diff --git "a/stack-exchange/math_overflow/shard_3.txt" "b/stack-exchange/math_overflow/shard_3.txt" deleted file mode 100644--- "a/stack-exchange/math_overflow/shard_3.txt" +++ /dev/null @@ -1,35970 +0,0 @@ -TITLE: Guessing a subset of {1,...,N} -QUESTION [11 upvotes]: I pick a random subset $S\subseteq\lbrace1,\ldots,N\rbrace$, and you have to guess what it is. After each guess $G$, I tell you the number of elements in $G \cap S$. How many guesses do you need to determine the subset? (If there is only one possibility left, then you can omit the last guess.) -There is an obvious strategy that requires only $N$ guesses. Guess $\lbrace1\rbrace$, then guess $\lbrace2\rbrace$, then guess $\lbrace3\rbrace$, and so on. But there is a clever strategy that requires only $\lceil 4N/5 \rceil$ guesses. -We know that the minimum number of guesses is at least $\left\lceil \frac{N}{\log_2{(N+1)}}\right\rceil$, because each guess reveals at most $\log_2(N+1)$ bits of information. I seek a proof or disproof of the conjecture that the number of guesses $g(N)$ is sublinear, i.e. $\lim_{N\to\infty} g(N)/N = 0$. -I will donate $100 to the American Red Cross if a proof or disproof is posted to this thread by April 30, 2011. For this purpose, I will accept an argument as correct if I believe it to be correct; or if a user with reputation above 1000 asserts that it is correct, and no user with reputation above 1000 denies that it is correct. Naturally, I would welcome improved upper bounds, even if they are linear. - -REPLY [18 votes]: This is a well-studied problem, sometimes phrased as a coin-weighing problem. It is known that $g(N)$ is $O(N / \log N)$. (We can even specify the guessing sets in advance, without knowing the previous answers.) I believe these three papers are the earliest to show this bound: -B. Lindstrom (1964), "On a combinatory detection problem I", Mathematical Institute of the Hungarian Academy of Science 9, pp. 195-207. -B. Lindstrom (1965), "On a combinatorial problem in number theory", Canadian Math. Bulletin 8, pp. 477-490. -D. Cantor and W. Mills (1966), "Determining a subset from certain combinatorial properties", Canadian J. Math 18, pp. 42-48. -There was a lot of work after these papers too (some with simpler constructions, some to solve more general problems). A book by Aigner covers this topic and more: -M. Aigner (1988), "Combinatorial search", John Wiley and Sons.<|endoftext|> -TITLE: Contour integration of $\zeta(s)\zeta(2s)$ -QUESTION [7 upvotes]: I have been looking at this for days and I am going insane. -I need to show that for a Dirichlet series equal to $\zeta(s)\zeta(2s)$, the sum of the coefficients less than $x$ is $x\zeta(2)+O(x^{(3/4)})$, and then expand that to the $\Pi \zeta(ks)$ for all $k$ in an effort to find the formula for the number of non-isomorphic abelian groups. -I know that using Perron's formula there is a simple pole at $s=1$ that gives a residue of $X\zeta(2)$, but I can't find a contour that converges or the exact error term. - -REPLY [6 votes]: If you havn't done so already, you might find it useful to look at the proof of theorem 12.2 on the divisor problem in Titchmarsh - The theory of the Riemann zeta function. Here, he goes through a detailed application of Perron's formula for the function $\zeta^k(s)$, which I believe to be very similar to your case. -Indeed, for $s=\sigma + it$ and $\sigma>1/2$, $\zeta(2s)$ is absolutley convergent and hence uniformly bounded with respect to $t$. So this will not contribute to the contours that you choose (as long as $\sigma>1/2$!). What you need then is good upper bounds for the order of the zeta function in the critical strip. -To get these, one normally finds the order of the function at two points, and then uses the Phragmén–Lindelöf principle for strips to get estimates for the function between these two points. For example, it is known that $\zeta(1/2 + it) = O(t^{1/4})$ (see The Lindelöf hypothesis), although there are much better bounds available than that. This is all done in Titchmarsh's book. -I hope this helps!<|endoftext|> -TITLE: Fano 3-fold of degree 4 -QUESTION [6 upvotes]: Let $X$ be the intersection of two quadrics in $P^5$. It is well known that the intermediate Jacobian $J(X)$ is isomorphic to $J(C)$ for a genus 2 curve, related to the pencil of quadrics whose base locus is $X$. -It seemed then natural to me to ask the following question: -Is there an explicit construction where $X$ is obtained as a smooth blow-up of $P^3$, or of a smooth quadric, or of a $P^2$ bundle over $P^1$, along a curve isomorphic to $C$? - -REPLY [5 votes]: The projection from a line $L_0$ is a birational isomorphism of $X$ onto $P^3$. It decomposes as the blow-up of the line $L_0$ followed by the contraction of a surface swept by lines intersecting $L_0$ onto a curve of genus 2 in $P^3$.<|endoftext|> -TITLE: Does every irreducible representation of a compact group occur in tensor products of a faithful representation and its dual? -QUESTION [17 upvotes]: (Previously posted on math.SE with no answers.) -Let $G$ be a compact Lie group and $V$ a faithful (complex, continuous, finite-dimensional) representation of it. Is it true that every (complex, continuous, finite-dimensional) irreducible representation of $G$ occurs in $V^{\otimes n} (V^{\ast})^{\otimes m}$ for some $n, m$? -The proof I know for finite groups doesn't seem to easily generalize. I want to apply Stone-Weierstrass, but can't figure out if the characters I get will always separate points (in the space of conjugacy classes). Ben Webster in his answer to this MO question seems to be suggesting that this follows from the first part of Peter-Weyl, but I don't see how this works. -Certainly the corresponding algebra of characters separates points whenever the eigenvalues of an element $g \in G$ acting on $V$ determine its conjugacy class (since one can get the eigenvalues from an examination of the exterior powers). Does this always happen? - -REPLY [21 votes]: You can mimic the standard finite group argument. Recall that, if $X$ and $Y$ are two finite-dimensional representations of $G$, with characters $\chi_X$ and $\chi_Y$, then $\dim \mathrm{Hom}_G(X,Y) = \int_G \overline{\chi_X} \otimes \chi_Y$, where the integral is with respect to Haar measure normalized so that $\int_G 1 =1$. The proof of this is exactly as in the finite case. -Now, let $W=1 \oplus V \oplus \overline{V}$. Your goal is to show that, for any nonzero representation $Y$, $\mathrm{Hom}(W^{\otimes N}, Y)$ is nonzero for $N$ sufficiently large. So you need to analyze $\int (1+\chi_V + \overline{\chi_V})^N \chi_W$ for $N$ large. Just as in the finite group case, we are going to split this into an integral near the identity, and an exponentially decaying term everywhere else. -I'm going to leave a lot of the analytic details to you, but here is the idea. Let $d=\dim V$. The function $f:=1+\chi_V + \overline{\chi_V}$ makes sense on the entire unitary group of $V$, of which $G$ is a subgroup. On a unitary matrix with eigenvalues $(e^{i \theta_1}, \cdots, e^{i \theta_d})$, we have $f=1+2 \sum \cos \theta_i$. In particular, for an element $g$ in the Lie algebra of $G$ near the identity, we have -$$f(e^g) = (2n+1) e^{-K(g) + O(|g|^4)}$$ - where $K$ is $\mathrm{Tr}(g^* g) = \sum \theta_i^2$. -And, for $g$ near the identity, $\chi_X(e^g) = d + O(|g|)$. So the contribution to our integral near the identity can be approximated by -$$\int_{\mathfrak{g}} \left( (2n+1) e^{-K(g) + O(|g|^4)} \right)^N (d+ O(|g|) \approx (2n+1)^N d \int_{\mathfrak{g}} e^{-K(\sqrt{N} g)} = \frac{(2n+1)^N d}{N^{\dim G/2}} C$$ -where $C$ is a certain Gaussian integral, and includes some sort of a factor concerning the determinant of the quadratic form $K$. You also need to work out what the Haar measure of $G$ turns into as a volume form on $\mathfrak{g}$ near $0$, I'll leave that to you as well. -For right now, the important point is that the growth rate is like $(2n+1)^N d$ divided by a polynomial factor. In the finite group case, that polynomial is a constant, which makes life easier, but we can live with a polynomial. -Now, look at the contribution from the rest of $G$. For any point in the unitary group $U(d)$, other than the identity, we have $|f| < 2n+1$. (To get equality, all the eigenvalues must be $1$ and, in the unitary group, that forces us to be at the identity.) So, if we break our integral into the integral over a small ball around the identity, plus an integral around everything else, the contribution of every thing else will be $O(a^N)$ with $a<2n+1$. -So, just as in the finite group case, the exponential with the larger base wins, and the polynomial in the denominator is too small to effect the argument. Joel and I discussed a similar, but harder, argument over at the Secret Blogging Seminar.<|endoftext|> -TITLE: Bounding the number of character degrees of a finite group in terms of the order of the group -QUESTION [8 upvotes]: Let $cd(G)$ be the set of degrees of irreducible complex characters of the finite group $G$ (so $cd(G) = \{\chi(1) | \chi\in Irr(G)\}$). - -What bounds are known of the form $|cd(G)|\leq f(|G|)$ (ie, what functions $f$ are known which satisfies such an inequality)? - -I can show that $|cd(G)|\leq \sqrt[3]{3|G|-7}$ and if $|G|$ is odd that $|cd(G)|\leq \sqrt[3]{\frac{12}{15}|G|}$. -On the other hand, one clearly has $|cd(G)|\leq d(|G|)-1$ (where $d(n)$ is the number of divisors of $n$), and this is better asymptotically. -There are two reasons for this question. -One is that I am looking at certain inequalities which guarantee that a group will be solvable, and having a good bound on $|cd(G)|$ in terms of the order of the group would help. -Also, the question is interesting when compared to the Taketa-inequality ($dl(G)\leq |cd(G)|$ which is conjectured to hold for all solvable groups), since clearly the derived length of a solvable group only grows logarithmically in the order of the group (being bounded by the sum of the exponents in the prime factorization of the order of the group). - -REPLY [4 votes]: The number of divisors at least roughly resembles the best achievable lower bound. For each prime $p$, there is a group $G_p$ with $p^3$ elements which has an irreducible representation of dimension 1 (the trivial representation) and an irreducible representation of dimension $p$ (because it's non-abelian). Now let $p_n$ be the $n$th prime. The number -$$P_n = p_1p_2\cdots p_n$$ -is a type of number with a lot of divisors. If you likewise let -$$H_n = G_{p_1} \times G_{p_2} \times \cdots \times G_{p_n},$$ -then $H_n$ has an irreducible representation for every $d$ that divides $P_n$, and it has $P^3_n$ elements. -Now, it is not quite true that $P_n$ has more divisors than any $N < P_n$. It is a good strategy for making a number with many divisors, but soon enough it is better to add more factors of $p_1$, then eventually more factors of $p_2$, etc., than to keep adding new prime factors. To understand this situation better, we can make many numbers (but not all numbers) that have more divisors than their predecessors with the "threshold method". The idea is to optimize the ratio $\log(d(N))/\log(N)$ globally by optimizing it locally (with respect to prime factorization). Pick a constant $t > 0$, the threshold, and say that $N$ should have at least $k > 0$ factors of a prime $p$ if and only if -$$\frac{\log(k+1) - \log(k)}{\log(p)} \ge t.$$ -Then I think that $d(M) < d(N)$ when $M < N$. -In fact, finding large values of $cd(G)$ (which I will use to mean the cardinality of the character degrees rather than the set) is a very similar problem when $G$ is nilpotent. A finite group is nilpotent if and only if it is the product of its Sylow subgroups. The main idea of the construction above is that in this case $cd(G)$ is multiplicative, i.e., the product of its values for $p$-groups. Following the comment by Frieder Ladisch, $cd(G)$ is maximized for $p$-groups by $C_{p^m} \ltimes C_{p^{m+1}}$. (In the first version of the answer I used other $p$-groups that aren't as good.) I.e., this group has character degrees $1,p,\ldots,p^m$, and no $p$ group with $p^{2m}$ or fewer elements can have an irrep with $p^m$ elements. So you can find many record values of $cd(G)$ for nilpotent groups using instead the threshold formula -$$\frac{\log(k+1) - \log(k)}{\min(4-k,2)\log(p)} \ge t.$$ -Let's incorporate the concept of a "record value" by defining $d'(N)$ to be the maximum of $d(M)$ with $M \le N$. Likewise define $cd'(N)$ to be the maximum of $cd(G)$ with $|G| \le N$. Then I think that the above constructions show that $d'(N)$ and $cd'(N)$ are at least similar functions, and that -$$d'(N) > cd'(N) > \sqrt[3]{d'(N)}$$ -when $N$ is large enough. In fact I think that the exponent of the second inequality climbs from $1/3$ to some higher value, although for nilpotent groups one also has -$$\sqrt{d'(N)} > cd'_{\text{nil}}(N).$$ -Let me also mention that the bound $O(\sqrt[3]{|G|})$ follows immediately from the fact that $|G|$ is the sum of the squares of the dimensions of the irreducible representations --- maybe that's what you have in mind with your bound.<|endoftext|> -TITLE: What is the subfactor planar algebra of type $\tilde{A}_n$, of index 4? -QUESTION [6 upvotes]: As I understand it, there is a subfactor whose principal graph is the affine Dynkin diagram $\tilde{A}_n$. Since every vertex has two neighbors, does that mean the space of 1-boxes is two dimensional? Is that allowed? - -REPLY [6 votes]: One concrete realization of this planar algebra goes as follows. - -Start with the (generalized) PA freely generated by an oriented strand. -Impose "$Z$-homology" relations: (a) oriented saddle moves, and (b) erase small loops (loop value = 1). -Now introduce an unoriented strand which is the formal direct sum of an upward pointing strand and a downward pointing strand. -If we now draw the principal graph of this PA from the point of view of the newly introduced unoriented strand type, we get the $\tilde{A}_\infty$ graph. -I just noticed that you want $\tilde{A}_n$, not $\tilde{A}_\infty$. For this case, start with $Z/n$ homology instead of $Z$ homology. If you want more details let me know.<|endoftext|> -TITLE: finite quotients of fundamental groups in positive characteristic -QUESTION [5 upvotes]: For affine smooth curves over $k=\bar{k}$ of char. $p,$ Abhyankar's conjecture (proved by Raynaud and Harbater) tells us exactly which finite groups can be realized as quotients of their fundamental groups. -What about complete smooth curves, or more generally higher dimensional varieties? Are there results or conjectural criteria (or necessary conditions) for finite quotients of their $\pi_1?$ (Definitely, not too much was known around 1990; see Serre's Bourbaki article on this.) -In particular, let $G$ be the automorphism group of the supersingular elliptic curve in char. $p=2$ or $3$ (see supersingular elliptic curve in char. 2 or 3 for various descriptions of its structure). Is there (and if yes, how to construct) a projective smooth variety in char. $p$ having $G$ as a quotient of its $\pi_1?$ Certainly there are lots of affine smooth curves with this property (e.g. $\mathbb G_m$), and I wonder if for some of them, the covering is unramified at infinity (so that we win!). - -REPLY [2 votes]: For a supersingular elliptic $E$ over an algebraically closed field of characteristic two or three there exists a smooth curve $C$ of higher genus such that $Aut_0(E)$ is a finite quotient of $\pi_1(C)$. -This is explained in section 3 of http://arxiv.org/abs/1005.2142v3 -This is an easy application of a general theory of finite quotients of fundamental groups of smooth curves as explained in the paper -Amilcar Pacheco and Katherine F. Stevenson. Finite quotients of the algebraic fundamental group of projective curves in positive characteristic. Pacific J. Math., 192(1):143–158, 2000 -In this paper, it is explained how to realize groups which have the property that their maximal $p$-Sylow subgroup ($p$ being the characteristic) is normal. The automorphism groups of supersingular elliptic curves satisfy this property.<|endoftext|> -TITLE: Clifford theory: behaviour of a very general irreducible representation under restriction to a finite index subgroup. -QUESTION [19 upvotes]: Let $G$ be a group and let $H$ be a subgroup of finite index. -Let $V$ be an irreducible complex representation of $G$ (no topology or anything: $V$ is just a non-zero complex vector space with a linear action of $G$ and no non-trivial invariant subs). -Now consider $V$ as a representation of $H$. Is $V$ a finite direct sum of irreducible $H$-reps? -I am almost embarrassed to ask this question here. It looked to me initially like the answer should be "yes and this question is trivial". If $G$ is finite it is trivial and Clifford theory tells you basically what can happen. Here is another case I can do: if $H$ has index two in $G$ then $V$ is indeed a finite direct sum of irreducibles. For either $V$ is irreducible as an $H$-rep, in which case we're done, or $V$ is reducible, so there's $0\not=W\not=V$ an $H$-stable sub. Say $g\in G$ with $g\not\in H$. One checks easily that $gW$ is $H$-stable, that $W\cap gW$ is $G$-stable, so must be zero, and that $W+gW$ is $G$-stable, so must be $V$. Hence $V$ is the direct sum of $W$ and $gW$. This implies that $W$ is irreducible as an $H$-rep---for if $X$ were a non-trivial sub then the same argument shows $V=X\oplus gX$ but this is strictly smaller than $W\oplus gW=V$. -I thought that this argument should trivially generalise to, say, the case where $H$ is a normal subgroup of prime index. But I can't even do the case where $H$ is normal and $G/H$ has order $3$, because I can't rule out $V$ being the sum of any two of $W$, $gW$ and $g^2W$, and the intersection of any two being trivial. -Either I am missing something silly (most likely!) or there's some daft counterexample. I almost feel that I would be able to prove something if I knew Schur's lemma [edit: by which I mean that if I knew $End_G(V)=\mathbf{C}$ then I might know how to proceed], but in this generality I don't see any reason why it should be true. Perhaps if I knew a concrete example of an irreducible complex representation of a group for which Schur's lemma failed then I might be able to get back on track. [edit: in a deleted response, Qiaochu pointed out that $G=\mathbf{C}(t)^\times$ acting on $\mathbf{C}(t)$ provided a simple example] [final remark that in the context in which this question arose, $G$ was a locally profinite group and $V$ was smooth and I could use Schur's Lemma, but by then I was interested in the general case...] - -REPLY [9 votes]: Since $V$ is irreducible, it is a finitely generated $\mathbb C[G]$-module, any non-zero element is a generator. Since $H$ is of finite index, $\mathbb C[G]$ is a finitely generated -$\mathbb C[H]$-module. Hence $V$ is a finitely generated $\mathbb C[H]$-module. Zorn's Lemma implies the existence of an irreducible quotient $W$. -Suppose that $H$ is normal in $G$. (It is probably enough to suppose that $H$ contains a subgroup of finite index, which is normal in $G$.) Let $K$ be the kernel of $V\rightarrow W$, for every $g\in G$ the quotient $V/g K$ is an irreducible $H$-rep, isomorphic to $W^g$. The kernel of the natural map -$$V\rightarrow \bigoplus_{g\in G/H} V/g K$$ -is $G$ invariant, and hence $0$. So we may inject $V$ into a finite direct sum of irreducible $H$-reps. Choose a smallest subset $X\subset G/H$, such that $\varphi: V\rightarrow \bigoplus_{g\in X} V/g K$ is injective, then $\varphi$ is also surjective. -Edit. Kevin pointed it out that $H$ always contains a subgroup of finite index, which is normal in $G$, and F.Ladisch finished off the general case, i.e without assuming that $H$ is normal in the comments below.<|endoftext|> -TITLE: Trivial fiber bundle -QUESTION [8 upvotes]: Suppose $(E,p,B;F)$ is a fiber bundle such that $E$ is homeomorphic to $B\times F$, is it true that the fiber bundle is trivial? -A non connected counter example has been provided, so I'll ask for E,B and F to be connected (hopefully low dimensional) manifolds. - -REPLY [13 votes]: Consider the pullback $\xi$ of $TS^2$ via the projection of $S^2\times\mathbb R$ onto the first factor. The bundle $\xi$ is a nontrivial $\mathbb R^2$-bundle over $S^2\times\mathbb R$ because its pullback under the inclusion $S^2\to S^2\times\mathbb R$ is $TS^2$, which is nontrivial. On the other hand, its total space is $\mathbb R\times TS^2$ which is diffeomorphic to $S^2\times\mathbb R^3$, which is the total space of the trivial $\mathbb R^2$-bundle over $S^2\times\mathbb R$.<|endoftext|> -TITLE: Compact open topology on $\mathrm{Homeo}(X)$ -QUESTION [29 upvotes]: Let $X$ and $Y$ be topological spaces. Define the compact open topology on the set $\mathrm{M}(X,Y)$ of continuous maps from $X$ to $Y$ via the subbase $[K,O]$ of all maps $f:X\rightarrow Y$ s.t. $f(K)\subset O$, where $K$ is any compact subset of $X$, and $O$ is any open subset of $Y$. So a basis of open sets is given by the following subsets: $[K_1,\dots,K_n,O_1,\dots,O_n]=[K_1,O_1 ]\cap\dots\cap [K_n,O_n]$, the collection of continuous maps $f:X\rightarrow Y$ that send each $K_i$ into $O_i$ for some specified collection of compact $K_i$'s and open $O_i$'s. -This topology has some nice properties: the exponential law holds under some hypotheses on the spaces $X$ and $Y$, and is certainly true if all spaces involved are locally compact Hausdorff spaces, as will be the case from now on. -My question is as follows: if $X$ is a locally compact Hausdorff space (or even a topological manifold), the compact open topology induces a topology on the set of homeomorphisms of $X$, which is a group. Does this topology turn $\mathrm{Homeo}(X)$ into a topological group? I can show that the product (composition) is continuous, but is the inverse too? $(f\rightarrow f^{-1})$ -I was able to prove continuity for compact spaces, where it is very easy to establish. I also managed to prove it for $X=\mathbb{R}$ because all homeomorphisms of $\mathbb{R}$ are monotone, but that's everything so far. -I tried looking it up in several textbooks on topology and algebraic topology where the C.O. topology is usually discussed, but couldn't find a discussion on this topic anywhere. - -REPLY [8 votes]: R. Arens, Topologies for homeomorphism groups, Amer. J. Math. 68 (1946) 593–610. -If $X$ is locally compact and locally connected (!), then $\mathrm{Homeo}(X)$ is a topological group.<|endoftext|> -TITLE: Why study Lie algebras? -QUESTION [137 upvotes]: I don't mean to be rude asking this question, I know that the theory of Lie groups and Lie algebras is a very deep one, very aesthetic and that has broad applications in various areas of mathematics and physics. I visited a course on Lie groups, and an elementary one on Lie algebras. But I don't fully understand how those theories are being applied. I actually don't even understand the importance of Lie groups in differential geometry. -I know, among others, of the following facts: - -If $G$ and $H$ are two Lie groups, with $G$ simply connected, and $\mathfrak{g,h}$ are their respective Lie algebras, then there is a one to one correspondance between Lie algebra homomorphisms $\mathfrak{g}\rightarrow\mathfrak{h}$ and group homomorphisms $G\rightarrow H$. -The same remains true if we replace $H$ with any manifold $M$: any Lie algebra homomorphism from $\mathfrak{g}$ to the Lie algebra $\Gamma(TM)$ of smooth vector fields on $M$ gives rise to a local action of $G$ on $M$. -Under some conditions like (I think) compactness, the cohomology of $\mathfrak{g}$ is isomorphic to the real cohomology of the group $G$. I know that calculating the cohomology of $\mathfrak{g}$ is tractable in some cases. -There is a whole lot to be said of the representation theory of Lie algebras. -Compact connected centerless Lie groups $\leftrightarrow$ complex semisimple Lie algebras - -How do people use Lie groups and Lie algebras? What questions do they ask for which Lie groups or algebras will be of any help? And if a geometer reads this, how, if at all, do you use Lie theory? How is the representation theory of Lie algebras useful in differential geometry? - -REPLY [3 votes]: I hope you may be interested in your questions restricted to Dynamical Systems (DS): - -How do people use Lie groups and Lie algebras in DS? -What questions do they ask for which Lie groups or algebras will be -of any help in DS? - -In DS people are interested in particular in: dense orbits and invariant measures for actions of groups on a manifold. A tool to study this is Ergodic Theory (ET). -There is an abstract ET of amenable groups actions (Ornstein-Weiss), I have seen that Lie groups provide: concrete examples, counter-examples and extensions of this abstract theory. I have also seen that one important reason why Lie groups allow to prove deep ET results is because they are well adapted to the theory of harmonic analysis, and harmonic analysis is one of the main machinery of ET. -The ET of Lie groups is a very broad theory, just a few examples: -Example 1. -Let $G\subset SL(\mathbb{Z},d)$ (such that $G$ acts irreducibly on $\mathbb{R}^d$ and $G$ does -not contain an abelian subgroup of finite index). Then the $G$-action on the d-torus is strongly ergodic (''the G-invariatn measure is the unique G-invariant mean''). Moreover, there is a characterization of the $G$-action with $G\subset SL(\mathbb{Z},d)$ on the d-torus that are strongly ergodic. And more in abstract, if $G$ is a connected non-compact simple Lie group with finite center, there is a characterization for the action to be strongly ergodic. -These are theorems of A. FURMAN and Y. SHALOM that generalize Rosenblatt: - -A. Furman and Y. Shalom. Sharp ergodic theorems for group actions and -strong ergodicity. -J. Rosenblatt. Uniqueness of invariant means for measure preserving transformations. Trans. Amer. -Math. Soc. 265 (1981), 623–636 - -Example 2. -Orbit-equivalence rigidity (Zimmer): -Let $G_1$ and $G_2$ be noncompact, simple Lie groups, with $R$-$rank(G_1)\geq 2.$ If $(G_1,X_1)$ is orbit equivalent to $(G_2,X_2)$, then $G_1$ is locally isomorphic to $G_2$, and, up to a group automorphism, the actions are isomorphic. -Example 3. -A. Avila, B. Fayad, and A. Kocsard provide in (On manifolds supporting distributionally uniquely ergodic diffeomorphisms) some counterexamples to a conjecture proposed by Forni in “On the Greenfield-Wallach and Katok conjectures in dimension three. -A Good references for ET of actions of Lie group is R. Zimmer. Ergodic Theory and Semisimple Lie Groups. Birkh¨auser, Boston, 1984. -You can find deep results by: A. Avila, V. Bergelson, G. Forni, J. Rosenblatt, A. Furman, H. Furstenberg, D. Y. Kleinbock, G. A. Margulis, Y. Shalom, A. Katok and Zimmer, between many others.<|endoftext|> -TITLE: A geometric series equalling a power of an integer -QUESTION [18 upvotes]: The following problem cropped up whilst considering generalised quadrangles with a product structure, and it boils down to a simple number theoretic problem. Let $s$ be an integer greater than 2 and suppose the geometric series $(s^r-1)/(s-1)$ is a nontrivial power of a positive integer. It seems the following is true: -If $r=3$, then $s= 18$. -If $r=4$, then $s = 7$. -If $r=5$, then $s = 3$. -If $r>5$, there are no solutions. -Does anyone know a proof of this curious property? - -REPLY [32 votes]: This is a well-investigated Diophantine equation known as Nagell--Ljunggren equation (they investigated this equation in the 1920s and 1940s, resp). Indeed, it is conjectured that the three solutions mentioned by the questioner are the only ones; however, it is not even known that the number of solutions is finite, though there are numerous partial result. -Below, I try to give some rough overview of some results that are known, and some references to (recent) articles. - -First, I restate the question to bring the notation in line with some sources I quote. -What are the solutions $(x,y,n,q)$ of the equation -$$ \frac{x^n - 1}{x - 1} = y^q $$ -with integers $x,y>1$, $n>2$, $q \ge 2$ ? -As mentioned, in the question, one finds three 'small' solutions -$(3,11,5,2)$, $(7,20,4,2)$, and $(18,7,3,3)$. -And, the remaining question is: -(A) Are these three solutions all the solutions ? -Or more modestly -(B) Is the number of solutions finite? - -As said, even (B) is open; but (A) is conjectured to be true. -By early works of Nagell and Ljunggren it is known that with any of the following conditions there are no other solutions: $q=2$, $n$ a multiple of $3$, $n$ a multiple of $4$, or ($q=3$ and $n$ not $5$ modulo $6$). -Shorey and Tijdeman proved (1976) that the number of solutions is finite with any of the following conditions: $x$ is fixed, $n$ has a fixed prime divisor, $y$ has a fixed prime divisor. Also, Shorey proved that the ABC-conjecture implies that the number of solutions is finite. -There are numerous additional results, imposing various conditions on $x,y,n$ or $q$ (due to Bennet, Bugeaud, Le, Mignotte, and others) for a survey of the state of the art around a decade ago see, e.g., a 2002 survey (in French) of Bugeaud and Mignotte (which was also the main bases for the above written part) available here. -The early results were obtained via passing to certain rings of algebraic integers; later results often used Baker's method (linear forms in logarithms) and results on Diophantine approximation. -Some years ago, the solution of Catalan's conjecture (which is on a somewhat similar equation), by Mihailescu that (as far as I understand, very surprisingly) avoided all these types of tools and used instead (only) results on cyclotomic fields/integers, provided a new impetus. -Specifically, it is now known, see Bugeaud and Mihailescu (2007), that -a. for any other solution (so not one of three known ones) the smallest prime divisor of $n$ is at least $29$ and $n$ has at most $4$ prime divisors (counted with multiplicity). Moreover, $n$ is prime if $q=3$. And, if $q\mid n$, then $q=n$. -b. to prove that there are no other solutions, it suffices to show that there is no solution with $n\ge 5$ an odd prime and $q$ an odd prime. -Moreover, related to the latter assertion Mihailescu recently proved (see here and here) various results in the case that $n$ and $q$ are odd primes (saying, in one of the abstracts that methods used in the cyclotomic approach to FLT are used, so Yemon Choi's intuition was very right). -This answer does certainly not give a complete picture (in this format, it would be difficult to give one, and no matter the format, it would be impossible for me); I am aware of various omissions I made, and I am afraid there are many of which I am not aware. -The references I mentioned should however allow to retrieve more complete information. -[Note: in case the tex is broken, it is not carelessness; at the moment, for technical reasons, I cannot test it myself.]<|endoftext|> -TITLE: Number of divisors of an integer of form 4n+1 and 4n+3 -QUESTION [6 upvotes]: Suppose $n$ is a large odd integer. Let $D_1(n)$ be the number of divisors of $n$ of the form $4k+1$ and let $D_3(n)$ be the number of divisors of the form $4k+3$. I would like to compute $(D_1(n),D_3(n))$. -As Joe Silverman points out, the number of representations of $n$ as a sum of two squares of integers is $4(D_1(n)-D_3(n))$. For example, $D_1(225)=6$ and $D_3(225)=3$, so there are $4(6-3)=12$ lattice points on the circle of radius $\sqrt {225}$ centered at the origin including $(0,15)$ and $(-9,-12)$. - -Is there a faster way to find $(D_1(n),D_3(n))$ than factoring $n$? - - -Original: -Hi, one way to do so is to list all the divisors of the integer and check each if it is of the form $4n+1$ or $4n+3$. -Is there any faster method to it, especially for large $n$? - -REPLY [12 votes]: Not quite what you're asking, but an interesting theorem of Legendre's says that the number of ways of writing an integer $N$ as a sum of two squares is $4D_1(N)-4D_3(N)$, where $D_1(N)$ is the number of positive divisors of $N$ that are congruent to 1 modulo 4 and $D_3(N)$ is the number of positive divisors of $N$ that are congruent to 3 modulo 4. There are undoubtedly also results proved via analytic methods that describe the distribution of $D_1(N)$ and $D_3(N)$. But I'd have to agree with the other posters that computing $D_1(N)$ and $D_3(N)$ for a specific $N$ sounds about as hard as factoring $N$. Indeed, if $N=pq\equiv 1 \pmod{4}$, computing $D_1(N)$ is equivalent to computing the first bit in the factors of $N$, which seems hard.<|endoftext|> -TITLE: Construction of the determinant line bundle on the degree $g-1$ Picard variety -QUESTION [15 upvotes]: Consider $J^{g-1}$, the variety of degree $g-1$ line bundles on a compact Riemann surface of genus $g$. Recall that $J^{g-1}$ is a torsor for the Jacobian, thus has dimension $g$. We can produce elements in $J^{g-1}$ by choosing $g-1$ points $p_1,\ldots,p_{g-1}$ and constructing the associated line bundle to the divisor $p_1+\cdots+p_{g-1}$. In this way, we obtain a $g-1$-dimensional family of elements of $J^{g-1}$, forming the so-called $\Theta$ divisor. Elements of the $\Theta$ divisor may be distinguished from other points in $J^{g-1}$ by the fact that they represent line bundles admitting sections, i.e. $h^0(L)\neq 0$. -For degree $g-1$ line bundles, the Riemann-Roch theorem gives $h^0(L)=h^1(L)$, so over $J^{g-1}$ we see that $h^0(L),h^1(L)$ are generically zero and jump along the $\Theta$ divisor. One might imagine that there are vector bundles $V,W$ over $J^{g-1}$ of the same rank, with fibres over $L\in J^{g-1}$ given by Dolbeault $0$- and $1$-forms with coefficients in $L$ respectively, together with a bundle map $\overline{\partial}:V\rightarrow W$ which is generically an isomorphism but drops rank along $\Theta$; then $\det \overline{\partial}$ would be a section of a line bundle $\det V^*\otimes\det W$ over $J^{g-1}$ which cuts out $\Theta$. This line bundle turns out to be a well-defined object called the determinant line bundle and it was introduced by Quillen. -Quillen's construction of the line bundle proceeds by replacing the family of (2-term) Dolbeault complexes parametrized by $J^{g-1}$ by a quasi-isomorphic family of finite-dimensional (2-term) complexes and taking the determinant line bundle of this. The finite-dimensional replacement for the Dolbeault complex is given by the Dolbeault operator acting on forms lying in the first few eigenspaces of the Laplacian, roughly speaking. Then as we move along $J^{g-1}$, some of the eigenvalues may hit zero and we have a jump of $h^0(L)$. -For a good list of references for the above, see determinant line bundle (nlab). -My question: What is a simple, modern construction of the determinant line bundle on $J^{g-1}$, perhaps one which uses the tools of algebraic geometry? It may well be tautological from some point of view, in which case I would probably be unsatisfied. Also, if you wish to rhapsodize on determinant line bundles, please feel free. -My motivation: Many of the classic papers by Beauville, Narasimhan, Ramanan etc. on moduli of vector bundles expend a lot of effort working with determinant line bundles and extracting information from them in quite ad-hoc and clever ways. I'd like to understand these better. - -REPLY [16 votes]: Let's say that your (compact!) Riemann surface is $X$ (of genus at least $1$). What does it mean to say that $J^{g-1}$ is the "variety of degree $g-1$ line bundles?" One way to formalize this is to say that there is a line bundle $\mathcal{L}$ on the product $X \times J^{g-1}$ with the property the rule $p \mapsto \mathcal{L}|_{X \times \{ p \}}$ defines a bijection between the ($\mathbb{C}$-valued) points of $J^{g-1}$ and the line bundles of degree $g-1$. (A stronger statement is that $J^{g-1}$ represents the Picard functor, but I'm going to try to sweep this under the run.) -Let $\pi \colon X \times J^{g-1} \to J^{g-1}$ denote the projection maps. Recall that the formation of the direct image $\pi_{*}(\mathcal{L})$ does not always commute with passing to a fiber. However, the theory of cohomology and base change describes how the fiber-wise cohomology of $\mathcal{L}$ varies. The main theorem states that there is a 2-term complex of vector bundles -$d \colon \mathcal{E}_0 \to \mathcal{E}_1$ -that computes the cohomology of $\mathcal{L}$ universally. That is, for all morphisms $T \to J^{g-1}$, we have -$\operatorname{ker}(d_{T}) = (\pi_{T})_{*}(\mathcal{L}_{X \times T})$ -and -$\operatorname{cok}(d_{T}) = (R^{1}\pi_{T})_{*}(\mathcal{L}_{X \times T}).$ -(The most important case is where $T = \operatorname{Spec}(\mathbb{C})$ and -$T \to J^{g-1}$ is the inclusion of a point.) -The complex $\mathcal{E}_{\cdot}$ is not unique, but any other complex of vector bundles with this property must be quasi-isomorphic. -In the literature, many authors construct a complex $\mathcal{E}_{\cdot}$ using an explicit procedure, but the existence is a very general theorem. I learned about this topic from Illusie's article "Grothendieck's existence theorem in formal geometry," which states the base change theorems in great generality. -In your question, you described how to convert the complex into a line bundle (take the difference of top exterior powers). But now we must check that two quasi-isomorphic complexes have isomorphic determinant line bundles. This can be checked by hand, but this statement has been proven in great generality by Mumford and Knudsen (see MR1914072 or MR0437541). The determinant of the complex $\mathcal{E}_{\cdot}$ is exactly the line bundle you are asking about. -One subtle issue is that the universal line bundle $\mathcal{L}$ is NOT uniquely determined. If $\mathcal{M}$ is any line bundle on $J^{g-1}$, then -$\mathcal{L} \otimes \pi^{-1}(\mathcal{M})$ also parameterizes the degree $g-1$ line bundles in the sense described above (and in a stronger sense that I am sweeping under the rug). However, one can show that $\mathcal{L}$ and $\mathcal{L} \otimes p^{*}(\mathcal{M})$ have isomorphic determinants of cohomology by using the fact that a degree $g-1$ line bundle has Euler characteristic equal to zero. -Added: Here is one method of constructing the complex. Fix a large collection of points $\{x_1, \dots, x_n\} \subset X$ (at least $g+1$ points will do), and let $\Sigma \subset X \times J^{g-1}$ denote the subset consisting of pairs $(x_i,p) \in X \times J^{g-1}$. The universal line bundle $\mathcal{L}$ fits into a short exact sequence: -$$ - 0 \to \mathcal{L}(-\Sigma) \to \mathcal{L} \to \mathcal{L}|_{\Sigma} \to 0 -$$ -Here $\mathcal{L}(-\Sigma)$ is the subsheaf of local sections vanishing along $\Sigma$ and $\mathcal{L}|_{\Sigma}$ is the pullback of $\mathcal{L}$ to $\Sigma$. -Associated to the short exact sequence on $X \times J^{g-1}$ is a long exact sequence relating the higher direct images under $\pi$. The first connecting map is a homomorphism -$d : \pi_{*}(\mathcal{L}|_{\Sigma}) \to R^{1}\pi(\mathcal{L}(-\Sigma))$ -This is a complex $K_{\cdot}$ of vector bundles that compute the cohomology of $\mathcal{L}$ universally in the sense described earlier. Thus, its determinant is the desired line bundle. -Why is this true? There are two statements to check: that the direct images are actually vector bundles and that the complex actually computes the cohomology of $\mathcal{L}$. Both facts are consequences of Grauert's theorem. Indeed, the hypothesis to Grauert's theorem is that the dimension of the cohomology of a fiber $\pi^{-1}(p)$ is constant as a function of $p \in J^{g-1}$. Using Riemann-Roch (and the fact that we chose a large number of points), this is easily checked. The theorem allows us to conclude that both $\pi_{*}(\mathcal{L}|_{\Sigma})$ and $R^1 \pi(\mathcal{L}(-\Sigma))$ are vector bundles whose formation commutes with base change by a morphism $T \to J^{g-1}$. -An inspection of the relevant long exact sequence shows that the cohomology of the complex -is $\pi_{*}(\mathcal{L})$ and $R^1\pi(\mathcal{L})$ (i.e. the complex compute the cohomology of $\mathcal{L}$). We would like to assert that this property persists if we base-change by a morphism $T \to J^{g-1}$ and work with the complex -$K_{\cdot} \otimes \mathcal{O}_{T}$. -But we already observed that the formation of the direct images appearing in the complex commutes with base-change, and so the base-changed complex fits into a natural long exact sequence. Again, an inspection of this sequence shows that base-changed complex compute the cohomology of $\mathcal{L}|_{X \times T}$, and so the original complex computes the cohomology universally.<|endoftext|> -TITLE: When does symmetry in an optimization problem imply that all variables are equal at optimality? -QUESTION [33 upvotes]: There are many optimization problems in which the variables are symmetric in the objective and the constraints; i.e., you can swap any two variables, and the problem remains the same. Let's call such problems symmetric optimization problems. The optimal solution for a symmetric optimization problem - like many of the ones that show up in calculus texts - frequently has all variables equal. To take some simple examples, - -The rectangle with fixed area that minimizes perimeter is a square. (Minimize $2x+2y$ subject to $xy = A$ and $x,y \geq 0$.) -The rectangle with fixed perimeter that maximizes area is a square. (Maximize $xy$ subject to $2x + 2y = P$ and $x,y \geq 0$.) -The difference between the arithmetic mean and the geometric mean of a set of numbers is minimized (and equals $0$) when all the numbers are equal. - -There are also more complicated symmetric optimization problems for which the variables are equal at optimality, such as the one in this recent math.SE question. -However, it is not true that every symmetric optimization problem has all variables equal at optimality. For example, the problem of minimizing $x +y$ subject to $x^2 + y^2 \geq 1$ and $x, y \geq 0$ has $(0,1)$ and $(1,0)$ as the optimal solutions. - -Does anyone know of general conditions on a symmetric optimization problem that guarantee the optimal solution has all variables equal? - -The existence of such conditions might be very nice. Unless the conditions themselves are ugly, they ought to vastly simplify solving a large class of symmetric optimization problems. -(Maybe convexity plays a role? My last example has a nonconvex feasible region.) - -REPLY [2 votes]: Suppose you have any optimization problem that is symmetric. I somehow weaker question is: How much of the original symmetry carries over to the solutions? For the symmetric group $S_n$ if the degree $d$ of the polynomials that describe the problem is low (compared to the number of variables) the "degree and half principle" says that one always finds minimizers contained in the set of points invariant by a group $S_{l_1}\times\ldots\times S_{l_d}$ where $l_1+\ldots+l_d=n$<|endoftext|> -TITLE: Torus based cryptography -QUESTION [5 upvotes]: In cryptography one needs finite groups $G$ in which the discrete logarithm problem is infeasible. Often they use the multiplicative group $\mathbb{G}_m(\mathbb{F}_p)$ where $p$ is a prime number of bit length $500$, say. -Rubin and Silverberg suggested (cf. [1]) to use certain tori instead, if the goal is to minimize the key size. In the easiest case, this comes down to using the group -$$T_2(p)=ker(Norm: \mathbb{F}_{p^2}^\times\to \mathbb{F}_p^\times).$$ -If I understood correctly, then the underlying philosopy seems to be: The group $T_2(p)$ should be as secure as $\mathbb{F}_{p^2}^\times$, but its size is only $p+1$. (So, if you use groups of type $T_2(p)$ instead of groups of type $\mathbb{G}_m(\mathbb{F}_p)$, then you can achive the same security with half the key size.) -Question. What are the reasons, be they heurisical or strictly provable, to believe -in this philosopy? -Denote by $\mathbb{G}'_m$ the quadratic twist of the algebraic group $\mathbb{G}_m$. -It is easy to see that $T_2(p)$ is isomorphic to $\mathbb{G}'_m(\mathbb{F}_p)$. -(This isomorphism is easy to compute). The philosophy predicts: The quadratic twist of the multiplicative group should be better than the multiplicative group itself. -(Compare with elliptic curves: If $E/\mathbb{F}_p$ is an elliptic curve, then I would certainly not expect its quadratic twist to be better than $E$ itself.) -Remark: I concentrated on the simplest case above. One also considers certain groups $T_n(p)$ which are expected to be as secure as $\mathbb{F}_{p^n}^\times$, while their size is only $\approx\varphi(n)p$. Lemma 7 in [1] is meant to explain this. However, I would be keen on a more detailed explanation. -[1] Lect. Notes in Comp. Sci. 2729 (2003) 349-365. (available at http://math.stanford.edu/~rubin/) - -REPLY [6 votes]: Can I refer you to my paper `On the Discrete Logarithm Problem on Algebraic Tori', Advances in Cryptology – CRYPTO 2005, Lecture Notes in Computer Science, 2005, Volume 3621/2005, 66-85, in which myself and Frederik Vercauteren studied this very problem. -In particular, we showed that the compression mechanism afforded by the birationality of some algebraic tori may be exploited to obtain a faster discrete logarithm algorithm for some cryptographically practical field sizes. In these instances, attacking the discrete logarithm in $\mathbb{F}_{p^n}^{\times}$ via its decomposition -$\prod_{d \mid n} T_d(\mathbb{F}_p)$ -is faster than using L[1/3] index calculus techniques. -Since then, other work has improved the L[1/3] index calculus techniques. However, our work demonstrates that it is naive to argue that the DLP in algebraic tori must be -hard purely because the DLP in the multiplicative group of the extension field is hard, precisely because an attack on the former provides an attack on the latter.<|endoftext|> -TITLE: Why is Casson's invariant worth studying? -QUESTION [20 upvotes]: Hi everybody! I am reading some papers about Casson's invariant for (integral) homology 3-spheres...as the wiki says "Informally speaking, the Casson invariant counts the number of conjugacy classes of representations of the fundamental group of a homology 3-sphere M into the group $SU(2)$". It seems to be something interesting to study, but this is "my first trip" in the realm of 3-manifold topology, so I don't get the deep meaning of this invariant. I mean why should one study this invariant? what should I be expecting from it? which contributions is it likely to bear to this field? In particular I came across the Casson invariant while studying Heegaard splittings...do you have any reading to suggest? -Thank you, -Lor - -REPLY [31 votes]: Wikipedia's description of the Casson invariant gives the first important reason to study it. As an invariant that comes from the $\text{SU}(2)$ representation variety of $\pi_1(M)$, it reveals in particular that $\pi_1(M)$ is non-zero. At the time, before Perelman's proof of the Poincaré conjecture and geometrization, there was a lot of mystery about potential counterexamples to the Poincaré conjecture. For instance, one speculation was that the so-called $\mu$ invariant could reveal a counterexample. Since the Casson invariant lifts the $\mu$ invariant, and since it proves that $\pi_1(M)$ is non-trivial when it is non-zero, it is one way to see that the $\mu$ invariant can never certify a counterexample to the Poincaré conjecture. (Of course, no we know that there are no counterexamples.) -A second fundamental reason to study the Casson invariant is that it is the only finite-type invariant of homology spheres of degree 1. Many interesting 3-manifold invariants are finite-type, or (conjecturally) carry the same information as a sequence of finite-type invariants. This is known more rigorously at the level of knots; for instance, the derivatives of the Alexander polynomial, the Jones polynomial, and many other polynomials at $1$ are all finite-type invariants. At the level of knots, the second derivative of the Alexander polynomial, $\Delta''_K(1)$, is known to be the only non-trivial finite-type invariant of degree 2, and there is nothing in degree 1. So it means that this invariant appears over and over again as part of the information of many other invariants; there are many different definitions of the same $\Delta''_K(1)$. The same thing should happen to the Casson invariant, and indeed there are already two very different-looking types of definitions: (1) Casson's definition; (2) either the first LMO invariant or the first configuration-space integral invariant. -A third fundamental reason is that Casson invariant has an important categorification, Floer homology, which is the objects in the theory whose morphisms come from Donaldson theory. One wrinkle of this construction is that it is only a categorification of one of the definitions of Casson's invariant, Casson's definition. If Casson's invariant has many definitions, then it might (for all I know) have many different categorifications. -If your question is meant in the narrow sense of what topology you can prove with the Casson invariant, then you can definitely prove some things but only (so far) a limited amount. However, if you are interested in quantum topological invariants in their own right, and not just as a tool for pre-quantum topology problems, then the Casson invariant is important because it is a highly non-trivial invariant that you encounter early and often.<|endoftext|> -TITLE: Is the unitary group of $l^2(A)$ with the strict topology contractible? -QUESTION [5 upvotes]: Let $A$ be a $C^*$-algebra with countable approximate unit. Let $\mathbb{K}$ denote the compact operators on a separable Hilbert space. Mingo and later Cuntz and Higson have shown that the unitary group in the multiplier algebra $M(A \otimes \mathbb{K})$ is contractible in the norm topology. It was then shown by Troitsky in a paper with the title -Geometry and Topology of operators on Hilbert $C^*$-modules -that $U(M(A \otimes \mathbb{K}))$ is also contractible, if it is equipped with the left strict topology, i.e. the topology generated by the semi-norms $\lVert xa \rVert$ for $x \in M(A \otimes \mathbb{K})$ and $a \in A \otimes \mathbb{K}$. Is the theorem still true, if we change from left strict to strict (which is the topology that includes the semi-norms $\lVert ax \rVert$)? - -REPLY [4 votes]: I would post this as a comment but as it just happens I can't do that. I do think that -the exercise that you mention proves strict contractibility. The same formula for the homotopy, -$$ -(u,t)\mapsto w_tuw_t^*+(1-w_tw_t^*), -$$ -is given in Proposition 12.2.2 of Blackadar's book on K-theory, although the statement -only says that the unitary group of $M(A\otimes K)$ is path connected. This formula goes back to Dixmier and Douady's paper on fields of Hilbert spaces, applied to $B(H)$.<|endoftext|> -TITLE: Ideals in smooth subalgebras of C*-algebras -QUESTION [12 upvotes]: Let $B$ be a $C^{*}$-algebra and $\mathcal{B}$ a dense *-subalgebra stable under holomorphic functional calculus and $C^{1}$-functional calculus for selfadjoint elements. Also, $\mathcal{B}$ is a Banach algebra in a norm $\|\cdot\|_{1},$ satisfying -$\|\cdot\|\leq\|\cdot\|_{1}$. -Also, there is a countable bounded approximate unit $u_{n}$ for $\mathcal{B}$ which is a contractive, increasing approximate unit for $B$. Let $\mathcal{I}$ be a closed two sided ideal in $\mathcal{B}$, and denote by $I$ its closure in $B$. -Is it true that $\mathcal{I}=I \cap \mathcal{B}$ ? -The pertinent examples are Lipschitz functions on the circle and on the real line, both with norm $\|f\|_{1}=\|f\|+\|\partial f\|$. - -REPLY [5 votes]: $\newcommand{\norm}[1]{\Vert#1\Vert}$ -In general, I think the answer to your question is no. Take ${\mathcal B}=C^1[-1,1]$ with the norm -$\norm{f}= \norm{f}_\infty+\norm{f'}_\infty$ -and let ${\mathcal I}$ be the closed ideal consisting of those $C^1$-functions which vanish at $x=0$ and whose 1st derivative vanishes at $x=0$. Then $I\cap {\mathcal B}$ contains the function $f(x)=x$ which is evidently not in ${\mathcal I}$. - -[Some general remarks follow, in a rambling style owing to lack of sleep. I may try to edit these later.] -In the commutative unital setting, taking $B=C(X)$, we know what the closed ideals of $B$ are (they are precisely the "kernels" of closed subsets of $X$, in the language of hulls and kernels). -If your subalgebra ${\mathcal B}$ also has maximal ideal space (homeo to) $X$, then your question is related to -- perhaps is equivalent to, I have not thought in detail -- the following one: -Can I find a closed two sided ideal in ${\mathcal B}$ which is not the kernel of its hull? -Without your restrictions on stability-under-func-calc, this kind of question has been much studied for commutative examples, and I think also for certain noncommutative examples related to group algebras. -For little Lipschitz algebras (on the circle) the answer is no -- this ought to be in a paper of Sherbert from the 1970s --- so I expect the answer to your original question is "yes". (For the "big" Lipschitz algebras my suspicion is that the counter-example I gave for $C^1[-1,1]$ would also work.)<|endoftext|> -TITLE: Approximate units from strictly positive elements in $C^{*}$-algebras. -QUESTION [6 upvotes]: The existence of a countable approximate unit in a $C^{*}$-algebra $B$ is equivalent to the existence of a strictly positive element $h\in B$. There are several ways to construct an approximate unit from $h$. My question is, does $h(h+\frac{1}{n})^{-1}$ constitute an approximate unit for $B$? - -REPLY [7 votes]: I think this works: Functional calculus shows that $h h (h+1/n)^{-1} \rightarrow h$. Then $h$ is strictly positive if and only if $Bh$ is dense in $B$ (See, for example, Jensen+Thomsen, "Elements of KK-Theory", Lemma 1.1.21). So for $b\in B$ and $\epsilon>0$, we can find $c\in B$ with $\|b-ch\|<\epsilon$, and for all $n$ sufficiently large, also $\|ch h(h+1/n)^{-1} - ch\| < \epsilon$. Thus -\begin{align*} &\| b - bh(h+1/n)^{-1} \| \\&< \epsilon + \| ch - chh(h+1/n)^{-1}\| -+ \|chh(h+1/n)^{-1} - bh(h+1/n)^{-1}\| \\ -&< 2\epsilon + \epsilon \|h(h+1/n)^{-1}\| < 3\epsilon. \end{align*} -Thus we're done.<|endoftext|> -TITLE: coherent sheaves on affine formal schemes -QUESTION [7 upvotes]: Let $\hat{X} = \text{Spf} \hat{A}$ be obtained as the formal completion of an affine scheme $X = \text{Spec} A$ where $A$ is an adic noetherian ring. Given a coherent sheaf $\mathfrak{F}$ on $\hat{X}$, is it always possible to find a coherent sheaf $\mathcal{F}$ on $X$ such that $\hat{\mathcal{F}} = \mathfrak{F}$? - -REPLY [6 votes]: Take $A=k[x,y]$, $\hat A=k[[x,y]]$, and suppose coherent sheaf corresponds to the $\hat A$-module $N=\hat A/(f)$, where $f\in k[[x,y]]$ is not "algebraic": say, $f=y-exp(x)$, assuming $k$ has characteristic zero. It is clear that $N$ does not come from completion of any f.generated $A$-module $M$. (Proof: Suppose $N=\hat M$. Since $N$ has no $A$-torsion, any torsion of -$M$ maps to zero, so we can replace $M$ with $M/(torsion)$ and assume $M$ is torsion-free. Then $M$ embeds in a locally free module (its second dual), but that would make $N$ embed in the free $\hat A$-module, which is false.)<|endoftext|> -TITLE: Curve integral of exponent of superharmonic function. -QUESTION [15 upvotes]: Let $\phi$ be a real smooth superharmonic function on unit disc $D$ in $\mathbb C$; i.e. $\triangle \phi\le 0$. - Then there is a curve $\gamma$ from the center of $D$ to its boundary such that - $$\int\limits_\gamma e^\phi<\infty.$$ - -The question came from my failed answer to this question. -I know that the answer is YES, but I do not see a direct proof. - -REPLY [5 votes]: I'll assume that $\varphi=-\psi$ where $\psi$ is subharmonic and not too weird (say, with isolated non-degenerate critical points; it seems like you can always add something bounded to achieve it but I haven't checked the details). Take any piece of some level curve of $\psi$ inside the disk parameterized by length $\ell$ and start the gradient accent from each point parameterized by the level $t$ of $\psi$. All but countably many of those escape to the boundary. Let $v(t)$ be the absolute value of the gradient and $S(t)$ be the "cross-section factor". Then $S(t)v(t)$ is non-decreasing (divergence of the gradient field is positive), the length element is $1/v(t)dt$ and the area element is $S(t)/v(t)d\ell dt$. Note that the total area of the disk is finite and the gradient curves cannot meet (we use the non-negativity of the divergence again here). Thus $\int S(t)/v(t)dt<+\infty$ most of the time. But then, since $Sv$ is non-decreasing, we also have $\int 1/v(t)^2dt<+\infty$ while we need just $\int e^{-t}/v(t)<+\infty$ and Cauchy-Schwartz ends the story.<|endoftext|> -TITLE: Cyclic Permutations - but not what you think -QUESTION [12 upvotes]: This question is not about elements of $S_n$ that consist of a single $n$-cycle, though naturally it's related. -Instead, consider permutations modulo the action of $(123\ldots n)$. That is, we want ABCD to be the same as BCDA and CDAB and DABC. (It's optional whether this also is the same as DCBA, but for now let's say it's not.) I am primarily interested in the graph that these generate, sort of like the Cayley graph for $S_n$ with generators $(12),(23),\ldots (n-1 n),(n1)$, but with vertices and edges identified. (I don't think this is a Cayley graph of a quotient of $S_n$; I don't even think this set is identifiable with a group since that subgroup isn't normal, if I recall correctly.) -What are these things called, and are there references to them in the literature? (Say to their symmetry groups, rep. theory, or whatever else.) I can't imagine there aren't, but because 'cyclic permutations' nearly always means something else, it's frustrating to look for this. I found pages of MathSciNet references to those terms, and none were about this. Not surprisingly! But presumably combinatorics experts have studied them - not just counted them, though Polya enumeration immediately comes to mind. -Edit: For a concrete example, imagine people around a dinner table, where you don't care which chair you sit in, you just care what the arrangement is. Maybe it's been thought of that way before? -Edit: Well, I have to say that Tilman and Mark Sapir both have been very helpful, but I guess Tilman answered the actual question. -Very oddly, I can only find ONE paper on MathSciNet that actually deals with the object I am interested in directly - Woodall's "Cyclic-order graphs and Zarankiewicz's crossing-number conjecture" proves some basic facts. Nearly every reference to such things is about using cyclic orders without considering all of them (in graph theory or queueing theory), is using them to create ribbon graphs, or is about extending partial cyclic orders to complete cyclic orders. - -REPLY [2 votes]: Here is one interpretation of this set in terms of Hochschild homology. Let $A$ be the group algebra of $\mathbb{Z}/n\mathbb Z$, and let $B$ be the group algebra of $S_n$ (say over $\mathbb C$ for simplicity). There is a map $A \to B$ taking the generator to the $n$-cycle $(1 2 \cdots n)$, and this makes $B$ an $A$-bimodule. Then the set you define is a basis for the Hochschild homology $HH_0(A, B)$. (Of course, this isn't combinatorial, so I don't know if this is useful for your purposes.) -http://en.wikipedia.org/wiki/Hochschild_homology<|endoftext|> -TITLE: A remark in Swinnerton-Dyer's paper in Cassels-Frohlich -QUESTION [14 upvotes]: In Swinnerton-Dyer's charming paper "An application of computing to classfield theory", in Cassels-Frohlich, he discusses the genesis of the Birch/Swinnerton-Dyer conjecture and numerical tests of it for the curves $y^2=x^3-dx$. At the end of the paper, he speculates on higher-dimensional analogues of the conjecture, linking Chow groups to L-functions, which in hindsight were essentially correct (Beilinson-Bloch made them precise). Regarding these conjectures, he says that Bombieri and himself had found some evidence for them in the special case of cubic threefolds and the intersection of two quadric hypersurfaces. I know that in the case of cubic threefolds $X$, codimension-two cycles can be related to zero-cycles in the Albanese $A_X$ of the Fano surface of lines in $X$, which (I assume) reduces the conjecture in this case to BSD for $A_X$. But what can be done for the intersection of two quadrics? Does anyone know what he was talking about here? Is this in print in more detail somewhere? - -REPLY [4 votes]: Maybe you already know this! But Miles Reid and Ron Donagi have shown that the intermediate Jacobian of the intersection of two quadrics is the Jacobian of a hyperelliptic curve. See -Donagi's old paper where there is a beautiful generalization of the group law on an elliptic curve. Recent results on such questions all use Nori's general results or Bloch-Srinivas's paper on the diagonall see for example Nagel's paper or Voisin's paper.<|endoftext|> -TITLE: Bounds on number of conjugacy classes in terms of number of elements of a group ? -QUESTION [7 upvotes]: What are bounds on number of conjugacy classes in terms of number of elements of a group ? -(I allowed myself to edit the question in spirit of remarkable answers given to it by Gerry Myerson and Geoff Robinson. Below is original text of the question. (Alexander Chervov) ). - -It's about the first step to find an upper bound to the order of a finite group with h conjugacy classes (right or left) that depends only on h. (h a natural non nul integer). -I have some doubts about the rigor of my proof that I am sharing with you so that you can help me find a likely error or an omited step. -I have attached the scan of my proof to this post. -Many thanks - -REPLY [4 votes]: In the meantime, Barbara Baumeister, Attila Maróti and Hung P. Tong-Viet -have obtained a better lower bound on the number of conjugacy classes of -a group of given order. -- Namely, in 2015 they have proved that for every -$\epsilon > 0$ there exists a $\delta > 0$ such that every group of order -$n \geq 3$ has at least -$$ - \frac{\delta \log_2(n)}{\log_2(\log_2(n))^{3+\epsilon}} -$$ -conjugacy classes. -- See here. In this paper, the authors further -cite a conjecture by Edward A. Bertram which asserts that the number of -conjugacy classes of a finite group is bounded below by the logarithm -for base $3$ of its order. This is basically the best one can expect -to be true, as the Mathieu group ${\rm M}_{22}$ has order -$443520 > 3^{11}$ and only $12$ conjugacy classes. The authors also -show that the conjecture holds for groups with trivial solvable radical.<|endoftext|> -TITLE: Ax–Grothendieck and the Garden of Eden -QUESTION [27 upvotes]: It's an obvious consequence of the pigeonhole principle that any injective function over finite sets is bijective. But there are some similar results in different areas of mathematics that apply to less-finite settings. -In algebraic geometry, the Ax–Grothendieck theorem states (if I have it correctly) that any injection from an algebraic variety over an algebraically closed field to itself is bijective; the standard proof involves some sort of local-global principle together with the same fact over finite fields. -In the theory of cellular automata, the Garden of Eden theorem states that any injective cellular automaton (over an integer grid of some fixed finite dimension, say) is bijective; the standard proof involves again the same fact for finite sets of cells together with a limiting argument that shows that for large enough bounded regions of an unbounded grid, the boundary of the region has negligible effect compared to the interior. -Is there some way of viewing these three injective-bijective statements (or others) as instances of a single more general phenomenon? - -REPLY [3 votes]: In the theory of von Neumann algebras, there is a similar phenomenon. -Let $M$ be a type $II_1$-factor. That is, it's an infinite dimensional von Neumann algebra with center $\mathbb C$, and with an (everywhere defined) trace $tr:M\to \mathbb C$. -Then there is a complete invariant of $M$-modules called the von Neumann dimension. -This invariant takes values in $\mathbb R_{\ge 0}\cup \{\infty\}$ and can take any value in that set. It has the property that any isometric map $H_1 \to H_2$ between modules of the same dimension is actually a unitary isomorphism (except if the von Neumann dimension is $\infty$, in which case, that's not true). -In particular, if $H$ is an $M$-module of finite von Neumann dimension, then any isometry (not assumed to be surjective) is actually a unitary isomorphism.<|endoftext|> -TITLE: Polynomial roots and convexity -QUESTION [45 upvotes]: A couple of years ago, I came up with the following question, to which I have no answer to this day. I have asked a few people about this, most of my teachers and some friends, but no one had ever heard of the question before, and no one knew the answer. -I hope this is an original question, but seeing how natural it is, I doubt this is the first time someone has asked it. -First, some motivation. Take $P$ any nonzero complex polynomial. It is an easy and classical exercise to show that the roots of its derivative $P'$ lie in the convex hull of its own roots (I know this as the Gauss-Lucas property). To show this, you simply write $P = a \cdot \prod_{i=1}^{r}(X-\alpha_i)^{m_i}$ where the $\alpha_i~(i=1,\dots,r)$ are the different roots of $P$, and $m_i$ the corresponding multiplicities, and evaluate $\frac{P'}{P}=\sum_i \frac{m_i}{X-\alpha_i}$ on a root $\beta$ of $P'$ which is not also a root of $P$. You'll end up with an expression of $\beta$ as a convex combination of $\alpha_1,\dots,\alpha_r$. It is worth mentioning that all the convex coefficients are $>0$, so the new root cannot lie on the edge of the convex hull of $P$'s roots. -Now fix $P$ a certain nonzero complex polynomial, and consider $\Pi$, its primitive (antiderivative) that vanishes at $0:~\Pi(0)=0$ and $\Pi'=P$. For each complex $\omega$, write $\Pi_{\omega}=\Pi-\omega$, so that you get all the primitives of $P$. Also, define for any polynomial $Q$, $\mathrm{Conv}(Q)$, the convex hull of $Q$'s roots. -MAIN QUESTION: describe -$\mathrm{Hull}(P)=\bigcap_{\omega\in\mathbb{C}}\mathrm{Conv}(\Pi_{\omega})$. -By the property cited above, $\mathrm{Hull}(P)$ is a convex compact subset of the complex plane that contains $\mathrm{Conv}(P)$, but I strongly suspect that it is in general larger. -Here are some easy observations: - -replacing $P$ (resp. $\Pi$) by $\lambda P$ (resp. $\lambda \Pi$) will not change the result, and considering $P(aX+b)$ will change $\mathrm{Hull}(P)$ accordingly. Hence we can suppose both $P$ and $\Pi$ to be monic. The fact that $\Pi$ is no longer a primitive of $P$ is of no consequence. -the intersection defining $\mathrm{Hull}(P)$ can be taken for $\omega$ ranging in a compact subset of $\mathbb{C}$: as $|\omega| \rightarrow \infty$, the roots of $\Pi_{\omega}$ will tend to become close to the $(\deg (P)+1)$-th roots of $\omega$, so for large enough $\omega$, their convex hull will always contain, say, $\mathrm{Conv}(\Pi)$. -$\mathrm{Hull}(P)$ can be explicitly calculated in the following cases: -$P=X^n$, $P$ of degree $1$ or $2$. There are only 2 kinds of degree $2$ polynomials: two simple roots or a double root. Using $z\rightarrow az+b$, one only has to consider $P=X^2$ and $P=X(X-1)$. The first one yields {$0$}, which equals $\mathrm{Conv}(X^2)$, the second one gives $[0,1]=\mathrm{Conv}(X(X-1))$. - -Also, if $\Pi$ is a real polynomial of odd degree $n+1$ that has all its roots real and simple, say $\lambda_1 < \mu_1 < \lambda_2 < \dots < \mu_n < \lambda_{n+1}$, where I have also placed $P$'s roots $\mu_1, \dots, \mu_n$, and if you further assume that $\Pi(\mu_{2j}) \leq \Pi(\mu_n) \leq\Pi(\mu_1) \leq\Pi(\mu_{2j+1})$ for all suitable $j$ (a condition that is best understood with a picture), then $\mathrm{Hull}(P)=\mathrm{Conv}(P)=[\mu_1,\mu_n]$: just vary $\omega$ between $[\Pi(\mu_n), \Pi(\mu_1)]$; the resulting polynomial $\Pi_{\omega}$ is always split over the real numbers and you get -$$[\mu_1,\mu_n]=\mathrm{Conv}(P)\subset\mathrm{Hull}(P)\subset -\mathrm{Conv}(\Pi_{\Pi(\mu_1)})\cap -\mathrm{Conv}(\Pi_{\Pi(\mu_n)}) = \\= [\mu_1,\dots]\cap -[\dots,\mu_n]=[\mu_1,\mu_n]$$ - -The equation $\Pi_{\omega}(z)=\Pi(z)-\omega=0$ defines a Riemann surface, but I -don't see how that could be of any use. - -Computing $\mathrm{Hull}(P)$ for the next most simple -polynomial $P=X^3-1$ has proven a challenge, and I can only conjecture what -it might be. -Computing $\mathrm{Hull}(X^3-1)$ requires factorizing degree 4 -polynomials, so one naturally tries to look for good values of $\omega$, -the $\omega$ that allow for easy factorization of $\Pi_{\omega}=X^4-4X-\omega$---for instance, the $\omega$ that produces a double root. All that remains to be done -afterwards is to factor a quadratic polynomial. The problem is symmetric, -and you can focus on the case where 1 is the double root (i.e., $\omega=-3$). -Plugging in the result in the intersection, and rotating twice, you obtain the following superset of -$\mathrm{Hull}(X^3-1)$: a hexagon that is the intersection of three similar isoceles -triangles with their main vertex located on the three third roots of unity $1,j,j^2$ -QUESTION: is this hexagon equal to $\mathrm{Hull}(X^3-1)$? -Here's why I think this might be. -Consider the question of how the convex hulls of the roots of $\Pi_{\omega}$ vary as $\omega$ varies. -When $\omega_0$ is such that all roots of $\Pi_{\omega_0}$ are simple, then the inverse function theorem shows that the roots of $\Pi_{\omega}$ -with $\omega$ in a small neighborhood of $\omega_0$ vary holomorphically $\sim$ -linearly in $\omega-\omega_0$: $z(\omega)-z(\omega_0)\sim \omega-\omega_0$. If however $\omega_0$ is such that $\Pi_{\omega_0}$ -has a multiple root $z_0$ of multiplicity $m>1$, then a small variation of $\omega$ -about $\omega_0$ will split the multiple root $z_0$ into -$m$ distinct roots of $\Pi_{\omega}$ that will spread out roughly as -$z_0+c(\omega-\omega_0)^{\frac{1}{m}}$, where $c$ is some nonzero coefficient. This -means that for small variations, these roots will move at much higher velocities -than the simple roots, and they will constitute the major contribution to the variation of -$\mathrm{Conv}(\Pi_{\omega})$; also, they spread out evenly, and (at least if the -multiplicity is greater or equal to $3$) they will tend to increase the convex hull -around $z_0$. Thus it seems not too unreasonable to conjecture that the convex hull -$\mathrm{Conv}(\Pi_{\omega})$ has what one can only describe as -critical points at the $\omega_0$ that produce roots with -multiplicities. I'm fairly certain there is a sort of calculus on convex sets that -would allow one to make this statement precise, but I don't know see what it could be. -Back to $X^3-1$: explicit calculations suggest that up to second order, the double root $1$ of $X^4-4X+3-h$ for $|h|<<1$ splits in half nicely (here $\omega=-3+h$), and the convex hull will continue to contain the aforementioned hexagon. -QUESTION (Conjecture): is it true that -$\mathrm{Hull}(P)=\bigcap_{\omega\in\mathrm{MR}}\mathrm{Conv}(\Pi_{\omega})$, where -$\mathrm{MR}$ is the set of all $\omega_0$ such that -$\Pi_{\omega_0}$ has a multiple root, i.e., the set of all $\Pi(\alpha_i)$ where the -$\alpha_i$ are the roots of $P$? -All previous examples of calculations agree with this, and I have tried as best I can to justify this guess heuristically. -Are you aware of a solution? Is this a classical problem? Is anybody brave enough to -make a computer program that would compute some intersections of convex hulls -obtained from the roots to see if my conjecture is valid? - -REPLY [12 votes]: This problem has been considered before: -The notes to chapter 4 of -Rahman/Schmeißer: Analytic Theory of Polynomials, Oxford University Press, 2002 -mention this problem, state that conv(P) is a proper subset of hull(P) in general, and give two references: -1) J. L. Walsh: The location of Critical Points of Analytic and Harmonic Functions, -AMS Colloquium Publications Volume 34, 1950, p. 72 -2) E. Chamberlin and J. Wolfe: Note on a converse of Lucas's theorem, Proceedings of the American Mathematical Society 5, 1954, pp. 203 - 205 -I had a look at both of them, the paper of Chamberlin and Wolfe can be obtained online via the AMS for free. -The relevant paragraph in Walsh's book is 3.5.1 (starts on page 71). I state the 4 theorems given there for convenience: -1) conv(P) = hull(P) if P is of degree 1 or 2 -2) conv(P) = hull(P) if P is of degree 3 and its zeros are collinear -3) There exists a P of degree 3, such that conv(P) is a proper subset of hull(P); example $P(z) = z^3 + 1$. -4) There exists a P of degree 4 with real zeros, such that conv(P) is a proper subset of hull(P); example $P(z) = (z^2 - 1)^2$. -Chamberlin and Wolfe prove, that -1) Any point on the boundary of hull(P) is on the boundary of conv($\Pi_\omega$) for some $\omega$. -2) If a side of any conv($\Pi_\omega$) contains a point of hull(P) and only two zeros of $\Pi_\omega$ , counting multiplicities, then P is of degree 1. (Here a side is equal to conv($\Pi_\omega$), if the zeros of $\Pi_\omega$ are collinear.) -3) Vertices of conv(P) need not lie on the boundary of hull(P), example - $\Pi(z) = z^2 (z + 1)(z^2 - 2az + 1 + a^2)$, where a is positive and sufficiently small. -4) Even if P is cubic, hull(P) need not be determined by its primitives with multiple zeros, example P(z) = $4z^3 + 9/2 z^2 + 2z + 3/2$. -While I have corrected what I considered a minor typo in the example $\Pi(z) = z^2 (z + 1)(z^2 - 2az + 1 + a^2)$, I did no thorough proofreading. -Walsh gives no special references to further work in section 3.5.1 and the only reference provided by Chamberlin and Wolfe is to Walsh's book cited above.<|endoftext|> -TITLE: Arnoldi method to compute the dominant eigenvector -QUESTION [6 upvotes]: Hi, everyone! -I have a problem of computing the dominant eigenvector. When I want to approximate the dominant eigenvector of a large sparse matrix via the famous Arnoldi method, I am wondering how to choose the reduced order $k$ (i.e., the number of Arnoldi iterations). I know that the approximation accuracy is relevant to the choice of $k$. As $k$ approaches to $n$ (the dimension of the full space) , the accuracy is very high. Now the problem is that given an accuracy $e$, can we find an approperate $k$ (depending on $e$) such that the difference between my approximate dominant eigenvector via Krylov subspace and the exact one is less than $e$? That's to say, I need to find an a-prior error bound, rather than an a-posterior error bound (however, I only found some a-posterior error bound results in some papers). Could you give me some suggestions? Thanks! - -REPLY [5 votes]: This is a very preliminary answer; my knowledge of these things is rather dated, but I got curious and found something of potential interest, so I thought I might as well share it. -There is a recent paper - -Bellalij, M.; Saad, Y.; Sadok, H. - Further analysis of the Arnoldi process for eigenvalue problems. - SIAM J. Numer. Anal. 48 (2010), no. 2, 393–407. - -(In case you do not have access to the journal there is a technical report that is similar to be found here http://www-users.cs.umn.edu/~saad/reports.html (year 2007). However, the discussion below referes to the article.) -At the end of Section 4.1 (where they analyse the algorithm under some assumptions) one can find the following: - -There does not seem to exist in the nonnormal case any formal theoretical results which - establish the convergence of an approximate pair obtained by the Arnoldi process - toward an exact eigenpair. Establishing the link with classical iterative methods - helps understand the nature of the convergence in some instances where the error - may be quite large early in the process. - -And at the beginning of the conclusions: - -The convergence analysis of the Arnoldi process for computing eigenvalues and eigenvectors is difficult, and results in the nonnormal case are bound to be limited in scope. - -So, it seems to me, that the a priori bounds that you seek in general might not exists/be known. In case you know something on the matrix (in particular, symmetric/hermitian) then the situaton seems to be better. -Of course, I quoted the 'negative' passages of the paper only, and it mainly contains 'positive' results (so, if you are not yet aware of it, might well be worth looking at), but as far as I could see not of the strength you seem to be hoping for.<|endoftext|> -TITLE: Non-degenerate alternating bilinear form on a finite abelian group -QUESTION [15 upvotes]: I asked this question on math.stackexchange yesterday, but nobody has helped so far, and only 44 people have seen it! So I hope people do not mind me asking it here... -Let $A$ be a finite abelian group, and let -$ \psi : A \times A \to \mathbb{Q}/\mathbb{Z} $ -be an alternating, non-degenerate bilinear form on $A$. Maybe I should say what I mean by these words; bilinear means it is linear in each argument separately; alternating means that $\psi(a,a) = 0$ for all $a$; non-degenerate means that, if $\psi(a,b) = 0$ for all $b$, then $a$ must be $0$. - -Why must $A$ have square cardinality? - -I believe it will follow from the following theorem in Linear algebra: -Theorem. Let $V$ be a finite dimensional vector space over a field $K$ that has an alternating, non-degenerate bilinear form on it (from $V \times V \to K$). Then dim $V$ is even. -My idea was to proceed as follows: If the size of $A$ is not square, then for some prime $p$, $A(p)$ is not square, where $A(p)$ means the $p$-primary part of $A$. The original $\psi$ induces a map on $A(p)$ that is non-degenerate, alternating and bilinear. I then wanted to say that $A(p)$ is an $\mathbb{F}_p$-vector space, and then applying the theorem I am done, but this is not true, e.g, $\mathbb{Z}/25\mathbb{Z}$ is not an $\mathbb{F}_5$-vector space. -Any pointers anyone? - -REPLY [27 votes]: Actually, one can show the following stronger result: - -Proposition. Assume that a finite abelian group $A$ admits a non-degenarate, bilinear alternating form $\psi$. Then $A$ has a lagrangian decomposition, i.e. there exists a subgroup $G$, isotropic for $\psi$, such that $$A \cong G \times \widehat{G},$$ where $\widehat{G}$ denotes as usual the group of characters of $G$. In particular, $|A|=|G|^2$. - -Therefore, the elements of $A$ can be written as $(x, \chi)$, with $x \in G$ and $\chi \in \widehat{G}$. Moreover, in such a presentation the form $\psi$ take the following form: $$\psi((x, \, \chi), \, (y, \, \eta))=\chi(y)\eta(x)^{-1}.$$ -An easy proof, by induction on the order of the group, can be found in Lemma 5.2 of A. Davydov, Twisted automorphisms of group algebras, arXiv:0708.2758 -Remark. It is interesting to notice the analogy with symplectic vector spaces. In fact, any symplectic vector space $(V, \omega)$ can be written as $V = W \oplus W^{*}$, where $W$ is a lagrangian (=isotropic of maximal dimension) subspace for $\omega$. In particular, $\dim V = 2 \dim W$. Moreover, with respect to this decomposition, $\omega$ has the following form: $$\omega(x \oplus \chi, \, y \oplus \eta) = \chi(y) - \eta(x).$$ -In the case of finite abelian groups the "dual role" is played by the group of characters, as usual.<|endoftext|> -TITLE: Degeneration of the Hodge spectral sequence -QUESTION [27 upvotes]: Let $f\colon X \to S$ be a smooth proper morphism of schemes. If $S$ is of characteristic zero (i.e., $S$ is a $\mathbb Q$-scheme), then Deligne has shown: - -$R^af_*\Omega^b_{X/S}$ is locally free for all $a,b \geq 0$. -The Hodge-De Rham spectral sequence -$E^{ab}_1 = R^af_*\Omega^b(X/S) \Rightarrow H_{\rm DR}^{a+b}(X/S)$ -degenerates in $E_1$. - -This is known to fail in positive characteristic. Mumford gave examples of smooth projective surfaces over algebraically closed fields. Nevertheless there are several interesting cases of schemes $X \to S$ in characteristic $p > 0$ where I know this to be true: -a. $X$ is an abelian scheme, a relative curve, a global complete intersection in projective space, or a K3-surface over $S$. -b. $X$ is a smooth projective toric variety over a field. -c. There is also a criterion of Deligne and Illusie which in particular shows 1. und 2. to hold if $\dim(X/S) < p$ and $X$ can be lifted to $W_2(S)$. -Question: What are other examples in positive characteristic, where 1. and 2. hold? -There is also a variant of the result of Deligne for logarithmic schemes. In particular I would be also interested for examples where the logarithmic analogue of 1. and 2. hold. -ADDITION: I am taking the risk to name two examples of smooth projective schemes over a field, where I would not be too surprised if (1. and) 2. hold, but where I know of no results: -d. $X$ is Calabi-Yau (i.e., its canonical bundle is trivial). -Edit: As Torsten Ekedahl pointed out below this definition of "Calabi-Yau" is not the "right" one (not even in char. $> 2$ as I wrote in an earlier edit) and does not imply in general that the Hodge-De Rham spectral sequence degenerates. -e. $X$ is $G$-spherical for a reductive group $G$ (i.e., $X$ carries a $G$-action such that there exists a dense $B$-orbit, where $B$ is a Borel subgroup of $G$). -Edit: Again one might to have exclude some small primes depending on the Dynkin type of $G$. - -REPLY [11 votes]: [I misunderstood Torsten Ekedahl's earlier comment. I'm reverting the lemma -to its original form which was a bit stronger.] -Since the question seemed to resonate with me, I've been thinking about this on and off (but mostly off) for a couple of days now. Here's what I've come up with. -What seems to make the example of complete intersections work is the fact that -the Hodge numbers can be computed by formulas independent of the characteristic -(a standard generating function can be found in SGA7, exp XI). -Here's the underlying principle. - -Lemma. - Suppose that $D$ is the spectrum of a mixed characteristic DVR with closed point $0$ and - generic point $\eta$. Let $\mathcal{X}\to D$ be a smooth projective family such that - $$\dim H^q(\mathcal{X}_0,\Omega_{\mathcal X_0}^p)= \dim H^q(\mathcal{X}_\eta,\Omega_{\mathcal X_\eta}^p)$$ - for all $p,q$. Then Hodge to De Rham degenerates on the closed fibre. - -Proof. It degenerates on $\mathcal{X_\eta}$ by Hodge theory. This plus semicontinuity -implies -$$\dim E_1(\mathcal{X}_0)\ge \dim E_\infty(\mathcal{X}_0)\ge \dim E_\infty(\mathcal{X}_\eta) -=\dim E_1(\mathcal{X}_\eta)=\dim E_1(\mathcal{X}_0)$$ -This can be used to check degeneration for the following cases: -Ex 1. Complete intersections in projective spaces as noted already. -Ex 2. Products of smooth projective curves and complete intersections. Use -Kuenneth and the fact that curves lift into characteristic $0$ (the obstruction -lies in $H^2$ of the tangent sheaf; or see Oort, Compositio 1971). -Ex 3. Certain cyclic branched covers of projective space, and more generally -certain hypersurfaces in weighted projective space. I'm too lazy to say -what "certain" means exactly. But a careful reading of Dolgachev's notes -on weighted projective spaces ought to yield something more precise.<|endoftext|> -TITLE: "Résumé de cours" by Jacques Tits -QUESTION [7 upvotes]: I have been reading a number of papers by Jacques Tits (mostly written in the second half of 1980s) and in them he frequently refers to following publications of his: - -Résumé de cours, Annuaire du collège de France, 81e année (1980-1981), 75-86. -Résumé de cours, Annuaire du collège de France, 82e année (1981-1982), 91-105. - -Unfortunately I have not been able to locate these two papers (internet, library, faculty). Any help with finding these two would be immensely appreciated. - -REPLY [5 votes]: The "résumés de cours" of Tits are now published by the SMF, see -here.<|endoftext|> -TITLE: Does de Branges's theorem extend to several variables? -QUESTION [12 upvotes]: Consider injective homolomorphic functions $f:\mathbb D\to \mathbb C$ on the unit disk $|z|\leq 1$, normalized by the conditions $f(0)=0$ and $f'(0)=1$. -Thus for $|z|\leq 1$ we have $ f(z)=\sum_{k=0}^{\infty} a_k z^k $ with $a_0=0$ and $a_1=1$. -Ludwig Bieberbach conjectured in 1916 and Louis de Branges proved in 1984 that for all $k \in \mathbb N$ the inequality $|a_k|\leq k$ holds. -Question Is there an analogue to de Branges's result for functions defined on a suitable domain (polydisk, ball,...?) in $\mathbb C^n$ for $n\geq 2$ ? - -REPLY [14 votes]: Such a result would have to be quite different in several variables, because the holomorphic automorphism group of $\mathbb{C}^n$ is very big when $n \geq 2$. For injectivity, we need to look at equidimensional mappings $F$ from the domain (whatever it may be), and into $\mathbb{C}^n$ say. For simplicity, choose $n = 2$, which already contains the main features of the general case. Any mapping $\psi$ of the form $(z,w) \mapsto (z,w + h(z))$ (a "shear") with $h$ an arbitrary entire function is holomorphic everywhere on $\mathbb{C}^2$ and injective. Thus -$\psi \circ F$ is also an injective holomorphic mapping from the domain into $\mathbb{C}^2$. Assuming the domain contains the origin and that $F$ is normalized (taking the origin to the origin and having the identity as derivative at the origin), we can ensure that $\psi \circ F$ is also normalized by choosing the entire function $h$ to have a double zero at the origin. The freedom of choosing $h$ implies that the power series coefficients of the mappings in the normalized family of injective holomorphic mappings have no universal bounds over the family; unlike in the one-dimensional case, the normalized family is not compact. -The key difficulty is this: In one variable the automorphism group of the target $\mathbb{C}^1$ consists of mappings of the form $z \mapsto az + b$, and we can mod out this automorphism group simply by fixing $f(0)$ and $f^{\prime}(0)$, and then it turns out that injectivity implies compactness. When $n \geq 2$ the automorphism group is enormous (a dense subgroup of it was determined explicitly by E. Andersén and L. Lempert, so it is known in a sense), thus a linear normalization is very inadequate, and we should normalize by the whole automorphism group. But it is not clear what it means in practice to mod out by the automorphism group, and whether injectivity would imply compactness after moding out when $n \geq 2$. If we obtain a compact family, there will exist sharp bounds on all coefficients, of course, though we might not be able to establish them. -A precise question (though in an unexplicit form) would be: When can we write the family of all injective holomorphic mappings from a domain $\Omega$ into $\mathbb{C}^n$ by composing $\mathrm{Aut}(\mathbb{C}^n)$ with a compact family? Trivially we can do it when -$\Omega = \mathbb{C}^n$ by choosing the compact family to consist of the identity map alone. -If one is willing to impose strong geometric conditions (e. g. starlikeness) on the injective mappings, one can get compactness. See the book Geometric function theory in one and several variables by I. Graham and G. Kohr for example. -ADDED: Having thought about this a little more, I believe it unlikely that for $n \geq 2$ compactness can hold for $F : \mathbb{B}_n \rightarrow \mathbb{C}^n$ even after moding out by the automorphism group of $\mathbb{C}^n$. For generic domains have only the identity automorphism, and so we could replace $\mathbb{C}^n$ by a slightly smaller domain $D$ with only the identity automorphism and look at injective holomorphic mappings $F : \mathbb{B}_n \rightarrow D$. Since $D$ is nearly as "roomy" as $\mathbb{C}^n$ the family should be noncompact, but obviously cannot be made compact by moding out by automorphisms. So moding out by automorphisms seems to be an "accidental" device, so to speak, and unlikely to work when $n \geq 2$. I would support this conclusion as follows: -One way to see that the family of injective holomorphic functions $f : \mathbb{D} \rightarrow \mathbb{C}$ with $f(0) = 0$ and $f'(0) = 1$ is compact is to apply the Schwarz Lemma to the inverse $f^{-1}$ to see that $f$ omits a point on the circle $|w| = 1$. Since the omitted set of $f$ on the unit sphere is a continuum containing $\infty$, $f$ also omits a point on the circle $|w| = 2$. A well known theorem states that if a family of holomorphic functions omits three points on the Riemann sphere, then it is precompact ("normal" in complex analysis parlance). Here we have three points, but two are "movable" (on the circles) and the third is fixed (at infinity) and a routine extension of the theorem mentioned shows that the family is still precompact, because the three omitted points stay uniformly away from each other. Since the normalized family of injective holomorphic functions $f$ is closed by a theorem of Hurwitz, the family is compact. -Now let us see what this yields for injective holomorphic mappings $F : \mathbb{B}_n \rightarrow D$ with $F(0) = 0$ and $F'(0) = I$ (we assume $0 \in D$) for $n \geq 2$. We can still apply the Schwarz Lemma (in several variables) to the inverse $F^{-1}$ to conclude that $F$ omits a point on the sphere $|w| = 1$, so that the omitted set of $F$ is an unbounded continuum with a point on that sphere. But this information is hardly strong enough to conclude that the family is precompact. The higher-dimensional analogue of the theorem on three omitted points is that if we remove $2n+1$ hyperplanes in general position from $\mathbb{C}^n$ then any family of holomorphic mappings into what remains is precompact. So it seems that we have removed much too little when $n \geq 2$.<|endoftext|> -TITLE: Polynomial with Galois Group $D_{2n}$ -QUESTION [8 upvotes]: How does one construct a polynomial with Galois Group $D_{2n}$? A general method would be preferable or if that's impractical then an example of it being done for any n would be appreciated. -Thanks! - -REPLY [8 votes]: One general method proceeds by making use of invariant polynomials. Let $G$ be a candidate Galois group for an irreducible polynomial of degree $n$ over a field $F$ so that in particular $G$ has a transitive permutation action on $n$ objects $r_{i}$ which we identify with the roots of the polynomial. Basic Galois theory then tells us that any polynomial in the $r_{i}$ which is invariant under the action of $G$ must then lie in $F$. Thus $G$ stabilizes the invariant polynomial ring $F[r_{1}, \cdots, r_{n}]^{G}$. -Now, since $G$ is a subgroup of $S_{n}$ the invariant ring includes the elementary symmetric polynomials $\sigma_{i}$ defined as -$ -\sigma_{1}=r_{1}+r_{2}+\cdots +r_{n}, -$ -$ -\sigma_{2}=r_{1}r_{2}+r_{1}r_{3}+\cdots +r_{n-1}r_{n}, -$ -$\cdots = \cdots ,$ -$\sigma_{n} = r_{1}r_{2}\cdots r_{n}. $ -On the other hand in the splitting field, a polynomial with roots $r_{i}$ can be completely factored as -$\prod_{i}(z-r_{i}) = z^{n}-\sigma_{1}z^{n-1}+\sigma_{2}z^{n-2}+\cdots + (-1)^{n}\sigma_{n}.$ -To construct a polynomial with Galois group $G$ we can then simply choose the $\sigma_{i}$ to be consistent with whatever relations occur in the invariant ring and write a polynomial as above. In general this leads to a polynomial whose Galois group is a subgroup of $G$, but provided we choose the invariants sufficiently generically the Galois group will in fact be $G$ itself. In general this method works well for groups of small order where the invariant rings are managable. -Now let us apply this to produce an example of a degree four polynomial over $\mathbb{Q}$ with Galois group $D_{4}$. Up to a shift in the indeterminate we may assume that this polynomial takes the form -$ -p(z)=z^{4}+\sigma_{2}z^{2}-\sigma_{3}z+\sigma_{4}. -$ -In other words without loss of generality we may assume that the sum of the roots of $p(z)$ vanishes. -An elementary problem in the theory of finite group representations shows that the invariant ring of $D_{4}$ acting as permutation on four objects $r_{i}$ subject to the constraint that $\sum_{i}r_{i}=0$ is generated by four objects $\alpha, \beta, \chi, \lambda$ subject to the single relation $\alpha \lambda =\chi^{2}$. Thus the relevant invariant polynomial ring is simply -$ -\mathbb{Q}[r_{1}, r_{2}, r_{3}, -r_{1}-r_{2}-r_{3}]^{D_{4}}\cong \mathbb{Q}[\alpha, \beta, \chi, \lambda]/\langle \alpha \lambda=\chi^{2}\rangle. -$ -Then the symmetric polynomials are expressed in terms of the generators of the invariant ring as -$ -\sigma_{2}=-\frac{1}{8}(\beta+\alpha), -$ -$\sigma_{3}= \frac{\chi}{16},$ -$\sigma_{4}= \frac{1}{256}((\alpha-\beta)^{2}-\lambda).$ -Thus any polynomial with Galois group $D_{4}$ can be written by choosing arbitrary $\alpha, \beta, \chi, \lambda$ subject to the single constraint in the invariant ring and plugging into the above. For example, one solution with integer coefficients is given by -$ -p(z)=z^{4}+z^{2}+2z+1. -$<|endoftext|> -TITLE: Advances and difficulties in effective version of Thue-Roth-Siegel Theorem -QUESTION [16 upvotes]: A fundamental result in Diophantine approximation, which was largely responsible for Klaus Roth being awarded the Fields Medal in 1958, is the following simple-to-state result: -If $\alpha$ is a real algebraic number and $\epsilon > 0$, then there exists only finitely many rational numbers $p/q$ with $q > 0$ and $(p,q) = 1$ such that -$$\displaystyle \left \lvert \alpha - \frac{p}{q} \right \rvert < \frac{1}{q^{2 + \epsilon}}$$ -This result is famous for its vast improvement over previous results by Thue, Siegel, and Dyson and its ingenious proof, but is also notorious for being non-effective. That is, the result nor its (original) proof provides any insight as to how big the solutions (in $q$) can be, if any exists at all, or how many solutions there might be for a given $\alpha$ and $\epsilon$. -I have come to understand that to date no significant improvement over Roth's original proof has been made (according to my supervisor), and that the result is still non-effective. However, I am not so sure why it is so hard to make this result effective. Can anyone point to some serious attempts at making this result effective, or give a pithy explanation as to why it is so difficult? - -REPLY [2 votes]: There have been some effective results on Roth's theorem -See this work by Luckhardt from 1989: -H. Luckhardt: Herbrand-Analysen zweier Beweise des Satzes von Roth: Polynomiale Anzahlschranken, - The Journal of Symbolic Logic, vol. 54 (1989), pp. 234–263. -https://www.cambridge.org/core/journals/journal-of-symbolic-logic/article/herbrandanalysen-zweier-beweise-des-satzes-von-roth-polynomiale-anzahlschranken/2C2B71C11AC25B06B0F1FD0E7F08D936<|endoftext|> -TITLE: Slices of infinity sheaves -QUESTION [5 upvotes]: I know from classical category theory that if $C$ is a small category and $X$ is a presheaf, that there is a canonical equivalence $$Set^{C^{op}}/X \simeq Set^{\left(C/X\right)^{op}},$$ where $C/X$ is the category of elements of $X$ (i.e. the Grothendieck construction of $X$). Moreover, if $C$ carries a Grothendieck topology, this statement is true for sheaves, where $C/X$ inherits a canonical topology from $C$. -Can I make a similar statement if I go to infinity sheaves, and if so, does anyone have a reference? Thanks. If someone knows a way of proving this model theoretically, that would also be nice. - -REPLY [2 votes]: This is Corollary 5.1.6.12 in HTT. Somehow I overlooked this.<|endoftext|> -TITLE: Examples of naturally occurring Quadratic forms or quadrics. -QUESTION [9 upvotes]: I am always fascinated when a quadratic form (or a quadric) arises naturally. I have -some elementary examples, but most of all, I want to learn more examples. I hope this question isn't considered too vague for MO. Most forms I list are really -elementary, and all are finite dimensional. -I got most of the following examples from M.Berger, Geometry I & II, and from the truly beautiful book "Eléments de géométrie : actions de groupes" by french author Rached Meinmné. -$(0)$ the discriminant on the affine space of unitary degre 2 polynomials -$(i)$ the determinant on endomorphisms of a 2 dimensional vector space, and -$\mathrm{Tr}^2-4\mathrm{det}$ -$(ii)$ the radical on the space of quadratic forms on a 2 dimensional vector space, -and the isotrope cone (not sure about the name, degenerate cone?). -$(iii)$ the family of hermitian forms (built from the Wronskian) on the solution -space of the discrete Schroedinger equation that allow one to show the existence of -right and left side $L^2$ solutions, and the Weyl m function. -$(iv)$ If $\Delta$ is any $2$ dimensional complex vector space, then -$\mathrm{Herm}(\Delta)$, the real vector space of hermitian forms on $\Delta$, -carries a natural quadratic form obtained by constructing an essentially unique -morphism $\rho$ from $\mathrm{Herm}(\Delta)$ to -$\mathrm{Hom}(\Delta\oplus\overline{\Delta})$ such that for all -$h\in\mathrm{Herm}(\Delta),~\rho(h)^2$ is proportional to $\mathrm{Id}$, the proportionality defining the quadratic form. Here, $\rho$ only depends on a choice of -a nonzero element $\omega\in\Lambda^2\Delta^*$. -$(v)$ If $V$ is a 4 dimensional vector space, then $\Lambda^2 V$ carries the natural -quadric $Q(v)=v\wedge v$ where $\Lambda^4 V$ is identified with the underlying -field, which vanishes exactly when $v$ comes from the canonical map -$\mathrm{Gr}(2,V)\rightarrow P\Lambda^2V$. -I remember reading about one on the space of circles, but I forgot the details. What other examples of natural quadratic forms are there? - -REPLY [2 votes]: Binary quadratic forms arise in nature as norm forms for a quadratic field. This point of view has various consequences in number theory. - -For a fixed negative discriminant (the definite case), Gauss discovered that the quadratic forms (or their $\mathrm{SL}_2(\mathbb{Z})$ equivalence classes) can be composed. This led him to the phenomenon of the ideal class group before ideals were invented by Kummer and Dedekind. Besides in the ideal class group for more general number fields, Gauss's composition law has found an extension in Bhargava's higher composition laws. These are based on the representation theory of arithmetic groups ($\mathrm{SL}_2(\mathbb{Z})$ and its generalizations), in which regard they are natural structures in themselves. They have striking applications to old problems regarding mean asymptotics of Selmer ranks of elliptic curves, the $3$-parts of class groups of quadratic fields, etc. -The Epstein zeta function takes the shape $\zeta_Q(s) := \sum_{\mathbf{n} \neq \mathbf{0}} Q(\mathbf{n})^{-s}$, for a given signature $(d,0)$ quadratic form $Q$. It has all the right analytical properties (meromorphic with simple pole at $s = 1$ and a functional equation relating $s \leftrightarrow 1-s$), allowing to decompose the zeta function of an imaginary quadratic field over a set of representatives $Q$ for the class group. This has consequences for the arithmetic of these fields, beautifully developed in Siegel's Lectures on Advanced Analytic Number Theory (Tata Institute lecture series, 1961). -For $d = 2$, $\zeta_Q(s)$ is in effect an Eisenstein series ($|mz+n|^2$ being a binary quadratic form in $m,n$), which is a natural structure all over mathematics, being a continuum of modular forms in the spectral resolution of the hyperbolic Laplacian. Siegel apparently had much interest in the conceptual role played in number theory by the higher rank quadratic forms and their Epstein zeta function. Much of his work was put on representation theoretic footing in Weil's 1964 paper Sur certains groupes d'operateurs unitaires. Michael Berg's book, The Fourier-Analytic Proof of Quadratic Reciprocity, is a terrific introduction to these ideas.<|endoftext|> -TITLE: semisimplicity of braid reps? -QUESTION [5 upvotes]: Here's something I really feel I should know, but do not: -Let $q$ be some sufficiently nice complex number (just pretend we're working over $\mathbb Q(q)$, for example), and $V$ some simple representation of $U_q(\mathfrak g)$. Then you have the usual representation of the braid group $B_n$ on $V^{\otimes n}$. The question is: is this representation semisimple? -I haven't been able to find a direct reference to the problem, but it really sounds like it should be known so I thought I'd check here. - -REPLY [3 votes]: A complete analysis of this is given in the paper by Orellana-Ram. Actually, they consider the action of the affine braid group on $M\otimes V^{\otimes n}$, but you can recover your case by taking $M$ to be the trivial module. I believe that this is is a semisimple representation (assuming you mean $V$ to be finite dimensional), but in any case it is all spelled out in the paper above.<|endoftext|> -TITLE: Generic filter over $V$ -QUESTION [5 upvotes]: I re-read Jechs chapter about forcing, and got a question. There he characterizes a (what he calls) modern way to make the forcing argument legitimate which (I think) goes like this: -It is pointed out there that, in order to establish the consistency of a statement $\varphi$ relative to ZFC, it is sufficient to exhibit a complete Boolean algebra $B$ such that the Boolean value of $\varphi$ in the Boolean-valued model $V^B$ is not zero, i.e. $|| \varphi|| = p \ne 0$. -This fact gives us an alternative approach to forcing, which avoids the assumption of a transitive, countable model $M$ of ZFC, in order to construct the generic extension $M[G]$. -Nevertheless one hardly sees a proof where the Boolean value of an interesting $\varphi$ is really evaluated because it is more easy to pretend that for the p.o. $P \quad$there exists a generic $G$ over $V$, then build $V[G]$ and show that $V[G] \models \varphi$, which is equivalent to the existence of a $p \in G$ such that $p \Vdash \varphi$, which implies that $||\varphi|| \ne 0$. -My problem with this approach is that a generic $G$ over $V$ cannot exist whenever $P$ satisfies the following property: For every $p \in P$ there exist $q \le p, r \le p$ such that $q$ and $r$ are incompatible. -(To see this it suffices to realize that for every filter $F$ on $P$ the set {$ p \in P \quad : p \notin F$} is dense, thus a generic $G$ leads to a contradiction). The existence of a generic $G$ does not change the Boolean value of $\varphi$, so I think that the argumentation in the break above remains valid although $G$ cannot exist. -However it seems strange to me that one assumes the existence of thing that must not exist, to faciliate calculation. One can say that this resembles the complex numbers but to me adding a new element to the reals is way weaker than adding an element that actually must not exist to the universe. -So my question is: is the just described strategy of showing the relative consistency of a statement really valid? - -REPLY [12 votes]: The existence of a generic $G$ over $V$ is indeed impossible in $V$ (for nontrivial forcing notions), but it has truth value 1 in appropriate Boolean-valued models. In more detail: If $P$ is a partially ordered set (to be used as a notion of forcing) and $B$ is the complete Boolean algebra of regular open subsets of $P$, then the following paragraph is true (i.e., has truth value 1) in the Boolean-valued model $V^B$: -There is a transitive class $\check V$ that contains all the ordinals and satisfies all the sentences true in the original ground model $V$ (so it can serve as a "copy" of $V$ in $V^B$). There is a subset $G$ of $P$ generic over $\check V$. Every set is the value of some forcing name in $\check V$ with respect to $G$. -Together, these say that, with truth value 1, the universe (of $V^B$) is a $P$-generic extension $\check V[G]$ of the ground model. So when people pretend to move to a (nonexistent) $P$-generic extension of $V$, a correct interpretation of this is that they move to $V^B$ and that whatever they assert about $V[G]$ is really asserted to have truth value 1 in this Boolean-valued model. -(Technicalities: "All the sentences true in the ground model $V$" should really be specified by a theory $T$, which should include a name (or at least a definition) for $P$ so that one can talk, in $V^B$, about the $P$ that lives in $\check V$.)<|endoftext|> -TITLE: Complex Hypersurface in Complex Projective Space -QUESTION [14 upvotes]: Apparently 2 smooth complex hypersurface in complex projective space that have the same degree -are diffeomorphic. Does anyone know where the proof of this can be found ? Is there a counter example for symplectic manifolds? - -REPLY [5 votes]: There is a proof of this result in Gompf and Stipsicz's book "4-Manifolds and Kirby Calculus": See Claim 1.3.11.<|endoftext|> -TITLE: What is the "Physically Consistent" proper subset of arithmetic? -QUESTION [10 upvotes]: Suppose 1st-order arithmetic is inconsistent along with Voevodsky http://video.ias.edu/voevodsky-80th. -It nevertheless remains true that when you have 2 apples and 2 apples, you have 4 apples. Preforming an experiment gives you the result of an experiment, which cannot be inconsistent. So there is a subset of arithmetic that is "necessarily" consistent, given the notion (maps) that arithmetic models reality. The question is, what is this "physically consistent" proper subset of arithmetic? -The second question is, what happens if the physical theory is quantum field theory, where quanta loose their individual identity or "primitive thisness"? - -REPLY [7 votes]: Presburger arithmetic which is the first order theory of natural numbers with addition has been proven to be consistent by Mojżesz Presburger. My reference for this is the wikipedia article on Presburger arithmetic.<|endoftext|> -TITLE: Is every field extension of an ultrafield an ultrafield? -QUESTION [10 upvotes]: Let $K=\lim(K_{i})$ be an ultrafield (over a non-principal ultrafilter), and let $K\hookrightarrow K'$ be a field extension of $K$. -When the field $K'$ is finite over $K$ it is also an ultrafield by Łoś's theorem. What can be said when the trascendence degree of $K'$ over $K$ is infinite? - -REPLY [5 votes]: I have mentioned in a comment that the accepted answer is not correct, although the argument is correct when the index set is countable. -Here's a result with no algebraic closedness assumption, which actually shows that the algebraic structure of the multiplicative group is enough to set up the argument, and also purports to clarify what we should allow on the index set, or even on the ultrafilter. I'm starting with the countable case. - -Proposition. Let $K$ be a field. Then the field $K(t)$ is not isomorphic to any nonprincipal countable ultraproduct of fields. Actually, the multiplicative group $K(t)^*$ is not isomorphic to any nonprincipal countable ultraproduct of groups. - -Since $K(t)^*\simeq K^*\times\mathbf{Z}^{(J)}$, where $\mathbf{Z}^{(J)}$ is the free abelian group on the nonempty set $J$ of monic irreducible polynomials, we have $\mathrm{Hom}_{\mathrm{Group}}(K(t)^*,\mathbf{Z})\neq\{0\}$. Hence the proposition follows from the following lemma. -Given a family $(G_n)$ of groups, I denote by $\prod^\star_nG_n$ the near product, i.e., the quotient of the product $\prod_n G_n$ by the finitely supported (aka restricted, aka finitely supported) product $\bigoplus_n G_n$. - -Lemma. Let $(G_n)$ be a sequence of abelian groups. Then $\mathrm{Hom}(\prod^\star_n G_n,\mathbf{Z})=\{0\}$. In particular, for every nonprincipal ultrafilter $\sigma$ on $\mathbf{N}$, we have $\mathrm{Hom}(\prod^\sigma_n G_n,\mathbf{Z})=\{0\}$, where $\prod^\sigma_n G_n$ is the ultraproduct with respect to $\sigma$. - -Proof: This is the classical Specker proof for $G_n=\mathbf{Z}$. The proof is an immediate adaptation. Consider a homomorphism $\prod^\star_nG_n\to\mathbf{Z}$, vanishing on $\bigoplus_nG_n$. For each $n$, choose a Bézout relation $2^na_n+3^nb_n=1$. -For $x=(x_n)_n\in\prod_nG_n$, write $p^\omega y=(p^ny)_n$. Denote by $x\mapsto \bar{x}$ the quotient map $\prod\to\prod^\star$. Since for every $x$ and for $p=2,3$, the element $\overline{p^\omega x}$ is $p$-divisible (i.e., has $p^n$-roots for all $n$), so is its image by $f$, which forces $f\big(\overline{2^\omega x}\big)=f\big(\overline{3^\omega x}\big)=0$. Hence, for every $y=(y_n)_n\in\prod_nG_n$, we have $y=2^\omega ay+3^\omega by$ and hence $f(\overline{y})=0$. So $f=0$. -[Note: this also follows from the $G_n=\mathbf{Z}$ case, by a simple composition argument.] -The second assertion follows from the first since $\prod^\sigma_n G_n$ is a quotient of $\prod^\star_n G_n$. - -Let me now considerably relax the countability assumption. Let me say that a set $I$ is reasonable if every ultrafilter on $I$, stable under countable intersections, is principal. (If there's a standard terminology I'd be happy to change.) -This only depends on the cardinal of $I$. It's consistent that every set is reasonable. If there's a non-reasonable cardinal, there's a minimal one, which is known as smallest measurable cardinal. All "reasonably small" cardinals are reasonable; obviously $\omega$ is, and notably $\alpha$ reasonable implies $2^\alpha$ reasonable and in particular sets of cardinal $\le \mathfrak{c}$ (continuum), $\le 2^{\mathfrak{c}}$, etc, are reasonable. -The point is that the proposition and lemma above still hold when the index set is supposed to have reasonable cardinal. - -Lemma'. Let $I$ be a reasonable set, and $(G_i)$ a family of abelian groups. Then $\mathrm{Hom}(\prod^\star_{i\in I} G_i,\mathbf{Z})=\{0\}$. -Remark: using a nonprincipal ultrafilter stable under countable intersections, conversely if $I$ is not reasonable then $\mathrm{Hom}(\prod^\star_{i\in I} \mathbf{Z},\mathbf{Z})\neq\{0\}$, since a nontrivial homomorphism is obtaining by taking the limit along such an ultrafilter. - -Proof: again, this follows from the case $G_i=\mathbf{Z}$: let $f:\prod_iG_i\to\mathbf{Z}$ vanish on $\bigoplus G_i$. If $f$ is nonzero, say on some element $(g_i)_i$, consider the homomorphism $\mathbf{Z}^I\to\prod_i G_i$ mapping $(n_i)_i$ to $(n_ig_i)_i$; it maps $\mathbf{Z}^{(I)}$ into $\bigoplus G_i$ and we can conclude from the fact that $\mathrm{Hom}(\prod^\star_{i\in I}\mathbf{Z},\mathbf{Z})=\{0\}$. -I don't know a reference, so let me provide a proof. -Choose, by contradiction, $f\neq 0$ as above. For every subset $J$ of $I$, denote by $R_J(f)$ the restriction of $f$ to $\prod_{j\in J}G_j$. Let $\mathbf{F}$ be the set of $J$ such that $R_J(f)\neq 0$. Clearly $J'\subset J\notin\mathcal{F}$ implies $J'\notin\mathcal{F}$. Let us say that $J\in\mathcal{F}$ is simple if it is not disjoint union of two elements of $\mathcal{F}$. -Claim: There exists a simple $J\in\mathcal{F}$. -Proof of claim: otherwise, one chooses $X'_0\in\mathcal{F}$, partition it as $X'_0=X_0\sqcup X'_1$ with $X_0,X'_1\in\mathcal{F}$, then partition $X'_1=X_1\sqcup X'_2$, etc, to obtain a partition $(X_n)$ of disjoint elements in $\mathcal{F}$. Then choosing $x_n$ with $f(x_n)\neq 0$, supported in $X_n$, for $u=(u_n)_n\in\mathbf{Z}^{\mathbf{N}}$, we define $f'(u)=f(\sum u_nx_n)$, we obtain a contradiction with the previously proved countable case. -Now choose $J$ simple, and define $\mathcal{U}=\{J'\in\mathcal{F}:J'\subseteq J\}$. Then $\mathcal{U}$ is an ultrafilter on $J$. Furthermore, it is stable under countable intersections: it is enough to show that the set of complements of elements of $\mathcal{U}$ (which is also the complement of $\mathcal{U}$) is stable under countable disjoint unions. This is, again, a straightforward consequence of $\mathrm{Hom}(\prod_{n\in\mathbf{N}}^\star\mathbf{Z},\mathbf{Z})=\{0\}$. - -Corollary: let $K$ be a field. Then the field $K(t)$ is not isomorphic to any nonprincipal ultraproduct of fields indexed by any reasonable (in the above sense) set. More precisely, the group $K(t)^*$ is not isomorphic to any nonprincipal ultraproduct of groups indexed by any reasonable set. - -Again, this is false if one allows non-reasonable cardinals. Indeed, if the ultrafilter is stable under countable intersections, and the field $K$ is countable, then the field $K(t)$ is isomorphic to its own ultrapower with respect to such an ultrafilter. - -Edit: - -Fact. Let $\sigma$ be an ultrafilter on a set $I$. Equivalences: - -Then $\sigma$ is not stable under countable intersections. -There exists a function $u:I\to\mathbf{N}$ such that $\lim_{i\to\sigma}u=\infty$ -There exists a strictly decreasing sequence $(I_n)$ of subsets of $I$, with empty intersection and $I_n\in\sigma$ for all $n$. - - -(An ultrafilter stable under countable intersections is usually called $\omega_1$-complete, or $\sigma$-complete.) -Proof: Suppose 1. So we have a sequence $(J_n)$ with $J_n\in\sigma$ and $J=\bigcap J_n\notin\sigma$. Setting $I_n=J^c\cap \bigcap_{i\le n}J_n$, we obtain 3. Supposing 3 (with $I_0=I$), we obtain 2 by defining $u(i)=\sup\{n:i\in J_n\}$. 2 implies 1 is clear, choosing $I_n=\{i:u(i)\ge n\}$. $\Box$ -Therefore, the ultrafilter being non-$\sigma$-complete (rather than non-principal) is the optimal assumption for the proposition to hold. Namely: - -Proposition: let $K$ be a field. Then the field $K(t)$ is not isomorphic to any non-$\sigma$-complete ultraproduct of fields; more precisely, the group $K(t)^*$ is not isomorphic to any non-$\sigma$-complete ultraproduct of groups. Conversely, if $K$ is a countable field, then the field $K(t)$ is isomorphic to any $\sigma$-complete ultraproduct of itself.<|endoftext|> -TITLE: What are the pillars of Langlands? -QUESTION [31 upvotes]: I had previously asked: -Narratives in Modular Curves -Since then, I've read quite a bit more (but not nearly enough) and I have a few follow up questions about the big picture. As you will soon see, I'm confused about how to think about things, and seeing the big picture will help me a lot in learning the specifics (learning in the dark is difficult!). -As I understand it, the story goes like this. First, one defined for every number field $\zeta_K(s)=\sum_{\mathfrak{a}} \frac{1}{(N\mathfrak{a})^s}$. One then defines a Dirichlet character, and for any such one defines $L(\chi,s)$. Further, for any $1$-dimensional Galois representation, $\rho: Gal(K/\mathbb{Q}) \rightarrow \mathbb{C}$, one defines $L(\rho,s)$. Now, in the $1$-dimensional case, the main two theorems that comprise class field theory are: if $K$ is abelian over $\mathbb{Q}$ with group $G$, then $\zeta_K(s)=\prod_{\rho \in \hat{G}} L(\rho,s)$; and for any such $\rho$ there exists a unique primitive Dirichlet character $\chi$ such that $L(\rho,s)=L(\chi,s)$. So far I follow the story perfectly. -There is also the issue of what if the base field is not $\mathbb{Q}$, which, admittedly, I don't fully have down. -Already in dimension $2$ I have a hard time figuring out what generalizes what corresponding thing from dimension $1$. For Galois representations, one continues to define $L(\rho,s)$ in a similar manner: as the product over $p$ of the characteristic polynomials of the action of the corresponding Frobenius (whenever defined for that $p$! It is still a little murky to me what happens at the bad primes). But now we have modular forms coming in to the picture, and the whole theory of modular curves. So how does this fit in as a generalization of the 1-dimensional case? Here's my best guess, you can tell me if I'm right. For a modular form $f$, one defines the $L$-function for it by the $q$-expansion of $f$: If $f(z)=\sum a(n)e(nz)$ then $L(f,s)=\sum \frac{a(n)}{n^s}$. Then various things that I do not fully understand come into play, claiming things like: $L(s,f)=\prod_{q|N}(1-a(q)q^{-s})^{-1} \prod_{p\not |N} (1-a(p)p^{-s}+f(p)p^{k-1-2s})^{-1}$ (probably just for $f$'s with some property, akin to being primitive). It seems (is this true?) that Hecke theory implies that these $L$'s are ``nice'' in the sense that they generalize Dirichlet $L$ functions. Is this the right way to see it? How? What is the $1$-dimensional analogue of modular functions, and modular curves? -Then I imagine that one has the modularity theorem, one of whose versions is(?) that for every $2$-dimensional Galois representation there's a modular function for which $L(\rho,s)=L(f,s)$. -You will notice that at no point did I talk about the adelic aspect. This is because I don't know where to put it. Is the adelic side easily equivalent to the (Dirichlet characters)-(modular functions) side?(are these two even on the same side?) Is it another pillar with which equivalence is far from trivial with both the Galois representations side AND the (Dirichlet characters)-(modular functions) side? In short -- I'm not sure what the pillars of Langlands are! -Further, let us assume that we have some version of Langlands. Is there a conjectural equivalent form of $\zeta_K(s)=\prod_{\rho \in \hat{G}} L(\rho,s)$ for $K/\mathbb{Q}$ not abelian? - -REPLY [36 votes]: In either the one or two dimensional case, there are two sides: Galois and automorphic. -Let me talk for a moment about the $n$-dimensional situation. -The Galois side involves studying continuous $n$-dimensional representations of the Galois group of -a number field $K$. -There are subtleties here about over what field these representations are defined, which I will return to. For now, let's imagine that they are representations into $GL_n(\mathbb C)$, -hence what are usually called Artin representations. -The automorphic side involves so-called automorphic representations of $GL_n(\mathbb A)$, -where $\mathbb A$ is the adele ring of $K$. These are irreduible representations which appear as Jordan--Holder factors in the -space of automorphic forms, which is, roughly speaking, the space of functions -on the quotient $GL_n(K) \backslash GL_n(\mathbb A)$. (To provide some orientation with regard to this notion: Note that, by Frobenius reciprocity, -any irreducible representation of a group $G$ appears in the space of functions on $G$ --- here I am ignoring topological issues --- and so any irreducible rep. of $GL_n(\mathbb A)$ will appear in the space of functions on $GL_n(\mathbb A)$. On the other hand, appearing in the space of functions on the quotient $GL_n(K)\backslash GL_n(\mathbb A)$ turns out to be a serious restriction --- most representations of $GL_n(\mathbb A)$ don't appear there.) -The idea --- roughly --- is that $n$-dimensional Galois representations will match -with automorphic representations. How does this happen? -Well, $n$-dimensional representations have traces of Frobenius elements, -so to each unramified prime we can attach a number. On the other hand, it turns out -that automorphic representations have essentially canonical generators, which can -be interpreted as Hecke eigenforms, and so (for primes not dividing the conductor -of the representation) we get a Hecke eigenvalue. The matching is given by the rule that traces of Frobenius elements should equal Hecke eigenvalues. -Let's consider the case of dimension $n = 1$: -First, any $1$-dim'l rep'n of the Galois group of $K$ factors through $G_K^{ab}$, while -the space of functions on the abelian group $K^{\times}\backslash \mathbb A^{\times}$ -(which is the adelic quotient considered above in the case $n = 1$) is (by Fourier -theory for locally compact abelian groups, and speaking somewhat loosely) the sum of spaces spanned by characters, so that automorphic representations are just characters of -$K^{\times}\backslash \mathbb A^{\times}$. -So our goal is to match characters of $G_K^{ab}$ with characters of $K^{\times}\backslash -\mathbb A^{\times}$. -Now global class field theory actually does something stronger: it directly relates -$G_K^{ab}$ to $K^{\times}\backslash \mathbb A^{\times}$. In particular, finite order -characters of the latter (which are just Dirichlet characters in the case $K = \mathbb Q$) -correspond to finite order characters of the former, the $L$-functions match, and so on. -Now consider the case $n = 2$: -The quotient $GL_2(K)\backslash GL_2(\mathbb A)$ is not a group, -just a space (and this is the same for any $n \geq 2$). Also, there is no quotient -of the Galois group $G_K$ akin to $G_K^{ab}$ with the property that all $2$-dimensional reps. of $G_K$ factor precisely through that quotient. -In short, unlike in the case $n = 1$, we can't hope to find groups that will be related in some direct manner (unlike in the case $n = 1$, where class field theory gives a direct relation between $G_K^{ab}$ and $K^{\times}\backslash \mathbb A^{\times}$); the relation -really has to be made at the level of representations. -Also, when you consider the quotient $GL_2(K)\backslash GL_2(\mathbb A)$, it is much harder -(in fact, impossible) to separate the archimedean and non-archimedean primes. If you play around and follow up some of the references suggested by others, you will see (in the case $K = \mathbb Q$) that this quotient is related to the upper half-plane and congruence subgroups, and the generating vectors for automorphic representations are precisely primitive Hecke eigenforms (either holomorphic modular forms or else Maass forms). The archimedean prime contributes the upper half-plane, and the finite primes contribute the level of the congruence subgroup and the Hecke operators. -Thus the analogue of a Dirichlet character in the $2$-dimensional case is a Hecke eigenform. -(But it is better to regard the latter as corresponding more generally to idele class characters --- also called Hecke characters or Grossencharacters; these are not-necessarily-finite-order generalizations of Dirichlet characters.) -If you really want to study just Galois represenations with finite image (i.e. Artin representations), on the automorphic side (for $K = \mathbb Q$ and $n = 2$) you should restrict to weight one holomorphic modular forms and Maass forms with $\lambda = 1/4$. (The former are proved to correspond precisely to two-dimensional reps. for which complex conjugation has non-scalar image, while the latter are conjectured to correspond to two-dimensional reps. for which complex conjugation has scalar image.) The other Maass forms -are not supposed to correspond to any Galois representations (just as non-algebraic idele class characters --- those that are not type $A_0$ in Weil's terminology --- don't correspond to anything on the Galois side in class field theory), while higher weight holomorphic modular forms are supposed to corresond to compatible systems of two-dimensional $\ell$-adic Galois representations with infinite image (just as infinite order algebraic Hecke characters correspond to -compatible systems of one-dimensional $\ell$-adic Galois representations --- see Serre's bok on this topic). -As the preceding summary shows, the big picture here is pretty big, and the technical details are quite extensive and involved. There is the added complication that the proofs of what is known in the non-abelian case are very involved, and often use methods that don't play a role in the general story, but seem to be crucial for the arguments to go through. (E.g. Mellin transforms don't play any role in the general theory of attaching $L$-functions to automorphic representations, but in the case $n = 2$, the fact that you can pass from a modular form to its $L$-function via a Mellin transform is often useful and important.) -What I would suggest is that you try to learn a little about Hecke characters beyond the finite order case, perhaps from Weil's original article and also from Serre's book. (Silverman's treatment of complex multiplication elliptic curves may also help.) This will help you become familiar with the crucial idea of compatible families of $\ell$-adic Galois representations, and see how it relates to objects on the automorphic side, in the fundamental $n = 1$ case. -Then, to begin to grasp the $n = 2$ case, you will want to understand how modular forms -give compatible systems of two-dimensional $\ell$-adic representations. The general statement here is due to Deligne, and can be learnt (at least at first) as a black-box. The case of weight 2, which includes the case of elliptic curves over $\mathbb Q$, is easier, and it's possible to learn the whole story (other than proof of the modularity theorem itself) in a reasonable amount of time. The case of weight one is special, and is treated in a beautiful paper of Deligne and Serre. The converse here (that every two-dimensional -Artin rep'n of appropriate type comes from a weight one form) is at least as difficult as the modularity of elliptic curves, and so learning the statement should suffice at the beginning. -And now you confront the difficulties in learning non-abelian class field theory: already -in the two-dimensional case, the theory has some aspects that remain almost entirely conjectural, -namely the conjectured relationship between certain Maass forms and certain two-dimensional -Galois representations. So at this point, you have to choose whether you just want to get some sense of the big picture and the expected story in general (say by looking over Clozel's Ann Arbor article and the recent preprint of Buzzard and Gee), or to begin actually working in the area. If you choose the latter, then the big picture is useful, but only in a limited way: to make progress, it seems that one has to then start learning many techniques that don't seem to be essential to the story from the big picture point of view, -but which are currently the only known methods for making progress (depending on what direction you want to pursue, these include the trace formula, Shimura varieties, the deformation theory of Galois representations, the Taylor--Wiles method, ... ).<|endoftext|> -TITLE: Injections to binary sequences that preserve order -QUESTION [5 upvotes]: Suppose we have a countable set S with a total order. Can we give an injection from S to the set of finite binary sequences that end in all zeros that preserves the ordering? The order on binary sequences is the dictionary ordering (e.g. 001001 <= 01). -For a finite set this is easy: arrange the set in order and assign an increasing sequence of binary sequences. -For the natural numbers this is also easy: send a number n to the sequence that starts with n ones (a similar solution works for negative numbers). -For the rationals this is already a bit more difficult. I believe the following works: Take the Stern-Brocot tree. Start at the root and walk down to the rational number. Every time you go left, write a 0. Every time you go right, write a 1. Finally write another 1. -So an equivalent formulation seems to be: can we arrange S into a binary tree such that the elements are arranged in order from left to right as in the Stern-Brocot tree. -My question is: can this be done for any countable set with a total order? -The question came up in a discussion whether radix sort can be used to sort any set (radix sort can sort binary sequences). - -REPLY [6 votes]: It is easy to prove that for any countable linearly ordered set there is an order preserving injection to the rationals. This can be proven by enumerating the base set and then specifying the values of the mapping by induction. -Since you have a solution for $\mathbb Q$, for other sets just compose the order preserving injection from above with this solution.<|endoftext|> -TITLE: Blow-ups at points in non-general position -QUESTION [5 upvotes]: Is it well known what happens if one blows-up $\mathbb{P}^2$ at points in non-general position (ie. 3 points on a line, 6 on a conic etc)? Are these objects isomorphic to something nice? - -REPLY [2 votes]: I'll just add to Francesco's answer by saying that general position of the points on the plane is equivalent to ampleness of the anticanonical sheaf $\omega_X^{\otimes -1}$. -The key observation is that on a del Pezzo surface, an irreducible negative curve ($C^2 < 0$) must be an exceptional curve (i.e. $C^2 = C\cdot K_X = -1$). This follow from the adjunction formula and the Nakai-Moishezon criterion. -If you blow up 3 colinear points, then the strict transform of the line containing these points will have self-intersection $-2$, which is not allowed by the key observation. Similiarly for the strict transform of a conic through 6 blown-up points. There is one more condition you have to impose: if you blow up 8 points, they cannot lie on a singular cubic with one of the points at the singularity. -If you relax the requirement that $\omega_X^{\otimes -1}$ be ample to just big and nef, then you can have some degenerate point configurations: this time $C^2 = -2$ is allowed, so 3 colinear points ar OK. However, 4 colinear points would not be OK.<|endoftext|> -TITLE: Torus minimizer of Willmore energy -QUESTION [10 upvotes]: It is my understanding that the torus that minimizes the Willmore energy -is not yet known -(this from a sentence in a 2005 paper by John Sullivan). -Willmore conjectured that the Willmore energy for any smooth, -immersed torus is $\ge 2 \pi^2$. -From what I can gather, -this energy is achieved by a standard rotationally round torus, -derived from the Clifford torus. -Assuming the problem remains open, my question is: - -Are there any viable candidates for different torus - Willmore minimizers, or does the evidence point to the one - just mentioned? - -I'm wondering if the problem remains open because all -strange, unlikely alternatives have not been ruled out, -or because there are actually some viable alternatives. -Likely the right reference I am not finding would suffice. -Thanks! - -Addendum (2Sep13). -As Renato Bettiol first pointed out below, the Willmore conjecture has -been solved by -Fernando Marques and -André Neves. They posted a 96-page paper to the arXiv: - -Fernando Marques and André Neves. - "Min-Max theory and the Willmore conjecture." - arXiv:1202.6036 (2012). Updated March 2013. - -        - -        -The Willmore Torus. Image by Tom Banchoff, cited in Morgan article. -The result was hailed in a Huffington Post article by -Frank Morgan: - -Frank Morgan. "Math Finds the Best Doughnut." - Huffington Post, 2 April 2012. - -And there is a very nice description by Dana Mackenzie -in an article that also describes the related resolution of -the Lawson conjecture by Simon Brendle: - -Dana Mackenzie. "What's Happening in the Mathematical Sciences." - Vol. 9. American Mathematical Soc., 2013. (AMS link) - -(7Sep13). One more remark. It remains open what might be the 2-holed or -3-holed Willmore surface of minimal bending energy. Dana says (p.28), - -..., it is difficult to know what even to conjecture about such surfaces. - -REPLY [10 votes]: Just a few days ago, Fernando C Marques and Andre Neves posted a preprint on the arxiv in which they (claim to) provide a complete proof of Willmore's Conjecture. I have no idea how much of it has been verified, especially given it is so recent, but the geometric ideas used (min-max theory for minimal submanifolds) seem very elegant.<|endoftext|> -TITLE: Generalization of plane geometric trees? -QUESTION [5 upvotes]: View a plane tree drawn in $\mathbb{R}^2$ as a joining of geometric (straight) segments at endpoints such that (a) they avoid intersecting one another (except where they share a vertex), and (b) they avoid creating a cycle, which would enclose a positive planar area. -I am interested in the generalization to $\mathbb{R}^3$ as follows. -Join together (flat) polygons, glued edge-to-edge, such that (a) they avoid intersecting one another (except where they share vertices and/or edges), and (b) they avoid enclosing (water-tightly) -a positive volume. -My question is: - -Is there a name for this construct? Has it been studied? - -I am mainly seeking references to any literature on this or related concepts. -Of course there is a generalization to $\mathbb{R}^d$, but I would be happy to learn -of work just generalizing plane trees to ??? in $\mathbb{R}^3$. -I cannot think of what it might be named: open panel structures? -It's come up in my work, and I would be delighted to christen it, but surely it has been -studied...? -Thanks in advance! -Addendum. As Greg Kuperberg kindly explained, the concept I described is -a collapsible complex. It is usually defined for simplicial complexes, but works -as well when the constituents are polytopes rather than simplices, e.g., polygons in $\mathbb{R}^3$. - -REPLY [3 votes]: Also check out the House with One Room notes form Allen Hatcher<|endoftext|> -TITLE: Is a non-compact Riemann surface an open subset of a compact one ? -QUESTION [10 upvotes]: Let $X$ be a non-compact holomorphic manifold of dimension $1$. Is there a compact Riemann surface $\bar{X}$ suc that $X$ is biholomorphic to an open subset of $\bar{X}$ ? -Edit: To rule out the case where $X$ has infinite genus, perhaps one could add the hypothesis that the topological space $X^{\mathrm{end}}$ (is it a topological surface?), obtained by adding the ends of $X$, has finitely generated $\pi_1$ (or $H_1$ ). Would the new question make sense and/or be of any interest? -Edit2: What happens if we require that $X$ has finite genus? (the genus of a non-compact surface, as suggested in a comment below, can be defined as the maximal $g$ for which a compact Riemann surface $\Sigma_g$ minus one point embeds into $X$) - -REPLY [3 votes]: Useful references for your question are Robert Brooks' "Platonic surfaces" and Dan Mangoubi's "Conformal Extension of Metrics of Negative Curvature" (both on arxiv). -I emailed Luca Migliorini requesting his paper. He told me it was basically his undergraduate thesis, published in a defunct italian journal, and that no copy of it remains. In otherwords, utterly useless. -The basic fact on compactifying a riemann surface is this: if $S$ is a finite area riemann surface, then there exists a compact riemann surface $S^c$ and a finite set of points $p_1, \ldots, p_k$ on $S^c$ such that $S^c \setminus {{p_1, \ldots, p_k}}$ is conformally equivalent to $S$. -In Brooks' paper, he states that this riemann surface $S^c$ is unique. However I'll admit to not be convinced of this uniqueness. The expression he uses throughout is "conformally filling punctures" -- a phrase which I think deserves more explanation than is given. -Lemma 1.1 in Brooks is interesting, and justifies the above claim. Of course we know what cusps on riemann surfaces look like. A cuspidal neighborhood $C$ of a Riemann surface can be taken isometric to the quotient of $\{ z\in \mathbb{H}^2: \Im(z)\geq 1/y \}$ by the isometry $z\mapsto z+1$, for some $y>0$. The parameter $y$ gives a measure on the size of the cusp, i.e. gives a geodesic loop homotopic to the puncture with hyperbolic length $y$. So the cusp $C$ is really isometric to the punctured ball of euclidean radius proportional to $1/y$ via the mapping $z\mapsto e^{2\pi i z}$ on the punctured open unit disk $D^\ast$ equipped with the metric $ds^*=\frac{-1}{r \log r} |dz|$. However $ds^*$ blows-up as $r\to 0$ like $1/r$. -Brooks (and afterwords Mangoubi more explicitly) gives, for any $\epsilon>0$, smooth bump functions $\delta$ concentrated at the origin on $D$ such that $e^\delta ds^*$ extends to a smooth metric past the origin and whose curvature remains pinched $-1 \pm \epsilon$. -I am going to include the details of this construction, together with some remarks relating to Donaldson's compactification of algebraic curves (from his book) shortly.<|endoftext|> -TITLE: Toy Models of Quantum Mechanics -QUESTION [33 upvotes]: Do toy models of quantum mechanics help us better understand "regular" quantum mechanics? For example, if we look at quantum mechanics over a finite field $F$ (e.g. $\mathbb{Z}_2$), can this lead to new insights for "regular" quantum mechanics? Or do these toy models just help clarify our understanding of "regular" quantum mechanics without actually providing any new insight? In other words, what is the utility of having toy models of quantum mechanics? - -REPLY [2 votes]: The theory of nonlocal boxes is quite illuminating to the beginner (or expert) trying to get a handle on entanglement. (Google "nonlocal box" or "box world".)<|endoftext|> -TITLE: Question on eigenvalue square root subadditivity -QUESTION [9 upvotes]: ORIGINAL QUESTION -Let $\lambda_{1}\left(\cdot\right)$ be the larger eigenvalue of a -$2\times2$ matrix and $\lambda_{2}\left(\cdot\right)$ the smaller -eigenvalue of a $2\times2$ matrix. Is it true that $$ -\left|\sqrt{\lambda_{1}\left(A+B\right)}-\sqrt{\lambda_{1}\left(B\right)}\right|+\left|\sqrt{\lambda_{2}\left(A+B\right)}-\sqrt{\lambda_{2}\left(B\right)}\right|\leq\sqrt{\lambda_{1}\left(A\right)}+\sqrt{\lambda_{2}\left(A\right)}$$ -for any two $2\times2$ positive-definite symmetric real matrices $A$ and $B$? -EDITED QUESTION (after Mikael de la Salle's original answer) -If $A$ is not positive definite, does one have $$ -\left|\sqrt{\left|\lambda_{1}\left(A+B\right)\right|}-\sqrt{\left|\lambda_{1}\left(B\right)\right|}\right|+\left|\sqrt{\left|\lambda_{2}\left(A+B\right)\right|}-\sqrt{\left|\lambda_{2}\left(B\right)\right|}\right|\leq\sqrt{\left|\lambda_{1}\left(A\right)\right|}+\sqrt{\left|\lambda_{2}\left(A\right)\right|}$$ -for any $2\times2$ symmetric real matrices $A$ and $B$, where $\lambda_{1}\left(\cdot\right)$ is the larger absolute value -eigenvalue and $\lambda_{2}\left(\cdot\right)$ -is the smaller absolute value eigenvalue? -Thanks for any helpful answers. - -REPLY [13 votes]: The answer to both your questions are yes. Let me start with the first question, which more straightforward. -A first remark: since $A$ is positive-definite, $\lambda_i(A+B) \geq \lambda_i(B)$ for $i=1,2$. -(to check this, use the formulas $\lambda_1(X) = \max_{\xi} \langle X\xi,\xi\rangle$ and $\lambda_2(X) = \min_{\xi} \langle X\xi,\xi\rangle$ where the min and max run over all unit vectors $\xi$. This formulas hold whenever $X$ is a symmetric $2 \times 2$ matrix.) -Your question is therefore whether $Tr(\sqrt{A+B})\leq Tr(\sqrt A)+ Tr(\sqrt B)$ for any symmetric positive definite matrices $A$ and $B$, or equivalently $Tr( \sqrt{X X^{T} + Y Y^{T} } ) \leq Tr(\sqrt{X X^{T} })+Tr(\sqrt{Y Y^{T} })$ for any matrices $X$ and $Y$, where $X^{T}$ denotes the transpose of $X$, or hermitian tranpose if you work with complex matrices. This inequality is true in any dimension (not just 2), and it is just the triangle inequality for the Schatten 1-norm given by $\|X\|_1 = Tr(\sqrt{X X^{T} })$. The expression $Tr(\sqrt{X X^{T} + Y Y^{T} })$ is indeed the 1-norm of the matrix $\begin{pmatrix}X&Y \\\\ 0&0\end{pmatrix}$. - -EDIT: It seems from the comments that my answer to your second question was far from clear. Let me try to explain differently the proof I had in mind. -For 6 real numbers $\alpha_1 \geq \alpha_2$, $\beta_1\geq \beta_2$ and $\gamma_1 \geq \gamma_2$, denote by $f(\alpha_1,\alpha_2,\beta_1 , \beta_2,\gamma_1,\gamma_2)$ the quantity $\sqrt{|\alpha_1|} + \sqrt{|\alpha_2|} - |\sqrt{\max(|\gamma_1|,|\gamma_2|)} - \sqrt{\max(|\beta_1|,|\beta_2|)}| - |\sqrt{\min(|\gamma_1|,|\gamma_2|)} - \sqrt{\min(|\beta_1|,|\beta_2|)}|$. -You are asking whether $f \geq 0$ provided that $\alpha,\beta,\gamma$ are the ordered eigenvalues of respectively $A,B,A+B$ for symmetric $2 \times 2$ matrices $A$ and $B$. The answer is yes, and I am sketching a proof. Denote by $D$ the possible values for $(\alpha_1,\alpha_2,\beta_1 , \beta_2,\gamma_1,\gamma_2)$. -$D$ is exactly described by Horn's inequalities. These inequalities are -$$\alpha_1 \geq \alpha_2 \ \ , \ \ \beta_1\geq \beta_2,$$ -$$\gamma_1 + \gamma_2= \alpha_1 + \alpha_2+\beta_1+\beta_2,$$ -$$\alpha_2+\beta_2 \leq\gamma_2 \leq \min(\alpha_1+\beta_2,\alpha_2+\beta_1).$$ -In particular, $D$ is a convex subset of dimension $5$ of $\mathbb R^6$, and one easily checks that its boundary corresponds to the case when $A$ and $B$ commute. Since the inequality is true when $A$ and $B$ commute (this is eay to check, see the other answer), your question reduces to whether $\inf_D f = \inf_{\partial D} f$. This transforms your eigenvalue question to a purely calculus question. -Notice now that $\beta,\gamma$ and $\alpha_1+\alpha_2$ being fixed, $f(\alpha,\beta,\gamma)$ decreases as $\min(|\alpha_1|,|\alpha_2|)$ decreases. Moreover, if you started with $\alpha,\beta,\gamma$ in the interior of $D$, you stay in $D$ if you make $\min(|\alpha_1|,|\alpha_2|)$ decrease, until you reach the boundary of $D$, or $\min(|\alpha_1|,|\alpha_2|)=0$. You are therefore left to prove that $f(\alpha,\beta,\gamma) \geq \inf_{\partial D} f$ if $(\alpha,\beta,\gamma) \in D$ with $\min(|\alpha_1|,|\alpha_2|)=0$. -In the same way, fixing $\alpha,\beta$ and $\gamma_1+\gamma_2$, you reduce the question to proving that $f(\alpha,\beta,\gamma) \geq \inf_{\partial D} f$ if $(\alpha,\beta,\gamma) \in D$ with $\min(|\alpha_1|,|\alpha_2|)=0$ and $\min(|\gamma_1|,|\gamma_2|)=0$. -Last, fixing $\alpha, \gamma$ and $\beta_1+\beta_2$ with $\min(|\alpha_1|,|\alpha_2|)=0$ and $\min(|\gamma_1|,|\gamma_2|)=0$, you see that $f(\alpha,\beta,\gamma)$ decreases as $\min(|\beta_1|,|\beta_2|)$ increases, until you reach the boundary of $D$. This proves that $\inf_D f = \inf_{\partial D} f$.<|endoftext|> -TITLE: More on universal homeomorphisms -QUESTION [6 upvotes]: I would like to understand this notion better; where could I find some examples? In particular, I am interested in the following questions (and references for the answers). - -Is a universal homeomorphism of connected regular (excellent finite dimensional) schemes an isomorphism if these schemes are not positive characterstic ones? -Suppose that a finite morphism $f:X\to Y$ of connected regular (excellent finite dimensional) schemes is generically purely inseparable. Does $f$ restrict to a universal homeomorphism of some open (non-empty) subschemes of $X$ and $Y$? - -REPLY [8 votes]: Yes. Let $f\colon X \to Y$ be a universal homeomorphism of locally noetherian schemes. Assume that $X$ and $Y$ are integral and $Y$ is normal, and the function field $k(Y)$ has characteristic 0. Then $k(X) = k(Y)$, and $f$ is an isomorphism by Zariski's main theorem. -Under these conditions $f$ is a universal homemorphism. In fact $X$ is the normalization of $Y$ in $k(X)$, again by Zariski's main theorem. But $k(X)$ is purely inseparable over $Y$, so it is obtained by a successive extraction of $p^{\rm th}$ roots of 1. If $F \colon Y \to Y$ is the absolute Frobenius, some power of $F$ will factor through $X$, and then it is easy to see that $X$ is universally isomorphic to $Y$, since $F$ is a universal isomorphism.<|endoftext|> -TITLE: Solvability in differential Galois theory -QUESTION [15 upvotes]: It is well known that the function $f(x) = e^{-x^2}$ has no elementary anti-derivative. -The proof I know goes as follows: -Let $F = \mathbb{C}(X)$. Let $F \subseteq E$ be the Picard-Vessiot extension for a suitable homogeneous differential equation for which $f$ is a solution. -Then one may calculate $G(E/F)$ and show it is connected and not abelian. -On the other hand, a calculation shows that if $K$ is a differential field extension of $F$ generated by elementary functions then the connected component of $G(K/F)$ is abelian, so it is impossible for an anti-derivative of $f$ to be contained in such a field $K$. -However, in classical Galois theory we can do much better, there, we know that a polynomial equation is solvable by radicals if and only if the corresponding Galois group is solvable. -So to my question - is an analog of this is available in differential Galois theory? Is there a general method to determine by properties of $G(F/E)$ if $F$ is contained in a field of elementary functions? - -REPLY [11 votes]: The analogue to "solvable by radicals" in differential Galois theory is "solvable by quadratures". The theorem says that a PV-extension is Liouvillian (adjoining primitives and exponentials) iff the connected component of the differential Galois group is solvable. See "A first look at differential algebra" by Hubbard and Lundell, for an expository account. - -I slightly misread the question at first, thinking you were looking for the analogue of solvbility by radicals in differential algebra. When it comes to determining if the primitive of a function is elementary or not, the characterization is given by Liouville's theorem. Now, for the general case of differential equations solvable in terms of elementary functions, there is a generalization of Liouville's theorem, that you can find in the article "Elementary and Liouvillian Solutions of Linear Differential Equations", by M.F. Singer and J. Davenport (link to Singer's papers here).<|endoftext|> -TITLE: Is there a table of (fibred knot) monodromies? -QUESTION [10 upvotes]: Background/motivation -I'm working on contact topology (in dimension three): a fundamental theorem of Giroux gives us a bijection between contact structures (up to isotopy) and open books (up to negative stabilisation). -Moreover, a stabilisation is a "hands on" operation both at the abstract level (i.e. surface with boundary and monodromy) and at the concrete level (i.e. fibred link in a three-manifold). -Whenever we have a fibred knot $K$ (say) in $S^3$, the corresponding fibration is an open book for $S^3$, which in turn supports a contact structure $\xi$ on $S^3$, where $K$ sits as a transverse knot. - -Now that we have a good theoretical framework, we'd like to put our hands on some examples, and we can start stabilising the open books coming from the Hopf bands and obtain many fibred knots/links and their monodromies, and these examples are not too hard to identify as knots/links in $S^3$. -My question is purely topological: what about the inverse approach? I want to recover the monodromy on the fibre, and I have an "algorithm" to find it, once I have compressing discs for the complement of a fibre (that is a Seifert surface of minimal genus), but to me it's rather crafty. I figure that someone must have done it at some point, so: - -Is there a table associating to each fibred knot the monodromy on its fibre? - -I could find only one source giving concrete examples, namely Burde and Zieschang's Knots, where monodromies for the trefoil and the figure-eight knot are computed. Also, I'm pretty sure that there's much material available for algebraic knots, and that's as far as I've got. - -REPLY [13 votes]: I've produced a table of monodromies for about 63% of the hyperbolic, fibred knots listed on knotinfo. This is available at: http://surfacebundles.wordpress.com/knot-complements/ https://bitbucket.org/Mark_Bell/bundle-censuses/overview (this link now contains significantly more data - for the origional data look under /source/Fibred Knots/). -This was done by producing a triangulation of every possible surface bundle over the circle for the surfaces $S_{1,1}, \ldots, S_{5,1}$ made from a composition of at most 15 Dehn twists about generators. Non-hyperbolic and non-knot complement manifolds were discarded and for each pair of isometric triangulations the short-lex later one was also discarded. Finally, for each hyperbolic, fibred knot complement listed on knotinfo, SnapPy was used to find a bundle on this list isometric to it if it existed. -As Sam points out, there is no canonical choice of generating set for $\mathop{Mod}(S_{g,1})$ so I used the Humphries generating set in each case. However, the monodromies obtained are the short lex earliest for each knot with respect to this generating set and the ordering of the generators shown at the bottom of the page. This ordering was chosen to minimise the running time; a different ordering can run several orders of magnitude slower. -I should point out that these results don't show the millions of knot complements that were also found but that don't (yet) appear on the knot tables. This simply comes from the fact that my tables are ordered by monodromy length whereas knotinfo's is ordered by crossing number.<|endoftext|> -TITLE: An example of a complex manifold without a finite open cover -QUESTION [6 upvotes]: Are there non-compact complex manifolds that -a) Don't embed in C^n (holomorphically) -and -b) Cannot be covered by a finite number of coordinate open sets? -If b) can be satisfied, then I think so can a) be by taking a product with a compact complex manifold. If one takes a Riemann surface of infinite genus, one does not have a "good" finite open cover, but I allow non-contractible open covers as well. Apologies in advance for this elementary question. - -REPLY [5 votes]: If $\widetilde X$ is a compact complex manifold of dimension $\geq 2$ and $x \in \widetilde X$ then $X = \widetilde X - \lbrace x \rbrace$ is a non-compact manifold that cannot be holomorphically embedded in $\mathbb C^N$. This is because, by Hartogs' Theorem, we have $\mathcal O(X) = \mathcal O(\widetilde X)$ and therefore global holomorphic functions on $X$ are constant, which is not the case for complex submanifolds of $\mathbb C^N$.<|endoftext|> -TITLE: The Monster Group uses in mathematical physics -QUESTION [19 upvotes]: I am doing a project on the inverse Galois problem, and am seeking to show that the monster group is realisable over the rationals. I have heard that the monster group has found uses in theoretical physics, and was wondering what those uses might be. Also, is there any practical significance to theoretical physics in the result I am aiming to prove? - -REPLY [32 votes]: There are presently no applications of the monster group in physics, though there is a lot of misleading speculation about this. However in the other direction there are some applications of ideas from physics to the monster group. In particular the no-ghost theorem in string theory is used to construct the monster Lie algebra acted on by the monster group. -The ideas from physics seem to have no direct connection with the problem of realizing the monster as a Galois group over the rationals. This was solved by Thompson, who showed by character-table calculations that the monster satisfies a sufficient condition ("rigidity") for it to be a Galois group.<|endoftext|> -TITLE: Occurrences of a simple reflection in the longest element of a Weyl group? -QUESTION [6 upvotes]: While looking at a preprint I've just bumped into a question about the longest element $w_0$ of a Weyl group $W$ (say irreducible of a Lie type $A$ - $G$ and of rank $n>1$, to simplify). Suppose this element is written as a not necessarily reduced product of the form $w s_n w'$, where both $w$ and $w'$ lie in the proper parabolic subgroup of $W$ generated by simple reflections other than $s_n$. My first impression is that this can occur only for type $A_n$, where for example when $n=2$ we get $w_0 = s_1 s_2 s_1 = s_2 s_1 s_2$. There is some relevant discussion of reduced expressions for $w_0$ in an earlier post here. But getting to a reduced expression from an arbitrary expression takes some work, as shown in the standard Tits algorithm. So I may be overlooking something. - -Is my impression stated above correct, and if so how can it best be made rigorous without case-by-case arguments? - -EDIT: As Ben points out, I need to avoid type A here, so my question is really about the remaining simple types. (I've edited the language above.) - -REPLY [4 votes]: First, I'll show that you can assume that $ws_nw'$ is a reduced product. -To see this, first note that if we multiply $s_n$ by a reduced expression for $w'$, the result will still be reduced, and this can then be extended to a reduced expression for the longest element by multiplying some reduced expression $s_{i_1} \cdots s_{i_k}$ on the left, where $k$ is $1 + l(w')$ less than the length of the longest element...although, as of yet, we don't know that none of the $i_p$ are equal to $n$. -But then $s_{i_1} \cdots s_{i_k} s_nw' = ws_nw'$ so that $s_{i_1} \cdots s_{i_k} = w$. By the nature of the braid relations, any simple reflection that appears in $s_{i_1} \cdots s_{i_k}$ must appear in any expression for $w$, so if $s_n$ appears in $s_{i_1} \cdots s_{i_k}$, then it must also appear in $w$, which is a contradiction. -So, if we have your expression $ws_nw'$, we then in fact have a reduced expression for the longest element that has only a single occurrence of $s_n$. -Now consider any positive root $\alpha$ that has a nonzero coefficient on the simple root $\alpha_n$ corresponding to $s_n$ and also satisfies $w_0 \alpha = -\alpha$. -Claim: $\alpha$ must go negative at the location of $s_n$ in our reduced expression. -Proof: As you apply the simple reflections in our reduced expression one by one to $\alpha$, the only simple reflection that can affect the coefficient of $\alpha_n$ is $s_n$, and so when you get to $s_n$, the simple reflection $s_n$ must negate the coefficient of $\alpha_n$ (for otherwise $w_0 \alpha$ could not be equal to $-\alpha$). (That is, if $s_n$ just nullified the coefficient on $\alpha_n$, because $s_n$ doesn't occur again, one could not end up at $-\alpha$ after all the simple reflections in the reduced expression for $w_0$ have been applied.) -In every case except for type A, there are at least two such $\alpha$...for instance, you can take the sum of the simple roots and the highest root. But they cannot both go negative at the occurrence of $s_n$ in our reduced expression for $w_0$! -Therefore, no such expression exists except in type A. -And then in type A, there's the expression 1 21 321 4321 ... n (n-1) (n-2) ... 321.<|endoftext|> -TITLE: equations defining a subvariety -QUESTION [6 upvotes]: The following question feels to me like a standard sort of 'fact' in birational geometry, but I can't seem to write down a correct set of details. Hopefully someone can point me to a reference and not a counter example! -Suppose $X$ is a variety (reduced and irreducible over an algebraically closed field, perhaps of characteristic zero) and suppose that there exist a very ample line bundle $L$ and a linear system $V \subset H^0(X,L)$ such that $Y = Bs(V)$ is the singular set of $X$ scheme theoretically, that $Y$ is smooth of codimension at at least 2, and that $\tilde X$, the blow up of $X$ along $Y$ is smooth. Further assume that $\phi_{|V|} X--> S$ birationally maps $X$ onto a smooth variety $S$. Let $\tilde \phi$ be the map from $\tilde X \to S$ induced by $V$. Further assume that, denoting by $f$ the map $\tilde X \to X$, that $f^{-1}(Y) = T$ surjects onto $S$. Let $v_1, \dots v_s$ be $s = \dim(S)$ general sections of $V$ so that the intersection $Z(v_1) \cap \dots Z(v_s) \cap S$ consists of finitely many smooth points say $p_1, \dots p_m $. -Also assume the $P = f( \tilde \phi^{-1}(\cup_{i=1:m} p_i))$ is a proper subset of $Y$. Then can one say that away from $P$, the sections $ v_1 \dots v_s$ generate the ideal of $Y$ in $X$ ? -The case I have in mind is where $Y$ is a smooth curve embedded in a sufficiently ample manner so that 1) $Y$ is defined by quadrics and 2) $X = Sec(X)$ is singular only along $Y$. Then $V$ would be the quadrics through $Y$. The point would be to use this sort of an argument to establish a minimum depth of $Sec(Y)$ along $Y$. -This is my first question, so please feel free to correct etiquette with this question as well as the mathematics. - -REPLY [2 votes]: Credit: This answer came out of trying to understand why auniket's answer (a.k.a. counterexample) works. -1) auniket is correct that for dimension reasons $T$ cannot surject onto $S$, so in particular my comment about $X$ being normal possibly helping is irrelevant. So is $T$. -2) It seems to me that there is a much more general problem with your desired statement. Namely I believe the following is true: -Claim: Under the conditions of the question, if in addition $\dim Y=0$ and $Y$ is reduced, then the desired statement cannot be true. -Proof: We may assume that $Y$ is a single point. Since by assumption $X$ is singular at $Y$, the local ring of $X$ at $Y$ is not a regular local ring. Therefore the ideal of $Y$ cannot be generated by $\dim X$ number of elements. On the other hand, by assumption $X$ is birational to $S$, so $\dim X=\dim S=s$. Therefore $v_1,\dots,v_s$ cannot generate the ideal of $Y$. $\square$ -Note: I think this actually covers both of auniket's examples and would definitely give an arbitrary number of normal examples. -3) It seems that this still leaves a sliver of hope for you as your $Y$ is a curve (and even in the zero-dimensional case if $Y$ is non-reduced, it could work out). However, if it is reduced then you are at the absolute minimal number of generators that the singularity condition allows.<|endoftext|> -TITLE: Where can I find a modern write-up of Heegner's solution of Gauss' class number 1 problem? -QUESTION [9 upvotes]: In a recent MO question someone mentioned Heegner's solution of the Gauss "class number 1" problem which takes the following form: - -When the class number of an imaginary quadratic form is 1 an elliptic curve is defined over $\mathbb{Q}$ and a modular function takes on integer values at certain quadratic irrationalities which leads to a collection of Diophantine equations: The solution of which finishes the theorem. - -I sadly can't read Heegner's original work (since I cannot read German) but also I don't think it's necessarily the best thing to read for this proof due to an alleged gap. So if anyone recognizes this proof sketch sketch and knows where I could read this in detail that would be wonderful! Thanks. - -REPLY [12 votes]: In his article On the "gap'' in a theorem of Heegner, Stark does a pretty thorough job of explaining where people thought the purported gap came from, to what extent it actually was a gap, and what you would need to fix such a thing if it existed. I'm paraphrasing, but he basically argues that the confusion stemmed from some errors (typos?) in some analytic results of Weber that Heegner had heavily used. So in a literal sense, Heegner had not proved it because he had cited faulty results, but Stark shows that he deserved credit for the theorem since using Heegner's argument with the correct versions of Weber results (which were indeed known to Weber), the job gets done. -Here's the mathscinet review of the article: -http://www.ams.org/mathscinet-getitem?mr=241384<|endoftext|> -TITLE: Question about the representation theory of SL(n,Z) -QUESTION [16 upvotes]: In this question, all representations are finite-dimensional representations over $\mathbb{C}$. -Fix some $n \geq 3$. Assume that $V$ is a representation of $\text{SL}(n,\mathbb{Z})$. Also, assume that $W$ is a subrepresentation of $V$ and set $V' = V/W$, so we have a short exact sequence -$$0 \longrightarrow W \longrightarrow V \longrightarrow V' \longrightarrow 0$$ -of $\text{SL}(n,\mathbb{Z})$ representations. Assume that the actions of $\text{SL}(n,\mathbb{Z})$ on $W$ and $V'$ extend to actions of $\text{SL}(n,\mathbb{C})$. Question : Must the action of $\text{SL}(n,\mathbb{Z})$ on $V$ extend to an action of $\text{SL}(n,\mathbb{C})$? -Margulis superrigidity says that the action virtually extends, but I can't construct representations where it doesn't extend on the nose. Of course, if it extends then $V = V' \oplus W$ as $\text{SL}(n,\mathbb{C})$-representations. - -REPLY [10 votes]: Consider the surjective map of $SL(n,\Bbb Z)$-modules $Hom_{\Bbb C}(V',V)\to Hom_{\Bbb C}(V',V')$. -Tim tells us that the identity map from $V'$ to $V'$ lifts to an $f:V'\to V$ which is invariant under a finite index subgroup $\Gamma $ of $SL(n,\Bbb Z)$. Then by averageing one can make it invariant under $SL(n,\Bbb Z)$.<|endoftext|> -TITLE: Can you flip the end of a large exotic $\mathbb{R}^4$ -QUESTION [28 upvotes]: Can you flip the end of a large exotic $\mathbb{R}^4$ - -Background -Definition (Exotic $\mathbb{R}^4$): -An exotic $\mathbb{R}^4$ is a smooth manifold $R$ homeomorphic but not diffeomorphic to $\mathbb{R}^4$, where $\mathbb{R}^4$ is equipped with its standard smooth structure. -Definition (Large exotic $\mathbb{R}^4$): -A large exotic $\mathbb{R}^4$ is an exotic $\mathbb{R}^4$ containing a four-dimensional compact smooth submanifold $K'$ that can not be smoothly embedded into $\mathbb{R}^4$. -Definition (End of a large exotic $\mathbb{R}^4$): -If $R$ is a large exotic $\mathbb{R}^4$ and $D^4$ is a four-dimensional disk topologically embedded into $R$ such that $K' \subset D^4$, then $R - D^4$ is an end of $R$. -Remark: -The previous definition varies slightly from the standard definition of "end", however it will be used for the remainder of this question. (See Gompf and Stipsicz Exercise 9.4.11 for the standard definition.) -Remark: -If $R - D^4$ is the end of a large exotic $\mathbb{R}^4$, then $R - D^4$ is a smooth manifold that inherits a smooth structure from $R$. -Definition (Flip of the end of a large exotic $\mathbb{R}^4$): -Given $R - D^4$, the end of a large exotic $\mathbb{R}^4$, a flip of $R - D^4$ is a diffeomorphism $f: R - D^4 \rightarrow R - D^4$ that maps the "inner region" of $R - D^4$, that "near" the removed $D^4$, to the "outter region", that "near infinity", and vica-versa. -Remark: -The previous definition is also non-standard. I am not aware of any standard definitions that carry, more-or-less, the same meaning. -So, at this stage the meaning of the question is hopefully clear. -Foreground -In our attempt to flip the end of a large exotic $\mathbb{R}^4$ an inconvenient truth stands in our way: -Theorem 1 (Uncountably many flips fail): -There are uncountably many large exotic $\mathbb{R}^4$'s that one can not flip the end of. -Quickly, let us see why this is true. Lemma 9.4.2 along with Addendum 9.4.4 of Gompf and Stipsicz state: -Lemma 1: -There exist pairs $(X,Y)$ and $(L,K)$ of smooth, oriented four-manifolds with $X$ simply connected, $Y$ and $K$ compact, $X$ and $L$ open (i.e. noncompact and boundaryless), $L$ homeomorphic to $\mathbb{R}^4$, and $X$ with negative definite intersection form not isomorphic to $n\langle-1\rangle$, such that $X - int(Y)$ and $L - int(K)$ are orientation-preserving diffeomorphic. -Theorem 9.4.3 of Gompf and Stipsicz states: -Theorem: -Any $L$ as appears in Lemma 1 is a large exotic $\mathbb{R}^4$. -The two statements above lead to: -Lemma: -One can not flip the end of any $L$ as appears in Lemma 1. -Proof: -Assume one could flip the end of $L$. Thus, one could use this flip to glue $L$ to the "end" of $X$ and obtain a simply connected closed smooth four-manifold with negative definite intersection form not isomorphic to $n\langle-1\rangle$. However, according to Donaldson's Theorem (Gompf and Stipsicz Theorem 1.2.30) there exists no such manifold. Thus, there exists no such flip. QED -Now we have shown that $L$ can not be flipped. Before we show how uncountably many large exotic $\mathbb{R}^4$'s can not be flipped, we need the definition: -Definition (Radial Family): -Let $R$ be an exotic $\mathbb{R}^4$. Thus, there exists a homeomorphism $h:\mathbb{R}^4 \rightarrow R$. Define $R_t$ as the image under $h$ of the open ball of radius $t$ centered at $0$ in $\mathbb{R}^4$. A radial family is a set of the form $\{R_t | 0 < t \le \infty \}$. -Remark: -If $R_t$ is a member of a radial family, then $R_t$ is a smooth manifold as it inherits a smooth structure from $R$ -Theorem 9.4.10 of Gompf and Stipsicz states: -Theorem: -If $\{L_t | 0 < t \le \infty \}$ is a radial family for an $L$ as appears in Lemma 1 and $r$ is such that $K \subset L_r$, then $\{L_t | r \le t \le \infty \}$ is an uncountable family of non-diffeomorphic large exotic $\mathbb{R}^4$'s. -This leads directly to a proof of Theorem 1. -Proof: -Assume one could flip the end of $L_t$ for $r \le t \le \infty$, where all notation is as in the previous theorem. Thus, one could use this flip to glue $L_t$ to the "end" of $X$ less the image of $L - L_t$ and obtain a simply connected closed smooth four-manifold with negative definite intersection form not isomorphic to $n\langle-1\rangle$. Again, according to Donaldson's Theorem (Gompf and Stipsicz Theorem 1.2.30) there exists no such manifold. Thus, there exists no such flip.QED -Things are seeming rather hopeless at this point. In fact, things are worse -than they seem! But, before we can revel in this despair, we must introduce two -definitions: -Definition (Simply Connected at Infinity): -Let $Z$ be a topological manifold. $Z$ is simply connected at infinity if for -any compact subset $C$ of $Z$ there exists a compact subset $C'$ of $Z$ that -contains $C$ and is such that the inclusion $Z - C' \rightarrow Z - C$ induces -the trivial map $\pi_1(Z - C') \rightarrow \pi_1(Z - C)$. -Definition (End Sum): -Let $Z_1$ and $Z_2$ be non-compact oriented smooth four-manifolds that are -simply connected at infinity. Choose two proper smooth embeddings $\gamma_i : [0, \infty) -\rightarrow Z_i$. Remove a tubular neighborhood of $\gamma_i((0, \infty))$ from -each $Z_i$ and glue the resulting $\mathbb{R}^3$ boundaries together respecting -orientations. The result is the end sum $Z_1 \natural Z_2$ of $Z_1$ and $Z_2$. -Remark: -The requirement that $Z_i$ is simply connected at infinity guarantees -that $\gamma_i$ is unique up to ambient isotopy and thus $Z_1 \natural Z_2$ -is unique up to diffeomorphism (Gompf and Stipsicz Definition 9.4.6). -Remark: -If $R_1$ and $R_2$ are exotic $\mathbb{R}^4$, then they are non-compact oriented -smooth four-manifolds that are simply connected at infinity and $R_1 \natural R_2$ -is a smooth manifold homeomorphic to $\mathbb{R}^4$. -Remark: -$X$ of Lemma 1 is simply connected at infinity. -Theorem 2: -If $\{L_t | 0 < t \le \infty \}$ is a radial family for an $L$ as appears in -Lemma 1 and $r$ is such that $K \subset L_r$, where $K$ is as in Lemma 1, then for -$R$ an exotic $\mathbb{R}^4$ and $t$ such that $r \le t \le \infty$ there exists -no flip of $R \natural L_t$. -Proof: -The proof is basically a slight variation on the above theme. Assume one could flip -the end of $R \natural L_t$ for $r \le t \le \infty$. Thus, one could use this flip -to glue $R \natural L_t$ to the "end" of $X$ less the image of $L - L_t$ end summed -with $R$, in other words with the flip glue $R \natural L_t$ to -$R \natural (X - (L - L_t))$, and obtain a simply connected closed smooth -four-manifold with negative definite intersection form not isomorphic to -$n\langle-1\rangle$. Again, according to Donaldson's Theorem -(Gompf and Stipsicz Theorem 1.2.30) there exists no such manifold. Thus, there -exists no such flip.QED -Now we can revel in this despair! -However, other ways of creating large exotic $\mathbb{R}^4$'s exist. For example, given a topologically slice knot that is not smoothly slice one can create a large exotic $\mathbb{R}^4$. (See, for example, Davis.) Such a large exotic $\mathbb{R}^4$, as far as I can see, might admit an end flip. But, I'm not sure. Thus, we end where we began. -Question - -Can you flip the end of a large exotic $\mathbb{R}^4$? - -REPLY [14 votes]: As stated, your question is equivalent to the existence of a large exotic 4-ball (a smooth $D^4$ which cannot be smoothly embedded into $\mathbb{R}^4_{std}$). -The existence of a flip would give rise to an exotic $S^4$, by gluing the $D^4$ at infinity using the flip diffeomorphism. Removing a (small) standard ball from this $S^4$ gives a large exotic $D^4$, since it contains a smooth submanifold $K'$ which cannot embed in $\mathbb{R}^4_{std}$. -Conversely, if you had a large exotic $D^4$, then you could adjoin a collar neighborhood of $S^3\times \mathbb{R}$ to get an exotic $\mathbb{R}^4$ which has a standard end (diffeomorphic to $S^3\times \mathbb{R}$), and therefore admits a flip. -Although I'm not an expert, I'm certain that the existence of a large exotic 4-ball is open (otherwise, the 4D smooth Schoenflies conjecture would imply the 4D smooth Poincare conjecture). -I realize that this does not answer the spirit of your question, which is whether there is a large exotic $\mathbb{R}^4$ which does not have a standard product end and which admits a flip.<|endoftext|> -TITLE: Will a ball fired through a focus of an ellipse eventually tend to a horizontal line? -QUESTION [8 upvotes]: A couple of years ago I came across this phenomenon which appears to be true although I am having difficulty proving it. -F and F' are foci of a billiard table in the shape of an ellipse. A ball is fired through one of the foci, I can prove that it will subsequently pass through the other focus (this is in fact the question that sparked off this idea). Now if we continue to follow the path of the ball then it seeems as though it will eventually tend to travel in a horizontal line. Some ideas I had were to create triangles and develop a recursive sequence of an angle and see if this tended to 0 or do a similar thing with the gradient of the path of the ball and see if this tended to 0 but couldn't quite get a nice formula either way. -This question is actually Exercise 4.3 in the following notes http://www.math.psu.edu/tabachni/Books/billiardsgeometry.pdf - -REPLY [8 votes]: Edit: sorry, there used to be a completely wrong solution here (I thought that a certain singular curve was a projective line). Now it is fixed. -There is another solution using algebraic geometry. Identify the ellipse with the projective line by sending the two points where the line through the foci meets the ellipse to $0$ and $\infty$. The map we get by starting from a point on the ellipse, getting the second intersection of the line through it and $F$ with the ellipse, then getting the second intersection of the line through that point and $F'$ with the ellipse is an invertible algebraic map sending $0$ to $0$ and $\infty$ to $\infty$, so it must have the form $x \mapsto cx$ for some constant $c$ (depending on the eccentricity). Thus, the billiard ball approaches the horizontal line at an exponential rate.<|endoftext|> -TITLE: Why can projective varieties just have abelian group operations? -QUESTION [10 upvotes]: I just started to read Shimura - Automorphic forms and number theory (Lecture notes in mathematics, 54). On page 20 or so, he mentions that every projective variety which is an algebraic group, is necessary abelian. -Why? - -REPLY [35 votes]: There are several different ways to see this. Here is one: -Let $G$ be our irreducible projective algebraic group variety over the field $k$, with identity element $e$. The group $G$ acts on itself by conjugation, and this action fixes $e$. Thus this induces -a $k$-linear action of $G$ on the local ring $\mathcal O_e,$ and hence on the (finite-dimensional) quotients -$\mathcal O_e/\mathfrak m_e^n$ for each $n$. Now any morphism from the irreducible projective variety $G$ to the affine variety $End_k(\mathcal O_e/\mathfrak m_e^n)$ must be constant, and so the $G$-action on each $\mathcal O_e/\mathfrak m_e^n$, and hence on $\mathcal O_e$ itself, must be trivial. -Since $G$ is irreducible, it is now easy to see, using the fact that the conjugation action -induces a trivial action on $\mathcal O_e$, that the conjugation action is in fact trivial on $G$ itself, and hence that $G$ is commutative. -(This argument breaks down if $G$ is not projective, because then $G$ can have non-trivial morphisms to the matrix rings $End_k(\mathcal O_e/\mathfrak m_e^n)$; it is instructive to think about this in the case $G = GL_n(k)$. It also breaks down if $G$ is not irreducible, e.g. if $G$ is a finite non-abelian group, then we can think of it as a zero-dimensional projective algebraic group. The point in this case is that one can't make the "analytic continuation" argument from the action on $\mathcal O_e$ to all of $G$.) - -REPLY [9 votes]: I borrow this proof from [Birkenhake-Lange, Complex Abelian Varieties, Lemma 1.1.1]. -Let $X$ be a projective variety having a group structure. I assume that we are working over $\mathbb{C}$. -Consider the commutator map $\Phi(x,y)=xyx^{-1}y^{-1}$, and let $U$ be a coordinate neighborhood of $1 \in X$. By the continuity of $\Phi$, and since $\Phi(x,1) =1 \in U$, for all $x \in X$ we can find open neighborhoods $U_x$ and $W_x$ such that $\Phi(U_x, W_x) \subset U$. -Since $X$ is compact, finitely many $V_x$ cover $X$. Calling $W$ the intersection of the corresponding subsets $W_x$, we get $\Phi(X, W) \subset U$. -Now $\Phi(1, y)=1$ for all $y \in W$. Since holomorphic functions on a compact variety are constant, it follows $\Phi(X, W)\equiv 1$. Being $W$ open and non-empty, this in turn implies $\Phi(X, X) \equiv 1$, which is our claim. -Notice that "projective" is not really necessary, in fact what we actually use in the proof is "compact complex". Indeed, pushing further this argument (by a straightforward use of the exponential map) one can show that any compact complex connected Lie group is a complex torus. - -REPLY [3 votes]: A very accessible proof (at the beginning of the book for the case over $\mathbb{C}$, and further on in the book for any characteristic) of that statement is present in Mumford's book Abelian varieties.<|endoftext|> -TITLE: Frobenius splitting and derived Cartier isomorphism -QUESTION [20 upvotes]: Let $X$ be a smooth algebraic variety over an algebraically closed field $k$ of characteristic $p>\dim X$. The motivation for my question comes from the following results. - -1. If $X$ is Frobenius split (the $p$-th power map $\mathcal{O}_X \to F_* \mathcal{O}_X$ admits n $\mathcal{O}_X$-linear splitting) then the Kodaira vanishing theorem holds for $X$. - -The proof uses nothing but Serre vanishing and the projection formula. - -2. If the complex $F_* \Omega^\bullet_X$ is quasi-isomorphic to a complex with zero differentials, then the Kodaira-Akizuki-Nakano vanishing theorem holds for $X$. - -The proof uses Cartier isomorphism, hypercohomology spectral sequences, Serre vanishing and the projection formula and is similar to that of 1. - -3 (Deligne-Illusie 1987). If $X$ lifts to $W_2(k)$, then the complex $F_* \Omega^\bullet_X$ is quasi-isomorphic to a complex with zero differentials. -4 (Buch-Thomsen-Lauritzen-Mehta 1995). If $X$ is strongly Frobenius split (that is, $0\to B_1\to Z_1\to \Omega^1_X\to 0$ splits, where $Z_i$ and $B_i$ are cocycles/coboundaries in $F_* \Omega^\bullet_X$), then $X$ and $F$ lift to $W_2(k)$ and the Bott vanishing theorem holds for $X$. - -My (maybe incorrect) feeling is that strong Frobenius splitting and lifting of the Frobenius to $W_2(k)$ are quite uncommon, Frobenius splitting is a common behavior "on the Fano side" and that lifting of $X$ to usually $W_2(k)$ exists. - -Question. Are there examples of Frobenius split varieties for which $F_* \Omega^\bullet_X$ is not quasi-isomorphic to a complex with zero differentials (for example, because the Hodge spectral sequence does not degenerate, see also this question on the Hodge spectral sequence)? If yes (that's my intuition here), does Frobenius splitting imply some weaker property of $F_* \Omega^\bullet_X$ which implies Kodaira vanishing? - -Edit. Note that Frobenius splitting just states that the complex $F_* \Omega^\bullet_X$ is quasi-isomorphic to a complex whose first differential $C^0 \to C^1$ is zero. - -REPLY [4 votes]: I just found a nice reference for this question: K. Joshi "Exotic Torsion, Frobenius Splitting and the Slope Spectral Sequence" (Canad. Math. Bull. Vol. 50 (4), 2007), section 9. -Theorem 9.1 (unpublished work of V. B. Mehta). Let X be a smooth, projective, F-split variety over an algebraically closed field $k$ of characteristic $p>0$. Then for all $i+j < p$, the Hodge to de Rham spectral sequence degenerates at $E^{i,j}_1$. In particular, for $i+j = 1$ we have the following exact sequence -$$ 0\to H^0(X, \Omega^1_X)\to H^1_{DR}(X/k)\to H^1(X, \mathcal{O}_X)\to 0. $$ -Moreover, any F-split variety with $dim(X) -TITLE: What elementary problems can you solve with schemes? -QUESTION [223 upvotes]: I'm a graduate student who's been learning about schemes this year from the usual sources (e.g. Hartshorne, Eisenbud-Harris, Ravi Vakil's notes). I'm looking for some examples of elementary self-contained problems that scheme theory answers - ideally something that I could explain to a fellow grad student in another field when they ask "What can you do with schemes?" -Let me give an example of what I'm looking for: In finite group theory, a well known theorem of Burnside's is that a group of order $p^a q^b$ is solvable. It turns out an easy way to prove this theorem is by using fairly basic character theory (a later proof using only 'elementary' group theory is now known, but is much more intricate). Then, if another graduate student asks me "What can you do with character theory?", I can give them this example, even if they don't know what a character is. -Moreover, the statement of Burnside's theorem doesn't depend on character theory, and so this is also an example of character theory proving something external (e.g. character theory isn't just proving theorems about character theory). -I'm very interested in learning about similar examples from scheme theory. - -What are some elementary problems (ideally not depending on schemes) that have nice proofs using schemes? - -Please note that I'm not asking for large-scale justification of scheme theoretic algebraic geometry (e.g. studying the Weil conjectures, etc). The goal is to be able to give some concrete notion of what you can do with schemes to, say, a beginning graduate student or someone not studying algebraic geometry. - -REPLY [3 votes]: In line with Felipe Voloch's remark "Spec $\mathbb{Z}$ are where schemes really shine", I thought I'd also add this beautiful (expository and very readable!) paper of Serre: -"How to use finite fields for problems concerning infinite fields" -which makes crucial use of being able to do algebraic geometry over (finitely generated algebras over) $\mathbb{Z}$ in order to prove geometric statements (many of which are easily understable without any background in algebraic geometry!) over fields like $\mathbb{C}$ and $\mathbb{Q}$.<|endoftext|> -TITLE: Bivariate polynomials with special properties -QUESTION [7 upvotes]: I recently came across some polynomials with some remarkable properties. -A polynomial $P(u,v) \in \mathbb{R}[u,v]$ in 2 variables is remarkable if -the set of solutions to the system $P(u,v)=P(v,u)=0$ is a finite number of points in $\mathbb{C}^2$ such that each point is of the form $(x, \overline{x}).$ -Here are some examples of such polynomials: -$v^2-u$ -$-2 u v+v^3+1$ -$u^2-3 u v^2+v^4+2 v$ -$3 u^2 v-4 u v^3-2 u+v^5+3 v^2$ -$-u^3+6 u^2 v^2-5 u v^4-6 u v+v^6+4 v^3+1$ -$-4 u^3 v+10 u^2 v^3+3 u^2-6 u v^5-12 u v^2+v^7+5 v^4+3 v$ -$u^4-10 u^3 v^2+15 u^2 v^4+12 u^2 v-7 u v^6-20 u v^3-3 u+v^8+6 v^5+6 v^2$ -A quite complicated algorithm is behind this sequence, any help identifying -a formula for these polynomials would be helpful. -Also, a proof that these polynomials are remarkable would be nice, -these are only checked by numeric computations in Mathematica. -Question: Can one classify the set of remarkable polynomials? - -REPLY [3 votes]: Your sequence $p_n (u,v)$ can be defined by $p_n (u,v) = vp_{n - 1} (u,v) - up_{n - 2} (u,v) + p_{n - 3} (u,v)$ with initial values $p_0 (u,v) = 1,p_1 (u,v) = v,p_2 (u,v) = v^2 - u.$ -In my above remark I have overlooked a term in the fifth polynomial. This is now the same as the formula given by ARupinski. -Added later: -Extend the sequence $p_n (u,v)$ to negative indices by $p_{ - 1} (u,v) = p_{ - 2} (u,v) = 0$ and $p_{ - n} (u,v) = p_{ n - 3} (v,u)$ for $n >2 .$ -Define a new sequence of polynomials $r_n (u,v)$ by the same recurrence and initial values $r_{ - 1} (u,v) = 1,r_0 (u,v) = 0,r_1 (u,v) = - u.$ Extend it to negative values by $r_{ - n} (u,v) = r_{n - 2} (v,u).$ -Let $A$ be the matrix with rows $(0,1,0),(0,0,1),(1, - u,v).$ -Then $A^n$ is the matrix with the following rows: $\left( {p_{n - 3 + j} (u,v),r_{n - 2 + j} (u,v),p_{n - 2 + j} (u,v} \right)$ for $0 \le j \le 2.$ -It seems that the sequence $r_n (u,v)$ or the sequence $r_{2n} (u,v)/(1 - uv)$ has analogous properties with respect to the zeroes. Is it also related to the group representation?<|endoftext|> -TITLE: Motivation for and history of pseudo-differential operators -QUESTION [54 upvotes]: Suppose you start from partial differential equations and functional analysis (on $\mathbb R^n$ and on real manifolds). Which prominent example problems lead you to work with pseudo-differential operators? -I would appreciate any good examples, as well as some historical outlines on the topic's development. (Shubin's classical book spends a few lines on history and motivation in the preface, but no "natural" examples. I am not aware of any historical outlines in the literature.) - -REPLY [3 votes]: Disclaimer. This answer stems from a draft of an entry on the symbol of a singular integral operator I am writing for the Wikipedia, thus it is by no means complete. -From the sources I have read until now (for example Gaetano Fichera's papers [1], p. 475 and [2], pp. 52-54 and Vladimir Maz'ya's book [5], p. 143), it seems that the motivation for the development of the theory of pseudodifferential operators lies in the development of the theory of singular integral operators, emerged as means for solving variable coefficients PDEs: precisely, Fichera and Maz'ya identify the starting point of the theory with the discover of the symbol of a singular integral operator made by Solomon Mikhlin. -Short answer -While studying the $3$-dimensional potential generated by a plane thin disk (of arbitrary shape), Francesco Tricomi succeeded for the firs time ever in giving a closed form expression for the composition of two $2$-dimensional (constant coefficient) singular integral operators (see [11], [12]). Tricomi's work led Solomon Mikhlin (see [5], pp. 2124-2125, [6], pp. 535-536 and for a brief but fairly complete historical survey see [8], §1.1.4 pp. 8-10), followed by Georges Giraud, to the discovery of the concept of the symbol of a singular integral operator. Be it noted that at that time, due to the form in which it was discovered, it was not clear that the symbol was strictly related to the Fourier transform the kernel respect to the integration variable, therefore it was not clear its relation with the Fourier transform of partial derivatives: however, its discovery made clear that the symbol is the key to uncover the algebraic structure of the composition of such operators and the structure of their spaces. Later on, Solomon Mikhlin ([2], p. 54 and [8], §1.1.7 p. 11) discovered that the symbol is the partial Fourier transform, respect to the integration variable, of the kernel of the singular integral operator. The use of the machinery of multidimensional Fourier integrals, developed by Alberto Calderón and Antoni Zygmund, became thus one of the strong points of the theory, and finally Kohn & Niremberg and Hörmader synthesized operator algebras which include both linear partial differential and singular integral operators: the theory of pseudodifferential operators was born. -A more detailed survey - -Background on PDE theory at the beginning of the twentieth century: reduction to integral equations. -One of the more extensively used techniques for the solution of PDEs which was developed at the beginning of the 20th century was the "Method of Potentials": it stemmed from the methods developed the in the nineteenth century for solving Laplace's equation and consist in representing the solution of a given problem as the sum of properly chosen potential type integrals (volume, single and double layer potentials) (see [8], Chapter III, p. 39 and §18, pp. 44-49 and [9], Chapter III, p. 49 and §18, pp. 54-59). These integral representations are obtained for example by using Hadamard's elementary solutions and Levi functions (now called parametrices): the set of integral equations obtained is hopefully analyzable by applying Fredholm's theory. -However, this is not always so. These integral equations are, as a rule, often singular i.e. do not exist in the ordinary sense but only in the sense of their principal values: in this case, Fredholm's theory cannot be applied directly. This happens, for example, for the Oblique Derivative Problem for Laplace's equation, as Henri Poincaré and Gaston Bertrand noted in their analysis of the problem of tides, de facto the first oblique derivative ever posed (see for example [3], pp. 251-252). Thus the time was ready for a theory of singular integral equations to be developed. - -The symbol of a singular integral: Tricomi, Mikhlin and Giraud. -Tricomi, while studying problem of determining the harmonic potential of a plane thin disk (of arbitrary shape) immersed in $\Bbb R^3$, found a formula for the explicit calculation of the composition of two $n$-dimensional (precisely $2$-dimensional) singular integral operators (see [11] and [12], §4 pp. 107-112 and also [8] §1.2, pp. 2-4). Building on the the work of Tricomi, Mikhlin discovered the concept of symbol in the form of a complex Fourier series (see Fichera [2], p. 53): let -$$ -Su(z)=\iint u(\zeta)K(z,z-\zeta)\mathrm{d}\xi\mathrm{d}\eta\quad z=x+iy,\;\zeta=\xi +i\eta,\;u\in L^p\label{1}\tag{1} -$$ -be a syngular integral operator whose kernel has the form -$$ -K(z,\zeta)=\frac{1}{|\zeta|^2}f\left(z\frac{\zeta}{|\zeta|}\right), -$$ -and let -$$ -f(z,\theta)=\sum_{h\in\Bbb Z\setminus \{0\}} c_h(z)e^{ih\theta}. -$$ -where the zero order term is missing since the condition -$$ -\int\limits_{-\pi}^{+\pi} f(z,e^{i\theta})\mathrm{d}\theta=0 -$$ -must be fulfilled in order for the integral \eqref{1} to exists if, for example, $u\in\operatorname{Lip}$. Then Mikhlin defines the symbol of $S$ as -$$ -\sigma(z,\zeta)=\sum_{h\in\Bbb Z\setminus \{0\}} c_h(z)\frac{-i^{|h|}}{|h|} - e^{ih\theta}. -$$ -In the paper [5] he shows how to extend the results of [4] to singular integrals defined on a closed surface. -Georges Giraud, who was the leading mathematician in the development of the method of potentials (to the point that, according to Carlo Miranda ([7], p. 44 and [8], p. 54) he provided "the most general and definitive contributions"), was consequently quite interested in the analysis of singular integrals. -Closely after the appearance of the works [4] and [5], Giraud published the note [3] in the Comptes rendus, giving (without proof) formulas for the symbol of a singular integral operator, expressed in the form of a series of harmonic polynomials, and for the composition of two singular operators that extends the work of Mikhlin to the $(n\ge 2)$-dimensional case: nor Giraud's nor Mikhlin's definitions of the symbol rely on multidimensional Fourier transform theory. -A proof of Giraud's formulas was given later by Mikhlin ([6], §1.4, p. 9), and only later, in 1956 ([2], p. 54 and [8], §1.1.7 p. 11) he also proved that the symbol is the partial multidimensional Fourier transform respect to the integration variable of the singular integral, connecting his theory to the one Alberto Càlderon and Antoni Zygmund were developing in the same years. - - -References. -[1] Fichera, Gaetano, "Francesco Giacomo Tricomi", Atti della Accademia Nazionale dei Lincei. Serie VIII. Rendiconti. Classe di Scienze Fisiche, Matematiche e Naturali 66, pp. 469-483 (1979), MR0606447, Zbl 0463.01022. -[2] Fichera, Gaetano, "Solomon G. Mikhlin (1908-1990)", Atti della Accademia Nazionale dei Lincei, Rendiconti Lincei, Matematica e Applicazioni, Serie XI, Supplemento 5, pp. 49-61, 1 plate (1994), Zbl 0852.01034. -[3] Giraud, Georges, "Équations à intégrales principales; étude suivie d’une application", Annales Scientifiques de l'École Normale Supérieure. (3) 51, pp. 251-372 (1934), MR1509344, Zbl 0011.21604. -[4] Giraud, Georges, "Sur une classe générale d’équations à intégrales principales", Comptes rendus hebdomadaires des séances de l'Académie des sciences, Paris 202, pp. 2124-2127 (1936), JFM 62.0498.01, Zbl 0014.30903. -[5] Maz’ya, Vladimir, Differential equations of my young years. Translated from the Russian by Arkady Alexeev, Cham: Birkhäuser/Springer (ISBN 978-3-319-01808-9/hbk; 978-3-319-01809-6/ebook), pp. xiii+191 (2014), MR3288312, Zbl 1303.01002. -[6] Mikhlin, Solomon G., "Equations intégrales singulieres à deux variables independantes", Recueil Mathématique (Matematicheskii Sbornik) N.S. (in Russian), 1(43) (4), pp. 535-552 (1936), JFM 62.0495.02, Zbl 0016.02902. -[7] Mikhlin, Solomon G., Complément à l’article “Equations intégrales singulières à deux variables indépendantes”, Recueil Mathématique (Matematicheskii Sbornik) N.S. (in Russian), 1(43) (6), pp. 963-964 (1936), JFM 62.1251.02, Zbl 62.1251.02." -[8] Mikhlin, Solomon G., Multidimensional singular integrals and integral equations. Translated from the Russian by W. J. A. Whyte. Translation edited by I. N. Sneddon. (International Series of Monographs in Pure and Applied Mathematics. Vol. 83), Oxford-London-Edinburgh-New York-Paris-Frankfurt: Pergamon Press, pp. XII+255 (1965). MR0185399, ZBL0129.07701. -[9] Miranda, Carlo, Equazioni alle derivate parziali di tipo ellittico, Ergebnisse der Mathematik und ihrer Grenzgebiete, Neue Folge, 2. Heft., Berlin- Göttingen-Heidelberg: Springer-Verlag. VIII, 222 S. (1955), MR0087853, ZBL0065.08503. -[10] Miranda, Carlo, Partial differential equations of elliptic type, Ergebnisse der Mathematik und ihrer Grenzgebiete, Band 2, Berlin-Heidelberg-New York: Springer-Verlag, pp. XII+370 (1970), MR0284700, Zbl 0198.14101. -[11] Tricomi, Francesco, "Formula d’inversione dell’ordine di due integrazioni doppie “con asterisco”"., Atti della Accademia Reale dei Lincei, Rendiconti della Classe di Scienze Fisiche, Matematiche e Naturali, (6) 3, pp. 535-539 (1926), JFM 52.0235.05. -[12] Tricomi, Francesco, Equazioni integrali contenenti il valore principale di un integrale doppio, Mathematische Zeitschrift 27, pp. 87-133 (1927), JFM 53.0359.02.<|endoftext|> -TITLE: Weierstrass' function and Brownian motion -QUESTION [5 upvotes]: Is there a known connection between Weierstrass' function -$W_\alpha (x) = \sum_{n=0}^\infty b^{- n \alpha} \cos(b^n x)$ -and Brownian motion? Specifically, when $\alpha = 1/2$, the Weierstrass function has same Holder continuity peoperties that Brownain sample paths do. Some crude Matlab experiments seem to suggest this function has linear quadratic variation for low values of $b$, too. -The expression reminds me a little of the Karhunen-loeve expansion of Brownian motion, but I don't see how the two might relate. -Many thanks. - -REPLY [3 votes]: Some quick Googling brought me to this paper. The idea is to take the coefficients in the summation to be suitable independent random variables according to a suggestion of Mandelbrot. I can't actually access the paper right now so I can't say if the authors were able to include the case of standard Brownian motion in their results. -``Convergence of the Weierstrass-Mandelbrot process to Fractional Brownian Motion'' Murad Taqqu and Vladas Pipiras. Fractals. 8 (2000) 369-384. - -REPLY [3 votes]: You might find this account useful. In particular, see the end of Section 1, page 7, and the end of Section 4.<|endoftext|> -TITLE: Looking for deterministic criteria to generate the symmetric group? -QUESTION [7 upvotes]: So let $S_N$ be the symmetric group of degree $N$. We think of it as a permutation group via its -natural action on the set $T=\{1,2,\ldots,N\}$. -Say that $H\leq S_N$ is a subgroup which acts transitively on $T$. However, I DONT'T WANT to assume necessarily that $H$ is primitive (that is the whole point of my question). Assume furthermore that there is an onto group homomorphism -$$ -f:H\rightarrow S_n -$$ -where $n=\lfloor{N/2}\rfloor$. In fact, as was pointed out by Schmidt, the existence of this onto group homomorphism implies that $H$ is imprimitive. -In general, one cannot rule out the existence of such an $H$. For example -one could have $H=S_n\ltimes\mathbf{F}_2^n$ where $N$ is even and $n=\frac{N}{2}$. -We let $H$ act on $T$ in the following way: We divide $T$ in $n$ disjoint blocks of size $2$. We let $S_n$ permute the $n$ blocks without swapping the pair in each block, and we let $\mathbf{F}_2^n$ permute (resp. acts like the identity) the two elements in the i-th block if the i-th coordinate of an element $\sigma\in \mathbf{F}_2^n$ is $\overline{1}$ (resp. $\overline{0}$). It thus follows that $H$ acts transitively (but imprimitively) on $T$. -Furthermore, suppose that I can produce " a lot of elements " in $H$ which contain a cycle of length $r$ in their cycle presentations (their writing as a product of disjoint cycles of $T$) for $r>n$. Then may I conclude that such an $H$ does not exist? -Q1: Is there some kind of results that would allow me to conclude that $H\supseteq A_N$, so that this would contradict the imprimitivity and therefore rule out the existence of such an $H$? -For example here is one key result which is good to know: if $H$ is assumed to be primitive and contains a cycle of length $\ell$ with $2\leq \ell\leq N-7$ ($\ell$ not necessarily prime) then combining classical results on permutation group theory one may show that $H\supseteq A_N$. However, since in my setting $H$ is imprimitive I cannot apply this result. -Q2: Do we have a good understanding of the tree of subgroups of $S_N$, especially -the maximal subgroups? -Q3: Is there some kind of probabilistic result that could be used in my context? - -REPLY [3 votes]: Embarrassingly, the posting below contains another incomplete proof, as was pointed out by Hugo Chapdelaine. I am working on a new (third) version, but after two wrong proofs, I want to hold off for a while on posting it until it is written up carefully and checked. What I am trying to prove is this: -THEOREM: Let $H$ be a transitive subgroup of $S_N$, and suppose -$M \triangleleft H$ and $H/M \cong S_n$, where $N/2 \le n < N$ and -$5 < n$. Then $n = N/2$ and either $M = 1$ or all orbits of $M$ have size $2$. - -Unfortunately, the proof in the original version of this post had -a gap that I do not see how to repair. The following weaker -result seems to be true, however. -THEOREM. Let $H \subseteq S_N$ and assume that $M \triangleleft H$ and $H/M \cong S_n$, where $n \ge N/2$. Then either $M$ is an elementary abelian $2$-group and $n = N/2$ and $H$ is transitive, or else there exists $K \subseteq H$ such that $H = MK$ and -$M \cap K = 1$. In particular, $K \cong S_n$. -Proof. Induct on $N$. Let $S$ be a point stabilizer in $S_N$. Write -$u = |H:(H \cap S)|$, so $u \le N$ with equality only if $H$ is transitive. -Write $v = |M:(M \cap S)|$, so $v = |M(H \cap S):(H \cap S)|$, and this divides $u$. Also, $u/v = |H:M(H \cap S)|$ is the index of a subgroup of -$H/M \cong S_n$, so this index is either $1$ or at least $n$. -Suppose $u/v = 1$. Then $M(H \cap S) = H$, and so -$(H \cap S)/(M \cap S) \cong H/M \cong S_n$. Also, $S \cong S_{N-1}$ and $(N-1)/2 < n$, so the inductive hypothesis applies with $H \cap S$ in place of $H$ and $M \cap S$ in place of $M$. Since $n$ is not $(N-1)/2$, there exists $K \subseteq H \cap S$ such that $K \cap (M \cap S) = 1$ and $H \cap S = (M \cap S)K$. Then $K \cap M = 1$ and -$H = M(H \cap S) = M(M \cap S)K = MK$ as required. -We can assume now that for every choice of point stabilizer $S$ we have -$u/v \ne 1$. Then $N/v \ge u/v \ge n \ge N/2$, and thus $v \le 2$. Thus all orbits of $M$ have size $1$ or $2$, so $M$ is an elementary abelian 2-group. If we always have $v = 1$, then $M = 1$ and we can take $K = H$. We can thus assume that $v = 2$ for some choice of $S$. Then $N/2 \ge u/2 = u/v \ge n \ge N/2$ and we have equality. Thus $n = N/2$ and $u = N$, and the latter equality shows that $H$ is transitive. QED - -In the case where $H = MK$ and $M \cap K = 1$, we have $K \cong S_n$, so we can ask what copies $K$ of $S_n$ are contained in $S_N$ if -$n \ge N/2$. If $n > 5$, It seems that the only possibility is that $K$ is the stabilizer of $N - n$ points. This can be proved using the fact that $S_n$ has no proper subgroup of index less than or equal to $2n$ except for point stabilizers. (At least I think that is a fact.)<|endoftext|> -TITLE: Bimodules over division rings -QUESTION [7 upvotes]: Inspired by other questions i have two questions about modules over division rings: given a division ring $D$ with center $Z(D)=K$. One has the notion of dimension for left modules (vector spaces) $V$ over $D$, which is well behaved like in linear algebra. -But does the following also hold: Given a left vector space $W$ of dimension $n$ and a right vector space $V$ of dimension $m$. Assume that at least one of them is even a $D$-bimodule, so that $V\otimes_D W$ has the structure of a $D$-module. -$Q1$: Do we have $dim_D(V\otimes _D W)=dim_D(V)dim_D(W)=nm$ in this case? -Now $D$ is naturally a $D$-bimodule, so is $D^{\*}=Hom_K(D,K)$. -$Q2$: Is there a $D$-bimodule isomorphism between $D$ and $D^{\*}$? -Both modules are one dimensional, so pick generators $x\in D$ and $y\in D^{\*}$. Now $x\mapsto y$ defines a map $D \rightarrow D^{\*}$, which satisfies $ax\mapsto ay$. Assume that $ax=xb$ for some $b\in D$, then we need to have $xb\mapsto yb$, which leads to $x^{-1}ax=y^{-1}ay$ for all $a\in D$. Can we find $x$ and $y$ such that the last relation is fulfilled? Or is there no such bimodule isomorphism? - -REPLY [12 votes]: The answer to both questions is positive if $D$ is finite-dimensional over its center $K$, and negative, in general, otherwise. -Q1. Suppose $V$ is a $D$-bimodule over $K$ (i.e., a $D\otimes_K D^{op}$-module), while $W$ is a left $D$-module of dimension $n$, as in your question. Then $W$ is isomorphic to a direct sum of $n$ copies of $D$, hence $V\otimes_D W$ as a left $D$-module is isomorphic to a direct sum of $n$ copies of $V$. So the $\dim_D (V\otimes_D W) = n\dim_D V$, where $\dim_D$ denotes the dimension of left $D$-modules. Now, to answer your original question, it remains to notice that the dimensions of $V$ as a left and a right $D$-module coincide, since both are equal to $\dim_KV/\dim_KD$. -It does not matter in this argument that $K$ is the whole center of $D$, only that $K$ is a field contained in the center of $D$ and $D$ is finite-dimensional over $K$. On the other hand, if $D$ is infinite-dimensional over $K$, it may happen that $D$ can be $K$-linearly embedded into itself as a proper subring. This would allow to define a $D$-bimodule structure on $V=D$ where $V$ is one-dimensional as a right module over $D$ and more than one-dimensional (possibly infinite-dimensional) as a left module over $D$. This would break your dimension formula. -Q2. Left $D$-module maps $D\to D^\ast$ are in one-to-one correspondence with elements of $D^\ast$, i.e., $K$-linear functions $D\to K$. A linear function $l\colon D\to K$ defines a $D$-bimodule map $D\to D^\ast$ if and only if one has $l(xy)=l(yx)$ for all elements $x$, $y\in D$. Taking $l$ to be the trace map (defined e.g. by tensoring $D$ with the separable closure of $K$ over $K$, identifying the result with matrices over the separable closure, and taking the traces of matrices), one can obtain a $D$-bimodule isomorphism between $D$ and $D^\ast$ when $D$ is finite-dimensional over its center $K$. -The above argument can be easily extended to the case when $K$ is just a subfield of the center of $D$ and $D$ is finite-dimensional over $K$ (one just composes the trace map with an arbitrary nonzero linear function from the center of $D$ to $K$). On the other hand, when $D$ is infinite-dimensional over $K$, it is no longer true that $D^\ast$ is a one-dimensional left $D$-module.<|endoftext|> -TITLE: A density on the natural numbers invariant with respect to the multiplication -QUESTION [9 upvotes]: The "classical Beurling density" of a subset of the natural numbers is $d(A)=lim_{n\rightarrow\infty}\frac{|A\cap[1,n]|}{n}$, when it exists. It defines a finitely additive probability measure on the natural numbers which is invariant with respect to the sum. Here is my question: does there exist a "nice formula" to describe a finitely additive probability measure on $\mathbb N$ which is invariant with respect to the multiplication? -A couple of remarks: I don't know if it is trivial that such a measure exists, but anyway it follows from the application of a general result of Vern Paulsen (on arxiv "Syndetic sets and amenability"). -Another problem would be that of finding the measure of particular sets. What about the measure of {$1!,2!,3!,4!...$}? Sets with measure differente from 0 and 1? For example the set of numbers whose first digit through 4 to 9 seems to have measure $=\log_{10}4$... any other? -Thanks in advance, Valerio - -REPLY [7 votes]: The natural thing to do here is to replace the intervals $[1,n]$ (and $n = |[1,n]|$) in the definition of $d(A)$ with a sequence $F_n$ of subsets of $\mathbb{N}$ which is multiplicatively asymptotically invariant (or, in other words, a Folner sequence for the semigroup $(\mathbb{N},\cdot)$). For an exploration of this idea, as well as applications, see for instance this article by Vitaly Bergelson: -Multiplicatively large sets and ergodic Ramsey theory, Israel Journal of Mathematics 148 (2005), 23-40. -EDIT: One particular example (mentioned in the article) is to take $F_n$ to be the set of all positive integers which can be written as a product of powers of the first $n$ primes, where the powers are allowed to be any non-negative integer which is less than or equal to $n$. The motivation for choosing $F_n$ this way is that just as $1$ generates the additive semigroup $(\mathbb{N},+)$, the primes generate the multiplicative semigroup. Think of balls in the corresponding Cayley graphs with radius getting larger and larger. The article contains many other examples of such $F_n$, and each of them gives a notion of "multiplicative density" by setting $d(A) = \lim_{n \to \infty} \frac{|A \cap F_n|}{|F_n|}$ (if the limit exists. Otherwise one usually considers the limsup and liminf).<|endoftext|> -TITLE: local class field theory (Norm map) -QUESTION [9 upvotes]: Let $K$ be a local field, for example the $p$-adic numbers. In Neukirch's book "Algebraic number theory", there is the statement: if $K$ contains the $n$-th roots of unity and if the characteristic of $K$ does not divide $n$, and we set $L=K(\sqrt[n]{K^{\times}})$, then one has $N_{L/K}(L^{\times})=K^{\times n}$. -My questions are the following, for a (finite) Galois extension $L$ of $K$:\ -(1) What happens if the characteristic of $K$ divides $n$? Can one obtain an explicit form of the image of the norm of $L^\times$?\ -(2) If we don't add the condition that $K$ contains the $n$-th roots of unity, what is the image of the norm operator? If $L = K(x)$, can one write it in terms of the primitive element $x$ and $K^{\times n}$?\ - -REPLY [7 votes]: Here is another interesting case where the image of the norm map can be written down explicitly. Let $F$ be a finite extension of $\mathbf{Q}_p$ containing a primitive $p$-th root $\zeta$ of $1$, and denote the filtration on the multiplicative group $F^\times$ by -$$ -\ldots\subset U_2\subset U_1\subset\mathfrak{o}^\times\subset F^\times. -$$ -We thus get the extensions $L_n=F(\root p\of{U_n})$. It can be checked that $L_{pe_1+1}=F$, where $e_1$ is the absolute ramification index of $F$ divided by $p-1$, and that $L_{pe_1}$ is the unramified degree-$p$ extension of $F$, so the image of the norm map $L_{pe_1}^\times\to F^\times$ is $\mathfrak{o}^\times F^{\times p}$. -What is the image of the norm map $L_{n}^\times\to F^\times$ for other $n$ ? Local class field theory and a certain orthogonality relation for the Kummer pairing (see for example the last section of arXiv:0711.3878) can be used to answer this question. Basically, for $n\in[1,pe_1]$, $N(L_n^\times)=U_mF^{\times p}$, where $m=pe_1+1-n$. -There are similar results for elementary abelian $p$-extensions of finite extensions of $\mathbf{F}_p((t))$. See for example the last section of arXiv:0909.2541. -These two papers have appeared in J. Ramanujan Math. Soc. 25 (2010), no. 1, 25–80, -and 25 (2010), no. 4, 393–417. -There are other instances where the image of the norm map can be computed explicitly. This happens for the cyclotomic extension $K_m$ of $\mathbf{Q}_p$ obtained by adjoining $\root{p^m}\of1$. It can be shown that $p\in N(K_m^\times)$, and that the image of the units of $K_m$ under the norm map down to $\mathbf{Q}_p$ is $1+p^m\mathbf{Z}_p$. See for example Artin, -Algebraic numbers and algebraic functions, p. 208, or Neukirch, -Class Field Theory, p. 45.<|endoftext|> -TITLE: Assigning positive edge weights to a graph so that the weight incident to each vertex is 1. -QUESTION [7 upvotes]: Let $\Gamma=(G,E)$ be a connected undirected graph, with no loops or multiple edges. $G$ is finite or countably infinite. For each edge $e=\{x,y\}\in E$, we assign a positive, symmetric edge weight $c_e := c_{\{x,y\}} = c_{xy} = c_{yx}$. I would like to know for which graphs $\Gamma$ it is possible to choose $(c_e)_{e\in E}$ so that for each $x\in G$, -\begin{equation*} -\sum_{y\sim x} c_{xy} = 1. -\end{equation*} -For example, this is possible on any $d-$regular graph if one sets $c_e \equiv 1/d$. The graph with vertex set $\{x,y,z\}$ and edges $\{x,y\}$ and $\{y,z\}$ shows that it is not always possible. - -REPLY [8 votes]: Here is a solution along the lines of JBL's answer. -First a couple of definitions: - -A disjoint cycle cover of a graph is a collection of cycles of our graph which are disjoint subgraphs, and contain all the vertices of our graph. A special case of a disjoint cycle cover is a perfect matching, for instance. -We will call the the permanent of a graph, the permanent of its adjacency matrix - -An easy fact is that the permanent of a graph counts its disjoint cycle covers. -Now, to our result: - -Theorem: A graph admits edge weights as in the problem if and only if every edge is contained in a disjoint cycle cover. Equivalently if and only if removing an edge decreases the permanent. - -proof: -Suppose the graph $G$ has such weights. Then the matrix $A$ with $a_{ij}$ being the weight of the edge connecting vertices $v_i$ and $v_j$, is doubly stochastic, and thus by the Birkhoff-von Neumann theorem can be written as a convex combination of permutation matrices. For every edge of $G$, there is a non zero term $a_{ij}$ in $A$ which means that there is a permutation matrix in our sum with a $1$, in the $ij$ entry, call this matrix $M_{ij}$. The first observation is that $M_{ij}$ has all zeros on the diagonal, and secondly that all it's non-zero entries correspond to edges in $G$. This collection of edges is of course a disjoint cycle cover. -Now for the other direction, each cycle cover can be assigned weights as in the problem (just assign $1/2$ to all edges in proper cycles and $1$ to all isolated edges). So taking an appropriate convex combination of all such covers gives us weights for $G$. -A special case is of course when all edges are contained in perfect matchings, but this property doesn't characterize all graphs as in the question, as the example I gave in the comment to JBL's answer shows (also just look at odd cycles). Which is why one must include more general cycle covers. - -Perhaps it is a bit more clear if we phrase it in the following way. When restricting to bipartite graphs, the property of each edge being contained in a disjoint cycle cover is equivalent to every edge being in a perfect matching (there are no odd cycles). Now the result above follows because weights on our graph induce weights on its bipartite double cover which sum to 1 at each vertex. A disjoint cycle cover of a graph is equivalent to a perfect matching of its bipartite double cover.<|endoftext|> -TITLE: Dual space of $\ell^\infty$ -QUESTION [8 upvotes]: Why can the elements of the dual space of $\ell^\infty(\mathbb N)$ be represented as sums of elements of $\ell^1(\mathbb N)$ and Null$(c_0)$? - -EDIT: As confirmed in the comments, the OP intended to ask about this sentence -"$f\in\ell_\infty^*$ is the sum of an element of $\ell_1$ and an element null on $c_0$" from the paper D. H. Fremlin and M. Talagrand: A Gaussian Measure on $l^\infty$ http://jstor.org/stable/2243023 (Which is different claim from what was in the original version of the question.) - -REPLY [6 votes]: Let's recall a simple, elementary, and general fact that hasn't been explicitly mentioned: a dual Banach space is always a splitting subspace in the isometric embedding into its double dual. -Let $i_X:X\to X^{**}$ denote the natural isometric embedding of $X$ in $X^{**}$. If we dualize, we have a transpose operator, $i_X^*:X^{***}\to X^*$ (that we may identify as the restriction map, which takes a linear form on $X^{**}$ to its restriction on $X$ as a subspace of $X^{**}$). On the other hand we also have the isometric embedding $i_{X^*}:X^*\to X^{***}$. It is a straightforward (though a bit formal) computation checking that $i_{X}^*$ is left-inverse to $i_{X^*}$, that is $i_{X}^*i_{X^*}=1_{X^*}.$ As a consequence of this, $P:=i_{X^*}i_{X}^*$ is a linear projector with $\operatorname{ker}P=\operatorname{ker}i_X^*=X^\perp$ corresponding to the splitting $X^{***}=X^*\oplus X^\perp$. -$$*$$ -Checking the identity $i_{X}^*i_{X^*}=1_{X^*}.$ This means $i_{X}^*i_{X^*}f=f$ for all $f\in X^*$, which also means $\langle i_{X}^*i_{X^*}f, x\rangle=\langle f, x\rangle$ for all $x\in X$ and $f\in X^*$. Indeed -$$\langle i_{X}^*i_{X^*}f, x\rangle_{X^*\times X}=\langle i_{X^*}f, i_{X} x\rangle_{X^{***}\times X^{**}}=\langle i_{X} x,f\rangle_{X^{**}\times X^*}=\langle f, x\rangle_{X^*\times X},$$ -by the definition of the transpose operator $i_{X}^*$, respectively by the definition of the embeddings $i_{X^*}$ and $i_{X}$.<|endoftext|> -TITLE: Galois Bicentennial? -QUESTION [20 upvotes]: The 200-th anniversary of the birth of Galois will be on October 25th, 2011. For Abel's bicentennial birth year in 2002, Norway had a big conference and initiated the Abel prize. A cursory web search doesn't reveal any major (or minor) conference to mark the bicentennial for Galois. Is there something being planned that hasn't been broadly announced yet? -[Edit: Before posting this question, I looked at the number theory conference page http://www.numbertheory.org/ntw/N3.html and found nothing there related to my question.] -I realize that is not a math research question, but I don't think it should be closed since it should interest many users of this site, it should have a definite answer, and it is not "argumentative", "too localized" (at what prime?) or "not a real question". -[Edit: Now that Emerton has answered the question well and nobody else has followed up after several days with other replies, I suspect he has given the best possible answer and I have designated it as such.] - -REPLY [3 votes]: There is also some events (exhibitions, large audience talks) in Bourg la Reine, the town where Galois grew up (near Paris). -Parlons d'Évariste Galois<|endoftext|> -TITLE: Reference Request: Independence of the ultrafilter lemma from ZF -QUESTION [10 upvotes]: I'm looking for references for the following facts concerning the ultrafilter lemma (~ "there exist non-principal ultrafilters"): - -The ultrafilter lemma is independent of ZF. -ZF + the ultrafilter lemma does not imply the Axiom of Choice. - -I would prefer an overview article / book that links to the original papers instead of the original papers themselves. That's because I don't rely on these facts; I only want to give some context for my readers. - -REPLY [22 votes]: In any of the formulations mentioned (so far) in the comments, the ultrafilter lemma is independent of ZF but weaker than AC. That the strongest form (all filters can be extended to ultrafilters) doesn't imply AC is a theorem of J.D. Halpern and A. Lévy ["The Boolean prime ideal theorem does not imply the axiom of choice" in Axiomatic Set Theory, Proc. Symp. Pure Math. XIII part 1, pp. 83-134]. That ZF doesn't prove even the weakest form (there exists a nonprincipal ultrafilter on some set) is a theorem of mine ["A model without ultrafilters," Bull. Acad. Polon. Sci. 25 (1977) pp. 329-331], building on S. Feferman's construction of a model with no non-principal ultrafilters on the set of natural numbers ["Some applications of the notions of forcing and generic sets," Fundamenta Mathematicae 55 (1965) pp. 325-345].<|endoftext|> -TITLE: Example of an infinite index subgroup of a non-amenable group whose normalizer is of non-zero finite index, and such that the Schreier graph is of subexponential growth -QUESTION [6 upvotes]: EDIT: In this question I forgot to put one of the assumptions, and the question was easier than it should be. Here is the revised question. Please vote to close this question as it is no longer relevant. - - -Question. Let $G$ be a finitely generated non-amenable discrete group, and $H$ be a subgroup of $G$ of infinite index. Can it happen that the index of the normalizer $N(H)$ of $H$ in $G$ is finite greater than $1$, and the Schreier graph of $G/H$ has subexponential growth? - -If the answer is yes, I would very much like to see an example. It would be especially nice if $G$ could be taken to be a property $(T)$ group. - -REPLY [9 votes]: The answer to the question is yes. Here are two examples, the first one answers the question as it is posted and in the second one I assume that $N(H)$ is the normal closure of $H$ in $G$. The second example seems to be more interesting, in the sense that one have to do more computation on Schreier graph, compering to the first example, where the computation of the growth of the Schreier graph of $G/H$ is straightforward. - -If $N(H)$ is normalizer of $H$, then the following example works. $G:=\mathbb{Z}\times\mathbb{F}_2$ and $H$ is a subgroup $(e,H_0)$, where $H_0$ is of finite index but not normal in $\mathbb{F}_2$. Then index of $N(H)=\mathbb{Z}\times N_{\mathbb{F}_2} (H_0)$ is finite and not equal to $1$, since $N_{\mathbb{F}_2}(H_0)\neq \mathbb{F}_2$, because $H_0$ is not normal in $\mathbb{F}_2$. -If $N(H)$ is normal closure of $H$ in $G$, then the following example works. Let $\mathbb{F}_2$ be the free group on two generators $a$ and $b$. Define a homomorphism $\phi:\mathbb{F}_2\rightarrow Aut(\mathbb{Z})=\mathbb{Z}_2$ by $\phi(a)=0,\phi(b)=1$. Then the question is valid for the semidirect product $G=\mathbb{F}_2\ltimes_{\phi} \mathbb{Z}$. The Schreier graph of $G/H$ has polynomial growth and the normal closure of $H$ is $\mathbb{F}_2\ltimes_{\phi} 2\mathbb{Z}$.<|endoftext|> -TITLE: Wiener Sausages in Riemann Surfaces -QUESTION [11 upvotes]: Let $M$ be a Riemann surface (or a higher dimensional manifold) and let's assume that it's geodesically complete. Let $W(t)$ be a Brownian motion on the surface accordingly to the manifold's Laplacian and let $r>0$. -Define the Wiener sausage as: -$$ -W_{r}(t):=\{ x\in M: d(x,W(s))\leq r\quad\text{for}\quad 0\leq s\leq t \}. -$$ -It is known that in $\mathbb{R}^{2}$ and for t sufficiently large and $r$ fixed -$$ -\mathbb{E}[\mathrm{vol}(W_{r}(t))]=\frac{2\pi t}{\log(t)}(1+o(1)). -$$ -Is there any analogue result for a general Riemann surface or at least the hyperbolic space? -Thanks! ---Gabriel - -REPLY [3 votes]: I just found out that the case $r$ fixed and $t\to\infty$ for simply connected symmetric manifolds of non-positive sectional curvature and dimension $d\geq 3$, and strictly negative curvature for dimension $d=2$, was solved by Chavel and Feldman in "The Wiener Sausages and a Theorem of Spitzer in Riemannian Manifolds", Probability and Harmonic Analysis, New York, pp. 45-60, 1986.<|endoftext|> -TITLE: Up to $10^6$: $\sigma(8n+1) \mod 4 = OEIS A001935(n) \mod 4$ (Number of partitions with no even part repeated ) -QUESTION [11 upvotes]: Up to $10^6$: -$\sigma(8n+1) \mod 4 = OEIS A001935(n) \mod 4$ -A001935 Number of partitions with no even part repeated -Is this true in general? -It would mean relation between restricted partitions of $n$ and divisors of $8n+1$. -Another one up to $10^6$ is: -$\sigma(4n+1) \mod 4 = A001936(n) \mod 4$ -A001936 Expansion of q^(-1/4) (eta(q^4) / eta(q))^2 in powers of q -$\sigma(n)$ is sum of divisors of $n$. - -sigma(8n+1) mod 4 starts: 1, 1, 2, 3, 0, 2, 1, 0, 0, 2, 1, 2, 2, 0, 2, 1, 0, 2, 0, 2, 0, 3, 0, 0, 2, 0, 0, 0, 3, 2 -sigma(4n+1) mod 4 starts: 1, 2, 1, 2, 2, 0, 3, 2, 0, 2, 2, 2, 1, 2, 0, 2, 0, 0, 2, 0, 1, 0, 2, 0, 2, 2 - -Update -Up to 10^7 -A001935 mod 4 is zero for n = 9m+4 or 9m+7 -A001936 mod 4 is zero for n = 9m+5 or 9m+8 -Question about computability - -REPLY [23 votes]: Let's call A001936(n) by $a(n)$. Here is a sketch of why $$a(n)\equiv \sigma(4n+1)\pmod{4}$$ -Firs note that the generating function of $a(n)$ is -$$A(x)=\sum_{n\geq 0}a(k)x^n=\prod_{k\geq 1}\left(\frac{1-x^{4k}}{1-x^k}\right)^2$$ for $\sigma(2n+1)$ the generating function is $$B(x)=\sum_{k\geq 0}\sigma(2k+1)x^k=\prod_{k\geq 0}(1-x^k)^4(1+x^k)^8$$ -So $$B(x)\equiv \prod_{k\geq 1}(1+x^{2k})^2(1+x^{4k})^2\equiv \prod_{k\geq 1}(\frac{1-x^{8k}}{1-x^{2k}})^2\equiv A(x^2)\pmod{4}$$ -Now the proof is complete once we know that $$B(x)\equiv \sum_{k\geq 0} \sigma(4n+1)x^{2n}\pmod{4}$$ this is an other way of saying $\sigma(4n-1)$ is divisible by $4$, which can be shown by pairing up the divisors $d+\frac{4n-1}{d}\equiv 0\pmod{4}$. -The proof for the other congruence is similar, but slightly longer, I might update this post later to include it. - -Let's prove that $\sigma(8n+1)\equiv q(n)\pmod{4}$, where $q(n)$ is the number of partitions with no even part repeated. The generating function is $$Q(x)=\sum_{n\geq 0}q(n)x^n=\prod_{k\geq 1}\frac{1-x^{4k}}{1-x^k}$$ -Since we know from above that $$\sum_{n\geq 0}\sigma(4n+1)x^{2n}\equiv \prod_{k\geq 1}(1+x^{2k})^2(1+x^{4k})^2 \pmod{4}$$ we conclude that $$L(x)=\sum_{n\geq 0}\sigma(4n+1)x^n\equiv Q(x)^2 \pmod{4}$$ -so that $$\sum_{n\geq 0} \sigma(8n+1)x^{2n}\equiv \frac{L(x)+L(-x)}{2}\pmod{4}$$ -So to finish off the proof we need the following -$$\frac{Q(x)^2+Q(-x)^2}{2}\equiv Q(x^2)\pmod{4}$$ which I will leave as an exercise Actually let me write the proof, just to make sure I didn't mess up calculations. This reduces to proving -$$\frac{\prod_{k\geq 1}(1+x^{2k})^4(1+x^{2k-1})^2+\prod_{k\geq 1}(1+x^{2k})^4(1-x^{2k-1})^2}{2}$$ $$\equiv \prod_{k\geq 1}(1+x^{4k-2})(1+x^{4k})^2 \pmod{4}$$ and since $$(1+x^{2k})^4\equiv (1+x^{4k})^2 \pmod{4}$$ this reduces to -$$\frac{\prod_{k\geq 1}(1+x^{2k-1})^2+\prod_{k\geq 1}(1-x^{2k-1})^2}{2}\equiv \prod_{k\geq 1} (1+x^{4k-2})\pmod{4}$$ but we can write $$\prod_{k\geq 1}(1-x^{2k-1})^2\equiv \left(\prod_{k\ geq 1}(1+x^{2k-1})^2\right) \left(1-4\sum_{k\geq 1}\frac{x^{2k-1}}{(1+x^{2k-1})^2}\right)\pmod{8}$$ therefore now we have to show -$$\prod_{k\geq 1}(1+x^{2k-1})^2\left(1-2\sum_{k\geq 1}\frac{x^{2k-1}}{(1+x^{2k-1})^2}\right)\equiv \prod_{k\geq 1}(1+x^{4k-2})\pmod{4}$$ Now everything is clear since $$\prod_{k\geq 1}(1+x^{2k-1})^2\equiv \prod_{k\geq 1}(1+x^{4k-2})\left(1+2\sum_{k\geq 1}\frac{x^{2k-1}}{(1+x^{2k-1})^2}\right)\pmod{4}$$<|endoftext|> -TITLE: Regge calculus: Questions of consistency resolved? -QUESTION [13 upvotes]: Hello, -Regge calculus is an approximation scheme for General Relativity, which has been introduced in early-sixties and has been adopted both in numerical relativity and numerical quantum relativity. In contrast to its widespread use in computational science, there does not seem to exist much theory on whether the Regge calculus is actually consistent - i.e. whether there is some degree of exactness (like mesh width) we can adapt arbitrarly to obtain a solution with arbitrarly small error (say in some Sobolev-norm). -In fact there has been debate about this: - -The Regge Calculus is not an approximation to General Relativity, 1995 -On the convergence of Regge calculus to general relativity, 2000 - -The question of consistency is a purely mathematical one, and therefore I do not expect it to be "debatable". I will have to work with this theory and hence I do wonder in how far there is a theoretical basis to tell apart "It works" and "It does not work". -Thank you very much! - -REPLY [12 votes]: The consistency is proved by Cheeger, M\"uller and Schrader in 1984, "On the Curvature of Piecewise Flat Sapces". Roughly speaking, given a smooth Riemannian manifold with a smooth metric, there exists a sequence of triangulation, on which Regge's definition converges to the smooth curvature as a measure. -At the linearized level, there is also a recent paper on the consistency: Christiansen 2011, "On The Linearization of Regge Calculus". One of the theorems in the paper is that we have consistency between linearized Regge and linearized Einstein equation as well. -That said, when you talk about the convergence of a numerical algorithm, it depends on a lot of other things as well, such as your formulation of the Einstein's equation. You will also need some form of stability to ensure convergence. Those questions remain to be solved (hopefully in my thesis:-).<|endoftext|> -TITLE: Is a flat, locally finite type morphism open? -QUESTION [8 upvotes]: Let $f: X\to Y$ be a morphism between arbitrary schemes. It is proved in EGA IV Prop 2.4.6 that if f is flat and locally of finite presentation, then f is open. If one replaces locally of finite presentation by locally of finite type, I don't think that the statement is still true (unless X and Y are locally noetherian) but I don't know of any counterexample. - -REPLY [4 votes]: Over a field $k$, another related (and somehow more algebraic) example is the "universal" $k$-algebra $A$ generated by countably many orthogonal idempotents. In other words $A$ is the $k$-algebra obtained as the quotient of the polynomial ring $k[X_1,X_2,...]$ in countably many variables, by the ideal generated by all polynomials $X_i^2-X_i$ and $X_iX_j$ for $i\ne j$. It is easy to see that the maximal ideal $m=(X_1,X_2,...)$ gives a nonisolated closed point of $Spec(A)$. Moreover the local ring $A_m$ is just $k$, so that the closed immersion $i:Spec(A/m)\to Spec(A)$ is flat (the extension of local rings is an isomorphism). Thus it is a flat, finite type, morphism which is not open.<|endoftext|> -TITLE: Fixing a mistake in "An introduction to invariants and moduli" -QUESTION [9 upvotes]: On page 13 of the book "An introduction to invariants and moduli" of Mukai -http://catdir.loc.gov/catdir/samples/cam033/2002023422.pdf there is a mistake, in the end of the proof of Proposition 1.9. It seems to me that this proof can not be fixed, without using the notion of Noetherian rings and Hilbert basis theorem. -The question is: Can this proof be fixed, without using commutative algebra -- i.e., by the elementary reasoning that Mukai is using there? -I reproduce here the proof from the book for completeness. $S$ is the ring of polynomials, $G$ a group, $S^G$ is the ring of invariants -Proposition. If $S^G$ is generated by homogeneous polynomials $f_1,...,f_r$ -of degrees $d_1,...,d_r$, then the Hilbert series of $S^G$ is the power -series expansion at $t=0$ of a rational function -$$P(t)=\frac{F(t)}{(1-t^{d_1})...(1-t^{d_r})}$$ -for some $F(t)\in \mathbb Z[t]$. -Proof. We use induction on $r$, observing that when $r=1$, the ring $S^G$ is just $\mathbb C[f_1]$ with the Hilbert series $$P(t)=1+t^{d_1}+t^{2d_1}+...=\frac{1}{1-t^{d_1}}.$$ -For $r>1$ consider the injective complex linear map $S^G\to S^G$ defined by $h\to f_rh$. Denote the image by $R\subset S^G$ and consider the Hilbert series for the graded rings $R$ and $S^G/R$. Since $R$ and $S^G/R$ are generated by homogeneous elements, we have -$$P_{S^G}(t)=P_{R}(t)+P_{S^G/R}(t).$$ -On the other hand, $dim(S^G\cap S_d)=dim(R\cap S_{d+d_r})$, so that $P_R(t)=t^{d_r}P_{S^G}(t)$, and hence -$$P_{S^G}(t)=\frac{P_{S^G/R}(t)}{1-t^{d_r}}.$$ -But $S^G/R$ is isomorphic to the subring of $S$ generated by the polynomials $f_1,...,f_{r-1}$, and hence by the induction hypothesis $P_{S^G/R}(t)=F(t)/(1-t^{d_1})...(1-t^{d_{r-1}})$ for some $F(t)\in \mathbb Z[t]$... -Mistake: It is not true that $S^G/R$ is isomorphic to the subring of $S$ generated by polynomials $f_1,...,f_{r-1}$. For example consider $\mathbb C^2$ with action $(x,y)\to (-x,-y)$. Then let $f_1=x^2$, $f_2=y^2$, $f_3=xy$. -Motiviation of this question. Of course this proposition is a partial case of Hilbert-Serre theorem, proven for example at the end of Atiyah-Macdonald. But the point of the introduction in the above book is that one does not use any result of commutative algebra. - -REPLY [3 votes]: This is not an answer to the question, I just decided to give for completeness a standard proof of the above statement that uses (a version of) Hilbert-Serre theorem. In this proof we need to use Hilbert basis theorem. In the above statement $S^G$ is clearly a finitely generated graded module over the ring of polynomials $\mathbb C[x_1,...,x_r]$, so it is sufficient to prove: -Theorem (Hilbert, Serre). Suppose that $S=\sum ^{\infty}_{j=0}S_j$ is a commutative graded ring with $A_0=\mathbb C$, finitely generated over $\mathbb C$ by homogeneous elements $x_1,...,x_r$ in positive degrees $d_1,...,d_r$. Suppose that -$M=\sum_{j=0}^{\infty} M_j$ is a finitely generated graded $S$-module (i.e., we have $S_iS_j\subset S_{i+j}$ and $S_iM_j\subset M_{i+j}$). -Then the Hilbert series $P(M,t)$ is of the form -$$\sum_{j=0}^{\infty}dim(M_j)t^j=P(M,t)=\frac{F(t)}{\Pi_{j=1}^r(1-t^{d_j})}, -\;\; F(t)\in \mathbb Z[t].$$ -Proof. -We work by induction on $r$. If $r=0$ then $P(M,t)$ is a polynomial with integer coefficients, so suppose $r>0$. Denote by $M'$ and $M''$ the kernel and cokernel of the multiplication by $x_r$, we have an exact sequence for each $j$ -$$0\to M'_j\to M_j \to^{x_r}M_{j+d_r}\to M''_{j+d_r}\to 0.$$ -Now $M'$ and $M''$ are finitely generated graded modules for $K[x_1,...,x_{r-1}]$, and so by induction their Hilbert series have the given form. From the above exact sequence we have -$$t^{d_r}P(M',t)-t^{d_r}P(M,t)+P(M,t)-P(M'',t)=0.$$ -Thus -$$P(M,t)=\frac{P(M'',t)-t^{d_r}P(M',t)}{1-t^{d_r}}$$ -has the given form. -Where did we use Hilbert basis theorem? We use it when we say that $M'$ is finitely generated.<|endoftext|> -TITLE: Hopfian property preserved by extensions with finite kernel? -QUESTION [7 upvotes]: Let $G$ be a group which is Hopfian and given a short exact sequence $1\to F \to H \to G \to 1$ -with $F$ a finite normal subgroup of $H$. Is $H$ Hopfian? - -REPLY [6 votes]: It seems that the answer is no: there exists an exact sequence -$$1\to F\to H\to Q\to 1$$ -with $F$ finite (central), $H$ non-Hopfian, and $Q$ Hopfian (with in addition, $H$ finitely generated solvable). -Mark Sapir's answer refers to a group constructed here (see 5.10), which is Abels' group over the ring $\mathbf{F}_p[t,1/t]$, and which probably be used to provide a negative answer to the question. Define the group $G$ as the group of matrices -$$\left(\begin{array}{rrrr} -1 & u_{12} & u_{13} & u_{14}\newline -0 & d_{22} & u_{23} & u_{24}\newline -0 & 0 & d_{33} & u_{34}\newline -0 & 0 & 0 & 1\newline -\end{array}\right) -$$ -where $u_{ij}\in\mathbf{F}_p[t,1/t]$, -and $d_{ii}\in\mathbf{F}_p[t,1/t]^\times=\langle t\rangle\mathbf{F}_p^\times$. -(Actually one can restrict to $d_{ii}\in\langle t\rangle$ but it's not important.) Let $e_{14}$ denote the map $x\mapsto\begin{pmatrix} 1&0&0&x\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}$. -As suggested by user "BS.", let $N$ be the central subgroup $e_{14}(\mathbf{F}_p[t^2])$ and $M$ the larger central subgroup $e_{14}(\mathbf{F}_p[t^2]\oplus \mathbf{F}_pt^{-1})$. (In a the initial post, $M$ and $N$ were chosen as other central subgroups but unfortunately $G/M$ failed to be Hopfian). So we have the central exact sequence with finite kernel -$$1\to M/N\to G/N\to G/M\to 1.$$ -Conjugation by the diagonal matrix $(t^2,1,1,1)$ is an automorphism of $G$, which maps $N=e_{14}(\mathbf{F}_p[t^2])$ strictly into itself and hence $H=G/N$ is non-Hopfian. -I haven't completely checked but here are some guidelines to show the group $Q=G/M$ is Hopfian. -Write the original group (given by $4\times 4$ triangular matrices) as $G=D\ltimes U$ with $D=\mathbf{Z}^2$ and $U$ its unipotent part. Set $U^2=[U,U]$ and $U^3=[U,U^2]$, which is central and equal to $e_{14}(\mathbf{F}_p[t,1/t])$. -Let $f$ be a surjective endomorphism of $Q$. - -check that the center of $G$ is precisely $U^3$. -It follows that $f$ induces a surjective endomorphism of $G/U^3$. Since this group is linear, it is Hopfian so this is an automorphism of $G/U^3$. - -Describe the group of automorphisms of $G/U^2 = \mathbf{Z}^2\ltimes F_p[t,1/t]^3$. (It should be reasonably easy to describe). - -Deduce a description of the group of automorphisms of $G/U^3$, or at least describe how these automorphisms act on $U^2/U^3$, showing that modulo something, the coefficient $12$ is multiplied by a monomial $w\cdot t^a$ ($w\in \mathbf{F}_p^*$) and the coefficient $24$ is multiplied by $vt^b$. So, taking a commutator (which should kill the "modulo something"), we obtain that in the "action of $f$ on $G$", the coefficient $14$ should be multiplied by a nonzero monomial. This multiplication should stabilize $M$ so this is multiplication by a scalar in $\mathbf{F}_p^*$, which implies that f actually induces an automorphism of $Q$.<|endoftext|> -TITLE: Generating finite simple groups with $2$ elements -QUESTION [43 upvotes]: Here is a very natural question: -Q: Is it always possible to generate a finite simple group with only $2$ elements? -In all the examples that I can think of the answer is yes. -If the answer is positive, how does one prove it? Is it possible to prove it without using the -classification of finite simple groups? - -REPLY [5 votes]: Carlisle King recently posted an arXiv preprint which (claims to) show that every finite simple group is generated by an involution, together with another element of prime order. -http://arxiv.org/abs/1603.04717 -The paper uses the classification, as most of these do. -Elements of prime (or even prime power) order seem to be particularly easy to work with for a number of kinds of arguments. See e.g. this mathoverflow question. -Now, if you want a result that doesn't use the classification, Paul Flavell gave an elementary proof that every non-solvable group has 2 elements that generate a non-solvable group. See here.<|endoftext|> -TITLE: Which primes can divide orders of Tate-Shafarevich groups? -QUESTION [10 upvotes]: Heuristic arguments due to (I believe) Delauney predict that every prime divides the order of the Tate-Shafarevich group of infinitely many elliptic curves over $\mathbf{Q}$. However, is it even known that for every prime $p$, there's at least one elliptic curve over $\mathbf{Q}$ whose Tate-Shafarevich group is finite and has order divisible by $p$? - -REPLY [10 votes]: Just to record an answer: yes, it is believed to be true that for each prime $p$ there is an elliptic curve $E_{/\mathbb{Q}}$ whose Shafarevich-Tate group contains an element of order $p$. But this is wide open. -One can however prove the following: -Theorem (R. Kloosterman): If you fix a prime $p$, then there is a number field $K_p$ -- i.e., depending on $p$ -- and an elliptic curve $E/K_p$ whose Shafarevich-Tate group contains an element of order $p$. -As Kloosterman remarked (in his 2005 paper on the subject, and also just now in the above comments) via a simple "Weil restriction" argument, this implies that there exists an abelian variety $A_{/\mathbb{Q}}$ whose Shafarevich-Tate group contains an element of order $p$. But now the dimension of $A$ goes to infinity with $p$. -Others have since worked out various improvements of Kloosterman's Theorem. Here are three different directions: -1) What is the minimal degree $d_p = [K_p:\mathbb{Q}]$ of a number field $K_p$ in above theorem? Kloosterman's argument gave $d_p = O(p^4)$. In a 2005 paper I showed that one can take -$d_p = 2p^3$. In a 2010 paper with Shahed Sharif, we showed that one can take $d_p = p$. (Note that whether $d_p = p$ is in fact best possible is an open question. At one point I thought I had an argument to show that it was, but this was wrong. I now suspect that this is only best possible by a method proceeding along the lines of our construction. For instance, I believe it is conceivable that $d_p = 2$ for all $p$, and this has something to do with how Sha behaves in quadratic twists...) -2) What kind of elliptic curves $E$ can be used to produce these elements in Sha? Sharif and I showed that in fact one can start with any elliptic curve $E_0$ over $\mathbb{Q}$ and take its base change to $K_p$ a degree $p$ number field. (Similarly, one can start with any elliptic curve $E$ over any number field and get order $p$ elements of Sha in an extension field of degree $p$.) And in fact $p$ does not need to be a prime number here: it holds for every integer $n$. And in fact you can get as many elements of order $n > 1$ as you want. -3) What kind of control can one get over the field extension $K_p/\mathbb{Q}$? Can one for instance choose it to be Galois, abelian, or cyclic? Matsuno proves that for any cyclic degree $p$ extension $K_p/\mathbb{Q}$ one can find elliptic curves $E_{\mathbb{Q}}$ such that the base change to $K_p$ has as many order $p$ elements in its Shafarevich-Tate group as one wants. There are similar results along these lines (with dihedral extensions and elements of the Selmer group) by Alex Bartel. -P.S.: As Tim Dokchitser points out, replacing $\mathbb{Q}$ by any fixed number field does not help to answer the OP's question, although the results of Sharif and myself hold equally well in this relative setting. (I was going to ask "In what way could an arbitrary but fixed number field be any better than $\mathbb{Q}$?" but then I remembered an amazing result about $2$-Selmer parity that only holds over certain imaginary number fields...due to Dokchitser and Dokchitser.)<|endoftext|> -TITLE: Importance of Poincaré recurrence theorem? Any example? -QUESTION [22 upvotes]: Recently I am learning ergodic theory and reading several books about it. -Usually Poincaré recurrence theorem is stated and proved before ergodicity and ergodic theorems. But ergodic theorem does not rely on the result of Poincaré recurrence theorem. So I am wondering why the authors always mention Poincaré recurrence theorem just prior to ergodic theorems. -I want to see some examples which illustrate the importance of Poincaré recurrence theorem. -Any good example can be suggested to me? -Books I am reading: -Silva, Invitation to ergodic theory. -Walters, Introduction to ergodic theory. -Parry, Topics in ergodic theory. - -REPLY [2 votes]: This is an old question but I don't see the obvious answer, so here we go. -A huge field of research in mathematical physics during the XIXe century revolved around giving explicit solutions to the equations of classical mechanics, using brute computation. At that time, finding a new first integral in the equations of motions of some physical system was a sure path to academic fame. If sufficiently many first integrals are found, the system is integrable. If moreover the motion is constrained to a bounded domain of the phase space , it is quasi-periodic and thus both regular and recurrent. The long term behavior of the system is thoroughly described. -There was certainly some hope at first to show the stability of the solar system, or at least the three body problem, by finding these first integrals. Huge efforts went into that line of research. A century later, we know that the three body problem is not integrable in general and integrability is a pretty rare property of dynamical systems. In particular it is not stable by perturbation. -But wait, we know that Earth won't suddenly fly towards Pluto and stay there forever. Actually, we are pretty sure that it will come back to its current position, and this follows from Poincaré recurrence theorem, once you note that the Liouville measure is left invariant by the motion. No need to explicitly solve the equations of motion, and the result is so general that it applies to all hamiltonian systems in restriction to a compact level of energy. And the proof is short and elegant! To be pedantic, this was a paradigm shift and the birth of a new method in the field of mathematical physics, best described by Poincaré himself in his numerous books. -We now consider the Poincaré recurrence theorem as marking the birth of a new mathematical discipline called ergodic theory, with striking applications to arithmetic, Lie groups, foliations, moduli spaces etc. To the point that people seem to have forgotten its celestial origins.<|endoftext|> -TITLE: Index theorem interpretation of the spectral flow for a pseudo holomorphic curve -QUESTION [9 upvotes]: Let $(M , \omega)$ be a symplectic manifold, $J$ a compatible almost complex structure. We call pseudo holomorphic strip a solution $u : \mathbb R \times I \to M$ of the equation $\partial_s u + J \partial_t u = 0.$ Given a pair of Lagrangian submanifolds $L_0, L_1$, such a strip is said to be bounded by the pair if $u(s, i) \in L_i, i = 0, 1.$ Under mild conditions, such a strip has limits as $s \to \infty$ that are intersection points between the Lagrangian submanifolds. -Robbin & Salamon proved that If the Lagrangian intersections are transverse, the Fredholm index between suitable Sobolev spaces of this linearized Cauchy-Gromov-Riemann operator coincides with the Maslov-Viterbo of the strip. Their proof involves general considerations for linear operators of the form $\partial_s + A_s$ defined on the space of paths $\mathbb R$ to some Hilbert space and rely on reducing the problem to a finite dimensional Hilbert space. -However the eventual result admits a purely intrinsic formulation : it states that the Fredholm index of a Dirac operator is given in terms of a characteristic class. This seems like a particular instance of the Atiyah-Singer index theorem, only on a manifold with boundary (the strip) and with totally real boundary conditions. -Can this particular result (index = Maslov class) be obtained through a less coordinate-bound and maybe more striking way ? - -REPLY [6 votes]: I try to elaborate on my own comment to Tom Mrowka's answer. -Assume that $f : \mathbb S^1 \to \mathbb S^1$ is càdlàg and has a unique discontinuity point at $1 \in \mathbb S^1$. This discontinuity is therefore a "jump" by a certain angle $\alpha.\pi$. This matches the behavior of the function $f_\alpha : z \mapsto (z-1)^\alpha,$ so that upon normalizing inside $\mathbb D^2$ by this holomorphic function, the problem reduces to the modelization proposed by Tom Mrowka. -By using multiple such normalizations, one gets a geometric description of the virtual dimension for discs with marked points on their boundary, each segment thus delimited abutting on a different "real condition", as relevant in the study of products in Floer theory.<|endoftext|> -TITLE: What is the cover time of a random walk on a cube? -QUESTION [11 upvotes]: I can't quite figure this problem yet. There is an ant at one vertex of a cube. The ant goes from one vertex to another by choosing one of the neighboring vertices uniformly at random. What is the average minimum time it takes to visit all vertices? - -REPLY [6 votes]: Yet another reference: Some sample path properties of a random walk on the cube, by Peter Matthews (1989). This covers the asymptotic distribution of the time $T$ taken to visit all vertices, the distribution of the number of vertices not visited at times near to $\mathbb{E}[T]$, and the expected time taken for the walk to come within a distance $d$ of all vertices.<|endoftext|> -TITLE: What relations exist among (quasi-) modular forms of different levels? -QUESTION [13 upvotes]: Given a (quasi-) modular form $f(\tau)$ for some congruence subgroup (say) $\Gamma(k)$, we know that $f(N\tau)$ is a (quasi-) modular form for $\Gamma(N k)$. Is there anything known about when we can do a partial reverse, that is, when we can take linear combinations of (quasi-) modular forms for some higher level subgroup to obtain one of stricktly lower level? -An example is the following: -Let $E(q) = \sum_{k=0}^\infty \sigma_1(2k+1)q^{2k+1}$ and $A(q) = \sum_{k=1}^\infty \sigma_1(k)q^k$. Then $E(q)$ is modular with respect to a non-trivial character, and both $A(q^2)$ and $A(q^4)$ are quasi-modular of level 2 and 4, respectively (though not of pure weight). -However: it turns out that -$$ -E(q) + 3A(q^2) - 2A(q^4) = A(q) -$$ -which shows that a linear combination of higher level terms (and one which is modular with respect to a non-trivial character) yields one of lower level. -Is this simply random chance? Are there known relations of this type? - -REPLY [2 votes]: There are (at least) two answers: there is a Galois theory of modular functions (as in G. Shimura's 1971 book, or in Lang's book on elliptic functions), and a theory of newforms" (Atkin-Lehner, and also Casselman in a representation-theoretic context). -Thinking in terms of general Galois theory, it ought not be so surprising that various sums of non-invariant things become invariant, I suppose, but the particulars are often non-trivial.<|endoftext|> -TITLE: Confidence intervals when the number of samples is random -QUESTION [6 upvotes]: I am interested in computing confidence intervals for the mean of a random variable $X$ given $\require{cancel}\xcancel{N \text{ i.i.d. samples}}$ an i.i.d. sample of $N$ copies of $X$, where $N$ is $\operatorname{Binomial}(n, p)$. Any time I read about confidence intervals for the mean it is assumed that the number of samplessize of the sample is fixed, which makes the asymptotic distribution of the sample mean Gaussian, and therefore allows for student-based confidence intervals and the like to be justified. However, if the number of samples size $N$ of the sample is a random variable itself, then the ratio -$$ \frac{\sum_i X_i}{N} $$ -will not be necessarily normal (see, for instance, http://en.wikipedia.org/wiki/Ratio_distribution#Gaussian_ratio_distribution). -What is the best way to deal with this scenario? Will bootstrapping be theoretically justified in this case? - -REPLY [2 votes]: Let $\hat{\mu}=\frac{\sum_i X_i}{N}$ and $\mu = \mathbf{E}(X)$ -$$\mathcal{P}(\mu \in [x,y]\,|\, a) = \sum_i \binom ni p^i(1-p)^{n-i}\mathcal{P}(\mu \in [x,y]\,|\,a,N=i)$$ -If $pN$ is relatively large, and $\hbox{var}(X) = \sigma^2$ you can represent the true mean as taken from a mixture of gaussian with mean $a$ and variance $\sigma^2/N$. This is not gaussian, but you can compute the pdf. -While the mixture isn't gaussian, you can compute its variance as -$$\sigma^2 \frac{np(1-p)^{n-1} F_{3,2}\left(1,1,1-n;2,2;\frac{p}{p-1}\right)}{1-(1-p)^n}$$ -(N.B this only valid when $X$ is normally distributed or approximately valid when $n >> (1-p)/p$) -$F_{3,2}$ is the hypergeometric function -For $p = 0.25$ and $N=100$ your average has about $4.13\%$ of the variance of $X$ -You can then use conservatively Chebyshev's inequality. - -REPLY [2 votes]: You should speak of the size of the sample, rather than of the number of samples. -You haven't said whether you can actually observe the sample size. Nor whether the probability distribution of the sample size in any way depends on the mean that you're trying to estimate. If you can observe the sample size and if you know it doesn't depend on that which you're trying to estimate, then it is an ancillary statistic (see http://en.wikipedia.org/wiki/Ancillary_statistic) and you can in effect treat it as non-random by conditioning on it.<|endoftext|> -TITLE: The ring of SL_2 invariants in sums of conjugation and tautological modules -QUESTION [5 upvotes]: Rings of Invariants -Consider $G=SL_2(\mathbb{C})$, and let $V$ be a finite-dimensional $G$-representation. Let $\mathbb{C}[V]$ denote the ring of polynomial functions on the space $V$; it is a free polynomial ring in $\dim(V)$-many variables. Then the ring of invariants of $V$ is the subring -$$\mathbb{C}[V]^G:=(f\in\mathbb{C}[V] s.t. \forall g\in G, f(gv)=f(v))$$ -The Conjugation Representation -Let $M_2(\mathbb{C})$ denote the space of $2\times 2$ complex matrices. The conjugation representation $V_C$ of $G$ on $M_2(\mathbb{C})$ is defined by $g\cdot m:= gmg^{-1}$. Thus, $V_C$ is a 4-dimensional $G$-representation. -The ring of invariants $\mathbb{C}[V_C]^G$ is well-known. It is freely generated by the functions $tr$ and $det$; this is a special case of the general fact that the only conjugation-invariant functions on $M_n(\mathbb{C})$ are symmetric functions of eigenvalues. -The Tautological Module -Let $V_T$ denote the $2$-dimensional $G$-representation where $G$ acts in the 'obvious' way, by the inclusion $G=SL_2(\mathbb{C})\subset GL_2(\mathbb{C})$. This is called the tautological $G$-representation. Since $SL_2(\mathbb{C})$ acts on $V_T$ with a dense orbit, the ring of invariants $\mathbb{C}[V_C]^G$ is boring; it is just $\mathbb{C}$. -Now, let $V_T^n$ denote the $n$-fold direct sum of $V_T$. This is a natural $2n$-dimensional $G$-representation. We can choose a $G$-invariant skew-symmetric bilinear form $\omega$ on $V_T$; this is big words for the usual scalar cross product $v_1\times v_2$. -This defines a natural $G$-invariant function on the space of pairs $(v_1,v_2)\in V_T^2$. -Then the ring of invariants $\mathbb{C}[V_T^n]^G$ is generated by $\omega(v_i,v_j)$ for $1\leq i j\leq n$. However, these do not (in general) freely-generate the ring of invariants, there are relations between them. The relations are all of the form -$$ \omega(v_i,v_k)\omega(v_j,v_l)=\omega(v_i,v_j)\omega(v_k,v_l)+\omega(v_l,v_i)\omega(v_j,v_k)$$ -for $1\leq ijkl\leq n$. -Both of these facts can be deduced by observing that $\mathbb{C}[V_T^n]^G$ can be identified with the homogeneous coordinate ring of the Grassmanian $Gr(2,n)$. Then the generators above are the Plucker coordinates, and the relations are the 3-term Plucker relations (there are no higher Plucker relations here). -The Question -Both of these examples have clever solutions and pretty answers. I am curious about the combination of both cases. -Let $m$ and $n$ be positive integers, and consider the direct sum $G$-representation $V^m_C\oplus V_T^n$. What is the ring of invariants $\mathbb{C}[V_C^m\oplus V_T^n]^G$? -I am aware of general procedures for producing these rings, for arbitrary finite-dimensional $G$-reps; see e.h. Sturmfel's Algorithms in Invariant Theory. However, I suspect that there is a clever solution to this particular problem. Not only because it is a combination of two problems with a clever solution, but because I have a guess as to what the answer is, and all the relations seem to be similar to the 3-term Plucker relations. I also suspect the answer is ancient (like much invariant theory), which is why I am trying to find an answer rather than try to prove my guess is correct by brute force. - -REPLY [2 votes]: I don't have the literature with me, but yes the answer to your question was well known to classical invariant theorists. My recollection is that it is one of the examples (sections) in Grace and Young (J.H. Grace and A. Young, The algebra of invariants, Cambridge Univ. Press, Cambridge, 1903). They give an answer without using polarization. All the generating invariants are the ones you describe or else invariants of degree 3 which are quadratic in (one or 2) of the tautological reps and linear in a conjugation rep. Describing the relations is harder but was understood classically. I am not sure of a reference though.<|endoftext|> -TITLE: Induction, the infinitude of the primes, and workaday number theory -QUESTION [15 upvotes]: There are various open problems in the subject of logical number theory concerning the possibility of proving this or that well-known standard results over this or that weak theory of arithmetic, usually weakened by restricting the quantifier complexity of the formulas for which one has an induction axiom. It particular, the question of proving the infinitude of the primes in Bounded Arithmetic has received attention. -Does this question make known contact with "workaday number theory" - number theory not informed by concepts from logic and model theory? I understand that proof of the infinitude of the primes in bounded arithmetic could not use any functions that grow exponentially (since the theory doesn't have the power to prove the totality of any such function). So especially I mean to ask: -1) If one had such a proof, would it have consequences about the primes or anything else in the standard model of arithmetic? -2) If one proves that no such proof exists, would that have consequences... -3) Do any purely number theoretic conjectures if settled in the right way, settle this question of its kin? - -As a side question, I'd be interested to know the history of this question. I first heard about it from Angus Macintyre and that must have been 25 years ago. - -REPLY [11 votes]: Two comments: - -Work of Paris, Wilkie and Woods shows that we can prove the existence of infinitely primes, and indeed that there is always a prime between $n$ and $2n$, assuming $I\Delta_0+\forall x -\exists y\ y=x^{|x|}$, where $|x|$ is the length of a binary expansion of $x$. So we know functions of exponential growth aren't necessary, but we are still using a function of super-polynomial growth. -Actually, they proved that this theory implies a weak Pigeon-hole Principle which Woods had shown earlier implied the infinitude of primes. -Another question in this spirit is whether $I\Delta_0$ proves that for any prime $p$ there is a non-square mod $p$. - -It is known that neither of these number theoretic results can be proved if the base theory is weakened to allow induction only for quantifier free formulas.<|endoftext|> -TITLE: Is the $\infty$-category of stable $\infty$-categories stable? -QUESTION [12 upvotes]: More generally, are there any remarkable properties enjoyed by the $\infty$-category of stable $\infty$-categories? - -REPLY [14 votes]: No. It's pointed by the zero category, but then taking loops of a stable category C (in the sense of the pullback of 0 --> C along itself) always gives the zero category, so loops is definitely not an equivalence. -One important structural feature of the category of stable categories along these lines is that it has some nice cofiber sequences (Verdier localization sequences), but I'm not sure the categorical properties these satisfy have been axiomatized (into a possible definition of ``stable (infty,2)-category''?)<|endoftext|> -TITLE: local Langlands and the Jacquet module -QUESTION [14 upvotes]: Let $G = GL_n(F)$ be the general linear group over a finite extension $F$ of $Q_p$. This question could be posed for a larger class of groups, but let us stay with $Gl_n$ for the moment. -Let $\pi$ be a smooth irreducible complex representation of $G$. Let $P \subset G$ be a parabolic subgroup and $P =MN$ a Levi-decomposition. The Jacquet module $\pi_N$ of $\pi$ is by definition the module of $N$-coinvariants in $\pi$. -Via the local Langlands correspondence $\pi$ corresponds to a Weil-Deligne representation $\sigma_\pi$. Furthermore, in the cases where $\pi_N$ is irreducible, the representation $\pi_N$ has a corresponding Weil-Deligne representation via local Langlands for $M$. -My question is, does the operation $\pi \to \pi_N$ from $G$-representations to $M$-representations has a "satisfactory" interpretation on the Galois side, via local Langlands for $G$ and $M$? -A point of caution is that $\pi_N$ need not be $M$-irreducible, so it does not go directly into local Langlands for $M$. - -REPLY [9 votes]: Let us consider the simple case: $G=GL_2(F)$, $n=2$. (cf. ''The local langlands conjecture for $GL_2(F)$'' C.J. Bushnell and G.Henniart) -In order to tell the story, first we need to give some definitions. Clearly we only need to consider the non-cuspidal case. Let $\chi=\chi_1\otimes \chi_2$ be the character of $T$, we denote $ \chi^{\omega}=\chi_2\otimes \chi_1$, we define $\pi_{\chi}=Ind_B^G(\delta_B^{-\frac{1}{2}}\otimes \chi)$ where $\delta_B$ is the modular function of the group $B$ i,e $\delta_B(tn)=||t_2t_1^{-1}||$ for $t=diag(t_1,t_2)$, $n\in N$, we write $\phi\circ det$, $\phi \cdot St_G$ two other kind of principal series for $GL_2(F)$. -Now we arrive to write the Jacquet functor $J: Rep(G) \longrightarrow Rep(T); (\pi, V) \longrightarrow -(\pi_N, V_N)$. -(1) For $\chi_1\chi_2^{-1}\neq ||.||^{\pm}$, $\pi=\pi_{\chi}$ is irreducible, then -$\pi_N=\delta_B^{-\frac{1}{2}}\otimes \chi \oplus \delta_B^{-\frac{1}{2}}\otimes \chi^{\omega}$. -(2) $\pi=\phi\circ det$, then $\pi_N=\phi\otimes \phi$. -(3) $\pi=\phi \cdot St_G$, then $\pi_N=||.||\phi\otimes ||.||^{-1}\phi$. -We recall some result about local langlands correspondance for general linear group. We denote $\mathcal{G}_2(F)$ to be the set of equivalence classes of 2-dimensional Frobenius semisimple, Deligne representation of the Weil group $\mathcal{W}_F$; also $\mathcal{A}_2(F)$ to be the set of equivalence classes of irreducible smooth representations of $GL_2(F)$. The local langlands correspondance tell us that there is a natural bijective map $l_2$ between $\mathcal{A}_2(F)$ and $\mathcal{G}_2(F)$. The naturality often involves some compatibility conditions. ( For detail one should see the article of Borel in Corvallis). -Assume $\pi$ is irreducible, lying in $\mathcal{A}_2(F)$, we denote $l_2(\pi)=(\rho,W,\mathbf{n})$. -(1) if $\pi=\pi_{\chi}$, then $\rho=\chi_1 \oplus \chi_2$ and $\mathbf{n}=0$, here we regard $\chi_i$ as the representation of Weil group $\mathcal{W}_F$. -(2) if $\pi=\phi\circ det$, then $\rho=||.||^{-\frac{1}{2}}\phi \oplus ||.||^{\frac{1}{2}}\phi$ and $\mathbf{n}=0$. -(3) if $\pi=\phi \cdot St_G$, then $\rho=||.||^{-\frac{1}{2}}\phi \oplus ||.||^{\frac{1}{2}}\phi$, but in this case -$\mathbf{n}\neq 0$. -Finally we comme to the question that Arno asks. We translate directly ''the Jacquet functor'' to the Galois side via the local langlands correspondence. -$J: \mathcal {G}_2(F) \longrightarrow \mathcal{G}_1(F)^{\otimes 2}$. More precisely, the result is outlined as follows: -(1) $J\big((\pi_{\chi}, \mathbf{n}=0)\big)=(\delta_B^{-\frac{1}{2}}\otimes \chi) \oplus (\delta_B^{-\frac{1}{2}}\otimes \chi^{\omega})$; -(2) $J\big((\phi\circ det, \mathbf{n}=0)\big)=\phi\otimes \phi$. -(3) $J\big((\phi \cdot St_G, \mathbf{n}=0)\big)= ||.||\phi\otimes ||.||^{-1}\phi$. -Remark: for general case, we take $\pi \in Irr_{\mathbb{C}}(G)$, one knows $\pi_N$ has finite length and is admissible as the representation over its Levi subgroup $M$, although we don't even know it is semi-simple or not.<|endoftext|> -TITLE: Geometry of Whitehead manifolds. -QUESTION [11 upvotes]: I'm currently studying some problems about the Whitehead manifold $W$ (the open 3-manifold which is contractible but not homeomorphic to $\mathbb{R}^3$). Does there exists some survey paper on its properties ? -I'm particularly interested in geometric results in the spirit of "Taming 3-manifolds using scalar curvature" by Chang, Weinberger and Yu (MathSciNet page) which shows that $W$ admits no metrics of uniformly positive scalar curvature. -Thanks. - -REPLY [15 votes]: McMillan proved that any contractible 3-manifold is obtained as a union of handlebodies, each of which is homotopically trivial in the next (this generalizes the method of construction of Whitehead). -Later he proved that there are uncountably many topologically distinct contractible 3-manifolds. -There is a survey of these and other results by McMillan. Even though these are old results, they are essentially state-of-the-art. The outstanding open problem was whether one of these contractible manifolds could cover a closed 3-manifold, but this is now resolved by the geometrization theorem (any closed aspherical 3-manifold is covered by $R^3$). Robert Myers obtained partial results on this problem, extended by David Wright, but this is superseded by geometrization. Myers also studied the space of end-reductions for such manifolds. -Of course, the paper you are interested in needs a very minimal amount of 3-manifold topology (aside from the Poincare conjecture!), which seems to be contained in Lemma 4.1 (although I'm not sure what 3-manifold results are used in papers which they cite, in particular reference 14).<|endoftext|> -TITLE: Quantitative measurement of infinite dimensionality -QUESTION [5 upvotes]: I recently encountered the metric mean dimension, which is a numerical metric invariant of (discrete time, compact space) dynamical systems that refines topological entropy for infinite-entropy systems. I am wondering if anything similar can be found in the literature for any metric notion of dimension (Let say that by ``metric'' means bi-Lipschitz invariant). -Put another way, I have a compact metric space $X$ that has infinite dimension for any sensible notion of dimension, and I would like to make this statement quantitative. I see two ways to do this. -The first one is to mimic the box-dimension, and consider the (extra-polynomial) growth rate of the smallest number of $\varepsilon$-balls needed to cover $X$ when $\varepsilon^{-1}$ goes to infinity. This is the simplest way to go, but I am concerned by the fact that box dimension have not as nice a behavior than Hausdorff dimension (for example countable spaces can have positive box dimension). -The second one, suggested by Greg Kuperberg, is to mimic Hausdorff dimension but replacing the family of "size functions" $(x\mapsto x^s)_s$ by another family with similar properties, like $(x\mapsto\exp(-\lambda/x)_\lambda)$. -My question is the following: do you know any example of such an invariant in the literature? Where is it used, in what purpose? - -REPLY [2 votes]: See my paper LINK -Centered densities and fractal measures, New York Journal of Mathematics 13 (2007) 33-87 -Some references are also at the end of it. In particular, Boardman, Goodey, and McClure.<|endoftext|> -TITLE: "Sums-compact" objects = f.g. objects in categories of modules? -QUESTION [29 upvotes]: Hello, -Let us call an object of an additive category sumpact (contraction of "sums" and "compact") if taking $Hom$ from it (considered as functor from the category to $Ab$) commutes with coproducts. Note that to be sumpact is weaker than to be compact (which means that $Hom$ from you commutes with filtered colimits). -Let us take, for our additive category, the category of left modules over some ring. It is known that compact objects in this category are exactly the finitely presented objects. What about sumpact objects? -It is clear that every finitely generated module is sumpact. When I try to prove the converse, I get into some pathological things. -Say, if a module has an increasing $\mathbb{N}$-sequence of submodules whose union is the whole module, and such that the union of every finite subsequence is not the whole module, then it is clear that this module is not a sumpact object (by considering the morphism from it to the direct sum of the quotients by members of our sequence). But it seems not clear (perhaps not true) that every non finitely generated module has such a sequence. -Also, when I check in the internet, it seems people put some condition: the ring is assumed to be perfect. Then indeed sumpact = f.g. -So my question is: for a general ring it is not true that sumpact implies f.g.? Can you give an example? Can you give an example when the ring is commutative? Can you indicate what perfect means and why then everything is OK? -Thank you - -REPLY [12 votes]: If it's considered bad form to resurrect year-old threads, then please slap my wrist (gently, please; I'm new here!) -A fairly simple explicit example of a "sumpact" module that is not f.g. is as follows. -Let $R$ be the ring of functions from an uncountable set $X$ to, say, a field $k$. Let $M$ be the ideal of functions with countable support. -Then it's very easy to show that $M$ isn't f.g., and fairly easy to show that it is "sumpact", using no set theory beyond the fact that a countable union of countable sets is countable. -Edit to add details requested in comments: -To show that $M$ is "sumpact", suppose that $\alpha:M\to\bigoplus_{i\in I}N_i$ is a homomorphism that doesn't factor through a finite subsum. I.e., for infinitely many $i$ the composition $\pi_i\alpha:M\to\bigoplus_{i\in I}N_i\to N_i$ of $\alpha$ with projection onto the summand $N_i$ is non-zero. Replacing $I$ with a countable collection of such $i$ we can assume that $I$ is countable and that $\pi_i\alpha$ is non-zero for all $i\in I$. -For each $i\in I$ choose $f_i\in M$ so that $\pi_i\alpha(f_i)\neq0$. Then the union of the supports $\text{supp}(f_i)$ is countable, so there is some $f\in M$ with $\text{supp}(f)=\bigcup_{i\in I}\text{supp}(f_i)$. -But then the ideal generated by $f$ contains every $f_i$, and so $\pi_i\alpha(f)\neq0$ for every $i$, contradicting the fact that $\alpha(f)\in\bigoplus_{i\in I}N_i$.<|endoftext|> -TITLE: Why is p=2 special, if we want to classify cplx. representation of GL2(Zp)? -QUESTION [8 upvotes]: I am currently reading Shalika's article "Representation of the two by two unimodular group over local fields" and various other related articles, which deal with the classification of complex representation of reductive groups over local rings. It is cumbersome, that some authors consider only local rings with residue fields of characteric $p \neq 2$. -Why is $p=2$ special here? -Perhaps some words about the strategy, which one should use - at least from my perspective: -Consider a finite extension $K$ of $\mathbb{Q}_p$. Let $\mathfrak{o}$ be the ring of integers in $K$ and $\mathscr{p}$ its maximal ideal. We want to classify all representation of $\mathrm{GL}_2( \mathfrak{o})$. Since we deal with a pro-$p$-group, the representations live on $\mathrm{GL}_2( \mathfrak{o}/\mathfrak{p}^n)$ for some $n>0$. We proceed by induction over $n$: -1) Classify all representation of $\mathrm{GL}_2(\mathbb{F}_q)$, where $\mathbb{F}_q$ is the residue field. -2) Use Mackey's formalism for the group extension (non split) -$$ 0 \rightarrow M_{2\times2}(\mathbb{F}_q) \rightarrow \mathrm{GL}_2( \mathfrak{o}/\mathfrak{p}^n) \rightarrow \mathrm{GL}_2( \mathfrak{o}/\mathfrak{p}^{n-1}) \rightarrow 0.$$ -Apparently the difficulties do already arise in step 1, since Piatesko-Shapiro in his lecture "Complex representations of $\mathrm{GL}_2$ over a finite field" only considers characteristic $\neq 2$. - -REPLY [4 votes]: A few more reasons/repetitions of reasons above in addition to the answers above : - -For $p = 2$, two different quadratic extensions can give rise to the same representation of $GL_2$ (this can happen for p not equal to 2, but only in the case of a Weil representation associated to a non-trivial character that has order 2 when restricted to the norm 1 elements of the extension). I don't know how bad this makes things, but criteria for this happening involves the discriminant (which can be pretty weird when $p = 2$). -For p = 2, the characters are hard to compute, even for the "Weil representations" Mander referred to above. This is because Weil representations are not so useful for character computation as with applying Frobenius formula to induction with a relatively simple representation of a subgroup of $GL_2(F)$ of the form $T B_n$, where T is an elliptic torus and B_n is something like a congruence subgroup. Such a representation is often a just a (quasi)character, and hence easier to handle. You see - the approach is not even to compute the character of a representation of $GL_2$ of the integers or an Iwahori but to start from a smaller subgroup times torus. The characters are computed on a torus-by-torus basis, and the presence of other tori in $T B_n$ can create havoc, when $p \neq 2$. Or so I hear. That the characters for $p \neq 2$ were computed enabled Sally and Shalika to compute Fourier transforms of orbital integrals and deduce Plancherel formula (for $SL_2$) from them.<|endoftext|> -TITLE: Completion of a category -QUESTION [37 upvotes]: For a poset $P$ there exists an embedding $y$ into a complete and cocomplet poset $\hat{P}$ of downward closed subsets of $P$. It is easy to verify that the embedding preserves all existing limits and no non-trivial colimits --- i.e. colimits are freely generated. $\hat{P}$ may be equally described as the poset of all monotonic functions from $P^{op}$ to $2$, where $2$ is the two-valued boolean algebra. Then we see, that $P$ is nothing more than a $2$-enriched category, $2^{P^{op}}$ the $2$-enriched category of presheaves over $P$ and that $y$ is just the Yoneda functor for $2$-enriched categories. -However, for a poset $P$ there is also a completion that preserves both limits and colimits --- namely --- Dedekind-MacNeille completion link text, embedding $P$ into the poset of up-down-subsets of $P$. -Is it possible to carry the later construction to the categorical setting and reach something like a limit and colimit preserving embedding for any category $\mathbb{C}$ into a complete and cocomplete category? - -REPLY [8 votes]: The following is what I wrote about six months ago during a private discussion (posted on request :-) -Let me start with a positive (obvious :-) result. -$\newcommand{\word}[1] { -\mathit{#1} -} -\newcommand{\catl}[1] { - \mathbb{#1} -} -\newcommand{\catw}[1] { - \mathbf{#1} -} -\newcommand{\mor}[3] { - {#1 \colon #2 \rightarrow #3} -}$ -1) Let $\catl{C}$ be a small category. Denote by $\word{Cont}(\catl{C}^{op}, \catw{Set})$ the full subcategory of presheaves $\catw{Set}^{\catl{C}^{op}}$ that consists of such $\mor{H}{\catl{C}^{op}}{\catw{Set}}$ that preserve small limits that exist in $\catl{C}^{op}$ (i.e. map existing colimits from $\catl{C}$ to limits in $\catw{Set}$). It follows from abstract nonsense that $\word{Cont}(\catl{C}^{op}, \catw{Set})$ has small (co)limits and, in fact, is a reflective subcategory of $\catw{Set}^{\catl{C}^{op}}$. -Of course, every presheaf preserves colimits, therefore the restricted Yoneda embedding $A \mapsto \hom(-, A)$ gives functor: $$\mor{y}{\catl{C}}{\word{Cont}(\catl{C}^{op}, \catw{Set})}$$ -This functor almost by definition preserves all limits that exist in $\catl{C}$. Observe, that it also preserves colimits. Let $\mor{F}{\catl{J}}{\catl{C}}$ be a functor from a small category, and assume that the colimit $\mathit{colim}(F)$ exists in $\catl{C}$. Consider any $H \in \word{Cont}(\catl{C}^{op}, \catw{Set})$. Then morphisms: -$$y(\mathit{colim}(F)) = \hom(-, \mathit{colim}(F)) \rightarrow H(-)$$ -by Yoneda, are tantamount to elements: -$$H(\mathit{colim}(F)) \approx \mathit{lim}(H \circ F)$$ -Similarly, morphisms: -$$y(F(J)) = \hom(-, F(J)) \rightarrow H(-)$$ -are tantamount to elements: -$$H(F(J))$$ -and one may easily verify, that the above exhibits $\hom(-, \mathit{colim}(F))$ as the limit $\mathit{lim}(y \circ F)$. -The above construction is described, for example, in "Basic concepts of enriched categories" of Max Kelly (Section 3.12). Moreover, as pointed out there, $\word{Cont}(\catl{C}^{op}, \catw{Set})$ inherits any monoidal (closed) structure from $\catl{C}$ via (restricted) convolution. -On the other hand, if $\catl{C}$ is not (monoidal) closed, then $\word{Cont}(\catl{C}^{op}, \catw{Set})$ generally will not be (monoidal) closed, due to the following fact. -2) There is no universal limit and colimit preserving fully faithful embedding of a small category to a cartesian closed complete and cocomplete category. -In fact, there is no universal embedding into cartesian closed category that preserves terminal object and binary coproducts. -For let us assume that $\catl{C}$ is non-degenerated and has a costrict terminal object (terminal object $1$ is costrict if whenever there exists a morphism $1 \rightarrow X$ then $X \approx 1$) and binary coproducts (you may take $\catl{C} = \catw{FinSet}^{op}$), and there is such an embedding $\mor{E}{\catl{C}}{\overline{\catl{C}}}$. -Because $1$ is costrict in $\catl{C}$ we have that $1 \sqcup 1 \approx 1$ in $\catl{C}$ and since coproduct and the terminal object are preserved by $E$, we have also $1 \sqcup 1 \approx 1$ in $\overline{\catl{C}}$. Let $\mor{f,g}{1}{A}$ be two morphisms in $\overline{\catl{C}}$. By the universal property of coproducts they induce the copairing morphism $\mor{[f,g]}{1 \sqcup 1}{A}$ that commutes with the coproduct's injections. However, since $1\approx 1 \sqcup 1$, the coproducts injections are identities and so $f = [f,g] = g$. Therefore, for every $A \in \overline{\catl{C}}$ there is at most one arrow $1 \rightarrow A$. If we take for $A$ an exponent $E(Y)^{E(X)}$, then by cartesian clossedness of $\overline{\catl{C}}$ and faithfulness of $E$: -$$\hom_\catl{C}(X, Y) \leq \hom_\overline{\catl{C}}(E(X), E(Y)) \approx \hom_\overline{\catl{C}}(1, E(Y)^{E(X)}) \leq 1$$ -which contradicts the fact that $\catl{C}$ is non-degenerated. -3) Generally, the completions obtained in any of the mentioned ways will not be the smallest completion. In fact, for a general $\catl{C}$, the smallest reflective subcategory of $\catw{Set}^{\catl{C}^{op}}$ containing representables may not be a smallest limit and colimit completion of $\catl{C}$. -Let us take for example $\mathcal{Z}_3$ group thought as of a category with a single object. Actually one may easily compute the category $\overline{\mathcal{Z}_3}$ induced by the idempotent monad associated to the monad $T$ on the Isbell cojugation for $\mathcal{Z}_3$ (by the theorem of Fakir this category can be computed by taking the objects that respects $T$-weak equivalences). Explicitly, category $\overline{\mathcal{Z}_3}$ is the full subcategory of $\catw{Set}^{\mathcal{Z}_3^{op}}$ on free permutations, together with the terminal permutation (up to the terminal permutation it is equivalent to the Kleisli resolution of the monad $\mor{(-) \times \mathcal{Z}_3}{\catw{Set}}{\catw{Set}}$, but I'm not sure if this is deep or meaningless...). -One may easily see that $\overline{\mathcal{Z}_3}$ is not self-dual, thus cannot be the smallest limit and colimit completion (since $\mathcal{Z}_3$ is self-dual, if $\overline{\mathcal{Z}_3}$ was the smallest, then $(\overline{\mathcal{Z}_3})^{op}$ would be the smallest). Moreover, $\overline{\mathcal{Z}_3}$ satisfies the following properties: - -its every object is a colimit over $\mathcal{Z}_3$, -its every object is a double-limit over $\mathcal{Z}_3$ - -Therefore (by the second property) $\overline{\mathcal{Z}_3}$ is the smallest reflective subcategory of $\catw{Set}^{\mathcal{Z}_3^{op}}$ containing representables. -4) The Dedekind-MacNeille construction as described in Todd's answer works when the monad induced by the Isbell conjugation is itself idempotent. For example, this is always true in the world of posets, in the world of Lawvere metric spaces, and more generally in the world of categories enriched over affine quantales (because in such a case every enriched monad is idempotent). -I actually think that only Dedekind-MacNeille completitions for categories enriched over (co)complete posets deserve the name. -The reason is that there is a subtlety here (which may look minor at first, but I think is crucial for the whole construction) that makes that the direct categorification of the original Dedekind-MacNeille requirements for completion have no sense --- the completion should be complete and cocomplete, which means that it should be closed under all limits/colimits. On the other hand, the term "a category is complete and cocomplete" means that the category is closed under small limits and small colimits (obviously, there is no (co)completion under all (co)limits in a $\catw{Set}$-enriched world). Therefore, the direct categorification of the requirement is possible only when the base of the enrichment has all limits/colimits, which in turn, implies that the base of the enrichment is a poset (at least in the classical mathematics). -As I said, the distinction between small and all may look minor at first, but in fact the distinction is not about quantity, but about quality. This distinction shows up on many occasions. For example, constructively if a category is complete (with respect to all cones) then it is also cocomplete (with respect to all cocones), and every colimit can be expressed as a limit; however, this is no longer true if the category is only small complete --- the cone from the canonical representation of a small cocone may be large, and there may be no reason for its limit to exists.<|endoftext|> -TITLE: Iwasawa mu-invariant for abelian extensions of quadratic number fields -QUESTION [12 upvotes]: Let K be a number field and $p$ an odd prime. Let $\mu$ be the Iwasawa $\mu$-invariant of the class group of the cyclotomic $\mathbb{Z}_p$-extension of $K$. If $K$ is abelian over $\mathbb{Q}$ then it is known that $\mu=0$ (Ferrero-Washinton, see Washington 7.5). Iwasawa conjectured that $\mu=0$ for all $K$. -Is something known for the case when $K$ is abelian over an imaginary quadratic field $k$ ? - -REPLY [2 votes]: If I am not mistaken, it proves that mu invariant of the Z_p times Z_p extension is 0 and this was Schneps's thesis. It is unfortunately not enough to show the conjecture of Iwasawa in this case even using the vanishing of anticyclotomic mu invariant proven by Hida. -Edit: In fact, what Schneps proves is that the mu invariant of the Z_p extension in which only one of the primes above p is ramified.<|endoftext|> -TITLE: A Question on Random Matrices -QUESTION [14 upvotes]: Consider the following $n\times n$ random matrix $V_{n}$ where the $(p,q)$ entry is given by -$$ -V_{n}(p,q):= \frac{1}{\sqrt{n}}\exp(2\pi i(p-1) x_{q}) -$$ -where $x_{1},x_{2},\ldots,x_{n}$ are iid random variables with uniform distribution on $[0,1]$. -It is not difficult to prove that $V_{n}^{*}V_{n}$ has the same eigenvalues as $X_{n}$ where -$$ -X_{n}(p,q)=\frac{\sin(n(x_p-x_q)/2)}{n\sin((x_p-x_q)/2)}. -$$ -This matrix is positive definite and invertible with probability one. The minimum eigenvalue, $\lambda_{1}(n)$, goes to zero as $n\to\infty$. I'm interested in the rate at which this eigenvalue goes to zero. Simulations suggest that -$$ -\mathbb{E}(\lambda_1(n))\sim \exp(-\alpha n), -$$ -the expected value decays exponentially. Using the Cauchy interlacing theorem I can only get the upper bound $O(\frac{1}{n^2}).$ -Any idea of what can work here? -Thanks! ---Gabriel - -REPLY [12 votes]: It is actually more like $e^{-\sqrt n}$. Let's look at the norm of the inverse matrix. The entries are $\pm\prod_{i:i\ne j}\frac 1{z_j-z_i}\sigma_m(z_1,\dots,z_{j-1},z_{j+1},\dots,z_n)$ where $z_k=e^{2\pi i x_k}$ is a random point on the unit circle and $\sigma_m$ is the $m$-th symmetric sum. Since $\log |Z-z_j|$ has zero mean and finite variance, you expect the first factor to be $e^{O(\sqrt n)}$ most of the time. The size of second factor is essentially the size of the random polynomial $\prod_i(z-z_i)$ on the unit circle. The typical value at one point is $e^{O(\sqrt n)}$ and we need about $n$ points to read the true maximum (plus we have $n$ rows to serve), so my educated guess (which I can try to convert into a proof if this subexponential dependence is of any value for you) would be $e^{-\alpha \sqrt{n\log n}}$ with some $\alpha$ (with high probability; the expectation may be a bit larger because there is a chance that the rare large values will still dominate). -I hope it makes sense but I'm in quite a hurry right now, so accept my apologies if I said some nonsense somewhere.<|endoftext|> -TITLE: square root of diffeomorphism of R: does it always exist? -QUESTION [23 upvotes]: Let $f:\mathbb R\to\mathbb R$ be a smooth, orientation preserving diffeomorphism of the real line. -Is it the case that there always exist another diffeomorphism $g:\mathbb R\to\mathbb R$ -such that $g\circ g = f$? -Note: it is relatively easy to show that a continuous $g$ exists, but I am not managing to find a smooth (i.e. $\mathcal C^\infty$) solution of the above equation. - -REPLY [2 votes]: Here this problem (and some more general) are studied for the case $f$ has only one fixed point. In general, one can get $C^1$ square roots, but for higher smoothness there are obstructions when the derivative at the origin is the identity which are studied there.<|endoftext|> -TITLE: Kähler metrics for projective space that are not the Fubini-Study metric -QUESTION [6 upvotes]: For projective $N$-space $CP^{N}$, there is a canonical Kähler metric called the Fubini-Study metric. Do there exist other Kähler metrics for $CP^N$. If so, is there any classification of such metrics? -More generally, how does this work for the Grassmanians, or even flag manifolds? - -REPLY [4 votes]: This is not a classification, but you can get a grip on the space of Kahler metrics on $CP^N$ using Bergman metrics and the Segre embeddings. -To explain this conside let $\{s_\alpha\}$ be a basis of homogeneous polynomials of degree k in N+1 variables. This gives an embedding $CP^N\to CP(H^0(O(k)))$. Now define a metric on H^0(O(k)) by declaring that the s_i are orthonormal. This defines a Fubini-Study metric on P(H^0(O(k)) and which pulls back to a metric on $\omega'$ on $CP^N$. -Now it can be proved that any Kahler metric $\omega$ on $CP^N$ is the limit of metrics of the form $k^{-1}\omega'$ for suitable bases $\{s_\alpha\}$ as $k$ tends to infinity (You can take the $\{s_\alpha\}$ to be orthonormal with respect to the $L^2$-metric induced by $\omega$). -In fact there is nothing special about $CP^N$ here. The above works for any projective manifold $X$ with ample line bundle $L$.<|endoftext|> -TITLE: Group theory required for further study in von Neumann algebra -QUESTION [12 upvotes]: After over half a year's study on operator algebra (especially on von Neumann algebra) by doing exercises in Fundamentals of the theory of operator algebras 1, 2 --Kadison, I was told that the current research focus is on the Ⅱ1 factor, and certain background on group theory is necessary, such as studying the free product, specific group construction and the ergodic action. Then, I want to know are there any good books on the group theory that might be necessary for further study on von Neumann theory? - -REPLY [15 votes]: As for a book on group theory that may be useful or interesting to read for further study of $II_{1}$ factors, I think that de la Harpe's book Topics in Geometric Group Theory is good for this. The reason I say this is that geometric group theory is concerned with the "large scale" structure of groups, and concerns ways that groups can be equivalent in ways that are weaker than group isomorphism. A lot of contemporary $II_{1}$ factor theory is also concerned with weak equivalence of groups and their measure-preserving actions. I'll say a bit more below, for context. -Before I do, though, let me mention that you should check out Sorin Popa's ICM talk, Deformation and rigidity for group actions and von Neumann algebras, a preprint listed on his website, and read all of it. This gives a really good intuition about a big part of what's going on in the subject right now, and says everything I'll say here and more. -One classical construction of a $II_{1}$ factor using a group $G$ is the (left) group von Neumann algebra, i.e. the commutant of the right regular represention of an i.c.c. (all nontrivial conjugacy classes are infinite) group G in $B(\ell^2 G)$. If two i.c.c. groups are isomorphic, then certainly their group von Neumann algebras are too. On the other hand, it is very difficult in general to tell if the group von Neumann algebra construction "remembers" the group used to construct it. For example, any two i.c.c. amenable groups have isomorphic group von Neumann algebras, so if you begin with an i.c.c. amenable group and whip up the group von Neumann algebra, it won't remember which group you used, but only will remember the amenability. It turns out that this construction is also sensitive to Kazhdan's Property (T), the Haagerup property (a weak amenability that is strongly non-(T)), in the sense that these properties are reflected in the structure of the von Neumann algebra. This construction is also sensitive to freeness, as in the freeness of the generators of a group. (See Gabriel's answer above.) Gromov's hyperbolicity is also reflected in the structure of the group von Neumann algebra, in that this "large scale" property severely governs the structure of the von Neumann algebra: this construction for Gromov hyperbolic groups give rise to solid factors. These things all seem to be "global" group properties, and so this is why I'm suggesting geometric group theory. -We're sort of listening for echos of the group in the von Neumann algebra built from it... -The broad question is: What properties of a group survive the construction of a $II_{1}$ factor using that group? -Another classical construction of a $II_{1}$ factor the group-measure space construction, which in modern terms is the way we build the crossed product von Neumann algebra from a discrete group and an ergodic measure-preserving action of that group on a standard Borel space. Check out Popa's above-mentioned ICM talk for more weak equivalences for groups surrounding this construction. -If you look at other ways of constructing $II_{1}$ factors, you can consider the same question. -Good luck with your study of von Neumann algebras!<|endoftext|> -TITLE: amenable equivalence relation generated by an action of a non-amenable group -QUESTION [8 upvotes]: Question. Give a (possibly elementary) example of a probability measure preserving action $\rho\colon G \curvearrowright X$ of a finitely-generated discrete group $G$ on a standard borel space $X$ with a probability measure, such that - -the equivalence relation generated by $\rho$ is ergodic and amenable, -the action $\rho$ is faithful, -the group $G$ is non-amenable. - - -A friend of mine asked me this question couple of days ago, which led us to another question, but perhaps there is an easier way to provide an example. - -REPLY [6 votes]: The answer is yes, such an action exists. -What is needed for the construction is the following very nice example of an action of a non-amenable group on $\mathbb Z$, which I just learned from Gabor Elek. -Consider a graph with vertices given by $\mathbb Z$ and unoriented edges between $n$ and $n+1$. -Pick a random labelling of the edges by the letters $a,b$ and $c$ with no $a$, $b$ or $c$ adjacent to the same letter. This defines an action of the group $G=\mathbb Z/2 \mathbb Z \ast \mathbb Z/2 \mathbb Z \ast \mathbb Z/2 \mathbb Z$. Indeed, just act according to existing labels or fix the element. -This action has the nice feature that it keeps invariant all counting measures on $\mathbb Z$, i.e. all $\mathbb Z$-Folner sequences sets are also Folner sequences for the $G$-action. -Now, the space of labellings (as above) of the graph is itself a probability measure space (a Bernoulli space), which carries an ergodic p.m.p. $\mathbb Z$-action by shifting. It is easy to see that $G$ acts on this space by measure preserving transformations (just by the method described above, done orbit by orbit) and induces an action as required. Indeed, the orbits are just the $\mathbb Z$-orbits, so its ergodic and amenable. Faithfulness follows the fact that you considered all labellings, so that with positive probability (on the space of labellings), an element will act non-trivially. Note also that $G$ is not amenable. -EDIT: As requested, more details on the action. The elements of the shift space are maps $f: \mathbb Z \to \lbrace a,b,c \rbrace$ with $f(n) \neq f(n+1)$. A letter shifts $f$ to the right if $f(1)$ equals that letter, it shifts to the left, if $f(0)$ is equal to the letter; otherwise you fix $f$. It is obvious that the orbits are just the orbits of the shift-action of $\mathbb Z$. Hence, the induced equivalence relation is just the one induced by the action of $\mathbb Z$.<|endoftext|> -TITLE: Computing squaring operations in the Adams spectral sequence -QUESTION [9 upvotes]: This question is about the classical Adams spectral sequence. Squaring operations are defined on its $E_2$ term. I'd like to know how to compute some of the non-trivial operations, such as $Sq^2 ( c_0 ) = h_0 e_0$. I feel like this ought to be doable in the May spectral sequence, but I don't know the details. -I'm aware of some work of Milgram on the subject, but there are some problems with his approach because of indeterminacies of Massey products. -Thanks! - -REPLY [2 votes]: This is an answer Bob's question after my previous answer. It doesn't fit in a comment. -As a spectrum, $\mathrm{End}(H\mathbb{Z}/2)$ is known to be a product of Eilenberg-MacLane spectra, so it is the classifying spectrum $\mathbb{H} C_{*}$ of a chain complex $C_{*}$, i.e. it is in the image of the Eilenbeg-MacLane functor $\mathbb{H}$ from chain complexes to spectra. The work of Shipley shows that this functor is well behaved with respect to multipliciative structures, since the symmetric monoidal model category of chain complexes is Quillen equivalent to that of $H\mathbb{Z}$-algebra spectra. In particular $\mathbb{H}$ takes chain algebras to ring spectra. However, we know that the ring spectrum $\mathrm{End}(H\mathbb{Z}/2)$ is not an $H\mathbb{Z}$-algebra, so it is the classifying spectrum of a chain complex, but not the classifying ring spectrum of a chain algebra. -Your question would be answered in full generality if we understood what a ring spectrum structure over the classifying spectrum of a chain complex is, i.e. what structure on a chain complex $C_{*}$ is equivalent to an $S$-algebra structure on $\mathbb{H} C_{*}$. -This is an achievable task, since there are nice models for the functor $\mathbb{H}$ taking values on symmetric spectra, and if $X$ is a symmetric spectrum, it is easy to describe a ring spectrum structure in terms of maps of simplicial sets $X_p\wedge X_q\rightarrow X_{p+q}$. The outcome should be some sort of non-additive multiplication on the chain complex $C_*$. -Now, let me remind you that Leif Kristensen constructed a long time ago a chain complex $C_{*}$ whose cohomology is the Steenrod algebra (reference below). This chain complex is endowed with a multiplication, which induces the product on the Steenrod algebra in cohomology. However, Kristensen's multiplication is not a chain algebra structure on $C_{*}$ since it is left distributive but not right distributive. This could be an approximation to a solution to the previous problem. -So far, I've been very lazy to pursue this project. There'd be a lot of work to do, but the outcome can be very valuable and surprising, and it looks like a feasible project. -MR0159333 (28 #2550) Kristensen, Leif On secondary cohomology operations. Math. Scand. 12 1963 57–82. (Reviewer: J. F. Adams)<|endoftext|> -TITLE: M12 simple sporadic group -QUESTION [6 upvotes]: I've spent quite a bit of time studying the Mathieu Groups, and I own the ATLAS. -My question is about M12. It is based on the ternary Golay Code, and is the automorphism -group of a Steiner S(5,6,12) system. Now, all of these Steiner systems are isomorphic -up to labelling. The order of M12 is 95040, which is 132 x 720. Since there are -132 blocks in this Steiner system, one can see that the 720 or S6 piece merely scrambles -the six elements of the hexad. Then, the 132 part is just sending the elements of -one hexad to another, of which there are 132 ways to do this. -Can someone give me an intuitive construction for this, not just generators...would it -make sense to say that the (sharply) quintuple transitive action might be to send -block 1 to 2, and 2 to 3, and perhaps another action to send block 1 to 3, 3 to 5 etc. or -something of this nature? Is there a hard and fast way to look at this action (M12) in -terms of the blocks? Or was I wrong about the 720 X 132 decomposition of the order of -the group...Thanks, Paul. - -REPLY [3 votes]: If you want an intuitive presentation of M12, also take a look at Curtis's construction.<|endoftext|> -TITLE: Karhunen–Loève approximation of Brownian motion and diffusions -QUESTION [12 upvotes]: The Karhunen–Loève theorem says that Brownian motion on the interval [0,1] can be represented as follows: -$W_t = \sum_{n=1}^\infty Z_n \frac{\sin((n-1/2)\pi t)}{(n-1/2)\pi},$ -where $Z_n \sim \mathcal{N}(0,1)$ \re i.i.d random variables. -Suppose we truncate this sum at some finite $N$, and define the process -$V_t = \sum_{n=1}^N Z_n \frac{\sin((n-1/2)\pi t)}{(n-1/2)\pi}.$ -Let $V^{(k)}$ be a vector of $k$ i.i.d copies of $V$. -We can define a new process -$X_t = \int_0^t \mu(X_t) dt + \int_0^t \sigma(X_t)dV^{(k)}_t,$ -where the second integral is defined pathwise as a Lebesgue-Stieltjes integral. As $N \rightarrow \infty$, will this process converge weakly to a diffusion in either the Ito or Stratonovich sense? -Karatzas and Shreve's book discusses the the one-dimensional case - one can show strong convergence, so clearly weak convergence follows too. They give a caveat about strong convergence of multidimensional processes, which is partly what prompted this question. -I'm interested in weak convergence for multidimensional processes, since this might give an interesting alternative to the Euler-Maruyama numerical approximation scheme. This is a fairly straightforward idea, so I'm sure there are results in the literature. I'd appreciate any references you could provide. -Many thanks. - -REPLY [4 votes]: Even in the multidimensional case, from the Wong-Zakai theorem, the sequence of processes which are solutions of the equation -$X^N_t=\int_0^t \mu(X^N_s) ds+\int_0^t \sigma(X_s^N)dV^N_s -$ -will converge uniformly in probability on the interval $[0,T]$ to the solution of the Stratonovitch stochastic differential equation -$X_t=\int_0^t \mu(X_s) ds+\int_0^t \sigma(X_s)\circ dW_s -$ -A survey on Wong-Zakai type theorems is -Wong-Zakai approximations for stochastic differential equations -By using the more recent rough paths theory, we can see that this convergence also holds in the $p$-variation topology for $2 -TITLE: Bounds on the size of sets not containing a given finite pattern -QUESTION [6 upvotes]: Recall the following version of Szemerédi's Theorem: let $r_k(N)$ be the largest cardinality of a subset of $[N]:=\{1,\ldots, N\}$ which does not contain an arithmetic progression of length $k$. Then, for any $k\ge 3$, $r_k(N)/N\to 0$ as $N\to\infty$. -It follows that the same is true for any finite pattern. i.e. if $A\subset\mathbb{N}$ is a finite set, then, if $r_A(N)$ is the largest cardinality of a subset of $[N]$ which does not contain a set of the form $t+n.A$, we again have $r_A(N)/N\to 0$ as $N\to\infty$. This is obvious since $A\subset [\max A]$. -For $k=3$, Tom Sanders has recently substantially improved the best known upper bound for $r_k(N)$, namely $O(N/\log^{1-o(1)}N)$. I believe for $k=4$ the current "world-record" is due to Green and Tao and for $k>4$ to Gowers (corrections welcome). -My question is about quantitative bounds for $r_A(N)$. Obviously, $r_A(N)\le r_{\max A}(N)$, but if $A$ is sparse this is likely to be far from optimal. Can one do better? Are there any quantitative results for more general sets $A$? -In particular, is it true that $r_A(N)$ ``behaves like'' $r_{|A|}(N)$? -To be concrete, what type of bounds can one get for $r_A(N)$ when $A=\{1,2,m\}$? Will they be more like $r_3(N)$, or more like $r_m(N)$ (or something strictly in between)? - -REPLY [4 votes]: Your question is quite broad, I won't be able to answer everything. -Let me start by answering your most concrete question regarding the pattern -$ \{1,2,m\} $. I prefer to use $\{0,1,k\}$, which is equivalent. -Recall that the existence of a three term arithmetic progression can be recast as asking for a soution to $x-2y+z=0$ with distinct $x,y,z$ in the subset. -Similarly, for the set you ask for one can recast the question whether there is such a pattern into the question whether there is a solution to the equation -$$(k-1)x - k y + z =0$$ -with distinct $ x, y, z$ in the subset. (Note that the sum of the coefficients is $0$.) -By contrast, one cannot encode the existence of a $k$-term arithmetic progressions, for $k>3$, by a single equation of the above form. Instead, one needs to consider a system of linear equations. -For all I know this is the crucial difference between $3$-term and $k$-term, with larger $k$, arithmetic progressions; at least, if a Fourier-analytic approached is used (as in the papers you mentioned); other approaches often yield only much weaker bounds or no explcit bounds at all. -So, the problem for $\{1,2,m\}$ is 'close' to $3$-term arithemtic progressions. -When comparing problems of this form a way to compare them is to compare the encoding of the 'pattern' as a solution to a (system) of linear equations. -Roughly, if you can encode it with one equation (and the sum of the coefficients is $0$) then it is close to $3$-term arithemtic progressions if not, then not. -In particualr, the number of variables in the equation is not the main problem (in some sence, it even helps to have more variables; however to guarantee that the variables all have distinct values can then become more of a problem), it is the number of equations. -See for example this paper by Liu and Spencer http://www.math.ksu.edu/~cvs/liu_spencer-roth_group.pdf -dealing with arbitrary linear equations (subject to some technical conditions) in finite abelian groups; but the integer problem is basically the problem for cyclic groups. -(There is also a second part of this paper.) -So, it really depends on the precise form of the pattern how hard the problem is; as you suspected its maximum is not a very good measure. - -To compare different systems of linear equations with respect to problems of this form one needs to compare the right parameters. -I am not the right person to explain this in detail; however, for example, the recent paper of Gowers and Wolf 'The true complexity of a system of linear equations' should give an impression of this. -And, there is a large variety of (recent) work related to this circle of ideas, also studying higher dimensional analogues or 'polynomial progressions' (e.g., Bergelson and Leibman, Polynomial extensions of van der Waerden's and Szemerédi's theorems). - -A final remark: as far as I know, the fact that the bounds for $r_3$ and $r_k$ -are so relatively far appart might well not be due to the true behavior of these functions, but due to the fact that in one case the bound was provable and in the other case it is not (yet) provable. For example, it is a conjecture of Erdős that for $S$ a subset of the natural numbers the divergence of $\sum_{s\in S}1/s$ implies the existence of $k$ arithmetic progression,for any $k$, in $S$, which roughly translates into a bound of $N/\log N$ for $r_k$ for any $k$. -Moreover, while the recent progress for $r_3$ that you mentioned is impressive, it is not (or at least not known to be) close to optimal; however, it is close to establishing the above mentioned conjecture for $k=3$. The best known lower bound (Behrend recently improved by Elkin) is roughly only of the form $N \exp(- c \sqrt{N})$ for a $c > 0$, which is smaller than $N/ (\log N)^D$ for any $D$; and as far as I know some people believe this might be closer to the truth than the upper bound. -Thus, if you ask about the size of your fucntions it is difficult to say close to which of the classical functions the functions you mentioned are, as even the classical ones are not completely understood (I believe not even conjecturally). However, how difficult it would be to prove bounds for your functions can be judged to some extent by what I said in the middle. -Much more could be said, but I have to stop.<|endoftext|> -TITLE: Applications of the group completion theorem -QUESTION [9 upvotes]: I've been reading Hatcher's exposition of the Madsen-Weiss theorem here. One of the key results needed is the group completion theorem. Hatcher gives several applications of the group completion theorem (the Barratt-Priddy theorem relating the stable cohomology of the symmetric group to the stable stems, the description of the loop space associated to the stable braid group, the Madsen-Weiss theorem, and Galatius's theorem on the stable cohomology of automorphism groups of free groups). Looking around the internet, I have not been able to locate other applications. The above theorems are all proven in kind of the same way (using "scanning"). Can people provide other interesting applications of the group completion theorem? -This should probably be community wiki, but I can't seem to find the box to check to make it so. - -REPLY [3 votes]: I use the group completion theorem quite a lot. -For example, I used it (along with Yang-Mills theory and work of Tyler Lawson) to study the spaces Hom$(\pi_1 S, U)$ and Hom$(\pi_1 S, U)/U$ when $S$ is an aspherical surface (or a product of aspherical surfaces) and $U =$ colim $U(n)$ is the infinite unitary group. These spaces turn out to be the homotopy group completions of the monoids $\coprod_n$ Hom$(\pi_1 S, U(n))$ and $\coprod_n$ Hom$(\pi_1 S, U(n))/U(n)$, respectively. Here the monoid structure comes from block sum of unitary matrices. This means, in particular, that both of these representation spaces are infinite loop spaces (because block sum is commutative up to coherent isomorphisms). For orientable surfaces, the picture is quite nice: - -Hom$(\pi_1 S, U)/U \simeq (S^1)^{2g} \times \mathbb{C} P^\infty$. - -The first factor can be seen entirely via the determinants of representations ($g$ is the genus). The second factor gives a (non-canonical) 2-dimensional cohomology class (or line bundle) over the moduli space of representations. This should be a reflection of Goldman's symplectic form, but I don't have any idea how to prove that. -The group completion story in this case is spelled out in quite a bit of detail in my paper "Excision for deformation K-theory..." (AGT, 2007). It goes back to the basic ideas of McDuff and Segal. The applications to surface groups show up in "The stable moduli space..." (Trans. AMS, 2011). You also can find these papers on my webpage or on the arXiv.<|endoftext|> -TITLE: Why does so much recent work involve K3 surfaces? -QUESTION [38 upvotes]: I've been noticing that a whole lot of papers published to the Arxiv recently involve K3 surfaces. Can anyone give me (someone who, at this point, knows little more about K3 surfaces than their definition) an idea why they are coming up so often? -Some questions that might be relevant: Are there particular reasons that they are so important? Are there special techniques that are available for K3 surfaces, but not more generally, making them easier to study? Are they just "in vogue" at the moment? Are they more like a subject of research (e.g., people are carrying out some sort of program to better understand K3 surfaces) or a testing ground (people with ideas in all sorts of different areas end up working the ideas out over K3 surfaces, because more general versions are much more difficult)? - -REPLY [11 votes]: K3 surfaces are also interesting from the point of view of complex dynamics. To quote from Curtis T. McMullen's introduction to his paper ``Dynamics on K3 surfaces: Salem numbers and Siegel disks", Journal fur die Reine und Angewandte Mathematik 2002(545): 201–233, -"The first dynamically interesting automorphisms of compact -complex manifolds arise on K3 surfaces. -Indeed, automorphisms of curves are linear (genus 0 or 1) or of finite -order (genus 2 or more). Similarly, automorphisms of most surfaces (including -$\mathbb{P}^2$, surfaces of general type and ruled surfaces) are either linear, finite -order or skew-products over automorphisms of curves. Only K3 surfaces, -Enriques surfaces, complex tori and certain non-minimal rational surfaces -admit automorphisms of positive topological entropy [Ca2]. The automorphisms of tori are linear, and the Enriques examples are double-covered by -K3 examples." -In this paper McMullen gives examples of K3 surfaces admitting automorphisms with Siegel disks (i.e., domains on which the automorphism is conjugate to a rotation). There are countably many such surfaces, all of them non-projective. The citation [Ca2] is to the paper by S. Cantat, Dynamique des automorphismes des surfaces -projectives -complexes. -CRAS Paris S ́er. I Math. -328 -(1999), 901–906, which grew into a larger work: S. Cantat : Dynamique des automorphismes des surfaces K3 ; Acta Math. 187:1 (2001), 1--57 -These papers are only a few examples (perhaps of landmark character); there has been a lot of study of dynamics on K3 surfaces going on indeed. - -REPLY [5 votes]: In the mathematical physics community, there was recently a resurgence of interest for K3-surfaces due to the observation of Egushi, Ooguri and Tashikawa that the elliptic genus of K3-surfaces seems to be built out of representations of the Mathieu group M24. This phenomenon has been dubbed the "Mathieu moonshine". There is still no definitive understanding of this relation, as far as I know.<|endoftext|> -TITLE: Why chain homotopy when there is no topology in the background? -QUESTION [21 upvotes]: Given two morphisms between chain complexes $f_\bullet,g_\bullet\colon\,C_\bullet\longrightarrow D_\bullet$, a chain homotopy between them is a sequence of maps $\psi_n\colon\,C_n\longrightarrow D_{n+1}$ such that $f_n-g_n= \partial_D \psi_n+\psi_{n-1}\partial_C$. I can motivate this definition only when the chain complexes are associated to some topological space. For example if $C_\bullet$ and $D_\bullet$ are simplicial chain complexes, then for a simplex $\sigma$, a homotopy between $f(\sigma)$ and $g(\sigma)$ is something like $\psi(\sigma)\approx\sigma\times [0,1]$, whose boundary is $f(\sigma)-g(\sigma)-\psi(\partial\sigma)$. -But chain homotopy also features in contexts where there is no topological space lurking in the background. Examples are chain homotopy of complexes of graphs (e.g. Conant-Schneiderman-Teichner) or chain homotopy of complexes in Khovanov homology. In such contexts, the motivation outlined in the previous paragraph makes no sense, with "the boundary of a cylinder being the top, bottom, and sides", because there's no such thing as a "cylinder". Thus, surely, the "cylinder motivation" isn't the most fundamental reason that chain homotopy is "the right" relation to study on chain complexes. It's an embarassing question, but: - -What is the fundamental (algebraic) reason that chain homotopy is relevant when studying chain complexes? - -REPLY [5 votes]: This was a comment, but I guess it was not so clear, although it is hiding in some of the above answers: -There is an object in $Ch^+(R)$ that behaves like an interval. Think of the simplicial interval, and you get a chain complex $I_*$ over $R$ with $I_0=Re_0 \oplus Re_1$, $I_1=Rf_0$ and all other groups 0, with differential $d(f_0)=e_1−e_0$. This plays the role of the unit interval. Now a chain homotopy of two chain maps $f,g:A_∗ \to B_∗$ really is $H:A_∗ \otimes I_* \to B_*$ where the tensor product takes place in R chain complexes. I am not sure how well this object behaves with respect to the Dold-Kan maps, but it certainly is the normalized chains on $\Delta^1$. You could similarly look at maps of $R$ complexes from $I_*$ into $Hom_R(A_*,B_*)$. -With this in hand you can form cylinder objects and cofiber sequences in the "obvious" way, and things start looking a bit more geometric.<|endoftext|> -TITLE: Is it true that no one-dimensional group variety acts transitively on $\mathbb{P}^1$? -QUESTION [6 upvotes]: This question may be trivial to people with the right background, but I do not see the answer. - -Let $\Bbbk$ be an algebraically closed field. Can any one-dimensional group variety (over $\Bbbk$) act transitively on $\mathbb{P}^1_{\Bbbk}$? - -Here is my reasoning so far: --Every $g \in G$ fixes a point of $\mathbb{P}^1$ since the associated linear map has an eigenvector. --Let $G_0$ be the identity component of $G$. If $G_0$ fixes a point $p$, then $G/G_0$ surjects onto the orbit of $p$, so $p$ has finite $G$-orbit. Hence $G$ does not act transitively. So, it suffices to assume $G$ is connected and show that it fixes a point. --I think (but am not confident here) that the only one-dimensional connected group varieties are $(\Bbbk, +)$, $(\Bbbk^*, \cdot)$, and abelian varieties (i.e., elliptic curves). The last are not an issue since any morphism $E \to PGL_2$ is constant, where $E$ is complete and $PGL_2$ is affine. --Since $G$ is one-dimensional irreducible and the isotropy subgroup of a point is closed, to show that $G$ fixes $p$, it suffices to show that infinitely many elements of $G$ fix $p$. -The last step, I think I can do for the multiplicative group, and for the additive group if $\Bbbk$ has characteristic zero, but my basic line of approach does not seem to work for the additive group in characteristic $p$. I also am less than completely confident of the classification I've stated for one-dimensional connected group varieties; if someone could point me to a good reference for this, I would appreciate it. - -REPLY [9 votes]: The case of a 1-dimensional connected affine group (which in particular is solvable) is settled most easily by invoking the Borel Fixed Point Theorem: a connected solvable (affine) group acting as an algebraic group on an irreducible projective variety has a fixed point. If the given "group variety" is an abelian variety, of course, you have to argue separately. Similarly for a non-connected affine group, in which the connected component of the identity has a fixed point. -ADDED: The question here involves only the 1950s work on algebraic groups ("group varieties") by Borel, Chevalley, Rosenlicht, ... By now parts of the theory have been streamlined, but anyway it's a little artificial to extract from the structure theory just the minimal amount of information you need for your purpose. However it's done you probably need to distinguish somewhere between characteristic 0 and characteristic $p$. Here is a more detailed outline of what might be argued: -1) It's useful to reduce the problem first to the case where $G$ is connected, which is an elementary step. -2) To handle the case when $G$ is not affine (thus an abelian variety of dimension 1), you implicitly rely on Chevalley's deep structure theorem on algebraic groups which was recently revisited by Brian Conrad: A modern proof of Chevalley’s theorem on algebraic groups, J. Ramanujan Math. Soc. 17 (2002), no. 1, 1–18. -3) When $G$ is an abelian variety, it has no everywhere-defined regular functions (like other complete varieties) and thus can't have a nontrivial algebraic group morphism to an affine group such as $PGL_2$, which is shown concretely to be the automorphism group of the projective line (6.4 in my book). -4) If $G$ is connected and affine of dimension 1, it is commutative: see 20.1 in my book, which doesn't yet get into the precise classification of such groups. -5) To finish the argument as Peter suggests you need something like Lemma 21.1 in my book, which is the key step toward the Borel Fixed Point Theorem in 21.2. -(That theorem is relatively easy to prove after some preparation, but is the key to further deep structure theory for reductive groups.) In Peter's sketch, you get only a bijective morphism from $G/H$ to the projective line, not necessarily an isomorphism; so 21.1 is needed in characteristic $p$. For me the best conceptual viewpoint is found in the fixed point theorem, which of course applies much more generally to actions of connected solvable groups on projective (or complete) varieties.<|endoftext|> -TITLE: When is a quasi-isomorphism necessarily a homotopy equivalence? -QUESTION [22 upvotes]: Under what circumstances is a quasi-isomorphism between two complexes necessarily a homotopy equivalence? For instance, this is true for chain complexes over a field (which are all homotopy equivalent to their homology). It's also true in an $\mathcal{A}_\infty$ setting. -Is it true for chain complexes of free Abelian groups? The case I'm particularly interested in is chain complexes of free $(\mathbb{Z}/2\mathbb{Z})[U]$ modules or free $\mathbb{Z}[U]$ modules, but I'm also interested in general statements. - -REPLY [21 votes]: If your complexes are bounded, this is always true for any ring more generally replacing free modules with projectives. The statement is that $\mathrm{D}^b(A\text{-}mod)$ is equivalent to $\mathrm{Ho}(Proj\text{-}A)$ and you can find it in Weibel Chapter 10.4. If your complexes are unbounded things are more tricky. Then your statement is true in over any ring of finite homological dimension. Basically you have two notions K-projective(which have the property that you want) and complexes of projectives. Bounded complexes of projectives are K-projective, but unbounded ones are not unless you have the finiteness hypothesis(see Matthias' answer). See this post for the injective version of this story Question about unbounded derived categories of quasicoherent sheaves. In the cases you are interested in there is no problem.<|endoftext|> -TITLE: Best way to find a closest vector in a lattice -QUESTION [11 upvotes]: Let $v_1,\dotsc,v_n$ be linearly independent vectors in $\mathbb{R}^n$, and let $\Lambda=\bigoplus_{i=1}^n \mathbb{Z}v_i$. The question is, given a vector $w$ in $\mathbb R^n$, find the element $v$ of the lattice $\Lambda$ closest to $w$. It is assumed that an inner product structure has been imposed on $\mathbb{R}^n$. This is called the CVP (for closest vector problem). -I've found various algorithms that can find the closest vector, but is there a particular algorithm that is especially simple to understand and to program? I want to test a few low dimensional examples, and easy to write computer code would be preferable. -Any suggestion? - -REPLY [7 votes]: A nice rahter-simple option would be Babai's Nearest Plane Algorithm. This is an approximation algorithm which outputs a vector of the lattice which is "close" the given target $w$. The accuracy of the approximation depends on the rank $n$ of the lattice. The good things are: the algorithm runs in polynomial-time and it is fairly easy to implement. - -Input: Basis $B\in\mathbb{Z}^{m\times n}, w\in\mathbb{Z}^{m}$ -Output: A vector $x\in \mathcal{L}(B)$ such that $\lVert x - w\rVert \leq 2^{\frac{n}{2}} \:\text{dist}(w,\mathcal{L}(B))$ - -Run $\delta$-LLL on $B$ with $\delta=3/4$. -$b \leftarrow w$ -for $j = n$ to $1$ do -$\qquad b=b-c_j b_j$ where $c_j = \lceil \langle b, \tilde{b}_j \rangle / \langle \tilde{b}_j, \tilde{b}_j \rangle\rfloor$ -Output $x:=w-b$ - - -Above, $\delta$-LLL denotes the Lenstra-Lenstra-Lovasz algorithm, used as a subroutine to obtain a $\delta$-LLL Reduced Basis $\lbrace b_1,\ldots,b_n \rbrace$ of your original basis $\lbrace v_j \rbrace$. The vectors $\lbrace \tilde{b}_j \rbrace$ are the Gram-Schmidt orthogonalization of the LLL basis. -It seems that this algorithm is normally the first to be taught in university courses. I read these algorithms and definitions in Oded Regev's course about Lattices in Computer Science. -For exacts algorithms, this short review might be helpful. Alternatively, this text is more technical but rather complete.<|endoftext|> -TITLE: CM for primary ideal -QUESTION [5 upvotes]: Let $R$ be a regular local ring, $I$ a prime ideal and $J$ an $I$-primary ideal in $R$. Is it true that if $R/I$ is CM then also $R/J$ is CM? -This question is in some way the inverse of this one. - -REPLY [6 votes]: A useful way to think about this issue is to consider $J=I^{(n)}$, the $n$-symbolic power of $I$, which by definition is the $I$-primary component of $I^n$. -When $R$ is a polynomial rings over $\mathbb C$, this is the ideal consisting of functions vanishing to order at least $n$ on $X = \text{Spec}(R/I)$. -It is then well-known that the depth of $R/I^{(n)}$ can go down. For example, take $I$ generated by the $2\times2$ minors of the generic $2\times 3$ matrix inside the polynomial rings of the $6$ variables, localized at the maximal ideal of those variables. Then $R/I$ is Cohen-Macaulay of dimension $4$, but $R/I^{(n)}$ would have depth $3$ eventually. For more general statements about ideals of maximal minors, see for example Section 3 of: - -Powers of Ideals Generated by Weak d-Sequences, C. Huneke, J, Algebra, 68 (1981), 471-509. - -EDIT: the example above looks specific, but such examples should be abound. I expect most Cohen-Macaulay ideals which are not complete intersections to give an example (it is known that $R/I^n$, the ordinary powers, are CM for all $n>0$ iff $I$ is a complete intersection). The $2\times 2$ minors gives is a generic situation of non-complete intersection but CM ideal. -A philosophical comment: it is unlikely that Cohen-Macaulayness will be preserved by basic operations on ideals. So if $R/I, R/J$ are CM, we do not expect $R/\sqrt{I}, R/I^n, R/I^{(n)}, R/P$ ($P$ an associated primes), or $R/(I+J), R/IJ$ etc. to be CM. -The reason is that to preserve depth one needs to control the associated primes, and these operations only allow you to control the support. However, finding an explicit example is usually not so obvious.<|endoftext|> -TITLE: A rank 3 geometry for the sporadic simple group of Suzuki -QUESTION [7 upvotes]: I am actually studying coset geometries (in the sense of Tits and Buekenhout) for the sporadic simple group of Suzuki. I came aware that Buekenhout found in 1979 a geometry over the following diagram - c 6 -O----------O----------O -1 4 4 - -However, I couldn't find any information about the maximal (or minimal) parabolic subgroups of this geometry. -Has anyone ever studied this geometry? Is there a paper where I could find the informations I am looking for? -As usual, thanks in advance! - -REPLY [7 votes]: I finally found the maximal parabolic subgroups of this geometry. Let us first denote the types of the elements with 0,1 and 2 when reading the diagram from left to right, and let us denote with $G_0$, $G_1$ and $G_2$ the stabilizer of an element of type 0, 1 and 2 respectively. Then we have: -$$ -G_0 = G_2(4),\quad G_1 = 2^{2+8}:(A_5 \times S_3),\quad G_2 = 2^{4+6} : 3 A_6 -$$ -which are all maximal subgroups of $Sz$, and the Borel is $$B = 2^{12}.3^2$$ -Historically, I read that this geometry was built using polar spaces (see Francis Buekenhout, Diagrams for geometries and groups, Journal of Combinatorial Theory A, 27, 121-151, 1979 doi:10.1016/0097-3165(79)90041-4). However, I have not studied yet how to build it geometrically.<|endoftext|> -TITLE: Maximal number of directed edges in suitable simple graphs on $n$ vertices without directed triangles. -QUESTION [8 upvotes]: We consider the class $C$ of directed simple (no multiple edges) graphs having the property that every vertex is reachable by a directed path from every other vertex. -Given an integer $k$, what is the maximal possible number of (directed) edges in a graph of $C$ with $n$ vertices -such that there are no directed cycles of length $\leq k$? -For $k=2$ this means simply -that the existence of an edge from $v$ to $w$ forbids the existence of an edge from $w$ to $v$ and one can thus choose arbitrary orientions (giving rise to a graph in $C$) on the edges of the complete unoriented graph. -For $k=3$, one has also to forbid oriented triangles which is not possible by orienting all edges of a complete graph on $n\geq 3$ vertices such that the result is in $C$. -On the other hand, there are of course no triangles by choosing arbitrary orientations -(giving rise to an element in $C$) of a complete bipartite graph. -There are thus such graphs having roughly $n^2/4$ directed edges. -There should be better upper and lower bounds. -Motivation: G. Higman (A finitely generated infinite simple group. J. London Math. Soc. 26, -(1951). 61--64) constructed finitely generated infinite simple groups -by considering quotients of the finitely presented group $$\langle g_1,\dots,g_n|g_{i-1}^{-1}g_ig_{i-1}=g_i^2\rangle$$ -where indices are modulo $n$. -This group is trivial for $n=2,3$ and infinite for $n\geq 4$. Given a directed graph, -one can consider the corresponding group-presentation with generators corresponding to vertices and directed edges corresponding to relations $a^{-1}ba=b^2$. The triviality -of the group constructed by Higman associated to $n=2,3$ implies that we want to avoid -oriented cycles of length $\leq 3$ when searching for interesting examples. - -REPLY [8 votes]: Updated 4/17/11: -(Originally, this answer contained a different proof of the result below for $k=3$. Not only did the proof not generalize, but it was wrong.) - -The maximum number of edges in a strongly-connected digraph with $n \geq k+1$ vertices and no cycles of length at most $k$ is $${\binom{n}{2}} - n(k-2) + \frac{(k+1)(k-2)}{2}.$$ - (A digraph where every vertex is reachable from every other vertex by a directed path is called strongly connected.) - -Gordon Royle conjectured this bound an gave an example achieving it for $k=3$. For general $k$ and $n$ the bound is attained by the following construction, almost identical to the one provided by Nathann Cohen in the comments: -Let vertices $x_1,x_2,\ldots,x_{n-k+2}$ form a transitive tournament with $x_i \to x_j$ being an edge for all $1 \leq i < j \leq n-k+2$. Now delete the edge $x_1 \to x_{n-k+2}$ and replace it with a path $x_{n-k+2} \to x_{n-k+3} \to \ldots \to x_n \to x_1$. (The vertices $x_{n-k+3},\ldots, x_n$ will have in-degree one and out-degree one in the resulting graph.) -It remains to prove that the above number is a valid upper bound. The proof is by induction on $n$. -Simple counting shows that the bound is valid if $G$ is a directed cycle. It is tight if $G$ is a cycle of length $k+1$. Assume now that $G$ is not a cycle. Then there exist $\emptyset \neq X \subsetneq V(G)$ such that $G|X$ is strongly connected. (For example, one can choose the vertex set of any induced cycle in $G$.) Choose $X$ maximal subject to the above. Let $u \to v_1$ be an edge of $G$ with $u \in X$, $v_1 \not \in X$, and let $P$ be a shortest path in $G$ from $v_1$ to $X$. Let $P=v_1 \to v_2 \to \ldots \to v_l \to w$. -Note that adding to $G|X$ any path starting and ending in $X$ produces a strongly connected digraph. It follows from the choice of $X$ that any non-trivial such path must include all the vertices in $V(G)-X$. In particular, if $l\geq 3$, $v_2,\ldots,v_{l-1}$ have no neighbors in $X$. -Let us further assume that $u$ and $w$ are chosen so that the directed path $Q$ from $w$ to $u$ in $G|X$ is as short as possible. (Perhaps, $w=u$.) Then $V(P) \cup V(Q)$ induces a cycle in $G$, and so $v_1$ and $v_l$ have at least $k-2$ non-neighbors on $V(P) \cup V(Q)$. At least $k-l$ of those non-neighbors are in $X$ if $l\geq 2$. Therefore there are at least $k-2$ non-edges (pairs of non-adjacent vertices) between $X$ and $V(G)-X$ if $l=1$, and at least -$$2(k-l)+(l-2)(k+1) \geq l(k-2)$$ -non-edges if $l \geq 2$. -By the induction hypothesis there are at least $|X|(k-2)- \frac{(k+1)(k-2)}{2}$ non-edges between vertices of $X$, and therefore at least -$$(l+|X|)(k-2)- \frac{(k+1)(k-2)}{2}=n(k-2) - \frac{(k+1)(k-2)}{2}$$ -non-edges in total, as desired.<|endoftext|> -TITLE: Characteristic classes of lifted bundles -QUESTION [6 upvotes]: Suppose $V$ is a vector bundle with structure group $SO(3)$, and suppose that it can be lifted to a $\text{Spin}(3) = SU(2)$ bundle (i.e. $w_2(V) = 0$). Let us call the lifted bundle $E$. Then it is stated on page 42 in The Geometry of Four-Manifolds by Donaldson and Kronheimer that we have the relation $p_1(V) = -4c_2(E)$. My question is: - -How does one show this? - -More generally, how does one compute the effect of going over to a lifted bundle on the characteristic classes? Generally one has $p_1(E) = c_1^2(E) - 2c_2(E)$. In our case the first term drops out, so that the claim in the book can also be written as $p_1(V) = 2p_1(E)$. This factor $2$ undoubtedly somehow comes from the covering homomorphism $SU(2) \rightarrow SO(3)$ which is 2 : 1, but how? -Probably related: on the same page he defines at the bottom for the so-called instanton number $\kappa = \frac1{8\pi^2}\int_M\text{Tr}(F^2)$, and then claims that this is $\kappa = c_2$ for $SU(r)$ bundles $E$ and $\kappa = -\frac14p_1$ for $SO(r)$ bundles $V$. There again is that factor 4; again, from the formula $p_1(E) = c_1^2(E) - 2c_2(E) = \left[\frac{-1}{4\pi^2}\text{Tr}(F^2)\right]$ one would expect this to be 2. I can see that this factor is chosen so that if one lifts the bundle that $\kappa$ does not change, but on the other hand, not every bundle is liftable, and for bundles which have both Chern classes and Pontryagin classes (such as complex $SU(r)$ bundles), I would expect that one would want the two formulae to give the same answer. As it is, they don't. -(I guess the real problem is I haven't managed to find any readable sources on $SO(r)$-bundles in the context of gauge theories. For special unitary groups there are sources in abundance, but for special orthogonal they are a lot harder to find.) - -REPLY [2 votes]: Let me rephrase your question so that I understand it: let $P \to X$ be an $SU(2)$-principal bundle. Then we get a $2$-dimensional complex vector bundle $E \to X$ by $P \times_{SU(2)} C^2$ (with the defining representation). And we get a $3$-dimensional real vector bundle $V:= P \times_{SU(2)} R^3$ (with the adjoint representation). -You (or Donaldson-Kronheimer) claim that $p_1(V)=-4c_2(E)$. -It is enough to consider the universal case $X=BSU(2)$. The map $BU(1) \to BSU(2)$ induced by the inclusion of the maximal torus is injective in cohomology, so it suffices to check your identity on $BU(1)$ (splitting principle). The pullback of $E$ to $BU(1)$ is $L \oplus L^{\ast}$, so its $c_2$ is a generator $\pm u$ of $H^4 (BU(1))$. The pullback of $V$ to $BU(1)$ is $L^2 \oplus \mathbb{R}$ (here the fact that $SU(2) \to SO(3)$ has degree comes in), hence the total Chern class of $V \otimes \mathbb{C}$ is $(1+2u)(1-2u)$, from which you can read off the Pontrjagin class. For the correct sign, you need a bit more care, though. -EDIT: The inclusion of the maximal torus is the group homomorphism $U(1) \to SU(2)$; $z \mapsto diag (z,z^{-1})$. Thus the defining rep. of $SU(2)$ restricts to a sum of the defining rep. of $U(1)$ and its dual, giving the pullback of $E$. -Consider the standard basis the Lie algebra $\mathfrak{su}(2)$ (http://en.wikipedia.org/wiki/Special_unitary_group#SU.282.29) and compute the action of $diag (z,z^{-1})$ via the adjoint rep in this basis. The result is that the adjoint rep. of $SU(2)$ restricts to a sum of the tensor square of the defining rep of $U(1)$ with the trivial real one-dim. rep. (this computation is basic in representation theory, it is the computation of the roots of $SU(2)$).<|endoftext|> -TITLE: Is this a characterization of Dedekind domain? -QUESTION [8 upvotes]: Let $R$ be an integral domain. Suppose that for any two nonzero ideals $I$ and $J$, we have $I \oplus J$ is isomorphic to $R \oplus IJ$ as $R$-modules. Does this implies $R$ is a Dedekind domain? - -REPLY [3 votes]: As Sampath has pointed out, we may assume that $R$ is local. Your hypothesis implies for any two non-zero ideals $I,J$, you have a surjection onto $R$. By Nakayama, this implies either $I\to R$ or $J\to R$ is surjective, since if neither is, then the both have images contained in the maximal ideal and so does their sum. But this means one of them is principal. -Having recognized the confusion I caused by my terseness, let me be more explicit. First, my definition of DD is: R a domain (not a field) and for any non-zero ideal $I$, there exists another ideal $J$ such that $IJ$ is isomorphic to $R$. Easy to see that $\text{Hom} (I,R)$ for any non-zero ideal $I$ can be identified naturally with $J=\{x\in K|xI\subset R\}$ where $K$ is the fraction field of $R$. Then, the definition of DD means that for any non-zero ideal $I$, defining $J$ as above, $IJ=R$. -Now assume that for any non-zero ideal $I$ of $R$, there exists a surjection $I\oplus I$ to $R$. Then, I claim that $R$ is a DD. The hypothesis implies, there exists $a,b\in I$, $x,y\in K$ with $xI,yI\subset R$ and $xa+yb=1$. Easy to check then that $I$ is generated by $a,b$. Thus all ideals are generated by atmost two elements and in particular $R$ is Noetherian. Now, by the above localization argument, $I$ is locally principal and the rest is clear. I hope this is clearer. -Even without localizing, letting $J$ as above, we have $x,y\in J$ and thus $1\in IJ\subset R$ and hence $IJ=R$.<|endoftext|> -TITLE: Is the group generated by two almost disjoint infinite cycles amenable? -QUESTION [6 upvotes]: Let $x$ and $y$ be two permutations of $\mathbb{Z}^2$ defined as follows. The permutation $x$ sends $(n,0)$ to $(n+1,0)$ and fixes all else while $y$ sends $(0,n)$ to $(0,n+1)$ and fixes all else. Is the group generated by $x$ and $y$ amenable? -I do know that the group does not contain a copy of the free group on two generators, so it is very likely to be amenable. I also know that if $y$ is defined, instead, by sending $(n,m)$ to $(n,m+1)$ then the group generated by $x$ and $y$ is amenable, in fact, it is a solvable extension of a locally finite group. - -REPLY [8 votes]: The derived subgroup of your group consists of permutations with finite support. Indeed, suppose that $w$ is a commutator word in $a$ and $b$ so the total exponent of $a$ (of $b$) is 0. Take a point $(m,n)$ where $m$ or $n$ are very large (comparing to $|w|$). Then $w(a,b)$ fixes that point. Therefore your group is an extension of a locally finite group by the Abelian group ${\mathbb Z}^2$, and is amenable.<|endoftext|> -TITLE: Why are viscosity solutions useful solutions? -QUESTION [23 upvotes]: I refer to definition of viscosity solution in user's guide to viscosity solutions of second order partial differential equations by Michael G. Crandall, Hitoshi Ishii and Pierre-Louis Lions. -Viscosity solutions are generalized solutions which can be implied if the Sobolev theory (or similar) doesn't provide "useful" solutions. A standard example is the problem -$|u'| = -1, u(-1) = 1, u(1)=1$ -All "zig-zag" functions with appropriate boundary conditions provide a solution, but $u(x)=|x|$ is the unique viscosity solution. -But except its formal beauty why do we regard a viscosity solution as useful, and what is the 'physical' or 'intuitive' interpretation of being a viscosity solution? - -REPLY [23 votes]: Viscosity solutions are the "appropriate" notion of solutions for second-order elliptic equations in nondivergence form, and for some classes of first-order equations. Here is a brief summary of why. -From the point of view of applications, the viscosity solution is almost always the right solution. For example, in optimal control theory, it has long been known that if the value function is smooth, it satisfies a certain PDE-- but the value is known to not be smooth in most cases. When Crandall and Lions invented viscosity solutions, it was clear immediately that the viscosity solution is precisely the value function. This story has played out over and over again, for many different applications, to the point that people in the field are completely shocked and stunned when there is some reason to consider a notion of solution other than viscosity solution. -From a mathematical point of view, viscosity solutions are natural. For equations in nondivergence form, energy methods are unavailable. Therefore, all one usually has is the maximum principle. Viscosity solutions are to weak solutions as the maximum principle is to energy methods. The term "viscosity solution" is rather unfortunate in this sense-- they should be called "comparison solutions" or something. The point is, these equations should satisfy a maximum principle. So it makes sense to define your weak solution as a one for which the maximum principle holds when you compare to smooth functions. -Finally, in the cases where there is overlap, like for the p-Laplacian, viscosity solutions are equivalent to (bounded) weak solutions in the integration-by-parts sense.<|endoftext|> -TITLE: Boundaries of the eigenvalues of a symmetric matrix (or of its Lapacian) -QUESTION [5 upvotes]: Given the adjacency matrix $A_{ij}$ of a graph with $N$ vertices and $M$ links (or any binary symmetric matrix of size $N \times N$), is it possible to establish lower and upper boundaries of its eigenvalues? I mean, do $N$ and $M$ determine the lowest and the largest possible eigenvalues of the matrix? -In other words, let $L$ be the Lapacian of $A_{ij}$. We know that its eigenvalues obbey the following: $\lambda_1 = 0 \leq \lambda_2 \leq \lambda_3 \leq \ldots \leq \lambda_N$. The lowest eigenvalue is always $\lambda_1 = 0$ and hence, the eigenvalues of $L$ have a lower bound. Is there an upper bound for $\lambda_N$ that depends only on the size $N$ of the graph and/or on $M$? -Stated yet in another manner. Does anyone know a graph $G'$ whose $\lambda'_N$ is largest than the $\lambda_N$ of any other graph of the same size $N$ and/or number of links $M$? -Thank you! - -REPLY [7 votes]: There are several notions of Laplacian for graphs. For instance, there is the normalized Laplacian, the classical Laplacian and Laplacians with boundary conditions as the Dirichlet Laplacian and so on. Assuming that you are taking about the classical Laplacian then its eigenvalues are -$$ -0=\lambda_1\leq \lambda_2\leq \ldots\leq\lambda_{n} -$$ -where $n$ is the number of nodes. Typically it is $\lambda_2$ the eigenvalue that gives the most information about the graph (like expanding properties, connectivity, isoperimetric properties, etc). For instance, $\lambda_2>0$ iff the graph is connected. -Here are a few known results: - -[Fiedler] $\lambda_2\leq \frac{n}{n-1}\min\{d(v):v\in V\}$ where $d(v)$ is the degree of the node $v$. -[Anderson and Moreley] The maximum eigenvalue satisfies -$$ -\lambda_n\leq\max\{ d(u)+d(v):uv\in E\}. -$$ -If the graph $G$ is connected then the equality holds iff the graph is bipartite. -[Kelmans] If the graph is simple (no multiple edges or loops) then $\lambda_n\leq n$ with equality iff the complement of $G$ is not connected. -$\sum_{i}^{n}{\lambda_i}=2m=\sum_{v}{d(v)}$ where $m$ is the number of edges in $G$. -[Fiedler] $\lambda_{n}\geq \frac{n}{n-1}\max\{d(v):v\in V\}$. - -The second and last bullet give you lower and upper bounds for the maximum eigenvalue in terms of the degree of the nodes. If the graph is regular of course much more can be said. -I hope it helps!<|endoftext|> -TITLE: A line bundle not big but with good intersection numbers -QUESTION [6 upvotes]: Let $X$ be a complex projective manifold of complex dimension $n$ and $A\to X$ an ample line bundle. Let $L\to X$ be a line bundle such that -$$ -c_1(L)^k\cdot c_1(A)^{n-k}>0,\quad k=1,\dots,n. -$$ -Is it true that then $L$ is big? -The answer is yes if $n\le 2$: for $n=1$ there is nothing to prove, and for $n=2$ the positivity of the top self intersection $c_1(L)^2>0$ says that $L$ or its dual is big. But then $c_1(L)\cdot c_1(A)>0$ implies that in fact $L$ is big. -The answer is again yes in all dimensions if $X$ is an abelian variety: in this case $L$ is moreover ample. This is because one can represent $c_1(L)$ and $c_1(A)$ by "constant" hermitian forms, the second being positive definite, and thus the intersection conditions simply tell that the elementary symmetric polynomials in the eigenvalues of the hermitian form representing $c_1(L)$ are all positive. Thus, $L$ is positively curved and hence ample. -I strongly suspect anyway that the result is false in general. -Could you give for instance a counterexample in dimension three? -Thanks in advance. - -REPLY [6 votes]: Here is a very straightforward contre-example. Let $X=\mathbb CP^2\times \mathbb CP^1$ blown up in one point. Denote by $E$ the exceptional divisor, and denote by $\pi$ the projection of $X$ to $\mathbb CP^2$, and take the following bundle: -$$L_n=\pi^*(O(n))\otimes O(E),$$ -where $n$ satisfies two conditions: $$c_1(O(n))^2\cdot c_1(A)>-c_1(O(E))^2\cdot c_1(A),\;\;\;\; -c_1(O(n))\cdot c_1(A)^2>-c_1(O(E))\cdot c_1(A)^2,$$ -it is obvious that such $n$ exists. -To that this bundle is what you want we just need the following two simple facts: $c_1(\pi^*(O(n)))\cdot c_1(O(E))=0$ and $c_1(O(E))^3=1$. $L_{n}$ is not big because the $H^0(kL_n)$ grows quadratically with $k$. -Idea of this example works in dimensions $2m+1$. We take a semi-ample line bundle $L_{sa}$ with Itaka dimension $2m$ http://en.wikipedia.org/wiki/Iitaka_dimension (in particular it is not big), and tensor it with a line bundle corresponding to an exceptional divisor $E$. We chose them so that $c_1(L_{sa})\cdot c_1(O(E))=0$, i.e., these bundles "don't interact". For large $n$ the class $(c_1(nL_{sa}))^k$ (provided -$k\le 2m$) is represented by a cycle of a high degree (with respect to $A$), so it "beats" $(c_1(O(E)))^k$. Finally - $E^{2m+1}=1$.<|endoftext|> -TITLE: whence commutative diagrams? -QUESTION [71 upvotes]: It seems that commutative diagrams appeared sometime in the late 1940s -- for example, Eilenberg-McLane (1943) group cohomology paper does not have any, while the 1953 Hochschild-Serre paper does. Does anyone know who started using them (and how they convinced the printers to do this)? - -REPLY [32 votes]: Eduard Study in Von den Bewegungen und Umlegungen, Math. Ann. 39 (1891) 441-566, writes on p. 508: - -Here $g, g^*, g'$ are rays in space with polar planes $\gamma, \gamma^*, \gamma'$, $\mathfrak P$ is the “polarity” taking rays to planes and conversely, the $\mathfrak t_i$ are commuting collineations acting on both rays and planes, and Study uses $\mathfrak t_i\mathfrak P$ to denote the composition we would write $\mathfrak P\circ\mathfrak t_i$. -(Essentially the same diagram is repeated in Study’s book Einleitung in die Theorie der Invarianten linearer Transformationen auf Grund der Vektorenrechnung (1923, p. 217), with credit to (1891).)<|endoftext|> -TITLE: Examples of "Unusual" Classifications -QUESTION [7 upvotes]: When one says "classification" in math, usually one of a handful of examples springs to mind: --Classification of Finite Simple Groups with 18 infinite families and 26 sporadic examples (assuming one believes the Classification is indeed complete) --Classification of finte-dimensional semisimple Lie algebras with 4 infinite families and 5 exceptional examples --Classification of (Simple, Formally Real) Jordan Algebras with 4 infinite families and 1 exceptional example -I'm sure there are other examples that non-algebraists would think of before these. All the examples I cited take the basic form of having several infinite families and some number of exceptional examples which do not fall into any of these families. Thus I am wondering: - - -Question: Does anyone know of examples of classifications of some mathematical objects such that the classification consists (A) only of infinite families or (B) only of a finite number of examples/ an infinite number of examples which do not seem to be closely related to one another (i.e. they do not "appear" to form any infinite families). - - -One example of case (B) that comes to mind would be finite dimensional division algebras over $\mathbb{R}$ of which there are 4. On the other hand, for case (B) I would like to rule out way too specific "classifications" such as "finite simple groups with an involution centralizer of such and such a form" since this is really a subclassification within the classification of FSG's. Although I am an algebraist, I would like to hear about examples from any branch of math, for comparison's sake. -(If anyone thinks of better tags for this, feel free to add or suggest them). - -REPLY [2 votes]: In extensions of number fields $E/k$, you can look at primes in the ring of integers of $k$ that either (a) split in the integers of $E$, (b) remain prime in the integers of $E$, or (c) ramify. The case (c) consists of finitely many exceptions.<|endoftext|> -TITLE: Kontsevich's formality theorem from an explicit homotopy -QUESTION [11 upvotes]: Suppose that $X$ is a smooth manifold, whose $C^{\infty}$-functions we denote by $A$. Let $D_{poly}^*(A):=\bigoplus_{n\geq -1}Hom(A^{\otimes n+1},A)$ be the Lie algebra of polydifferential operators with the Hochschild differential and the Gerstenhaber bracket. A version of the Hochschild-Kostant-Rosenberg theorem shows the cohomology of $D_{poly}^*(A)$ is isomorphic to $\bigoplus_{n\geq -1}\wedge^{n+1}T_X$, polyvector fields. The graded vector space of polyvector fields is also a Lie algebra with the Schouten-Nijenhius bracket. However, the HKR-isomorphism is not a morphism of Lie algebras. The formality theorem of Kontsevich shows the HKR-isomorphism can be corrected to an $L_{\infty}$-morphism whose first term is the HKR-isomorphism. The upshot is this proves star-products on $C^{\infty}(X)$ correspond to formal Poisson structures. -In the article: M. de Wilde and P. B. A Lecomte: An homotopy formula for the Hochschild cohomology, Compositio Mathematica, tome 96, no. 1 (1995), the authors construct an explicit homotopy for the HKR-isomorphism. From this explicit homotopy the authors construct a star-product on $\frak g^*$, the dual of the Lie algebra $\frak g$. My question is the following: can one do the same thing for the general case of a Poisson manifold. In particular, is the explicit homotopy which induces Kontsevich's $L_{\infty}$-quasiisomorphism known? I believe that one has to exist since two $L_{\infty}$-algebras are quasiisomorphic if and only if they are homotopy equivalent. But, can you write down the explicit homotopy from the quasiisomorphism. - -REPLY [5 votes]: The only way I know to construct a formality quasi-morphism for poy-vector fields out of an homotopy is via Tamarkins approach (i.e. $G_\infty$-formality). What Tamrakin does is - -prove that there is a suitable $G_\infty$-structure on Hochschild cochains -prove that the obstruction to construct a $G_\infty$-formality step by step is unobstructed. At each step you have to make some choice, and the homotopy for the Hochschild complex of De Wilde-Lecomte gives you a way to do such choices. - -But Tamarkin construction (part 1) involves the choice of an associator (i.e. choice of appropriate weights in Kontsevich's $L_\infty$-qausi-isomorphism)... so I think that this is hopeless to construct the formality out of an homotopy. -Moreover, even the other way I don't see how you could associate an homotopy for the Hochschild complex to an $L_\infty$-quasi-isomorphism from $T_{poly}$ to $D_{poly}$. -I might be wrong but I have the feeling that you are mixing two different notions of homotopy: that of higher homotopies in the $L_\infty$-morphism, and that of homotopy retract for the Hochschild cochain complex. In particular, what do you mean by a "homotopy equivalence" between $L_\infty$-algebras (whatever you mean, a homotopy for the Hochschild complex in the sens of De Wilde-Lecomte won't produce an example).<|endoftext|> -TITLE: Unbalancing lights in higher dimensions -QUESTION [16 upvotes]: In ''The Probabilistic Method'' by Alon and Spencer, the following unbalancing lights problem is discussed. Given an $n \times n$ matrix $A = (a_{ij})$, where $a_{ij} = \pm 1$, we want to maximise the quantity -$x^T A y = \sum_{i=1}^n \sum_{j=1}^n a_{ij} x_i y_j$ -over all $n$-dimensional vectors $x$, $y$ such that $x_i,y_j = \pm 1$. The name of the problem comes from interpreting $A$ as a grid of lights that are on or off, and $x$ and $y$ as sets of light switches, each associated with a row or column (respectively); flipping a switch flips all lights in that row or column, and the goal is to maximise the number of lights switched on. -Let $m(A)$ be the maximum of $x^T A y$ over $x$ and $y$ such that $x_i,y_j = \pm 1$. As Alon and Spencer discuss, for any $A$ it is possible to show that $m(A) \ge C n^{3/2}$ for some constant $00$? Or even just improved so that the dependence on $d$ is subexponential? Conversely, can the upper bound be reduced? -Background -Finding a lower bound on $m(A)$ in the more general case where $A$ is an arbitrary bilinear form was considered by Littlewood back in 1930. The bound above for the $d$-dimensional case is a special case of a bound for general $d$-linear forms which was proven later by Bohnenblust and Hille. In the functional analysis literature, the quantity $m(A)$ is known as the injective tensor norm of $A$; this norm, and the above results, are discussed extensively in the book Analysis in Integer and Fractional Dimensions by Ron C. Blei. However, I could not find any information about whether the bounds can be improved. - -REPLY [7 votes]: Belatedly answering my own question: Pellegrino and Seoane-Sepulveda have shown the lower bound that $m(A) \ge n^{(d+1)/2}/\text{poly}(d)$. As far as I know, it is still open whether the $\text{poly}(d)$ term can be replaced with a universal constant.<|endoftext|> -TITLE: $\partial \bar{\partial}$ lemma for contractible domains -QUESTION [10 upvotes]: Question. Is every $(p, \, p)$ closed form ($p\geq1$) in a contractible open set of $\mathbb{C}^n$ $\partial \bar{\partial}$ exact? - -We know that every $d$-exact $(p, \,p)$-form on a compact Kahler manifold is $\partial \bar{\partial}$ exact (by the Hodge theorem), but unfortunately, that can't be applied here... - -REPLY [8 votes]: If in addition you assume that your domain is pseudoconvex then by a theorem of -A.Aeppli what you want is true.The paper is titled :On the cohomology structure of Stein -manifolds 1965 (Proc.Conf.Complex Analysis(Minneapolis Minn 1964) pages 58 to 70 Springer, -Berlin. - David's example indicates why one might need a hypothesis like pseudoconvexity.<|endoftext|> -TITLE: Which torsion classes in integral cohomology are Chern classes of flat bundles? -QUESTION [13 upvotes]: Chern-Weil theory tells us that the integral Chern classes of a flat bundle over a compact manifold (i.e. a bundle admitting a flat connection) are all torsion. Given a compact manifold $M$ whose integral cohomology contains torsion, one can then ask which (even-dimensional) torsion classes appear as the Chern classes of flat bundles. What is known about this question? I would be interested both in statements about specific manifolds and about general (non)-realizability results. -One specific thing that I know: if $S$ is a non-orientable surface, then there is a flat bundle $E\to S$ whose first Chern class is the generator of $H^2 (S; \mathbb{Z}) = \mathbb{Z}/2$. This shows up, for example, in papers of C.-C. Melissa Liu and Nan-Kuo Ho. As Johannes pointed out in the comments, this also shows that the fundamental class of a product of surfaces can be realized by a flat bundle. -However, I suspect that for a product of 3 Klein bottles, not all the 4-dimensional torsion classes can be realized as second Chern classes of flat bundles. In fact, I think I know a proof of this if one restricts to unitary flat connections: the space of unitary representations has too few connected components. - -REPLY [3 votes]: The answer to Tom's formulation is no. It's possible if you restrict to finitely generated groups that my argument falls apart, but I doubt this is essential. -Take a group $\Gamma$ so that $B\Gamma^+=K(Q/Z,2n-1)$, ie, $H^k(\Gamma;Z)=H^k(K(Q/Z,2n-1);Z)$. Since $Ext(Q/Z,Z)=\hat Z$, there lots of interesting classes in $H^{2n}(\Gamma;Z)$. If we could lift them to flat bundles over $B\Gamma$, then after applying the plus construction and profinite completion, we would have split $K(\hat Z;2n)$ off of $BU^{\hat{}}$. But the torsion homology of the Eilenberg-MacLane space cannot be a retract of the torsion-free homology of $BU$. -I wanted to work one prime at a time, but $Ext(Q_p/Z_p,Z)=0$.<|endoftext|> -TITLE: Characterization of algebraic points on Shimura varieties? -QUESTION [5 upvotes]: Is there any (conjectural) characterization of $\overline{\bf{Q}}$-points -on Shimura varieties? -The question of course does not always make sense -for ${\bf{Q}}$-points: a theorem of Shimura shows that a quaternionic Shimura curve has -no ${\bf{R}}$-points, and a theorem of Mazur shows that the modular curve -$Y_0(N)$ has no ${\bf{Q}}$-points for $N$ sufficiently large. The question does -however seem to make sense for certain abelian extensions of CM fields. (For -instance, in the setting of a quaternionic Shimura curve $M$ defined over a totally -real field $F$, if a totally imaginary quadratic extension $K/F$ embeds -into the underlying quaternion algebra, then there is a supply of CM points -on $M$ defined over certain abelian extensions of $K$). In particular, I should -like to know more about the following questions: -(i) Over which number fields $k$ does a given Shimura variety $S(G, X)$ -have a $k$-rational point? -(ii) For which such number fields $k$ will $S(G, X)(k)$ be Zariski dense? -(iii) To what extent are such $k$-rational points accounted for by CM points (or similar constructions)? -Sorry if these questions are imprecise or wrongly formulated, I would be happy to at least have an indication of -where to look in the literature if someone has already thought about this. - -REPLY [10 votes]: If you haven't, you should first think about these questions just for modular curves, which are the simplest Shimura varieties. Then there are only finitely many $N$ for which the modular curve of level $N$ has genus $< 2$. Once the genus is at least $2$, there are only finitely many points over any fixed number field (by Mordell's conjecture/Faltings theorem). -For higher dimensional Shimura varieties, once the level gets large enough the variety will become of general type, and Lang's conjecture will (presumably) apply, so as to give restrictions on the rational points. In particular, the answer to (ii) will presumably be never if the level is large enough. The $\overline{\mathbb Q}$-points are Zariski dense by the Nullstellensatz, - and most of them won't be CM points (unless the Shimura variety is associated to a torus, so that every point is CM!), so eventually (i.e. if you make $k$ large enough) you will get points that are not CM.<|endoftext|> -TITLE: The relationship between Crofton formula and Radon transform. -QUESTION [9 upvotes]: The famous Crofton formula says that the length of a curve can be calculated by integral of the `line crossing' over the space of all oriented lines. My question is, is there a way to treat this formula as a special case or corollary of the Radon transform theory? If so, how can we express the relationship precisely? - -REPLY [3 votes]: You should definitely check these notes generated by three bright undergraduates for an REU project that I supervised a few years ago. I promise you, it will be worth your time.<|endoftext|> -TITLE: Bounded denominators for modular forms -QUESTION [8 upvotes]: I recently saw a conjecture that a modular form is a modular form for a congruence subgroup of $SL_2(Z)$ if and only if it has bounded denominators. Are both directions conjectures, or is one already known to be true? - -REPLY [11 votes]: As Ramsey pointed at, it is known that forms on congruence subgroups have bounded denominators. The other direction is still a conjecture, though some partial progress has been made. You might try looking at some of the papers by Ling Long ( http://orion.math.iastate.edu:80/linglong/ ). She has been interested in this problem for a while and has made some progress towards the conjecture. I'll briefly state one of her results (the others involve a bit more notation). All of her papers in this area are available on the arxiv. -Let $\Gamma$ be a finite index subgroup of $SL_2(\mathbb Z)$. -Unbounded denominators conjecture: Any meromorphic modular form on $\Gamma$ which is holomorphic on $\mathfrak h$ and has algebraic Fourier coefficients has bounded denominators if and only if it is a cusp form and $\Gamma$ is a congruence subgroup. -In their paper "Fourier Coefficients of Noncongruence Cusp Forms" Winnie Li and Ling Long prove the following theorem: -Theorem: Suppose the modular curve $X_\Gamma$ has a model defined over $\mathbb Q$ such that the cusp at $\infty$ is $\mathbb Q$-rational, $k\geq 2$ and $S_k(\Gamma)$ is one dimensional. Then a form in $S_k(\Gamma)$ with rational Fourier coefficients has bounded denominators if and only if it is a congruence modular form. -The other two papers by Long (joint with Chris Kurth) that I mentioned are available here and here. The idea of these papers is to prove the unbounded denominator property for certain classes of noncongruence subgroups. -Lastly, you may find this survey article a bit helpful.<|endoftext|> -TITLE: Question on the interpretation of a presheaf category as a co-completion -QUESTION [6 upvotes]: The category of presheaves $Pre(C)$ on a small category $C$ is the category of functors $C^{op}\to Sets$. Since the category of sets is co-complete and every presheaf is a colimit of representable ones (i.e. presheaves of the form $Hom(-,c)$ for an object $c$ of $C$) $Pre(C)$ may be interpreted as the co-completion of $C$. -I wonder how $Pre(C)$ relates to $C$ if $C$ is already co-complete. More generally, if $D$ is a subcategory of $Pre(C)$ containing all the representable presheaves, how is the relation between $Pre(C)$ and $Pre(D)$? -As a special instance consider $C=\Delta$. Then $Pre(C)$ is the category $sSet$ of simplicial sets. Let $D$ be the full subcategory of finite simplicial sets, is $Pre(D)\cong sSet$? -Thank you. - -REPLY [17 votes]: You should think of $Pre(C)$ as a formal (i.e., free) cocompletion of $C$. Even if $C$ is cocomplete, by passing to $Pre(C)$ there will be "new" colimits that were not in $C$ (are not in the essential image of the Yoneda embedding). A simple example of this is the empty presheaf (which returns the empty set when evaluated at any object of $C$); there is no way this is represented by an object of $C$, since $\hom(-, c)$ has at least one element when evaluated at $c$. -Part of the problem is nomenclatural: normally when people use the word "completion" (as when completing a metric space, or completing an integral domain to its field of fractions), the embedding of an complete object into its completion is an isomorphism. Not so in the case of free completion or free cocompletion! There is an interesting discussion of this in the nLab; roughly speaking, the traditional sorts of completions are covered by the concept of "idempotent monad", especially when the unit of the monad is an embedding, but free cocompletion is far from idempotent. -If $C$ is cocomplete in a very nice sense (called "totally cocomplete", or just "total" for short), then even though the Yoneda embedding will certainly not be an equivalence of categories, there will be something like the next best thing to an inverse: the Yoneda embedding will have a left adjoint. (This is actually a definition of total cocompleteness.) Most of the large cocomplete categories that arise in practice as categories of structured sets, such as categories algebraic over $Set$ for instance, or $Top$, have this property. -As for the last example, $Pre(C)$ and $Pre(D)$ are not equivalent. You can add certain colimits to $C$ without changing the free cocompletion, but there are precious few. In a nutshell, $C$ and $D$ have equivalent presheaf categories if they are Morita equivalent, which in the case of categories enriched in $Set$ means that they have equivalent Cauchy completions or Karoubi envelopes -- the category you get by adjoining coequalizers of pairs $1_x, e: x \to x$ where $e$ is idempotent. (This type of completion is idempotent!) So you can enlarge $C$ by splitting idempotents and get the same presheaf category. In the case of categories enriched in $Ab$, the analogous notion of Karoubi envelope is obtained by adjoining direct sums and splitting idempotents. The general rule in $V$-enriched category theory is that you can adjoin to $C$ what are called "absolute colimits" -- colimits which are preserved by any $V$-enriched functor, and still get the same presheaf category $V^{C^{op}}$ up to equivalence (provided that idempotents split in $V$!), but that's the limit of what you can do if you want the same presheaf category.<|endoftext|> -TITLE: The Circle Method and the binary Goldbach Problem -QUESTION [10 upvotes]: I've been learning about the Circle Method (at the level of the book -"An Invitation to Modern Number Theory," by Miller and Takloo-Bighash). The arguments in the book show how the -Circle method can be applied successfully to the ternary Goldbach problem, but fails for the binary problem (since the minor arc -contribution can't be bounded sufficiently well). Is this the final word on the Circle method's application to this problem? Is there any hope -that more sophisticated arguments/estimates might make it work? -I have seen some numerical experiments that seem to indicate that the minor arc contribution is small - for example, see the undergraduate project at -http://web.williams.edu/go/math/sjmiller/public_html/mathlab/public_html/jr02spring/goldbach/goldbach5.pdf . -Also, what book and/or set of notes would be good for me to continue studying the Circle Method? I have looked at -Vaughn's book, it is too dense for me, without enough motivating material. Is the book by Nathanson ("Additive Number Theory: The Classical Bases") a -good place to continue? -Thanks, -Tom - -REPLY [4 votes]: I am a little nervous writing this, because it is far from my field. But it was my understanding that there is a much more basic problem in proving the binary Goldbach conjecture by analytic methods. -Let $P$ be any set of positive integers and let $f_P(x) = \sum_{p \in P} x^p$. Then the behavior of $f_P$ near $1$ depends on the rate of growth of $\pi_P(n) := \# \{ k \in P: \ k \leq n \}$. Similarly, the behavior of $f_P$ near $e^{2 \pi i \ell/m}$ depends on the rates of growth of the functions $\pi_P(n: r,m) := \# \{ k \in P : \ k\leq n,\ k \equiv r \mod m \}$. The cricle method is all about making this dependence precise, and about approximating $f_P(e^{i \theta})$ by $f_P(e^{2 \pi i \ell/m})$ for some rational approximation $\ell/m$ to $\theta/(2 \pi)$. -However, there is a set of integers $P$ for which all the $\pi_{P}(n: r,m)$ have the same growth rates as for the primes, yet binary Goldbach is false for $P$. So, rather than improving the details of the circle method, one is left with a fundamentally new problem: Thinking of new properties of the primes to make use of. - -So, what is $P$? I'll first do it with $\pi_P(n)$ having the correct growth, then I'll add in the more refined $\pi_P(n, r,m)$'s. -Divide $\mathbb{Z}_{\geq 0}$ up into intervals $[a_0, a_1) \sqcup [a_1, a_2) \sqcup \cdots$ where $a_0=1$ and $a_{k+1} = a_k + \sqrt{a_k} + O(1)$ where the $O(1)$ is some absolute cnstant bound for how much error we will permit (I think $2$ would work). Choose a subsequence $b_j$ of the $a_i$ such that $b_1=a_1$ and $b_{i+1} \geq 2 b_i$. We will make sure that none of the $b_i$ are in $P+P$. -Don't put any elements of $[a_0, a_1)$ into $P$. That ensures that $b_1=a_1$ is not in $P+P$. We now inductively construct $[a_k, a_{k+1}) \cap P$. Suppose that we have already constructed the part of $P$ in $\bigcup_{i < k} [a_i, a_{i+1})$, and that $P+P$ does not contain any $b_j$. Let $b_j$ be the unique $b$ such that $b_{j-1} \leq a_k < a_{k+1} \leq b_j$. When adding elements of $[a_k, a_{k+1})$ to $P$, we can't make a sum of the form $b_{j'}$ with $j' < j$, because $b_{j'}< a_k$. And we can't make a sum $b_{j''}$ with $j' > j$, because $b_{j''} > 2 b_j \geq a_{k+1} + a_{k+1}$. So our only concern is that we might create the sum $b_j$. -Choose $\sqrt{a_k}/\log a_k+O(1)$ elements of $[a_k, a_{k+1})$ to put into $P$, subject to the sole condition that we don't make $b_j$ land in $P+P$. Again, the $O(1)$ is a once and for all global choice. In order to avoid creating this sum, we must avoid putting in $b_j - p$, where $p$ is an element of $P$ lying in an interval of length $[\sqrt{a_k}]$. The size of $P \cap [x, x+L]$ is $\approx L/\log(x+L) \leq L/\log L$. So we have to avoid $\sqrt{a_k}/\log \sqrt{a_k} = 2 \sqrt{a_k}/\log a_k$ things, while choosing $\sqrt{a_k}/\log a_k$ things, all from an interval of length $\sqrt{a_k}$. Since $\log a_k$ is eventually bigger than $3$, this is not a problem. -Now, let's do this while, at the same time, making sure that the $\pi_P(n, r,m)$ grow correctly. This time, make sure to take the $b_j$ to be even, since you won't be impressed if they are odd. The $O(1)$ in the defining recursion for the $a$'s gives me enough room to make sure lots of $a$'s are even, so that's fine. This time, we are going to require that, in addition, for every $m$, we have $(\sqrt{a_k}/\log a_k)/\phi(m!)+O(1)$ elements of $P \cap [a_k, a_{k+1})$ in each invertible residue class modulo $m!$. Note that, for $m$ large, the $O(1)$ swamps the first term, so this condition is trivial. So for each $k$, this is in fact finitely many conditions to obey. -And, since $\log a_k$ does eventually get larger than $3 \phi(m!)$, this should be doable. -There is a lot of unchecked material here; I welcome corrections and citations.<|endoftext|> -TITLE: Square-free diophantine approximation -QUESTION [9 upvotes]: Given an irrational algebraic number $\alpha$ (and maybe I want to add: of degree greater than $2$?), do there exist infinitely many relatively prime and square-free $p$,$q$ with -$$|\alpha - p/q | < \frac{1}{q^2}\ .$$ -I would guess "yes" based on a combination of Dirichlet's approximation theorem and the positive density of square-free integers among all the integers, but I can't even think of one example where I can prove this. - -REPLY [12 votes]: I believe this is an open problem. Heath-Brown (1984) proved that there are infinitely many solutions to $|\alpha-p/q| < q^{-5/3+\epsilon}$ in square-free numbers. Harman (1984) proved that for almost all $\alpha$ there are infinitely many solutions to $|\alpha-p/q| < q^{-2+\epsilon}$ in square-free numbers. See paper1 and paper2.<|endoftext|> -TITLE: Question about a Limit of Gaussian Integrals and how it relates to Path Integration (if at all)? -QUESTION [11 upvotes]: I have come across a limit of Gaussian integrals in the literature and am wondering if this is a well known result. -The background for this problem comes from the composition of Brownian motion and studying the densities of the composed process. So if we have a two sided Brownian motion $B_1(t)$ we replace t by an independent Brownian motion $B_2(t)$ and study the density of $B_1(B_2(t))$. If we iterate this composition n times we get the iterated integral in (**) below as an expression for the density of the n times iterated Brownian motion. The result I am interested in is derived in the following paper: -The original reference is "Fractional diffusion equations and processes with randomly varying time" Enzo Orsingher, Luisa Beghin http://arxiv.org/abs/1102.4729 -Line (3.14) of Orsingher and Beghins paper reads for $t > 0$ -$$(**) \qquad\lim_{n \rightarrow \infty} 2^{n} \int_{0}^{\infty} \ldots \int_{0}^{\infty} \frac{e^{\frac{-x^2}{2z_1}}}{\sqrt{2 \pi z_1}} \frac{e^{\frac{-{z_1}^2}{2z_2}}}{\sqrt{2 \pi z_2}} \ldots \frac{e^{\frac{-{z_n}^2}{2t}}}{\sqrt{2 \pi t}} \mathrm{d}z_1 \ldots \mathrm{d}z_n = e^{-2 |x|} $$ - -How do you prove this result without using probability? -Edit: there has been a solution posted to 1) using saddlepoint approximation but I am still not clear on how to make the argument rigorous -https://physics.stackexchange.com/q/7552/2757 -I have been studying a slight generalization of ** from the probability side of things and have been trying to use dominated convergence to show the LHS of ** is finite but I am having problems finding a dominating function over the interval $[1,\infty)^n$. Is dominated convergence the best way to just show the LHS of (**) is finite? -Is this a type of path integral (functional integral)? Or is this integrand some kind of kinetic plus potential term arsing in quantum mechanics? Do expressions like (**) ever come up in physics literature? - -(I tried using the change of variable theorem for Wiener measure to transform (**) into a Wiener integral with respect a specific integrand and have had some success with this.. I think this shows how to compute a Wiener integral with respect to a function depending on a path and not just a finite number of variables but did not see how to take this any further - The change of variable theorem for Wiener Measure was taken from "The Feynman Integral and Feynman's Operational Calculus" by G. W. Johnson and M. L. Lapidus.) - -REPLY [3 votes]: The expression you are interested in is of the form $\lim_n T^n\psi_t$ where $T$ is the integral operator -$$Tf(x)=2\int_0^\infty \frac{e^{-\frac{x^2}{2y}}}{\sqrt{2\pi y}} f(y)dy$$ -and $\psi_t(x)= \frac{e^{-\frac{x^2}{2t}}}{\sqrt{2\pi t}}$. Note that $T$ is an operator with a positive kernel $T(x,y)=2 I[y>0] \frac{e^{-\frac{x^2}{2y}}}{\sqrt{2\pi y}}$ which satisfies $\int_{-\infty}^\infty T(x,y)dx =1.$ That is, $T$ is much like a stochastic matrix and the limit you wish to obtain could only hold if $e^{-2|x|}$ is the principle eigenvector (with eigenvalue $1$). -To realize $T$ as something like a stochastic matrix, we should present it as a compact operator. It is not compact on $L^2(\mathbb{R})$, but if we think of it as an integral operator on the space $L^2(e^x dx)$ of functions with -$$\int_{-\infty}^\infty |f(x)|^2 dx <\infty,$$ -i.e. -$$Tf(x)=\int_{-\infty}^\infty K(x,y) f(y) e^y dy$$ -with $K(x,y)= 2I[y>0] \frac{e^{-\frac{x^2}{2y} -y}}{\sqrt{2\pi y}}$, then -$\int\int K(x,y)^2 e^{x+y}dxdy <\infty$ so $T$ is Hilbert-Schmidt on $L^2(e^x dx)$, hence compact. Because $K(x,y)>0$ for all $x,y$ the Perron-Frobenius theorem (suitably generalized to compact operators of this type) shows that $T$ has a unique positive eigenvalue $\lambda_0$ with a positive eigenfunction and all other eigenvalues $\lambda$ are of modulus $|\lambda|<\lambda_0$. I claim that $T\phi=\phi$ where $\phi(x)=e^{-2|x|}$, so the unique positive eigenvalue is one! (Certainly this has to do with the origins of the problem in probability theory.) To see that $T\phi=\phi$ it is most convenient to take a Fourier transform in $x$ to get -$$ \widehat{T\phi}(k)=2 \int_0^\infty e^{-\frac{k^2}{2}z}e^{-2z}dz=\frac{4}{k^2 +4}=\widehat{\phi}(k).$$ -The other thing we need is the left principle eigenvector -- the eigenvector of $T^\dagger$ with eigenvalue $1$. Here the adjoint must be on $L^2(e^x dx)$ so -$$ T^\dagger f(x) =\int_{-\infty}^\infty K(y,x)f(y)e^y dy= 2 e^{-x} I[x>0] \int_{-\infty}^\infty \frac{e^{-\frac{y^2}{2x}}}{\sqrt{2\pi x}} e^y f(y).$$ -Observe that $T^\dagger \widetilde{\phi}(x)=\widetilde{\phi}(x)$ where $\widetilde{\phi}(x)=2 e^{-x}I[x>0]$. The factor of $2$ enforces the normalization $\langle \phi,\widetilde \phi \rangle =1$ (with the inner product in $L^2(e^xdx)$). It now follows that -$$T^n f =\phi \langle \widetilde{\phi},f\rangle +o(1)$$ -for any function $f\in L^2(e^xdx)$. Since -$$\langle \widetilde{\phi},\psi_t \rangle =\int_{0}^\infty \widetilde{\phi}(x)\psi_t(x)e^x dx=2 \int_0^\infty \psi_t(x)= $$ -your identity follows.<|endoftext|> -TITLE: If a representation has enough reductive stabilizers, is it a direct sum of characters? -QUESTION [8 upvotes]: Suppose $G\to GL(V)$ is a linear representation of an irreducible algebraic group over a field $k$. - -Suppose $C\subseteq V$ is a $G$-invariant closed cone that spans $V$, and that the stabilizer of any point of $C$ is linearly reductive. Must $V$ be a direct sum of 1-dimensional representations? - -[Edit] I mostly care about the case where $k$ is algebraically closed and of characteristic zero. In this case, all instances of "linearly reductive" may be changed to "reductive." -[Edit] Remark 0: If they answer is "yes," then the image of $G$ in $GL(V)$ is a torus. -Remark 1: In my situation, $C$ is actually the closure of a $G$-orbit, so I'm happy to assume $C$ contains a dense open copy of $G$. -Remark 2: I can prove this result in the case $C=V$ as follows. -Taking $v=0$, we see that $G$ is linearly reductive. A subgroup $H\subseteq G$ is linearly reductive if and only if $G/H$ is affine (I can't recall the name associated to this result). It follows that for every $v\in V$, the orbit $G\cdot v \cong G/Stab(v)$ is affine. Every non-zero orbit is either a $k^\times$-torsor or a $\mu_n$-torsor (for some $n$) over its image in $\def\P{\mathbb P}\P(V)$, so the image of each orbit in $\P(V)$ is also affine (this uses that $k^\times$ and $\mu_n$ are linearly reductive). Choose a non-zero $v\in V$ so that the image $Z$ of $G\cdot v$ in $\P(V)$ is minimal dimensional. Minimality of dimension implies that $Z$ is closed. A closed subscheme of $\P(V)$ is affine if and only if it is finite. Since $G$ is irreducible, we have that $Z$ is a single point, so $G$ fixes the 1-dimensional subspace $k\cdot v$. $G$ is linearly reductive, so you can find a complement to $k\cdot v$ and repeat this argument until you've decomposed all of $V$ into 1-dimensional $G$-invariant subspaces. -In general, this proof produces $\dim(C)$ many $G$-invariant 1-dimensional spaces (contained in $C$), but the span of these subspaces need not contain $C$. - -REPLY [4 votes]: This is meant to be a slightly cleaned up version of Peter's answer above. If I'm not mistaken, it is not necessary to assume the characteristic of $k$ is zero. However, I don't know the theory of reductive groups in positive characteristic—even in characteristic 0, I've only dealt with semi-simple groups—so I may be making wrong assumptions about how the representation theory works. - -Let $B\subseteq G$ be a borel with torus $T$ and unipotent radical $U$. Let $V=\bigoplus V_\lambda$ be the decomposition of $V$ into weight spaces with respect to $T$. -Lemma: Suppose $v_\lambda\in V_\lambda$, $v=\sum v_\lambda$ is a point of $C$, and $\mu\in T^\vee$ is extremal among weights in the support of $v$ (i.e. weights for which $v_\mu\neq 0$). Then $C$ contains $v_\mu \def\l{\langle}\def\r{\rangle}$. -Proof: Since $\mu$ is extremal among weights in the support of $v$, there is a 1-parameter subgroup $\alpha\in \hom(\mathbb G_m,T)\cong \hom(T^\vee,\mathbb Z)$ such that $\l\alpha,\mu\r > \l\alpha,\lambda\r$ for any $\lambda\neq \mu$ in the support of $v$. Consider the map $f\colon\mathbb A^1\to V$ given by $f(t)= \sum t^{\l\alpha,\mu\r-\l\alpha,\lambda\r}v_\lambda$. Away from $t=0$, we have that $f(t)=t^{\l\alpha,\mu\r}\alpha(t^{-1})\cdot v$ is in $C$. Since $C$ is closed, we get that $f(0)=v_\mu$ is in $C$. $\square$ -Now suppose $\mu$ is an extremal highest (with respect to $B$) weight appearing in the decomposition of $V$. Since $C$ spans $V$, there is a point of $C$ with a component in $V_\mu$, so by the lemma, there is a non-zero vector $v\in V_\mu\cap C$. Since $\mu$ is a highest weight, we have that $U$ stabilizes $v$. Since the stabilizer of $v$ is reductive, it must contain the derived subgroup of $G$, so the irreducible representation with highest weight $\mu$ is a character, so $\mu$ pairs to zero with every coroot. It follows that every weight in the decomposition of $V$ pairs to zero with every coroot, so $V$ is a direct sum of characters.<|endoftext|> -TITLE: How true are theorems proved by Coq? -QUESTION [52 upvotes]: Less tongue in cheek, is it known what the relative consistency is for theorems proved with an automatic theorem prover? Of course this depends somewhat on what assumptions one makes with respect to predicativity and so on. But for concreteness take one of the popular packages with its standard installation. -Perhaps this is a can of worms, or a piece of string of indeterminate length, but the recent surge of interest in Voevodsky's univalent foundations raises questions about the consistency strength of the system HoTT he (and others) propose. - -REPLY [5 votes]: I just learned about the part of the Coq FAQ titled What do I have to trust when I see a proof checked by Coq?. To quote from the Apr 24, 2018 revision: - -You have to trust: - -The theory behind Coq: The theory of Coq version 8.0 is generally admitted to be consistent wrt Zermelo-Fraenkel set theory + inaccessible cardinals. Proofs of consistency of subsystems of the theory of Coq can be found in the literature. -The Coq kernel implementation: You have to trust that the implementation of the Coq kernel mirrors the theory behind Coq. The kernel is intentionally small to limit the risk of conceptual or accidental implementation bugs. -The Objective Caml compiler: The Coq kernel is written using the Objective Caml language but it uses only the most standard features (no object, no label ...), so that it is highly improbable that an Objective Caml bug breaks the consistency of Coq without breaking all other kinds of features of Coq or of other software compiled with Objective Caml. -Your hardware: In theory, if your hardware does not work properly, it can accidentally be the case that False becomes provable. But it is more likely the case that the whole Coq system will be unusable. You can check your proof using different computers if you feel the need to. -Your axioms: Your axioms must be consistent with the theory behind Coq. - - -So I guess Coq theorems are true modulo the above.<|endoftext|> -TITLE: What holomorphic functions are limits of polynomials? -QUESTION [15 upvotes]: Let $\Omega$ be a connected open set in the complex plane. What is the closure of the polynomials in $\mathcal{H}(\Omega)$ the set of holomorphic functions on $\Omega$? The topology is the usual compact convergence topology. Take, for instance, an annulus such as $D(r,R)$, the set of all complex $z$ such that $r<|z|< R$, you cannot recover the function $z\mapsto \frac{1}{z}$ because of the residue at $0$, so what holomorphic functions are limits of polynomials? - -REPLY [18 votes]: Let $\Sigma\supset\Omega$ be the union of $\Omega$ and all bounded components of ${\mathbb C}\setminus \Omega$. The algebra you get is the algebra of all holomorphic functions on $\Sigma$. -First, every $f\in{\cal H}(\Sigma)$ is a locally uniform limit of polynomials as a consequence of Runge's Theorem, see Corollary 1.15 in John Conway's "Functions of one complex variable I", 2nd Ed. -Second, if $p_n$ is a sequence of polynomials converging locally uniformly on $\Omega$ and if $K$ is a bounded component of ${\mathbb C}\setminus \Omega$, then $p_n$ also converges uniformly on a neighborhood of $K$ by Cauchy's integral Theorem, as there exists a path around $K$ with winding number 1. (Take the positively oriented boundary of an $\varepsilon$-neighborhood of $K$, where $\varepsilon$ is chosen so small that it does not hit any other components.) - -REPLY [9 votes]: Dear Olivier, let's call Runge an open subset $\Omega \subset \mathbb C $ such that polynomials are dense in $\mathcal H(\Omega)$ . A hole of $\Omega$ is a compact connected component of $ \mathbb C \setminus \Omega $. We then have the equivalent statements, for the open subset $\Omega \subset \mathbb C$ (not assumed connected). -a) $\Omega$ is Runge -b) $\Omega$ has no hole -c) Every connected component of $\Omega$ is simply connected -In the general case, when these equivalent conditions are not fulfilled, Runge's theorem says that if you choose one point in each hole of $\Omega$, then the rational functions with poles only in these points are dense in $\mathcal H(\Omega)$. Beware that you can have a non-denumerable set of holes : take for $\Omega$ the complement of a Cantor set in $\mathbb R \subset \mathbb C$. -In a related vein, Mergelyan's (difficult) theorem says that if you take a compact subset $K \subset \mathcal H (\Omega)$ with connected complement in $\mathbb C$, then any continuous function on $K$ which is holomorphic in the interior of $K$ can be uniformly approximated by polynomials. -Bibliography: Remmert's book is probably the best reference for this question ( and many others...) The equivalence of the statements quoted above is proved in Chapter 13, section 2. -Mergelyan's theorem is not in Remmert's book but is proved on page 386 of Rudin's well known Real and complex analysis, of which you can find a review here -Remark: These results are somewhat astonishing. Take $\Omega=\mathbb C \setminus \{x\in \mathbb R| x \leq -1\}$ and for $f \;$ the holomorphic branch of the logarithm $f(z)=log (1+z) $ which is zero at the origin. Its Taylor series $\sum_{k=0}^{\infty} (-1)^k \frac {z^k}{k}$ diverges for -$|z|\gt 1$ and the partial sums of the series are polynomials which definitely don't converge to $f$. However, since $\Omega$ has no holes, there does exist some sequence of polynomials converging to $f$ uniformly on compact subsets of $\Omega$.<|endoftext|> -TITLE: Horst Knörrer's Permutation Cancellation Problem -QUESTION [18 upvotes]: The Problem: -The following question of Horst Knörrer is a sort of toy problem coming from mathematical physics. -Let $x_1, x_2, \dots, x_n$ and $y_1,y_2,\dots, y_n$ be two sets of real numbers. -We give now a weight $\epsilon_\pi$ to every permutation $\pi$ on {1,2,...,n} as follows: -1) $\epsilon_\pi =0$ if for some $k \ge 1$, $x_k \ge y_{\pi(1)}+y_{\pi(2)}+\cdots +y_{\pi(k)}$. -2) Otherwise, $\epsilon_\pi=sg(\pi )$. ($sg (\pi )$ is the sign of the prrmutation $\pi$.) -Problem: Show that there is a constant $C>1$ such that (for every $n$ and every two sequences of reals $x_1,\dots,x_n$ and $y_1, \dots, y_n$), -$$\sum_\pi \epsilon_\pi \le C^n \sqrt{n!}.$$ -Origin and Motivation from Mathematical Physics -1) The problem was proposed by Horst in a recent Oberwolfach's meeting as a combinatorial problem that arises (as a toy problem) from mathematical physics. -The context of this question is explained in Section 4 of -J.Feldman, H.Kn\"orrer, E.Trubowitz: -"Construction of a 2-d Fermi Liquid", Proc. XIV. International Congress on Mathematical Physics. Editor: Jean Claude Zambrini. World Scientific 2005 -"In this section, we formulate an elementary question about permutations that may be connected with implementing the Pauli exclusion principle in momentum space." -The problem and some variations are directly related to "cancellations between -Fermionic diagrams". -The wider picture (See the Eleven Papers by J.Feldman, H.Knörrer, E.Trubowitz) is toward mathematical understanding and formalism for highly successful physics quantum theories. (In a very very wide sense this is related to Clay's problem on Yang-Mills and Mass gap.) -Remarks and more Motivation -2) This remarkable cancellation property seems similar to cancellations that we often encounter in probability theory, combinatorics and number theory. -3) It look similar to me even to issues that came in my recent question on Walsh functions. -So this question about permutations is analogous to questions asserting that for certain +1,-1,0 functions on ${-1,1}^n$ there is a remarkable cancellation when you sum over all $\pm 1$ vectors. This is true (to much extent) for very "low complexity class functions" (functions in $AC^0$) by a theorem of Linial-Mansour-Nisan, So maybe we can expect remarkable cancellation for "not too complex" functions defined on the set of permutations. -4) We can simplify in the question and replace condition 1) by -1') $\epsilon_\pi =0$ if for some $k \ge 1$, $x_k \ge y_{\pi(k)}$. -I don't know if this makes much difference. -5) An affarmative answer seems a very bold statement, so, of course, perhaps the more promising direction is to find a counter example. But I think this may be useful too. - -REPLY [8 votes]: The question was beautifully solved affirmatively by Nikola Djokic. -His paper an upper bound on the sum of signs of permutations with a condition on their prefix sets is now posted on the arxive.<|endoftext|> -TITLE: Hilbert transforms of measures -QUESTION [9 upvotes]: Given a finite measure $\mu$ on the real line $\mathbb R$, one definition of its Hilbert transform is $(H\mu)(y) =\frac{1}{\pi}(PV)\int \frac{d\mu(x)}{x-y}$ which is known to exist almost everywhere on $\mathbb R$. Another way is to define the Borel transform ${\mathcal H}(z) = \frac{1}{\pi}\int\frac{d\mu(x)}{x-z}$ for $z\in \mathbb C\setminus \mathbb R$, and then take the limit of $\Re({\mathcal H}\mu)(x+iy)$ as $y\downarrow 0$, such limit existing almost everywhere. In the case that $\mu$ is absolutely continuous it is stated $explicitly$ in the literature that these agree (almost everywhere), and it is $implicit$ in the use of the term `Hilbert transform' in the literature that the two definitions agree (almost everywhere) for general finite $\mu$. Does anyone know where one can find explicit proof(s)? (Proofs for the $L^p$ case are easy to find.) - -REPLY [2 votes]: Showing that the two definitions agree almost everywhere is easy! Using the truncated transform -$$ -\mathcal{H}\_\epsilon\mu(x)=\frac1\pi\int_{\lvert y-x\rvert > \epsilon}\frac{d\mu(y)}{x-y} -$$ -then, by definition, $\mathcal{H}\mu(x)=\lim_{\epsilon\to0}\mathcal{H}\_\epsilon\mu(x)$ for all $x$ at which the limit exists. Convolve the identity -$$ -\Re\left(\frac{1}{x+ih}\right)=\frac{x}{x^2+h^2}=\int_0^11_{\left\lbrace\lvert x\rvert > h\sqrt{t/(1-t)}\right\rbrace}\frac1x\\,dt -$$ -with $\frac1\pi d\mu$ to obtain, -$$ -\Re\left(\mathcal{H}\mu\right)(x+ih)=\int_0^1\mathcal{H}_{h\sqrt{t/(1-t)}}\mu(x)\\,dt. -$$ -The integrand on the right hand side tends to $\mathcal{H}\mu(x)$ as $h\to0$, whenever this is defined, so bounded convergence gives $\Re(\mathcal{H}\mu)(x+ih)\to\mathcal{H}\mu(x)$. - -The question also asks how to show that $\mathcal{H}\mu$ is defined almost everywhere. I don't have a reference for this, but the standard proof can be modified without too much difficulty. The maximal operator $\mathcal{H}^*\mu(x)\equiv\sup_{\epsilon > 0}\lvert\mathcal{H}_\epsilon\mu(x)\rvert$ is weak (1,1) continuous, -$$ -\left\lvert\left\lbrace x\colon\mathcal{H}^*\mu(x) > \lambda\right\rbrace\right\rvert\le C\lVert\mu\rVert/\lambda, -$$ -for all $\lambda > 0$ and a fixed constant $C$ (I'm using $\lvert\cdot\rvert$ to denote the Lebesgue measure). I'm working from the notes Interpolation, Maximal Operators, and the Hilbert Transform, by Michael Wong (Theorem 8.7). These notes look at the case where $\mu$ is absolutely continuous, but the proof carries across to the general case with no big changes. -By weak continuity, if there exists measures $\mu_n$ with $\lVert\mu-\mu_n\rVert\to0$ such that $\mathcal{H}\mu_n$ all exist almost everywhere, then $\mathcal{H}\mu$ exists almost everywhere. -If $\mu$ is absolutely continuous with differentiable density, then $\mathcal{H}\mu$ exists everywhere in the standard way. As the differentiable functions are dense in $L^1$, this extends to all absolutely continuous measures. -Only the case for singular measures $\mu$ remain. So, there exists a measurable $S\subseteq\mathbb{R}$ with zero Lebesgue measure with $\mu(S^c)=0$. Then we can choose compact sets $K_n\subseteq S$ with $\mu(S\setminus K_n)\to0$. Note that the measures $\mu_n\equiv 1_{K_n}\cdot\mu$ are supported on the compact sets $K_n$ and $\lVert\mu-\mu_n\rVert\to0$. It follows that $\mathcal{H}\mu_n(x)$ is defined everywhere outside of $K_n$ (in fact, $\mathcal{H}\_\epsilon\mu_n(x)$ is a constant function of $\epsilon$ for $\epsilon$ small enough that $B_\epsilon(x)\cap K_n=\emptyset$). So, $\mathcal{H}\mu_n$ is defined almost everywhere and, hence, so is $\mathcal{H}\mu$.<|endoftext|> -TITLE: Connes's unpublished manuscript on correspondences, anyone? -QUESTION [13 upvotes]: There exist unpublished notes on correspondences of von Neumann algebras due to Connes. This is often cited, but I've never seen a copy. It would be nice to have this, say, to maybe look further into the point of view discussed in his book Noncommutative geometry. Particularly, the stuff involving half-densities. -Does anyone have a scanned copy of this they wouldn't mind sending/posting a link to? -Thanks, in advance! - -REPLY [6 votes]: Aren't you thinking of Popa's unpublished manuscript on Correspondences, available on his website: -http://www.math.ucla.edu/~popa/popa-correspondences.pdf<|endoftext|> -TITLE: Mathematics seminar for "non-mathematicians" -QUESTION [9 upvotes]: Next term I am leading a seminar for students, who will become teachers for elementary school i.e. for kids of age 6-10. The students in the seminar will have no mathematical background beyond the "Abitur". They are supposed to give a 45 minutes talk on a (not too difficult) mathematical topic, and they have to write an exposition of a few pages. My first ideas cover geometry of triangles, basics around Fibonacci sequences ect. Did somebody on MO teach a similar class already? -In brief: I would appreciate further suggestions for suitable topics very much! - -REPLY [3 votes]: Have a look at the exhibition on knots on http://www.popmath.org.uk . I have used this and other material for talks advertised as "How mathematics gets into knots" for ages 8-80: I was then criticised for suggesting one should stop at 80! -The material I used in a recent general lecture in Kansas (see my my preprint page) on the Dirac string trick, and the related belt trick and Phillipine wine trick, and on the pentoil, can be done for any age; it is essential to involve the audience in doing the tricks, or demos. One can also do addition of knots, and analogies between knots and numbers, getting over the point that while knots and numbers are quite different, the relations between knots can be analogous to the relations between numbers. Note also that my apparatus for the Dirac String Trick was home made: two piece of board, coloured ribbon, bulldog clips to attach the ribbon (in case it gets too tangled). My pentoil was made in a science workshop from copper tubing. -In giving such a presentation to children and teachers once, a teacher came up to me afterwards and said:"That is the first time in my mathematical career that anyone has used the word "analogy" in relation to mathematics!" Thus a real concern is that there is so little discussion about mathematics, and its context.<|endoftext|> -TITLE: Configuration space of little disks inside a big disk -QUESTION [27 upvotes]: The space of configurations of $k$ distinct points in the plane -$$F(\mathbb{R}^2,k)=\lbrace(z_1,\ldots , z_k)\mid z_i\in \mathbb{R}^2, i\neq j\implies z_i\neq z_j\rbrace$$ -is a well-studied object from several points of view. Paths in this space correspond to motions of a set of point particles moving around avoiding collisions, and its fundamental group is the pure braid group. It is not hard to prove that this space is homotopy equivalent to the configuration space of the unit disk -$$F(D^2,k)=\lbrace(z_1,\ldots , z_k)\mid z_i\in D^2, i\neq j\implies z_i\neq z_j\rbrace$$ -In real life however, particles, or any other objects which move around in some bounded domain without occupying the same space, have a positive radius, and so would be more realistically modelled by disks rather than points. This motivates the study of the spaces -$$F(D^2,k;r)= \lbrace(z_1,\ldots , z_k)\mid z_i\in D^2, i\neq j\implies |z_i - z_j|>r\rbrace$$ -where $r>0$. The homotopy type of this space is a function of $r$ and $k$. Fixing $k$ and varying $r$ gives a spectrum (is this the right word?) of homotopy types between $F(\mathbb{R}^2,k)$ and the empty space. It seems like an interesting (and difficult) problem to study the homotopy invariants as functions of $r$. For instance, for $k=3$ what is $\beta_1(r)$, the first Betti number of $F(D^2,k;r)$? - -Question: Have these spaces and their homotopy invariants been studied before? If so, where? - -Of course one can also ask the same question with disks of arbitrary dimensions. - -REPLY [31 votes]: I have a number of results on hard disks in various types of regions, and preprints are in progress. The terminology "hard spheres" (or "hard disks" in dimension 2) comes from statistical mechanics, and I believe Fred Cohen is following my lead on this. (See for example the hard disks section of Persi Diaconis' survey article.) ---- With Gunnar Carlsson and Jackson Gorham, we did numerical experiments and computed the number of path components for 5 disks in a box, as the radius varies over all possible values. This is quite a complicated story already, as the number is not monotone or even unimodal in the radius. (This preprint is almost done, a rough copy is available on request.) ---- With Yuliy Baryshnikov and Peter Bubenik we develop a general Morse-theoretic framework and proved that in a square, if $r < 1/2n$ then the configuration space of $n$ disks of radius $r$ is homotopy equivalent to $n$ points in the plane. On the other hand this is tight: if $r> 1/2n$ then the natural inclusion map is not a homotopy equivalence. (Again this preprint is getting close to be being posted to the arXiv...) There is a much more general statement here. ---- I also have some results with Bob MacPherson about hard disks in a square and also in (the easier case of) an infinite strip. We have been talking to Fred Cohen about this a bit lately, who believes there may be connections to more classical configuration spaces. -I am slightly self-conscious about claiming results here, without first having posted the preprints to the arXiv, but I just wanted to state that there are a number of things known now, and I am working hard to get everything written up in a timely fashion. In the meantime I have some slides up from a talk I recently gave at UPenn. -Here is a concrete example, since that might be more satisfying. -It turns out for $3$ disks in a unit square: -For $0.25433 < r$, the configuration space is empty, -for $0.25000 < r < 0.25433$ it is homotopy equivalent to $24$ points, -for $ 0.20711 < r < 0.25000$ it is homotopy equivalent to $2$ circles, -for $ 0.16667 < r < 0.20711 $ it is homotopy equivalent to a wedge of $13$ circles, and -for $ r < 0.16667$ it is homotopy equivalent to the configuration space of $3$ points in the plane. -For $4$ disks in a square it looks like the topology changes $9$ or $10$ times, and for $5$ disks it looks like the topology might change $25$-$30$ times or more. The general idea is that certain types of "jammed" configurations act like critical points of a Morse function, and mark the only places where the topology can chance. - -Update: two preprints have been posted to the arXiv. -Min-type Morse theory for configuration spaces of hard spheres (Baryshnikov, Bubenik, Kahle): -http://arxiv.org/abs/1108.3061 -Computational topology for configuration spaces of hard disks ( Carlsson, Gorham, Kahle, and Mason): -http://arxiv.org/abs/1108.5719 - -Update (January 2022): A lot has happened in the past year or two. Here are some more preprints that have been recently posted to the arXiv. -Configuration spaces of disks in an infinite strip (Alpert, Kahle, MacPherson) -https://arxiv.org/abs/1908.04241 -Configuration spaces of disks in a strip, twisted algebras, persistence, and other stories -(Alpert and Manin) -https://arxiv.org/abs/2107.04574 -Homology of configuration spaces of hard squares in a rectangle (Alpert, Bauer, Kahle, MacPherson, Spendlove) -https://arxiv.org/abs/2010.14480<|endoftext|> -TITLE: Quantum mathematics? -QUESTION [21 upvotes]: "Quantum" as a term/prefix used to be genuinely physical: what was supposed to be physically continuous turned out to be physically quantized. - -What sense does this distinction make inside mathematics? - -Especially: Is "quantum algebra" a well-chosen name? (According to Wikipedia, it's one of the top-level mathematics categories used by the arXiv, but it's not explained any further.) - -REPLY [15 votes]: I think that the basic intuition relating quantum algebra and quantum physics is something like: -quantum stuff = classical stuff + $\hbar$ (something complicated) -where $\hbar$ is a "small" formal variable. In other words, the point is to consider that the mathematical objects everybody knows are only approximations of more complicated objects. Hence, quantum mathematics has something to do with perturbation theory, because most of the interesting objects in quantum mathematics are perturbations of trivial solutions of some problems/equations. Here, perturbation means that these objects are formal power series in $\hbar$ whose constant term is a trivial solution (eg: 1 :) ) of some equation (eg: the Yang Baxter equation). -Hence, as John pointed out, quantum algebra involves the study of objects for which classical properties (eg: commutativity) are "almost" true (ie: true modulo $\hbar$).<|endoftext|> -TITLE: Geometric interpretation of the Pontryagin square -QUESTION [18 upvotes]: The Pontryagin square (at the prime 2) is a certain cohomology operation -$$ -\mathfrak P_2: H^q(X;\Bbb Z_2) \to H^{2q}(X;\Bbb Z_4) -$$ -which has the property that its reduction mod 2 coincides with $x\mapsto x^2$. Furthermore, -If $x\in H^q(X;\Bbb Z_2)$ is the reduction of an integral class $y$, then $\mathfrak P_2(x)$ -is the mod 4 reduction of $y^2$. -For a definition of $\mathfrak P_2$, see e.g.: -Thomas, E.: A generalization of the Pontrjagin square cohomology operation. Proc. Nat. Acad. Sci. U.S.A., 42 (1956), 266–269. -Suppose now that $X = M^{2q}$ is a closed smooth manifold of dimension $2q$. -Then $x\in H^q(X;\Bbb Z_2)$ is Poincaré dual to a class $x' \in H_q(X;\Bbb Z_2)$. -By Thom representability, $x'$ is represented by a map $f: Q^q\to M$ in which $Q$ is a (possibly unorientable) closed $q$-manifold (by taking the image of the fundamental class). -By transversality, we can even assume that $f$ is an immersion. -Question: Is there an interpretation of $\mathfrak P_2(x)$ as a geometric operation on -$f$? -(Note: By the properties of the Pontryagin square, if $Q$ and $M$ are orientable, then -then $\mathfrak P_2(x)$ is represented by the self-intersection of $f$ reduced mod 4.) - -REPLY [11 votes]: This is just a guess, based on your comment about the case when $Q$ and $M$ are both oriented. -Claim: If $M$ is oriented and $Q$ is unoriented (and perhaps nonorientable), then the self-intersection of $Q$ is well defined in $\mathbb Z$ if $q$ is odd and well-defined in $\mathbb Z_4$ is $q$ is even. -Proof: Let $Q'$ be a parallel copy of $Q$ which intersects $Q$ transversely. Choose local orientations of $Q$ and $Q'$ near their intersection points so that the local orientations of corresponding points of $Q$ qnd $Q'$ agree. Define an intersection number (dependent on these choices) by summing the signs of the intersections. There are two types of intersections of $Q$ and $Q'$: (a) those coming from the non-triviality of the normal bundle of $Q$, and (b) those coming from intersections of $Q$ with itself (if $Q$ if immersed but not embedded). Changing a local orientation does not affect the contribution of type (a) intersection points to the total intersection, since both the $Q$ and $Q'$ local orientations change together. Type $b$ intersections come in pairs. If $q$ is odd then the signs of these pairs always cancel, so changing local orientation has no effect. If $q$ is even then changing a local orientation flips the signs of both points, so the total intersection changes by $\pm 4$. -So perhaps in the case when $M$ is oriented the Pontryagin square is given by the above mod 4 intersection number, just as in the case when $Q$ is oriented.<|endoftext|> -TITLE: Recent developments on the Artin conjecture -QUESTION [14 upvotes]: Hi to everybody! I am studying right now the work of Serre and Deligne about the modularity of 2-dimensional complex Galois representations. I know that if $\rho \colon G_{\mathbb Q} \to GL_2(\mathbb C)$ is an irreducible Galois representation s.t. $\pi(\rho(G_{\mathbb Q}))$ is dihedral or isomorphic to $S_4$ or $A_4$ (where $\pi \colon GL_2(\mathbb C) \to PGL_2(\mathbb C)$ is the projection onto the quotient), then the Artin conjecture holds for $L(s,\rho)$, namely such function has an holomorphic continuation to the entire complex plane. So my question is: has the conjecture been proven (or disproven) in the case $\pi(\rho(G_{\mathbb Q})) \cong A_5$? -And moreover, I read that it is expected a more general result to be true, something called "strong Artin conjecture" which deals with cuspidal representations. Where can I find something introductory to this topic? - -REPLY [5 votes]: Rogawski's article "Functoriality and the Artin Conjecture", available from his website, is a good introduction. It was published in 1997, but I think that Khare-Wintenberger is the only advance since then. -A bit more is known -- the Strong Artin Conjecture holds for $n$-dimensional representations of nilpotent Galois groups by Arthur-Clozel ("Simple algebras, base change, and the advanced theory of the trace formula").<|endoftext|> -TITLE: Computing factorials modulo p^N -QUESTION [20 upvotes]: Can one compute $(p-1)!$ modulo $p^2$ in time polynomial in $\log p$? I can do it modulo $p$! (The last one is an exclamation point, not a factorial.) -More generally, I would like to be able to compute $d!$ modulo $p^N$ in time polynomial in $N \log p$. - -REPLY [12 votes]: I believe that the question whether $(p-1)! \bmod p^2$ is computable in polynomial time is still open. The best I am aware of is $O(p^{1/2+\epsilon})$, but I am not a specialist on this issue. -The computation of $(p-1)! \bmod p^2$ plays a role in certain p-adic algorithms to compute zeta functions of hypersurfaces in $\mathbb{P}^n$. -There is an algorithm of David Harvey for zeta functions of hypersurfaces, where your question plays a role. I.e., the fact that this is not known that $(p-1)! \bmod p^2$ can be computed in less than $O(p^{1/2+\epsilon})$ is an obstruction to improve the complexity of this algorithm. (See the slides "Computing zeta functions of projective surfaces in large characteristic" on http://www.cims.nyu.edu/~harvey/.)<|endoftext|> -TITLE: Reference in Riemann Surfaces -QUESTION [11 upvotes]: Can any one recommend me a good introductory book in Riemann Surface? - -REPLY [4 votes]: I enjoyed Simon Donaldson's book "Riemann Surfaces", Oxford Graduate Texts in Mathematics (2011). Extremely well-written and recommendable for beginners.<|endoftext|> -TITLE: Rational points à la Chabauty-Coleman -QUESTION [17 upvotes]: I have been trying to learn the method of Chabauty and Coleman to find rational points on curves; I have been reading an exposition by McCallum and Poonen which was pointed out to me by Emerton in this question. -Let $X$ be a curve of genus $g$ over $\mathbb{Q}$ with jacobian variety $J$, let $p$ be a prime of good reduction, and let $\overline{J(\mathbb{Q})}$ be the $p$-adic closure of the Mordell-Weil group $J(\mathbb{Q})$ in $J(\mathbb{Q}_p)$. Denote by $r'$ the dimension of the $p$-adic manifold $\overline{J(\mathbb{Q})}$. -The main assumption of the approach is that $r' < g$. This is automatic if $r < g$, where $r$ is the rank of $J$, because in general one has $r' \leq r$. This last inequality needn't be equality, "since $\mathbb{Z}$-independent points in log $J(\mathbb{Q})$ need not be $\mathbb{Z}_p$-independent". - -How do I compute $r'$? - -I wrote down a toy example, that is, $X : y^2 = x^5 + 17$. Here $r = 2$, and the method might work if $r'$ was 0 or 1, but I don't know how to check this. -I suspect that $r' = 2$, in which case the method is not even applicable, and I must think harder, but my question is not about this example, rather the general approach. - -Is there an example of a curve $X$ with $r = g = 2$ but with $r' = 0$ or 1? - -REPLY [7 votes]: If $r=2$ then $r'>0$. For an example where $r'=1$, take a curve such that the jacobian has a nontrivial endomorphism $f$ -and such that the group of points in the jacobian is generated by $P,f(P)$ for some point $P$ -Now find a prime $p$ splitting in $\mathbb{Q}(f)$ so that $f(P)=\alpha P$ for some $p$-adic number $\alpha$. -Then $r'=1$. -Added later: In your toy example, the endomorphism ring contains the fifth roots of unity so it may fall in my example above for those primes that split in that ring. A reasonable conjecture would be that if the endomorphism ring of the jacobian is $\mathbb{Z}$, then $r' = \min \{ g,r \}$. This is likely to be very hard to prove, as it is an abelian variety analogue of Leopoldt's conjecture. -Added much, much later: There are problems with my construction and what I say is probably not right. M.D. had a good reason to be skeptical in the comments. This was pointed out to me by Bjorn Poonen in a recent email exchange. His question is discussed further in a paper of M. Waldschmidt On the p-adic closure of a subgroup of rational points on an Abelian variety, Afr. Mat. 22 (2011), no. 1, 79–89.<|endoftext|> -TITLE: Semisimple Weil-Deligne representations -QUESTION [7 upvotes]: I've just realized that I don't understand something important and basic about the Weil-Deligne group and its representations. (I'm not very surprised by this). -Following Deligne's article, Section 8 of "Les Constantes des Equations Fonctionnelles Des Functions L" from the 1972 Antwerp volume, one is led to consider representations of the Weil-Deligne group in the following algebraic sense: for any local nonarchimedean field, there is a group scheme $W'$, defined over $Q$, which is a semidirect product of the Weil group scheme $W$ and the additive group scheme $G_a$. This is a non-affine group scheme, and the Weil subgroup scheme $W$ is obtained as a countable disjoint union of affine subschemes (the cosets of inertia). -So, as is now standard, we consider algebraic representations of the group scheme $W'$, over various fields $E$ of characteristic zero, as such representations provide a unified framework (thanks to results of Grothendieck, Deligne, Serre) for $\lambda$-adic representations that arise from arithmetic. -A crucial piece of this is to restrict attention (or semisimplify) to the semisimple representations of the Weil-Deligne group. And presumably, such semisimple representations form a Tannakian category (over any base field of characteristic zero). -And so to my question... what is the algebraic group associated to this Tannakian category? Or am I just confused? And how does the (non-affine) Weil-Deligne group scheme relate to this (affine) algebraic group obtained by restricting attention to these semisimple representations? Does this involve one of these awfully large group schemes like $Spec(E[E^\times])$, where $E$ is a characteristic zero field (something like the semisimple algebraic hull of the discrete group $Z$)? Any references? - -REPLY [4 votes]: Is this really a sensible question to ask, I wonder? -Here's a guess as to what the answer might look like. The Weil-Deligne group comes in three pieces. First there's inertia. Then there's a copy of $\mathbf{Z}$. And finally there's your $N$. Now the $N$ piece works fine: that will contribute something like an affine line (considered as additive group) to the situation. But the other two pieces are surely monstrous. The inertia action is via a finite group, by definition, and so you'll take something like the affine group schemes corresponding to these finite groups and then take some monstrous projective limit. And the $\mathbf{Z}$ part, because it's semisimple by definition, is basically a grading of your vector space by non-zero elements of the ground field, so it too will be monstrous: you seem to be already aware of the monsters that occur when you try and write these things down explicitly: a $\mathbf{G}_m$ gives a $\mathbf{Z}$-grading, some funny projective limit of such things gives a $\mathbf{Q}$-grading, now raise this to some uncountable cardinal $I$ to get a $\mathbf{Q}^I$-grading, because as an abelian group $\mathbf{C}^\times$ is quite close to $\mathbf{Q}^I$ for some uncountable $I$, but then there's some torsion so add some $\mu_n$'s for the $n$-torsion etc etc and you get the sort of group you mention in the question. -And finally glue them all together carefully.<|endoftext|> -TITLE: Lovasz theta function and independence number of product of simple odd-cycles -QUESTION [9 upvotes]: Lovasz theta function $\vartheta(G)$ of a graph $G$ provides an upper bound for the independence number of a graph, $\alpha(G)$ and $\Theta(G) = \lim_{k\rightarrow \infty}\sqrt[k]{\alpha(G^{k})}$. That is, $\Theta(G) \le \vartheta(G)$. -If the graph is a pentagon ($G=C_{5}$), then $\Theta(C_{5}) = \vartheta(C_{5})$. -$\Theta(C_{2k+1}) \ne \vartheta(C_{2k+1})$ if $k > 2$ since $\vartheta(C_{2k+1})^{r}$ fails to be integral for any $r \in \mathbb{Z}^{+}$. However, are there known lower and upper bounds for for $\vartheta(C_{2k+1}) - \Theta(C_{2k+1})$ that depends on $k$? - -REPLY [3 votes]: The theta function of odd cycles can be calculated explicitly: -$$\vartheta(C_{2n+1})=\frac{(2n+1)\cos(\frac{\pi}{2n+1})}{1+\cos(\frac{\pi}{2n+1})}=n+\frac{1}{2}-O(1/n)$$ -while computing $\Theta(C_{2n+1})$ for any odd $2n+1\geq 7$ is an open problem. So your question is about bounds on $\Theta(C_{2n+1})$. The best upper bound is the one we know for all graphs, $\Theta(G)\le \vartheta(G)$, there haven't been any improvements when $G$ is an odd cycle. -As for lower bounds, a first improvement on $\Theta(C_{2n+1})\geq \alpha(C_{2n+1})=n$ is given by -$$\Theta(C_{2n+1})\geq \sqrt{\alpha(C_{2n+1}^2)}=\sqrt{n^2+\lfloor\frac{n}{2}\rfloor}=n+\frac{1}{4}-O(1/n)$$ -and further by considering a lower bound on $\alpha(C_{2n+1}^3)$, Bohman, Ruszink and Thoma proved in "Shannon capacity of large odd cycles" that $$\Theta(C_{2n+1})\geq n+\frac{1}{3}-O(1/n)$$ -I believe the best known lower bounds are the ones one gets from extracting the exact counting given in "A limit theorem for the Shannon capacities of odd cycles, I" and II, by T. Bohman. Bohman didn't write the explicit lower bound but instead just gave an estimate which is enough to prove -$$\lim_{n\to \infty} \left(n+\frac{1}{2}-\Theta(C_{2n+1})\right)=0$$<|endoftext|> -TITLE: A Perturbation problem for U(n) -QUESTION [9 upvotes]: Let G be a finite subgroup of U(n), the unitary group acts on $\mathbb{C}^n$. If there is a unit vector $x$ in $\mathbb{C}^n$ such that g(x) is almost orthogonal to x, for all $g\in G$ except the identity, can we perturb x so that g(x) is exactly orthogonal to x, for all $g\in G$ except the identity? More precisely, can we find a very small number $\epsilon>0$, so that if there exist a unit vector $x$ and the inner product $|(g(x),x))|<\epsilon$ for all $g\in G$ \ {1}, then we can find another unit vector $y$, such that $(g(y),y)=0$ for all $g\in G$ \ {1}? Is it possible to further require that $||x-y||$ be small too? - -REPLY [10 votes]: If gx is a bounded distance away from x (which in particular occurs when gx is nearly orthogonal to x), then g is a bounded distance away from the identity. Since U(n) is compact, this and the pigeonhole principle forces the group G to have bounded cardinality; in particular, the set of all such groups is compact (if one chooses closed conditions for properties such as "bounded distance away from origin") in the Hausdorff distance topology, as the limit of a sequence of finite groups with bounded cardinality in the Hausdorff metric is again a finite group with bounded cardinality. For any single group, the claim is true for some epsilon by continuity (and the compactness of the unit sphere), so the claim is true in general by compactness of the space of groups. -With a bit more effort one can extract an explicit value of epsilon by making the compactness arguments quantitative, though the bounds are likely to be somewhat poor. -(More generally, for studying finite subgroups of compact linear groups, a useful fact to know here is Jordan's lemma, which says that one can always find a bounded index subgroup of such a group which is abelian (the bound can depend on the ambient dimension of the linear group). Here, of course, much more is true, because we are able to exclude group elements from getting too close to the origin, but Jordan's lemma is useful in situations in which we do not have this luxury.)<|endoftext|> -TITLE: Ramification formula for orbifolds -QUESTION [6 upvotes]: It's well known for smooth curves that if $\pi:X\to Y$ is a finite map, $K_X=\pi^*K_Y+Ram(\pi)$, this is just the Riemann-Hurwitz formula at the level of line bundles. I've been told that this formula is true for a finite morphism of smooth varieties over $\mathbb{C}$, at least. -First, is the above true? -Second, how must the formula be corrected if $X$ and $Y$ are both smooth orbifolds/DM-stacks? -In particular, what if I want to use that $X$ and $Y$ are of the forms $\tilde{X}/G$ and $\tilde{Y}/H$ for groups $G,H$, and express the formula in terms of objects on $\tilde{X},\tilde{Y}$? - -REPLY [5 votes]: In order for $K_X$ to exist you need some restriction on the singularities of $X$. See this MO answer for some thoughts on that. The most usual assumption to make is that $X$ is normal which implies that it is non-singular in codimension $1$. -In order to pull-back $K_X$ and still get a divisor, you need something more, the usual assumption is that it should be $\mathbb Q$-Cartier. See this MO answer for more on that. -If you have that $X$ and $Y$ are non-singular in codimension $1$, then you can forget about the second condition and just consider the formula on the non-singular part. (Strictly speaking you would want to restrict to the complement of the image of the singular part of $X$ on $Y$ so you'd still have a finite map). -So, let's say that both $X$ and $Y$ are smooth and $\pi$ is finite. Then localizing at height $1$ primes (or codimension $1$ points) reduces the statement to curves. Albeit not over the original field, but it's still OK. -I think this should cover two of your questions.<|endoftext|> -TITLE: Helix translates as geodesics -QUESTION [13 upvotes]: I believe one can fill $\mathbb{R}^3$ with -horizontal translates of the helix -$(\cos t, \sin t, t) \;,\; t \in \mathbb{R}$, -so that every point of $\mathbb{R}^3$ -lies in exactly one helix. -I am wondering if it is possible to assign a metric to $\mathbb{R}^3$ -so that these helices are all geodesics? -(I am imagining these as world lines of particles stationary in the plane.) - -REPLY [6 votes]: I'm pretty sure that Nil geometry works, but I don't know a reference. I seem to remember that one may think of Nil geometry as fibering over the plane, and that geodesics connecting different points in the same fiber had projections to circles. I think then these give helices. -A more general criterion was given by Dennis Sullivan for when a foliation may be realized as the geodesics of a Riemannian metric.<|endoftext|> -TITLE: Cobordism categories that don't involve manifolds -QUESTION [16 upvotes]: In order to capture the various flavors of cobordism into one concept, the notion of a "cobordism category" is introduced. This is an essentially small category $C$, together with finite coproducts, an initial object, and an additive functor $\partial$ satisfying some properties. Of course, the idea is that one should think of the objects of $C$ as manifolds of one sort or another and $\partial$ as taking the boundary. Indeed, in Tom Weston's notes on the subject he immediately restricts his attention to cobordism categories of "$(B,f)$" manifolds, which are a special type of manifold. -The question is: What are some examples of cobordism categories $(C, \partial, i)$, where the objects of $C$ are not manifolds of some kind? -The only example I know of is motivic cobordism. I'm hoping there are others! -(For the reader's convenience, here is the definition: -A cobordism category is an essentially small category $C$ with finite coproducts (including an initial object $0$), equipped with a coproduct-preserving functor $\partial: C \to C$ and a natural transformation $i: \partial \to 1_C$, such that $\partial\partial c \cong 0$ for every object $c$.) - -REPLY [7 votes]: One famous (in my field!) example is Witt space bordism. Witt spaces are not manifolds but rather pseudomanifolds (which aren't so far off from manifolds, but they can have singularities). A pseudomanifold is a Witt space if certain local rational intersection homology groups vanish. The bordism group of Witt spaces is important because it turns out to be a geometric model for ko-homology after inverting 2. The original reference is Paul Siegel's thesis: http://www.jstor.org/stable/2374334 -There's a slightly fancier version due to Pardon using "IP spaces" which satisfy integral Poincare duality: http://www.springerlink.com/content/6m5j386lr5hx2444/<|endoftext|> -TITLE: (Non?)-linearity of the consistency strength ordering in ZF -QUESTION [11 upvotes]: Much of the research taking place in set theory, is related to the classification of sentences according to their consistency strength relative to ZF. In order to be more specific, we say that for all sentences $\sigma,\tau\in Sent$, $\sigma$ has less consistency strength than $\tau$ (or $\sigma\leq_{cons}\tau$) iff $cons(ZF+\tau)\rightarrow cons(ZF+\sigma)$. My main question (which I believe should have a negative answer) is the following: -$$\mbox{ Is } \leq_{cons}\mbox{ linear? If not is there a natural counterexample?}$$ -In fact I have several similar questions concerning $\leq_{cons}$ and I would be glad if someone could provide me with some references. Thanks! - -REPLY [6 votes]: Unlike - seemingly - all the answers above, and since strangely he seems to have missed this question (as not so often), Pr. J.D. Hamkins claims he has natural statements that contradict the pre-linearity of this pre-ordering! -He will talk about them on 25 January at 7pm UTC: -http://jdh.hamkins.org/natural-instances-of-nonlinearity-in-the-hierarchy-of-consistency-strength-uwm-logic-seminar-january-2021/<|endoftext|> -TITLE: How to construct a maximally difficult NxN labyrinth ? -QUESTION [12 upvotes]: This question is just for fun; I don't know if it has a nontrivial answer. Let $\Gamma$ be an $n\times n$ square grid, thought of as a graph where the edges of the graph are the edges of the squares of the grid and the vertices of the graph are the vertices of the squares. Vertices of $\Gamma$ have coordinates $(m,k)$, $0\leq m,k \leq n$. Take a subgraph $\Lambda$ of $\Gamma$. So it looks like this: -alt text http://www.ich-liebt-du.de/wp-content/uploads/2008/09/labyrinth.jpg -Let $\Delta$ be the "frame", i.e. the subgraph of $\Gamma$ of points with at least one coordinate equal to $n$, and all the edges between them. Let $A=(x,0)$ , $0 < x < n$, be a point on the lower edge (the entrance) and likewise $\Omega =(y,n)$, $0 < y < n$, a point in the upper edge (the exit). -We say that $\Lambda$ is an $n\times n$ $(A,\Omega)$-labyrinth if it is connected and $A,\Omega\in\Lambda$. -Theseus enters the labyrinth at $A$ and has to reach $\Omega$. He only knows what I said above (except he does not know the value of $n$) and of course ignores the structure of $\Lambda$ and the position of $A$ and $\Omega$ in the lower (respectively upper) edge. At each point (vertex) $v$, just by looking, he has only the following information: - -Whether $v$ is equal to $A$, or to $\Omega$. -How long are the maximal forward, backward, left and right straight paths $S$ starting from $v$ contained in $\Lambda$. Some lengths may be $0$. -Whether or not the "end" of $S$ different from $v$ lies in $\Delta$. -Whether or not the "end" of $S$ different from $v$ is $\Omega$, and whether it's $A$. - -Since he enters at $A$, Theseus has a strategy. A strategy is a function that, at each vertex, tells him whether to go forward (resp. bacwards, left or right) by one step depending on the path that has already been walked and on the information gathered so far. -Let $D(\Lambda)$ be the smallest number of steps he must do in order to get to $\Omega$, where the minimum is taken over all the possible strategies. -The question is: - -Given $n$, $A$ and $\Omega$, how to construct a labyrinth $\Lambda$ that maximizes $D(\Lambda)$ ? - -I'm interested in aswers, if there can be, that do not involve listing all possible strategies and all possible graphs with a "computer". - -REPLY [6 votes]: UPDATE: the answer below works fine for tree-like labyrinths. The previous last paragraph (now struck out) is wrong, though, because the tree-like labyrinth you simulate depends on the random coins you flip. -Suppose the labyrinth is a tree. Further, suppose you do a random depth-first search on a tree, and stop when you reach a specific leaf called Ω. I claim that the expected number of times you traverse each edge is exactly once. -Proof: suppose you're at a node which is an ancestor Ω. There is one subtree of this node that contains Ω, and possibly several other subtrees that don't. For each subtree that doesn't, you explore it with probability ½, and if you do explore it, you traverse every edge twice (this is a property of depth-first-search). There is also one edge leading out of this node that leads to Ω. You will only go down this edge once. QED -Now, for any labyrinth that has a dead-end path, you need never go down the last segment of that path; thus the strategy of depth-first search, and turning back when you see a dead end, takes time strictly less than n2. Thus, there is only one type of tree-like labyrinth that requires maximum expected time to traverse: the type where there are no choices, but just a single path that takes you to from the start to the goal. -If the labyrinth is not tree-like, I believe you can pretend that it is tree-like by pretending there are walls whenever a step will take you back to a section you've already visited, so the expected n2 time upper bound will hold for this kind of labyrinth as well. I don't have a solution.<|endoftext|> -TITLE: "The complex version of Nash's theorem is not true" -QUESTION [29 upvotes]: The title quote is from p.221 of the 2010 book, -The Shape of Inner Space: String Theory and the Geometry of the Universe's Hidden Dimensions -by Shing-Tung Yau and Steve Nadis. "Nash's theorem" here refers to the Nash embedding theorem -(discussed in an earlier MO question: "Nash embedding theorem for 2D manifolds"). -I would appreciate any pointer to literature that explains where and why embedding fails. -This is well-known in the right circles, but I am having difficulty locating sources. Thanks! - -REPLY [10 votes]: In his 1952 Annals paper, Calabi characterised those Kähler manifolds $(X,\omega)$ which admit an isometric embedding into a complex space form. Here, a complex space form is a Kähler manifold of constant holomorphic sectional curvature which is complete with respect to the metric, hence either complex Euclidean space, complex hyperbolic space or complex projective space. -Here, we describe the case of a Kähler manifold $(X,\omega)$ admitting an isometric embedding into complex Euclidean space $\mathbb{C}^N$ for some $N$, as requested. -Firstly, the Kähler form $\omega$ has to be real-analytic. -Associated to a real-analytic Kähler manifold $(X,\omega)$, Calabi introduced a diastasis function $D_\omega$ (coming from the Greek word for distance) in order to describe his characterization. This function $D_\omega$ is a partially defined on on $X\times X$ and is given locally as follows: Let a point $u \in X$ be given and choose a local real-analytic Kähler potential $\Phi : U \to \mathbb{R}$ defined in an open neighbourhood $U$ of $u$ in $X$ (i.e. $\sqrt{-1} \partial\bar{\partial} \Phi = \omega$ on $U$). Let $\tilde{\Phi} : U \times \overline{U} \to \mathbb{C}$ denote a polarization of $\Phi$ (i.e. $\tilde{\Phi}$ is a holomorphic function on $U \times \overline{U}$ satisfying $\tilde{\Phi}(x, \overline{x}) = \Phi(x)$ for all $x\in U$, where $\overline{U}$ is the conjugate complex manifold of $U$). Calabi defined -$$D_\omega(x , y) := \tilde{\Phi}(x, \bar{x}) + \tilde{\Phi}(y, \bar{y}) - \tilde{\Phi}(x, \bar{y}) - \tilde{\Phi}(y, \bar{x}) \quad {\text{for }} (x,y)\in U\times U,$$ -and showed that the right-hand side is independent of the choice of Kähler potential $\Phi$. -The diastasis function detects isometries, in the sense that a holomorphic map $f : (X, \omega) \to (X^\prime, \omega^\prime) $ between real-analytic Kähler manifolds is an local isometry near a point $x \in X$ if and only if -$$D_{\omega^\prime}(f(x), f(y)) = D_\omega (x, y) \quad \text{for all } y\in U$$ -in some open neighbourhood $U$ of $x$ in $X$. -Since the diastasis of the flat metric $\omega_0 $on $\mathbb{C}^N$ from the origin is given by -$$D_{\omega_0}(z, 0) = |z_1|^2 + \cdots + |z_N|^2 \quad \text{for } z\in \mathbb{C}^N,$$ -this leads to the following result. - -Theorem [Calabi]. A real-analytic Kähler manifold $(X,\omega)$ admits an local isometric map near a point $x \in X$ to complex Euclidean space $\mathbb{C}^N$ if and only if there exists local holomorphic functions $f_1,\ldots, f_N$ defined near $x$ such that - $$D_\omega(x, y) = |f_1(y)|^2 + \cdots + |f_N(y)|^2.$$ - for points $y$ in some open neighbourhood of $x$. - -Regarding embedding (i.e. injective immersion), which is a global question, other than checking the above criterion at some point $x\in X$, there is an added condition, namely that the diastasis function $D_\omega$ extends to a total function on $X\times X$ such that $D_\omega (x, y) \ge 0$ for all $(x, y)\in X\times X$ and equality holds if and only if $x = y$. This added condition ensures injectivity. -Finally, one should note the perhaps surprising observation that if a (connected) real-analytic Kähler manifold $(X,\omega)$ admits an local isometric map near a point $x \in X$ to some complex Euclidean space $\mathbb{C}^N$, then it admits a local isometric map near every point $x \in X$ to $\mathbb{C}^N$, as Calabi showed. This explains why to check the existing of an embedding, it is sufficient to check the sum of squares condition in the highlighted block at some point $x\in X$ plus the global positive-valued extension condition on the diastasis function $D_\omega$.<|endoftext|> -TITLE: Gaussian processes, sample paths and associated Hilbert space. -QUESTION [10 upvotes]: Given a Gaussian process on some topological space $T$, with a continuous covariance kernel -$C(\cdot,\cdot)\colon T\times T\to R$, we can associate a Hilbert space, which is the reproducing kernel Hilbert space of real-valued functions on $T$, with $C$ as kernel function. This contruction is given in, for instance, R J Adler & J E Taylor: "Random Fields and Geometry", and surely a lot of other places. We can suppose the topological space $T$ is separable. -A very rapid review of the construction is: Define an inner product space $H_0$ as consisting of all real-valued functions on $T$ of the form $f(x) = \sum_{i=1}^n a_i C(x_i,x)$, -for real numbers $a_i$ and points in $T$, $x_i$. We can define an inner product on $H_0$ by -$\left\langle\sum a_i C(x_i,\cdot), \sum b_j C(y_j,\cdot)\right\rangle = \sum \sum a_i b_j C(x_i,y_j)$. -Then the reproducing kernel Hilbert space associated with our gaussian process is the completion $H$ of $H_0$. -Now, this strongly suggests (to be usefull, and by the Karhunen-Loéve theorem, which is based on this construction) that sample paths of our Gaussian process belongs to H with probability 1. This must be proved somewhere, but where? Anybody knows a reference? - -REPLY [2 votes]: Short answer: - -If the RKHS $\mathcal{H}_C$ associated to the kernel of your GP is infinite dimensional, then sample paths are in $\mathcal{H}_C$ with probability 0. - -Nevertheless, one can show that sample paths are contained in a "larger" RKHS, and Steinwart showed how to construct such an RKHS as a power of the original RKHS. - -Sections 4.2 and 4.3 of Gaussian Processes and Kernel Methods: A Review on Connections and Equivalences by Kanagawa et al. give a nive and brief account of sample path properties of gaussian processes and of how to construct an RKHS containing the paths of a GP.<|endoftext|> -TITLE: are there infinitely many triples of consecutive square-free integers? -QUESTION [17 upvotes]: The title says it all ... Obviously, any such triple must be of the form -$(4a+1,4a+2,4a+3)$ where $a$ is an integer. Has this problem -already been studied before ? The result would follow from Dickson's -conjecture on prime patterns, which implies that there are -infinitely many integers $b$ such that $4(9b)+1,2(9b)+1$ and $4(3b)+1$ are all prime -(take $a=9b$). -A related question : Question on consecutive integers with similar prime factorizations - -REPLY [7 votes]: The question of the number of positive integers $n \leq x$ for which all members of an associated fixed pattern are squarefree (or r-free) was studied by Leon Mirsky: -L. Mirsky, Note on an asymptotic formula connected with r-free integers. -Quart. J. Math., Oxford Ser. 18 (1947), 178-182. -L. Mirsky, Arithmetical pattern problems relating to divisibility by rth powers. -Proc. London Math. Soc. (2) 50 (1949), 497–508. -As I remember it, Mirsky proved that the number is $cx + O(x^{2/3})$ for patterns of squarefrees, where $c$ is a constant depending on the pattern, and is positive if the pattern is not excluded by certain necessary congruential conditions.<|endoftext|> -TITLE: Gromov's Hyperbolicity and Positive Cheeger Constant in Planar Graphs -QUESTION [13 upvotes]: It is known that for metric graphs the concepts of Gromov's hyperbolicity and strictly positive Cheeger constant are related. Let us first recall the definition of the Cheeger constant. Let $G$ be a graph with vertex set $V(G)$ and edge set $E(G)$. For a collection of vertices $A \subseteq V(G)$, let $\partial A$ denote the collection of all edges going from a vertex in $A$ to a vertex outside of $A$, -$$ - \partial A := \{ (x, y) \in E : x \in A, y \in V(G) \setminus A \}. -$$ -Then the Cheeger constant of G, denoted $h(G)$, is defined as -$$ -h(G) := \inf \left\{ \frac{|\partial A|}{|A|} : A \subseteq V(G)\quad\text{and}\quad |A|<\infty \right\}. -$$ -However, neither Gromov's hyperbolicity nor strictly positive Cheeger constant implies the other. For instance, the graph $\mathbb{Z}$ is hyperbolic in the sense of Gromov but its Cheeger constant is zero. On the other hand, the Cayley graph of the group $\mathbb{F}_{2}\times\mathbb{F}_2$ has positive Cheeger constant but it is not hyperbolic since $\mathbb{Z}\times \mathbb{Z}$ sits as a subgroup. -Does there exist a planar graph with positive Cheeger constant but not hyperbolic? - -REPLY [10 votes]: If I add the assumption that the given graph has bounded degree, and the same holds for the dual graph, then the answer to your question is no. A positive Cheeger constant implies a linear isoperimetric inequality which in turn implies Gromov hyperbolicity. -However, if you allow the dual graph to have unbounded degree (say), there are examples. Let $T$ be the regular degree $3$ tree. Note that $h(T) > 0$. Properly embed $T$ in the plane. Let $C_i$ be an ordering of the complementary regions. Now add a single edge $e_i$, in the interior of $C_i$, so that $C_i$ is divided into two regions, one compact and the other not compact. We add the edge so that the compact region is bounded by a cycle of length $i$. -Let $T' = T \cup (\cup e_i)$. The cycles created above prevent $T'$ from being Gromov hyperbolic. However $h(T') \geq h(T)$ as adding edges does not decrease the Cheeger constant.<|endoftext|> -TITLE: What structure is needed to define a Gaussian distribution on a given space? -QUESTION [9 upvotes]: In most textbooks, the normal distribution is defined on $\mathbb{R}^n$ by specifying its probability density function. This works perfectly well, but it isn't really amenable to generalisation. -I'm wondering what the minimal structure is that one must have on a given space $S$ before one can define an analogue of the normal distribution. If $S$ has a Riemannian metric defined on it, then one can define a Brownian motion on $S$ using the Laplace-Beltrami operator. The family of normal distributions on $S$ could then be defined as the one-dimensional marginals of the Brownian motion. -Alternatively, the normal distribution could be characterised as the distribution that maximises entropy when its mean and variance are known. This characterisation only seems to require a notion of "mean" (which suggests that $S$ must be a metric space). -Is there a more general construction of the normal distribution? - -REPLY [5 votes]: Let $X$ be a Banach space with its Borel $\sigma$-algebra $\mathcal B(X)$, and let $\mathbb P$ be a Radon measure on $X$. The continuous linear functionals on $X$ are measurable functions, hence random variables. Let's assume that for all $f \in X^*$, $$\mathbb E|f|^2 = \int_X |f(x)|^2 \ \mathrm{d}\mathbb P(x) < \infty.$$ This assumption implies a powerful structure theorem (cf. Theorem III.2.1 of [1]): -Theorem. There exists an element $\mu \in X$ (the mean element) and a continuous, symmetric operator $K : X^* \to X$ (the covariance operator) such that for all $f, g \in X^*$, $$f(\mu) = \mathbb E f \qquad \mathrm{and} \qquad f(Kg) = \mathbb E(fg) - f(\mu)g(\mu).$$ -Furthermore, the support of the measure $\mathbb P$ is contained in the affine subspace spanned by the mean and covariance operator: $$\operatorname{supp} \mathbb P \subseteq \mu + \overline{KX^*},$$ -and this space is separable. -As Mark Meckes said, we say that the measure $\mathbb P$ is Gaussian if each $f \in X^*$ is a real-valued Gaussian random variable (hence with mean $f(\mu)$ and variance $f(Kf)$. In this case, the support is the full affine subspace $\mu + \overline{KX^*}$. Furthermore, the measure is completely characterized by its characteristic functional $\varphi : X^* \to \mathbb R$: -$$\varphi(f) = \int_X \mathrm e^{\mathrm i f(x)} \ \mathrm d\mathbb P(x) = \mathrm e^{\mathrm i f(\mu) - f(Kf)/2}.$$ -For more on this topic, I suggest the reference: -[1] Probability Distributions on Banach Spaces by Vakhania, Tarieladze and Chobanyan.<|endoftext|> -TITLE: Model category structure on categories enriched over quasi-coherent sheaves -QUESTION [15 upvotes]: Gonçalo Tabuada has shown that there is a Quillen model category structure on the category of small dg-categories, i.e. the category of small categories enriched over chain complexes (for a fixed commutative ring k). Julia Bergner has shown that the category of simplicial categories, i.e. categories enriched over simplicial sets, also has a model category structure. -The above two results are connected by the fact that chain complexes and simplicial sets are both monoidal model categories (in the sense of Hovey Chapter 4). Toen in the introduction to his paper "Homotopy theory of dg-categories and derived Morita theory" mentions that work has been done by Tapia to put a model structure on categories enhanced over very general monoidal model categories. Unfortunately there seem to be no mentions of this work since. -I'm very interested in the following case of the above statement. Let $X$ be a nice scheme, say noetherian, separated, and finite dimensional. Even assuming regular would be ok. The category $\mathcal{C}$ of chain complexes of quasi-coherent $X$-modules is a monoidal model category by Gillespie; see Theorem 6.7 of the paper http://arxiv.org/abs/math/0607769. -Question: Does anyone know of a reference, or general machinery, that shows that there is a model structure on small categories enriched over $\mathcal{C}$ in which the weak equivalences are the functors that induce an equivalence after taking homology? I suppose one could go through and try to mimic Tabuada's proof, but that would not be ideal. - -REPLY [4 votes]: There were two preprints posted on the arXiv this week that seem to answer the question. They are Dwyer-Kan homotopy theory of enriched categories by Fernando Muro and On the homotopy theory of enriched categories by Clemens Berger and Ieke Moerdijk.<|endoftext|> -TITLE: Are the nontrivial zeros of the Riemann zeta simple? -QUESTION [19 upvotes]: A few years ago, I found on arXiv an article in which the authors (I think they were at least two to write it) claimed to have proven that the non trivial zeros of the Riemann zeta function were all simple using the concept of Riemann surfaces. But unfortunately, I just can't find it back. Does someone know if such a result has been published and widely accepted by the mathematical community? -Thank you in advance. - -REPLY [4 votes]: I finally managed to find back the article I was talking about. Just click on the green link in the first message of the following link: -link text<|endoftext|> -TITLE: What do whitehead towers have to do with physics? -QUESTION [26 upvotes]: First let me say something that I don't completely understand, since I do not know enough physics. If I say anything wrong, someone please tell me: -For the spinning particle, there is a sigma-model, which is a type of quantum field theory, which describes how a spinning particle can propagate. The input data for this sigma model is an oriented pseudo-Riemannian manifold $X$ equipped with a line bundle with connection. The condition for "quantum anomaly cancellation", can be shown to be that the classifying map $X \to BSO(n)$ corresponding to the tangent bundle has a lift through the map $BSpin(n) \to BSO(n)$. Such a lift is called spin-structure. -The sigma-model for the spinning string, starts similarly, but the role of the line bundle is replaced by that of a bundle-gerbe (that is a gerbe with band U(1)). - I believe what is going on here is that a line bundle is the same data as a principal $U(1)$-bundle, and a bundle-gerbe is the same data as a principal bundle for the $2$-group $[U(1)\to 1]$. Anyhow, for this new quantum-field theory, the condition for "quantum anomaly cancellation" is that the classifying map $X \to BSO(n)$ has a lift through $BString(n) \to BSpin(n) \to BSO(n)$. In fact, $String(n)$ does not exist as a Lie group, but it does exist as a (weak) group object in differentiable stacks, which are in particular sheaves (over the category of manifolds) in homotopy 1-types. -Apparently this can be taken even further, and one can talk about a sigma-model for the so-called $5$-brane, and the condition "quantum anomaly cancellation" is that the classifying map $X \to BSO(n)$ has a lift through $BFiveBrane(n) \to BString(n) \to BSpin(n) \to BSO(n)$, and $Fivebrane(n)$ at least exists as a group object in sheaves (over the category of manifolds) in homotopy $5$-types. (Is this a sigma-model with bundle-gerbe replaced by a principal bundle for $U(1)$ promoted to a $3$-group?) -Note: We should really be starting with $O(n)$ since a lift of $X \to BO(n)$ through $BSO(n) \to BO(n)$ is the same as equipping $X$ with an orientation. -Anyway, my question is, what exactly is "quantum anomaly cancellation" (perhaps in layman's terms) and what does it have to do with the whitehead tower of $O(n)$? -Also, is there more after $5$-branes? - -REPLY [9 votes]: Hi Dave, -I only just saw this question now. Maybe I can still react anyway. -The anomalies that we are talking about here mean the following: the action functional of a given QFT may turn out to be not quite a function on the configuration space, but instead a section of some line bundle with connection over configuration space. Hence for the theory to make sense, that line bundle with connection must be trivialized, and hence first of all must be trivializable. The Chern class of that line bundle is the the global anomaly, the obstruction to there existing a trivialization of the underlying bundle. It's curvature is the local anomaly, a measure for the obstruction for it to trivialize also as a bundle with connection. -That such anomalous contributions to the action functional appear for heterotic-type super-branes ("spinning branes") comes from the fact that for these the fermionic path integral is not a function on the bosonic configurations, but is a section of the Pfaffian line bundle of the given Dirac operator. -So anomaly cancellation is the process where we identify those constraints on the field content which make these anomaly line bundles become trivialized. This, and the standard references on it, is reviewed here: -http://ncatlab.org/nlab/show/quantum+anomaly -That the anomaly of the spinning particle vaishes if the target space has Spin structure is classical. That the anomaly of the heterotic string vanishes if the target has String structure is due to Killingback and Witten, originally, by an argument that was recently made rigorous by Bunke. See the references at -http://ncatlab.org/nlab/show/differential+string+structure -That the cancellation of the anomaly of the super 5-brane is similarly related to higher topological structures, and the introduction of the term "Fivebrane group" and "Fivebrane structure" is originally due to -Hisham Sati, Urs Schreiber, Jim Stasheff, Fivebrane structures Rev.Math.Phys.21:1197-1240 (2009) (arXiv:0805.0564) -That article also includes a review of the whole story of anomaly cancellation in its section 3. -In the successor -Hisham Sati, Urs Schreiber, Jim Stasheff, Twisted differential string and fivebrane structures Communications in Mathematical Physics (2012) (arXiv:0910.4001) -we develop the full differential geometry corresponding to this. Various related articles with further developments are listed here -http://ncatlab.org/nlab/show/Geometric+and+topological+structures+related+to+M-branes -http://ncatlab.org/schreiber/show/differential+cohomology+in+a+cohesive+topos#Subprojects<|endoftext|> -TITLE: reference containing the list of irreducible finite dimensional representation of real general linear group -QUESTION [8 upvotes]: It seems that it is not easy to find a reference containing a classification and construction of finite dimensional irreducible representations of $GL_n(\mathbb{R})$. One way to look at it is via $(\mathfrak{g},K)$ module, and we do have a classification theorem in terms of highest weight, but it is not obvious to get the explicit constructions from this,especially in higher rank case. -So I'm wondering is there some good reference containing explicit constructions of all irreducible finite dimensional representations of $GL_n(\mathbb{R})$? - -REPLY [5 votes]: To expand a little on Allen's useful answer: -1) The question doesn't specify over which field (presumably $\mathbb{R}$ or $\mathbb{C}$) the f.d. irreps of $G=GL_n(\mathbb{R})$ should be studied, but in this particular case it doesn't matter much. From the viewpoint of Steinberg's 1967-68 Yale lectures, the special linear group is a Chevalley group; so its structure and representations in characteristic 0 can be studied first over $\mathbb{Q}$ followed by extension of scalars. -2) The f.d. representation theory of semisimple, or more generally reductive, Lie groups has been well understood for a century, going back to work of Elie Cartan and Hermann Weyl. But most of the difficulty involves simple groups (semisimple with finite center), while passage to general linear groups over $\mathbb{C}$ just involves easy modifications by powers of the determinant. So the textbook literature has mostly concentrated on simple Lie groups, though books for physicists usually focus more on $GL_n(\mathbb{C})$ and the combinatorics of partitions along with Schur-Weyl duality. Adapting this to the real group $G$ is fairly straightforward, as Allen indicates. -3) Anyway, it's not easy to specify a standard source for the specific question raised here. That's partly because modern research efforts have focused mostly on the less-understood world of infinite dimensional representations. -4) The question asks for a "list" and a "construction". Allen basically gives a list, taking for granted the classical parametrization of f.d. irreps of the simple groups over $\mathbb{C}$ (or restrictions to $\mathbb{R}$) by highest weights in general or partitions in this special case. But constructions are not so easy even for general or special linear groups, where the best concrete approach seems to be via Schur-Weyl duality.<|endoftext|> -TITLE: The generalization of Brouwer's fixed point theorem? -QUESTION [8 upvotes]: Let $X$ be a contractible compact [edit: locally connected] topological space -(Hausdorff and second countable). Let $f\colon X\to X$ -be a continuous map. Then (I suppose) $f$ has a fixed -point. Personally, I cannot think of a better generalization -of Brouwer's fixed point theorem, but is it true? - -REPLY [24 votes]: No. I believe the first counterexample is from: -Kinoshita, S. On Some Contractible Continua without Fixed Point Property. Fund. Math. 40 (1953), 96-98 -which I unfortunately can't find online. Kinoshita's example is described on page 127 in this excellent article by Bing, however. -EDIT: The question has been revised to add the local connectivity condition; as stated, I think the question is open. If "contractible" is replaced with "acyclic" there are counterexamples dating back to Borsuk, referenced e.g. here; Borsuk's paper, which I can't find online, is: -K. Borsuk, Sur un continua acyclique qui se laisse transformer topologiquement en lui-meme sans points invariants, Fund. Math. 24 (1935), pp. 51-58. -This suggests that if the OP's conjecture is true, it is unlikely to be open to homological attacks.<|endoftext|> -TITLE: Are GIT's good categorical quotients just locally ringed space coequalizers? -QUESTION [12 upvotes]: Introduction: The definition of "good categorical quotient" in geometric invariant theory (given below) seems fairly ad hoc to me, except that it looks very similar to the coequalizer of the action in the category of locally ringed spaces. I'm trying to understand this similarity. -Some background: Suppose we have a group action $G\times X \to X$ in the category $\bf Sch$ of schemes. The category $\bf LRS$ of locally ringed spaces is cocomplete (see propositiion 1.12 in A. Vezzani's thesis), so in particular, $G\times X \rightrightarrows X$ has a $\bf LRS$-coequalizer; call it $Q$. One can construct it as the coequalizer in the category of topological spaces equipped with the sheaf of $G$-invariant functions to make it a locally ringed space, as Vezzani's proposition explains more carefully. -If $G\times X \rightrightarrows X$ has a $\bf Sch$-coequalizer $Y$, then GIT calls $Y$ a "categorical quotient"; see, for example, R. Birkner's Introduction to GIT. Note that here we get a universal map $u: Q\to Y$ in $\bf LRS$. -$Y$ is called a "good categorical quotient" if it satisfies 3 properties, also found in Birkner's notes, which I prefer to re-express here in terms of the universal map $u$: -i) $u^\sharp : {\cal O}_Y \to u _* {\cal O}_Q$ is an isomorphism. -ii) $u: Q\to Y$ is a closed map (at the level of topological spaces). -iii) $u$ takes disjoint closed sets to disjoint closed sets. -(These together imply that $X\to Y$ is a categorical quotient, i.e. $\bf Sch$-coequalizer; see Mumford's GIT I.2 remark 6.) -All of these properties are clearly satisfied if $u$ is an isomorphism. That is, if the $\bf LRS$-coequalizer is a scheme, that scheme is a good categorical quotient. -Towards a converse, (i) implies that $u$ must be dominant, and hence by (ii) it is surjective. And (iii) implies it is injective on closed points, so it's looking a lot like an isomorphism. Can anything more be said? - -1) When is a good categorical quotient an $\bf LRS$-coequalizer of the group action? I.e., are there nice and general conditions where $u$ must be an isomorphism? -2) If the answer is not always, what's a counterexample? (Answered by Anton) - -Thanks! - -REPLY [4 votes]: The answer to the title question is NO. If $X\rightrightarrows Y\to Z$ is a coequalizer in $\bf LRS$, then the underlying topological space of $Z$ is the coequalizer of the underlying topological spaces (and the structure sheaf is the equalizer of pushforwards of the structure sheaves). On the other hand, good categorical quotients identify any two points whose orbit closures intersect. So one of my favorite good quotients, $\mathbb A^1/\mathbb G_m$, provides a counterexample. The good categorical quotient in schemes has a single point as its underlying topological space, but the quotient in $\bf LRS$ has two points. -I think the bad behavior on underlying topological spaces is probably the only problem. That is, a good categorical quotient is a coequalizer in $\bf LRS$ precisely when it is a geometric quotient (i.e. when we impose the additional condition that each fiber is a single orbit).<|endoftext|> -TITLE: Rationality of intersection of quadrics -QUESTION [20 upvotes]: Let $X \subset \mathbb{P}^n$ be a complete intersection of two quadrics. It is classical that, if $X$ contains a line, then it is rational. The proof is very simple and basically it is given by taking the projection off the line. -It is also stated in a paper by Colliot-Thélène, Sansuc and Swinnerton-Dyer that if $X$ contains a curve of odd degree, then it is rational as well. Is there a geometric proof of this? -Are there known examples of complete intersections of two quadrics that contain a curve of odd degree and not a line? -And finally: are there other sufficient conditions for the rationality of such a variety? - -REPLY [7 votes]: I've thought about this and discussed about this with someone else. A big part from what will follow is not my own contribution (in particular I hadn't thought about invoking Amer's theorem). -In fact it suffices to prove that any complete intersection of two quadrics containing a curve of odd degree must contain a line (hence the answer to your second question is affirmative). -By a theorem of Amer, an intersection of quadrics $f = g = 0$ contains a linear space of dimension $r$ over $k$ iff the quadric given by $f + tg = 0$ contains a linear space of dimension $r$ over $k(t)$. Hence it suffices to prove that any quadric which contains a curve of odd degree actually contains a line. -To do this, note that such a quadric must have a rational point: if you cut by a generic hyperplane, you will get a finite set of points, at least one of which has odd degree. But Springer's theorem says that if a quadric contains a point in an odd degree extension of the base field, then it must have a rational point. So the quadric is isotropic and we can split off a hyperbolic plane. With the "rest" of the quadric you can repeat the argument, although you probably have to be careful with configuration issues - I'd have to write down this part more carefully. Hence you will get two hyperbolic planes, hence a line.<|endoftext|> -TITLE: A question on classification of almost complex structures on $4$-manifolds -QUESTION [16 upvotes]: I have a (basic?) question in topology. -Question 1. Is it possible to characterise compact $4$-manifolds $M^4$, such that almost complex structures on $M^4$ are uniquely defined up to homotopy by their first Chern classes? Or is there at least a large class of $4$-manifolds where this is true? Maybe there is a reference? -Recall the theorem of Wu. Denote by $\tau$ the signature of $M^4$ and by $e$ the Euler characteristics. Then for any element $c\in H^2(M^4,\mathbb Z)$ with $c^2=3\tau+2e$ -and such that $c\; {\rm mod} \;\mathbb Z_2= w_2\in H^2(M^4,\mathbb Z_2)$ -there is an almost complex structure $J$ on $M^4$, such that $c_1(M,J)=c$ (in particular, as Paul says $e+\tau=0\; {\rm mod} \;4$). However this $J$ might be non-unique, up to homotopy among almost complex structures with $c_1=c$, as the following example shows. -Example. Let $M^4=S^1\times S^3$, then the tangent bundle is trivial, so almost complex structures on $M^4$ can be identified up to homotopy with homotopy classes of maps $M^4\to S^2$ (see page 11 in The geometry of Four-manifolds Donaldson Kronheimer). Maps $S^3\to S^2$ can have different Hopf invariants, but $c_1=0$ since $S^1\times S^3$ has no second cohomology... -Added Question 2. Is there at least one manifold that satisfies condition of Question 1? -In McDuff-Salamon (footnote on page 120) it is written, that if $M^4$ is spin then -there are precisely two homotopy classes of $J$ with given $c_1$ (this is said to be related to $\pi_4(S^2)=\mathbb Z_2$). But since $S^1\times S^3$ is spin this statement from McDuff-Salamon seem to contradict to my conclusion (that for $S^1\times S^3$ homotopy classes are can have different Hopf invariants). So, where is the mistake?...:) - -REPLY [22 votes]: This can be answered by obstruction theory for the fibration -$$ F=SO(4)/U(2) \to BU(2) \to BSO(4) $$ -where the fibre is actually a 2-sphere: $F=S^2$. Start with the tangent bundle of an oriented 4-manifold $M$ and ask for existence respectively uniqueness of a lift of its Gauss-map $M\to BSO(4)$, that is, existence respectively uniqueness for an almost complex structure on $M$. -The obstructions for existence lie in $H^{i+1}(M;\pi_i(F))$ and the obstructions for uniqueness lie in $H^{i}(M;\pi_i(F))$. Assume for simplicity that $M$ is simply connected and closed and let $X$ be its 2- (and hence) 3-skeleton. The same obstruction theory tells you that the tangent bundle of $X$ has a complex structure and that it's uniquely determined by the first Chern-class. If such a structure extends to $M$, there is a single uniqueness obstruction in -$$ H^{4}(M,X;\pi_4(F)) = \pi_4(S^2) = Z/2$$ -By naturality, this obstruction is realized by a second almost complex structure on $M$ (with the same first Chern-class) if and only if the quotient map $M \to M/X=S^4$ induces a nontrivial map on cohomotopy: -$$ \pi_4(S^2) = [S^4,S^2] \to [M,S^2] $$ -It is well known that this map is nontrivial if and only if $M$ is spin, see for example the cohomotopy preprint with Kirby and Melvin on my homepage. -This explains the footnote in McDuff-Salamon (where they probably assume that $M$ is simply connected, otherwise there is another uniqueness obstruction in $H^3(M)$ that you found for $M = S^1 \times S^3$). It also answers your question for simply connected almost complex manifolds $M$ as follows: -The first Chern-class characterizes almost complex structures on $M$ if and only if $M$ is not spin. Good examples are those Hirzebruch surfaces which are nontrivial $S^2$-bundles over $S^2$.<|endoftext|> -TITLE: When is the different in a number field a principal ideal? -QUESTION [10 upvotes]: Q1: Do you known examples, where the different is not a principal ideal? -Q2: Is there a good interpretation for the reason, why this happens? -See e.g. Neukirch, Proposition 2.4, page 197. -The reason why I ask: the definition of the canonical additive character $\psi:x \mapsto \mathrm{e}^{2 \pi i (\mathrm{Tr}_{F / \mathbb{Q}} x \mod \mathbb{Z})}$ is somewhat unsatisfactory, if I want to identify the Pontryagin dual of the additive group of $\mathfrak{o}$ with the additive group of $F/\mathfrak{o}$, which simplifies some of notation involved when computing some $p$ adic integrals or Gauss sums. -Btw. with the canonical additive character, the Pontryagin duality is of the following form: -$$F / \mathfrak{D}^{-1} \cong \mathrm{Hom}_{ab.group} ( \mathfrak{o} , \mathbb{C}^\times).$$ -where $\mathfrak{D}$ is the different and the isomorphism is given by -$$ \xi \in F \mapsto \psi( \xi \cdotp).$$ - -REPLY [12 votes]: A partial answer: -Regarding Q1: An example for this is the number field generated by third root of $175$ . -See e.g. a comment by KConrad on this question - Which number fields are monogenic? and related questions -or also Ex 4.15 in these notes -http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/different.pdf -which contain several explicit examples related to differents. -Regarding Q2: I can give at least a good reason when it does not happen (that is when it is principal). Namely, when the ring of integers is generated by a single element. If it is generated by a single element $u$, then the different is generated by $f'(u)$ where $f$ is the minimal polynomial of $u$. -A proof of this can be found in the above mentioned notes (Thm 4.3) as well as a discussion of the correct generalization of this result in the general case (Rem 4.5). -Thus, for example, for quadratic fields it is always principal. And, more generally, one does not have to look for examples for non-principal in too 'nice' fields (i.e., those where the ring of integers is generated by one element, such as cyclotomic fields).<|endoftext|> -TITLE: Use of traces in physics -QUESTION [9 upvotes]: This is borderline physics, but I'll post here first as it relates to a mathematical concept. -I was wondering if anyone could clarify what the physical significance is of traces, whether of matrices or more general operators, and are there more general notions of the same kind? -For example, there is what seems (as far as I can follow it) quite a promising "precursor" quantum theory called Trace Dynamics, developed by Stephen Adler and summarized in http://www.worldscinet.com/ijmpd/19/1914/free-access/S0218271810018335.pdf -Also, I have seen traces used in many other physics contexts. -I guess the first thing, more on topic here, would be to explain what mathematical significance this apparently slightly artificial (and coordinate dependent?) notion has. -Feel free to waffle in generalities - The reply needn't be impeccably formal, as I am more interested in gaining intuition than results that could easily be looked up ;-) - -REPLY [4 votes]: There was a similar question recently posed (and often answered) here: -Geometric Interpretation of Trace<|endoftext|> -TITLE: Given an odd integer N find the smalletst prime p > N such that (p-1,N)=1 -QUESTION [7 upvotes]: So the title says it all, - -> - Q: Given a large odd integer $N>>0$, what can we prove about the smallest prime - $p>N$ such that $gcd(p-1,N)=1$? - -Note that such a prime exists: Given an integer $a$ coprime to $N$ we know that there are infinitely many primes $p$ -with $p\equiv a\pmod{N}$. In particular, since $N$ is odd we may take $a=2$ and thus we know that there are infinitely many primes $p\equiv 2\pmod{N}$ and thus infinitely many primes -$p$ such that $gcd(p-1,N)=1$ so the question makes sense. -I would be extremely happy if we could always prove the existence of a prime -$N < p < \frac{3N}{2}$ (for $N$ a large odd integer) such that $(p-1,N)=1$. -The first question seems to be difficult so here is a more tracktable question: - - -Q: So let $N$ be a large odd integer. Is it always possible to find two prime numbers $p,q$ in the interval $(N,\frac{3N}{2})$ - such that $(p-1,q-1,N)=1$? - -REPLY [9 votes]: Your first question seems to follow fairly easily from the Bombieri-Vinogradov theorem; actually it seems only to need Renyi's original result. -By Moebius inversion, it should suffice to find an asymptotic formula of the form $\sum_{d|N} \mu(d) \psi(x;d,1) = cx + O(x/\log{x})$ for some $c > 0$ and -$N < x \leq 3N/2$. -If $d>z$ with $z$ some fixed small power of $x$ (say $z=x^{1/100}$) then the trivial bound $\psi(x;d,1) \ll x/d$ gives a satisfactory error term. Otherwise, by Bombieri-Vinogadov we have $$\sum_{d|N, d < z} |\psi(x;d,1) - x/\phi(d)| = O(x (\log{x})^{-100}).$$ Finally, compute $\sum_{d | N , d < z} \mu(d)/\phi(d)$.<|endoftext|> -TITLE: Randomly contracting edges of a graph - expected number of vertices? -QUESTION [7 upvotes]: Let $G'$ be a graph obtained from $G$ after contracting each edge with probability $p$. Let $n = |V(G)|, e = |E(G)|$. -I would like to compute (or at least obtain a lower bound) for $E[|V(G')|]$ in terms of some known graph invariants (number of edges, degree sequence, connectivity,..) -I am sure I am not the first one that studied such a probabilistic space and since I couldn't find any estimates for $E[|V(G')|]$ in my textbook I am asking: is there any simple identity/estimate for $E[|V(G')|]$ ? Is there any reference to a paper studying this quantity? -Edit: I have removed the completely wrong attempt to estimate $E[|V(G')|]$. - -REPLY [5 votes]: The expected number of vertices of $G'$ is given by a sum (over all sets of edges) that bears a certain resemblance to the Tutte polynomial $T(x,y)$ of (the graphical matroid associated to) $G$, as defined, for example, at http://en.wikipedia.org/wiki/Matroid#Tutte_polynomial. A quick calculation (maybe too quick --- I don't guarantee the following and I apologize for any errors here, but something like this should be true) indicates that, if we let $v,e,c$ be the number of vertices, edges, and components of $G$, then the expected number of vertices of $G'$ is -$$c+p^{v-c}(1-p)^{e+c-v}\frac{\partial T(x,y)}{\partial x}\left(x=\frac1p,y=\frac1{1-p}\right).$$ -EDIT: -In answer to the question about how I got this, and in the interest of correcting missing factors of $p$ and $1-p$ in the answer above, here are some details. -I'll use the notation $v,e,c$ as above. In addition, for any subset $S$ of the edges, I'll write $z(S)$ for the number of points after the edges in $S$ are contracted; it's also the number of connected components of the graph with the same vertices as $G$ but only the edges in $S$. Note that the rank of $S$ in the graphical matroid is $v-z(S)$. Obviously, $z(S)$ is always at least $c$; it turns out to be convenient to work with the difference, which I'll abbreviate $q(S)=z(S)-C$. Taking the definition of the Tutte polynomial $T(x,y)$ from the Wikipedia page linked above, and using the connection between ranks and $z$, we get -$$ -T(x,y)=\sum_{S\subseteq E}(x-1)^{z(S)-c}(y-1)^{|S|-v+z(S)}= -\sum_{S\subseteq E}(x-1)^{q(S)}(y-1)^{|S|-v+c+q(S)}. -$$ -Compare this with the expectation of $q$, namely -$$ -\sum_{S\subseteq E}q(S)p^{|S|}(1-p)^{|E|-|S|}= -(1-p)^{|E|}\sum_{S\subseteq E}q(S)\left(\frac p{1-p}\right)^{|S|}. -$$ -We can convert $T(x,y)$ into this expectation in a few steps, as follows. First, if we differentiate $T(x,y)$ with respect to $x$ and multiply the result by $x-1$, the effect is to bring down a factor $q(S)$ from the exponent without changing anything else. That factor $q(S)$ matches part of what we want in the expectation of $q$. Next, we can make the $x-1$ and $y-1$ stuff in $T$ match the $p$ and $1-p$ stuff that we want. First set $y-1=p/(1-p)$; equivalently, $y=1/(1-p)$. Then the terms in $T$ will have the factor $(p/(1-p))^{|S|}$ that we want. They also have other factors that we don't want, namely $(x-1)^{q(S)}$ and $(p/(1-p))^{-v+c+q(S)}$. We can get rid of the $q(S)$ powers here by setting $x-1=(1-p)/p$; equivalently $x=1/p$. What remains is an unwanted factor $(p/(1-p))^{-v+c}$ that doesn't involve $S$ and can therefore be pulled out of the sum. Collecting all this stuff (and hoping that I'm copying everything correctly --- copying is sometimes the hardest part of mathematics), I get that the expectation of $q$ is -$$ -(1-p)^e\cdot \left[\left((x-1)\frac{\partial T(x,y)}{\partial x}\right) -(\frac1p,\frac1{1-p})\right]\cdot\left(\frac p{1-p}\right)^{v-c}, -$$ -which simplifies to -$$ -p^{v-c-1}(1-p)^{e+c-v+1}\frac{\partial T}{\partial x}\left(\frac1p,\frac1{1-p}\right). -$$ -Finally, add $c$ to get the expectation of $z$.<|endoftext|> -TITLE: Surprising and Useful Physical Intuition for Mathematical Objects -QUESTION [39 upvotes]: I believe I.M. Gelfand said that when beginning to learn a new subject, one should learn it like a physicist. -In this spirit, what are some helpful and surprising physical intuitions accompanying mathematical objects? -I especially am interested in intuitions that suggest nice proofs...and helpful new ways of thinking. - -REPLY [6 votes]: The Jacobi triple product formula is -$$\prod_{m=1}^\infty -\left( 1 - x^{2m}\right) -\left( 1 + x^{2m-1} y^2\right) -\left( 1 +\frac{x^{2m-1}}{y^2}\right) -= \sum_{n=-\infty}^\infty x^{n^2} y^{2n} -$$ -Borcherds found an elegant combinatorial argument for this formula based on counting the number of ways it is possible for a certain physical system to achieve particular energy levels. The physical model was based on the Dirac sea model of the vacuum. -Many combinatorial formulae find interpretations this way. For example the appearance of $n-\frac{1}{24}$ in Rademacher's formula for the partition function is less of a surprise for physicists familiar with vacuum energies.<|endoftext|> -TITLE: Quotient Surface of A Hyperelliptic Involution -QUESTION [5 upvotes]: Let $X$ be a hyperelliptic Riemann surface, and let $J$ be the hyperelliptic involution. Then consider the quotient surface $X/ < J > ,$ my question is whether $X/ < J > $ is a Riemann surface? -On the standard Riemann surface textbook, the answer is yes, $X/< J >$ is the Riemann sphere $S^{2}$. More generally, let $H$ be a subgroup of the automorphism group of a Riemann surface $\Sigma,$ then $\Sigma/H$ is also a Riemann surface, and the quotient map is holomorphic, and the fixed points of $H$ are the branch points of the quotient map. -However, when we consider the problem in another way, the above mentioned $X/< J >$ should be an orbifold. The fixed points of $J$ are the singular points of orbifold. An orbifold is not a manifold, and if $X/< J >$ is even not a manifold, how could it be a Riemann surface? We know that only when an group $G$ act freely and properly discontinuously on a manifold $M$, the quotient space $M/G$ is a manifold. Here the question is that the action of automorphism group of a Riemann surface on itself has the fixed points. -A simple example is the American football model of an orbifold. Let $\tau$ be the $\pi$ rotation around $z$-axis, then we can get an orbifold with two singular points located at north pole and south pole of the sphere. However, under the Riemann surface view, $S^{2}$ is a Riemann surface, and $\tau$ is a holomorphic involution with two fixed points, so $S^{2}/<\tau>$ should be a Riemann surface. It seems we get a contradiction here. Where is the mistake or confusion in my above statement? - -REPLY [3 votes]: I think the confusion stems from a common misconception, that orbifold points should somehow be considered as singular. The quotient in question - whether as a complex variety or as an orbifold - is not singular. In fact you will never be able to produce something singular by taking the quotient of a compact Riemann surface (i.e. a smooth complex algebraic curve) wrt the action of a finite group (because smoothness in dimension 1 is equivalent to normality, which is preserved by quotients under finite group actions).<|endoftext|> -TITLE: Computing a polynomial product over roots of unity -QUESTION [7 upvotes]: I'm trying to compute the coefficients of the following polynomial, where $\omega$ is a primitive $p$-th root of unity, for $p$ prime: -$$a(x) = \prod_{i=0}^{p-1} f(\omega^ix).$$ -It turns out that the $i$-th coefficient is always an integer, and non-zero only when $i$ is a multiple of $p$. So it seems to me like there should be an elementary expression for $a$. -So far I've got this expression for the $i$-th coefficient: -$$a_i = x^i\sum_{k_0 + \ldots + k_{p-1} = i}f_{k_0} \cdots f_{k_{p-1}} \omega^{k_1 + 2k_2 + \ldots + (p-1)k_{p-1}}$$ -where each $k_i$ is non-negative and bounded by the degree of $f$. -Clearly the roots of unity all cancel out somehow, but I can't figure out how to get a 'nice' expression out. Any suggestions? - -REPLY [3 votes]: A slightly more general result is shown in Theorem A here -https://www.maths.tcd.ie/pub/ims/bull47/R4701.pdf -which states: -If $r$ is relatively prime to $n$ then the coefficient of $x^r$ in $\prod_{i=1}^n f(\omega^i x)$ is 0. -(Here $\omega$ is a primitive $n$-th root of unity, $f$ has rational coefficients.)<|endoftext|> -TITLE: Why is the integral of the second chern class an integer? -QUESTION [16 upvotes]: I'm currently reading the paper "Holonomy, the Quantum Adiabatic Theorem, and Berry's Phase" by Barry Simon. -Imagine a vector bundle with a connection $\nabla$. For simplicity, we assume that this is a $U(1)$ vector bundle. Parallel transport gives rise to the holonomy group, which assigns to each curve $C$ a number $e^{i\gamma(C)}$ that indicates how a vector is "rotated" when transporting it along the curve. In turns out that the phase change $\gamma(C)$ can be expressed as an integral of the curvature form over any surface $S$ that delimits the curve, $C = \partial S$, -$$ \gamma(C) = \int_{S} F^{\nabla} .$$ -I am interested in the integral of the curvature form over the whole manifold, which turns out to be an integer multiple of $2\pi$, -$$ \int_{M} F^{\nabla} = 2\pi k, k\in\mathbb{Z}$$ -Simon notes that this "standard fact" is a consistency condition on the holonomy group. I can understand that: integrating over the whole manifold is like taking the holonomy of the constant path, which must be the identity. -What I would like to understand is the generalization to higher Chern classes. For instance, - -Why is the integral of the second Chern form an integer multiple of $4\pi^2$? - -$$ \int_{M} F^{\nabla}\wedge F^{\nabla} = 4\pi^2 k, k\in\mathbb{Z}$$ -I have a pedestrian proof for special cases, but I would like to understand a general reason behind this phenomenon. Is there a "higher holonomy" at work here? -Obviously, my knowledge of vector bundles and characteristic classes is rather limited. I can find my way around the book "From Calculus to Cohomology", but have by no means absorbed all the material. Basically, my question is why the Chern classes defined via connections are normalized with a factor of $1/(2\pi)^k$. - -REPLY [10 votes]: Greg, -I realize that I may as well write an answer rather than a series of comments. -Although Jessica has given a good answer, I'll try to say this as concretely as possible, -since I now think I understand the question more clearly. The question was actually -about the integrality of -$$\frac{1}{4\pi^2}\int_M F\wedge F$$ -where $F$ is the curvature of line bundle $V$ on a $4$-manifold $M$. This is what mathematicians -(I'm assuming you're a physicist) would call $c_1(V)^2$. -The first thing is observe that -$c_1(V)\in H^2(M,\mathbb{Z})$, and that it's image in de Rham cohomology is - given by $1/(2\pi i)[F]$. To see this in explicit terms, note that the classifying space -for line bundles in $\mathbb{C}\mathbb{P}^\infty$. This implies that the $V$ is the pull -back of the tautological bundle under a $C^\infty$ map $f:M\to \mathbb{C}\mathbb{P}^N$, -for $N\gg 0$. Working on projective space, we can check integrality of the class -$1/(2\pi i)[F]$ by doing a direct calculation to see that this integrates to $1$ over a complex line (aka $2$-sphere). This suffices because the line generates $H_2(\mathbb{C}\mathbb{P}^N)$. After this, $c_1(V)^2=-1/4\pi^2[F]^2$ is automatically integral. That's it. -Postscript: If you are unhappy with the last part, you can replace $f$ with its composition with a generic projection to obtain $f:M\to \mathbb{C}\mathbb{P}^2$. Then your integral -becomes the degree of $f$ which is certainly an integer. Hopefully, you can take it from -here.<|endoftext|> -TITLE: Does the 2-category of toposes admit pseudo-colimits? -QUESTION [12 upvotes]: In Johnstone's book Topos Theory, he mentions an unresolved problem as to whether the 2-category $\mathbf{\text{Topos}}$ of Grothendieck toposes and geometric morphisms admits pseudo-colimits. $\mathbf{\text{Topos}}$ does admit lax colimits (including, famously, examples of Artin-Wraith gluing), and he remarks that insofar as -$$\mathbf{\text{Topos}}^{op} \to Cat$$ -(which you can think of as the forgetful 2-functor from from Grothendieck toposes and left exact left adjoints to categories and functors) is represented by the object classifier $Set^{Fin}$, hence takes pseudo-colimits in $\mathbf{\text{Topos}}$ to pseudo-limits in $Cat$, what we are really asking is whether a pseudo-limit in $Cat$ of toposes and lex left adjoints gives back a topos. -Johnstone's book was written a long time ago, and I was wondering whether there has been progress on this problem since then. - -REPLY [13 votes]: The answer is yes; it seems to be due to Ieke Moerdijk in "The classifying topos of a continuous groupoid, I" (1988). There is also an exposition in section B3.4 of Johnstone's more recent book Sketches of an Elephant. -The idea is: a pseudo-limit in Cat of toposes and lex left adjoints "obviously" satisfies all the exactness conditions of Giraud's theorem, since lex left adjoints preserve all of that structure. Thus one "only" has to construct a small generating family in the limit category—but in general that is nontrivial!<|endoftext|> -TITLE: Can the Sobolev norm of order 1/2 detect "jumps"? -QUESTION [5 upvotes]: We are given a function $f: \mathbb R^d \to \mathbb R$. For simplicity we can assume that $f$ is smooth and compactly supported. Is the Sobolev norm of order $\frac{1}{2}$ strong enough to prove an inequality like this -$$|f(a)-f(b)| \leq \| f \|_{H^{\frac 1 2}}$$ -or a variant, where the left side depends only on the difference in function values and the right hand side on the norm. -"Jumps" is not the exact term, since $H^{\frac 1 2}$-functions can't have jump discontinuities, but is the norm able to measure, by how much the function values change? -For $H^s$ and $0 \leq s < \frac 1 2$ this is not the case. Characteristic functions of bounded sets lie in $H^s$ and when the measure of the set goes to 0, so does the $H^s$-norm. On the other hand, this is true for the $H^1$-norm, because of $f(b) - f(a) = \int_a^b f(t)dt$. -At which order does the behaviour switch? - -REPLY [3 votes]: You need $s>\frac{d}{2}$ for this to hold. You will find this in any text on Sobolev spaces.<|endoftext|> -TITLE: For which rings $R$ is $\mathrm{SL}_n(R)$ generated by transvections? -QUESTION [16 upvotes]: Let $R$ be a commutative ring $R$ with $1$, and $n \geq 2$ an integer. - -Under which conditions is the group $\operatorname{SL}_n(R)$ generated by transvections? - -(A transvection is a matrix with $1$ everywhere on the diagonal and exactly one other non-zero entry.) -This is certainly the case if $R$ is a field, or if $R$ is a Euclidean domain, but I'm wondering whether there is a complete answer to the question. - -REPLY [7 votes]: The goal of my answer is only to provide recent references. -I warmly recommend these two bits of T. Y Lam's book [2]: - -§I.8, for examples where transvections fail to generate $SL_n(R)$ -the second to last paragraph of §VIII.12 for other interesting examples of rings $R$ satisfying $SL_2(R) = E_2(R)$ or its negative. - -And also B. Magurn's latest article on generalized Euclidean group rings [4]. -Update. -Here are newer references focussing on the instances of $SL_2(R) \neq E_2(R)$ for $R$ a quadratic order in a totally imaginary quadratic field. The state of the arts is to be found in [3] and [6], while [5] gives a nice geometric insight on the set $SL_2(R)/E_2(R)$. -An older, but in my humble opinion, important paper is [1], where the structure of $SL_2(R)$ as an amalgamated product with factor $E_2(R)$ is described for $R$ the ring of integers of a totally imaginary quadratic field (with few exceptions), see Theorem 2.4. - -[1] C. Frohman and B. Fine, "Some amalgam structure for Bianchi groups", 1988. -[2] T. Lam, "Serre's problem on projective modules", 2006. -[3] B. Nica, "The unreasonable slightness of $E_2(R)$ over imaginary quadratic rings", 2011. -[4] B. Magurn, "On a note from Oliver concerning generalized Euclidean group rings", 2014. -[5] K. Stange, "Visualizing the Arithmetic of Imaginary Quadratic Fields", 2017. -[6] A. Sheydvasser, "A Corrigendum to Unreasonable Slightness", 2017.<|endoftext|> -TITLE: Talagrand's concentration inequality with limited independence -QUESTION [10 upvotes]: Is there a version of Talagrand's concentration inequality known when the variables have limited independence. More precisely, Let $F:\mathbb{R}^n \rightarrow \mathbb{R}$ be a $1$-Lipschitz convex function. Then, we know that if $X$ is drawn u.a.r. from the $n$-dimensional hypercube, then $\Pr[ |F(X)-M(F)|>t ] \le 2e^{-t^2}$ where $M(F)$ is the median of $F$. If instead $X$ is sampled from a $k$-wise independent distribution over the hypercube, does one get a similar measure concentration result? - -REPLY [2 votes]: It won't be true for small $k$. For example, flip a fair coin: if it's heads, set $X = 0$, and if it's tails, set $X = (1,\dots,1)$. This measure is $1$-wise dependent, but there is certainly no concentration. (Edit: This is a trivial counterexample, but it illustrates the point: When $k$ is small, then each component can have an outsized influence on the function $F(X)$, and this prevents the concentration phenomenon from occurring). -For general Lipschitz functions, I am not familiar with any extension to a $k$-wise setting. -Hoeffding's inequality is the deviation estimate you stated above (but for the mean), applied to the special case $F(X) = \sum_{i=1}^n X_i$: $$\mathbb P( |F(X) - \mathbb E X| > t) \le 2 e^{-2t^2}.$$ -There is a generalization of Hoeffding's inequality to the setting of $k$-wise independent random variables, for $k$ sufficiently large. This was proved by Schmidt, Siegel and Srinivasan in [1]. -Theorem. (cf. Theorem 4.21 on page 66 of [2]) -Let $X_i$ be random variables on $[0,1]$ with $\mathbb E(X_i) = p_i$. Let $X = \sum_{i=1}^n X_i$, and write $\mu := \mathbb E X$ and $p := \mu/n$. Let $\delta > 0$, and let $k_*$ be the first integer greater than $\mu \delta / (1-p)$. If $X_1, \dots, X_n$ are $k$-wise independent for $k \ge k_*$, then $$\mathbb P( X \ge \mu(1+\delta) ) \le \binom{n}{k_*} p^{k_*} \Big/ \binom{\mu(1+\delta)}{k_*}$$ -[1] Chernoff-Hoeffding Bounds for Applications with Limited Independence, Schmidt, Siegel and Srinivasan, 1995. -[2] Concentration of Measure for the Analysis of -Randomised Algorithms, Dubhashi and Panconesi, 2006.<|endoftext|> -TITLE: Co-ends as a trace operation on profunctors -QUESTION [12 upvotes]: The n-lab site on profunctors (http://ncatlab.org/nlab/show/profunctor) describes profunctor composition as using a co-end to "trace out" the connecting variable: -$F\circ G := \int^{d\in D} F(-, d) \times G(d, -)$ -Naturally, one could picture for more general profunctors, $\psi : A \times B^{op} \times C \times D \rightarrow Set$, $\phi : B \times E^{op} \rightarrow Set$ a more generalised composition as co-end'ing together an input from one with an output from the other: -$\xi := \int^{b \in B} \psi(-,b,-,-) \times \phi(b,-)$ -This then looks a lot like contraction of tensors: -$\xi_{e}^{a,c,d} := \sum_{b\in B} \left( \psi_{b}^{a,c,d} \phi_{e}^{b} \right)$ -Even in the use of the language "trace out" (and the fact that both operations form abstract trace operations $Tr(-)$ in their respective traced monoidal categories), this analogy seems to be implied. This also seems to be a useful way to think about profunctor composition, and it appears quite feasible that tensor contraction could be described as a co-end of suitably-enriched profunctors. However, it doesn't seem obvious how to go about unifying these two operations. So, my question is: - -To what extent can the analogy between tensor contraction and profunctor composition be made precise? - -REPLY [13 votes]: The connection can be made precise via the notion of compact closure for symmetric monoidal structures. -Recall that a symmetric monoidal category $C$ is compact closed if every object $c$ has a right adjoint $c^\ast$, meaning that there are unit and counit arrows $\eta: I \to c^\ast \otimes c$ and $\varepsilon: c \otimes c^\ast \to I$ satisfying triangular equations ($I$ is the monoidal unit; the idea is that $c^\ast \otimes -$ is right adjoint to $c \otimes -$). The classical example is of course the category of finite-dimensional vector spaces: the counit $V \otimes V^\ast \to k$ is given by evaluation, and the unit $k \to V^\ast \otimes V$ takes the unit element $1 \in k$ to $\sum_i f^i \otimes e_i$ for any chosen basis $e_i$ and dual basis $f^i$. -In such a situation, there are various equivalent ways of considering morphisms $f: b \to c$. By the adjunction, they are in natural bijection with morphisms $I \to b^\ast \otimes c$. (I'll call this the 'right' picture.) By taking advantage of the symmetry of the tensor, it's also true that one can switch things around and give an adjunction $c^\ast \dashv c$, and thus morphisms $f: b \to c$ will also be in natural bijection with morphisms $b \otimes c^\ast \to I$ (the 'left' picture). Also in this situation, one can define an abstract trace of an endomorphism $f: b \to b$ by the formula -$$Tr(f) = (I \stackrel{unit}{\to} b \otimes b^\ast \stackrel{f \otimes 1_{b^\ast}}{\to} b \otimes b^\ast \stackrel{counit}{\to} I)$$ -which returns the classical trace for endomorphisms on a finite-dimensional vector space. Continuing with this, the composition of two morphisms in the right picture, say $f: I \to b^\ast \otimes c$ and $g: I \to c^\ast \otimes d$, is obtained by 'tracing out': -$$I \stackrel{f \otimes g}{\to} b^\ast \otimes c \otimes c^\ast \otimes d \stackrel{1_{b^\ast} \otimes \varepsilon \otimes 1_d}{\to} b^\ast \otimes d$$ -which is to say composing with the counit in a tensor sandwich. -Now, the bicategory of small categories and profunctors is a compact closed bicategory, meaning that it is a symmetric monoidal bicategory (the tensor being given at the object level by taking cartesian products), and every object $C$ has a right biadjoint, which turns out to be $C^{op}$. If we think of a profunctor from $C$ to $D$ as essentially the same thing as a cocontinuous functor -$$Set^{C^{op}} \to Set^{D^{op}},$$ -then the unit $1 \to C^{op} \otimes C$ in $Prof$ corresponds to the unique (up to iso) cocontinuous functor -$$Set \to Set^{C^{op} \times C}$$ -that sends the terminal object $1$ to $\hom_C$. The counit $C \otimes C^{op} \to 1$ corresponds to a cocontinuous functor -$$Set^{C \times C^{op}} \to Set$$ -which sends a functor $F: C \times C^{op} \to Set$ to the coend $\int^c F(c, c)$. Thus, composition of profunctors in the right picture will involve a tracing out by applying a coend operation to the middle two factors.<|endoftext|> -TITLE: Find the point on the Stiefel Manifold that is closest to a matrix -QUESTION [5 upvotes]: I don't have much background on high-dimensional geometry, so I dare to ask it. -For a given point in $x\in\mathbb{R}^n$, assume that I want to find the point on the unit sphere that is closest to the point $x$. It's easy -- this can be computed by constructing a line through $x$ and the origin, and find the point where this line intersects with the sphere. I'd like to generalize this to $\mathbf{X}\in\mathbb{R}^{n\times k}$ where the sphere is replaced with a Stiefel manifold $\mathbf{Z}^T\mathbf{Z}=\mathbf{I}_{k\times k}$. -Originally the columns of $\mathbf{X}$ constitute the orthonormal basis of $k$-dimensional subspace of $\mathbb{R}^n$, which means $\mathbf{X}$ is on the Stiefel manifold. But the matrix is perturbed by some random matrix $\mathbf{E}$ such that $\tilde{\mathbf{X}}=\mathbf{X}+\mathbf{E}$. I want to find the closest point of $\tilde{\mathbf{X}}$ that is on the Stiefel manifold. In this case, the distance between two matrices $\mathbf{X}$ and $\mathbf{Y}$ must be measured by comparing their orthonormal projectors $||\mathbf{X}\mathbf{X}^T - \mathbf{Y}\mathbf{Y}^T||_2$ to avoid technical issues with ordering and rotations among the basis vectors. -I think there exists only one point $\mathbf{Y}$ on the Stiefel manifold that is closest to $\tilde{\mathbf{X}}$, but I'm not sure how to find it. One thing I can try is employing Lagrange multipler and try to minimize $||\tilde{\mathbf{X}}\tilde{\mathbf{X}}^T - \mathbf{Y}\mathbf{Y}^T||_2$ subject to $\mathbf{Y}^T\mathbf{Y}=\mathbf{I}$, but I'm not sure whether I have to compute derivative of a spectral norm of a matrix. Is there any 'natural' computation to find it? - -REPLY [2 votes]: Theorem IX.7.2 of Bhatia's Matrix Analysis generalizes Bart's answer to any unitarily invariant norm, although it doesn't apply in an obvious way to your distance measure.<|endoftext|> -TITLE: Normality of a locus of points in projective space -QUESTION [7 upvotes]: Let $U_{d,n}\subseteq(\mathbb{P}^d)^n$ denote the locus of $n$-distinct points in projective space $\mathbb{P}^d$ that lie on a rational normal curve of degree $d$, and let $V_{d,n}$ denote its topological closure. I would like to know if, for $d \le n-3$, the spaces $V_{d,n}$ are normal. If they are, this implies that their GIT quotients by $\text{SL}_{d+1}$ are normal, and hence that the birational morphisms from $\overline{M}_{0,n}$ to these GIT quotients (constructed in http://arxiv.org/abs/1012.4835) are contractions. This would clarify some results in this preprint and would be useful in future work as well. -Note that these loci $V_{d,n}$ can be constructed in other ways than the topological closure one given above. For instance, one can take the Kontsevich stable map space $\overline{\mathcal{M}}_{0,n}(\mathbb{P}^d,d)$ and use the product of evaluation maps $\overline{\mathcal{M}}_{0,n}(\mathbb{P}^d,d) \rightarrow \mathbb{P}^d$ to cut out this locus. Similarly, if $\mathcal{H}$ denotes the closed component of the Hilbert scheme parametrizing rational normal curves of degree $d$ then there is an incidence locus in $\mathcal{H}\times(\mathbb{P}^d)^n$ and the projection to $(\mathbb{P}^d)^n$ also defines $V_{d,n}$. Perhaps one of these constructions helps address normality? - -REPLY [4 votes]: Hi Noah, -Consider the forgetful morphism $(\mathbb P^d)^{n+1}\to (\mathbb -P^d)^n$. This restricts to a forgetful morphism $\pi:V_{d,n+1}\to V_{d,n}$. This is -kind of like the map $\overline{\mathscr M}_{g,n+1}\to \overline{\mathscr -M}_{g,n}$. The general fiber of $\pi$ is a $\mathbb P^1$ (the original degree $d$ -rational normal curve). -Now let's see normality. Usually the best way to prove normality is via Serre's -criterion: normal is equivalent to $R_1$ and $S_2$. So, you want to prove these -conditions. -For $n=d+3$ it is easy since $V_{d,n}=(\mathbb P^d)^n$. So it is normal (actually smooth of -course). In particular it is both $R_1$ and $S_2$, in fact it is Cohen-Macaulay -(CM). -Now consider $\pi:V_{d,n+1}\to V_{d,n}$. If $V_{d,n}$ is $R_1$, then I think -(likely) so is $V_{d,n+1}$: the singular set of $V_{d,n+1}$ is contained in the union -of the pre-image of the singular set of $V_{d,n}$ and the locus where $\pi$ is not -smooth. -The former is of at least codimension $2$ by induction. I am not entirely certain -about the latter, but in any case the non-smooth locus has two parts: the locus of -fibers with multiple components and the locus of singular set of the reduced singular -fibers. -The latter of these is definitely of at least codimension $2$ since the locus of -these fibers is of at least codimension $1$ and the singular part is at least -codimension $1$ in that. So, you need that the locus of non-reduced fibers is at -least codimension $2$. I think this seems OK, but this is one of those things I did -not carefully work out. It seems to me that (any component of this) ought to be -contained in (a component of) the reduced singular locus. The reason I think that is -that it seems that there should be a partial "smoothing" of a non-reduced fiber when -you just separate the components. If this is true, then the locus is at least of -codimension $2$. Then again, if this fails then normality fails, so at least you got a necessary condition. -This should take care of condition $R_1$. Now on to $S_2$. -In some sense this is harder, because we need this condtion at the singular set as -well while above we just needed to show that the singular set is not too big. -For $d=2$ this is easy, because a conic (including degenerate ones) is defined by a single equation. So we can -prove that $V_{d,n}$ is actually CM (=Cohen-Macaulay) and hence $S_2$. This follows -by induction: If $V_{d,n}$ is CM, then so is $V_{d,n}\times \mathbb P^d$ and -$V_{d,n+1}$ is a Cartier divisor in $V_{d,n}\times \mathbb P^d$ and hence itself is -CM. -The same proof does not work for $d>2$. Although a smooth curve in a smooth total space is a local complete intersection, as pointed out by mdeland in the remarks this does not remain true for all degenerations. (I should have realized this as this is a famous example...) -At the same time I feel that one might still be able to do something like this. After all we do not need the individual curves to be lci or even S_2. (If we did, this example would lead to non-normality). -It seems this definitely works for $d=2$ but not for other $d$'s. However, at least this might give you some ideas on how one might go about proving something like this. I would add -that if this does not turn out to be normal, then perhaps this is not the "right" -compactification to consider. At the least you should assume that your degenrate curves have no embedded points, but I could imagine that the best to do is to demand that degenerations be stable, i.e., consider the pointed curve degenerating to a pointed degenerate curve -that is stable. This would lead to a partial resolution of $V_{d,n}$ and the above -proof would essentially prove that it is normal. Of course, this may be what you're -doing in general or it does not work for some reason, I am not familiar with the -relevant results, so this is just the first thing I would try.<|endoftext|> -TITLE: A question related to the abc conjecture -QUESTION [9 upvotes]: The abc conjecture asserts that whenever $a,b,c$ are pairwise coprime positive integers such that $a + b = c$ and $\epsilon > 0$, there exists a constant $C_\epsilon > 0$ (which depends on $\epsilon$ but not on $a,b,c$) such that if $N(a,b,c) = \displaystyle \prod_{p | abc} p$ is the radical of $a,b,c$, we have -$$\displaystyle c \leq C_\epsilon N(a,b,c)^{1 + \epsilon}.$$ -Now, if we define $R(n)$ to be the number of ways of writing $n$ as the sum of two positive integers $a,b$ such that $a,b,n$ are pairwise coprime, then infinitely often (when $n$ is prime) we have $R(n) = n -1$. What if we defined $R_{\epsilon, C}(n)$ to be the number of ways of writing $n = a + b$ and $n > C N(a,b,n)^{1 + \epsilon}$? If the $abc$-conjecture is true then $R_\epsilon(n)/n$ should tend towards 0 (in fact, if the conjecture is true, then $R_\epsilon(n) = 0$ for all $n$ sufficiently large). Of course, this is a much weaker statement (one 'almost' version of $abc$ conjecture if you will) than the full conjecture. Is anything of this sort accomplished? - -REPLY [10 votes]: For any $n$ the number of relatively prime $a,b ({\rm rad}(ab))^{1+\epsilon}$ is $o(n)$, indeed $o(n^{1-\epsilon'})$ -for any $\epsilon' < \epsilon / (1+\epsilon)$. Therefore this bound -is true a fortiori of the number of such $a,b$ for which $a+b=n$. -We use the following lemma: -For all $\delta > 0$ there exists $C_\delta$ such that for every $r,n$ -the number of solutions of ${\rm rad}(x) = r$ with $x\leq n$ -is at most $C_\delta n^\delta$. -Proof: We may assume $r$ squarefree, else the number of solutions is zero. -Then (even without the condition $x \leq n$) we compute -$$ -\sum_{{\rm rad(x)} = r} x^{-\delta} - = \prod_{p|r} \, (p^{-\delta} + p^{-2\delta} + p^{-3\delta} + p^{-4\delta} + \cdots) - = \prod_{p|r} \frac{p^{-\delta}}{1 - p^{-\delta}}. -$$ -All summands are positive, and each solution with $x \leq n$ contributes -at least $n^{-\delta}$, so the number of $x \leq n$ terms in the sum -is at most $n^\delta \prod_{p|r} p^{-\delta} / (1 - p^{-\delta})$. -But each factor in this product is less than $1$ except for the -finitely primes $p \leq 2^{1/\delta}$. This proves the lemma with -$$ -C_\delta = \prod_{p \leq 2^{1/\delta}} \frac{p^{-\delta}}{1 - p^{-\delta}}. -\qquad \Box -$$ -Now, given $\epsilon$, the number of pairs $(r,r')$ such that -$n > (rr')^{1+\epsilon}$ is asymptotically proportional to -$n^{1/(1+\epsilon)} \log n$. By the lemma, each one arises as -$({\rm rad}(a), {\rm rad}(b))$ at most $C_d^2 n^{2\delta}$ times -with $a,b -TITLE: Cusp width for an arbitraty Fuchsian group -QUESTION [5 upvotes]: In Shimura's Intro to Arithmetic Theory of Automorphic Forms, he defines a cusp of a Fuchsian group $\Gamma$ as a point $s \in \mathbb{R} \cup \{ \infty \}$ that is fixed by a parabolic element of $\Gamma$. -I know the definition of the width of a cusp in the case $\Gamma = SL_2(\mathbb{Z})$ or a congruence subgroup (namely, it is the only positive integer $h$ such that $\rho^{-1} \overline{\Gamma_s} \rho$ is generated by the matrix $\begin{pmatrix} 1 & h \\\\ 0 & 1 \end{pmatrix}$, where $\rho$ is any matrix in $SL_2(\mathbb{Z})$ such that $\rho(s) = \infty$) but I was wondering if we can also define a notion of cusp width for any Fuchsian group. - -REPLY [11 votes]: For subgroups $ \Gamma \subset SL_2(\mathbb Z)$, every cusp is a covering space of the single cusp for the quotient of the upper half plane by $SL_2(\mathbb Z)$ (that is, the (2,3,infinity orbifold), and the index of the covering is a natural concept that equals the width. There is no such structure in general: in fact, -(as is well-known and I'm sure is described in Shimura's book) subgroups -of $SL_2(\mathbb Z)$ are conjugate (within $GL_2(\mathbb Q)$) to other subgroups where this index changes: for example, let $\Gamma$ be the subgroup of $SL_2(\mathbb Z)$ -that when conjugated by -the diagonal matrix with diagonal entries $2,1$ is still integral. -There is another, related concept, however, that is universally applicable: the set of areas of maximal horoball neighborhoods of cusps. For any cusp in the quotient of upper half space by a Fuchsian group, there is some maximal neighborhood of the cusp of the form horoball mod parabolic subgroup (stabilizer of cusp) that is embedded in the quotient. Its area (which equals the length of its boundary curve) is an invariant. If the quotient orbifold has $k$ cusps, then there is moreover a convex subset of $\mathbb R^k$ consisting of the set of $k$-tuples of logs of areas of these horoball neighborhoods such that the union is embedded. It is bounded by hyperplanes of the form $\log(A_i) < C_i$ and $\log(A_i) + \log(A_j) < K_{i,j}$. -For the 3-dimensional generalization of this (having to do with discrete subgroups of -$PSL(2,\mathbb C)$ act on hyperbolic 3-space, Jeff Weeks' program snappea is very good for getting a sense of how these horoball neighborhoods interact: you can see the pictures change as you move sliders. Unfortunately the best version of the original snappea (you can easily find it by googling) only runs on obsolete versions of the macintosh operating system, but there is a modern version, SnapPy, modernized using Python by Marc Culler and Nathan Dunfield, that will give the same information, but requires more keyboard and less GUI interaction. The geometry of the maximal horoball neighborhoods of cusps has considerable significance for 3-dimensional topology, and has been studied quite a lot.<|endoftext|> -TITLE: Kind of submultiplicativity of the Frobenius norm: $\|AB\|_F \leq \|A\|_2\|B\|_F$? -QUESTION [10 upvotes]: Let $\|\cdot\|_F$ and $\|\cdot\|_2$ be the Frobenius norm and the spectral norm, respectively. -I'm reading Ji-Guang Sun's paper 'Perturbation Bounds for the Cholesky and QR Factorizations' from BIT 31 in 1991. -While deriving the perturbation bound on Cholesky factorization $A=LL^T$ with $A\in\mathbb{R}^{n\times k}$ of rank $k$, he begins with the perturbed matrix $A+E$ and its Cholesky factorization: -$$A+E=(L+G)(L+G)^T.$$ -Expanding the RHS and subtracting $A$ from both sides, we have $E = GL^T + LG^T + GG^T$. Then he claims in (2.12) of his paper that -$$\|E\|_F\leq 2\|L\|_2\|G\|_F + \|G\|_F^2\,.$$ -If I read $\|L\|_F$ instead of $\|L\|_2$, everything seems perfectly normal to me by the following three properties of any matrix norm: - - $\|X^T\|=\|X\|$ for any matrix norm - subadditivity: $\|X+Y\|_p\leq\|X\|_p + \|Y\|_p$ - submultiplicativity: $\|XY\|_p\leq \|X\|_p\cdot\|Y\|_p$ for $p=2$ or $F$ - -But I'm not sure why $\|L\|$ is a spectral norm in this equation even though all other norms are Frobenius norms. Does $\|XY\|_F\leq \|X\|_2\|Y\|_F$ always hold? - -REPLY [11 votes]: A simpler, more direct proof that requires no SVD: let $Y_j$ be the $j$th column of $Y$ and $Z_j$ that of $Z=XY$. Then, -$$\|Z\|_F^2 = \sum_j \|Z_j\|_2^2 = \sum_j \|XY_j\|_2^2 \leq \sum_j \|X\|_2^2\|Y_j\|_2^2 = \|X\|_2^2\|Y\|_F^2.$$<|endoftext|> -TITLE: Is there a concept of Combined Teichmuller space for surfaces with both geodesic boundary and punctures/cusps -QUESTION [5 upvotes]: If we take a sequence of compact hyperbolic Riemann surface with k geodesic boundary components such that the lengths of the geodesic boundary components go to zero, then in the "limit", we should get a surface with k punctures/cusps. Is there a concept of distance ( in Teichmuller theory/ Riemannian geometry ) which would realize this convergence ? I know that there is a concept of Teichmuller space with genus g and k geodesic boundary components and also there is a concept of Teichmuller space with genus g and k cusps with finite area metrics. But is there a concept of Teichmuller space which will unify both of them, i.e. is distance will realize the above convergence and will contain both of the above Teichmuller spaces as subspace ? -Should we use Gromov-Housdorff convergence to realize the above convergence ? - -REPLY [8 votes]: If you have a compact hyperbolic surface with geodesic boundary $\Sigma$, then you may double the surface along its boundary to get a closed hyperbolic surface $D\Sigma=\Sigma\cup_{\partial\Sigma}\Sigma$, which has an involution $\tau:D\Sigma\to D\Sigma$ which exchanges the two sides and fixes $\partial \Sigma$. One may then identify the Teichmuller space parameterizing hyperbolic structures on $\Sigma$ with geodesic boundary with the subspace of the Teichmuller space of $D\Sigma$ which is fixed under the involution $\tau$ (which as an element of the mapping class group acts by an isometric involution on Teichmuller space of $D\Sigma$). -The Weil-Petersson metric on Teichmuller space is an incomplete metric. -Its completion is obtained by appending the Teichmuller spaces of Riemann surfaces with nodes, where some of the curves on the surface have been pinched (one may think of these as hyperbolic metrics with a double cusp, by letting the length of the curve approach zero). The Weil-Petersson metric extends to this completion, as proved by Masur. -Consider the Teichmuller space of $D\Sigma$, and the corresponding subspace associated to $\Sigma$ (the fixed point set of $\tau$). If one pinches some of the curves associated to the fixed point of $\tau$ on $D\Sigma$, then $\tau$ acts as an involution on the noded surface, and one may append the Teichmuller space of this noded surface invariant under $\tau$ to the Teichmuller space of $\Sigma$ with the Weil-Petersson completion. This gives you the moduli space you're describing of surfaces with geodesic boundary and punctures, and the Weil-Petersson metric gives a well-defined distance function on this space.<|endoftext|> -TITLE: What monsters does the "growth condition" required of holomorphic modular functions bar? -QUESTION [7 upvotes]: Even though the title of this question pretty much captures what I'd like to know, I'll add -two side questions: -1) Is it difficult to get a handle on the totality of functions that arise if one drops -the growth condition? -2) Presumably these functions have no known number theoretic interest; is there a theoretical reason to explain why they shouldn't? -Edit: In light of the early responses (thank you all!), particularly Scott Carnahan's, I'm particularly interested to know about modular functions with essential singularities and I suppose non-zero weight. As Scott points out, for quite arbitrary $g$, $g(f(z))$ will be $\Gamma$-invariant if $f$ is $\Gamma$-invariant ($\Gamma\subset SL_2{\Bbb Z}$). Then -I suppose one could multiply such a $g(f(z))$ by a classical modular function of -non-zero weight. -Does this observation admit some sort of converse, something in the spirit of the Weierstrass factorization theorem for entire functions, say? - -REPLY [9 votes]: There are several growth conditions at infinity that you can impose: - -If $f(z) \to 0$ as $z \to i \infty$ (or in higher level, all cusps), then $f$ is a cusp form, and if it is weight $2k$, then it descends to a section of $\Omega^{\otimes k}$ on the compact modular curve. -If $f(z)$ is bounded as $z \to i \infty$ (this is the condition you mention in the comments), then $f$ is often called a holomorphic modular form, and it descends to a section of $\Omega^{\otimes k}(\text{cusps})$, i.e., these are pluricanonical forms with logarithmic poles at cusps. -If $f(z)$ grows at most exponentially as $z \to i \infty$, then we can call $f$ a meromorphic modular form, and it descends to a section of $\Omega^{\otimes k}(\infty \cdot \text{cusps})$. Common examples of number theoretic interest include the $j$-invariant, other principal moduli of genus zero modular curves, partition functions of conformal vertex algebras, and forms of negative weight that enumerate colored partitions. You might want to rephrase your claim of "no number theoretic interest" to a statement that there is no obvious connection to Galois representations through Langlands correspondence. -If $f(z)$ has no growth condition at all, then it does not necessarily descend to a section of a bundle on an algebraic curve, and it only exists as an analytic object. For example, you can pull back the exponential function by $j$ to get the holomorphic $SL_2(\mathbb{Z})$-invariant function $e^{j(z)}$ on the upper half-plane that grows like $e^{e^z}$ as $z \to i\infty$ along some trajectories. I don't know of any number theoretic applications for modular functions with essential singularities at cusps. - -REPLY [7 votes]: The coefficients of functions with poles at cusps do have great number theoretic interest, and have been heavily studied. A classical example is the coefficients of $1/\eta(\tau)$, whose coefficients are partitions of a number. It is also fairly easy to describe all such functions that are meromorphic at cusps: these are more or less just sections of various line bundles over a compact Riemann surface, so the number of such functions is given by Riemann-Roch, just as for modular forms that are bounded at cusps.<|endoftext|> -TITLE: Efficient presentations for finite groups -QUESTION [11 upvotes]: A finitely presented group which has more generators than relations has an infinite abelianization and so is an infinite group. Therefore, for a finite group, all presentations must have at least as many relations as generators. A presentation for a finite group is called efficient if the number of generators equals the number of relations. -I know the following examples of efficient presentations for finite groups with $2$ generators: -$< x,y | x^{2} = y^{2} = (xy)^{n}>$, for any $n \geq 2$, which is a group of order $4n^{2}-4n$ in which $xy$ has order $2n^{2}-2n$ and conjugating through $x$ or $y$ sends $xy$ to $(xy)^{2n-1}$. (When $n=2$, this is the quaternion group of order $8$.) -$< x,y | x^{3} = y^{3} = (xy)^{2}>$, which is a double cover of $A_{4}$ isomorphic to $SL(2,3)$. This group is isomorphic to the group of quaternion units, together with $\pm \frac{1}{2} \pm \frac{i}{2} \pm \frac{j}{2} \pm \frac{k}{2}$. -$< x,y | x^{4} = y^{3} = (xy)^{2}>$, which is a double cover of $S_{4}$ not isomorphic to $GL(2,3)$. This is isomorphic to the preceding group, together with every quaternion of the form $\pm \frac{u}{\sqrt{2}} \pm \frac{v}{\sqrt{2}}$, where $u$ and $v$ are two distinct elements of the set $\{ 1, i, j, k \}$. -$< x,y | x^{5} = y^{3} = (xy)^{2}>$, which is a double cover of $A_{5}$ isomorphic to $SL(2,5)$. This is isomorphic to the group of unit quaternions representing $SL(2,3)$ as above, together with a set of icosians explicitly described in Conway and Sloane's SPLAG, which I am too lazy to describe here. -All of these groups have a central element of order $2$, and except for the members of the infinite family that have $n$ being odd, a subgroup isomorphic to the quaternion group of order $8$. -How does this generalize to larger numbers of generators, or, at least, what are some examples of efficient presentations of finite groups using $3$ generators? (I need not explain why I know all examples of efficiently presented groups with a single generator.) Is there a group analogous to the quaternion group of order $8$ for each number of generators? Is there a group analogous to $SU(2)$, which has subgroups isomorphic to the smallest member of the infinite family and all three exceptional examples? Is the center of a nontrivial finite group possessing an efficient presentation always nontrivial? (And what is a good reference for discussing this?) - -REPLY [8 votes]: You asked about the center of a group with an efficient presentation: the relation of "central" to "efficient presentation" is probably strongest in the Schur multiplier. In particular, a perfect group can only have an efficient presentation if its Schur multiplier is trivial (so it is the universal perfect central extension of its inner automorphism group). If the inner automorphism group has a non-trivial Schur multiplier, then the perfect group with an efficient presentation must have a non-trivial center. -However, many perfect groups have trivial Schur multipliers and centers. For instance, the Mathieu group M11 has an efficient presentation on two generators with two relations, and of course has a trivial center. -Efficient presentations are somewhat poorly named (balanced might be better) since they are typically awful for computational group theory. However, people have taken them as a challenge, and so you can find scores of papers dealing with efficient presentations of (covering groups of) finite simple groups. One paper with nice tables is: - -Campbell, Colin M.; Havas, George; Ramsay, Colin; Robertson, Edmund F. - Nice efficient presentations for all small simple groups and their covers. - LMS J. Comput. Math. 7 (2004), 266–283. - MR2118175 - Online version. - -Note that efficient is sometimes defined in terms of the rank of the Schur multiplier. So be careful that your usage may contradict some of the papers (which give "efficient" presentations for groups with non-trivial Schur multiplier; they are allowed to use one extra relation per independent generator of the Schur multiplier). -A related concept is an asymptotic version that seeks to describe how complicated presentations of finite simple groups get. In fact it seems they are very, very simple: - -Guralnick, R. M.; Kantor, W. M.; Kassabov, M.; Lubotzky, A. - Presentations of finite simple groups: a quantitative approach. - J. Amer. Math. Soc. 21 (2008), no. 3, 711–774. - MR2393425 - DOI:10.1090/S0894-0347-08-00590-0 - -where it is shown that there is an absolute constant C so that a total C generators and relations is needed to define a presentation of a finite quasi-simple group, and indeed the total word length of the relations is bounded by C*(log(n)+log(q)) where n is the rank of the BN-pair and q is the size of the field of definition (and alternating and sporadic groups just get less than C). They left out the case of Ree groups (the characteristic 3 kind), but probably it will be taken care of at some point. -If you want to explore these things in GAP, part of the documentation for GAP is dedicated to this problem: - -http://www.gap-system.org/Doc/Examples/balanced.html<|endoftext|> -TITLE: Examples for non-naturality of universal coefficients theorem -QUESTION [30 upvotes]: Does anyone have good examples of a space $X$ and a map $f: X \to X$ so that $f_*: H_*(X) \to H_*(X)$ is the identity but (e.g.) $f_*: H_*(X; \mathbb{F}_2) \to H_*(X; \mathbb{F}_2)$ is not the identity? -Edit: As mentioned in the comments, $f_*$ is an isomorphism on the mod-2 homology, but I don't see why it needs to be the identity. More precisely, the exact sequence -$$ -C(X; \mathbb{Z}) \overset{\times 2}{\longrightarrow} C(X; \mathbb{Z}) \longrightarrow C(X;\mathbb{Z}/2) -$$ -gives an exact triangle in homology, which in turn induces a 2-step filtration on $H_*(X; \mathbb{Z}/2)$ (where one step is the image of the map $H_*(X;\mathbb{Z}) \to H_*(X;\mathbb{Z}/2)$). The assumption that $f_*$ is the identity on integral homology implies that it is the identity on the associated graded space to this filtration, but that still doesn't imply it is the identity. -I came across a similar phenomenon in the context of Heegaard Floer homology, with the rings $(\mathbb{Z}/2)[U]$ and $\mathbb{Z}/2$ playing the roles of $\mathbb{Z}$ and $\mathbb{Z}/2$. - -REPLY [8 votes]: Here's another solution; it's really the same as the ones Sam and Tom give, but with a more "geometric" flavor. -Start with the unit 2-sphere $S^2\subset \def\R\mathbb{R}\R^3$, and let $r: S^2\to S^2$ be given by reflection across the $xz$-plane ($r(x,y,z)=r(x,-y,z)$). This passes to a self-map $f: \R P^2\to \R P^2$ on the quotient; it carries the subspace $\R P^1\subset \R P^2$ to itself, which I'm thinking of as the quotient of the circle in the $xy$-plane. In other words, $f$ is a cellular map, with respect to the "usual" CW-filtration $\R P^0\subset \R P^1\subset \R P^2$. On the "cellular chain complex" of $\R P^2$, the map $f$ is degree $-1$ on the cells in dimension $1$ and $2$. -Now let $X=\R P^2\cup_{\R P^1} \R P^2$, obtained by gluing two projective planes along a circle. Let $g:X\to X$ be the self-map which (i) sends the first $\R P^2$ to the second $\R P^2$ by the map $f$, (ii) sends the second $\R P^2$ to the first $\R P^2$ by the map $f$, and thus (iii) sends the common $\R P^1$ to itself by the map $f|\R P^1$. -Since $g$ is a cellular map, its easy to compute its effect on the cellular chain complex of $X$: it's degree $-1$ on the $1$-cell, and it switches the two $2$-cells with each other by degree $-1$ maps. So it's identity on $H_*(X;\def\Z\mathbb{Z}\Z)$, since $H_1(X;\Z)=\Z/2$ and the "cellular chain" $(1,-1)$ which generates $H_2(X;\Z)=\Z$ is clearly fixed; but it's not the identity on $H_2(X;\Z/2)=\Z/2\oplus \Z/2$, as the switching of the $2$-cells is visible here. -(The generator of $H_2(X;\Z)$ is actually the image of a map $q:S^2\to X$, which you get by gluing the characteristic maps of the two $2$-cells together, and $fq$ differs from $q$ exactly by a 180-degree rotation of the sphere (and so $q$ and $fq$ are homotopic). This also shows that $X$ is stably equivalent to $S^2\vee \R P^2$.)<|endoftext|> -TITLE: Identifying poisoned wines -QUESTION [66 upvotes]: The standard version of this puzzle is as follows: you have $1000$ bottles of wine, one of which is poisoned. You also have a supply of rats (say). You want to determine which bottle is the poisoned one by feeding the wines to the rats. The poisoned wine takes exactly one hour to work and is undetectable before then. How many rats are necessary to find the poisoned bottle in one hour? -It is not hard to see that the answer is $10$. Way back when math.SE first started, a generalization was considered where more than one bottle of wine is poisoned. The strategy that works for the standard version fails for this version, and I could only find a solution for the case of $2$ poisoned bottles that requires $65$ rats. Asymptotically my solution requires $O(\log^2 N)$ rats to detect $2$ poisoned bottles out of $N$ bottles. -Can anyone do better asymptotically and/or prove that their answer is optimal and/or find a solution that works for more poisoned bottles? The number of poisoned bottles, I guess, should be kept constant while the total number of bottles is allowed to become large for asymptotic estimates. - -REPLY [4 votes]: I have a solution that uses just 28 rats. The idea is to use a 28x1000 2-separable testing matrix $M$, where $M_{ij}$ is 1 if rat $i$ drinks bottle $j$ and 0 if he doesn't drink it. A matrix is 2-separable if the boolean OR of any of its two columns is unique. You can read more about such matrices here: https://en.wikipedia.org/wiki/Disjunct_matrix -You can find the actual matrix $M$ here: https://pastebin.com/4yDd1XvG -I plan to write a short report describing the method I used to find $M$. -UPDATE: -A 27x1065 2-separable matrix has been found so it is possible to solve this problem with 27 rats. See https://oeis.org/A054961<|endoftext|> -TITLE: Transcendental numbers: yet another classification -QUESTION [13 upvotes]: Let $\mathbb{A^+}$ be the set of non-negative algebraic numbers. Consider the set of "polynomials" : $$\mathbb{P} = \lbrace a_0 + a_1x^{r_1} + a_2x^{r_2} + a_3x^{r_3} +\cdots + a_nx^{r_n}| a_0, a_i, r_i \in \mathbb{A}, r_i > 0, i= 1,2,\cdots,n\rbrace$$ We call $\alpha \in \mathbb{R}, \alpha \geq 0$ extra-algebraic if there exists a polynomial in $\mathbb{P}$ satisfying $f(\alpha)=0$. Denote the set of all extra-algebraic numbers by $\mathbb{A}_E$. So, $\mathbb{A} \subset \mathbb{A}_E$.(The strict inclusion is because of numbers like $2^\sqrt2$ which are extra-algebraic but not algebraic and more by the Gelfond–Schneider theorem). We call $\beta \in \mathbb{R}, \beta > 0$ extra-transcendental if it is not extra-algebraic. Candidates for examples of extra-transcendental numbers are $e^\pi$ and $e^\frac{-\pi}{2}$. -Question: - -Do extra-transcendental numbers exist? -Is $\mathbb{R^+} - \mathbb{A}_E$ uncountable? - -Many thanks. - -REPLY [20 votes]: $\mathbb P$ is countable. Moreover, any $f\in\mathbb P$ is analytic, hence it has only countably many zeros. Thus $\mathbb A_E$ is countable, and in particular, extra-transcendental reals exist, and $\mathbb R^+\smallsetminus\mathbb A_E$ has the power of continuum. -For a concrete example, the Lindemann–Weierstrass theorem implies that $e$ is extra-transcendental. -EDIT: To tie up a loose end, every nonzero $f\in\mathbb P$ has only finitely many positive real roots. Since $f(x)$ is eventually dominated by its nonzero term with the highest exponent, the roots are bounded. Similarly, $f(x)$ is dominated by the term with the smallest exponent when $x\to0+$, hence the roots are bounded away from $0$, i.e., they are contained in a compact subset of $(0,+\infty)$. However, choosing a branch of logarithm makes $f$ holomorphic in $U=\mathbb C\smallsetminus(-\infty,0]$, therefore it can have only finitely many roots in any compact subset of $U$. -In fact, if $f(x)=a_0x^{r_0}+a_1x^{r_1}+\cdots+a_nx^{r_n}$ (with $r_i\in\mathbb R$ pairwise distinct, $a_i\in\mathbb R\smallsetminus\{0\}$), then $f$ has at most $n$ positive real roots. -We can prove this by induction on $n$. If $n=0$, then $f(x)=a_0x^{r_0}$ has no positive real root. Assume the statement holds for $n-1$. Put $g(x)=a_0+a_1x^{r_1-r_0}+a_2x^{r_2-r_0}+\cdots+a_nx^{r_n-r_0}=f(x)/x^{r_0}$. Then every positive root of $f$ is also a root of $g$. Moreover, between each two consecutive roots of $g$, there is a root of its derivative $g'$. Since $g'$ has at most $n$ nonzero terms (the derivative of the constant $a_0$ vanishes), it has at most $n-1$ positive real roots by the induction hypothesis, thus $f$ has at most $n$ such roots. -I guess that one could also prove a variant of Decartes' rule of signs for these generalized polynomials along similar lines.<|endoftext|> -TITLE: Additive Subgroups of the Reals. -QUESTION [24 upvotes]: Does anyone know if there is a classification of the subgroups of the real numbers taken under addition? If not can anyone point me in the directiong of any papers/materials which discuss properties of or interesting facts about these subgroups? - -REPLY [4 votes]: Slightly off-topic, a weird subgroup in two dimensions is constructed in this paper: -Ryuji Maehara. On a connected dense proper subgroup of ${\bf R}^2$ whose complement is connected . Proc. Amer. Math. Soc. 97 (1986) 556-558. MR 840645. -That subgroup is constructed as a graph of a wild group homomorphism $f:\mathbf{R}\to \mathbf{R}$, so it is also a subgroup: -$\{ (x,f(x)) \mid x\in \mathbf{R}\}$<|endoftext|> -TITLE: Global dimension and localization -QUESTION [7 upvotes]: Is there any condition on a commutative ring $R$ so that the global dimension of $R$ coincides with the supremum of the global dimensions of the localizations $R_{\mathfrak{m}}$ at all maximal ideals $\mathfrak{m}\subset R$? I'm looking (if possible) for conditions which are easy to verify. - -REPLY [7 votes]: This problem is discussed at length in T.Y. Lam's Lectures on Modules and Rings. The hyperlink should take you to the Theorem in question (5.92 in section 5G). The point is that for a commutative noetherian ring $R$ you get the result you wanted and also more: - -For a commutative noetherian ring $R$ gl.dim$(R_m)=$pd$_R(R/m)$ for all maximal ideals $m$. This implies gl.dim$(R)=\sup($gl.dim$(R_m)) = \sup($pd$_R(S))$ where the last supremum runs over all simple $R$-modules. - -The proof Lam gives avoids the machinery of Ext, using instead the fact that the global dimension of a commutative noetherian local ring is the injective dimension (also the projective dimension) of its residue field. -Note that the noetherian assumption really is necessary. On page 197, Lam points out that B. Osofsky has constructed some interesting examples (he gives details) which I suspect would show this theorem fails without the noetherian hypothesis.<|endoftext|> -TITLE: Is this long sequence of Hom's exact ? -QUESTION [8 upvotes]: Let $\mathcal{A}$ be an abelian category with enough projectives and let $\underline{\mathcal{A}}$ be its stable category with loop functor $\Omega: \underline{\mathcal{A}} \to \underline{\mathcal{A}}$ (for definitions see remark 3 below). Define $\Omega^0 := id_{\underline{\mathcal{A}}}$ and $\Omega^{n+1} := \Omega \circ \Omega^n$ and denote the hom-groups of $\underline{\mathcal{A}}$ by $[-,-]$. Then, if -$$ 0 \to B' \to B \to B'' \to 0$$ -is a short exact sequence in $\mathcal{A}$, there is a sequence (starting with $n=0$ from the left): - -$$ ... \to [\Omega^n(A),B'] \to -> [\Omega^n(A),B] \to -> [\Omega^n(A),B''] \to -> [\Omega^{n+1}(A),B'] \to ...$$ - -and the composition of two consecutive maps is zero. - -Is this sequence exact, or - equivalently, is $[\Omega^n(A),-]_{n -> \ge 0}$ a delta-functor ? - -Remark 1: I know that the following is a long exact sequence (ending with $n=0$ at the right): -$$ ... \to [A, \Omega^n(B')] \to [A, \Omega^n(B)] \to [A, \Omega^n(B'')], \to [A, \Omega^{n-1}(B')] \to ...$$ -Therefore is guess that the sequence above is also exact. -Remark 2: There is a natural epimorphism -$$Ext_\mathcal{A}^n(A,B) \to [\Omega^n(A),B].$$ -If $\mathcal{A}$ satisfies $Ext_\mathcal{A}^n(-,P)=0$ for all projectives $P$ and all $n > 0$ then the epimorphism is actually an isomorphism (for $n >0$) and the -exactness of the sequence follows from the long exact $Ext$-sequence. -Remark 3: The stable category $\underline{\mathcal{A}}$ is defined as follows: It has the same objects as $\mathcal{A}$ and the hom's are given by -$$[A,B] := Hom_{\underline{\mathcal{A}}}(A,B) := Hom_\mathcal{A}(A,B) / P(A,B)$$ -where $P(A,B)$ is the subgroup of homomorphisms that factor through a projective. The endo-functor $\Omega$ is obtained by taking fixed projective presentations in $\mathcal{A}$: -$$0 \to \Omega(A) \to P \to A \to 0.$$ - -REPLY [10 votes]: It's not exact in general. Take $\mathcal{A}$ to be the category of finitely generated abelian groups, so that $\underline{\mathcal{A}}$ is the category of finite abelian groups and $\Omega=0$. Take the short exact sequence -$$0\rightarrow \mathbb{Z}/2 \rightarrow \mathbb{Z}/4 \rightarrow \mathbb{Z}/2\rightarrow 0$$ -and $A=\mathbb{Z}/2$. Then the long sequence is -$$\mathbb{Z}/2\stackrel{1}\rightarrow \mathbb{Z}/2\stackrel{0}\rightarrow \mathbb{Z}/2\rightarrow 0\rightarrow \cdots,$$ -which is not exact.<|endoftext|> -TITLE: Sum of log p/p for p equivalent to l mod D -QUESTION [12 upvotes]: It's fairly classical that for $D>1$ and $(D,l)=1$ one has -$$\sum_{\stackrel{p\leq x}{p\equiv l\; (mod \; D)}}\frac{\log p}{p} = \frac{\log x}{\phi(D)} + \textrm{O}(1)$$ -where if I understand correctly the dependence on $D$ in the $\textrm{O(1)}$ is captured by something like -$$\frac{1}{\phi(D)}\sum_{\textrm{non-principal} \; \chi}\frac{-L'(1,\chi)}{L(1,\chi)}$$ -which is $\ll_{\epsilon}D^{\epsilon}$ for any $\epsilon>0$. If this is correct, suppose I am interested in the sum -$$\sum_{\stackrel{p\leq x}{\left(\frac{-D}{p}\right)=1}}\frac{\log p}{p}$$ -where $\left(\frac{-D}{p}\right)$ is the Legendre symbol. There are $\phi(D)/2$ residue classes mod $D$ that $p$ can lie in, and so then this is just my previous sum $\phi(D)/2$ times, and I get that it's $1/2 \log x +\textrm{O}(1)$, where the dependence on $D$ in $\textrm{O}(1)$ is now something like $\textrm{O}(D^{1+\epsilon})$. Is this the best one can do? I was hoping I could get it to be $\textrm{O}(D^{\epsilon})$ but if that's not even correct perhaps I should stop trying. - -REPLY [15 votes]: You're right that your proposed method for estimating the new sum, while correct, gives a worse error bound than we can obtain otherwise. Rather than quoting the classical result itself, I suggest going back to the proof of that classical result and modifying it to address the new sum. -Somewhat more precisely: the Legendre symbol $\big( \frac{-D}p \big)$, or rather its multiplicative generalization the Jacobi symbol $\big( \frac{-D}n \big)$, is a Dirichlet character (mod $D$)—call it $\chi_1(n)$. The sum you are interested in is just -$$ -\sum_{p\le x} \frac{\log p}p \bigg( \frac{1+\chi_1(p)}2 \bigg). -$$ -(Not exactly, since this expression counts primes dividing $D$ with weight $1/2$, but that's easily dealt with.) Therefore you're left with needing to understand $\sum_{p\le x} (\log p)/p$ (no problem) and $\sum_{p\le x} (\chi_1(p)\log p)/p$. The latter sum is going to be $O(1)$ in the same way that the error term in the classical problem is $O(1)$; it is going to be related to $-L'(1,\chi_1)/L(1,\chi_1)$. So in fact you don't have all the different $-L'(1,\chi)/L(1,\chi)$ error terms to deal with, only the single one where $\chi=\chi_1$.<|endoftext|> -TITLE: Starting PhD at the age of 25 -QUESTION [10 upvotes]: For over an year I have otherwise been pretty active on MathOverflow but for this question I would like to remain anonymous. -I would be starting my grad school in Fall 2011 in the US while in the 25th year of my life. I will be joining a grad school which is ranked by most lists within their top 10. -I have been working and hope to continue to work in areas in theoretical physics which have a strong interface with mathematics, especially geometry. -I understand that most people start their PhD at the age of 22 (or even below!). -I would like to know how does it affect my career, now that I will be getting my PhD around the age of 30. I am very worried and extremely depressed that this is possibly too late to start a PhD. I guess most scientists become faculty by the age of 30 when I would be getting my PhD! -I would be happy to get any feedback/advice about starting PhD so late in life. - -Also can one get a PhD in theoretical physics/mathematics in less than 5 years? -I have made this question community wiki. - -REPLY [5 votes]: The difference between 25 and 22 may seem like a lot at your age, but honestly it's nothing in the context of a life span of 70 years or more. You're certainly not too old to start graduate study and have a very productive career. - -REPLY [2 votes]: Something that can be very satisfying is to acheive a series of long-term goals you have set. Something that can be unsettling is the sense of achieving something unintended or finding that what was previously considered a good goal is not. -The issue of starting a Ph.D. at age 25 versus 22 or some other age is less important (I think) than the issue of whether this fits in with your other personal goals. For example, I can't say that starting a Ph.D. at 25 kept me from starting a family over a decade later, but it probably contributed to the delay. Hindsight shows me the advantages of starting a family earlier than later. There are also issues of employment/financial support, degree to which you spend time not doing academics, etc. If you think you have done a full self evaluation and found that Ph.D. studies fit, good for you; my bet is that you will need to reevaluate your goals on a yearly (if not more frequent) basis. -I do not want to discourage you from a Ph.D. I think that a few hours (days, weeks) spent -doing serious thinking and planning now will ensure years of satisfaction to come. -Gerhard "Ask Me About System Design" Paseman, 2011.03.29 - -REPLY [2 votes]: In France most of students (in pure mathematics) start their PhD at the age 24. Indeed they get a A level at 18, then they need to study during 5 years to get a master. They could start a PhD just after the master but most of the time they take one year more to pass the "agregation" diplom (a diplom for teaching). So you get 18+5+1=24.<|endoftext|> -TITLE: Trees in groups of exponential growth -QUESTION [28 upvotes]: Question: Let $G$ be a finitely generated group with exponential growth. - Is there a finite generating set $S \subset G$, such that the associated Cayley graph $Cay(G,S)$ contains a binary tree? - -Some background: - -The existence of such a tree clearly implies exponential growth. -Kevin Whyte showed in Amenability, Bilipschitz Equivalence, and the Von Neumann Conjecture, Duke Journal of Mathematics 1999, p. 93-112, that such trees exist if $G$ is non-amenable. So the question is only open for amenable groups of exponential growth. -One good reason for such a binary tree to exist is the existence of a free semigroup inside $G$. In fact, if $G$ is solvable, then the existence of such a semigroup is known to be equivalent to exponential growth (and equivalent to being not virtually nilpotent). This is part of some version or extension of the Tits alternative. Grigorchuk constructed an amenable torsion group with exponential growth, which does not contain such a semigroup, but it contains a binary tree. - -EDIT: Al Tal pointed out in an answer below that Benjamini and Schramm covered the non-amenable case (this is 2. from above) already in Benjamini and Schramm "Every Graph With A Positive Cheeger Constant Contains A Tree With A Positive Cheeger Constant", GAFA, 1997. - -REPLY [4 votes]: This (so nice) question seems to be equivalent with the notorious and old problem of constructing a supraamenable (or superamenable) group of exponential growth. Recall that a supraamenable group is one such that every non-empty set is not G-paradoxical, or equivalently, given any non-empty subset of G, there exists a finitely additive invariant measure on G that assigns measure one to the set. All groups of sub-exponential growth are supraamenable, but no other example is known. -Basic defintions are here (before Exercise 8 and Remark 3) http://terrytao.wordpress.com/2009/01/08/245b-notes-2-amenability-the-ping-pong-lemma-and-the-banach-tarski-paradox-optional/ -The question is stated eg. as Question 70 here but it is much older http://www.unige.ch/math/folks/delaharpe/articles/18-CGH.pdf -Non supraamenability seems to be equivalent with the existence of a generating set such that the Cayley graph contains the binary tree. Indeed, if such a generating set $S$ exists then the set $V$ containing the vertices of the tree is paradoxical using elements from $S^{-1}$: take the subset of the left sons and the one of the right sons, they will both cover their mothers. Conversely, if $X=X_1 \sqcup X_2$ is a paradoxical subset of $G$ and $S$ is the (finite) set of elements involved in the paradox, then $S^{-1}$ will do, as long as one gets sure identity is not in $S$ (which can be assumed). Start with any $x \in X$ as initial root. A paradox is nothing else than a $2$ to $1$ piecewise $G$- map $f:X \rightarrow X$ so one can choose the left son the counter image of $x$ in $X_1$ and for the right son the counter image in $X_2$ and continue by induction (there is a little problem, at exactly one point in the process there might be a cycle ocurring (due to the root trying to connect), but then one can delete the whole infected half...)<|endoftext|> -TITLE: A (nameless?) product in the category of groups, and its properties -QUESTION [7 upvotes]: Suppose I have a group $G,$ with normal subgroups $K, L$ such that $G=KL$ but $K\cap L \neq \{1\}.$ If the intersection were trivial, we would say that $G$ is the direct product of $K$ and $L,$ but this operation to direct product is like free product with amalgamation is to free product. Any names for it out there? -I am really more interested in the properties of this more than the name (though the latter might help me find the former) -- for example, given three groups $H_1, H_2, H3,$ is there a way to classify $G$ where $K\simeq H_1, L\simeq H_2, K\cap L \simeq H_3?$ What if $H_1, H_2$ are finitely generated free groups, and $H_3$ is an infinitely generated free group? - -REPLY [2 votes]: As already mentioned above, the group in question is an extension -$$1\rightarrow K\cap L \rightarrow KL \rightarrow K/K\cap L \times L/K\cap L \rightarrow 1,$$ -where we abbreviate the rightmost (a priori) non trivial group by $H$. -In particular $K\cap L$ is normal in $KL=G$, so $G$ is isomorphic to a subgroup of the wreath product $K\cap L \wr H$. I don't know if this point of view is helpful, but I for myself would try to use wreath products here.<|endoftext|> -TITLE: Estimates for Symmetric Functions -QUESTION [19 upvotes]: Let $z_1,z_2,\ldots,z_n$ be i.i.d. random variables in the unit circle. Consider the polynomial -$$ -p(z)=\prod_{i=1}^{n}{(t-z_i)}=t^n+a_{1}t^{n-1}+\cdots+a_{n-2}t^2+a_{n-1}t+a_n -$$ -where the $a_i$ are the symmetric functions -$$ -a_{1}=(-1)\sum_{i=1}^{n}{z_{i}}\hspace{0.3cm},\quad a_{2}=(-1)^2\sum_{1\leq i< j\leq n}{z_{i}z_{j}}\hspace{0.3cm} \quad\ldots\quad a_{n}=(-1)^{n}z_{1}z_{2}\ldots z_{n}. -$$ -How can we estimate the random variable $Z$ defined as -$$ -Z=\sum_{j=1}^{n}{|a_{j}|} -$$ -asymptotically as $n\to\infty$? -It is not very difficult to estimate $|\sum_{j=1}^{n}{a_{j}}|$ by estimating $\log p(1)$ via the CLT. However, $Z$ seems to be much more difficult. Any idea of what can work here? - -Update: If we look at the term at the central symmetric random variable - $a_{\lfloor n/2 \rfloor}$ $$ -> a_{\lfloor n/2 \rfloor}=\text{sum of -> the products of $\lfloor n/2 \rfloor$ -> of different $z_{i}$'s} $$ - it is not hard to see that it - has uniform distributed phase in - $(-\pi,\pi]$. -However, its magnitude is blowing up extremely fast! -Does anyone knows how to compute the limit distribution of - $|a_{\lfloor n/2 \rfloor}|$ under the appropriate normalizations? - -Thanks! - -REPLY [12 votes]: OK, here is my argument (sorry for the delay). -First of all, $Z$ is essentially the maximum of the absolute value of the polynomial $P(z)=\prod_j(1-z_jz)$ on the unit circumference (up to a factor of $n$, but it is not noticeable on the scale we are talking about). -Second, the maximum of the absolute value of a (trigonometric) polynomial of degree $K$ can be read from any $AK$ uniformly distributed points on the unit circumference $\mathbb T$ (say, roots of unity of degree $AK$) with relative error of order $A^{-1}$. -Now let $\psi(z)=\log(1-z)$. We want to find the asymptotic distribution of the $\max_z n^{-1/2}Re\sum_j \psi(zz_j)$ where $z_j$ are i.i.d. random variables uniformly. Since it is the logarithm of $|P(z)|$, the maximum can be found using $10n$ points. -Decompose $\psi(z)$ into its Fourier series $-\sum_{k\ge 1}\frac 1kz^k$. Then, formally, we have $n^{-1/2}\sum_j \psi(zz_j)=-\sum_k\left(n^{-1/2}\sum_j z_j^k\right)\frac{z^k}{k}$. It is tempting to say that the random variables $\xi_n,k=n^{-1/2}\sum_j z_j^k$ converge to the uncorrelated standard complex Gaussians $\xi_k$ by the CLT in distribution and, therefore, the whole sum converges in distribution to the random function $F(z)=\sum_k\xi_k\frac{z^k}k$, so $n^{-1/2}\log Z$ converges to $\max_z\Re F(z)$ (the $-$ sign doesn't matter because the limiting distribution is symmetric). This argument would be valid literally if we had a finite sum in $k$ but, of course, it is patently false for the infinite series (just because if we replace $\max$ by $\min$, we get an obvious nonsense in the end). Still, it can be salvaged if we do it more carefully. -Let $K$ run over the powers of $2$. Choose some big $K_0$ and apply the above naiive argument to $\sum_{k=1}^{K_0}$. Then we can safely say that the first $K_0$ terms in the series give us essentially the random function $F_{K_0}(z)$ which is the $K_0$-th partial sum of $F$ when $n$ is large enough. -Our main task will be to show that the rest of the series cannot really change the maximum too much. More precisely, it contributes only a small absolute error with high probability. -To this end, we need -Lemma: Let $f(z)$ be an analytic in the unit disk function with $f(0)=0$, $|\Im f|\le \frac 12$. Then we have $\int_{\mathbb T}e^{\Re f}dm\le \exp\left(2\int_{\mathbb T}|f|^2dm\right)$ where $m$ is the Haar measure on $\mathbb T$. -Proof: By Cauchy-Schwartz, -$$ -\left(\int_{\mathbb T}e^{\Re f}dm\right)^2\le \left(\int_{\mathbb T}e^{2\Re f}e^{-2|\Im f|^2}dm\right)\left(\int_{\mathbb T}e^{2|\Im f|^2}dm\right) -$$ -Note that if$|\Im w|\le 1$, we have $e^{\Re w}e^{-|\Im w|^2}\le \Re e^w$. So the first integral does not exceed $\int_{\mathbb T}\Re e^{2f}dm=\Re e^{2f(0)}=1$. Next, $e^s\le 1+2s$ for $0\le s\le\frac 12$, so $\int_{\mathbb T}e^{2|\Im f|^2}dm\le 1+4\int_{\mathbb T}|\Im f|^2dm\le 1+4\int_{\mathbb T}|f|^2dm$. Taking the square root turns $4$ into $2$ and it remains to use that $1+s\le e^s$ -The immediate consequence of Lemma 1 is a Bernstein type estimate for $G_K(z)=\sum_{k\in (K,2K]}\left(n^{-1/2}\sum_j z_j^k\right)\frac{z^k}{k}$ -$$ -P(\max|\Re G_K|\ge 2T)\le 20Ke^{-T^2K/9} -$$ -if $0\le TK\le \sqrt n$, say. -Indeed, just use the Bernstein trick on the independent random shifts of $g_K(z)=\sum_{k\in (K,2K]}\frac{z^k}{k}$: -$$ -E e^{\pm t\Re G_K(z)}\le \left(\int_{\mathbb T}e^{\Re tn^{-1/2}g_K}dm\right)^n\le e^{2t^2/K} -$$ -for every $t\le \sqrt n/2$ (we used the Lemma to make the last estimate) and put $t=\frac{TK}{3}$. After that read the maximum from $10K$ points with small relative error and do the trivial union bound. -Choosing $T=K^{-1/3}$, we see that we can safely ignore the sum from $K=K_0$ to $K=\sqrt n$ if $K_0$ is large enough. Now we are left with -$$ -G_K(z)=\sum_{k\ge \sqrt n}\left(n^{-1/2}\sum_j z_j^k)\right)\frac{z^k}{k} -$$ -to deal with. Recall that all we want here is to show that it is small at $10n$ uniformly distributed points. Again, if $g(z)=\sum_{k\ge \sqrt n}\frac{z^k}{k}$, we have $|\Im g|\le 10$, say so we can use the same trick and get -$$ -P(\max_{10n\text{ points}}|\Re G|\ge 2T)\le 20n e^{n^{-1/2}t^2-tT} -$$ -if $0\le t\le \sqrt n/20$, say. -Here we do not need to be greedy at all: just take a fixed small $T$ and choose $t=\frac{2\log n}T$. -Now, returning to your original determinant problem, we see that the norm of the inverse matrix is essentially $Z/D$ where $D=\min_i\prod_{j:j\ne i}|z_i-z_j|$. We know the distribution of $\log Z$ and we have the trivial Hadamard bound $D\le n$. This already tells you that the typical $\lambda_1$ is at most $e^{-c\sqrt n}$. The next logical step would be to investigate the distribution of $\log D$.<|endoftext|> -TITLE: holomorphic K-theory -QUESTION [19 upvotes]: Topological K-theory is usually defined by setting $K(X)$ to be the groupification of the monoid $Vect_\mathbb{C}(X)$ of complex vector bundles over $X$ (with addition given by Whitney sum). However, we can alternatively declare that $[B]\sim [A]+[C]$ whenever $0\rightarrow A \rightarrow B\rightarrow C\rightarrow 0$ is a short exact sequence of vector bundles over $X$ (morphisms are required to have locally constant rank): certainly if $B\cong A\oplus C$ then we have such a sequence, and in the other direction we can take a metric on $B$ and identify $C$ with $A^\perp \subseteq B$. -You can take the $K$-theory of any abelian category using this second definition. So, I'm curious to know if people do this for the category of holomorphic vector bundles over a complex manifold. The above splitting construction no longer works since it uses partitions of unity, so assuming we use this more general definition we'd get more equivalence relations. On the other hand, there's all this funny business going on with vector bundles topologically but not holomorphically isomorphic, which means that $K_{hol}(X)$ wouldn't just be a subquotient of $K(X)$. So in the end, I'm not sure whether I should expect this to be a more or less tractable sort of object. -I'm told that the Chow ring might have something to do with this, but the wikipedia page seems to indicate that it's more analogous to singular cohomology than anything else. - -REPLY [27 votes]: Grothendieck proved that there is an analytification functor $X \mapsto X^{an}$ from schemes locally of finite type over $\mathbb C$ to the category of (non-reduced!) analytic spaces, which is fully faithful when restricted to proper schemes. This induces isomorphisms from $K-$ groups in the algebraic sense on $X$ to $K-$ groups in the holomorphic sense on $X^{an}$ . This is just a mild generalization of Serre's GAGA principle proved for reduced, projective varieties. So this settles your problem in the compact algebraizable case, by telling algebraic geometers to solve it ( and they actually know quite a lot of the K-theory of schemes !) -In the diametrically opposed case of Stein manifolds, a landmark theorem of Grauert also answers your request. Namely, given a complex manifold there is an obvious forgetful functor $Vecthol(X) \to Vecttop(X)$ from isomorphism classes of holomorphic vector bundles -on $X$ to isomorphism classes of topological vector bundles -on the underlying topological space $X^{top}$. If $X$ is Stein, Grauert proved that the functor is an isomorphism of monoids : every topological vector bundle has a unique holomorphic structure. ( Results of this nature fit into what is called the "Oka principle". ) There are no extension problems for short exact sequences $0 \to \mathcal E \to \mathcal F \to \mathcal G \to 0$ because they all split: in the Stein case thanks to theorem B and in the topological case because of partitions of unity (theorem B in disguise, actually: fine sheaves are acyclic). So in the Stein case too you can relax and ask topologists to do your work . -Finally, there are complex manifolds between these extreme cases. I am not aware of a general theory there ( of course that proves nothing but my ignorance) . This looks like an interesting topic of investigation, especially in view of Winkelmann's theorem ( link to survey here) that on every compact holomorphic manifold of positive dimension $n $ there exists a non-trivial holomorphic vector bundle of rank $\leq n$.<|endoftext|> -TITLE: Double coset spaces of reductive groups and integral representations of L-functions -QUESTION [12 upvotes]: Let $G$ be a reductive group over a number field $k$, with center $Z$. Let $P$ be a parabolic subgroup. Let $H$ be a reductive subgroup of $G$. To what extent can we understand the double coset space $P_k\backslash G_k/H_k$? I'll give some examples, then my motivation for the question, and then I'll refine the question. -Let's look at some simple examples. -Example 1: $G=GL_2$, $P$ is the standard Borel subgroup, $H=Z\cdot GL_1$, embedded as the Levi component of $P$. Using the Bruhat decomposition, $P_k\backslash G_k/H_k=1\cup w\cup wn$, where $n$ is any nontrivial element of the unipotent radical of $P$, and $w$ is the nontrivial element of the Weyl group of $G$. -Example 2: $G=GL_2$, $P$ is the standard Borel subgroup, $H$ is the multiplicative group of a quadratic extension $k_1$ of $k$. Identifying $P_k\backslash G_k$ with $k^2-(0,0)=k_1^\times$, we see that $P_k\backslash G_k/H_k=1$. -Example 3: $G=GL_2\times GL_2$, $P$ is $B\times B$, the product of the Borel subgroups of $GL_2$, $H$ is $GL_2$ embedded diagonally in $G$. Then -$P_k\backslash G_k/H_k=B_k\backslash H_k/B_k=1\cup w$. -Example 4: $G=GL_2$ over a quadratic extension $k_1$ of $k$ (considered as a $k$-group), $P$ the standard Borel subgroup, $H=GL_2$ over $k$. Identifying $P_k\backslash G_k$ with $k_1^2-(0,0)$, we see that $P_k\backslash G_k/H_k=1\cup wn_\alpha$, where $n_\alpha=\bigg(\matrix{1&\alpha\cr 0&1}\bigg)$ and $\alpha$ generates $k_1$ over $k$. - -So why do I care? -One of the primary sources of integral representations of automorphic $L$-functions is integrating the restriction of an Eisenstein series on a group $G$, associated to a parabolic subgroup $P$, against a cusp form on a subgroup $H$, i.e. -$$Z(s,f)=\int_{Z_{\mathbb A} H_k\backslash H_{\mathbb A} } E_s(h)f(h)\ dh$$ -In order for this to converge, we need $Z\backslash Z_H$ to be anisotropic (so, e.g., Example 1 won't work without modification). Unwinding the Eisenstein series, we have -$$Z(s,f)=\sum_{\xi\in P_k\backslash G_k/H_k}\int_{Z_{\mathbb A} \Theta^\xi_k\backslash H_{\mathbb A} }\varepsilon_s(\xi h)f(h)\ dh$$ -where $\Theta^\xi=\xi^{-1}P\xi\cap H$. Let's consider a fixed term in the sum. -$$\int_{Z_{\mathbb A} \Theta^\xi_k\backslash H_{\mathbb A} }\varepsilon(\xi h)f(h)\ dh=\int_{Z_{\mathbb A} \Theta^\xi_{\mathbb A} \backslash H_{\mathbb A} }\int_{Z_{\mathbb A} \Theta^\xi_k\backslash\Theta^\xi_{\mathbb A} }\varepsilon_s(\xi\theta h)f(\theta h)\delta_{\Theta_{\mathbb A} }(\theta)^{-1}\ d\theta\ dh$$ -$$=\int_{Z_{\mathbb A} \Theta^\xi_{\mathbb A} \backslash H_{\mathbb A} }\int_{Z_{\mathbb A} \Theta^\xi_k\backslash\Theta^\xi_{\mathbb A} }\varepsilon_s(\xi\theta\xi^{-1}\xi h)f(\theta h)\delta_{\Theta_{\mathbb A} }(\theta)^{-1}\ d\theta\ dh$$ -$$=\int_{Z_{\mathbb A} \Theta^\xi_{\mathbb A} \backslash H_{\mathbb A} }\varepsilon_s(\xi h)\int_{Z_{\mathbb A} \Theta^\xi_k\backslash\Theta^\xi_{\mathbb A} }\varepsilon_s(\xi\theta\xi^{-1})f(\theta h)\delta_{\Theta_{\mathbb A} }(\theta)^{-1}\ d\theta\ dh$$ -Note that if $\varepsilon_s(\xi\theta\xi^{-1})$ is trivial on $\Theta^\xi_{\mathbb A} $ and $\Theta^\xi$ has a normal subgroup that is a unipotent radical of a parabolic subgroup of $H$, the term vanishes by the cuspidality of $f$. -Under certain conditions, the inner integral (and hence the entire integral) factors into a product of local integrals, which we can hopefully compute to be local factors of an automorphic $L$-function. But we must pass over this in silence. (See Garrett's Euler Factorization of Global Integrals for more.) - -In order to make further progress, we would need to understand the nature of $\Theta^\xi$, as well. Let's calculate it for the above examples. -Example 1: $\Theta^1=H$, $\Theta^w=H$, $\Theta^{wn}=Z$. -Example 2: $\Theta^1=Z$. -Example 3: $\Theta^1=B$, $\Theta^w=w^{-1}Bw$ (the "opposite" Borel subgroup). -Example 4: $\Theta^1$ is the Borel subgroup of $H$, $\Theta^{wn_\alpha}=k_1^\times$. -Another very interesting example is the set-up for the triple product $L$-function, found in either section 3.10 of Bump's book or in PS-Rallis, Rankin triple L functions. -It would be really nice to know, given $G$, $P$, and $H$, what automorphic $L$-function (if any, since most such triples won't produce anything interesting) is represented by the above integral. Hypothetically, we could make a list of the $L$-functions, like there is for the Langlands-Shahidi method. Understanding the double coset decomposition is the first step. -This is probably way too much to hope for; I'm interested in understanding why it is so hard. -It seems that, over $\mathbb C$, we know when the double coset is finite (at least when $H_{\mathbb C}$ is an open subgroup of the fixed points of an involution of $G_{\mathbb C}$). I can't find a good statement of what is known over $\mathbb Q$. Is anything known if $G$ and $H$ are split? Or if $G$ is split and $H$ is anisotropic? Or if $G=GL_n$? Etc . . . - -REPLY [8 votes]: This is a very nice question! -All of your examples are spherical quotients. Please take a look at -http://andromeda.rutgers.edu/~sakellar/rs.pdf and Yiannis' other papers -for connections between spherical quotients and integral representations for L functions.<|endoftext|> -TITLE: volume of the unit ball of the Banach space $\ell_1^n\otimes_{\epsilon}\ell_1^n$? -QUESTION [10 upvotes]: We denote by $\otimes_{\epsilon}$ the injective Banach tensor product. -Which is the asymptotic volume of the unit ball of the Banach space $\ell_1^n\otimes_{\epsilon}\ell_1^n$? - -REPLY [5 votes]: You might want to have a look at the work "On the volume of unit balls in Banach spaces" by C. Schuett (Lemma 3.2). He computes -$vol_n(B_{\ell_p^n\otimes_{\varepsilon}\ell_q^n})^{1/n^2}$ up to constants depending only on $p$ and $q$. -In your case, the estimate is -$$vol_n(B_{\ell_1^n\otimes_{\varepsilon}\ell_1^n})^{1/n^2} \asymp n^{-3/2}.$$<|endoftext|> -TITLE: Can curves differentiate vector bundles on P^2? -QUESTION [9 upvotes]: Not much is known about vector bundles on $\mathbb{P}^2$ but I wonder if the following is a tractable question: -If $E,E'$ are non-isomorphic vector bundles on $\mathbb{P}^2$, then is there always a smooth curve $C \subset \mathbb{P}^2$ such that $E|_C$ and $E'|_C$ are still non isomorphic? -A related and perhaps easier question: Can a vector bundle restrict to the trivial bundles on every curve without itself being trivial on $\mathbb{P}^2$? - -REPLY [21 votes]: Any curve of large enough degree will do. Set $F:= E'\otimes E^{\vee}$; if $d$ is a very large integer, then $\mathrm H^1(F(-d)) = 0$. Take any curve $C$ of degree $d$, and suppose that $E\mid_C$ and $E'\mid_C$ are isomorphic; this isomorphism is given by a section of $F\mid_C$. Since $\mathrm H^1(F(-d)) = 0$, this section extends to a global section of $F$, which yields a homomorphism $f \colon E'\to E$ which is an isomorphism when restricted to $C$. This $f$ is injective. Obviously $E$ and $E'$ have the same rank, so the cokernel of $f$ is torsion, and the same degree, so it in fact concentrated in finitely many points. But then the cokernel must be 0, because a sheaf concentrated in codimension 2 can't have projective dimension 1, and this concludes the proof.<|endoftext|> -TITLE: Connectivity of the Erdős–Rényi random graph -QUESTION [54 upvotes]: It is well-known that if $\omega=\omega(n)$ is any function such that $\omega \to \infty$ as $n \to \infty$, and if $p \ge (\log{n}+\omega) / n$ then the Erdős–Rényi random graph $G(n,p)$ is asymptotically almost surely connected. The way I know how to prove this is (1) first counting the expected number of components of order $2, 3, \dots, \lfloor n/2 \rfloor$, and seeing that the expected number is tending to zero. Then (2) showing the expected number of isolated vertices is also tending to zero. -This approach also allows more precise results, such as: if $p = (\log{n}+c) / n$ with $c \in \mathbb{R}$ constant, then Pr$[G(n,p)$ is connected] $\to e^{-e^{-c}}$ as $n \to \infty$, which follows once we know that in this regime the number of isolated vertices is approaching a Poisson distribution with mean $e^{-c}$. - -I am wondering if it is possible to - give an easier proof (of a coarser result) along the - following lines. There are $n^{n-2}$ - spanning trees on the complete graph, - and $G$ is connected if and only if - one of these trees appears. So the - expected number of spanning trees - is $n^{n-2}p^{n-1}$. One might expect that if this - function is growing quickly enough, - then with - high probability $G(n,p)$ is connected. - -I think I remember reading somewhere that this approach doesn't quite work --- for example the variance is too large to apply Chebyshev’s inequality. What I am wondering is if there is some way to fix this if we are willing to make $p$ a little bit bigger. In particular, what about $p = C \log{n} / n$ for some large enough constant $C > 1$, or even $p = n^{-1 + \epsilon}$ for fixed but arbitrarily small $\epsilon >0$? - -REPLY [7 votes]: this is not really an answer to your question but just a related matter. As you point out the expected number of trees in a random graph is 1 already when p=c/n and is very large when p=logn/n so the hope is that this can be used to show that with a large probability the random graph contains a tree. There is a collection of conjectured by Jeff Kahn and me trying to suggest a very general connection of this type. These conjectures are presented in the pape Thresholds and expectation thresholds by Kahn and me and are mentioned in this MO question. If true these conjectures will imply that the threshold for connectivity will be below logn/n (of course, we do not need it for this case...), and the proof will probably will at best be much much more complicated then existing proofs. -I should mention that the sharp threshold property which was proved by Erdos and Renyi for connectivity can be proved (with harder proofs) from more general principles: One is the Margulis-Talagrand theorem which applies to the threshold for random subgraphs of highly edge connected graphs and one is Friedgut's result which identify graph properties with coarse thresholds.<|endoftext|> -TITLE: Can Hölder's Inequality be strengthened for smooth functions? -QUESTION [24 upvotes]: Is there an $\epsilon>0$ so that for every nonnegative integrable function $f$ on the reals, -$$\frac{\| f \ast f \|_\infty \| f \ast f \|_1}{\|f \ast f \|_2^2} > 1+\epsilon?$$ -Of course, we want to assume that all of the norms in use are finite and nonzero, and $f\ast f(c)$ is the usual convolved function $\int_{-\infty}^{\infty} f(x)f(c-x)dx$. The applications I have in mind have $f$ being the indicator function of a compact set. -A larger framework for considering this problem follows. Set $N_f(x):=\log(\| f \|_{1/x})$. Hölder's Inequality, usually stated as - $$\| fg \|_1 \leq \|f\|_p \|g\|_q$$ -for $p,q$ conjugate exponents, becomes (with $f=g$) $N_f(1/2+x)+N_f(1/2-x)\geq 2N_f(1/2)$. In other words, Hölder's Inequality implies that $N_f$ is convex at $x=1/2$. The generalized Hölder's Inequality gives convexity on $[0,1]$. -It is possible for $N_f$ to be linear, but only if $f$ is a multiple of an indicator function. What I am asking for is a quantitative expression of the properness of the convexity when $f$ is an autoconvolution. - -Examples: The ratio of norms is invariant under replacing $f(x)$ with $a f(cx-d)$, provided that $a>0$ and $a,c,d$ are reals. This means that if $f$ is the interval of an indicator function, we can assume without loss of generality that it is the indicator function of $(-1/2,1/2)$. Now, $f\ast f(x)$ is the piecewise linear function with knuckles at $(-1,0),(0,1),(1,0)$. Therefore, $\|f\ast f\|_\infty=1$, $\|f \ast f\|_1 = 1$, $\|f \ast f \|_2^2 = 2/3$, and the ratio of norms is $3/2$. -Gaussian densities make another nice example because the convolution is easy to express. If $f(x)=\exp(-x^2/2)/\sqrt{2\pi}$, then $f\ast f(x) = \exp(-x^2/4)/\sqrt{4\pi}$, and so $\|f\ast f\|_\infty = 1/\sqrt{4\pi}$, $\|f\ast f\|_1=1$, and $\|f \ast f\|_2^2=1/\sqrt{8\pi}$. The ratio in question is then just $\sqrt{2}$. -This problem was considered (without result) by Greg Martin and myself in a series of papers concerning generalized Sidon sets. We found this ``nice'' example: $f(x)=1/\sqrt{2x}$ if $0 < x < 1/2$, $f(x)=0$ otherwise. Then $f\ast f(x) = \pi/2$ for $0 < x < 1/2$ and $f\ast f(x) = (\pi-4\arctan(\sqrt{2x-1}))/2$ for $1/2 < x < 1$, and $f\ast f$ is 0 for $x$ outside of $(0,1)$. We get $\|f \ast f\|_\infty = \pi/2$, $\|f \ast f\|_1 = 1$, $\|f \ast f \|_2^2 = \log 4$, so the norm ratio is $\pi/\log(16) \approx 1.133$. -In this paper, Vinuesa and Matolcsi mention some proof-of-concept computations that show that $\pi/\log(16)$ is not extremal. - -REPLY [2 votes]: Not an answer, but rather an extended comment. -Consider the following problem. -Given a set of integers $A\subset [1,N]$, denote by $\nu(n)$ the number of representations of $n$ as a sum of two elements of $A$. Thus, $\nu=1_A\ast1_A$ up to normalization, and, trivially, we have - $$ \sum_n \nu^2(n) \le |A|^2 \max_n \nu(n). $$ -Does there exist an absolute constant $\varepsilon>0$ such that if - $$ \sum_n \nu^2(n) > (1-\varepsilon) |A|^2 \max_n\nu(n), $$ -then $\alpha:=|A|/N\to 0$ as $N\to\infty$? (The flavor of this question to me is as follows: we want to draw a conclusion about a finite set, given that its ``additive energy'' is large -- but not as large as in Balog-Szemeredi-Gowers.) -What is the relation between this and the original problem? Although I cannot establish formally equivalence in both directions, it is my understanding that the two problems are ``essentially equivalent''; at least, if in the original problem we confine ourselves to indicator functions of open sets.<|endoftext|> -TITLE: Braid group analogue for signed symmetric group? -QUESTION [6 upvotes]: This is probably something well known (either in the affirmative or in the negative) but I couldn't get this information easily: -Braid group:Symmetric group::?:Signed symmetric group -By "signed symmetric group" I mean the wreath product of the cyclic group of order two by the symmetric group with its usual action as a symmetric group. Equivalently, it is the group of n by n matrices under multiplication where every row and every column has exactly one nonzero entry and that entry could be +1 or -1. -I don't see an obvious way to generalize, nor do I see a reason why no analogue can exist. - -REPLY [13 votes]: There are braid groups attached to every Coxeter group which are obtained by forgetting that the generators in the standard presentation square to the identity but keeping the other relations. I believe the signed symmetric group is the Coxeter group of type $B_n$. - -REPLY [7 votes]: There is an analog of the braid group for each Coxeter group. The signed symmetric group is the Coxeter group of type B: as for the usual braid group, a presentation of it can be obtained from the standard Coxeter presentation by removing the torsion relation. Hence, it is generated by $\tau, \sigma_1,\dots, \sigma_{n-1}$ and relations -$ \tau \sigma_1 \tau \sigma_1 = \sigma_1 \tau \sigma_1 \tau $ -$ \tau \sigma_i=\sigma_i\tau$ if i > 1 -$\sigma_i \sigma_{i+1} \sigma_i = \sigma_{i+1} \sigma_{i} \sigma_{i+1}\ i=1,\dots,n-2 $ -$ \sigma_i \sigma_j = \sigma_j \sigma_i \text{ if } |i-j| \geq 2 $ -There is also a topological definition of this group: Coxeter groups are finite reflection groups, that is finite subgroups of $GL_n(\mathbb{R})$ generated by reflections. For the type B, the corresponding hyperplane are defined by equation $z_i-z_j=0$ (as for the symmetric group), $z_i+z_j=0$ and $z_i=0$. -Hence let $X_n=\{(z_1,\dots,z_n) \in (\mathbb{C}^{\times})^n, z_i \neq \pm z_j\}$, and define the pure braid group of type B as the fundamental group of $X_n$. The full braid groupe of type B is the fundamental group of the quotient space $X_n/(Z_2^n \rtimes S_n)$. - -REPLY [5 votes]: To expand a bit on Qiaochu's answer: The signed symmetric group has a presentation with generators $s_i=(i,i+1)$ and $t$, which is the element that is -1 in the first coordinate and 1 in the others (in terms of signed permutation matrices, this is $\operatorname{diag}(-1,1,1,\dots,1)$). -The $s_i$'s satisfy the usual braid relations $$s_is_{i+1}s_i=s_{i+1}s_is_{i+1}$$ as well as $s_i^2=1$. You can check that there are two different ways of writing $\operatorname{diag}(-1,-1,1,\dots,1)$ in terms of these generators: $$s_1ts_1t=ts_1ts_1;$$ you also have that $t^2=1$. -If you forget about the relations that say that the squares of generators are 1, you get the braid group of type B. This braid group also has a topological intepretation as braids in a cylinder: see this paper.<|endoftext|> -TITLE: Checking whether a variety is normal -QUESTION [13 upvotes]: I am looking to check whether the hypersurface in $A^{n}$ defined by $x_{1}^{2} + x_2^{2} + .... + x_n^{2} = 0$ is a normal variety.....In general, are there any nice sufficiency conditions to prove normality? - -REPLY [3 votes]: Given a Cohen-Macaulay affine ring $R$ over a perfect field, it satisfies $S_k$ for all $k$. Since $R$ is equidimensional, Jacobian criterion says that the singular locus is cut out by the Jacobian ideal $J$. It is then sufficient and necessary to have codim $J \ge 2$ in $R$ for $R$ to satisfy $R_1$. -Assuming characteristic is not 2, the coordinate ring $S = k[x_1,\cdots,x_n]/(x_1^2+\cdots+ x_r^2)$ is cut out by a non-zerodivisor and is thus Cohen-Macaulay. Its singular locus is cut out by $I = (x_1,\cdots, x_r)$ which has codim $r-1$ in $S$. Therefore $S$ is normal iff $r\ge 3$.<|endoftext|> -TITLE: Density of Irreducible Polynomials in $\mathbb{Z}[x]$ -QUESTION [13 upvotes]: Recently I was thinking about some questions concerning $\mathbb{Z}[x]$ and realized that they might be a bit easier if I knew the relative densities of reducible polynomials. -Let $P_d$ denote the set of all elements of $\mathbb{Z}[x]$ of degree $\leq d$. My basic question is: - - -Question: What fraction of elements of $P_d$ factor in $\mathbb{Z}[x]$? - - -The fraction $f_d$ of elements of $P_d$ which factor satisfies $f_d\geq 1-\zeta(d+1)^{-1}$ (with $\zeta$ the Riemann Zeta Function) since this is the count of elements which factor as a constant times an element of $P_d$. When $d = 0,1$ one obviously has equality, but what is not clear to me is whether $f_d = 1-\zeta(d+1)^{-1}$ holds for all $d$. -I thought about the analogous problem in $\mathbb{F}_p[x]$ (where one ignores constant factors), but in this case as $d\rightarrow\infty$ one has $f_d\rightarrow 1$. If this behavior carries over to $\mathbb{Z}[x]$ then as $d$ grows large, almost every polynomial of degree $d$ factors which seems absurd; therefore looking at this question in $\mathbb{F}_p[x]$ doesn't seem to help much. -So, letting $f_d = 1-\zeta(d+1)^{-1}+\varepsilon(d)$, I am wondering if the correction term $\varepsilon$ equals zero or, if not, what is $\varepsilon(d)$ and how would one derive it? - -REPLY [5 votes]: Actually, a lot more is true: a random polynomial has Galois group the full symmetric group (a result of van der Waerden) and this is true under various restrictions, and with various error terms. For a summary and some related results, see -Igor Rivin, -Walks on groups, counting reducible matrices, polynomials, and surface and free group automorphisms. Duke Math. J. 142 (2008), no. 2, 353–379. MR2401624 -EDIT Concerning the result mentioned in @quid's answer: experiment shows that this is not quite sharp (at least for monic polynomials) -- the truth seems to be that the fraction of reducible polynomials is asymptotic to $1/t.$<|endoftext|> -TITLE: Is being a coequalizer a target-local property in schemes? (answered: no, and no) -QUESTION [5 upvotes]: This question is aimed at a better understanding of GIT's "categorical quotients", which are defined as coequalizers of group actions $G\times X\rightrightarrows X$ in the category of schemes. See also Anton's currently unanswered question about surjectivity of coequalizers, also answered by Laurent Moret-Bailly. -Suppose $f,g:W\rightrightarrows X$ and $h:X\to Y$ are scheme maps such that $hf=hg$. Let $Y_i$ be a Zariksi cover of $Y$, and let $X_i$ and $W_i$ be their pullbacks to $Y_i$ (i.e. the preimages of the open sets $Y_i$). - - (a) local to global: Is it true that if $W_i\rightrightarrows X_i\to Y_i$ is a coequalizer in the category of schemes for every $i$, then $Y$ is a coequalizer in schemes? - (b) global to local: How about the converse? - -Summary of answer by Laurent Moret-Bailly: -(a) local to global: answer is no, but yes if the maps on intersections $h_{ij}:X_{ij}\to Y_{ij}$ are epic (for example if $h$ is schematically surjective, or just universally epic). -(b) global to local: answer is simply no. -Remarks -1) The analogous statements (a) and (b) for coequalizers in the category of locally ringed spaces are true, which can be seen from the construction of coequalizers in LRS (coequalize the topological spaces, and take rings of invariants). -2) The analogous statements for coequalizers in the category of affine schemes is true: That $C\to B\rightrightarrows A$ is an equalizer is equivalent to the exactness of the $C$-module sequence $0\to C \to B \stackrel{f-g}{\to} A \to 0$, which can be checked in the localizations at prime (or maximal) ideals of $C$. -3) The analogous statements for good geometric quotients of schemes is true. That is, working in Schemes/$S$, if we take $W=G\times_S X$, then $X\to Y$ is a good geometric quotient iff $Y_i$ is a good geometric quotient of $W_i\rightrightarrows X_i$ for all $i$. -4) The analogous statements for equalizers of schemes is true, because fibred products can be checked/constructed on open covers, as is essentially proved in Hartshorne chapter II.3. In fact in any category, pulling back along a morphism preserves all limits, but not colimits, and in particular not coequalizers. -5) If $W=Spec(A),X=Spec(B)$ and $Y$ is their scheme coequalizer, then $Y$ is usually not affine (e.g. when gluing along opens), but $Spec(\cal{O}_Y(Y))$ is the coequalizer in the category of affine schemes. That is, $\cal{O}_Y(Y)$ is canonically isomorphic to the equalizer $C$ of $f^\sharp, g^\sharp:B\rightrightarrows A$ in rings, whose underlying set is the equalizer in sets. -6) If in (5) $B$ is a local ring, then $Y$ is affine, $Y=Spec(C)$, $C$ is local, and $C\to A$ is a local map. - -REPLY [7 votes]: Let me start with a remark [EDITED for clarity after Andrew's comments]. Given $h:X\to Y$, the following are equivalent: -(1) $h$ is the coequalizer of some $W\rightrightarrows X$, -(2) $h$ is the coequalizer of $X\times_Y X\rightrightarrows X$. -In other words, being a coequalizer is equivalent to being an effective epimorphism (This works in any category with fiber products). -Back to the questions. Question (b) asks whether if $h$ is a coequalizer, then its restriction $h^{-1}(V)\to V$ also is, for each open $V\subset Y$. Let me recall the example I gave to answer this question, which provides a counterexample where $h^{-1}(V)$ is empty (and $V$ isn't): take $Y=\mathrm{Spec}\,k[[t]]$ ($k$ a field), $X=$ the disjoint sum of all subschemes $\mathrm{Spec}\,(k[[t]]/(t^n))$ ($n\geq1$), $V=$ generic point of $Y$. -For question (a), assume each $h_i:X_i\to Y_i$ is a coequalizer and let $s:X\to S$ be a morphism such that $sf=sg$. Then for each $i$, the restriction of $s$ to $X_i$ descends uniquely to $t_i:Y_i\to S$. The question is whether $t_i$ and $t_j$ coincide on $Y_i\cap Y_j$. Composing them with (the restriction of) $f$ (or $g$) gives the same result, hence: -$\bullet$ gluing is automatic (and we get a positive answer) if we know that for each open $V\subset Y$, the restriction $h^{-1}(V)\to V$ is an epimorphism of schemes; -$\bullet$ but the above example shows that this is not true in general, and in fact we get a (nonseparated) counterexample to the question by taking two copies $X_i\to Y_i$ ($i=1,2$) of that example and putting $X=X_1\coprod X_2$, $Y=$ gluing of $Y_1$ and $Y_2$ along the generic points: here the coequalizer of $X\times_Y X\rightrightarrows X$ is $Y_1\coprod Y_2$.<|endoftext|> -TITLE: Describing the kernel of the exponential map as a homology group -QUESTION [5 upvotes]: I am reading Deligne: Hodge III, and am puzzled by a certain statement in section 10. If anyone could give a reference or a hint for how to prove this, I would be grateful. Maybe it is obvious and I just don't see why. -We consider an extension $G$ of an abelian variety by a torus. Then Deligne claims that the kernel of the exponential map $Lie(G) \to G$ can be identified with $H_1(G, \mathbb{Z} )$. Why is this true? - -REPLY [16 votes]: The exponential map realizes Lie(G) as the universal covering space of G, and the kernel is group of covering transformations. Thus the kernel is $\pi_1(G)$, which, being commutative in this case, equals $H_1(G,Z)$. (I assume we are over the complex numbers.)<|endoftext|> -TITLE: Occurrences of (co)homology in other disciplines and/or nature -QUESTION [140 upvotes]: I am curious if the setup for (co)homology theory appears outside the realm of pure mathematics. The idea of a family of groups linked by a series of arrows such that the composition of consecutive arrows is zero seems like a fairly general notion, but I have not come across it in fields like biology, economics, etc. Are there examples of non-trivial (co)homology appearing outside of pure mathematics? -I think Hatcher has a couple illustrations of homology in his textbook involving electric circuits. This is the type of thing I'm looking for, but it still feels like topology since it is about closed loops. Since the relation $d^2=0$ seems so simple to state, I would imagine this setup to be ubiquitous. Is it? And if not, why is it so special to topology and related fields? - -REPLY [9 votes]: To add a touch of very belated whimsy, cohomology has manifestations in art. Here are two: - -A Penrose triangle (an "impossible figure") can be viewed as a Čech $1$-cycle associated to an open covering of the image plane, taking values in the sheaf of positive functions under pointwise multiplication. -In more detail, place your eye at the origin of Cartesian space, let $P$ be a region in some plane (the screen) not through the origin, and let $C \simeq P \times (0, \infty)$ (for cone) be the union of open rays from your eye to a point of $P$. -A spatial scene in $C$ is rendered in the plane region $P$ by radial projection, discarding depth (distance) information. Recovering a spatial scene from its rendering on paper amounts to assigning a depth to each point of $P$, i.e., to choosing a section of the projection $C \to P$. -Roger Penrose's three-page article On the Cohomology of Impossible Figures describes in detail (copiously illustrated with Penrose's inimitable drawings) how a two-dimensional picture can be "locally consistent" (each sufficiently small piece is the projection of a visually-plausible spatial object) yet "globally inconsistent" (the entire picture has no visually-plausible interpretation as a projection of a spatial object). -For example, each corner of a Penrose triangle has a standard interpretation as a plane projection of an L-shaped object with square cross-section, but the resulting local depth data cannot be merged consistently in a way conforming to visual expectations. -Of course, the depth data can be merged consistently (from a carefully-selected origin!) by circumventing visual expectations. There are at least two "natural" approaches: use curved sides, or break one vertex. -M. C. Escher's Print Gallery is a planar, "self-including" rendition of the complex exponential map, viewed as an infinite cyclic covering of the punctured plane. (Contrary to mathematical convention, traveling clockwise around the center in Escher's print "goes up one level".) I know of no better reference that Hendrik Lenstra's and Bart de Smit's 2003 analysis of the Droste effect.<|endoftext|> -TITLE: contractible manifolds -QUESTION [12 upvotes]: Where can I find the proof of the following fact: If $M$ is a contractible manifold of dimension $n\ge 5$, then the direct product of $M$ and $\mathbb{R}^{n+1}$ is homeomorphic to $R^{2n+1}$ ? - -REPLY [15 votes]: This was proved in the PL-setting in: -McMillan, D. R.; Zeeman, E. C. -On contractible open manifolds. -Proc. Cambridge Philos. Soc. 58 1962 221–224. -From MathReviews: -"An open manifold is defined to mean a non-compact space that is triangulable by a countable complex which is a combinatorial manifold without boundary. The main theorem is that if $M^n$ is a contractible open manifold, then $M^n\times E^2$ is piecewise linearly homeomorphic to $E^{n+2}$. The principal tool used in the proof is the theorem due to Zeeman which implies that if $M^n$ is a contractible open manifold and $X$ is a subcomplex of codimension $\geq 3$, then $X$ lies in an $n$-cell in $M^n$. The results of this paper have been improved by Stallings for $n>3$ [same Proc. 58 (1962), 481--488] to show that $M^n\times E^1$ is piecewise linearly homeomorphic to $E^{n+1}$. Lemma 3 of the present paper is of independent interest and has been used to study cellularity of sets in products." (Reviewed by M. L. Curtis) - -REPLY [15 votes]: Expanding a little on Andreas Thom's answer, the easiest way to prove this is to use Stallings's theorem that says that a high-dimensional contractible PL manifold that is simply connected at infinity is homeomorphic to $\mathbb{R}^n$ (the key point being that if $M$ is a contractible PL manifold, then $M \times \mathbb{R}$ is clearly simply connected at infinity). There is a really beautiful account of this theorem in chapter 10 of Steve Ferry's notes on geometric topology, available here.<|endoftext|> -TITLE: Unwritten Rule Of Writing Own Name -QUESTION [10 upvotes]: In mathematics, there is an unwritten rule that one is not supposed to write his or her own name (using initials or "the author" if necessary) or attach his or her own name to something (e.g., to have something named after you, someone else should first call it that). -A few questions: -First, is this rule, in some form, in fact written down somewhere? If so, where? -Second, what circumstances, if any, is it appropriate to use your own name? -Lastly, does the rule extend to other people referring to the originator of an idea and the idea? For example, is it appropriate to say "X proved the X Theorem."? Or does this make it seem as though (to a small degree) that X broke the unwritten rule? While less explicit, would it be preferred to say "X proved his/her theorem regarding Y on Z."? -I apologize for the "soft" question, but the discussion and answers are certainly appreciated. - -REPLY [3 votes]: Some additional remarks: -All you said are certainly only unwritten rules, not formal ones; just as there is no written rule or law that one has to be modest or polite. (With the possible exception that some journals might have a specific style regarding certain of these matters; however, if this is so a technical editor will take care of it or you will be made aware of it explicitly.) -I would not say 'X proved the X Theorem' but just as it feels redundant to mention X twice; -but either 'X proved the following theorem' or 'the X theorem' are perfectly fine (in the latter case it should already be common to refer to this result as such or you should be in a position to feel comfortable to suggest naming conventions). -One additional point where I think it can make sense to deviate from the not mentioning the own name in a paper is when it would start to be complicated. What I mean is this, say you write all references in the form Authornames [number]. And the current paper is written by A,B,C who need to quote a paper by A,E,F and another one by C,G,H. Then avoiding the present authors name by saying 'E,F and the first named author [5]' and 'G,H and the third named author [7]' seems sufficiently inconvenient to warrant an exception.<|endoftext|> -TITLE: What is the stalk of a stack? -QUESTION [8 upvotes]: When we study sheaves of sets (on a space X or a site C) we are often interested in the stalks of the sheaf (at either a point $p:1\to X$ or a left exact, cover-preserving functor $a:C\to Sets$). I wonder how one generalizes this notion to stacks? -Since we can always pass from a stack to an equivalent sheaf, one option might be to say the stalk of a stack is (any category equivalent to) the stalk of the associated sheaf of categories. This seems a little roundabout, though? Is there a more intrinsic definition of these categories? - -REPLY [10 votes]: Here is the categorical way to think about the stalk: -Let $X$ be space, and $x \in X$ a point. Regard $x$ as a map $$x:pt \to X$$. -Then $x$ induces a geometric morphism $$x_*:Sh(pt) \to Sh(X)$$ $$Sh(pt) \stackrel{}{\longleftarrow} Sh(X):x^*$$ -(so $x^*$ is left-adjoint to $x_*$ and left-exact). -Now notice that $Sh(pt) \cong Set$. Under this identification, the stalk of $F \in Sh(X)$ at $x$ is the set $x^*\left(F\right)$. -This can be done in the $2$-categorical setting with stacks with no adjustment: -$x$ induces an adjoint pair of $2$-functors -$$x_*:St(pt) \to St(X)$$ $$St(pt) \stackrel{}{\longleftarrow} St(X):x^*$$ -and under the identification $St(pt) \cong Gpd,$ the stalk of $\mathscr{X} \in St(X)$ at $x$ is the groupoid $x^*\left(\mathscr{X}\right)$. - -A way of viewing this in a geometric way is by using the etale-realization of the stack $\mathscr{X}$, (explained in arxiv.org/abs/1011.6070 ) which is a topological stack $$L\left(\mathscr{X}\right)$$ equipped with a local homoemosphism to $X$. The stalk can then be viewed as the fiber of this map over $x$.<|endoftext|> -TITLE: If Spec(A) has a G-fixed point and a dense G-orbit, is Spec(A) a cone? -QUESTION [5 upvotes]: [Edited to include a dense orbit] - -Let $X=Spec(A)$ be a normal affine scheme over an algebraically closed field $k$, with an action of a linearly reductive group $G$. Suppose $x\in X$ is a $G$-invariant $k$-point and that $X$ contains a dense open $G$-orbit. Note that this implies that every $G$-orbit contains $x$ in its closure (the good quotient $Spec(A^G)$ is $Spec(k)$). Must $X$ by $G$-equivariantly isomorphic to a cone inside a representation of $G$? - -The natural cone in the picture is the tangent cone of $X$ at $x$, $\def\m{\mathfrak m}Spec(gr_\m A)$, where $\m$ is the maximal ideal corresponding to $x$ and $gr_\m A = A/\m\oplus \m/\m^2\oplus \m^2/\m^3\oplus \cdots$. This tangent cone is a closed cone inside the tangent space at $x$, $Spec(Sym^*(\m/\m^2))$, which is a representation of $G$. -Since $G$ is linearly reductive, there is a $G$-equivariant isomorphism of vector spaces $A\cong gr_\m A$. The question is whether this can be made into an isomorphism of rings. -Associated graded doesn't have a universal property, which suggests that it should be hard (or impossible) to construct such an isomorphism, but I can't think of a counterexample. -Remark 1: The normality assumption is necessary. Otherwise consider the action of $\mathbb G_m$ on the cuspidal cubic $Spec(k[x^2,x^3])$ given by $t\cdot x^n=t^nx^n$. The tangent space at the fixed point is 2-dimensional, so if the cuspidal cubic were isomorphic to a cone, it would be a cone it $\mathbb A^2$, but all 1-dimensional cones in $\mathbb A^2$ are unions of lines. -Remark 2: Since $X$ contains a dense orbit, a natural place to look for ideas for a proof or counterexample is in literature on spherical varieties. I haven't been able to understand very much of it yet. If $X$ is an affine normal spherical variety with a $G$-invariant $k$-point, must it be a cone? In my situation, $X$ actually contains a dense open copy of $G$. -Remark 3: I should probably also impose the condition that $X$ is reduced, though I don't see why that should make much of a difference. - -REPLY [6 votes]: If you by "cone" mean exactly that $A$ should be isomorpic to -$\mathrm{gr}_{\mathfrak m}A$ it seems that the following is counterexample: Let -$G=\mathbb G_m$, $A=k[x,y,z]/(x^2+y^3+z^5)$ with $tx=t^{15}x$, $ty=t^{10}y$ and -$tz=t^{6}z$ (exponents chosen more or less at random). Then the tangent cone at -the origin (the fixed point) has affine algebra $k[x,y,z]/(x^2)$ and hence is -not isomorphic to $A$. This is just raising your cusp example one dimension so -that it becomes normal. -Addendum: Turning to the modified question let's try for a toric example: $G=\mathbb G_m^2$ so we should look at a monomial subring of $k[x,x^{-1},y,y^{-1}]$ with monoid of monomials saturated (equivalently being the set of integer vectors in a rational cone). Unless I am mistaken $k[xy,x^2y,xy^2,x^3y,xy^3]$ is such a ring (if I have missed some monomial in the saturation it won't matter for my argument). We then have that $x^2y$ and $xy^2$ are elements in $\mathfrak m\setminus\mathfrak m^2$ (this is clear irrespective if I have missed some elements in the saturation) but their product $x^2y\cdot xy^2=(xy)^3\in \mathfrak m^3$ which means that the associated graded is not a domain and hence not isomorphic to $A$. (I think the general condition for being a cone is that the generators of the monoid of monomials should lie on an affine hyperplane which is anyway how I arrived at this example.)<|endoftext|> -TITLE: Decidability in groups -QUESTION [15 upvotes]: This is not my area of research, but I am curious. Let $G=\left< X|R \right>$ be a finitely presented group, where $X$ and $R$ are finite. There are many questions which are undecidable for all such $G$, for example whether $G$ is trivial or whether a particular word is trivial in $G$. Is there any non-trivial question (by trivial I mean that the answer is always yes or always no) which is decidable? For instance, is there a class $S$ (non-empty and not equal to all the finitely presented groups) such you can always decide whether $G$ is in this class? - -REPLY [25 votes]: The problem whether $G$ is perfect, that is $G=[G,G]$ is decidable because you need to abelianize all relations (replace the operation by "+" in every relation) and solve a system of linear equations over $\mathbb Z$. For example, if the defining relations are $xy^{-1}xxy^5x^{-8}=1, x^{-3}y^{-2}xyx^5 = 1 $, then the Abelianization gives $-5x+4y=0, 3x-y=0$. Now you need to check if these two relations "kill" $\mathbb{Z^2}$. That means for some integers $a,b,c,d$ we should have $x = a(-5x+4y)+b(3x-y), - y = c(-5x+4y)+d(3x-y) $. This gives 4 integer equations with four unknowns: $1=-5a+3b, 0=4a-b, 1=4c-d, 0=-5c+3d - $. This systems does not have an integer solution (it implies $7a=1 $), so $G\ne [G,G]$. - -REPLY [12 votes]: Many of the undecidable properties of finitely presented groups are verifiable i.e. if they are true, then they can be proved true. Such properities include trivial, finite, abelian, nilpotent, free, automatic, hyperbolic, isomorphic to some other specified finitely presented group. -As a follow-up question, are there any properties of finitely presented groups that are known to be neither verifiable nor for there negation to be verifiable? Might solvability be such a property?<|endoftext|> -TITLE: Name of a polytope -QUESTION [16 upvotes]: What is the name of the polytope $\Sigma\cap (-\Sigma)$ for $\Sigma$ a $d-$simplex with barycenter at the origin? -In dimension $2$, one gets a hexagon, in dimension $3$ an octahedron (given by the $6$ midpoints of edges), in dimension $4$ -a polytope with 30 vertices (given for example by all permutations of $(0,1,1,-1,-1)$). -(More generally, this polytop has ${2n\choose n}$ vertices in dimension $d=2n-1$ -and $(2n+1){2n\choose n}$ vertices in dimension $d=2n$.) - -REPLY [5 votes]: I do not know of any specific name for this family of polytopes, but their Coxeter diagrams are of a recognizable form. -$\qquad\qquad\qquad\qquad\qquad\qquad$ -In fact, every second one is a hyper-simplex. In general, you can look them up on Wikipedia's list of uniform polytopes. -The intersection $\Sigma \cap(-\Sigma)$ of $d$-dimensional simplices can be constructed as -$$\Big\{\sum_{i=1}^{d+1} \lambda_i e_i \in\Bbb R^{d+1} \;\Big|\; \sum_{i=1}^{d+1}\lambda_i=1,\lambda_i\in\big[0,\frac{2}{d+1}\big]\Big\}.$$ -This is almost the convex hull of the $e_i$, but the coefficients of the convex combinations are more restricted. -With this, it is possible to prove what Vince's suspected about the coordinates of the vertices.<|endoftext|> -TITLE: Cartan subgroups of p-adic groups. -QUESTION [6 upvotes]: Does anyone know about any reference describing the structure of Cartan subgroups in the case of connected p-adic reductive (or let's say semi-simple) groups? -I would like to know how different is this from the real case. -Thanks in advance, -Jim Riel - -REPLY [6 votes]: just stumbled upon this old thread and decided to contribute my 2 cents (though I completely agree with Moshe's answer): J.-L. Waldspurger has a very explicit description of conjugacy classes of semisimple elements in the classical groups, in his Asterisque volume "Integrales orbitales nilpotentes et endoscopie pour les groupes classiques non ramifies", I believe in Section I.7. This is, essentially, Moshe's examples 1 --3 but done in a systematic way for every classical group. One can get a parametrization of tori from this description of semisimple conjugacy classes by looking at centralizers.<|endoftext|> -TITLE: Factorization of a real matrix into Hermitian x Hermitian. Is it stable ? -QUESTION [25 upvotes]: It is known (see Theorem 4.1.7 in R. Horn & C. Johnson) that every matrix $A\in M_n(\mathbb R)$ (real entries) can be written as the product $HK$ of two Hermitian matrices (complex entries). Of course, the pair $(H,K)$ is far from being unique, because the real dimension of $\mathbb H_n\times\mathbb H_n$ is $2n^2$, much larger than $n^2=\dim M_n(\mathbb R)$. The question is whether this factorization can be done in a stable manner: - -Does there exist a finite constant $c_n$ such that, for every $A\in M_n(\mathbb R)$, the pair $(H,K)\in\mathbb H_n\times\mathbb H_n$ can be chosen so that $A=HK$ and $\|H\|\cdot\|K\|\le c_n\|A\|$ ? - -Of course the answer does not depend on the choice of the matrix norm. Only the constant does. -Edit. I must mention, to my shame, that at the beginning of Chapter 6 of my book on matrices (Springer-Verlag, GTM 216), I pretend that $\mathbb H_n\times\mathbb H_n$ equals $M_n(\mathbb C)$; without proof of course. Thanks to Jean Gallier, who pointed it out. - -REPLY [17 votes]: Surprisingly (at least to me) the answer is no when $n\ge 3$. This was proved by Yves Benoist and me after I mentioned the problem in a talk at MSRI and Yves came up with a great idea. -It is enough to show that there is no uniform bound when $n=3$. -Here is an elementary argument that involves little computation. I don't pay attention to getting exact constants. -For small positive $\epsilon$ define -\begin{equation} - x_1 = e_1 \quad \quad - x_2 = e_1 + \epsilon e_2 \quad \quad - x_3 = e_1 + \epsilon e_3 -\end{equation}` -For certain distinct non zero real $\lambda_i$, which will depend on $\epsilon$, let $T\in M_n(\mathbb{R})$ be defined by $Tx_i = \lambda_i x_i$. The dual basis to $(x_i)$ is defined by -\begin{equation} - f_1=e_1 - {1\over\epsilon} (e_2 + e_3) \quad \quad - f_2={1\over\epsilon} e_2 \quad \quad - f_3={1\over\epsilon} e_3 -\end{equation} -So the transpose and adjoint of $T$ is defined by $T^*f_i = \lambda_i f_i$. If $S$ implements a similarity between $T$ and $T^*$, it is more or less clear that $\|S\|\cdot \|S^{-1}\| \to \infty$ as $\epsilon \to 0$ uniformly over all permissible choices of $\lambda_i$. (For me the easy way to see this is to semi-normalize the $f_i$ as -\begin{equation} - \tilde{f}_1=\epsilon e_1 - (e_2 + e_3) \quad \quad - \tilde{f}_2= e_2 \quad \quad - \tilde{f}_3= e_3 -\end{equation} -The similarity $S$ must be given by $Sx_i = a_i \tilde{f}_i$ for some non zero $a_i$ since the $\lambda_i$ are distinct, and if $\|S\| \vee \|S^{-1}\| \le (1/3) C$, then all $|a_i|$ and all $1/|a_i|$ are less than $C$. But $\|x_2-x_3\|=\sqrt{2}\epsilon$ and -$\|a_2 \tilde{f}_2 - a_3 \tilde{f}_3 \| =\sqrt{|a_2|^2+|a_3|^2} > \sqrt{2} /C$, which forces -$\|S\| >1/(C\epsilon)$.) -Next a trivial but (it seems) important point. For fixed $\epsilon$, you can choose the $\lambda_i$ close enough to one so that $T$ is as close to the identity as you want. So we can choose $\lambda_i$ so that $\|T\|=1$ and $\|T^{-1}\|< 1+\epsilon$; denote such a $T$ by $T_\epsilon$. -Write $T_\epsilon = H_\epsilon K_\epsilon$ with $H_\epsilon$, $K_\epsilon$ (complex) Hermitian and $\|H_\epsilon\|=1$. We want to see that $\|K_\epsilon\| \to \infty$ as $\epsilon \to 0$. Notice that $H_\epsilon$ and $K_\epsilon$ are non singular since no $\lambda_i$ is zero. So we have - $H_\epsilon^{-1}T_\epsilon H_\epsilon = K_\epsilon H_\epsilon = T_\epsilon^*$ and hence, by the first part of the proof, $\|H_\epsilon^{-1}\| \to \infty$ as $\epsilon \to 0$. But $H_\epsilon^{-1} = K_\epsilon T_\epsilon^{-1}$, so - $$\|H_\epsilon^{-1}\| \le \|K_\epsilon\| \|T_\epsilon^{-1}\| \le (1+\epsilon) \|K_\epsilon\|. - $$<|endoftext|> -TITLE: Cones, monoids, and the space of (very) ample divisors -QUESTION [7 upvotes]: An interesting and useful tool to study a projective variety is its ample cone. Understanding the structure of this cone reveals information about the variety, and it is an isomorphism-invariant so these cones sometimes can be used to distinguish non-isomorphic varieties. -Another interesting class of divisors are the very ample divisors. -Each ample divisor has a multiple that is very ample, so there is no such thing as the "cone" of very ample divisors: by tensoring with $\mathbb{R}$ or $\mathbb{Q}$ and looking only at rays, we lose the information of when an ample divisor becomes very ample. Thus one must look at the monoid of very ample divisors. Though it seems to be difficult in general, there are some criteria in some cases for what power of an ample line bundle one must take for it to become very ample. However, I am curious about the following: -Are there applications to studying the structure of the very ample monoid, similar to those of the ample cone appearing in the theory of surfaces and higher-dimensional geometry? -A more specific question is: -What is an example of two varieties such that their Picard groups are isomorphic and their ample cones coincide, but their very ample monoids do not? -And also: -For a variety to be a Mori dream space imposes rather strict conditions on the cones associated to the variety, e.g., polyhedrality of the effective and nef cones. Does it impose further restrictions on the monoids of basepoint-free divisors, very ample divisors, etc., beyond the restrictions imposed on the cones themselves? -Basically, I understand what information is lost about a specific divisor by passing from it to the ray it spans, but I would like to know examples of the information that is lost about the space of divisors in this process. This is admittedly somewhat vague, but any suggestions in this direction would be appreciated. - -REPLY [7 votes]: Regarding your second question, take $A$ and $X$, where $A$ is a general principally polarized Abelian surface and $X$ is a very general surface of degree $\geq 4$ in $\mathbb{P}^3$. -Then the Néron-Severi of both these varieties is cyclic of rank $1$, generated by the class $\Theta$ of a Theta divisor in the former case and by the class $H$ of a hyperplane section in the latter case. -Therefore $\textrm{NS}(A) \cong \textrm{NS}(X) \cong \mathbb{Z}$. -However, the very ample monoids are -$M_A:=\{n \Theta | n \geq 3 \}\cup \{0\} \quad \textrm{and} \quad M_X:=\{n H | n \geq 1 \} \cup \{0 \}$, -respectively. -These are not isomorphic monoids since $M_A$ is not cyclic whereas $M_X$ is cyclic. -EDIT. In order to give an example with the Picard groups instead of the Néron-Severi group, take a general $K3$ surface $Y$ which is a double cover of $\mathbb{P}^2$ branched along a sextic curve. Therefore the Picard group of $Y$ is generated by $D$, where $D$ is the pull-back in $Y$ of a line $\ell \subset \mathbb{P}^2$. Therefore -$\textrm{Pic}(Y) \cong \textrm{Pic}(X) \cong \mathbb{Z}$, -and consequently the ample cones are also isomorphic. However, the very ample monoid of $Y$ is -$M_Y:=\{n D | n \geq 2 \}\cup \{0\}$, -which is not cyclic, hence not isomorphic to $M_X$.<|endoftext|> -TITLE: Density of numbers not divisible by a large prime power -QUESTION [5 upvotes]: Although the answer to my question is probably implicit in the answers to the question asked here: Density of numbers having large prime divisors (formalizing heuristic probability argument), I can't extract it. -Problem: find decent bounds on the number of positive integers $n$, such that, for all primes $p$ dividing $n$, if $p^k$ exactly divides $n$, then $n > p^{k+1}$. -My idea for a first upper bound: if $n$ is divisible by a prime larger than $n^{\frac{1}{2}}$, it is immediately exluded that $n$ is of the above form, so the density can never be larger than $1 - \log{2}$ -My idea for a first lower bound: if $n$ has two prime divisors between $n^{\frac{1}{3}}$ and $n^{\frac{1}{2}}$, then $n$ is of the above form. But I don't know the density of these numbers. -I am probably very happy with a (reference to) a proof/theorem that implies that we have a positive lower density, but asymptotics would be great. -EDIT (after the first response of GH, for which I'm thankful!): assume $n$ lies in some moduloclass, say $a \pmod{b}$. Can we still show a positive lower density, whatever the values of $a$ and $b$? - -REPLY [4 votes]: Your numbers have positive lower density. To see this let $z$ be a positive integer to be fixed later, and denote -$$ c:=\prod_{p \leq z}(1-1/p). $$ -Consider all square-free integers $x < n \leq 2x$ which are composed of primes $z < p \leq \sqrt{x}$. Note that these numbers satisfy the requirements. Their number, by a crude estimate, is at least -$$ cx+O(1)-\sum_{\sqrt{x} < p \leq 2x}(cx/p+O(1)) - \sum_{z < p \leq \sqrt{2x}}(cx/p^2 +O(1)), $$ -which is at least -$$ c(1-\log 2-1/z+o(1))x. $$ -That is, the lower density is at least -$$ c(1-\log 2-1/z)/2.$$ -For $z:=5$ the left hand side exceeds $0.0142$, while for $z:=17$ it exceeds $0.0223$. -EDIT: I improved slightly my original argument.<|endoftext|> -TITLE: Classifying representation through extensions -QUESTION [5 upvotes]: Let $G$ be a group (you preferred type: finite, compact, ...): -Mackey has a machinery to classify all irreducible representations of a locally compact group $G$ in terms surjective group homomorphism: -$$ \sigma : G \rightarrow N$$ -by some irreducible representations of subgroups of $N$ and some projective representations $G/ kern(\sigma)$, where $kern(\sigma)$ is a so called type-1 subgroup (i.e. the von Neumann group algebra is a direct integral type $1$ factors), e.g. take a finite, abelian, compact or amenable group. -Q1: Is there a nice reference for a condensed treatment of these results? - -REPLY [2 votes]: Re: Q1, until real answers appear. -There is a survey article: "Projective representations and the Mackey obstruction - a survey", Contemporary Mathematics, v. 449 (2008), pp. 345-378. -Besides this article, there are a lot of interesting tidbits in the above-mentioned volume. (Group Representations, Ergodic Theory, and Mathematical Physics: A Tribute to George W. Mackey) One of the papers in here (the one by Kirillov) contains a simplified proof of the imprimitivity theorem in the Lie group case. -There is also an article by V.S. Varadarajan (posted on his website) about the work of Mackey, but the UCLA website seems to be misbehaving right now, so you'll have to check this later. -BTW: There is another closely related question this morning. Do you guys know each other?<|endoftext|> -TITLE: What is a Lagrangian submanifold intuitively? -QUESTION [68 upvotes]: What are good ways to think about Lagrangian submanifolds? -Why should one care about them? -More generally: same questions about (co)isotropic ones. -Answers from a classical mechanics point of view would be especially welcome. - -REPLY [3 votes]: As mentioned in previous answers - Lagrangian submanifolds encode vast amount of information on the symplectic geometry of the ambient symplectic manifold $M$ they live in (like sheaves on an algebraic manifold) - so studying them is really an essential feature of symplectic geometry. -As of themselves, Lagrangian submanifolds also admit, for instance, some special intersection properties - which are typically proved by pseudo-holomorphic techniques or the more "modern machinery" of Floer homology. If you are interested in the subject a great place to start is Biran's Lagrangian non-intersections (2005). -But, even if your interest lies in algebraic geometry, in general, Lagrangian submanifolds can still prove very interesting to you - due to the homological mirror symmetry conjectures of Kontsevich - which suggest that sheaves (or more generally, elements of $\mathcal{D}^b(X)$) on one manifold W (e.g Calabi-Yau) should literally -translate to Lagrangian submanifolds (or elements of Fukaya category) in another manifold $M$ - via a mirror functor. This led to some fascinating developments in both symplectic and algebraic geometry.<|endoftext|> -TITLE: Partitioning a polygon into convex parts -QUESTION [18 upvotes]: I'm looking for an algorithm to partition any simple closed polygon into convex sub-polygons--preferably as few as possible. -I know almost nothing about this subject, so I've been searching on Google Scholar and various computational geometry books, and I see a variety of different methods, some of which are extremely complicated (and meant to apply to non-simple polygons). I'm hoping there's a standard algorithm for this, with a clear explanation, but I don't know where to find it. -Can anyone point me to a source with a clear explanation of how to do this? - -REPLY [2 votes]: I don't know, whether this is an already known algorithm and I also don't know how optimal the resulting partioning will be, but its description and implementation are simple: - -identify the set of inflex points, i.e. where a right-turn is made when going around the polygon counter clockwise -calculate the nested convex hull of the inflex point set -cut the polygon along the edges of the nested convex hull, that are inside the polygon<|endoftext|> -TITLE: Density of fields with a decomposition condition -QUESTION [5 upvotes]: Let $p$ be a fixed prime number. Roughly speaking, I am interested in the following ratio -$$ -\frac{ |\{\text{ all CM number fields of degree }2g}|} -{ |\{\text{CM fields of degree 2g, such that p splits completely in K}\}|} -$$ -A possible definition could be the following: -let $d_{K}$ be the discriminant of K, -then we can define this ratio as -$$ \lim_{d \to \infty} \frac -{|\{ \text{all CM fields of degree 2g and}\ d_{K} \le d\}|} -{|\{ \text{CM fields of degree 2g such that p splits completely and}\ - d_{K} \le d \}|}.$$ -Was it studied by anyone? -I would appreciate any reference. - -REPLY [3 votes]: Let $K$ be a CM field with maximal real subfield $k$. Most of the time the normal closure -of $k$ will have the symmetric group $S_g$ as Galois group; thus the number of fields -$k$ in which $p$ splits completely has density $1/g!$ among all of them by density theorems due to Kronecker, Frobenius and/or Chebotarev. In about half of the cases, $p$ will also split in $K$, giving $1/(2 \cdot g!)$ as a rough estimate. -It is not necessarily true, however, that the density of fields with the symmetric Galois group is $1$ when the fields are ordered by discriminant. It is my impression that the -corresponding problems are still open for $g > 5$, although there might be conjectural densities in the articles by Malle etc. -Thus you should be able to get a definitve answer for $g \le 4$ using known results -(check the articles by Cohen et al. on the distribution of Galois groups), and -perhaps $g = 5$ using recent advances e.g. by Bhargava.<|endoftext|> -TITLE: Chern character of the index bundle for a family of Dirac operators -QUESTION [6 upvotes]: Suppose we have a family of compact oriented even dimensional spin manifolds $\{Y_x\}$ parameterized by a compact even dimensional manifold $X$. The $Y_x$'s -are all diffeomorphic to some $Y$, of dimension $n$, and fit together to form a fiber bundle $\pi : Z \rightarrow X$ with fiber $Y_x=\pi ^{-1}(x)$. $TZ$ has the subbundle -$V:=\text{ker }\pi_*$ which is tangent to the fibers. -There may be a family of coefficient bundles also and we obtain a family of twisted Dirac operators $D_x:\Gamma(S^+_x\otimes E_x)\rightarrow \Gamma (S^-_x\otimes E_x)$. -The index of the family gives rise to an element $\text{ind} D \in K(X)$, which is the virtual vector bundle $[\text{ker } D_x]-[\text{coker }D_x]$ when the dimension of both spaces are constant. -Finally, there is a map $\text{H} ^{*}(Z,\mathbb{R})\rightarrow \text{H} ^{*-n}(X,\mathbb{R})$ known as the Gysin homomorphism or integration over the fibers map. We'll use the latter terminology writing the map $\int_Y$ and regarding cohomology classes as living in de Rham cohomology. -The Atiyah-Singer index theorem gives -$$\text{ch }(\text{ind } D)= \int _Y \hat A (V) \text{ch}(E)$$ - - -What general results exist regarding the components of the Chern character of the index bundle, or equivalently the results of the integration over the fibers map, for twisted Dirac operators? - -To illustrate, an immediate answer is that the zero cohomology (virtual rank) is the index of the Dirac operator on $Y$. A more interesting answer is that in some cases that might be all one obtains: -it is a result of Borel-Hirzebruch that the signature is strictly multiplicative in all bundles where $\pi_1$ of the base acts trivially on the rational cohomology of the fibers. -The signature is the index of a certain twisted Dirac operator. If we have a family of these operators such that $Z\rightarrow X$ satisfies the condition involving the fundamental group, -then the strict multiplicativity gives $\text{ch}(\text{ind }D)=\int_Y \hat A (V)ch(E)=\text{sign }(Y)$. A priori one could expect higher degree cohomology classes. It seems interesting that these vanish. -If the question is too vague or broad, I would be happy knowing - -Are there any instances in which there are known relations between the Chern character of the index bundle and the Chern classes of $X$? - -REPLY [4 votes]: Your question is not so clearly stated; so I take the opportunity to interpret it and write a little essay. -As far as I understand your question, you found out that the multiplicativity result by "Borel-Hirzebruch" (I think it is actually due to Hirzebruch-Chern-Serre) implies that the higher Chern character of the index bundle of the signature operator are zero and now you want to see a more conceptual explanation. You should have a look -into Atiyah's "The signature of fibre bundles". -The following is definitly in the spirit of that paper, though I did not find the statement there. -Theorem: "Let $Z \to X$ be an oriented smooth fibre bundle with fibre $Y$, closed, of dimension $4k$. -Pick a fibrewise Riemann metric and let $D$ be the fibrewise signature operator. Assume that $\pi_1 (X)$ acts trivially on $H^{2k}(Y; \mathbb{R})$. Then -the index bundle of the signature operator $ind(D)$ is trivial of virtual rank $sign(Y)$. Moreover, -if the image of $\pi_1 (X)$ in $Gl(H^{2k} (Y); \mathbb{R})$ is finite, then the components of $ch(ind(D))$ in positive degree are zero." -Proof: - -A linear algebra fact. Given a finite-dimensional real vector space $V$ with a nondegenerate symmetric bilinear form $b$; of signature $(p,q)$. -Consider the space $Q(b)$ of all pairs $(V_{+},V_{-})$, such that $\pm b|_{V_{\pm}} $ is positive definite and $W_{-} \oplus W_{+}=V$. -$Q$ is a subspace of the product $Gr_p (V) \times Gr_q(V)$ of two Grassmannians. $Q(b)$ is diffeomorphic to the homogeneous space $O(p,q)/(O(p)\times O(q))$, -and $O(p) \times O(q) \subset O(p,q)$ is a -maximal compact subgroup. Therefore $Q$ is contractible (there should be a more direct proof, though). -Assume that the $\pi_1$-action on $V:=H^{2k} (Z_x; \mathbb{R})$ is trivial. There is the intersection form $b$ on $V$; pick a decomposition -$V=W_{+} \oplus W_{-}$ as before. -Now consider the bundle $E:=H^{2k} (Z/X) \to X$; the fibre over $x$ is the -cohomology $H^{2k} (Y_x; \mathbb{R})$. -Because the $\pi_1$-action is trivial, the bundle is -trivial and the splitting above gives a splitting $E=F_{+} \oplus F_{-}$ into two trivial subbundles. -Now pick a Riemann metric on the fibres. The Hodge theorem identifies $E$ with the bundle of harmonic $2k$-forms and the -Hodge star $\ast$ is an involution and it splits $E= E_{+} \oplus E_{-}$ into the sum of eigenspaces. -More or less by definition, $ind (D) = [E_{+}]-[E_{-}]$. Now I claim that $E_{\pm}\cong F_{\pm}$. This is because both decompositions can be viewed as sections to -the bundle $X \times Q(b)$ and since $Q(b)$ is contractible, they are homotopic. -If the image of the monodromy is a finite group, say $\mu: \pi_1 (X) \to G$, you pass to the finite cover $\tilde{X} \to X$ corresponsing to the kernel of $\mu$. -The pullback of $Z$ to $\tilde{X}$ has trivial monodromy and so the signature index bundle is trivial. But the induced homomorphism $H^{\ast} (X; \mathbb{Q}) \to H^{\ast}(\tilde{X}, \mathbb{Q})$ is injective, which allows to reduce the argument to the case of a trivial action. QED - -After I have written all this, I realize that the main point is that the bundle $E$ has structural group $O(p,q)$ and it is flat as such a bundle. -It allows a reduction of the structural group to $O(p) \times O(q)$ (by the contractibility of $Q(b)$), but not as a flat bundle. If you find a flat reduction, then $ch(Ind(D))=0$ (in positive degrees) by Chern-Weil theory. The index theorem is not really relevant here; it identifies $ch(ind(D))$ with the fibre integral of the $L$-class of the vertical tangent bundle.<|endoftext|> -TITLE: Are extensions of nuclear Fréchet spaces nuclear? -QUESTION [15 upvotes]: Consider the category of Fréchet spaces, the morphisms being -continuous linear maps with closed image. Suppose that we -have a short exact sequence in that category: -$0 \rightarrow V_1 \rightarrow V_2 \rightarrow V_3 \rightarrow 0$. -Of course $V_1$ and $V_3$ are nuclear if $V_2$ is. I recently asked -myself if the converse might be true. I haven't found anything useful -in the standard literature (Treves, Schaefer) but that might -be just me being too ignorant to see the obvious. I'm grateful if someone could -shed some light on this. -Cheers, -Ralf - -REPLY [8 votes]: You question was answered even for locally convex spaces by S. Dierolf and W. Roelcke -in proposition 3.8 of the article "On the three-space-problem for topological vector -spaces". Collect. Math. 32, p. 13-35 (1981). -The splitting theory for Frechet spaces is nowadays very well understood by results of D. Vogt and others. This can be found in my "Derived functors in functional analysis", Springer Lecture Notes in Mathematics 1810 (2003).<|endoftext|> -TITLE: Overview of the interplay of Harmonic Analysis and Number Theory -QUESTION [25 upvotes]: I'm kind of disappointed that the question here was never sharpened. -The Laplacian $\Delta$ on the upper half-plane is $-y^{2}(\partial^{2}/\partial x^{2}+\partial^{2}/\partial y^{2}))$. Suppose $D$ is the fundamental domain of, say, a congruence subgroup $\Gamma$ of $Sl_{2}(\mathbb{Z})$. Eigenfunctions of the discrete spectrum of $\Delta$ are real analytic solutions to $\Delta (\Psi)=\lambda \Psi$ that are $\Gamma$-equivariant functions in $L^{2}(D, dz)$, where $dz$ is the Poincare measure on the upper half-plane. These eigenfunctions evidently carry quite a bit of number theoretic information. Frankly, this point of view on number theory sounds incredibly interesting... - -Question: Would someone please suggest a readable introductory account that tells this story? - -(I imagine that answers will include the words Harish-Chandra, Langlands, etc...) -Also, if experts are inclined to write a short overview as an answer, that would also be much appreciated. - -REPLY [5 votes]: I think it is also fair to say that things like hyperbolic $n$-spaces (and other symmetric spaces), and arithmetic quotients of them, are primordial, one-of-a-kind (well, not quite) objects. The opposite of "generic" mathematical objects. Partly because of the abruptly greater technical complexity in discussing the harmonic analysis on them, they are much less familiar than Euclidean spaces or their familiar quotients, circles and products thereof. -Apart from Langlands' program and direct, intentional discussion of $L$-functions, I find it provocative that the basic harmonic analysis of $SL_2(\mathbb Z)\backslash H$ is not merely far subtler than that of $\mathbb R^2$ or $\mathbb R^2/\mathbb Z^2$, but that those subtleties are directly related to profound unsolved problems. As a well-known example, while we easily understand the sup norm of exponentials in the harmonic analysis on the real line, sharp estimates on pointwise behavior of eigenfunctions for the Laplacian on the upper half-plane give Lindelof: e.g., the value of the Eisenstein series $E_s$ at $z=i$ is the zeta of the Gaussian integers (divided by $\zeta(2s)$). -Continuing, unlike the fact that the product of two exponentials is an exponential (that is, the tensor product of two one-dimensional irreducibles is still irreducible), decomposition of tensor products of waveforms, or of the repns they generate, is very tricky, and, again, is connected to serious outstanding problems. (Iwaniec' book mentions such examples.) -That is, apart from "big conjectures" about automorphic forms and L-functions by themselves, even-more-primitive number-theoretic things just arise unbidden when we try to do innocent, ordinary things that would be trivial in Euclidean space.<|endoftext|> -TITLE: Why don't ideals and quotients work well for categories? -QUESTION [39 upvotes]: Ideals are intimately related to quotients and congruence relations. They clearly play a very important role in ring theory and order theory. So do normal subgroups in group theory. (Enriched) category theory could be regarded as a common generalization of all these settings. Why is it that such important structures don't work well for categories? -I am aware that there is a categorical notion of congruence relation. However, this doesn't seem to take the spirit of multiple objects to heart: all it does is keep the same objects and relate morphisms within homsets. For one thing, the accompanying notion of quotient category doesn't correspond to coequalizers in $\mathbf{Cat}$ (of which there are many more). -It is not even clear how to define an ideal of a category. To allow for proper ideals, it probably shouldn't simply be a subcategory. Naively one thinks of a subset $I(X,Y)$ of each $\mathrm{Hom}(X,Y)$ that is invariant under composition with arbitrary morphisms, or just of subsets $I(X)$ of each $\mathrm{Hom}(X,X)$, or of $I(X)$ just for some objects; but this doesn't really take objects into account. Thinking of an appropriate definition is even more perplexing for higher categories. - -Question: are there related notions of ideal and quotient for categories that have interesting consequences but are not trivial on the level of objects? - -It is left open what roles left (postcomposition) or right (precomposition) ideals should play; a related question is if there is a notion of commutativity for categories with interesting consequences. -A convincing explanation why one shouldn't consider such questions would also be a good answer. - -REPLY [5 votes]: I don't think I saw anyone give this construction above and if so I apologize for the repetition. -One could consider the orbit category by an endofunctor. The example I have in mind is the cluster category $\mathcal{C}$ important in the categorification of cluster algebras. One starts with the category of finite-dimensional representations $mod_\mathbb{C}~Q$ of a Dynkin quiver $Q$ over the field $\mathbb{C}$ of complex numbers. Then the cluster category is the orbit category of the bounded derived category $D^b(mod_\mathbb{C}~Q)$ by the endofunctor $F=\tau^{-1}[1]$ where $\tau$ is the Auslander-Reiten translation of $mod_\mathbb{C}~Q$ and $[1]$ is the shift functor. The paper at http://de.arxiv.org/pdf/math.RT/0402054.pdf was one of the first on cluster categories. -I guess to answer your main question: the objects in these categories are very different. For example $D^b(mod_\mathbb{C}~Q)$ has infinitely many isomorphism classes of indecomposable representations, for example the shifts of indecomposable objects in $mod_\mathbb{C}~Q$. However $\mathcal{C}$ has only finitely many indecomposable objects, namely the indecomposable representations of $Q$ considered as complexes concentrated in degree 0 and the shifts $P[1]$ of the projective representations of $Q$. -I suppose this construction has a slightly different flavor than the quotient of a ring by an ideal.<|endoftext|> -TITLE: When is non amenablity witnessed by a single non measurable set? -QUESTION [15 upvotes]: Suppose $G$ is a finitely generated discrete group and that there is a subset $E$ of $G$ such that if -$\mu$ is a finitely additive probability measure on $G$, then there is a -$g$ in $G$ such that $\mu(E \cdot g) \ne \mu(E)$. -Certainly $G$ is non amenable. -Can more be said about $G$? -Must $G$ contain $\mathbb{F}_2$? -It should be noted that the above situation can happen: -let $E$ be all elements of $\mathbb{F}_2$ which ``begin'' with $a$ or $a^{-1}$. -Then both $E$ and its complement have infinitely many disjoint translates (by powers of $b$ and $a$, respectively). - -REPLY [12 votes]: The answer is that the above is equivalent to non amenability. -Fix a group $(G,*)$. -Since $(G,*)$ is non amenable if and only if every finitely generated subgroup is non amenable, -we may assume that $G$ is finitely generated. -If $\mu$ and $\nu$ are finitely supported probability measures on $G$, -define -$$ -\mu * \nu (Z) = \sum_{x * y \in Z} \mu (\{x\}) \nu (\{y\}) -$$ -Observe that $g * \nu (E) = \nu ( g^{-1} * E)$. -If $S$ is a subset of $S$, let $P(S)$ denote all probability measures on $S$ -(which are identified with probability measures on $G$ which are supported on $S$). -I will identify $G$ with the point masses in $P(G)$. -If $A$ and $B$ are subsets of $G$ and $A$ is finite, we say that -$B$ is $\epsilon$-Ramsey with respect to $A$ if for every $E \subseteq B$, -then there is a $\nu$ in $P(B)$ such that $P(A) * \nu \subseteq P(B)$ and -$$ -|\mu * \nu (E) - \nu (E)| < \epsilon -$$ -for all $\mu$ in $P(A)$. -Notice that in some sense $E$ is defining a partition of $P(B)$ and we are -postulating the existence of a copy of $P(A)$ in $P(B)$ which is homogeneous for $E$ up to -an error of $\epsilon$. -It can be shown with an argument similar to the one below that if $B$ is $\epsilon$-Ramsey with respect to $A$, then for every $f:B \to [0,1]$ -there is a $\nu$ in $P(B)$ such that -$$ -|f(\mu * \nu) - f(\nu)| < \epsilon -$$ -where $f$ has been extended linearly to $P(B)$. -We say that $(G,*)$ is Ramsey if for every finite subset $A \subseteq G$ and every $\epsilon > 0$, -there is a finite subset $B$ of $G$ with is $\epsilon$-Ramsey with respect to $A$. -Notice that if $B$ satisfies that for every $E \subseteq B$ there is a $\nu$ in $P(B)$ such that -$$ -|g * \nu (E) - \nu (E)| < \epsilon -$$ -for all $g$ in $A$, then $B$ is contained in a finite set which is $\epsilon$-Ramsey -(we need only to replace $B$ by $A * B \cup B$). -To connect this to the question, suppose that $G$ is not Ramsey, as witnessed by a finite -$A \subseteq G$ and $\epsilon > 0$. -I claim there is a set $E \subseteq G$ such that for every $\mu \in P(G)$, there is a $g \in A$ -such that $|\mu(E \cdot g) - \mu (E)| \geq \epsilon/2$. -Let $B_n$ $(n < \infty)$ be an increasing sequence of finite sets covering $G$. -Let $T_n$ be the set of all subsets $E$ of $B_n$ which witness that -$B_n$ is not $\epsilon$-Ramsey with respect to $A$. -Observe that if $E$ is in $T_{n+1}$, then $E \cap B_n$ is in $T_n$. -Otherwise there would be a $\nu$ in $P(B_n)$ such that $g * \nu$ is in $P(B_n)$ for each -$g$ in $A$ and -$$ -|g * \nu (E \cap B_n) - \nu (E \cap B_n)| < \epsilon -$$ -Such a $\nu$ would also witness that $E$ is not in $T_{n+1}$. -Define $T = \bigcup_n T_n$ and order $E \leq_T E'$ if $E = E' \cap B_m$ where $E$ is in $T_n$. -This order makes $T$ into an infinite finitely branching tree. -By König's lemma, $T$ has an infinite path whose union is -some $E \subseteq G$. -If there were a measure $\mu$ which was $\epsilon/2$-invariant for $E$ with respect to translates by elements of $A$, -there would be a finitely supported $\nu$ which was $\epsilon$-invariant for $E$ with respect to translates -in $A$. -But this would be a contradiction since then the support of $\nu$ would be contained in some $B_n$ -and $\nu$ would witness that $E \cap B_n$ was not in $T$. -Now the claim is that the Ramsey property of a discrete group is equivalent to its amenability. -That amenability implies the Ramsey property follows from Følner's characterization of amenability. -Also observe that $G$ is amenable provided that for every $\epsilon > 0$, every -finite list $E_i$ $(i < n)$ of subsets of $G$, and -$g_i$ $(i < n)$ in $G$, there is a finitely supported $\mu$ such that -$$ -|\mu (g_i * E_i) - \mu (E_i) | < \epsilon. -$$ -Set $B_{-1} = \{1_G\} \cup \{g^{-1}_i :i < n\}$ and construct a sequence -$B_i$ $(i < n)$ such that $B_{i+1}$ is $\epsilon/2$-Ramsey with respect to $B_i$. -Now inductively construct $\nu_i$ $(i < n)$ by downward recursion on $i$. -If $\nu_j$ $(i < j)$ has been constructed, let $\nu_i \in P(B_i)$ be such that -$$ -|\mu * \nu_{i} * \ldots * \nu_{n-1} (E_i) - \nu_i * \ldots * \nu_{n-1} (E_i)| < \epsilon/2 -$$ -for all $\mu$ in $P(B_{i-1})$. -Set $\mu = \nu_0 * \ldots * \nu_{n-1}$. -If $i < n$, then since $\nu_0 * \ldots * \nu_{i-1}$ and -$g_i^{-1} * \nu_0 * \ldots * \nu_{i-1}$ are in $P(B_{i-1})$, -$$ -|g_i^{-1} * \mu (E_i) - \nu_i * \ldots * \nu_{n-1} (E_i)| < \epsilon/2 -$$ -$$ -|\mu (E_i) - \nu_i * \ldots * \nu_{n-1} (E_i)| < \epsilon/2 -$$ -and therefore -$|\mu (g_i * E_i) - \mu (E_i)| < \epsilon$.<|endoftext|> -TITLE: Information about publishing and citations -QUESTION [32 upvotes]: In the context of some discussions we are having at my university, it has become evident that some statistical information regarding publishing practices in the various areas of mathematics would be necessary to proceed---you know, facts. In particular, I would be immensely happy to know - -are there measurable and measured differences in the number of papers published by people working in different areas (think PDEs v. Algebraic Geometry v. Number theory v. Combinatorics; top level MSC groups, say)? Here I mean papers published by individual authors as well as collectively. -are there measurable and measured differences in the number of citations gotten by papers in each area?; similar question about the out-degree in the citation digraph? -what is the time profile of the citations to a typical math paper (ideally, broken by area again), whereby I mean: how are the citations to papers typically distributed in time? - -Google has pointed to studies in which such comparisons are made between different disciplines (mathematics v. chemistry, say) but not at all between areas of mathematics. - -Can anyone point to such information? - -I would love to get hold of MathSciNet's raw tables (only papers, authors, subject area, citations) which would allow me to compute such things... (MathSciNet only has citation information since 2000, and I do not really know if that would make a representative sample for all-time statistics. It would be very interesting to have these kind of information diachronically, but I don't expect that data to be available) - -REPLY [7 votes]: In addition to the previous responses: - -Pure mathematicians are a lot worse about citing each other than applied mathematicians. Similarly, the threshold for being an author is quite high (I have seen (many) papers by X where the key lemma was proved by Y. In applied math or CS there is no question that Y would be a coauthor). -In applied mathematics it is much more common for the advisor to be co-author on papers written by his/her grad students/postdocs (perhaps because of the greater influence of hard sciences, where the lab director's name appears on ALL papers published by people working in the lab. In the hard sciences the lab director's name is typically not the first, but in applied mathematics people use the pure math practice of alphabetical name order [at least more often than not]). This leads to a much greater average number of authors per paper, and so more publications and more citations. Unfortunately, the system is that a co-author of a five author paper gets full "citation credit". - -If you look at the mathscinet list of top cited journals, you will see that the best pure math journals (Annals, Acta) are doing at best as well as some fairly generic applied math journals, which supports the above.<|endoftext|> -TITLE: Approximate search space on a 5x5x5 cube with 3 different possible classes? -QUESTION [5 upvotes]: Hey all, -I read the meta, and I realize this question might be pretty elementary for this site, but I'm having trouble computing this, and I know it won't take too much insight for someone to give me an approximation. -Say I have a 5x5x5 tic-tac-toe board (noughts and crosses), where each of the 125 spaces on the cube can belong to one of 3 different classes (X, O, empty). Now obviously the naive observation is that there are 3^125 possible 'boards', but after taking the following eliminating criteria in mind, can someone please give me a general idea of the order of the space complexity? ---EDITED TO ADDRESS #3-- - -Eliminate duplicate boards (equivalent after rotation, reflection) -Eliminate all boards that are not valid 'game boards'. That is, eliminate all boards where there is 5-in-a-row of one class (excluding the 'empty' class) in either horizontal, vertical, or diagonal directions, in all dimensions. -Similar to (2), but going one step further and eliminating all boards where there 4 in a row in any direction/dimension and that four in a row can possibly lead to a win. So exclude all boards that contain at least one row, column, or diagonal with 4 of one class, and the 5th being empty. -Because TTT (N&C) is a turn-based, ZS game, we should also eliminate possible boards where the difference in classes is greater than one. - -As mentioned above, I'm certainly not looking for any kind of precise number, just looking for a broad estimate. I've tried determining this for 2-dimensional boards and simple 3-dimensional boards, but I'm quite unsure of how these would scale to 5x5x5. -Thanks in advance for the help, this has been gnawing at me for a few days now. - -REPLY [3 votes]: The set of boards which have any symmetry is microscopic compared with the total number of boards, so the number of boards up to symmetry is roughly $3^{125}/48 \approx 9.1\times 10^{57}$. -The probability that a random board has a particular $5$ in a row is $2/(3^5)$, and if I count correctly there are $94$ places to win. So, the expected number of wins is $188/243 \lt 1$. At least $1-\frac{188}{243}$ of the boards have no $5$ in a row. -As I commented, you may want to force #crosses-#noughts to be $0$ or $1$, but that is not a very strong condition on a random board, only reducing the number by a factor of about $11.5$. -These are not independent conditions, but the interactions are of lower order. There are at least $2\times 10^{56}$ boards up to symmetry with no win yet so that the players could reach the position by alternating.<|endoftext|> -TITLE: linear order in a group of functions -QUESTION [6 upvotes]: Let $A$ be the smallest set of the following functions on the positive reals: - -The identity function is in $A$, -for a function $f\in A$ also the inverse $f^{-1}$ is in $A$, -for two functions $f,g\in A$ also the sum $f+g$, the product $f\cdot g$ and the composition $f\circ g$ is in $A$. - -In each generation the following properties are conserved: each function is strictly increasing and maps the positive reals surjectively to the positive reals. That's why we always can take the inverse above. The set $A$ is a group with respect to the composition operation. -The set $A$ contains all the polynomials with positive integer coefficients and zero constant term. Each non-trivial polynomial has finitely many fixpoints. Having finitely many fixpoints also implies being linearly orderable by "orders of infinity" (Hardy), i.e. by $f<_\infty g$ if there is an $x_0$ such that $f(x)\lt g(x)$ for all $x\gt x_0$ . -Do all the functions of $A$, except the identity function, have finitely many fixpoints? - -REPLY [9 votes]: Yes. Every function $f\in A$ is definable by a first-order formula in the structure $(\mathbb R,+,\cdot,\le)$. It follows that the set $F$ of its fixpoints is also definable. Since the structure is o-minimal, $F$ is a finite union of intervals (possibly degenerate). (The o-minimality follows from the Tarski–Seidenberg theorem: $\mathbb R$ admits elimination of quantifiers in the language of ordered rings, hence every definable set is a Boolean combination of sets defined by $p(x)\ge0$ for some polynomial $p\in\mathbb R[x]$.) Moreover, every $f\in A$ is an analytic function, and so is $f(x)-x$, hence if $F$ contains a nondegenerate interval, it contains the whole of $\mathbb R^+$, and $f$ is the identity. Otherwise, $F$ only contains degenerate intervals, i.e., it is finite. -The same property will hold if you also include in $A$ all functions $ax$, where $a\in\mathbb R^+$, and the exponential function (modified so that the property of being increasing and mapping $\mathbb R^+$ onto itself is maintained, e.g., $e^x-1$). (The o-minimality of the real field with exponentiation is a result of Alex Wilkie.) -Note that if you are only interested in the functions’ being linearly ordered by $<_\infty$, you don’t need the number of fixpoints to be finite, it suffices if they are bounded, and o-minimality gives this right away. That is, if you have any collection $A$ of functions $f\colon(a_f,+\infty)\to\mathbb R$, each of which is first-order definable in $(\mathbb R,+,\cdot,\exp,\le)$ with parameters (but they do not have to be injective, continuous, or anything like that), then for every $f,g\in A$, either $f<_\infty g$ or $f=_\infty g$ or $g<_\infty f$.<|endoftext|> -TITLE: Mr. G.P.K.'s questions -QUESTION [35 upvotes]: WARNING: An acquaintance of mine, Mr.Goosepond Prhklstr Kratchinabritchisitch, has requested permission to post his questions under my username. When I asked him why he didn't do it under his own name, he gave me a convoluted answer of which I didn't understand a word (as usual with him). So, as a pure courtesy, I'm letting him ask his questions but I want to emphasize that I do not in the least endorse anything he might say. As a precaution I'm making his post community wiki, but I confess that I'm still more than a little apprehensive... Georges Elencwajg - -Mr. Goosepond Prhklstr Kratchinabritchisitch's questions: -1) Why do you people continually talk of "elliptic curves"? They should just be called ellipses ( a special sort of comics) and are very easy to draw with a piece of string (I suppose that's what is nowadays bombastically called "string theory"). You can also easily study them with analytic geometry: their equation is simply $\frac {a^2}{x^2}+ \frac {b^2}{y^2}=1$. -2) Why do you hide all posts on descriptive geometry, compilation of seven decimal logarithmic tables, remarkable points of a triangle, etc: in other words the hard core of Mathematics? -3) Conversely, why so many posts on categories? Don't you people understand that they are just a fad, like the yo-yo, the hula-hoop or the computer? -4) I see many people refer to some GAGA article by a Jean-Paul Serre. Why don't grown-ups here tell this teenager to stop mentioning extravagant pop-singers ? What the FAC is coming next? -5) Why do people always talk of Dedekind's character? What's wrong with this guy? -6) I saw mentions of the Tango bundle and of the Serre twist (him again!) What's coming next: the Cauchie boogie-woogie, the Nagata salsa or the Cartan can-can ? -7) How come I couldn't find any financial statement in Forbes or the Wall Street Journal on a multinational like MO and the moguls running it? Why do investors (I'm talking to you, Warren Buffett!) tolerate this? Take the C.E.O, Mr. Geraschenko, for example: I couldn't even find out under which flag his yacht sails. -8) Why are you so afraid of acknowledging the top specialists in mathematics ? -G.P. Kratchinabritchisitch - -REPLY [10 votes]: Happy April Fool's?<|endoftext|> -TITLE: Status of an open problem about semilinear sets -QUESTION [19 upvotes]: In his book "The Mathematical Theory of Context-Free Languages" (1966), Ginsburg mentioned the following open problem: - -Find a decision procedure for determining if an arbitrary semilinear set -is a finite union of linear sets, each with stratified periods. - -Does anyone know if any progress has been made on this? I have searched, but not found any information. I did find that at least one of the other open problems mentioned by Ginsburg was solved already in the 1960s. -In case this has been done, but using different terminology, here are the definitions of the terms in the problem: -A linear set is a set of tuples of nonnegative integers of the form -$L = \{c + \sum_{i=1}^n \alpha_i p_i \mid \alpha_i\in \mathbb{N}_0\}$, where $\mathbb{N}_0$ denotes the nonnegative integers and $c,p_1,\ldots,p_n$ are fixed elements of $\mathbb{N}_0^r$. The set of periods of $L$ is $P = \{p_1,\ldots,p_n\}$. (The set of periods is not uniquely determined.) -A semilinear set is a union of finitely many linear sets. -For $p\in\mathbb{N}_0^r$, we denote the $i$-th component of $p$ by $p(i)$. -A subset $P$ of $\mathbb{N}_0^r$ is stratified if it satisfies the following conditions: - -each $p\in P$ has at most two non-zero components, and -there do not exist $i -TITLE: Is there a trivial construction of the trace on the Jones basic construction? -QUESTION [5 upvotes]: Let $N$ be a type $II_{1}$-factor with trace $\tau$, and $B$ a von Neumann subalgebra. The existence of the semifinite trace on the Jones basic construction $\langle N, e_{B} \rangle$ is reasonably easy to establish if $B$ is a subfactor of $N$, but appears not to be so easy in general. - -Question: What is the shortest known proof of the existence of the trace on the basic construction for a von Neumann subalgebra inclusion? - -Two complete proofs of this appear in the excellent book Finite von Neumann algebras and Masas by Sinclair and Smith. I (and others far more adept than I) am curious if any other proofs of this exist in the literature. -EDIT: The previous question, as asked, is embarrassingly silly. What seems less than obvious is how to construct the trace and verify that what you have written down is actually a faithful, NORMAL, semifinite trace. If someone could indicate how to do this, I'd very much appreciate it. - -REPLY [5 votes]: Perhaps I am missing some hypothesis, but I think the proof is just about the same whether or not $B$ is a factor. Here is the proof from Jones' original paper, and I believe it does not use factoriality of B (and not even facoriality of $N$?) -Lemma. Let $J:L^2(N)\to L^2(N)$ be the modular conjugation. Then $\langle N, e_B\rangle$ is $J B' J$. -Indeed, clearly $N=JN'J \subset JB'J$ and $e_B=Je_BJ\subset JB'J$ since $L^2(B)$ is $B$-invariant; thus $\langle N,e_B\rangle \subset JN'J$. On the other hand, $J\langle N, e_B\rangle J ' \subset JNJ ' \cap \{e_B\}' =\{e_B\}'\cap N$ since $JNJ'=N$ and $Je_BJ=e_B$. If $x\in N$ commutes with $e_B$ then, denoting by $1\in L^2(N)$ the trace vector, $x e_B 1 = e_B x 1 \in L^2(B)$ so that $x\in L^2(B)$. Thus $x=E_B(x)$ (where $E_B$ is the trace-preserving conditional expectation onto $B$) and so $x\in B$. Thus $\{e_B'\}\cap N = B$ and as a result $J\langle N, e_B\rangle J' \subset B$. By the bicommutant theorem you then get that $\langle N, e_B\rangle \subset JB'J$. -Now, given the Lemma, it is clear that $JB'J$ has a semi-finite trace, since $B'$ has a semi-finite trace (since $B$ has a finite trace).<|endoftext|> -TITLE: Distribution of a function in an arithmetic progression -QUESTION [6 upvotes]: I am going to have to borrow the opening passage from Bombieri, Friedlander, Iwaniec${}^*$ since they state this idea so well. In the following $\|f\|$ means $\big(\sum_{n\leqslant x} |f(n)|^2\big)^{1/2}$. - -Given an arithmetic function $f(n)$, it is natural to study its distribution in residue classes $a\: (\text{mod }q)$. One focuses on the classes $a$ with $(a,q)=1$, without restricting the generality, and expects that among these classes a reasonable function $f$ will be uniformly distributed, such uniformity being measured by upper bounds for the magnitude of - $$ -\Delta_f(x;q,a) = \sum_{\substack{n\leqslant x, \\ n\equiv a(\text{mod }q)}} f(n) - \frac{1}{\phi(q)} \sum_{n\leqslant x, \\ (n,q)=1} f(n). -$$ - A not unreasonable goal is the estimate - $$ -\Delta_f(x;q,a) \ll \frac{1}{\phi(q)} (\log x)^{-A}x^{1/2} \|f\|, \qquad \qquad (1) -$$ - for any $A>0$, the implied constant depending only on $A$, the result valid uniformly in $q$ in a range as large as possible. In view of Cauchy's inequality it is natural to regard (1) as saving $(\log x)^A$ from the "trivial" estimate. - -I have two queries. -1. Just a slight niggle. I only got $\ll \frac{1}{\sqrt{q}} x^{1/2} \|f\|$ for the "trivial" estimate using Cauchy's inequality. Is anyone able to get $\ll \frac{1}{\phi(q)} x^{1/2} \|f\|$? -2. More philosophically, how does one know when (or where does the intuition come from) that an estimate like (1) is actually true, or can be improved on? Normally my first instinct would be resort to some computations as a check that time isn't being wasted proving something that has an easy counterexample, but I cannot really see how this is done here. There are so many variables; one would presumably have to fix some $q$ and then let $x$ vary, then see what happens with different values of $q$ --- and this doesn't even take into account the infinite choices for $f$. - -${}^*$ Primes in arithmetic progressions to large moduli, Acta Math. 156 (1986). - -REPLY [5 votes]: For individual $a$'s the best universal bound you can get is $\ll \frac{1}{\sqrt{q}} x^{1/2} \|f\|$, as shown by the example of the characteristic function of a reduced residue class mod $q$. I think the authors meant the following. We are interested in the sums $\sum_{\substack{n\leqslant x, \ n\equiv a(\text{mod }q)}} f(n)$, but only when they are close to each other, i.e. close to their average $\frac{1}{\phi(q)} \sum_{n\leqslant x, \ (n,q)=1} f(n)$. So we are interested in a situation when the "error term" $\Delta_f(x;q,a)$ is smaller than the "main term" $\frac{1}{\phi(q)} \sum_{n\leqslant x, \ (n,q)=1} f(n)$, for which we know the bound $\ll \frac{1}{\phi(q)} x^{1/2} \|f\|$ by Cauchy. -As to your philosophical question, the sum $\sum_{\substack{n\leqslant x, \ n\equiv a(\text{mod }q)}} f(n)$ can be understood as the average of $\bar\chi(a)\sum_{n\leq x}f(n)\chi(n)$ as $\chi$ varies over the mod $q$ Dirichlet characters. In some fortunate situations, the contribution of the principal character, i.e. the term $\frac{1}{\phi(q)} \sum_{n\leqslant x, \ (n,q)=1} f(n)$ considered above, is much bigger then the total contribution of all the other characters. -This is closely tied to the following phenomenon: the Dirichlet series $\sum f(n)n^{-s}$ is regular in a right half-plane with a pole on its boundary. Replacing $f(n)$ by $f(n)\chi(n)$ for a nonprincipal $\chi$ often destroyes the pole, and makes $\sum f(n)\chi(n)n^{-s}$ converge in a larger right half-plane. This is certainly expected to happen for $f(n)=\Lambda(n)$ which measures the distribution of primes mod $q$, as it is a consequence of the Generalized Riemann Hypothesis. -So I think the philosophy is this: there is a miracle that Riemann and Dirichlet discovered. We don't understand it, but we want to understand it, try to understand it.<|endoftext|> -TITLE: Most efficient checking algorithm for Pell's Equation -QUESTION [9 upvotes]: What is the most computationally efficient way to check, given $x,y,D$ that they satisfy Pell's equation (positive or negative) ($x^2-Dy^2=1$)? (Obviously the question is concerned with very large values of $x,y,D$.) -I know (I think) that it'll have to be checking mod $p$ but I just can't find the right balance between computation time required for factorizing $x,y$ or $D$ and brute-force calculations. -Update: What I said about mod $p$ had to do with the fact that I thought that with Pell especially there might be some computationally efficient way to get a distinguished set of primes (and then the question was how best to balance the 'finding' with the moding) - but as Carnahan indicates the case might not be more special than that of evaluating any binomial. I think I'm pretty much covered by what Carnahan says, though I would be interested to know how these bounds on the number of operations are obtained. Also does it really follow that there is no more efficient way to evaluate a 'close' possible triple (i.e. one that has passed a mod-small-prime or log test) than to use multiplication algorithms? Is that obvious? - -REPLY [7 votes]: Based on the comments, it looks like this is not a question specific to Pell's equation, and that you just want to evaluate a single binomial with big inputs as quickly as possible. -If you check the equation directly using fast multiplication algorithms (e.g., Schönhage-Strassen), you can expect the calculation to require about $(\log z )\cdot (\log \log z)$ operations, where $z = \max \{ x, y, D \}$. -If you want a way to quickly find a negative answer, you can check relative sizes and leading digits, then try reduction modulo small integers. If you start with a random negative answer, you can expect it to be eliminated after at most a few divisions (i.e., about $\log z$ operations). -To find a positive answer using modular arithmetic, you can use the Chinese remainder theorem. To prove that the identity holds, it suffices to check it modulo $n$, for $n$ ranging over a collection of positive integers whose least common denominator is larger than $x^2$ and $Dy^2$. It is common to check modulo a large collection of small primes, and this will require about $\log z$ primes and $(\log z)^2$ operations. Another natural choice with a binary computer is Fermat numbers, of the form $2^{2^n}+1$, since the division-with-remainder can be optimized - this ends up looking a lot like direct calculation. -In summary, the advantage of checking modulo small primes is that it lets you quickly eliminate negative answers, and the disadvantage is that (if I'm not mistaken) it is roughly quadratically slower than direct calculation when you have a positive answer. You can choose your method depending on exactly what sort of calculation you plan to do.<|endoftext|> -TITLE: Modern reference for maximal connected subgroups of compact Lie groups -QUESTION [19 upvotes]: What's the nicest place to see a list of the maximal connected subgroups of compact Lie groups? Is there anything on-line? -I looked at Tits' Bourbaki talk on Dynkin's and others' work, but he admits early on that the theory leads to huge tables, which he isn't going to include. Those are the sort of thing I'd like to see. - -REPLY [14 votes]: I guess you are referring to the Tits Bourbaki seminar talk #119 here in the 1950s on subalgebras of semisimple Lie algebras (which translates the original question about compact Lie groups)? That's freely available online, but Dynkin's earlier papers probably aren't (though AMS published them long ago in translation volumes). Apparently Dynkin's tables contained some errors, as tables often do, but there is a useful AMS/IP collection here with corrections and commentaries which may be the best current source for the full story. -I suspect there has been too little incentive for anyone to rework and publish all of Dynkin's arguments and tables, but Gary Seitz (along with Martin Liebeck) did publish a number of long papers generalizing this work to the setting of semisimple algebraic groups (giving a lot of details in a modern style). -ADDED: There are two important (and long) papers by Dynkin in 1952, published by AMS in Series 2, Volume 6 of their translation series (1957), pages 111-378: Semisimple subalgebras of semisimple Lie algebras and Maximal subgroups of the classical groups. The papers contain lots of tables and are related to each other, as discussed by Tits in his Bourbaki talk. The main results are recovered by Seitz (and Liebeck) in their large AMS Memoir papers (such as No. 365 in 1987) treating maximal subgroups of semisimple algebraic groups. But this is a more elaborate framework, getting into prime characteristic as well. -To apply Dynkin's results to compact Lie groups one has to use Weyl's approach: complexify the Lie algebra or in reverse take a compact real form of a complex Lie algebra. Since a compact connected group is semisimple (or trivial) modulo its center and the latter lies in a torus, the semisimple case is crucial for maximal subgroup problems. The relevant maximal subalgebras of an associated complex Lie algebra are then semisimple. Unfortunately, compact groups don't seem to be treated explicitly in the literature (an exposition with some low rank groups as examples would be useful). But studying the group structure directly is too difficult, so the Lie algebra technology over $\mathbb{C}$ is most natural here, including some finite dimensional representation theory.<|endoftext|> -TITLE: References on Lie groups and dynamical systems -QUESTION [9 upvotes]: I'm interested in Lie theory and its connections to dynamical systems theory. I am starting my studies and would like references to articles on the subject. - -REPLY [9 votes]: The connections between Dynamics and Lie Groups (or Algebraic groups) comes mainly in two flavours: - -Smooth dynamics, like others have stated Hamiltonian dyanmics and differential equations. -Applications of Ergodic theory and Topological dynamics to Lie groups (or more generally, homogeneous spaces), or as Lindenstrauss' calls it - homogeneous dynamics. - -The homogeneous dynamics realm is again divided into two main areas: - -"Geometric Applications", i.e. most of Margulis works (rigidity and such). Those are problems that deal directly with these settings. -"Other applications", mainly number theoretical applications, which basically can be modelled on such spaces and dynamical methods (such as orbit classification or measure classification) come to use. - -I find myself more expert on 2.2 side, and I don't know anything about smooth dynamics, so I'll leave you with just one reference - Katok-Introduction to the Modern Theory of Dynamical Systems, which is some sort of general encyclopedia, and might be a good place to start your journey. -About homogeneous dynamics. -The area doesn't have a usual reference, and to be exact, there are hardly any references at all. -A good place to start would be - Einsiedler,Ward - Ergodic Theory With a View Towards Number Theory; this relatively a new book in the GTM, which is well written, and give you an introduction to ergodic theory, and in the later part he proves Ratner theorems for SL_2 (Furstenberg, Danni, Danni-Smilie). He also discusses some of the dynamics of nilpotent systems, such as the Heisenberg group (which is the starting point toward the Green-Tao theorem). -For the more advanced reader, the best place would be Elon's own notes - Lindenstrauss notes from a prev. course in HU, another good place would be the Clay Pisa proceedings, containing lecture notes of Eskin regarding Ratner theorems, and a paper by Lindenstrauss and Einsiedler about their work on diagonalizable actions. -If one is particularly interested in Ratner theorems, one can look in - Dave Morris' book about Ratner's theorems. -For those who are interested in Margulis works (Arithmeticity and such), I know only of two references - Margulis-Discrete subgroups of semisimple Lie groups which is out of print and extremely hard to find (and also hard to understand, one needs some familiarity with Lie groups and algebraic groups), and the other one is a book by Zimmer - Ergodic Theory and Semisimple Groups. -Dave Morris has a draft of a book about Arithmetic groups which might be of interest as well - Morris - Introduction to Arithmetic Groups. -Well I'm hoping I gave you enough reading for some time.<|endoftext|> -TITLE: lefschetz hyperplane section theorem -QUESTION [12 upvotes]: I am supposed to do a presentation on Lefschetz hyperplane section theorem via Morse theory (following Milnor's Morse Theory) for my algebraic geometry class...I more or less understand the proof, but I am really at a loss what could be good and easily presentable applications of the theorem. I am a beginner in algebraic geometry, so I can't do anything highly involved. Griffiths and Harris has a couple of examples....I was wondering if you people could suggest something really interesting and elegant... -thanks a lot. - -REPLY [39 votes]: Dear anonymous, let $ X \subset \mathbb P^n(\mathbb C)$ be a smooth hypersurface of degree $d$ in projective space. Let me show how Lefschetz directly calculates the homology $H_i(X)=H_i(X,\mathbb Z)$ of $X$, except for $H_{n-1}(X)$. -There is an embedding $v:\mathbb P^n(\mathbb C) \to \mathbb P^N(\mathbb C)$, called the Veronese embedding, mapping the original $P^n(\mathbb C)$ to a huge projective space $\mathbb P^N(\mathbb C)$ of dimension $N=\binom{n+d}{d} -1$ -The image of $v$ is then an isomorphic copy $P$ of $\mathbb P^n(\mathbb C)$ and the charm of Veronese's embedding is that $v(X)$ is now a hyperplane section of $P $ i.e. $v(X)=P \cap H$ for some hyperplane $H \subset \mathbb P ^N(\mathbb C)$ . -You thus get $H_i(X) \simeq H_i(P)$ for $i\leq n-2$ by applying Lefschetz for homology. For $i\geq n$ - you deduce (by applying Poincaré duality and Lefschetz for cohomology ) - $H_i(X) \simeq H^{2n-2-i}(X)\simeq H^{2n-2-i}(\mathbb P^n(\mathbb C))$. Since $P\simeq P^n(\mathbb C)$ all the above homology groups are easy to calculate. Indeed, for $0 \leq i \leq 2n$, projective space has homology $H_i(P^n(\mathbb C))$ alternately $0$ and $\mathbb Z$. Ditto for cohomology. -The only group which has partly escaped us is $H_{n-1}(X)$, for which Lefschetz only tells us that it surjects onto $H_{n-1}(P^n(\mathbb C))$. As a reality check, notice that for $n=2$ we know that the rank of -$H_{1}(X)$ is $2g=(d-1)(d-2)$, where $g$ is the genus of our curve $X$ . This is indeed not equal to the rank of any (co)homology group of $P^2(\mathbb C)$ for $d \geq 3$ -Optional complement: Allow me to show how concrete the Veronese embedding is. It maps the point -$(x_0:x_1:x_2:\cdots:x_n)\in \mathbb P^n$ to the point $(\cdots:M_\delta (x):\cdots) \in \mathbb P^N$ where $\delta = (d_0,\cdots,d_n) \quad d_0+\cdots d_n=d$ and $M_\delta (x)=x_0^{d_0}\cdots x_n^{d_n}$. -So that if for example $n=d=2$, the conic $x^2+2y^2+3y^2-4xy-5xz-6yz=0$ in $\mathbb P^2$ gets mapped isomorphically by Veronese onto the hyperplane section of the Veronese surface $v(\mathbb P^2)$ by the hyperplane $X+2Y+3Z-4U-5V-6W=0$ of $\mathbb P^5$ . What could be simpler?<|endoftext|> -TITLE: What kind of completion is this? -QUESTION [11 upvotes]: Let $X$ be a compact Hausdorff space, and $C(X)$ the unital commutative C*-algebra of continuous functions on it. The double Banach dual $C(X)^{**}$ is a commutative von Neumann algebra and hence has a compact Hausdorff space $X^{**}$ as Gelfand spectrum again. What is $X^{**}$, in terms of $X$? -This gives an (idempotent?) endofunctor (monad?) on the category of compact Hausdorff spaces, that I don't recognize as any of the usual ones like Stone-Cech. What completion is it? Is it related to the functor taking a compact Hausdorff space to the $\sigma$-algebra generated by its opens? -Accounts of enveloping von Neumann algebras of (commutative) C*-algebras in terms of double Banach duals seem hard to find in the literature, and any references are welcome. What is the von Neumann algebra $C(X)^{**}$, in the first place? - -REPLY [7 votes]: For what it's worth, I found a lot of information in [Dales, Lau & Strauss, "Second duals of measure algebras", Dissertationes Mathematicae 481:1-121, 2012]. The assignment $X \mapsto X^{\ast\ast}$ is functorial, and called the hyper-Stonean cover. It loses information: if $X$ is countable, then $X^{\ast\ast} \cong \beta\mathbb{N}$. -If $X$ is metrizable and uncountable, a lot of the structure of $X^{\ast\ast}$ is known -- it is characterised as follows: - -$X^{\ast\ast}$ is hyper-Stonean; -the set $D$ of isolated points of $X^{\ast\ast}$ has cardinality $2^{\aleph_0}$, its closure $Y$ is a clopen subspace homeomorphic to $D_d$; -$X\setminus Y$ contains a family of $2^{\alpha_0}$ pairwise disjoint, clopen subspaces, each homeomorphic to $\mathbb{H}$; -the union $U$ of the above sets is dense in $X \setminus Y$, and $\beta U = X \setminus Y$. - - -In general, there exist a continuous projection $p \colon X^{\ast\ast} \to X$ and a (not necessarily injective) injection $i \colon X \to X^{\ast\ast}$ with $i \circ p = 1_{X^{\ast\ast}}$. Moreover, $X$ consists of the isolated points of $X^{\ast\ast}$, and is therefore open.<|endoftext|> -TITLE: applications of the sphere theorem -QUESTION [12 upvotes]: I am looking for interesting applications of the 1/4-pinched sphere theorem. The theorem says: A compact, simply connected riemannian manifold whose sectional curvature K satisfies $1/4 < K \leq$ 1 (possibly after multiplying the metric by a constant) is homeomorphic (recently extended to "diffeomorphic") to the sphere. I just wanted to know: is it just a beautiful theorem or can you use it in concrete situations to derive some conclusions difficult to see otherwise? I am interested in this just because I am curious, I do not have any specific purpose in mind. - -REPLY [5 votes]: An application occurs in the study of asymptotic behavior of complete manifolds with certain curvature decay. -Let M be a n-dimensional complete non-compact manifold. -Suppose that - -M is simply-connected at infinity, -the sectional curvatures of M go to zero at infinity, -there exists a foliation of (n-1)-dimensional sub-manifolds on the ends of M -these sub-manifolds have controlled second fundamental form, - -then you may use Gauss equation and the differential sphere theorem to say that these sub-manifolds are diffeomorphic to the sphere.<|endoftext|> -TITLE: When are certain group C*-algebras exact? -QUESTION [9 upvotes]: This is somewhere between a "reference request" and "ask an expert", but I hope it is not too trivial or off-topic. -Anyway. There has been a lot of attention given to showing that for certain discrete groups $G$, the reduced group C*-algebra $C_r^*(G)$ is an exact C*-algebra in the sense of Kirchberg/Wassermann. For something I am working on, I want to consider certain classes of nondiscrete groups. (I am actually thinking of the case where the group von Neumann algebra is finite, but it's not clear to me how much that helps.) -This leads me to my question: does anyone know of results for when either of the following group C*-algebras are exact? -1) $C_r^*(G)$ where G is not discrete . If $G$ is amenable or connected then this algebra will be nuclear, hence in particular exact. -2) $C_d^*(G)$, which is defined to be the norm-closed subalgebra of ${\mathcal B}(L^2(G))$ generated by all left translation operators $\lambda_t$, with $\lambda_t(\xi)(s) = \int_G \xi(t^{-1}s)\\,ds$. -The second case would be of most interest to me right now (the problem is that for the groups I am interested in, $C^*(G_d)$ is too big to be exact, as $G_d$ will contain a nonabelian free subgroup). -I suspect that I am simply not looking in the right places or at the right parts of the standard literature, so a pointer to the right places would be enough. - -Update October 2012: In a partial vindication for the old-school way of doing things, involving not this new-fangled InterWeb2.0 stuff but "giving a conference talk and raising a question": I have been informed by Narutaka Ozawa that for many examples of compact non-abelian connected Lie groups $G$, the algebra denoted above by $C^*_d(G)$ fails to be exact. In cases such as $SO(n)$ for $n$ large enough this follows from the existence of dense subgroups that have Property (T) as discrete groups; in the case $G=SU(2)$ one needs some spectral gap results that have been recently been obtained in hard work of Bourgain--Gamburd. -(I am going from memory of our conversation here, so if anything in the above is faulty, then it is almost surely a mistake on my part and not on Ozawa's.) - -REPLY [3 votes]: This might not be directly relevant, but it is known that $C^*_d(G)$ is nuclear if and only if $G_d$ is amenable: see Theorem 3 in Erik Bedos, ON THE C*-ALGEBRA GENERATED BY THE LEFT REGULAR REPRESENTATION OF A LOCALLY COMPACT GROUP, Proceedings of the American Mathematical Society, Vol. 120, No. 2 (Feb., 1994), pp. 603-608. The references in that paper might be interesting too.<|endoftext|> -TITLE: Generalization or Improvement of Cheeger inequality on Graphs -QUESTION [9 upvotes]: Let $G=(V,E)$ be an undirected graph with vertex set $V$ and edge set $E$. Let $A$ denote the adjacency matrix of $G$ and $D$ denote the diagonal matrix such that $D_{i,i}$ equals to the degree $d_i$ of vertex $i$. Then the Laplacian of $G$ is defined to be $L:=I-D^{-1/2}AD^{-1/2}$, which has $n$ nonnegative real eigenvalues, say, $0=\lambda_1\leq \lambda_2\leq \cdots\leq \lambda_n$. -Now for any vertex subset $S\subseteq V$, let $e(S,\bar{S})$ denote the number of edges between $S$ and its complement $\bar{S}$. Let $vol(S):=\sum_{v\in S}deg_v$ be the sum of degrees of vertices in $S$. Then the conductance (or Cheeger constant) $h_G$ of graph $G$ is defined to be -$$h_G:=\min_{S\subseteq V}\frac{e(S,\bar{S})}{\min \{vol(S), vol({\bar{S}})\}}.$$ -The Cheeger inequality relates the second smallest eigenvalue $\lambda_2$ of $L$ to the conductance $h_G$ as follows: -$$2h_G\geq\lambda_2\geq \frac{h_G^2}{2}.$$ -The above inequality is known to be tight. For example, the left side of the inequlity is tight on the $d$-dimensional cube and the right side is tight on the $n$-vertex cycle. Thus, we do not hope to improve the inequality that applies to every graph. -My question is: - -Is there any result which gives that for some special class of graphs, the Cheeger inequality has an improved form, say, $2h_G^{1.2}\geq\lambda_2\geq \frac{h_G^{1.5}}{2}$ for any $G\in\mathcal{C}$, where $\mathcal{C}$ is a set of graphs. - -Ideally, we would hope that there is a nice tradeoff between the generality of the class $\mathcal{C}$ and the tightness of the Cheeger-type inequality. In another word, $\mathcal{C}$ contains a wide class of graphs and the upper bound is close to the lower bound. - -REPLY [3 votes]: There other classes of graphs such as generalizations of the $G(n,p)$ model with a given degree distribution. See for example Chung and Vu's paper on the spectrum of these graphs. In particular, they analyzed the case when the graph has a power law degree distribution.<|endoftext|> -TITLE: Relation between Theta series and Eisensteinseries -QUESTION [7 upvotes]: In "Mackey - Unitary Group Representation in Physics, Probability and Number Theory" on page 326, George Mackey mentions a result of Ludwig Siegel, which was later generalized to semi-simple Lie groups by André Weil. In his words: -"Now in one formulation Siegel's main result takes the form of an identity between an theta series and an Eisenstein series." -My first guess is that this relation goes through the constant term of the Eisenstein series, which is an L function. This L function is the Mellin transform of a Theta series, right? -Q1: What is the exact statement? -Q2: Is there a nice result for $\mathrm{GL}_2$ (in terms for intertwiner for the parabolic induced representation) for these kind of results? -Q3: How does this generalize to $\mathrm{GL}_n$? - -REPLY [10 votes]: Conceivably it is useful to imbed Siegel's result in a somewhat larger context, so that it is a bit of a special case of something. Namely, (and this is still a special case...), even-sized orthogonal groups $O(Q)$ defined over a number field $k$ "pair" with $Sp(2n,k)$ for all choices of size $2n$, as mutual commutators inside a two-fold cover $Mp(2n\cdot \dim Q)$ of $Sp(2n\cdot \dim Q)$. The "Segal-Shale-Weil/oscillator" repn of the adele group (restricted from a repn of the metaplectic Mp) gives a well-defined mapping from repns (local or global) from irreducibles of $O(Q)$ to irreducibles of $Sp(2n)$ for $2n\gg \dim Q$, and in the other direction for the opposite inequality. (The precise cut-offs are about "first-occurrence", as in work of Kudla-Rallis and others.) -In a rather degenerate situation, for $\dim Q\gg 2n$, the constant $1$ on the adelic orthogonal group is mapped to some sort of automorphic form on $Sp(2n)$. It is not profoundly difficult, but non-trivial, to see that the image is the Siegel-type Eisenstein series. With sufficiently many decades of hindsight, and with the benefit of working on adele groups rather than classically, as Siegel did, various people have found simplified arguments: there is a bibliography and an example of a simplified argument on-line in http://www.math.umn.edu/~garrett/m/v/siegel_weil.pdf, also in the Eisenstein Series volume from AIM Harris-Skinner-Li have a wider-scope discussion in "A simple proof of rationality of Siegel-Weil" (http://www.math.jussier.fr/~harris/resarticles/SW.pdf) -The translation into classical language engenders the discussion of lattices in a genus and cardinalities of automorphism groups, much as comparison of idele class groups to ideal class groups and unit groups gives rise to classical details regarding the latter.<|endoftext|> -TITLE: Structure of nonaveraging sets of integers -QUESTION [5 upvotes]: A set of integers is said to be nonaveraging if it contains no three-term arithmetic progression. I call a nonaveraging subset of $\lbrace 1,2, \ldots ,n \rbrace$ optimal when it has maximal cardinality. -There is a regularly updated website on nonaveraging sets records at http://www.math.uni.wroc.pl/~jwr/non-ave/DATABASE.TXT. Most of the research done on upper bounds for nonaveraging subsets of $\lbrace 1,2, \ldots ,n \rbrace$ (by Roth, Bourgain, Gowers, Tao, Green and others) involves randomness one way or another (be it in the form of Fourier analysis, extremal graph theory or ergodic theory) , and Behrend's lower bound is nonconstructive. -Despite the randomness, the optimal nonaveraging sets display some structure : -When $n=20$, there are two optimal solutions, which are $B \cup (B+5) \cup \lbrace 18 \rbrace$ and $B' \cup (B'+5) \cup \lbrace 3 \rbrace$, where $B=\lbrace 1,2,9,15 \rbrace$ and $B'=\lbrace 1,7,14,15 \rbrace$. When $n=30$, there is a unique optimal solution, $B \cup (B+19)$, where $B=\lbrace 1,3,4,8,9,11 \rbrace$. Looking at larger examples from the abovementioned website, the decomposition "two copies+error term" seems to persist, which inspires me the following list of (increasingly strong) conjectures : -** Conjecture 1. ** There is a function $E(n)$ tending to $+\infty$ when $n$ tends to infinity, such that for any optimal nonaveraging subset $X$ of $\lbrace 1,2, \ldots ,n \rbrace$ we can write $X=B \cup (B+t) \cup C$ for some nonzero integer $t$ and some $B,C \subseteq \lbrace 1,2, \ldots ,n \rbrace$ and $|B| \geq E(n)$. -** Conjecture 2. ** There is a function $\varepsilon(n)$ tending to $0$ when $n$ tends to infinity, such that for any optimal nonaveraging subset $X$ of $\lbrace 1,2, \ldots ,n \rbrace$ we can write $X=B \cup (B+t) \cup C$ for some nonzero integer $t$ and some $B,C \subseteq \lbrace 1,2, \ldots ,n \rbrace$ with $|C| \leq n\varepsilon(n)$. -** Conjecture 3. ** There is a function $\varepsilon(n)$ tending to $0$ when $n$ -tends to infinity, such that for any optimal nonaveraging subset $X$ of $\lbrace 1,2, \ldots ,n \rbrace$ we can write -$X=B \cup (B+t) \cup C$ for some nonzero integer $t$ and some $B,C \subseteq \lbrace 1,2, \ldots ,n \rbrace$ with $|C| \leq |X|\varepsilon(n)$. -Note that the two copies $B$ and $B+t$ are necessarily disjoint since -$X$ is nonaveraging. Also, for each conjecture we have a weaker variant -where "any optimal $X$" is replaced with "at least one optimal $X$". -Conjectures 2 and 3 may be out of reach but conjecture 1 seems really easier because containing no two copies of a set of size at least $k$ is a much stronger requirement than being nonavering, so that the corresponding optimal sets should be much smaller. Can anyone supply a proof? - -REPLY [3 votes]: As to Conjecture 3, I strongly doubt it is true. If it were true, one could have decomposed any optimal progression-free subset of $[1,n]$ as $B\cup(B+t)\cup C$, where $|C|$ is small as compared to $|B|$. Now, $B$ is a progression-free subset of $[1,n-t]$ with the property that $t\notin B-B$ and $2t\notin\pm(B+B-2\ast B)$, with $2\ast B:=\{2b\colon b\in B\}$. My feeling is that these requirements are quite restrictive, forcing $|B|$ to be much smaller than (roughly) one half of the size of an optimal progression-free subset of $[1,n]$.<|endoftext|> -TITLE: Parametrization of unit varieties -QUESTION [6 upvotes]: Let $K$ be a number field with integral basis $\{\omega_1,\ldots,\omega_n\}$. Then -$$ \Phi(X_1, \ldots, X_n) = N_{K/{\mathbb Q}}(\omega_1 X_1 + \ldots + \omega_n X_n) $$ -is a homogeneous polynomial of degree $n$ with integral coefficients, and the integral points on the affine variety -$$ \Phi(X_1,\ldots,X_n) = 1 $$ -correspond to units with norm $+1$ in the ring of integers of $K$. -For quadratic extensions, this "unit variety" is defined by $X_1^2 - mX_2^2 = 1$ -(a Pell conic) whenever $m \equiv 2, 3 \bmod 4$ is squarefree; for other extensions of small -degree it is similarly easy to write down explicit equations. -It is well known that the rational points on the Pell conic can be parametrized. The same thing holds for general cyclic extensions: the -proof via Hilbert 90 that Pell conics can be parametrized generalizes easily. This suggests the following question: - Can unit varieties of number fields be parametrized by rational functions with rational coefficients? - -REPLY [8 votes]: Thanks to Dror for pointing out a nice geometrical way to think of such varieties. -This paper shows that such varieties are not rational in general: -http://www.math.jussieu.fr/~florence/norm_one.pdf -However, any algebraic torus over a number field $K$ of dimension one or two is rational over $K$. Indeed, the case of dimension one is easy, and the case of dimension two is a theorem of Voskresenskiĭ, but algebraic tori of dimension three and above do not have to be rational over the ground field in general.<|endoftext|> -TITLE: Which Banach algebras are group algebras? -QUESTION [9 upvotes]: Given a locally compact Hausdorff group $G$, one can construct several Banach star-algebras using $G$ (and its associated Haar measure): $L^1 (G)$, $M(G)$ (regular complex measures on $G$), $L^{\infty} (G)$, $C^* (G)$, $C^*_r (G)$, $W^* (G)$, etc. (see this Wiki article for some details). Then one can ask, for instance, which Banach *-algebras can be represented (i.e. are isometrically *-isomorphic) to/as $L^{\infty} (G)$. In this case every such algebra is an abelian Von-Neumann algebra, and every abelian Von-Neumann algebra can be represented as $L^ \infty (X,\mu)$ for some set $X$ with measure $\mu$`, so the question reduces to the question "which measures are Haar measures (on some LCH group)?". However, I'm mostly interested in this type of question for the algebras $L^1 (G)$. It is known that such a Banach *-algebra is semisimple and symmetric for any abelian LCH group $G$, so obviously one can't represent all commutative Banach *-algebras as $L^1 (G)$. So, to summarize, I have two questions: - -Is there an operator-theoretic characterization of the Banach algebras which are isometrically *-isomorphic to $L^1 (G)$ for some $G$? In particular, is there an abelian, symmetric, semisimple Banach *-algebra which isn't $L^1 (G)$ for some $G$? -Are there classes of Banach *-algebras to which one can associate an LCH group $G$ in a canonical way (up to an isomorphism of topological groups), so that the algebra is naturally isomorphic to a group algebra $L^1 (G)$? - -Also, if there is some known reference for this kind of problems, I'll be glad to know of it. - -REPLY [3 votes]: [This is too long to be a comment, but is at best a partial answer; I just thought I'd write it down while I have spare time.] -Regarding the 2nd part of Question 1: you need to impose some extra hypotheses to capture the "real" question I suspect you are interested in. Otherwise, just take $C[0,1]$, which is certainly abelian, symmetric as a *-algebra, and semisimple. This cannot be isomorphic to $L^1$ of anything, even as a Banach space. (To give a bit more detail: every map from $C[0,1]$ to $L^1$ is weakly compact, so if the two were isomorphic as Banach spaces then the identity map on $C[0,1]$ would be weakly compact, implying $C[0,1]$ is reflexive which is not the case.) If you want non C*-examples then $C^1[0,1]$ will also do, for much the same reasons. -If you're after a counter-example where the underlying Banach space looks like $L^1$ or $\ell^1$, then I suggest the convolution algebra $\ell^1(S)$ where $S$ is an infinite semilattice (=commutative semigroup in which all elements are idempotent); the involution is just complex conjugation. Semisimplicity is proved in an old paper of Hewitt and Zuckerman (I am not sure of the spelling of the 2nd author's name, and as I am out of the office can't check right now). Off the top of my head, the quickest, if not the most direct, way to show that such an algebra is not isomorphic to an abelian group algebra, is to use the fact (Duncan & Namioka, late 1970s) that $\ell^1(S)$ for such $S$ is never amenable, while every abelian $L^1$-group algebra is amenable (Johnson, 1972). There ought to be a simper way to distinguish the two classes of algebra, though. -Update. If you are interested only in isometric isomorphism of Banach algebras then there is a much easier way to show $\ell^1(S)$ is not a group algebra. For if $\theta: \ell^1(S)\to L^1(G)$ is an isometric algebra isomorphism, then by examining where extreme points of the unit ball go, you find that $G$ must be discrete and for each $x\in S$, $\theta(\delta_x)=\delta_{\phi(x)}$ where $\phi:S \to G$ is an isomorphism of semigroups. Now $S$ has lots of idempotents, $G$ has only one, and we have reduced to absurdity as they say.<|endoftext|> -TITLE: Langlands conjectures in higher dimensions -QUESTION [5 upvotes]: Geometric class field theory (curves over a finite field) has been generalized -to higher dimensional varieties over a finite field (and other arithmetical fields). Some of the key names here are Lang, Serre, Bloch, Kato-Saito... -Question: Does there exist a precise formulation of Langlands's conjectures which constitutes a non-abelian generalization of these results? -Probably relevant: Kazhdan's talk here at the Panel on Open Questions held at Gross's 60th Birthday Conference. - -REPLY [6 votes]: From what I remember, Kapranov's paper "Analogies between the Langlands correspondence and topological quantum field theory" has some speculations about this. A wise old friend once called it the most speculative paper he'd ever seen in print.<|endoftext|> -TITLE: Properties of Some Random Graphs -QUESTION [14 upvotes]: Working in a problem the following family of graphs appears naturally. Consider the set $A_{n}=\{1,2,3,\ldots,n\}$ and let $\mathcal{C_{n}}$ be the set of all permutations of $A_{n}$ of order $n$ (cycles of order $n$). Let $\sigma_{1},\sigma_{2},\ldots,\sigma_{d}$ be random elements chosen uniformly and without repetition from $\mathcal{C}_{n}$. -Now we construct the random graph $\mathcal{G}$ where the node set is $A_{n}$ and there is an edge between each node $i$ and $\sigma_{p}(i)$ for every $p\in\{1,2,\ldots,d\}$. It's clear that every node in the graph $\mathcal{G}$ has degree at most $2d$ (we ignore multiple edges and loops). - -My questions are: - -Did anybody studied these graphs before? -Is it known what is the asymptotic diameter of $\mathcal{G}$ for fixed - $d$ as $n$ increases with high - probability? -Estimates on the Cheeger constant? Laplacian? - -REPLY [4 votes]: For those interested in further reading about the contiguity of the above mentioned models for random regular graphs and generalization of the above: -Catherine Greenhill, Svante Janson, Jeong Han Kim and Nicholas C. Wormald, Permutation pseudographs and contiguity, Combinatorics, Probability and Computing 11 (2002), 273 - 298. -Link: http://web.maths.unsw.edu.au/~csg/papers/gjkw-revised.pdf<|endoftext|> -TITLE: Is $\mathbb R^3$ the square of some topological space? -QUESTION [213 upvotes]: The other day, I was idly considering when a topological space has a square root. That is, what spaces are homeomorphic to $X \times X$ for some space $X$. $\mathbb{R}$ is not such a space: If $X \times X$ were homeomorphic to $\mathbb{R}$, then $X$ would be path connected. But then $X \times X$ minus a point would also be path connected. But $\mathbb{R}$ minus a point is not path connected. -A next natural space to consider is $\mathbb{R}^3$. My intuition is that $\mathbb{R}^3$ also doesn't have a square root. And I'm guessing there's a nice algebraic topology proof. But that's not technology I'm much practiced with. And I don't trust my intuition too much for questions like this. -So, is there a space $X$ so that $X \times X$ is homeomorphic to $\mathbb{R}^3$? - -REPLY [17 votes]: The Euler characteristic with compact support $\chi_c(X)$ is a very robust topological invariant available for any reasonable space (such as a subanalytic set). The key properties here are that $\chi_c(X)$ is a real number and $\chi_c(A\times B)$ is multiplicative: -$$ -\chi_c(X)\in \mathbb{R},\quad \quad \chi_c(A\times B)=\chi_c(A)\cdot \chi_c(B). -$$ -It follows that the Euler characteristic with compact support of a topological square is always non-negative: -$$\forall X: \quad \chi_c(X\times X)\geq 0.$$ -Thus, a space with negative Euler characteristic with compact support cannot be a topological square. -In particular, since $\chi_c(\mathbb{R}^3)=-1$, $\mathbb{R}^3$ is not a topological square. -More generally, since $\chi_c(\mathbb{R}^n)=(-1)^n$, for any $k\in\mathbb{N}$, $\mathbb{R}^{2k+1}$ is not a topological square. -For an introduction to the topological Euler characteristic with compact support, I would recommend the following notes by LIVIU NICOLAESCU.<|endoftext|> -TITLE: Do coproducts in categories of algebras preserve monos? -QUESTION [8 upvotes]: Even if the answer is no, I am interested in a more specific question. -Let $\Sigma$ be a set of operations of finite arity, $E$ be a set of equations over $\Sigma$ and $\mathcal{A}(\Sigma,E)$ be the respective category of algebras and algebra morphisms. Also denote the free algebra functor by $F: \mathsf{Set} \to \mathcal{A}(\Sigma,E)$. -If $f : A \to B$ is a monomorphism in such a category i.e. an injective algebra morphism, and also $X$ is set, does it follow that $FX + f : FX + A \to FX + B$ is also injective? -Any help much appreciated. - -REPLY [8 votes]: Well, I have a counterexample. -Let $\Sigma$ contain two unary operations $a$ and $b$. -Further suppose $E$ contains the equations $a(p) = a(q)$ and $b(p) = b(q)$. -Then $F0 = \emptyset$ because $\Sigma$ contains no nullary operations and we also have the terminal algebra $\mathsf{1}$ with singleton carrier $\{*\}$. Furthermore the free algebra on omega generators $F\omega$ has carrier $\omega + 1 + 1$, the omega corresponding to the generators, the remaining two elements being the equivalence class containing $a$-prefixed terms and the equivalence class containing $b$-prefixed terms. -Now we certainly have the injective algebra morphism $\iota : \emptyset \hookrightarrow \mathsf{1}$. -However $F\omega + \emptyset \cong F\omega$ so it has carrier $\omega + 1 + 1$, whereas $F\omega + \mathsf{1}$ has carrier $\omega + 1$. This follows because in $F\omega + \mathsf{1}$ we can deduce: -\[ -a(x) = a(*) = * = b(*) = b(x) -\] -for any $x \in U(F\omega + \mathsf{1})$. -Since $F\omega + \iota$ is surjective but not injective, the respective functor $F\omega + Id$ does not preserves monos.<|endoftext|> -TITLE: Non-vanishing of p-adic L-functions -QUESTION [18 upvotes]: In Non-vanishing of L-series of modular forms (easy case?) it was answered that for a cuspidal newform $f$ of weight strictly greater than 2, then $L(f,1)$ is non-zero. (Here the $L$-series is normalized so that the center of the critical strip is given by $s=k/2$.) In particular, for such modular forms, their associated $p$-adic $L$-functions are non-zero. As far as I know the non-vanishing of $p$-adic $L$-functions in the weight 2 case is a highly non-trivial result and relies upon a non-vanishing theorem of Rohrlich on twisted $L$-values. Further, from the non-vanishing of the $p$-adic $L$-function, one can deduce that $L(f,\chi,j)$ is non-zero for all but finitely many pairs $(\chi,j)$ where $\chi$ is a Dirichlet character of $p$-power conductor and $j$ is an integer between $1$ and $k-1$, as long as $p$ is an ordinary prime for $f$. -My questions: -1) Is there a direct argument to prove the non-vanishing of $L(f,\chi,j)$ for all but finitely many $\chi$ and $j$ in the ordinary and weight greater than 2 case (which doesn't use $p$-adic $L$-functions). -2) Is this result known in the non-ordinary case? - -REPLY [5 votes]: These arguments on the non-vanishing of $p$-adic $L$-functions are great! I had never seen them before. -What I'm writing here is neither an answer to your question nor an actual proof of any sort. But I think it at least follows the general theme of what you are asking. -Namely, I tried to use 2-variable p-adic L-functions and the non-vanishing of $p$-adic $L$-functions of higher weight modular forms to deduce Rohrlich's theorem in the weight 2 case (i.e. the non-vanishing of $L(f,\chi,1)$ for $f$ a form of weight 2 for all but finitely many $\chi$ of $p$-power conductor). It didn't actually work as I need to assume the non-vanishing of some mu-invariant which is deep stuff, but I think the argument is amusing enough to present in any case. -Here's the argument: put the original weight 2 form $f$ into a Hida family, and write down the corresponding 2-variable $p$-adic $L$-function. For simplicity, let me assume that the ordinary Hecke algebra in this case is just $\Lambda = {\bf Z}_p[[S]]$. Here one sets $S=\gamma^k-1$ to specialize to weight $k$ where $\gamma$ is some topological generator of ${\bf Z}_p^\times$. -Then the two-variable $p$-adic $L$-function can be thought of as a power series in ${\bf Z}_p[[S,T]]$. Say -$$ -L_p(S,T) = a_0(S) + a_1(S)T + a_2(S)T^2 + \dots -$$ -First let me point out that this power series is non-zero. Indeed, it interpolates the $p$-adic $L$-functions of each classical form in the Hida family which have already been observed to be non-zero in weight greater than 2 (without invoking Rohrlich's theorem). -Now let's assume that at least one form in the Hida family has zero $\mu$-invariant. This means there is some weight k such that -$$ -L_p(f_k,T) = L_p(\gamma^k-1,T) = a_0(\gamma^k-1) + a_1(\gamma^k-1)T + a_2(\gamma^k-1)T^2 + \dots -$$ -has non-zero $\mu$-invariant. In particular, for some $i \geq 0 $, $a_i(\gamma^k-1)$ is not divisible by $p$. This implies that $a_i(S)$ is a unit in ${\bf Z}_p[[S]]$, and in particular is non-zero. Thus, the $p$-adic $L$-function in weight 2 -$$ -L_p(f_2,T) = L_p(\gamma^2-1,T) = a_0(\gamma^2-1) + a_2(\gamma^2-1)T + \dots + a_i(\gamma^2-1)T^2 + \dots -$$ -is non-zero as $a_i(\gamma^2-1)$ is non-zero. -Let me point out that one needs to confront this $\mu$ issue in some way. Possibly the two-variable $p$-adic $L$-function could have looked like -$$ -L_p(S,T) = (S - (\gamma^2-1)) + 0T + 0T^2 + \dots -$$ -The specialization of this power series to weight 2 then vanishes. But note, this would mean that every form in this Hida family has positive $\mu$-invariant, and moreover, these $\mu$-invariants blow up as you approach weight 2. (Possibly there is some easy reason why this can't happen, but I can't see one.)<|endoftext|> -TITLE: Square roots of $\mathbb R^{2n}$ -QUESTION [34 upvotes]: Recently, Richard Dore asked us if $\mathbb R^3$ is the cartesian square of some space, and Tyler Lawson answered beautifully in the negative. -The even powers of $\mathbb R$ were left out in that question because, well, it is obvious that they are squares. Now: - -Are they squares in a unique way? In other words, if a space $X$ is such that $X\times X\cong\mathbb R^{2n}$, must $X$ be homeomorphic to $\mathbb R^n$? - -One can consider other values of $2$ in Richard's question or here, as well as look for factors instead of only square roots (but if I recall correctly $\mathbb R^5$ has all exotic $\mathbb R^4$s as factors, so the last variant might be «trivial»...) - -REPLY [28 votes]: I'm pretty certain the answer is no provided $n \geq 3$. Let $X$ be the Whitehead manifold: http://en.wikipedia.org/wiki/Whitehead_manifold -It's a contractible open 3-manifold which is not homeomorphic to $\mathbb R^3$. -$X^2$ I claim is homeomorphic to $\mathbb R^6$. I don't have a slick proof of this. The idea is, ask yourself if you can put a boundary on $X^2$ to make $X^2$ the interior of a compact manifold with boundary. Larry Siebenmann's dissertation says if $X^2$ is simply-connected at infinity, you're okay. But the fundamental-group of the end of $X^2$ has a presentation of the form $(\pi_1 Y * \pi_1 Y) / <\pi_1 Y^2>$, where $*$ denotes free product and angle brackets "normal closure", and $\pi_1 Y$ is "the fundamental group at infinity for $X$". So $\pi_1 X^2$ is the trivial group. -Once you have it as the interior of a compact manifold with boundary, the h-cobordism theorem kicks in and tells you this manifold is $D^6$.<|endoftext|> -TITLE: Can Chern class/character be categorified? -QUESTION [19 upvotes]: The Chern character sends the class of a locally free sheaf to the cohomology ring of the underlying variety X. And it is a ring homomorphism from K to H^*. I saw people write its source as the bounded derived category too, which make sense if the underlying variety is smooth (sending a bounded complex to the alternating "sum" of the Chern characters of its cohomology sheaf). -My question is, if I want to think $D^b(X)$ as a certain categorification of $K_0(X)$, is it possible to categorify the chern character map? What will be a good candidate of the target category? (Or is there a heuristic showing this is not likely to be true?) - -REPLY [10 votes]: The paper http://arxiv.org/abs/0804.1274 of Toën-Vezzosi is about categorifying the Chern character. Let me try to summarize their strategy. -First of all they introduce a triangulated $2$-category $Dg(X)$ of derived categorical sheaves on a (derived) scheme $X$. It is based on a the idea that a categorification of the theory of modules on a commutative ring $k$ is given by $k$-linear categories: they argue that dg-categories can be used in order to categorify homological algebra in a similar but better way (better in the sens that the non-dg setting seems to be too rigid to allow push-forwards). -The second step is to use, for a given (derived) scheme $X$, the pull-back along the projection $LX\to X$. For a categorical sheaf $F$ on $X$ on consider its pull-back $p^*F$, which naturally come equipped with a self-equivalence $u$. The rough idea to see this is to consider the pull-back (a-k-a >transgression) along the evaluation map $S^1\times LX\to X$, and then to observe that categorical sheaves on $S^1\times LX$ are categorical sheaves on $LX$ together with a $\mathbb{Z}$-action. -Finally, they conjecture the existence of an $S^1$-equivariant trace $Tr^{S^1}(u)\in D^{S^1}(LX)$. Its $K_0$ would be a candidate for the (categorified) Chern character of $F$. -Why does this categorify the Chern character ? -If we do the same construction starting with a sheaf of $X$, then we get in the end an element in $\pi_0(\mathcal O_{LX}^{S^1})=HC_0^{-}(X)$ (while the non-$S^1$-invariant trace takes values in $\pi_0(\mathcal O_{LX})=HH_0(X)$). -One can show that this constructs the ususal Chern character. The main difficulty is the (conjectural) existence of the $S^1$-invariant trace. -Follow-up -A complete treatment of this approach (together with a proof of the conjecture) has been done by the above mentioned authors in a long paper in french.<|endoftext|> -TITLE: Are Pappus Theorems generalized? -QUESTION [6 upvotes]: Pappus' Centroid Theorems provide a slick way of computing the center of mass for plane curves and plane areas. -The first theorem states that the surface area $A$ of a surface of revolution generated by rotating a plane curve $\Gamma$ about an axis external to $\Gamma$ and on the same plane is equal to the product of the arc length $s$ of $\Gamma$ and the distance $d$ traveled by its geometric centroid.$$A=sd$$ -The second theorem states that the volume $V$ of a solid of revolution generated by rotating a plane figure F about an external axis is equal to the product of the area $A$ of F and the distance $d$ traveled by its geometric centroid. $$V=ad$$ -In both theorems, $d=2 \pi y_{c}$, where $y_{c}$ is the required centroid. When we were taught in class the techniques of evaluation of the centroids of three dimensional figures via integration, I remember that the evaluation of the centroid of the hollow hemisphere was particularly difficult for me. So I ask if there are Pappus-like theorems which one could apply for three dimensional bodies? - -REPLY [5 votes]: There is a neat way of finding the centroid of surfaces/volumes of revolution. Suppose you have a planar figure $\Gamma$ and an axis $\ell$ in the plane that is disjoint from $\Gamma$. Now suppose that the centroid of $\Gamma$ projects on $\ell$ as $O$, and that it's distance from $\ell$ is $r$. Suppose we rotate $\Gamma$ around $\ell$ with an angle of $\alpha$, and call the resulting body $\Gamma(\alpha)$. -To find the centroid of $\Gamma(\alpha)$, $G_{\alpha}$, it is enough to measure it's distance from $O$, since $OG_{\alpha}\perp \ell$ and $G_{\alpha}$ is on the dihedral bisector of the angle of revolution. Let's denote $|OG_{\alpha}|$ by $f(\alpha)$, so that by cutting $\Gamma(\alpha)$ into two $\Gamma (\alpha/2)$'s, and using Archimedes lemma that the centroid of the union of two objects is in the line joining the separate centroids, we get the functional equation $$f(\alpha)=f(\alpha/2)\cos (\alpha/4)$$ with initial condition $f(0)=r$, so that by continuity the solution is $f(\alpha)=2r\frac{\sin (\alpha/2)}{\alpha}$. This method takes care of all examples like the hollow hemisphere etc.<|endoftext|> -TITLE: What information Hilbert polynomial encodes other than dimension, degree and arithmetic genus? -QUESTION [18 upvotes]: Consider the Hilbert polynomial for a projective scheme. -The degree, dimension and arithmetic genus extract information from the lowest term and the highest term in the polynomial. What about all other terms? It would seem they encode some more info about our scheme. I could not find any reference to these coefficients, though. -So my question is what else can we learn about our scheme from the Hilbert polynomial? -Thanks - -REPLY [19 votes]: Another way to look at this is the following: In my opinion, the important invariant is the Hilbert polynomial. It is not a single number, but it is still an invariant. -Actually one should be careful with what one means by the Hilbert polynomial. It is really the Hilbert polynomial with respect to an ample line bundle or what's the same an embedding. -Anyway, my point is that the Hilbert polynomial is an invariant that subschemes of projective space have to share in order to be deformation equivalent. (It is not sufficient for that though!) -This implies that when you are constructing moduli spaces you fix the Hilbert polynomial first. The resulting moduli space is still possibly disconnected, but it will be of finite type. (Insert here a longer discussion of Hilbert schemes.) -The fact, as Donu has already pointed out, that for curves the Hilbert polynomial is equivalent to the dimension, degree, and arithmetic genus is sort of a special case. These are obvious invariants that had been studied independently of Hilbert polynomials. It is a reassurance of their importance that they make up the Hilbert polynomials of curves. One can imagine that instead of defining Hilbert polynomials, one could define the various coefficients along the way Donu explains and then get similar results, but I think it would become pretty clumsy that way as if you consider the coefficients individually their transformation rules become pretty complicated.<|endoftext|> -TITLE: Are Chow groups generated by local complete intersections? -QUESTION [14 upvotes]: Let $X$ be a smooth projective variety over an algebraically closed field. The Chow group $\mathbb Q\mathrm{CH}^d(X)$ is $\mathbb Q$--linearly generated by irreducible subvarieties $Z \subseteq X$ of codimension $d$, modulo rational equivalence. -I am interested in the linear subspace of $\mathbb Q\mathrm{CH}^d(X)$ which is generated by the subvarieties $Z\subseteq X$ of codimension $d$ which are locally complete intersections, so those which are locally the zero set of exactly $d$ regular functions. Let us denote this subspace by $\mathbb Q\mathrm{CH}^d_{\mathrm{lci}}(X)$. Then the question is: - -Are Chow groups generated by local complete intersections? I.e. does equality - $$\mathbb Q\mathrm{CH}^d_{\mathrm{lci}}(X) = \mathbb Q\mathrm{CH}^d(X)$$ - hold? - -If for instance $d=1$, equality holds indeed, as $X$ is smooth. I suspect this not so in general for $d\geq 2$... but where to look for a counter example? - -REPLY [7 votes]: This is not an answer to the original question. Instead I will argue that the SUBRING of the Chow ring generated by lci subschemes is the whole ring. -Indeed, let us first check that Chern classes of vector bundles generate the Chow ring. Indeed, the structure sheaf of any subvariety $Z$ has a locally free resolution, hence its Chern classes (in particular the class of $Z$ itself) can be expressed as a liner combination of Chern classes of vector bundles. -So, it remains to check that Chern classes of vector bundles can be expressed as linear combinations of products of lci subschemes. Let us take a vector bundle $E$ of rank $r$. Let $O(h)$ be a very ample line bundle. Then for $n \gg 0$ the bundle $E(nh)$ is globally generated. Hence its top Chern class is represented by the zero locus of a generic section of $E(nh)$ which is lci. Considering different twists one can deduce that for any $i$ the class $c_i(E)h^{r-i}$ is represented by a linear combination of lci. -Now instead of zero loci of sections, consider degeneration schemes of morphisms $O^k \to E(nh)$. Again, the class of the degeneration scheme is $c_{r-k+1}(E(nh))$ and for generic morphism it is a smooth (and hence lci) subvariety. Taking different twists we conclude that $c_i(E)h^{r-k+1-i}$ is represented by a linear combination of lci.<|endoftext|> -TITLE: An effective way to tell if the saturation of a homogeneous ideal is the irrelevant ideal -QUESTION [7 upvotes]: Let $\Bbbk$ be an algebraically closed field, let $R$ denote the graded ring $\Bbbk[x_0, \dotsc, x_N]$, and let $f_1, \dotsc, f_n \in R_m$ be nonconstant homogeneous polynomials. Then the common vanishing locus $V(f_1, \dotsc, f_n) \subset \mathbb{P}^N$ is empty if and only if the saturation of the homogeneous ideal $I = (f_1, \dotsc, f_n)$ is the entire irrelevant ideal $R_+$. This is true iff for some $d > 0$, $I_d = R_d$ (the degree-$d$ parts are equal), in which case $I_e = R_e$ for all $e \geq d$. -It is not hard to compute the vector subspace $I_d \subset R_d$ for successive values of $d$: if $I_d$ is generated as a $\Bbbk$-module by $a_1, \dotsc, a_k \in R_d$, then $I_{d+1}$ is generated by the $x_i a_j$, plus any of the $f_i$ that are of degree $d+1$. -If you want to show that $I$ does, in fact, have saturation equal to the entire ideal $R_+$, you can start computing the vector subspaces $I_d$; if you're right, then sooner or later you'll get $I_d = R_d$ and have your answer. But suppose you go on and on, and $I_d$ remains stubbornly a proper subspace of $R_d$. Is there some point--when $d$ is a thousand, a million, $10^{100}$—at which you can say, "If $I_d$ does not contain all $R_d$ by now, it never will"? - -Does there exist $D$, depending only on the degrees of the $f_i$, sufficiently large that if $I_d = R_d$ for any $d$, then $I_D = R_D$? - -I'm reasonably confident that the answer to that question is yes, based on the following sketch: Look at the space $V$ of all $n$-tuples of homogeneous polynomials $(f_1, \dotsc, f_n)$ with fixed degrees $d_1, \dotsc, d_n$. Let $S_d \subset V$ be the subset of those for which $I_d = R_d$. Since the condition on $S_d$ comes down to the condition that some linear map of vector spaces is surjective, $S_d$ is Zariski-open. Thus, $S_d \subset S_{d+1} \subset S_{d+2} \subset \dotsb$ is an increasing union of Zariski-open sets, and consequently must stabilize at some $D$. -Unfortunately, this argument is entirely non-effective. We have no idea what the value of $D$ is, and so if we actually want to show that $I^{sat} \neq R_+$, we're out of luck. This motivates the following question: - -Assuming an affirmative answer to the previous question, what is a (preferably computable) function $$D = D(d_1, \dotsc, d_n)$$ that works? - -REPLY [4 votes]: Yes, there exists an effective bound on $D$. I am not sure who first found such a bound, but here is a nice reference : -MR2198324 : Jelonek, Z. On the effective Nullstellensatz. Invent. Math. 162 (2005), no. 1, 1--17. -The minimal number $e=e(I)$ such that $I \supset (\sqrt{I})^e$ is called the Noether exponent of $I$. The above article gives an effective bound for $e(I)$. -More precisely, in the situation at hand, one may assume $n>N$ and also $d_1 \geq d_2 \geq \cdots \geq d_n$. Then Jelonek proves that $e(I) \leq (d_1 \cdots d_N) \cdot d_n$ (see Corollary 1.4 with $X=\mathbf{P}^N$). -Thus the function $D(d_1,\ldots,d_n)=(d_1 \cdots d_N) \cdot d_n$ (with $d_1 \geq \cdots \geq d_n$) works.<|endoftext|> -TITLE: How does the group algebra look as a Lie algebra -QUESTION [11 upvotes]: It's probably a well known question, so it is just a reference question. -Let $G$ be a finite group and let $C[G]$ be a group algebra. Then we can define a bracket on $C[G]$ by $[f,h]=f*h-h*f$. What does $C[G]$ look like as a Lie algebra? When is it solvable? - -REPLY [5 votes]: Concerning the situation in characteristic $p$: when $p$ divides the order of $G$, the case not covered by Maschke's theorem, the group algebra $KG$ is no longer semisimple. There is, however, a whole array of papers, started from J.D. Donald and F.J. Flanigan, A deformation theoretic version of Maschke's theorem for modular group algebras: the commutative case, J. Algebra 29 (1974), 98-102, DOI:10.1016/0021-8693(74)90114-8, aiming to prove a conjecture which can be considered as a modular analog of Maschke's theorem: the group algebra KG is deformed to a semisimple algebra. Most of these papers have a group-theoretic flavor, arguing in terms of blocks and other group-representation-theoretic data. Murray Gerstenhaber and Anthony Giaquinto have claimed in: Compatible deformations, Trends in the Representation Theory of Finite Dimensional Algebras (ed. E.L. Green and B. Huisgen-Zimmerman), Contemp. Math. 229 (1998), 159-168, that there is a counterexample to this conjecture: a 8-element quaternion group over a field of characteristic 2. This was believed to be true for a decade or so, after it has been proved wrong (N. Barnea and Y. Ginosar, A separable deformation of the quaternion group algebra, Proc. Amer. Math. Soc. 136 (2008), 2675-2681, DOI: 10.1090/S0002-9939-08-09480-X, arXiv:0704.1556). As far as I know, the Donald-Flanigan conjecture is still open.<|endoftext|> -TITLE: Classifications of finite simple objects -QUESTION [7 upvotes]: I'm curious to know if other classifications are known of "finite simple structures" in the same spirit of the monumental classification of finite simple groups. Here I mean "classification" in the informal sense of the term, but also answers that take into account a more sophisticated viewpoint (as in this mo question) are welcome, as well as answers that consider reasonable weaker notions of classification (as e.g. this mo question, that asks about a classification of finite simple groups up to finitely many exceptions). -The (apperarently incomplete) case of finite commutative rings has already been discussed here. The finite p-groups have been considered here. -Also answers/remarks involving the classification of "finite simple objects" of some category (or higher category) are considered in topic (provided that a reasonable definition of "finite" and "simple" is suggested in that context). - -REPLY [3 votes]: Somewhat related to Igor Pak's comment is the classification of the finite irreducible Coxeter groups. Of course they are not "simple" as groups, but the irreducibility seems the natural replacement for simplicity; here "irreducible" means that the Coxeter diagram is connected, or equivalently, that the Coxeter system does not split as the direct product of two Coxeter systems. -The outcome is the famous list $A_n$, $B_n = C_n$, $D_n$, $E_6$, $E_7$, $E_8$, $F_4$, $G_2$, $H_3$, $H_4$, $I_{(n)}$, where the last three items are maybe less well known people only familiar with Lie groups and Lie algebras and/or algebraic groups since they don't survive there. -See also http://en.wikipedia.org/wiki/Coxeter_group.<|endoftext|> -TITLE: Is there a convenient differential calculus for cojets? -QUESTION [24 upvotes]: I understand the basics of exterior differential geometry and how to do calculus with exterior differential forms. I know how to use this to justify the notation dy/dx as a literal ratio of the differentials dy and dx (by treating x and y as scalar-valued functions on a 1-dimensional manifold and introducing division formally). I would like to extend this to second derivatives. Ideally, this would justify the notation d2y/dx2 as a literal ratio. -I can't do this with the exterior differential, since both d2y and dx ∧ dx are zero in the exterior calculus. It occurs to me that this would work if, instead of exterior differential forms (sections of the exterior bundle), I used sections of the cojet bundle (cojet differential forms). In particular, while degree-2 exterior forms may be written in local coordinates as linear combinations of dxi ∧ dxj for i < j (so on a 1-dimensional manifold the only exterior 2-form is zero), degree-2 cojet forms may be written in local coordinates as linear combinations of d2x and dxi · dxj for i ≤ j (so on a 1-dimensional manifold the cojet 2-forms at a given point form a 2-dimensional space). -I know some places to read about cojets (and more so about jets) theoretically, but I don't know where to learn about practical calculations in a cojet calculus analogous to the exterior calculus. In particular, I don't know any reference that introduces the concept of the degree-2 differential operator d2, much less one that gives and proves its basic properties. I've even had to make up the notation ‘d2’ (although you can see where I got it) and the term ‘cojet differential form’. I can work some things out for myself, but I'd rather have the confidence of seeing what others have done and subjected to peer review. -(Incidentally, I don't think that it is quite possible to justify d2y/dx2; the correct formula is d2y/dx2 − (dy/dx)(d2x/dx2); we cannot let d2x/dx2 vanish and retain the simplicity of the algebraic rules. It would be better to write ∂2y/∂x2; the point is that this is the coefficient on dx2 in an expansion of d2y, just as ∂y/∂xi is the coefficient of dy on xi when y is a function on a higher-dimensional space. The coefficient of d2y on d2x, which would be ∂2y/∂2x, is simply dy/dx again.) - -REPLY [4 votes]: As far as I can tell, your example computation in the comments is a computation in the Hasse-Schmidt ring of a polynomial algebra. Given a commutative ring $A$ and $A$-algebras $f:A \to B$ and $A \to R$, an order $k$ Hasse-Schmidt differential from $B$ to $R$ is a $k+1$-tuple $(D_0,\ldots,D_k)$ of $A$-module maps from $B$ to $R$ satisfying: - -$D_i(f(a)) = 0$ for all $i \geq 1$ and all $a \in A$. -$D_i(b_1 \cdot b_2) = \sum_{j=0}^i D_j(b_1) D_{i-j}(b_2)$. - -We write $Der^k_A(B,R)$ for the set of order $k$ differentials from $B$ to $R$. There is a Hasse-Schmidt algebra $HS^k_{B/A}$ with universal $k$-derivation that represents the functor $Der^k_A(B,-)$, and its relative spectrum over $\operatorname{Spec} B$ is the relative $k$th jet space of $B/A$. For example, $HS^0_{B/A} = B$, and $HS^1_{B/A} = Sym_B^*(\Omega_{B/A})$ yields the tangent bundle. You can find this information in Vojta's EGA-style exposition. -Concretely, here is your example: Let $A$ be a ring such as $\mathbb{R}$, and let $B = A[x]$. It is not hard to show that $HS^k_{B/A} \cong B[x^{(1)},x^{(2)},\ldots,x^{(k)}]$, with canonical maps $y \mapsto y^{(i)}$. In terms of ordinary differentials, we have $x^{(i)} = \frac{1}{i!}d^ix$, and in particular, if we were to write the higher Leibniz rule with differentials, we would need some $\binom{i}{j}$ factors. At any rate, repeated use of the Leibniz relation yields $d^2(x^3-3x) = (3x^2-3)d^2x + 6x(dx)^2$. -If you want to differentiate something twice, you use the fact that for any $y$, $dy$ is equal to $d^0 y' dx$ for some $y' \in B$, then apply the quotient rule: $d\left(\frac{dy}{dx}\right) = \frac{dx\cdot d^2 y - dy \cdot d^2 x}{(dx)^2}$. In particular, you find that $d\left(\frac{dy}{dx}\right)/dx \neq \frac{d^2 y}{(dx)^2}$, because $\frac{d^2x}{(dx)^2} \neq 0$ in this ring. If you really want the somewhat misleading notation $\frac{d^ky}{dx^k}$ to literally denote a $k$th derivative, you have to mod out by the ideal generated by $d^ix$ for $i \geq 2$.<|endoftext|> -TITLE: Hirzebruch's motivation of the Todd class -QUESTION [61 upvotes]: In Prospects in Mathematics (AM-70), Hirzebruch gives a nice discussion of why the formal power series $f(x) = 1 + b_1 x + b_2 x^2 + \dots$ defining the Todd class must be what it is. In particular, the key relation $f(x)$ must satisfy is that -($\star$) the coefficient of $x^n$ in $(f(x))^{n+1}$ is 1 for all $n$. -As Hirzebruch observes, there is only one power series with constant term 1 satisfying that requirement, namely -$$f(x) = \frac{x}{1-e^{-x}} = 1 + \frac{x}{2}+\sum_{k\geq 2}{B_{k}\frac{x^{k}}{k!}} = 1 + \frac{x}{2} + \frac{1}{6}\frac{x^2}{2} - \frac{1}{30}\frac{x^4}{24} + \dots,$$ -where the $B_k$ are the Bernoulli numbers. -The only approach I see to reach this conclusion is: - -Use ($\star$) to find the first several terms: $b_1 = 1/2, b_2 = 1/12, b_3 = 0, b_4 = -1/720$. -Notice that they look suspiciously like the coefficients in the exponential generating function for the Bernoulli numbers, so guess that $f(x) = \frac{x}{1-e^{-x}}$. -Do a residue calculation to check that this guess does satisfy ($\star$). - - -My question is whether anyone knows of a less guess-and-check way to deduce from ($\star$) that $f(x) = \frac{x}{1-e^{-x}}$. - -REPLY [5 votes]: A pure topological derivation of the characteristic power series of the Todd class can be obtained by looking at pushforward maps in complex oriented cohomology theories. This is an outline of the argument: basically, since both topological complex $K$-theory and even $2$-periodic rational singular cohomology are complex oriented cohomology theories, it is possible to define integration maps (for stably complex $X$): -$$ -\int_X^{K}:K(X) \to K(\mathrm{pt}) \cong \mathbb{Z} -$$ -$$ -\int_X^{HP_{\mathrm{ev}}\mathbb{Q}}: \prod_{i\in \mathbb{Z}}H^{2i}(X;\mathbb{Q}) \to \prod_{i\in \mathbb{Z}}H^{2i}(\mathrm{pt};\mathbb{Q}) \cong \mathbb{Q}. -$$ -It is easy to see, by doing some consideration on how integration maps are defined, that these integration maps are non natural with respect to the Chern character $\mathrm{ch}$, and that there exists a class $\mathrm{td}_X \in \prod_{i\in \mathbb{Z}}H^{2i}(X;\mathbb{Q})$ (which is defined to be the Todd class of $X$) such that -$\require{AMScd}$ -\begin{CD} -K(X) @>{\mathrm{ch}_X(-)\cdot \mathrm{td}_X}>> \prod_{i \in \mathbb{Z}}H^{2i}(X;\mathbb Q)\\ -@V{\int_X^K}VV @VV{\int_X^{HP_{\mathrm{ev}}\mathbb{Q}}}V \\ -\mathbb{Z} @>{\mathrm{ch}_{\mathrm{pt}}}>> \mathbb{Q} -\end{CD} -is commutative. The integration maps are defined by means of the Thom isomorphisms (multiplications by the Thom class) for complex vector bundles in a complex oriented cohomology theory. Now, by using the properties of these Thom classes and an application of the splitting principle, it is possible to show that the Todd class $\mathrm{td}_X$ can be expressed as a product of pullbacks of single cohomology class, namely -$$ -\dfrac{c_1(\mathcal{O}(1))}{1-e^{-c_1(\mathcal{O}(1))}} \in \prod_{i \in \mathbb{Z}}H^{2i}(\mathbb{P}^\infty;\mathbb{Q}) \cong \mathbb{Q}[[c_1(\mathcal{O}(1))]], -$$ -which shows that the characteristic power series of the Todd class is exactly $f(t)=t/(1-e^{-t})$.<|endoftext|> -TITLE: Lie algebra semisimple if and only if perfect? -QUESTION [6 upvotes]: If $L$ is a semisimple lie algebra then $L=[L,L]$. Is the opposite true? - -REPLY [29 votes]: No. A Lie algebra satisfying that property is called perfect. For an example of a perfect Lie algebra that isn't semisimple, take a semisimple $L$ and a nontrivial irreducible representation $V$ of $L$, and define a bracket on $L \times V$ by -$$ [(X,v),(Y,u)] := ([X,Y],Xu-Yv). $$ -This turns $L \times V$ into a perfect Lie algebra with $\text{Rad}(L \times V) = V$. -(Note (YCor): this is a semidirect product, hence usually denoted $L\ltimes V$.)<|endoftext|> -TITLE: p-adic Langlands in Families -QUESTION [6 upvotes]: Let $E/\mathbb{Q}_p$ be a finite extension. Let $\rho$ be a continuous irreducible representation of $Gal(\overline{\mathbb{Q}}_p/\mathbb{Q}_p)$ on a two dimensional $E$-vector space. To $\rho$, via the p-adic Langlands correspondence, there is an associated $E$-Banach vector space representation of $GL_2(\mathbb{Q}_p)$, which we denote by $\Pi(\rho)$. -Is it known if this construction behaves well in $p$-adic families? More precisely, if $S$ is a reduced affinoid algebra and $\rho_S$ is a continuous $S$-linear representation of $Gal(\overline{\mathbb{Q}}_p/\mathbb{Q}_p)$ on a free rank two $S$-module is there an associated $S$-Banach module, $\Pi(\rho_S)$, equipped with a contiuous $S$-linear action of $GL_2(\mathbb{Q}_p)$, such that for all $x \in Sp(S)$, $S/x \otimes\Pi(\rho_S) \cong \Pi(\rho_{S,x})$? - -REPLY [8 votes]: There is a discussion of the behaviour of $\Pi(\rho_S)$ in Colmez's Asterisque paper which says that the answer is essentially yes. (Probably it deals with the case of a formal family rather than a rigid analytic one, though.) See also the discussion in section 3 of my -paper on local-global compatibility, which presents the deformation-theoretic formulation -of $p$-adic local Langlands as originally suggested by Kisin. (Again, this will handle formal families rather than rigid ones.)<|endoftext|> -TITLE: Top degree local cohomology under action by a non-zerodivisor -QUESTION [8 upvotes]: Let $R$ be a noetherian commutative ring of dimension $n$, and let $M$ be a faithful finite $R$-module. Let $I$ be a proper ideal of $R$, and let $x\in I$ be a non-zerodivisor on $M$. -When does multiplication by $x$ induce an injection $H^n_I(M)\hookrightarrow H^n_I(M)$? - -REPLY [6 votes]: Graham was right, the map is not necessarily $0$ as I wrote in the first comment. However, it is true that $H_I^n(M)$ is $I$-torsion, so it will be injective if and only if $H_I^n(M)=0$. -Amusingly, I will observe that the map is actually surjective. -Apply $\Gamma_I(-)$ to the sequence: -$$ M \stackrel{x}{\to} M \to M/xM$$ -to get $$ \to H_I^n(M) \stackrel{x}{\to} H_I^n(M) \to H_I^n(M/xM) \to $$ -Now note that $\dim M/xM < \dim M = n$ because $x$ is $M$-regular, so $H_I^n(M/xM) = 0$. (In general, $H_I^n(N) =0 $ for $n>\dim N$). So the multiplication by $x$ map is surjective, as claimed (may be this is what you had in mind anyway). -For completeness, the question of when $H_I^n(M) =0$ is rather subtle. It will be true, for example, if $I$ can be generated up to radical by at most $n-1$ elements, because you can calculate local cohomology using the Cech complex on those elements. -Another instance is when $R$ is a complete local domain, and $\dim R/I>0$ (this is known as the Hartshorne-Lichtenbaum vanishing theorem). I do not know an easy equivalent condition off the top of my head.<|endoftext|> -TITLE: Non-algebraic curve visualisation -QUESTION [7 upvotes]: Is there any software which can automatically visualise a non-algebraic -complex curve, I mean the structure of it's ramification points and sheet? -I think a good test example would be the Lambert curve $y\exp y =x$ -(what I really need is a bit more complicated family of the curves). - -REPLY [3 votes]: I have found a good and simple paper Graphing Elementary Riemann Surfaces by Robert M. Corless and David J. Jeffrey with an explanation how to use Maple for graphing Riemann surfaces. In particular the Lambert curve is amnong the examples they consider.<|endoftext|> -TITLE: Subwords of the Fibonacci word -QUESTION [19 upvotes]: The Fibonacci word is the limit of the sequence of words starting with "$0$" and satisfying rules $0 \to 01, 1 \to 0$. It's equivalent to have initial conditions $S_0 = 0, S_1 = 01$ and then recursion $S_n= S_{n-1}S_{n-2}$. -I want to know what words cannot appear as subwords in the limit $S_\infty$. At first I thought $000$ and $11$ were the only two that could not appear. Then I noticed $010101$. Is there any characterization of which words can or cannot appear as subwords of the Fibonacci word? -Loosely related, this word appears as the cut sequence of the line of slope $\phi^{-1}$ through the origin where $\phi = \frac{1 + \sqrt{5}}{2}$. - -REPLY [5 votes]: There is a simple algorithm to determine the set of words that do not appear as factors (subwords) in the Fibonacci word $f$ (which comes essentially from the characterization Ale De Luca pointed out in his question). -Take the Fibonacci sequence $(F_n)_{n\geq 0}=1,1,2,3,5,8,\ldots$. For every $n$ larger than $1$ take the word of length $F_n-2$ appearing at the beginning of the Fibonacci word $f$ (i.e., the prefix of $f$ of length $F_n-2$). This gives you the sequence of words $\lambda,0,010,010010,\ldots$ Now, for each of these words $v$, if $v$ is followed by symbol $1$ then take $0v0$; otherwise, if $v$ is followed by symbol $0$, then take $1v1$. This gives you the set of words of minimal length that do not appear as factors in $f$, that is, the minimal forbidden factors of $f$: $11,000,10101,00100100,\ldots$ -The words that do not appear as subwords in the Fibonacci word $f$ are precisely the words over $\{0,1\}$ that contain a minimal forbidden factor.<|endoftext|> -TITLE: When is the torsion subgroup of an abelian group a direct summand? -QUESTION [20 upvotes]: For an abelian group $G$, let $G[\operatorname{tors}]$ be its torsion subgroup. -Consider the torsion sequence: -$0 \rightarrow G[\operatorname{tors}] \rightarrow G \rightarrow G/G[\operatorname{tors}] \rightarrow 0$. - -For which torsion abelian groups $T$ is it the case that for all abelian groups $G$ with $G[\operatorname{tors}] \cong T$, the torsion sequence splits? - -I know some sufficient conditions: -1) $T$ is divisible. Indeed, this holds iff $T$ is injective as a $\mathbb{Z}$-module iff any short exact sequence $0 \rightarrow T \rightarrow G \rightarrow G/T \rightarrow 0$ splits. -Thus divisibility is necessary and sufficient if one considers arbitrary short exact sequences, but in the special case $T = G[\operatorname{tors}]$ divisibility is not necessary. The torsion sequence also splits if: -2) $T$ has bounded order: $T = T[n]$ for some $n \in \mathbb{Z}^+$. (For this see e.g. see Corollary 20.14 of these notes of K. Igusa.) -I do know some examples where the torsion sequence does not split, e.g., when $G = \prod_{n=1}^{\infty} \mathbb{Z}/p^n \mathbb{Z}$. -But in fact I am interested in the case in which $T$ has "cofinite type", i.e., $T$ can be injected into $(\mathbb{Q}/\mathbb{Z})^n$ for some $n \in \mathbb{Z}^+$. (I am making up the terminology here; if I ever knew what the infinite abelian group people call this, it's not coming to mind at the moment.) -So for instance the simplest case that I don't know at the moment would be something like $T = \mathbb{Z}/p\mathbb{Z} \oplus \mathbb{Q}_p/\mathbb{Z}_p$. -Not that it makes any difference as to what the answer is, but I would be very pleased to hear that the torsion sequence splits whenever $G[\operatorname{tors}]$ has "cofinite type". If you care why, see Theorem 5 here. - -REPLY [22 votes]: These are the (torsion) cotorsion groups. The following follows from a theorem of Baer: - -A torsion abelian group is cotorsion if and only it is direct sum of a divisible torsion abelian group and an abelian group of bounded exponent. - -The original paper of R. Baer is "The subgroup of the elements of finite order of an abelian group", Ann. of Math. 37 (1936), 766-781. (See in particular Theorem 8.1.) -[I have made Baer's paper available here. --PLC]<|endoftext|> -TITLE: Can always a family of symmetric real matrices depending smoothly on a real parameter be diagonalized by smooth similarity transformations? -QUESTION [15 upvotes]: This question is related to another question, but it is definitely not the same. -Is it always possible to diagonalize (at least locally around each point) a family of symmetric real matrices $A(t)$ which depend smoothly on a real parameter $t$? The diagonalization is supposed to be done by similarity transformations with a family of invertible matrices $S(t)$ depending smoothly on $t$. -An equivalent formulation is whether, given a smooth vector bundle $E\to \mathbb R$ over the one-dimensional differentiable manifold $\mathbb R$, and a smooth symmetric bilinear form $b\in E^*\otimes_M E^*$ on the vector bundle, we can always find a local frame of smooth sections of $E$ in which the symmetric bilinear form $b$ is diagonal. -Some results about such diagonalizations are known, for example from this article. There it is proven that we can smoothly choose the eigenvalues of a family of hermitian matrices depending smoothly on a real parameter, given that there are no points where the roots of the characteristic polynomial meet of infinite order. In particular, this works for the analytic case. But because the eigenvectors are computed with respect to the natural basis, it is implicitly assumed that they are orthogonal with respect to this basis. -My need is to find a smooth diagonalization by similarity transforms, which therefore are not necessarily orthogonal in respect to the natural basis, only invertible. Therefore, the restriction is weaker than that of having smooth eigenvectors. -Is it always possible to find such a diagonalization? If not, what are the conditions under which it can be done? Can you provide some references? -Thank you. -Update with the conclusions: -The counterexample of the type given by Michael Renardy and Denis Serre (who also explains it) answers my question (negatively). Initially I though of such examples as being valid only for the problem of eigenvectors, being of the type in the article to which I referred in my question, and I hoped that allowing the transformations to be more general than those orthogonal may avoid this problem. -But I understand now that the two problems are in fact equivalent. I think that the essence of the counterexample is to have the basis made of two rotating vectors on which the quadratic form corresponding to $b(t)$ has opposite signs. In this case, if by absurd we can diagonalize the matrices, even if $S(t)$ are not orthogonal, then in the new basis it is like solving the eigenvalue problem, and this possibility is contradicted by the counterexamples. For the positive definite case, Johannes Ebert pointed that the diagonalization is possible. -Thank you all for your help. - -REPLY [4 votes]: The answer is yes for real analytic or quasi-analytic parametrizations, see the following paper, also for an overview on different types of parametrizations: -Andreas Kriegl, Peter W. Michor, Armin Rainer: Denjoy-Carleman differentiable perturbation of polynomials and unbounded operators. Integral Equations and Operator Theory 71,3 (2011), 407-416. -pdf<|endoftext|> -TITLE: Usefulness of using TQFTs -QUESTION [35 upvotes]: What is a topological feature, that a (some) TQFT (e.g. in 3 or 4 dim) sees but homology/cohomology/homotopy groups don't? Or: what is an example where using classical theories is hard, but using a TQFT is comparatively easy? - -REPLY [27 votes]: All the answers so far have focused on 3 dimensions, but the answer is much more striking in 4 dimensions. Freedman's theorem tells you that classical homology invariants give you complete information about topological, simply-connected 4-manifolds. These classical invariants cannot, however, distinguish between distinct smooth structures on the same topological 4-manifold, and essentially our only technique for distinguishing smooth 4-manifolds is Donaldson's invariant or the Seiberg-Witten invariant or their relatives. These do not quite form a TQFT, but are related to TQFTs. -Edit: On request, a little about how the 4-manifold invariants are related to a TQFT. This is all nicely explained in the beginning of Kronheimer and Mrowka's book Monopoles and 3-manifolds. -There are actually three different theories, denoted $\widehat{\mathit{HM}}$ ("HM-from"), $\check{\mathit{HM}}$ ("HM-to", unfortunately typeset badly here), and $\overline{\mathit{HM}}$. All are close to satisfying axioms for a TQFT assigning a vector space to a 3-manifold and maps to a 4-manifold, at least for connected manifolds. (The vector spaces are infinite dimensional, but finite in each graded piece.) Unfortunately, however you slice it, in each case the invariant associated to a closed 4-manifold in the usual TQFT way (when defined) is zero. -Instead, you use the fact that there is an exact triangle -$$ -\cdots \longrightarrow \widehat{\mathit{HM}} \longrightarrow \overline{\mathit{HM}} \longrightarrow \check{\mathit{HM}}\longrightarrow \cdots -$$ -(with right mapping to left), and the map $\overline{\mathit{HM}}(W)$ is $0$ for $b_2^+(W) \ge 1$. -If you have a 4-manifold $W$ with $b_2^+(W) \ge 2$, you factor it as two cobordisms $W = W_1 \cup_Y W_2$ for some 3-manifold $Y$, with $b_2^+(W_i) \ge 1$. Then the properties above let you map from $\check{\mathit{HM}}(S^3)$, to $\check{\mathit{HM}}(Y)$, backwards in the exact triangle to $\widehat{\mathit{HM}}(Y)$, and then forwards to $\widehat{\mathit{HM}}(S^3)$. The resulting map (from $\check{\mathit{HM}}(S^3)$ to $\widehat{\mathit{HM}}(S^3)$) gives the interesting Seiberg-Witten invariants of $W$.<|endoftext|> -TITLE: A torsionfree group with infinite cohomological dimension and no infinitely generated free abelian subgroup -QUESTION [10 upvotes]: Recently I've been reading about cohomological finiteness conditions for groups, my main source being Brown's book "Cohomology of Groups". -One of the first things one learns is that a group with finite cohomological dimension necessarily is torsionfree. It is easy to come up with an example of a torsionfree group with infinite cohomological dimension: non-finitely generated free abelian group immediately springs to mind. A little bit of googling revealed an example of an infinite-dimensional torsionfree $FP_{\infty}$ group.$^1$ (A group is said to be of type $FP_{\infty}$ if there exists a projective $ZG$-resolution {$P_i$} of $Z$ such that each $P_i$ is finitely generated.) This example is Thompson's group $F$. However, it is well-known that $F$ contains a non-finitely generated free abelian group. -So, the question is: what is an example of a torsionfree group with infinite cohomological dimension and no inifinitely generated free abelian subgroup (that is, if there exists one)? -$^1$ In case anyone is interested: K. S. Brown, R. Geoghegan, Invent. Math. 77, 367--381. - -REPLY [10 votes]: You can take a union of torsion-free word-hyperbolic groups with cohomological dimension approaching infinity to get such an example. -Word-hyperbolic groups are rank one, so one can show that any -abelian subgroup must intersect each subgroup in a cyclic group, and therefore the -subgroup cannot be an infinitely generated free abelian subgroup. -Here's an explicit sequence. Take a sequence of cocompact torsion-free lattices in $\Gamma_n< Isom(\mathbb{H}^n)$, such that $\Gamma_n < \Gamma_{n+1}$ from the natural embedding $Isom(\mathbb{H}^n) < Isom(\mathbb{H}^{n+1})$. The limit $\Gamma_{\infty}=\underset{n\to\infty}{\lim} \Gamma_n$ is a group of cohomological dimension $\infty$ since $\Gamma_n$ is of cohomological dimension $n$. -For example, consider the quadratic form over $\mathbb{Q}(\sqrt{5})$ which is $q_n(x_0,x_1,\ldots,x_n)=((1-\sqrt{5})/2 )x_0^2+x_1^2+\cdots+x_n^2$. This gives a quadratic form over $\mathbb{R}$ of signature $(n,1)$. Consider the group $O(q_n,\mathbb{Z}[\phi])< GL(n+1,\mathbb{Q}(\sqrt{5}))$, where $\phi=(1+\sqrt{5})/2$, the group of integral matrices preserving this quadratic form. Then there exists a prime ideal $\mathcal{P}<\mathbb{Z}[\phi]$ such that $\Lambda_n=Ker\{ GL(n+1,\mathbb{Z}[\phi])\to GL(n+1,\mathbb{Z}[\phi]/\mathcal{P})\}$ is torsion-free for all $n$. Let $\Gamma_n=O(q_n,\mathbb{Z}[\phi])\cap \Lambda_n$. Then $\Gamma_n$ is a torsion-free hyperbolic group acting cocompactly on $\mathbb{H}^n$ (by considering the Lorentzian model of hyperbolic space), and clearly $\Gamma_n<\Gamma_{n+1}$.<|endoftext|> -TITLE: Is an affine "G-variety" with reductive stabilizers a toric variety? -QUESTION [12 upvotes]: Let $X=Spec(A)$ be a reduced normal affine scheme over an algebraically closed field $k$ of characteristic $0$, with an action of a connected reductive group $G$. Suppose - -$x\in X$ is a $G$-invariant $k$-point, -$X$ contains a dense open stabilizer-free $G$-orbit (i.e. a dense open copy of $G$), and -the stabilizers of all points of $X$ are reductive. - -Must $G$ be a torus (and therefore $X$ a toric variety)? - -Thoughts so far -I'll include the ideas I've had so far in the hope that somebody might see a way to use them to answer the question. -Remark 1: Choosing a set of generators $f_1,\dots, f_r\in \mathfrak m_x$ for $A$ as an algebra, there is a finite-dimensional $G$-invariant vector space $V^\ast\subseteq \mathfrak m_x$ that contains them. The surjection $Sym^\ast(V^\ast)\to A$ induces $G$-equivariant a closed immersion $X=Spec(A)\to Spec(Sym^\ast(V^\ast))\cong V$ sending $x$ to the origin. So we may assume $X$ is a $G$-invariant closed subvariety of a faithful representation $V$ and that $x$ is the origin. -To fix our ideas, choose a decomposition $G=T_0\cdot H$, where $T_0$ is a central torus and $H$ is semisimple. Choose a borel $B\subseteq H$ and a maximal torus $T\subseteq B$, and let $V = \bigoplus_{\mu\in T^\vee}V_\mu$ be the weight decomposition of $V$ as a representation of $T$. To answer the question affirmatively, it suffices to show that $V$ is trivial as a representation of $H$. That is, that only the weight $\mu=0$ appears in the decomposition. -Remark 2: If $X$ contains a maximal highest weight vector $v$, then $v$ is stabilized by the unipotent radical of $B$. Since points of $X$ have reductive stabilizers, $v$ must be stabilized by all of $H$. Since $v$ was a maximal highest weight vector in $V$, all of $V$ must be stabilized by $H$. -Peter McNamara came up with a trick for cooking up a maximal highest weight vector in the case where $X\subseteq V$ is a cone. However, Torsten Ekedahl showed that we cannot hope that $X$ will be a cone in general. -Remark 3: Another possible way to get an affirmative answer is to find a vector $v\in X$ which is generic (i.e. has non-trivial component in each weight space, or at least in enough weight spaces) such that the torus orbit $T\cdot v$ is not closed. If such a $v$ exists, then there is a non-trivial 1-parameter subgroup $\gamma:\mathbb G_m\to T$ such that $\gamma(t)\cdot v$ has a limit. This means that all (or enough of) the weights of $V$ lie on one side of the hyperplane in $T^\vee$ determined by $\gamma$. Since the character of $V$ is symmetric under the action of the Weyl group of $H$, all the weights must be zero. - -This is sort of a combination of my two previous questions, both of which have gotten wonderful answers: -If a representation has enough reductive stabilizers, is it a direct sum of characters? -If Spec(A) has a G-fixed point and a dense G-orbit, is Spec(A) a cone? -In particular, they helped me pin down this question, which is what I think I'm really after. - -REPLY [7 votes]: Vera Serganova showed me the following (affirmative) answer. I'll use the setup from Remark 1. Note that Remark 2 shows that $X$ cannot contain a positive highest weight vector. So the following result does the job. - -Proposition: Let $V$ be a representation of a reductive group $G$ and let $X$ be the closure of an orbit of $G$ so that $0\in X$. If $X$ is not contained in a direct sum of 1-dimensional representations, then it contains a positive highest weight vector (with respect to some borel). - -Proof. We have that $X=\overline{G\cdot v}$, and we are given that $v$ is not contained in a sum of 1-dimensional representations (otherwise $X$ would be as well). Since $0\in \overline{G\cdot v}$, the Hilbert-Mumford criterion (Proposition 2.4 of GIT) tells us that there is a 1-parameter subgroup $\gamma$ so that $\gamma(t)\cdot v$ contains $0$ in its closure. Let $V=\bigoplus_{i\in \mathbb Z}V_i$ be the weight space decomposition of $V$ with respect to $\gamma$. Then $v=v_p+v_{p+1}+\cdots$ with $v_i\in V_i$, $v_p\neq 0$, and $p>0$. -Let $T$ be a maximal torus of $G$ which contains $\gamma$, and let $B\subseteq G$ be a borel containing $T$ so that $\gamma$ pairs non-negatively with all the positive roots. Since $V$ is finite-dimensional, we can tweak $\gamma$ so that it pairs positively with all the positive roots. -If $v$ is a highest weight vector, we are done. Otherwise there is some positive root $\alpha$ so that $e_\alpha\in \mathfrak g_\alpha$ does not annihilate $v$. Let $\exp(t\cdot e_\alpha)\cdot v = \sum_{i\ge p} f_i(t)$, where $f_i(t)\in k[t]\otimes V_i$, and let $m_i=\deg(f_i)$. Since $e_\alpha\cdot V_i\subseteq V_{i+\langle \gamma,\alpha\rangle}$ and $\langle \gamma,\alpha\rangle>0$, we have that $m_p=0$. Since $e_\alpha\cdot v\neq 0$, we have that $m_i>0$ for some $i>p$. -Let $a/b\in \mathbb Q$ be the positive rational number so that $m_i\cdot b \le i\cdot a$ for all $i$, with $m_j\cdot b = i\cdot a$ for some $j>p$. Define the morphism $g: \mathbb A^1 \to V$ by the formula $g(t) = \sum_{i\ge p} t^{i\cdot a}f(t^{-b})$. The condition that $m_i\cdot b \le i\cdot a$ implies this is a well-defined morphism, and the condition that $m_j\cdot b = j\cdot a$ for some $j>p$ implies that $g(0)\neq 0$. But for $t\neq 0$, we have that $g(t) = \gamma(t^a)\exp(t^{-b}e_\alpha)\cdot v\in X$, so since $X$ is closed, we have that $g(0)\in X$. -Note that $g(0)$ is non-zero, it lies in $\bigoplus_{i>p} V_i$, and it is not contained in a sum of 1-dimensional representations since it is in the image of $e_\alpha$ (which annihilates 1-dimensional representations). Since $V$ is finite dimensional, this procedure can only be repeated a finite number of times, so we get a highest weight vector after a finite number of iterations.<|endoftext|> -TITLE: Weil Conjectures for Number Fields -QUESTION [22 upvotes]: Let $K$ be a number field with integral basis $\{\omega_1,\ldots,\omega_n\}$. -The affine variety $A_K$ defined by -$$ N_{K/{\mathbb Q}}(X_1 \omega_1 + \ldots + X_n \omega_n) = 1 $$ -is an algebraic group, the group structure coming from multiplication -of units with norm $1$; in fact, $A_K$ is a norm-1 torus. For pure cubic extensions ${\mathbb Q}(\sqrt[3]{m})$ with $m \not \equiv \pm 1 \bmod 9$, the unit variety is defined by -$$ X_1^3 + mX_2^3 + m^2X_3^2 - 3mX_1X_2X_3 = 1, $$ -for example. -The affine part of the variety $A_K$ is smooth; the affine part of its -reduction modulo $p$ is smooth if and only if $p \nmid \Delta$, where -$\Delta$ denotes the discriminant of $K$. -For each prime $p \nmid \Delta$ let $N_r$ denote the number of -${\mathbb F}_q$-rational points on $A_K$, where $q = p^r$. Define the -Hasse-Weil zeta function -$$ Z_p(T) = \exp\bigg( \sum_{r=1}^\infty N_r \frac{T^r}r \bigg). $$ -This zeta function has the following properties: - -The zeta function $Z_p(T)$ is a rational function of $T$; the degrees -of numerator and denominators are equal. More exactly, $Z_p(T)$ can -be written in the form -$$ Z_p(T) = \begin{cases} - \frac{P_0(T) P_2(T) \cdots P_{n-1}(T)}{P_1(T)P_3(T) \cdots P_{n}(T)} - & \text{ if $n$ is odd}, \\\ - \frac{P_1(T)P_3(T) \cdots P_{n-1}(T)}{P_0(T) P_2(T) \cdots P_{n}(T)} - & \text{ if $n$ is even}, - \end{cases} $$ - where $P_j(T)$ is a product of terms of the form $1 - \zeta p^{j}T$ for suitable roots of unity $\zeta$. The actual factors $P_j(T)$ essentially - depend only on the prime ideal factorization of $p$ in $K$. -Moreover, $Z_p(\infty) = \lim_{T \to \infty} Z_p(T)$ exists and - satisfies $Z_p(\infty) = \epsilon_p$, where $\epsilon_p = \chi(p)/p$ for - Pell conics (unit varieties for quadratic extensions) and - $\epsilon_p = \pm 1$ in general. -The zeta function $Z_p(T)$ admits a functional equation of the form -$$ Z_p\Big(\epsilon_p \frac1{p^nT}\Big) = \eta_p Z_p(T)^{(-1)^n} $$ -for some constant $\eta_p$ depending only on $p$. -The global zeta function $Z_K(s)$ is constructed as follows: set -$L_p(s) = P_{n-1}(p^{-s})$ and $Z(s) = \prod_p L_p(s)$. Then, up -to Euler factors at the ramified primes, $Z(s) = \zeta_K(s+n-2)/\zeta(s+n-2)$, - where $\zeta_K$ is the Dedekind zeta function of $K$. - -My impression is that the case of Pell conics is slightly different from -the general case because Pell conics are smooth even at infinity. -I am unaware of almost any of the results on norm-1 tori obtained in the last -30 years, and my main question is: - Is all of this a special case of known results on algebraic tori, and if yes, what are the relevant references? -BTW, the rationality and the functional equation seem to be -known in quite general situations. So a more precise question would be whether these unit varieties have received any special attention. - -REPLY [10 votes]: Some of the relevant results can be found in the book "Algebraic Groups and their birational invariants" by V. E. Voskresenskii (Translations of Math Monographs 179, American Math Society 1998). -Specifically, Chapter 4, Section 9 is all about tori over a finite field, number of rational points, and zeta function; this proves the (local) results you mention for a general torus. Section 13 (same chapter) shows how the global zeta function of a general torus is expressible in terms of Artin L-functions. The case of norm-one tori is actually a simple but important case. Here is a link: -http://books.google.com/books?id=O-R8m2qRS04C&lpg=PP1&ots=jnOsvxLq_B&dq=algebraic%20groups%20and%20their%20birational%20invariants&pg=PP1#v=onepage&q&f=false<|endoftext|> -TITLE: Link of singularities -QUESTION [6 upvotes]: For an isolated plane curve singularity, given by homogeneous equation $F=0 \subset \mathbb{C}^2$, one consider the curve $(F=0) \cap S^3 \subset S^3$, and we call it the link of singularity. some properties of the singularity are embedded in the corresponding link and one can start studying the singularity by studyng the invariants of the corresponding link, like Jones polynomial ... -Question: Is there any similar construction for more general singularities? Like when $X$ is an irreducible singular variety and the singular locus $X^{sing}$ is isomorphic to a smooth variety $V$? - -REPLY [5 votes]: I think you will be interested in reading about the so-called Milnor map, and the so-called Minor Fibration Theorem. -The standard reference is John Milnor's book "Singular points of complex hypersurfaces" from 1968. -Although people ususally assume that the hypersurface has an isolated singularity. That means that the singularity has finite multiplicity. The non-isolated case (where the singular set is a smooth variety) is much more complicated. -Nevertheless, as a starting point, I recommend looking at the Milnor Map and the Milnor Fibration Theorem. -There are lots of results on the topology of the Milnor fibre. For example if $f : (\mathbb{C}^n,0) \to (\mathbb{C},0)$ is a holomorphic map germ with an isolated critical point at $0 \in \mathbb{C}^n$ then the Milnor fibre is homotopy equivalent to the bouquet of $\mu$-spheres, where $\mu$ denotes the Milnor number. The Milnor number is given by the absolute value of the Poincaré-Hopf index of the gradient vector field $\nabla f$ at $0 \in \mathbb{C}^n.$ -Take a look at the introduction to this paper for a little more detail and some references. Take a look at this paper for the non-isolated case. Both papers include information about the links.<|endoftext|> -TITLE: Is there a "motivic Gromov-Witten invariant"? -QUESTION [17 upvotes]: I recently attended an interesting seminar, where the concept of motivic Donaldson-Thomas invariants was explained (0909.5088). -Very roughly, the DT invariant is a generating function $\sum q^k e(M_k)$ of a numerical invariant $e(\cdot)$ of a sequence of moduli spaces $M_k$. The motivic DT invariant is obtained by considering $\sum q^k [M_k]$ where $[M_k]$ is the image in $K(Var)$. -This contains more info than the ordinary DT invariant. -Can this idea be applied for, say, the GW invariant of Calabi-Yau 3-folds, to get a finer invariant? -(Sorry for my vague question.) - -REPLY [8 votes]: Let me just add a bit to what Balazs said. The fact that the moduli spaces of sheaves on a CY3 have this symmetric obstruction theory is a reflection of properties of the category they live in, namely the derived category of coherent sheaves on the CY3. Indeed, Kontsevich and Soibelman's general construction of motivic DT invariants applies to a general class of CY3 categories and the motivic invariants live in a Hall algebra associated to this category. -My point is that the natural category associated to Gromov-Witten invariants is the Fukaya category and so I would expect any sort of "motivic" version of GW theory to live in something associated to the Fukaya category. In particular, I don't think these invariants will live in the Grothendieck group of varieties the way the motivic DT invariants do. GW theory (A-model) is inherently analytic (or symplectic) whereas DT theory (B-model) is inherently algebraic. -It would be really interesting to figure out what the analog of the Hall algebra is in the Fukaya category. Maybe the symplectic geometers already know it?<|endoftext|> -TITLE: Gromov-Witten invariants of singular spaces -QUESTION [7 upvotes]: I wonder if there is any situation where one can talk about Gromov-Witten invariants -or quantum multiplication for singular varieties. Ideally, I would like have a situation -where for a singular variety $X$ one can define quantum multiplication operators -by elements of ORDINARY cohomology of $X$ on -the INTERSECTION cohomology of $X$ (I have some examples where I know what I want the answer -to be, but I don't know how to ask the question). -In fact, I will be ready to start with the following simple example: assume that $X$ just -has quotient singularities, i.e. locally it looks like $Y/G$ where $Y$ is smooth -and $G$ is a finite group. In this case the intersection cohomology coincides with the -ordinary cohomology, so my question is whether in this case one can define quantum -multiplication. One warning: I am talking about quantum cohomology of $X$ itself, not -about what is called "orbifold quantum cohomology" (which in many cases -coincides with the quantum cohomology of a good resolution of $X$). - -REPLY [2 votes]: Since no one else has said anything, let me make two naive comments. - -The only case I know of where there is a well developed notion of Gromov-Witten invariants for a singular variety is the very special case when the target admits a gluing $X \cup_D Y$ where X and Y are smooth and projective and D is a smooth divisor. -You write that you are not interested in the orbifold Gromov-Witten invariants, but perhaps it is possible to define something similar to what you want in terms of the orbifold invariants. Let $X = Y/G$ and set $\mathscr X = [Y/G]$. The Gromov-Witten invariants of $\mathscr X$ are given by maps -$$ I_{g,n,\beta} : H^\ast(\overline I \mathscr X)^{\otimes n} \to H^\ast(\overline M_{g,n}),$$ -where $\overline I \mathscr X$ is the rigidified inertia stack of $\mathscr X$. However there is also a natural map $H^\ast(X) \to H^\ast(\overline I \mathscr X)$ induced by: (i) the rigidification morphism $I \mathscr X \to \overline I \mathscr X$, in particular the isomorphism it induces on cohomology; (ii) the forgetful map $I \mathscr X \to \mathscr X$; and (iii) the coarse moduli space map $\mathscr X \to X$. So by precomposition we get a collection of maps $H^\ast(X)^{\otimes n} \to H^\ast(\overline M_{g,n})$, and even though it seems they will not satisfy the axioms required of GW invariants maybe they are what is needed in your situation.<|endoftext|> -TITLE: Is $\zeta(3)/\pi^3$ rational? -QUESTION [33 upvotes]: Apery proved in 1976 that $\zeta(3)$ is irrational, and we know that for all integers $n$, -$\zeta(2n)=\alpha \pi^{2n}$ -for some $\alpha\in \mathbb{Q}$. Given these facts, it seems natural to ask whether we can have -$\zeta(n)=\alpha \pi^n$ for all $n$ (I'm mainly interested in the case $n=3$). The proofs I've seen of the irrationality of $\zeta(3)$ don't seem to give this information. -My gut feeling is that the answer is no, but I can't find any reference proving this fact. I know that the answer hasn't been proven to be yes ($\zeta(3)$ isn't even known to be transcendental), but ruling out this possibility seems an easier problem. - -REPLY [2 votes]: In this article, Takaaki Musha proves that $\zeta(2n+1) \notin (2\pi )^{2n+1} \mathbb{Q}$. I haven't read it so I can't say more. -See this question.<|endoftext|> -TITLE: Clifford PBW theorem for quadratic form -QUESTION [13 upvotes]: $\DeclareMathOperator\Cl{Cl}$Update Feb 3 '12: now with a question 2 which is much more elementary (and should be well-known!). -Let $k$ be a commutative ring with $1$. Let $L$ be a $k$-module, and $g:L\to k$ be a quadratic form, i. e., a map for which the map $L\times L\to k,\ \left(x,y\right)\mapsto g\left(x+y\right)-g\left(x\right)-g\left(y\right)$ is $k$-bilinear and which satisfies -$$g\left(\lambda x\right)=\lambda^2 g\left(x\right)\quad\text{for all $x\in L$ and $\lambda\in k$.}$$ -We define the Clifford algebra $\Cl\left(L,g\right)$ as the tensor algebra $\otimes L$ (all tensors are over $k$) divided by the two-sided ideal generated by $x\otimes x-g\left(x\right)$ for all $x\in L$. -Question 1: Is $\Cl\left(L,g\right)$ isomorphic to ${\bigwedge}L$ as a $k$-module? Is the associated graded object of $\Cl\left(L,g\right)$ isomorphic to ${\bigwedge}L$ as a $k$-algebra? -Remark: The answer is yes if the quadratic form $g$ comes from a bilinear (not necessarily symmetric!) form. As a consequence, the answer is yes if $L$ is a free $k$-module or, more generally, a direct sum of quotient $k$-modules of $k$ (in fact, it is easy to see that in this case, every quadratic form on $L$ comes from a bilinear form). (I think that in the case when $L$ is a finite-free $k$-module, the answer "yes" can also be proven by the diamond lemma, though I have not checked.) I am interested in the "perverse" cases when none of these holds, but I suffer from a lack of perversion: I can't name any such case. So here is the question which obviously needs to be addressed first: -Question 2: Find a commutative ring $k$ with $1$ and a $k$-module $L$ with a quadratic form (this is defined as above) which doesn't come from any bilinear (symmetric or not) form on $L$. (A quadratic form $g$ is said to come from a bilinear form $h$ if we have $g\left(v\right)=h\left(v,v\right)$ for every $v\in L$.) -Note: The Clifford algebra of a quadratic form is, in some sense, a "little sister" of the universal enveloping algebra of a Lie or pseudo-Lie algebra. ("Little sister" not in the historical sense, but in the sense of partly having the same properties, but them being easier to prove in the Clifford case than in the universal enveloping algebra case.) The above question asks for a kind of Poincaré–Birkhoff–Witt (PBW) theorem for the Clifford algebra. (Note that the PBW theorem itself requires some niceness conditions such as $L$ being finite-free or $k$ being a $\mathbb Q$-algebra, so it wouldn't surprise me if the Clifford case also doesn't work in full generality. But the opposite case wouldn't surprise me either, because PBW is substantially harder than PBW for Clifford algebras, even in characteristic $0$.) - -REPLY [3 votes]: If someone could check the below I'd be very indebted. -Found the counterexample (to Question 1 and thus also to Question 2). It is inspired by the counterexample to ring-theoretical PBW in P. M. Cohn, A remark on the Birkhoff-Witt theorem. J. London Math. Soc. 38 1963 pp. 197-203, MR0148717 (though I still don't know whether Cohn's counterexample is true for all primes $p$ - but here I only need $p=2$). -Let $k$ be the commutative ring $\mathbb F_2 \left[\alpha,\beta,\gamma\right] / \left(\alpha^2,\beta^2,\gamma^2\right)$. -Let $L$ be the $k$-module $\left\langle x,y,z\right\rangle / \left\langle \alpha x - \beta y - \gamma z\right\rangle$. -Define a map $q:L\to k$ by $q\left(\overline{ax+by+cz}\right) = a^2+b^2+c^2$ for all $a,b,c\in k$. This map $q$ is easily seen to be well-defined and a quadratic form; it also satisfies $q\left(\overline x\right)=q\left(\overline y\right)=q\left(\overline z\right)=1$. -In the Clifford algebra $\mathrm{Cl}\left(L,q\right)$, we have $\left(\overline{\alpha x-\beta y-\gamma z}\right)\cdot\overline x = \alpha \underbrace{\overline{x}\cdot\overline{x}}_{=q\left(\overline x\right)=1} - \beta \overline y \overline x - \gamma \overline z \overline x = \alpha - \beta \overline y \overline x - \gamma \overline z \overline x$. Since the left hand side of this equality is $0$ (because $\overline{\alpha x-\beta y-\gamma z}=0$), we thus have $ \alpha - \beta \overline y \overline x - \gamma \overline z \overline x = 0$. Multiplied with $\beta\gamma$, this becomes $\alpha\beta\gamma = 0$ (because $\beta^2=0$ and $\gamma^2=0$ make the other two terms disappear). This must hence hold in $\mathrm{Cl}\left(L,q\right)$. But this does not hold in $k$, and thus not in $\wedge L$ either, so we cannot have an isomorphism.<|endoftext|> -TITLE: Flow on Infinite Graphs -QUESTION [9 upvotes]: Assume you have a simple, infinite graph $G$ with bounded degree (there is an upper bound for the degree of the nodes). Choose an arbitrary vertex $x\in V(G)$ and consider -$$ -G_{n}:=\{x\in G:d(x_0,x)\leq n\} -$$ -with the graph metric (hop metric). Assume that each pair of nodes is communicating a unit load of information and the load goes through the minimum path between nodes (if there is more than one minimum path we choose one arbitrarily). The total traffic in $G_{n}$ is equal to $\frac{N(N-1)}{2}$ where $N=N(n)=|G_{n}|$. -Given a node $v\in G_{n}$ we define $T_{n}(v)$ as the total traffic generated in $G_{n}$ passing through $v$. In other words, $T_{n}(v)$ is the sum off all the geodesic paths in $G_{n}$ which are carrying traffic and contain the node $v$. -If the graph $G$ is planar and it has exponential growth $|G_{n}|=K\exp(\lambda n)$ for $n$ sufficiently large, then it is not difficult to prove that there are nodes in $G_{n}$ such that -$$ -T_{n}(v)\geq C\frac{N^2}{\log(N)} -$$ -for $n$ sufficiently large. -My question is the following - -Is the same true if we remove the - planar condition but we keep the - exponential growth? My intuition is - that the answer is no but I can't find - a counterexample. - -REPLY [5 votes]: I think the following provides a counterexample. -The idea is to use the fact that on a graph $G$ with maximum degree $\Delta$ and diameter $D$ reasonably close to the natural lower bound $\log_{\Delta-1}|V(G)|$ the traffic is almost uniformly distributed. -Explicitly, for a vertex $v$, let $S_k(v):=\{x \in V(G) \: | \: d(x,v)=k\}$. Then $|S_k(v)| \leq \Delta(\Delta-1)^{k-1}$ and the traffic through $v$ can be estimated as $$T_G(v) \leq \sum_{k+l \leq D} |S_k(v)||S_l(v)| < D^2\Delta^2(\Delta-1)^{D-2}.$$ -Bollobas and de la Vega in a paper "The diameter of random regular graphs" (last reference for Lecture 1 at the link.) show that for sufficiently large $N'$ a random $r$-regular graph on $N'$ vertices has diameter at most $\log_{r-1}(N') + \log_{r-1}(\log_{r-1}(N'))+C$, where $C$ is a (small) constant depending on $r$. By adjusting $C$, we may assume that an $r$-regular graph on $N'$ vertices satisfying this bound on diameter exists for any $N'$. -Finally, we construct $G$ by, first, taking an infinite $r$-regular tree, except that for simplicity of calculations we let the degree of $x_0$ be $r-1$. Secondly, we put an $r$-regular graph with the diameter bound listed above on the set of vertices $S_k(x_0)$, that is on each level of our infinite tree, considered as rooted at $x_0$. -Note that $|S_k(x_0)|=(r-1)^k$ and the diameter of $G|S_k(x_0)$ (i.e. $G$ restricted to $S_k(x_0)$) is at most $k + \log_{r-1}(k)+C$. It follows that the diameter of $G_n$ is at most $$\max_{k \leq l \leq n}( \mathrm{diam}(G|S_k(x_0)) + l-k) \leq n + \log_{r-1}(n)+C$$ -The graph $G_n$ has maximum degree $\Delta = 2r$ and by the bound on the traffic load above we have -$$ T_n(v) \leq n^2(2r-1)^{n+\log_{r-1}(n)+C'}$$ -for any $v \in V(G_n)$ and some choice of $C'(r)$ independent on $n$. -On the other hand $$|V(G_n)|=N =\sum_{k=0}^n (r-1)^k > (r-1)^n$$ -It follows that, for any $\epsilon >0$ we can choose large enough $r$ so that for large $n$ we have $T_n(v) \leq N^{1 + \epsilon}$ for every $v \in G_n$.<|endoftext|> -TITLE: Applications for p-Sylow subgroups theorem -QUESTION [25 upvotes]: I have searched for such a question and didn't find it. I recently had a presentation in which I introduced $p$-Sylow subgroups and proved Sylow's theorems. I will have another one soon, concerning applications of Sylow's theorem. -My question is: - -Are there any spectacular applications - of Sylow's theorem in group theory and - other fields of mathematics (which are - of course related to groups)? - -REPLY [3 votes]: Even the cyclicity of the groups of order 15, or the existence of a normal Sylow 5-subgroup in any group of order 100, is not merely a toy example. -The fact that Sylow p-subgroups of a finite group are always conjugate is one way to prove that normal implies characteristic for a Sylow p-subgroup. (So if a group has simple subgroups of index 100 which generate it, and no normal subgroup of order 25, the group itself is simple. Hence, the Higman-Sims group is simple because the Mathieu group $M_{22}$ is simple. This is done in Wilson's "The Finite Simple Groups".) Two consequences of this are that if $P$ is a Sylow p-subgroup of a finite group $G$ and $K$ is a subgroup satisfying $N_{G}(P) \leq K \leq G$, then $[K:N_{G}(P)] \equiv [G:K] \equiv 1 \mod{p}$ and $K$ is self-normalizing in $G$. In particular, the maximal subgroups of $G$ containing $N_{G}(P)$ are constrained by these results. -The cyclicity of groups of order 15 is more than just a toy example, since the cyclicity of groups of order 299 = 13*23 (which is provable the same way) is used in Thompson's original proof of the simplicity of the Conway group $Co_{1}$. (This proof also gives an example of the use of the Frattini argument.) -If you want to prove the Burnside $p^{a}q^{b}$-Theorem, you need to exploit the existence of Sylow subgroups. This is one of the few commonalities of the character-theoretic and character-free proofs of the theorem. Via character theory, the basic group-theoretic result is that a finite group with a conjugacy class whose size is a power of a prime cannot be simple -- but you can only get a conjugacy class of size equal to a power of a prime in a group of order $p^{a}q^{b}$ by choosing a nontrivial central element of a Sylow subgroup (unless you made a bad choice and it's in the center of the whole group, in which case nonsimplicity of the group is immediate unless the group is cyclic of prime order). -Eschewing character theory, Sylow subgroups are indispensable, whether you use the Glauberman $ZJ$-theorem or any other local-analytic tools to do the heavy lifting in the proof. They are also essential even for much lighter lifting which happens in these proofs. -When using the transfer to prove a finite group satisfying certain conditions is not perfect, it's good to have a subgroup from which this fact is visible. It is good to have a subgroup $H$ such that one knows $\phi : G \to A$ is nontrivial because its restriction to $H$ is nontrivial. If p is a prime dividing $| \phi(G) |$, then any subgroup whose index is a nonmultiple of p will work. A Sylow p-subgroup of $G$ fits the bill perfectly, and often comes with a fair amount of information about its own structure, to boot. -It is possible to build off of the Burnside $p^{a}q^{b}$-Theorem to prove the that the existence of Sylow systems characterizes finite solvable groups. Sylow system normalizers are all conjugate in a finite solvable group, and these facts form the starting point of the theory of finite solvable groups (which is substantial in its own right, as one can read in "Finite Soluble Groups", by Doerk and Hawkes).<|endoftext|> -TITLE: Strong group ring isomorphisms -QUESTION [25 upvotes]: Background/Motivation -Let $R$ be a commutative ring with unit. If $G$ is a finite (or in general, discrete) group, let $R[G]$ be the group $R$-algebra associated to $G$. The isomorphism problem for group rings asks for condiions on groups $G$ and $H$ such that $R[G]\simeq R[H]$. -Group rings are not a complete invariant, even of finite groups; in a 2001 Annals paper, Hertweck discovered two finite groups $G$ and $H$ with $\mathbb{Z}[G]\simeq \mathbb{Z}[H]$ and $G\not\simeq H$. In general, group algebras over a field are even weaker invariants; for example, if $G$ and $H$ are any two finite abelian groups of order $n$, then $\mathbb{C}[G]\simeq \mathbb{C}[H]\simeq \mathbb{C}^n$, by e.g. the Chinese Remainder Theorem or Artin-Wedderburn. -I am curious about a slight strengthening of the isomorphism problem for group rings. Namely, if $(S, +, \cdot)$ is a (not necessarily commutative) ring with unit, let the opposite ring $S^{op}=(S, +, \times)$ be the ring whose underlying Abelian group under addition is the same as that of $S$, but with the multiplicative structure reversed, i.e. $a\times b=b\cdot a$; the formation of the opposite is clearly functorial. Note that if $R[G]$ is a group ring, it is naturally isomorphic to its opposite through the map $\phi_G: g\mapsto g^{-1}$. -The Problem -Now if $G, H$ are groups and $\psi: R[G]\to R[H]$ is an isomorphism of group rings, we may ask if it is compatible with the formation of the opposite ring---that is, does $\phi_H\circ \psi=\psi^{op}\circ \phi_G$? Say that $G, H$ have strongly isomorphic group rings if such a $\psi$ exists. - -What is known about groups with strongly isomorphic group rings over commutative rings $R$? Are there non-isomorphic finite groups $G, H$ with $\mathbb{Z}[G]$ strongly isomorphic to $\mathbb{Z}[H]$, for example? More weakly, when is $\mathbb{C}[G]\simeq \mathbb{C}[H]$? - -Strong Isomorphism is Strong -Just to convince you that strong group ring isomorphism is in fact a stronger condition that group ring isomorphism, note that $\mathbb{C}[\mathbb{Z}/4\mathbb{Z}]$ is isomorphic to $\mathbb{C}[\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}]$ but not strongly isomorphic. This is because $\phi_{\mathbb{Z}/4\mathbb{Z}}$ is not the identity, but $\phi_{\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}}$ is the identity on the underlying set of $\mathbb{C}[\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}]$. -Addendum (4/6/2011) -Andreas Thom points out in his excellent answer that the case of finite abelian groups over $\mathbb{C}$ is not much harder than the case of usual group ring isomorphisms. Unfortunately the question over e.g. $\mathbb{Z}$ is likely to be extremely difficult, since the usual isomorphism problem over $\mathbb{Z}$ is apparently quite hard---I don't yet understand Hertweck's construction well enough, for example, to tell if the groups he constructs have strongly isomorphic group rings. In any case, I would accept as an answer a summary of the current state of the art for strong isomorphism over $\mathbb{Z}$ (for example, does Hertweck's construction admit a strong isomorphism?), or any relatively recent reference addressing the more general question (as Qiaochu Yuan points out in a comment, the question is equivalent to asking when the group rings of $G, H$ are isomorphic as $*$-algebras, which suggests to me that the question must have been studied by someone). - -REPLY [11 votes]: If $G$ is a finite abelian group, then $\mathbb C[G] = \lbrace f \colon \hat G \to \mathbb C \rbrace$, where $\hat G$ is the Pontrjagin dual of $G$. The isomorphism $g \mapsto g^{-1}$ translates into the same map on the Pontrjagin dual (basically multiplication by $-1$ on $\hat G$), but now it is a bit easier to analyze. Note also, that there is a non-canonical isomorphism $G \cong \hat G$. -Hence, in order to find two non-isomorphic abelian groups which have strongly isomorphic complex group rings, we just have to analyze (in addition to the cardinality of the group) the orbit structure of multiplication by $-1$ on the group. Indeed, we are now just talking about an algebra of function on a set with some $\mathbb Z/2 \mathbb Z$-action. -Example: In $\mathbb Z/8\mathbb Z \times \mathbb Z/2\mathbb Z$, there are precisely $4$ elements, which are fixed under multiplication by $-1$, namely $(0,0),(4,0),(0,1)$ and $(4,1)$. The same is true for $\mathbb Z/4\mathbb Z \times \mathbb Z/4\mathbb Z$. Here, one has $(0,0),(2,0),(0,2)$ and $(2,2)$. Hence, there exists an isomorphism between $\mathbb C[\mathbb Z/8\mathbb Z \times \mathbb Z/2\mathbb Z]$ and $\mathbb C[\mathbb Z/4\mathbb Z \times \mathbb Z/4\mathbb Z]$, which respect the isomorphism which is induced by $g \mapsto g^{-1}$. -I do not know about an example with coefficients in $\mathbb Z$.<|endoftext|> -TITLE: Can a certain sum converge to 0? -QUESTION [15 upvotes]: Let $\{r_i\}_{i \in \aleph}$ be sequence of integers such that, for some $t \in \mathbb{N}$ and all $i \in \mathbb{N}$, we have $r_i = r_{i+t}$. My question: -Can $\displaystyle \sum_{i=1}^n \dfrac{r_i}{i}$ converge to $0$ if $n \rightarrow \infty$ for a non-trivial choice of the $r_i$ and $t$? Or does $\displaystyle \sum_{i=1}^\infty \dfrac{r_i}{i} = 0$ imply $r_i = 0$ for all $i \in \mathbb{N}$? - -REPLY [22 votes]: Yes. All the $r_i$ must equal $0$ if the period is prime, however. Consider for example $$f(s)=(1-p^{1-s})^2 \zeta(s),$$ which is periodic with period $p^2$, at $s=1$. -I should probably expand on this answer a bit. The case where $t$ is prime is an old conjecture of Chowla, which was resolved by Baker, Birch, and Wirsing (all the $r_i=0$ in this case) in the paper I link to in the first word of this answer. They give the Dirichlet series for $f(s)$ above as a counterexample when $t$ is not prime. -To see that $f(s)$ has the desired properties, I'll work it out in a bit more detail for $p=2$. Expanding $f$ out as a Dirichlet series gives $$f(s)=\sum_{n=0}^\infty \frac{1}{(4n+1)^s}-\frac{3}{(4n+2)^s}+\frac{1}{(4n+3)^s}+\frac{1}{(4n+4)^s}$$ as Woett remarks in the comments. On the other hand, $(1-2^{1-s})^2$ has a double zero at $s=1$, whereas the zeta function $\zeta(s)$ has a simple pole at $s=1$; so $f(1)=0$. So taking the limit as $s\to 1^+$ gives that the OP's series converges to $f(1)=0$ for $r_1=r_3=r_4=1,~ r_2=-3, ~t=4$, as desired.<|endoftext|> -TITLE: Kazhdan-Lusztig graph for the Springer fiber of the minimal special unipotent class? -QUESTION [9 upvotes]: This graph was determined in the case of simply-laced root systems by Igor Dolgachev and Norman Goldstein here. For other root systems the original question should be modified, leading to a precise conjecture which remains unproved. It takes some preparation to state. -Briefly, the fibers in the Springer resolution of singularities of a unipotent class in a simple algebraic group (say over $\mathbb{C}$ or a field of good prime characteristic) are projective varieties, arising from the fixed points of a unipotent element on the flag variety $G/B$. (It's equivalent to study nilpotent orbits of the Lie algebra.) The varieties, as well as their irreducible components (all of equal dimension), have known dimensions (Springer, Steinberg, Spaltenstein), but it remains challenging to describe geometrically the fibers or the patterns of component intersections. The best understood nontrivial case involves the subregular class, where the fiber is a "Dynkin curve" having copies of $\mathbb{P}^1$ as components. The incidence graph in types $A,D,E$ is just the corresponding Dynkin diagram, which coincides in the subregular case with the associated "Kazhdan-Lusztig graph". The latter is defined at the end of the 1979 KL Inventiones paper and has one vertex for each component, with vertices joined by an edge just when the components intersect in codimension one. -Dolgachev and Goldstein studied the opposite extreme of the minimal (nontrivial) unipotent class and worked out its KL graph for types $A,D,E$. Their theorem is that the graph is the same as that for the subregular class. Their study of the minimal class for $G_2$ yields a different graph. Their consultation with Spaltenstein reported in the paper suggests substitution of the minimal special class. (I asked Dolgachev about that a year ago when he was here, but he hadn't seen it pursued further.) The special classes came up in Lusztig's work on Springer representations of Weyl groups, but have no geometric characterization so far. Anyway, Lusztig-Spaltenstein duality for the partial ordering of special classes (generalizing transpose duality for partitions in type $A$) leads in types $B, C, F_4$ to a minimal special class defined by the short root groups, but in type $G_2$ to the subregular class. Comparing data for the Langlands dual types -$B, C$ suggests strongly this formulation: -CONJECTURE: In all types, the KL graph for the subregular unipotent class agrees with the graph for the Langlands dual minimal special unipotent class. -A bit of numerical evidence in favor: the number of components of the Springer fiber for the minimal special class in Langlands self-dual type $F_4$ is 6, as for the subregular class of $F_4$, whereas for $B_3$ and $C_3$ the relevant numbers 4, 5 get switched and similarly for types $B_n, C_n$ in general via Springer theory. (For other pairs of classes related in this way, more than a graph would be needed to get a good comparison. Another problem.) Aside from wanting more geometric evidence related to the stated conjecture, my question here is: - -Is there a good reason for Langlands duality to play any role in this essentially geometric question? - -[EDIT: The paper is now freely available online; link added. But I haven't yet found an answer to my question.] -[UPDATE: Concerning the conjecture itself, an MIT graduate student Dongkwan Kim has followed the suggestion of Roman Bezrukavnikov below to provide a proof here using the notion of folding. (It then seems plausible to look for similar behavior whenever two special nilpotent orbits are related by Lusztig-Spaltenstein duality.) But my question seems to remain open.] - -REPLY [5 votes]: This is not quite an answer but an observation which perhaps may clarify the situation. In the simply-laced case the coincidence of Kazhdan-Lusztig graphs for the minimal and the subregular orbit is known, as pointed out in the posting, so we want to reduce the general case of your conjecture to the simply-laced one. -Let $\mathfrak{g}$ be a nonsimplylaced simple Lie algebra. I claim that (as can be observed e.g. from the case by case treatment in Slodowy's LNM volume) the Springer fiber for the subregular element in $\mathfrak{g}$ coincides with the subregular Springer fiber in the unfolding of the Langlands dual Lie algebra. -[One point this is related to is the action of the affine braid group of the dual Lie algebra on the derived category of coherent sheaves on the resolution of the slice -- since we have an embedding of the affine braid group for $G$ to that of its unfolding, the above statement is consistent with that action]. -In particular, it doesn't come from any direct geometric map between the Lie algebras or related spaces. Now one can expect a geometric relation between the minimal special Springer fiber for $\mathfrak{g}$ and the minimal Springer fiber -for its unfolding (no Langlands duality this time), or at least a relation between their KL graphs: this would provide a proof of the Conjecture in view -of the result by Dolgachev and Goldstein. In fact, Paul Levy's answer to: Uniform proof of dimension formula for minimal special nilpotent orbit? seems to point in the direction of such a relation.<|endoftext|> -TITLE: In what sense is the étale topology equivalent to the Euclidean topology? -QUESTION [34 upvotes]: I have heard it said more than once—on Wikipedia, for example—that the étale topology on the category of, say, smooth varieties over $\mathbb{C}$, is equivalent to the Euclidean topology. I have not seen a good explanation for this statement, however. -If we consider the relatively simple example of $\mathbb{P}^1_\mathbb{C}$, it seems to me that an étale map is just a branched cover by a Riemann surface, together with a Zariski open subset of $\mathbb{P}^1_\mathbb{C}$ that is disjoint from the ramification locus. (If there is a misconception there, small or large, please let me know) The connection to the Euclidean topology on $\mathbb{P}^1_\mathbb{C}$, however, is not obvious to me. -What is the correct formulation of the statement that the two topologies are equivalent, or what is a good way to compare them? - -REPLY [8 votes]: Here is another take on your question, in the direction of (ramified) covers. -1. Riemann There is an analytification functor $X\mapsto X^{an}$ from the category of $\mathbb C$-schemes locally of finite type to that of (non-reduced) complex analytic spaces. Its introduction is due principally to Riemann, Chow, Serre and Grothendieck. It has all the desirable properties and of course the set of points of $X^{an}$ is $X(\mathbb C)$. Riemann's existence theorem states that this functor induces an equivalence of categories between the finite étale covers of $X$ and the finite étale analytic covers of $X^{an}$. -This is the deep result (alluded to in Angelo's fine answer) which, in particular, yields the identification of the topological fundamental group of $X^{an}$ with a completion of the scheme-theoretic fundamental group. -2. Grauert-Remmert In a sense the classification of algebraic covers has been reduced to that of analytic ones. We can then apply the following result, due to Grauert and Remmert : -Let $X$ be a normal analytic space and $U\subset X$ an open subset with analytic complement. Then any (ramified) finite normal cover of $U$ uniquely extends to a normal finite cover of $X$. -(Grothendieck, in SGA 1, gave a slick proof of this theorem by invoking Hironaka's resolution of singularities)<|endoftext|> -TITLE: There is no lattice in PSL(2,R) which contains PSL(2,Z) properly? -QUESTION [15 upvotes]: How can I see that there is no lattice in $G=\mathrm{PSL}_2( \mathbb{R})$ which contains $\Gamma_1=\mathrm{PSL}_2( \mathbb{Z})$ properly, or equivalently, that $X_1 =\mathrm{PSL}_2(\mathbb{Z}) \backslash \mathbb{H}$ is not a finite sheated nontrivial cover of a Riemann surface. -As a motivation, the eigenvalues of the Laplace Beltrami operator on $\mathrm{PSL}_2(\mathbb{Z}) \backslash \mathbb{H}$ is conjectured to have only eigenvalues of multiplicity one. So now, if there exists $\Gamma \supset \Gamma_1$, an eigenvalue on $X_1$ can be associated to an irreducibel representation $\pi$ of $\mathrm{Ind}_{\Gamma_1}^{\Gamma} $, since $Ind_{\Gamma}^{G}$ $ \mathrm{Ind}_{\Gamma_1}^{\Gamma} 1 $ and $\mathrm{Ind}_{G}^{\Gamma}1 $ are isomorphic, and the eigenvalues appear with multiplicty being square of the dimension of $\pi$ (eigenfunctions =matrix coefficients). -Hence, if there would exists such a thing with a higher dimensional representation, the conjecture would be wrong. That's why the intuition that something should be known here. -What about $G$ reductive (see comment of David Loeffler)! - -REPLY [10 votes]: The existing answers all seem like huge overkill. The area of of the modular orbifold (quotient of the hyperbolic plane by $PSL(2, \mathbb{Z})$ is $\pi/3,$ so if it covers something, it should be an orbifold whose area divides that, so it should be a quotient by a triangle group, but it is easy to see that this is not possible by simple arithmetic.<|endoftext|> -TITLE: Subspaces of a Subfactor -QUESTION [5 upvotes]: Is the following true? - -Let $\mathcal N \subset \mathcal M$ be a subfactor. There is a bijective correspondence between the ultraweakly closed subspaces of $\mathcal M$ that are bimodules over $\mathcal N'\cap \mathcal M$, and the ultraweakly closed subspaces of $\mathcal N$. - -If the statement is false, is there a simple way to modify it to make it true? I am particularly interested in the case that $\mathcal M$ is of type $\mathrm I$. -If $\mathcal V \subseteq \mathcal N$ is ultraweakly closed, then $\mathcal V (\mathcal N' \cap \mathcal M)$ is a bimodule over $\mathcal N'\cap \mathcal M$. If the subfactor admits a conditional expectation, then this function is injective. (Edit. Jesse points out that the conditional expectation should be ultrweakly continuous.) -Edit. Steven points out that the statement as written is trivially true by a counting argument. Of course, I'm asking about the function $\mathcal V \mapsto \mathcal V \mathcal (\mathcal N'\cap \mathcal M)$, or something similarly natural. He also notes that irreducible subfactors are a counterexample to the bijectivity of $\mathcal V \mapsto \mathcal V(\mathcal N' \cap\mathcal M )$ in general. This leaves a single concrete question: - -Let $\mathcal N \subseteq \mathcal B (\mathcal H)$ be a factor. Is the function $\mathcal V \mapsto \mathcal V \mathcal N'$ a bijection between the ultraweakly closed subspaces of $\mathcal N$ and the ultraweakly closed subspaces of $\mathcal B(\mathcal H)$ that are $\\mathcal N'$ bimodules? - -REPLY [2 votes]: I suppose the statement is true because the set of ultraweakly-closed subspaces of $N$ has the same cardinality as the set of $N^\prime\cap M$-bimodular subsets of $M$. You probably want an explicit description of the correspondence, preferably such that the $N^\prime\cap M$-bimodule that corresponds to $V\subset N$ is exactely $V(N^\prime\cap M)$. -This is certainly not true in general. For an irreducible subfactor $N\subset M$, i.e. $N^\prime\cap M=\C$, the condition that $V\subset M$ is an $N^\prime\cap M$-bimodule is empty, while there are strictly more subsets of $M$. -Of course, the bicommutant theorem shows that there are no irreducible subfactors of type I factors. I do not immediately see a counterexample in the type I case.<|endoftext|> -TITLE: Presenting the Hyperfinite II_1 Factor -QUESTION [5 upvotes]: It's well known that all hyperfinite $\mathrm{II}_1$ factors are isomorphic. I risk the wrath of MathOverflow elders to ask if a particular isomorph is easier than others to handle. In particular, is there a presentation for which it's obvious that the commutant is also the hyperfinite $\mathrm{II}_1$ factor (presented in the same way)? - -REPLY [8 votes]: Another classical construction is to see the hyperfinite $\mathrm{II}_1$ factor as the infinite tensor product of the two by two matrices -$$ -\mathcal{R}=\otimes_{n=1}^{\infty}{M_{2}(\mathbb{C})} -$$ -acting on its $L^{2}$ closure $L^{2}\Big(\otimes_{n=1}^{\infty}M_{2}(\mathbb{C})\Big)$ by right multiplication. -Then its commutant is given by the left multiplication.<|endoftext|> -TITLE: Does $\mathbb{K}[G]\simeq\mathbb{K}[H]$ for some field $\mathbb{K}$ of characteristic $p$, imply $\mathbb{F}_p[G]\simeq\mathbb{F}_p[H]$? -QUESTION [15 upvotes]: Due to the first (and very helpful) answer I received, I've reformulated the question a little: $G$ and $H$ are now assumed to be $p$-groups. -Let $p$ be a prime, and let $\mathbb{F}_p$ be the field of $p$ elements. Let $G,H$ be finite $p$-groups, and let $\mathbb{k}[G]$ denote the group algebra. -Does $\mathbb{K}[G]\simeq\mathbb{K}[H]$ for some field $\mathbb{K}$ of characteristic $p$, imply $\mathbb{F}_p[G]\simeq\mathbb{F}_p[H]$? - -REPLY [2 votes]: I believe I have a counter-example in $p$-groups. I'm hoping someone who really knows non-commutative deformation theory can fill in the last half. -$\def\FF\mathbb{F}$Write $\FF_q$ for the field with $q$ elements. Let $p$ be an odd prime. -I first present my two groups, $G_1$ and $G_2$. For $i=1$, $2$, fix a short exact sequence $0 \to Z_i \to W_i \to V_i \to 0$ of $\FF_p$ vector spaces, with $\dim Z_i = 2$, $\dim V_i = 4$. We write $w \mapsto \overline{w}$ for the quotient map $W_i \to V_i$. Let $\phi_i : \bigwedge^2 V_i \to Z_i$ be a linear map. We extend $\phi_i$ to a map $W_i \times W_i \to Z_i$ by $(w,w') \mapsto \phi_i(\overline{w} \wedge \overline{w})$, and denote this map by $\phi_i$ as well. -The underlying set of $G_i$ will be $W$. Given an element $w$ in $W_i$, we write $a^w$ for the corresponding element of $G_i$. The relations in $G_i$ are - -$a^z$ is central, for all $z \in Z_i$ -$a^{w} \cdot a^{w'} = a^{w+w'+\phi_i(w,w')}$. - -Now, I must tell you how to choose the $\phi_i$. Dualizing $\phi_i$, we get a linear map $Z_i^{\ast} \to \bigwedge^2 V_i^{\ast}$. In other words, we get a two dimensional subspace of $\bigwedge^2 V_i^{\ast}$ or, in other words, a projective line in $\mathbb{P}\left( \bigwedge^2 V_i^{\ast} \right)$. Now, $\mathbb{P}\left( \bigwedge^2 V_i^{\ast} \right)$ contains the Grasmannian $G(2,4)$ of skew-symettric tensors of rank $1$. We will choose $\phi_1$ to meet $G(2,4)$ at two distinct points defined over $\FF_p$, and choose $\phi_2$ to meet $G(2,4)$ at two Galois conjugate points defined over $\FF_{p^2}$. -Explicit coordinates are as follows: Let $e_1$, $e_2$, $e_3$, $e_4$ be a basis for $V_i$. Take $\phi_1$ to be $(e_1^{\ast} \wedge e_2^{\ast}, e_3^{\ast} \wedge e_4^{\ast})$. Letting $\FF_{p^2} = \FF_p[\sqrt{D}]$, take $\phi_2 = (e_1^{\ast} \wedge e_3^{\ast} + D e_2^{\ast} \wedge e_4^{\ast}, e_1^{\ast} \wedge e_4^{\ast} + e_2^{\ast} \wedge e_4^{\ast})$; note that $(e_1^{\ast} \pm \sqrt{D} e_2^{\ast}) \wedge (e_3^{\ast} \pm \sqrt{D} e_4^{\ast})$ is a linear combination of the components of $\phi_2$. -The point of these choices is that there does NOT exist an $\FF_p$-linear map $W_1 \to W_2$ carrying $\phi_1$ to $\phi_2$, but there DOES exist such a map once we tensor with $\FF_{p^2}$. - -Now, we need to show that our construction is reflected in properties of the group rings. Let $R_i = \FF_p[G_i]$. Our first goal is to show that $R_1 \not \cong R_2$. We must show how to canonically recover $V_i$, $Z_i$ and $\phi_i$ from $R_i$. -The only $1$-dimensional representation of $G_i$ over $\FF_p$ is the trivial rep, since $G_i$ is a $p$-group. Therefore, $\FF_p$ is an $R_i$ module in only one way. Let $I_i$ be the kernel of the unique map $R_i \to \mathrm{End}(\FF_p)$. Explicitly, $I_i$ has $\FF_p$ basis given by the elements $g-1$, for $g \in G_i$. -The center of $R_i$ is $\FF_p[Z_i]$. For $z$ and $z' \in Z_i$, we have $(a^{z+z'}-1) - (a^z-1) - (a^{z'}-1) = (a^z-1)(a^{z'}-1)$, so $(a^{z+z'}-1) \equiv (a^z-1)+(a^{z'}-1) \bmod (Z(R_i) \cap I_i)^2$. We see that the $\FF_p$ vector space $Z_i$ is canonically isomorphic to $(Z(R_i) \cap I_i)/(Z_i \cap I_i)^2$, by $z \mapsto a^z-1$. (Expressions like $(Z_i \cap I_i)^2$ mean the square as a two-sided ideal.) -We claim similarly that $V_i \cong I_i /I_i^2$ by the map $v \mapsto a^w-1$, where $w$ is an arbitrary lift of $v\in V_i$ to $W_i$. To this end, we first must prove the formula is well defined. If $w$ and $w+z$ are two different lifts, then we must show that $a^{w+z} - a^w = (a^z-1) a^w \in I_i^2$. In other words, we must show that $a^z-1 \in I_i^2$ for $z \in Z_i$. Now, if $z = \phi_i(w, w')$, then $(a^w-1)(a^{w'}-1) - (a^{w'}-1)(a^w-1) = a^{w+w'+\phi_i(w,w')} - a^{w+w'+\phi_i(w',w)} = (a^{2\phi_i(w,w')}-1) a^{w+w'-\phi_i(w,w')} \in I_i^2$. Therefore, $a^{2 \phi_i(w,w')}-1 \in I_i^2$. Letting $w$ and $w'$ vary, we can obtain $a^z-1 \in I_i^2$ for any $z \in Z_i$. -We have now checked that $v \mapsto a^w-1$ is a well defined map of sets $V_i \to I_i/I_i^2$. To see that it is a map of $\FF_p$ vector spaces, note that -$$(a^{w+w'}-1) - (a^w-1)- (a^{w'}-1) = (a^w-1)(a^{w'}-1) + a^{w+w'} (1-a^{\phi_i(w,w')}).$$ -Both summands on the right hand side are in $I_i^2$. A bit more work shows that this map of $\FF_p$ vector spaces is an isomorphism. -Now, we must show that we can recover $\phi_i$. Let $\langle Z(R_i) \cap I_i \rangle$ be the two sided ideal of $R_i$ generated by $Z(R_i) \cap I_i$. Fix a section $V_i \to W_i$. Using this section, we can write any element $c$ of $R_i$ uniquely as $\sum_{V \in V_i} c_v a^v$. We have $c \in \langle Z(R_i) \cap I_i \rangle$ if and only if all the $c_v$ are in $Z(R_i) \cap I_i$. For $c \in \langle Z(R_i) \cap I_i \rangle$, define $\sigma(c) = \sum_{v \in V} c_v$. We check that $\sigma(c) \bmod (Z(R_i) \cap I_i)^2$ is independent of the choice of section. So $\sigma$ gives a well defined map $\langle Z(R_i) \cap I_i \rangle \longrightarrow (Z(R_i) \cap I_i)/(Z(R_i) \cap I_i)^2 \cong Z_i$, which we will also denote $\sigma$. -For any $w$ and $w' \in W_i$, we have $a^w a^{w'} - a^{w'} a^w = (a^{\phi_i(w,w')} - a^{\phi_i(w',w)}) a^{w+w'} \in \langle Z(R_i) \cap I_i \rangle$. Thus, it makes sense to talk about $\sigma(x x' - x' x)$ for any $x$ and $x' \in R_i$. Suppose furthermore that $x \in I_i$ and $x' \in I_i^2$. Then I claim that $\sigma(x x' - x' x)=0$. By linearity, we may assume that $x=a^w-1$ and $x' = (a^{w'}-1) (a^{w''}-1)$. So -$$x x' - x' x = (a^{\phi_i(w,w'+w'')} - a^{-\phi_i(w,w'+w'')}) a^{w+w'+w''+\phi_i(w', w'')} - (a^{\phi_i(w,w')} - a^{-\phi_i(w,w')}) a^{w+w'} - (a^{\phi_i(w,w'')} - a^{-\phi_i(w,w'')}) a^{w+w''}$$ -and $\sigma(x x'-x' x) = \phi_i(w,w'+w'') - \phi_i(w,w') - \phi_i(w,w'')=0$. This proves the claim. -Therefore, $(x,x') \mapsto \sigma(x x' - x' x)$ gives a well defined map $I_i/I_i^2 \times I_i/I_i^2 \to (Z(R_i) \cap I_i)/(Z(R_i) \cap I_i)^2$. Tracing through definitions, this map is $2 \phi_i$. - -Whew! That's only half the work. We now want to show that $\FF_{p^2}[G_1] \cong \FF_{p^2}[G_2]$. The idea is to write each side as a noncommutative $\FF_{p^2}$ algebra generated by $V_i$, where we choose some basis $v_1$, $v_2$, $v_3$, $v_4$ for $V_i$ and send these basis elements to $a^{w_1}-1$ etcetera, where $w_s$ is a lift of $v_s$ to $W_i$. We have an $\mathbb{F}_{p^2}$ linear map taking $V_1 \otimes \FF_{p^2}$ to $V_2 \otimes \FF_{p^2}$ in a way that respects $\phi$. The low order terms match up. The intuition is that, with enough knowledge of deformation theory, we should be able to lift this low order match to an isomorphism $\FF_{p^2}[G_1] \cong \FF_{p^2}[G_2]$. -Anyone want to finish the computation? - -ADDED I've been thinking more about this, and I am no longer so optimistic, although I'd still love to know what an expert thinks. Quillen associates a restricted Lie algebra $L_i$ to a $p$-group $G_i$. Writing $U_i$ for the restricted enveloping algebra, he shows that the associated graded of $\FF_p[G_i]$ with respect to $I_i$ (the augmentation ideal) is $U_i$. In our case, we get $\mathbb{F}_{p^2} \otimes L_1 \cong \mathbb{F}_{p^2} \otimes L_2$ and so $\mathbb{F}_{p^2} \otimes U_1 \cong \mathbb{F}_{p^2} \otimes U_2$. Now, can we lift this back to $\FF_{p^2}[G_1] \cong \FF_{p^2}[G_2]$? -The most obvious reason this should be true is if $U \cong \FF_p[G]$ as algebras. That's true if $G$ is an abelian $p$-group. I think it is also true for the noncommutative $p$-torsion group of order $p^3$. But I don't think it should be true in our case. More generally, let's talk about $2$-step $p$-torsion nilpotents in general. -So, let $G$ be the group generated by $X_i$ and $Z_{ij}$ with the relations that everything is $p$-torsion, the $Z_{ij}$ are central, $Z_{ij} = Z_{ji}^{-1}$ and $X_i X_j = X_j X_i Z_{ij}$. We might, in addition, impose some further commutative relations between the $Z_{ij}$. (In our example, $X_i$ runs from $i=1$ to $4$, and the $Z_{ij}$ generate a group of rank $2$.) Let $\mathfrak{g}$ be the corresponding restricted Lie algebra on $x_i$ and $z_{ij}$, where all $[p]$-powers are $0$, the $z_{ij}$ are central, $z_{ij} = - z_{ji}$ and $[x_i, x_j] = z_{ij}$. We'd like to know whether there is an isomorphism $\FF_p[G] \to \FF_p\langle \mathfrak{g} \rangle$, where the right hand side is the restricted enveloping algebra. Let $A$ be the subring $k[z_{ij}]/z^{ij}^p=0$ of the right hand side. We want the $Z_{ij}$ to go to central elements on the right, so it seems plausible that the $Z_{ij}$ should map into $A^{\times}$. -For low degree terms in the $x_i$, we should clearly send $X_i$ to the initial terms of $\sum_s x_i^s/s!$. But this runs into trouble in degree $p$. Suppose we try fixing this by writing $X_i = \sum_{s -TITLE: Reference for Mathematical Economics -QUESTION [17 upvotes]: I'm looking for a good introduction to basic economics from a mathematically solid(or, even better, rigorous) perspective. I know just about nothing about economics, but I've picked up bits and pieces in the course of teaching Calculus for business and social sciences, and I'd like to know more, both for my personal culture and to incorporate into my courses. My Platonic ideal of such a book would be along the lines of T W Körner's Naive Decision Making, but I'll take what I can get. - -REPLY [9 votes]: Let me first tell you that there are three types of economic theory: - -price theory -general equilibrium -game theory - -Nowadays, the last type of theory is by far the most popular, but maybe you need some familiarity with the other two to appreciate game theory as an economist does. I guess you could say the mathematics is very easy judged from what mathematicians are used to. What is difficult is the economic interpretation. For that you need a bit economic intuition and a bit of culture about classic economic models. -My recommendations for a mathematically mature neophyte would be: - -george stigler's price theory for the first -debreu's classic for general equilibrium; there is also a book by JWS Cassels -Fudenberg & Tirole or Vega Redondo for game theory<|endoftext|> -TITLE: The conformal group of $S^n$. -QUESTION [12 upvotes]: Is there any explicit computation of Conf($S^n$, $g_{std}$), the group of conformal diffeomorphisms of the standard $n$-sphere? - -REPLY [4 votes]: Try the book A Mathematical Introduction to Conformal Field Theory by Martin Schottenloher. Chapters 1 and 2 go over some of the proofs you are looking for and the book is example driven.<|endoftext|> -TITLE: Missing mass conjecture -QUESTION [6 upvotes]: Let $n,t$ be positive integers and $p_1,p_2,\ldots,p_n$ positive numbers summing to 1. Conjecture: -$$ -\sum_{i=1}^n p_i (1-p_i)^t \le \frac{n(1-1/n)^n}{t} -$$ -always holds. -The motivation comes from my missing mass question; the quantity $\sum_{i=1}^n p_i (1-p_i)^t$ is precisely the expected unseen mass after $t$ draws from the distribution $(p_i)$ on $n$ objects. - -REPLY [7 votes]: Since the single-variable optimization that David mentions still requires some work, I will present another solution. Let $f(p) = p(1-p)^t$. Define the function $g$ that is equal to $f$ on $[0, 1/t]$, and on $[1/t, 1]$ is a linear interpolation between the points $(1/t, f(1/t))$ and $(1, 0)$. Then we can check that $g$ is concave on all of $[0, 1]$ and $g \ge f$ on all of $[0, 1]$. (I learned this concept of "concave majorants" or "convex minorants" from Steele's book called The Cauchy-Schwarz Master Class.) Applying Jensen's inequality, we have $\sum f(p_i) \le \sum g(p_i) \le n g(1/n)$. -We now split into two cases, $t \le n -1$ or $t \ge n$. First, suppose that $t \le n - 1$. Then $g(1/n) = f(1/n) = (1/n) (1 - 1/n)^t$. So we need to show that $(1 - 1/n)^t$ is at most $n (1 - 1/n)^n / t$. That's equivalent to showing $t(1 - 1/n)^t \le n (1 - 1/n)^n$. We can check that the left side is an increasing function of $t$ for $t \le n - 1$, and when $t = n - 1$ we have an equality. So we have established the inequality in this case. -Next suppose that $t \ge n$. Then by linear interpolation, we find $g(1/n) = (1 - 1/t)^{t-1} (1 - 1/n) / t$. So we need to show that $(1 - 1/t)^{t - 1} (n - 1) \le n(1 - 1/n)^n$. That's equivalent to $(1 - 1/t)^{t-1} \le (1 - 1/n)^{n - 1}$. By taking reciprocals, that's equivalent to $(1 + 1/(t - 1))^{t - 1} \ge (1 + 1/(n-1))^{n - 1}$. The left side is an increasing function of $t$, and we have an equality when $t = n$. So we have established the inequality in the second case too.<|endoftext|> -TITLE: Simple connectedness via closed curves or simple closed curves? -QUESTION [5 upvotes]: I've recently read some papers and books involving simply connected domains in Euclidean space (dimension at least 2), where domain is an open connected set. The usual definition is a (connected) set for which every continuous closed curve is (freely) contractible while some authors only require that every continuous simple closed curve is contractible. The authors who define simple connectedness using simple closed curves do so in order to use Stokes' Theorem or the Jordan curve theorem somewhere in the sequel; however, they never mention (not even with a reference) that their definition is equivalent to the usual one! My question is if there is a proof written down somewhere (with all the details) proving the equivalence (for domains in $\mathbb{R}^n$ with $n \geq 2$)? If not, does someone know of an "easy" proof using a minimal amount of knowledge, say that of a first course in topology? - -REPLY [14 votes]: As pointed out by Pierre and Paul in comments, there are several standard ways to deal with this kind of issue. A good answer really depends what you're assuming you start from, and where you're trying to go to. The Jordan curve theorem and Stoke's theorem are both fairly sophisticated and difficult for beginners to grasp, so it's a bit hard to see how only analyzing embedded curves is streamlining anything, except perhaps helping with people's intuitive images---but even so, it may do more harm than good. -Perhaps it's worth pointing out that this statement is false in greater generality, for instance for closed subsets of $\mathbb R^3$. Here's an example in $\mathbb R^3$: -consider a sequence of ellipsoids that get increasingly -getting long and thin; to be specific, they can have axes of length $2^{-k}$, $ 2^{-k}$ and $2^k$. Stack them in $\mathbb R^3$ with short axes contained in the $z$-axis, so each one touches the next in a single point with long axes parallel to the $x$-axis, and let -$X$ be their union together with the $x$-axis. -Any simple closed curve in $X$ is contained in a single ellipsoid, since to go from one to the next it has to cross a single point, so every simple closed curve is contractible. -However, a closed curve in the $yz$-cross-section that goes down one side and back up the other sides is not contractible. The fundamental group is in fact rather large and crazy. -Anyway, here are some lines of reasoning that can overcome whatever hurdle needs to be ovrcome: - -PL approximation, as suggested by Pierre: this is easy, the keyword is "simplicial approximation". I'll phrase it for maps of a circle to Euclidean space as in the question, even though essentially the same construction works in far greater generality. Given an open subset $U \subset \mathbb{R}^n$ and given a map $f: S^1 \to U$, -then by compactness $S^1$ has a finite cover by neighborhoods that are components of $f^{-1}$ of a ball. If $U_i$ is a minimal cover of this form, there is a point $x_i$ that is in $U_i$ but not in any other of elements of the cover; this gives a circular ordering to the $U_i$. There is a sequence of points $y_i \in U_i \cap U_{i+1}$, indices taken mod the number of elements of the cover. The line segment between $y_i$ and $y_{i+1}$ is contained in $U_i$, since balls are convex. (This generalizes readily to the statement that for any simplicial complex, there is a subdivision where the extension that is affine on each simplex has image contained in $U$. It also generalizes readily to the case that $U$ is an open subset of a PL or differentiable manifold). -Raising the dimension: if you take the graph of a map of $S^1$ into a space $X$, it is an embedding. If you're (needlessly) worried about integrating differential forms on non-embedded curves, pull the forms back to the graph, where the curve is embedded. If you want to map to a subset of Euclidean space with the same homotopy type, just embed the graph of the map -(a subset of $S^1 \times U$ into $\mathbb R^2 \times U$. (There's a very general technique to do this, if the domain is a manifold more complicated than $S^1$, even when it's just a topological manifold, using coordinate charts together with a partition of unity to embed the manifold in the product of its coordinate charts). -The actual issue for integration, using Stoke's theorem etc., is regularity --- to make it simple, restrict to rectifiable curves, and don't worry about embededness. Any continuous map into Euclidean space is easily made homotopic to a smooth curve, by convolving with a smooth bump function---the derivatives are computed by convolving with the variation of the bump, as you move from point to point. -Similarly, you can approximate any continuous map by a real-analytic function, if you convolve with a time $\epsilon$-solution of the heat equation (a Gaussian with very small variance, wrapped around the circle). This remains in $U$ if $\epsilon$ is small enough. A real analytic map either has finitely many double points, or is a covering space to its image; in either case you reduce simple connectivity to the case of simple curves. -Sard's theorem and transversality, as mentioned by Paul. Sard's theorem is nice and elegant and has many applications, including the statement that a generic smooth map of a curve into the plane is an immersion with finitely many self-intersection points, as is any generic smooth map of an $n$-manifold into a manifold of dimension $2n$. If the target dimension is greater than $2n$, then a generic smooth map is an embedding.<|endoftext|> -TITLE: Splitting of the double tangent bundle into vertical and horizontal parts, and defining partial derivatives -QUESTION [9 upvotes]: Let $M$ be a manifold and $g$ a metric on $M$. -Let $TM$ denote the tangent bundle of $M$, and denote points in $TM$ by $(x,v)$ where $v \in T_xM$. -The Levi-Civita connection of $(M,g)$ induces a splitting of the double tangent bundle $TTM = V \oplus H$, where $V$ is the vertical distribution, defined by $V_{(x,v)} = T_{(x,v)}T_xM$ (i.e. the tangent space to the fibre), and $H_{(x,v)}$ is the horizontal distribution, which is determined by the connection. -Suppose $A:TM \rightarrow TM$ is a map such that $A(x,v)\in T_xM$ (so the map $A(x,\cdot)$ is a map from $T_xM$ to itself for all $x \in M$). -How does one use the splitting described above to define "partial derivatives" $\nabla_xA$ and $\nabla_vA$, which should be maps: -$(\nabla_xA)(x,v):T_xM \rightarrow T_xM$, -$(\nabla_vA)(x,v):T_xM \rightarrow T_xM$. -These should have the property that if $\gamma(t)$ is a curve on $M$ and $u(t)$ is a vector field along $\gamma$ (so $u(t) \in T_{\gamma(t)}M$ for all $t$), and $\nabla_t$ denotes the covariant derivative along $\gamma$, then -$\nabla_t(A(\gamma,u)) = (\nabla_xA)(\gamma,u) \cdot \dot{\gamma} + (\nabla_vA)(\gamma,u) \cdot \nabla_t u$ -(here on the LHS, $A(\gamma,u)$ is itself a vector field along $\gamma$, so the notation $\nabla_t(A(\gamma,u))$ is meaningful). -The expression above "makes sense" intuitively, but I can't get the formalism to work properly. - -REPLY [4 votes]: Denote $E=TM$. We have a vertical projection $P_{(x,v)}:T_{(x,v)}E\to E_x$ determined by the splitting. The covariant derivative of a vector field $(\gamma,u)$, regarded as a curve in $E$, is the vertical projection of its derivative. We also have linear maps $d\pi:T_{(x,v)}E\to T_xM$ (where $\pi:E\to M$ is the bundle projection), the horizontal lift $h:T_xM\to H_{(x,v)}$ such that $h\circ d\pi=id_H$, and the natural isomorphism $i:T_xM=E_x\to V_{(x,v)}$ (which is the only thing depending on the fact that $E$ is the tangent bundle and not an arbitrary vector bundle). In this notation, we have $\xi=h\circ d\pi(\xi)+i\circ P_{(x,v)}(\xi)$ for every $\xi\in T_{(x,v)}E$. -Define $\nabla_x A=P_{A(x,v)}\circ (dA)\circ h$ and $\nabla_v A=P_{A(x,v)}\circ (dA)\circ i$ where $dA:T_{(x,v)}E\to T_{A(x,v)}E$ is the ordinary derivative. Then, for a vector $\xi\in T_{(x,v)}E$ (the derivative of our vector field $(\gamma,u)$) we have -$$ -\begin{aligned} - \nabla_t (A(\gamma,u)) &= P_{A(x,u)}\circ dA(\xi) = P_{A(x,u)}\circ dA \circ h\circ d\pi(\xi) +P_{A(x,u)}\circ dA \circ i\circ P_{(x,v)}(\xi) \cr - &=(\nabla_xA)(d\pi(\xi)) + (\nabla_vA)(P_{(x,v)}(\xi)) =(\nabla_xA)\cdot\dot\gamma + (\nabla_vA)\cdot\nabla_t(\gamma,u) . -\end{aligned} -$$<|endoftext|> -TITLE: Tameness for the Galois closure of a map of curves -QUESTION [7 upvotes]: Say we are working over some $K=\overline{K}$, of characteristic $p>0$. Let $\phi: Y\rightarrow X$ be a nonconstant map of smooth projective curves. To this map we can associate a map $\psi: Z\rightarrow X$, where on the level of fields this is the Galois closure of $k(X)\subseteq k(Y)$. I would like to know about the tameness of this map. -Let $e_P$ denote the ramification indices (with the maps understood to be either $\psi$ or $\phi$ depending on where $P$ lives). Now obviously if $p|e_P$ and if $Q$ lies above $P$, $p|e_Q$ as well, so $\psi$ has wild ramification at $Q$. I am wondering when we can ensure this map is (everywhere) tamely ramified. For instance if $d=deg(\phi) < p$, then the degree of the Galois closure of $k(Y)$ over $k(X)$ has degree dividing $d!$, and hence $\psi$ remains tame. -My question is this: Suppose we can show for each $P\in Y$ such that $e_P \geq p$ that each point above $P$ is tamely ramified. Can we conclude that $\psi$ is (everywhere) tamely ramified? It seems to me that this isn't true but I cannot produce a counterexample. It would be fortuitous if it were true, however. Any help is greatly appreciated. - -REPLY [3 votes]: For the first glance -it should follow form the Abhaynkar's lemma (see "Algebraic Function Fields" by Stichtenoth, Theorem 3.9.1) and the fact the Galois closure is the composite of all the different embeddings of L over K into fixed algebraic closure of K (so each of them has the same properties of tame ramifications). -Then we just apply the lemma and get the result that $p=char(K)$ does not divide $e_P$ for any place P in K.<|endoftext|> -TITLE: Profinite completion of a semidirect product -QUESTION [10 upvotes]: If we have two finitely generated residually finite groups $G$ and $H$, is there are relation between -the profinite completions $\hat{G},\hat{H}$ and the profinite completion of a semidirect -product $\hat{G \rtimes H}$ -(and analogous question for pro-p completions) - -REPLY [2 votes]: Let $\mathscr{P}$ be any property -such that whenever a group has $\mathscr{P}$ then all its subgroups also have $\mathscr{P}$. -In [1] Theorem 3.1, K. W. Gruenberg has proved that if the wreath product -$W= A \wr B$, is residually $\mathscr{P}$, -then either $B$ is $\mathscr{P}$ or $A$ is abelian. -Consider $W= S_3 \wr \mathbb{Z}$, where $S_3$ is the symmetric group of -degree 3. -Since $S_3$ is not abelian, $\mathbb{Z}$ is not finite, and the subgroup of any finite group is finite, -the group $W$ is not RF. -Clearly, $W= \prod_{i \in \mathbb{Z}} S_3 \rtimes \mathbb{Z}$, where $\mathbb{Z}$ and $\prod_{i \in \mathbb{Z}} S_3$ -are residually finite. -[1] K. W. Gruenberg, Residual properties of infinite soluble groups}, -Prec. London Math. Soc., Ser. 3, 7 (1957), 29--62.<|endoftext|> -TITLE: Projecting the unit cube onto subspaces -QUESTION [8 upvotes]: Given a set $S$ of non-zero vectors in $\mathbb{R}^n,$ and a subspace $L,$ consider $f(S,L)=\max_{s \in S}\frac{\|Ps\|_2}{\|s\|_2}$ where $Ps$ is the orthogonal projection of $s$ onto $L.$ -Specifically, consider the set $X$ consisting of the $2^n-1$ (nonzero) vectors with all coordinates $0$ or $1$. -A recent question concerns criteria which might show, for a given subspace $L$, that $f(X,L)$ is small. Here I am concerned with choosing the subspace: - -For each $n$ and $d1$ case as a new question. -Here is a somewhat selective review, all for the case $d=1.$ - -Gerhard mentioned that $(1,-1,0,0,\cdots)$ attains $\sqrt{1/2}.$ -Seva suggested that $(1,1/\sqrt{2},1/\sqrt{3},\cdots)$ is asymptotically optimal attaining $O(\sqrt{1/\log n}).$ This is indeed better but the smallest $n$ for which it beats $\sqrt{1/2}$ is (by my calculations) $n=1203.$ The nice optimality proof references $(1,\sqrt{2}-1,\sqrt{3}-\sqrt{2},\cdots)$ This is actually a better choice, it first beats $\sqrt{1/2}$ for $n=56.$ The two choices are comparable (up to scalar multiple) because $\sqrt{k+1}-\sqrt{k} \approx 1/2\sqrt{k}$. The approximation is pretty good, except for $k=0.$ As this is the largest entry, it takes longer to beat $\sqrt{1/2}$. -Denis essentially does mention $(1,\sqrt{2}-1,\sqrt{3}-\sqrt{2},\cdots)$ and, at least in one edit, suggests reflecting in the center: So for $n=6$ use $(a_1,a_2,a_3,-a_3,-a_2,-a_1)$ where $a_k=\sqrt{k}-\sqrt{k-1}.$ This is superior at $n=4$ and even at $n=3$ with the right rule for the center: It seems obvious that there should be symmetry and entries summing to $0$, but actually $(1,\sqrt{2}-1,-1)$ is the best choice for $n=3$. -Emil nicely lays this out in a comment. Here is my somewhat informal explanation (with some minor details skipped) , in case another perspective is useful. I suggest at the end that this should remind one of linear programming.: - -There is no harm in assuming that the subspace is generated by a vector $v=(v_1,v_2,\cdots,v_n)$ with $v_1 > v_2 > \cdots > v_n$ (actually, only $\ge$ is clear, but I will ignore that in the interests of brevity.). By a mild abuse of notation, for $j>0$ let $x_{j}$ be the vector whose first $j$ entries are $1$ and the rest $0$ while $v_{-j}$ has first $n-j$ entries $0$ and the last $j$ equal to $1$. -Given $v$, Let $B \subset X$ be the set of vectors in $X$ which attain the maximum of $\frac{\|Px\|}{\|x\|}.$ I make two claims about $B$: - -Every vector in $B$ is $x_j$ for some $j$ with $v_j > 0$ or else $x_{-j}$ for some $j$ with $v_{n-j} < 0$ (think about why any other vector with $j$ entries equal to $1$ will be worse.) -Unless $B$ is a basis of $\mathbb{R}^n$, There is a $v$ which gives a better ratio. (Because otherwise we should be able perturb the entries of $v$ in such a way as to lower $\frac{\|Px\|}{\|x\|}$ for the vectors already in $B$ (simultaneously keeping them equal and maximal) while raising that ratio slightly for some $x_j$ not yet in $B$. - -Together these tell us that $B=\lbrace x_{1},x_{2},\cdots x_n\rbrace$ or else $B=\lbrace x_{1},x_{2},\cdots x_q;x_{q-n},\cdots,x_{-2},x_{-1}\rbrace$ where $v_q>0>v_{q+1}.$ Since the various ratios must all be the same, we deduce that $v=(\alpha a_1,\alpha a_2, \cdots,\alpha a_q; -\beta a_{n-q},\cdots,-\beta a_2,-\beta a_1)$ for the $a_k$ as above. Here $\alpha$ and $\beta$ are positive constants. A bit of reflection shows that they must be equal and hence can be taken to be $1$. Finally, $q$ should be $ n/2 $ (rounded if needed). -Comment: (disclaimer: I am an optimist but not an optimizer so the following may inexact.) Here we have the convex optimization problem of choosing $v$ so as to minimize the largest of $\frac{\|Px\|_2}{\|x\|_2}.$ Had it been $\frac{\|Px\|_1}{\|x\|_1}$ we would have been able to use linear programming. My argument above has the feel of linear programing, perhaps using duality. Perhaps a similar method could uncover optimal subspaces of dimension $d \ge 2.$ - -REPLY [4 votes]: My answer concerns with the case $d=1$ only. Without loss of generality, we can focus on the subspaces, generated by a vector with all coordinates non-negative. It is easy to verify that for the subspace $L$, generated by the vector $(1,1/\sqrt{2},...,1/\sqrt{n})$, the projection onto $L$ of any non-zero vector $\epsilon\in\{0,1\}^n$ has lenght at most $\frac{2}{\sqrt{\log n}}\,\|\epsilon\|$. This is essentially the worst case as, on the other hand, for any non-zero vector $z\in R^n$ with non-negative coordinates there exists a non-zero vector $\epsilon\in\{0,1\}^n$ such that - $$ \langle z,\epsilon \rangle \ge \frac{2}{\sqrt{\log n+4}}\,\|z\|\|\epsilon\|. $$ -To see this, write $z=(z_1,...,z_n)$ and, without loss of generality, assume that - $$ z_1 \ge \dotsb \ge z_n \ge 0\quad \text{and}\quad \|z\|=1. $$ -Let $\tau := 2/\sqrt{\log n+4}$. We will show that there exists $k\in[n]$ -with $z_1+...+z_k\ge\tau\sqrt k$; choosing then $\epsilon$ to be the vector -with the first $k$ coordinates equal to $1$ and the rest equal to $0$ -completes the proof. -Suppose, for a contradiction, that $z_1+...+z_k<\tau\sqrt{k}$ for $k=1,...,n$. Multiplying this inequality by $z_k-z_{k+1}$ for each $k\in[n-1]$, and by $z_n$ for $k=n$, adding up the resulting estimates, and rearranging the terms, we obtain - $$ z_1^2+...+z_n^2 < \tau \big(z_1+(\sqrt2-1)z_2 - +...+ (\sqrt n-\sqrt{n-1})z_n \big). $$ -Using Cauchy-Schwarz now gives - $$ 1 < \tau \Big( \sum_{k=1}^n - \big(\sqrt k-\sqrt{k-1}\big)^2 \Big)^{1/2} \|z\| - < \tau \sqrt{\log n +4}/2, $$ -a contradiction. - -REPLY [3 votes]: Edited. Conjecture for $d=1$: Define the sequence $v_1,\ldots,v_n$ by $v_1=1$ and -$$v_{k+1}=\left(\sqrt{1+\frac1k}-1\right)(v_1+\cdots v_k).$$ -Then the $\min\max$ equals $a$ where -$$a^2\sum_1^nv_j^2=1.$$ -It corresponds to the projection on the line spanned by $u:=(a_1,\ldots,a_n)$ where $a_j:=av_j$. -In this construction, the equality $\|Pe_I\|=\|e_I\|$ ($I$ a subset of indices) is achieved for every subset $I=(1,\ldots,p)$ with $1\le p\le n$. -We have $v_k=\sqrt{k}-\sqrt{k-1}\sim\frac{1}{2\sqrt k}$. Asymptotically, we have $a\sim\frac{2}{\sqrt{\log n}}$.<|endoftext|> -TITLE: The minimal model program and symplectic resolutions -QUESTION [9 upvotes]: I've been reading some papers of Namikawa lately, and have on several occasions come across a claim I would really like someone to expand on. -On page 4 of Poisson deformations of affine symplectic varieties, he says: - -According to Birkar-Cascini-Hacon-McKernan, we can take a crepant partial resolution $\pi: Y \to X$ in such a way that $Y$ has - only $\mathbb{Q}$-factorial terminal singularities. This $Y$ is called a $\mathbb{Q}$-factorial terminalization of $X$. - -Here $X$ is assumed to be affine and symplectic (as defined on the first page of Namikawa's paper). -So, I have 2 questions: - -Which results in this paper is this supposed to follow from? I can see some things along these lines, but with a lot of hypotheses I'm not used to dealing with (like "Kawamata log terminal"), and Namikawa doesn't say a word more than what is above for why this works. -How much control does one have on the ample divisors on the resolution? This is very vague, so let me lay out what I'm hoping for; Namikawa proves that $X$ is homotopy equivalent to a generic deformation $Y'$ of $Y$ in a reasonably canonical way (there are some choices involved, but they're controlled). For any isomorphism, you can ask if a class in $H^2(Y';\mathbb{R})$ is in the nef cone of $NS(X)$ under the induced isomorphism on cohomology. What I'm hoping is that there's a way of resolving $X$ and then choosing a homotopy equivalence to $Y$ that makes this so, which is unique if the class is in the interior. - - -Is there any hope of such a picture existing? I'm having too much trouble parsing the BCHM paper to tell whether such a story is in there or not. - -EDIT: Let me expand a little bit on what I am hoping for: In another paper of Namikawa (look in section (P.2)), he describes an approach to classifying symplectic resolutions which sounds a bit like my 2. above. You - -start with a line bundle $L$, which you want to make into the ample line bundle on a different resolution (you imagine it is the proper transform of that line bundle). -attempt to do a flop which makes this line bundle closer to being nef; that is, you find a curve $L$ is negative on, contract it, and then find a symplectic resolution of the contraction for which the proper transform $L^+$ is relatively ample. -rinse and repeat until $L$ is genuinely ample. - -In the paper mentioned above, this is done for nilpotent orbits, using very specific known facts about how these orbits and their resolutions work. What I was really hoping for was some indication of whether this story can be run on a general symplectic resolution (probably with a $\mathbb{C}^*$ action to keep everything nice). - -REPLY [4 votes]: The answer to the first question: if $X$ has klt singularities, then there exists a $\mathbb{Q}$-factorial variety $Y$ with a birational morphism $\pi: Y\to X$, such that if you write $\pi^*K_X=K_Y+\Delta$, then $\Delta$ is effective, and for any exceptional divisor $E$ of $Y$, we have the discrepancy $a(E,Y,\Delta)>0$, i.e., $(Y,\Delta)$ is terminal. This implies $Y$ itself is terminal. If you start with $X$ with only canonical singularities (This is stronger than klt singularities. But if $K_X$ is Cartier and $X$ is klt, then $X$ has canonical singularities. I mention this since I know in some cases from the representaion theory, indeed $K_X$ is trivial.), then $\Delta=0$. This follows from Corollary 1.4.3 of [BCHM]. In fact, it was known before that certain part of MMP would imply the existence of terminalization. The cases of MMP established in [BCHM] contain this part.<|endoftext|> -TITLE: Generalization of Moise's theorem -QUESTION [5 upvotes]: I am looking for a generalization of Moise's theorem, which the few professors that I asked treat as a "known geometric fact" but none could find a reference to an article proving it. -The claim is like this: -Let $M$ be a compact 3 manifold (Riemannian but I do not think that helps), and let $X,Y$ be compact 2 manifolds (Riemannian aswell) that intersect transversally, then there is a triangulation of $M$ such that $X,Y$ are sub-complexes. -Thank you in advance, -Ethan - -REPLY [9 votes]: In fact it helps immensely that $M$, $X$, and $Y$ are all Riemannian, so much so that the question is both true and not at all a generalization of Moise's theorem. Instead, you are looking for a smooth triangulation of $M$ that supports $X$ and $Y$. A much better citation is to Goresky's theorem that any smooth stratification of a smooth manifold is supported by a smooth triangulation. This theorem is not specific to dimension 3; and stratifications in this sense are much more general than transversely intersecting submanifolds. On the other hand, the special case of a transverse intersection was known long before. Actually all of Goresky's theorem was sort-of previously known; the merit of the paper was to clean up and unify previous knowledge. -In general, in geometric topology, the hard direction is to go from continuous structures, to piecewise linear structures like triangulations, to smooth structures. Moise's theorem is about triangulating a topological manifold with no smoothness in sight --- so if the manifold is secretly smooth, the simplices of the produced triangulation could well be extreme fractals. On the other hand, Moise's theorem can be adapted to a variation of your question, in which $X$ and $Y$ are both collared and intersect transversely. This is in a way what you requested, but since your manifolds are smooth, not really what you want. Note that if $X$ and $Y$ are not collared, they could have Alexander horns, and then the supporting triangulation would not exist.<|endoftext|> -TITLE: Diagonalizable subgroups of a connected linear algebraic group -QUESTION [13 upvotes]: Let $G$ be a connected linear algebraic group -over an algebraically closed field $k$ of characteristic 0. -Let $D\subset G$ be a closed diagonalizable subgroup of $G$ -(a subgroup of multiplicative type). -Is it true that $D$ is contained in some torus $T\subset G$? -This is so for $G=\mathrm{GL}_n$. -Is this true for any connected linear $G$ (or any connected reductive $G$)? -I am stuck with this simple question... -Edit. The answer to the original question is NO, see Angelo's answer. -However, is it true that any cyclic finite diagonalizable subgroup $C$ of $G$ -is contained in some torus $T\subset G$? - -REPLY [4 votes]: Here is another example similar to Angelo's construction of a non-toral diagonalizable subgroup of a reductive group. I'll suppose that the characteristic is not 2. -Let $G = SO(V) = SO(V,\beta)$ for $\dim V > 2$, and write $V$ as an orthogonal sum -$V = U \perp W$ for $0 < \dim U < \dim V$ with $\dim U$ even, -such that the restriction of $\beta$ to $U$ and $W$ is non-degenerate. -Let $t \in G$ act as the identity on $W$ and as $-1$ on $U$. Then the -centralizer $M=C_G(t)$ identifies with the subgroup -{$(x,y) \in O(U) \times O(W) \mid \det(x) = \det(y)$}. In particular, -this centralizer is not connected: $M/M^0$ has order 2. -One can evidently choose an involution $s \in M \setminus M^0$, and then -$D = \langle t,s\rangle$ is a diag. subgroup of $G$ which is contained -in no maximal torus. -Part of this construction can be made in char. 2. Instead of $t$, you have -to take a non-smooth subgroup $\mu \simeq \mu_2$, essentially given by -the action of a semisimple element $X \in \operatorname{Lie}(G)$ ($X$ should -act as $1$ on $U$ and $0$ on $W$). Then $M=C_G(\mu) = C_G(X)$ is again -disconnected (well, now you can't argue by determinants) with component -group of order $2$. But this doesn't seem to lead to a non-toral diagonalizable -subgroup (any finite order element representating the non-trivial -coset of $M/M^0$ has a non-trivial unipotent part).<|endoftext|> -TITLE: Relation between cohomology of ordered and unordered configuration spaces? -QUESTION [12 upvotes]: For any manifold $M$, the unordered configuration space of $k$ points is obtained as a quotient of ordered configuration space of $k$ points by the group action of symmetric group on $k$ letters. Does it induce some relation between the cohomology algebras of the two spaces? - -REPLY [8 votes]: Oscar Randal-Williams already mentioned the transfer isomorphism $H^*(C_n(M);\mathbb{Q}) \approx H^*(F_n(M);\mathbb{Q})^{\Sigma_n}$. At the risk of self-promotion, one place you can see this transfer in action is in my paper "Homological stability for configuration spaces of manifolds", arXiv:1103.2441. In that paper I used an analysis of $H^*(F_n(M);\mathbb{Q})$ and the action of $\Sigma_n$ on it to conclude that the cohomology of the unordered configuration space $C_n(M)$ is eventually independent of the number of points $n$: $$H^k(C_n(M);\mathbb{Q})\approx H^k(C_{n+1}(M);\mathbb{Q})\text{ for }n\gg k.$$ -The key idea is that we can relate $F_{n+1}(M)$ with $F_n(M)$, even when we cannot relate $C_{n+1}(M)$ with $C_n(M)$ directly, and then the transfer map lets us push information from $F_n(M)$ down to $C_n(M)$. The basic framework of this approach is explained in the introduction. -There are also some explicit computations (some going back to Bödigheimer–Cohen–Taylor) of $H^*(C_n(M);\mathbb{Q})$ for various manifolds $M$ in Section 4.2 that might be of interest to you. -Also, you should check out the papers on the cohomology of configuration spaces written by the other participants in this discussion, Oscar Randal-Williams and Giacomo d'Antonio, which should provide a somewhat different perspective.<|endoftext|> -TITLE: Are there Steenrod operations on Hochschild cohomology of the group algebra of a finite group? -QUESTION [7 upvotes]: I know that the steenrod algebra acts on the group cohomology. I would like to know whether this action extends to the Hochschild cohomology of the group algebra, as the latter contains group cohomology as a subalgebra. - -REPLY [2 votes]: Yes, there is an action of the Steenrod algebra on HH^*(k[G]; k[G]), G a discrete group and k = Z/2 for the mod 2 Steenrod algebra. In my paper "A Comparison of Products in Hochschild cohomology," (on arXiv.org), I show how Steenrod's cup-i products act on the Hochschild cochain complex Hom_k(k[G]^*, k[G]) for an arbitrary coefficient ring k. The cup-i products can then be used to define an action of the Sq^i operators.<|endoftext|> -TITLE: Simple example of renormalization -QUESTION [13 upvotes]: As far as I understand, the RG theory, or functional RG theory is a mathematical tool for moving in the "scale dimension". The tool can be used for calculation of Feigenbaums constant (e.g. mentioned here). Can the theory be given a simple example of how to move one "step" in the "scale dimension" ? - -REPLY [9 votes]: The simplest and earliest example I know regarding the -renormalization group idea is the following. -Suppose we -want to study some feature $\mathcal{Z}(\vec{V})$ of some object $\vec{V}$ -which is in a set $\mathcal{E}$ of similar objects. -Suppose that unfortunately this question is too hard. What can one do? -The renormalization group philosophy is try to find a -"simplifying" transformation $RG:\mathcal{E}\rightarrow\mathcal{E}$, such that $\mathcal{Z}(RG(\vec{V}))= -\mathcal{Z}(\vec{V})$, and $\lim_{n\rightarrow \infty} RG^n(\vec{V})=\vec{V}_{\ast}$ -with $\mathcal{Z}(\vec{V}_{\ast})$ easy to understand. -Example (Landen-Gauss, late 1700's): -Let $\vec{V}=(a,b)\in\mathcal{E}=(0,\infty)^2$ and consider -$$ -\mathcal{Z}(\vec{V})=\int_{0}^{\frac{\pi}{2}} -\frac{d\theta}{\sqrt{a^2 \cos^2\theta+b^2\sin^2\theta}}\ . -$$ -A good choice of renormalization transformation here is $RG(a,b)=\left(\frac{a+b}{2},\sqrt{ab}\right)$, as discovered by Gauss. -A recent example now. -Example (Kadanof-Wilson, late 1960's early 1970's): -Take $\mathcal{E}$ to be the set of Borel probability measures on $\mathbb{R}^{\mathbb{Z}^d}$. Let $\mathcal{Z}(\vec{V})$ be equal to $1$ if the two-point function decays exponentially and $0$ otherwise. -Then define $RG$ as the transformation which gives the law of the block-spinned/coarse-grained field as a function of the law of the original field. -Note that the feature $\mathcal{Z}$ that one would like to preserve can be defined a bit more loosely. One could, e.g., "define" it as the long-distance behavior of the random spin field with probability law $\vec{V}$ (or in physics jargon: the low energy effective theory). -Also in the dynamical systems context, it could be the chaotic behavior or not of a map $\vec{V}$. Then $RG$ could be a doubling transformation, i.e., composition of the map with itself together with some rescalings and reversals of orientation.<|endoftext|> -TITLE: Asymptotic Distribution of Primes -QUESTION [5 upvotes]: Given an integer $n$ and let $1\leq m\leq n$ be such that $n$ and $m$ are coprimes define -$$ -\mathcal{N_{n,m}}:=\text{the set of primes $p$ such that $p\equiv{m}\hspace{0.1cm}\mathrm{mod}(n)$}. -$$ -Let $\mathcal{P}$ be the set of all primes. I seem to recall that the following result is true: -$$ -\varphi(n)^{-1}=\lim_{k\to\infty}{\frac{|\mathcal{N_{n,m}}\cap\{1,2,\ldots,k\}|}{|\mathcal{P}\cap\{1,2,\ldots,k\}|}}. -$$ -where $\varphi$ is the Euler's function. -My question is two fold: - -Does anyone have a reference for the previous fact? I was unsuccesful finding it. -Are there finer results along these lines? Second order results? - -Thanks! - -REPLY [10 votes]: A good way to find the result you mentioned is to search for Dirichlet's (prime number) theorem; while Dirichlet only proved the infinitude of the set in question, nowadays one will frequently find the more precise assertion you mentioned when this result is discussed. -A more common way to state it is that the number of primes congruent to $m$ modulo $n$ smaller than $x$ is asymptotically equal to $\varphi(n)^{-1} x/log (x) $ (assuming coprimeness as you did), which in combination with the prime number theorem implies what you are looking for. -There are a variety of results related to finer aspects of this problem; -key words e.g. Bombieri-Vinogradov Theorem or Siegel-Walfiz Theorem. -See for example the wikipedia article on Dirichlet's theorem here which also links to the keywords I mentioned for a quick overview. -Other than that as Gerry Myerson said any typical book on Analytic Number Theory will contain something on this subject (how much depends of course on the book). - -REPLY [5 votes]: It's just the prime number theorem for primes in arithmetic progression, no? Should be in any analytic number theory text that does the prime number theorem.<|endoftext|> -TITLE: Removable Singularities for Elliptic Equations -QUESTION [6 upvotes]: The following fact is quite standard and does not have a very long proof: -$(\*)$ If $u$ is harmonic on $B_1(0)\setminus \{0\}$ and uniformly bounded, then $u$ in fact extends to a harmonic function on the whole ball. -Some googling reveals that such statements are in fact true for large classes of elliptic operators with much more general singularity sets and growth conditions. There appears to be a vast literature on the subject. -I would like to know if the exact analogue of $(\*)$ has a "simple" proof for linear elliptic operators. -"Simple" is slightly ambiguous but is meant to mean tools present in standard elliptic theory textbooks, i.e. Gilbarg and Trudinger or Jost. - -REPLY [5 votes]: The proof I had in mind was actually simpler and appears to work only in dimension greater than 2. Here is a sketch for a second order self-adjoint elliptic operator $Pu = \partial_i(a^{ij}\partial_ju)$. Suppose $u$ satisfies $Pu = 0$ on $B\backslash\{0\}$. We want to show that $u$ is a weak solution on $B$. It suffices to show that if $\phi$ is a smooth compactly supported function on $B$, then -$$ -\int \phi Pu = 0. -$$ -Let $\chi$ be a smooth nonnegative compactly supported function that is equal to $1$ on a neighborhood of $0$ and, given $\epsilon > 0$, let $\chi_\epsilon(x) = \chi(x/\epsilon)$. Then -$$ -\int \phi Pu = \int [\phi(1-\chi_\epsilon) + \phi\chi_\epsilon]Pu = \int \phi\chi_\epsilon Pu -= \int P(\phi\chi_\epsilon)u. -$$ -If we now expand out $P(\phi\chi_\epsilon)$, we see that it is $O(\epsilon^{-2})$. Therefore, if $u \in L_\infty$, then -$$ -\int P(\phi\chi_\epsilon)u = O(\epsilon^{n-2}) -$$ -Therefore, taking the limit $\epsilon \rightarrow 0$ shows that $u$ is a weak solution on $B$. Elliptic regularity tells you that it is in fact a strong smooth solution.<|endoftext|> -TITLE: What is the "correct" generalization of operator norms for nonlinear operators? -QUESTION [11 upvotes]: I have been recently wondering what is a (or even the) "correct" generalization of the notion of an operator norm to nonlinear operators? -Please excuse the naivete of my question; if you think that question will benefit from being made more precise, then I will appreciate help towards making it so. - -Because I lack formal education in mathematics, I might be missing something obvious or well-known here. Could somebody point me in the right direction, and let me know what are the key concepts to think about when defining operator norms for nonlinear operators? -Some vague ideas that occurred to me: - -Linearizing the operator (locally), so the essentially traditional operator norms of the linearized operator could be considered? This sounds very unsatisfactory though. -If $A$ is a nonlinear operator for which we can sensibly define $\log A$, maybe that helps in tackling the nonlinearity. - -REPLY [8 votes]: Probably the answer depends on the context, where you need the generalized concept. -Unfortunately, usually the linearized operator is not bounded anymore, because it contains differential operators. -As pointed out by Mikael, the Lipschitz constant may be one possibility. For example, in the Crandall-Liggett theory of nonlinear semigroups the Hille-Yosida generation theorem on linear contraction semigroups is generalized to nonlinear contractions. Here, definitely the Lipschitz constant replaces the role of the operator norm. Or in the Banach fixed point theorem the convergence of the geometric series is generalized to iterations of nonlinear maps. -But there might be other answers, I am really curious.<|endoftext|> -TITLE: Equations for an algebraic gömböc -QUESTION [6 upvotes]: A gömböc is a $3-$dimensional convex body (having uniform density) which has exactly one stable and one instable equilbrium position (see http://en.wikipedia.org/wiki/G%C3%B6mb%C3%B6c). -Such a convex body keeps its properties under tiny perturbations. There exists thus a gömböc given by algebraic equations. What is the simplest (small degree and small coefficients) polynomial $P(x,y,z)\in\mathbb R^3$ such that $P(x,y,z)\leq 0$ defines a gömböc? -Added after Stopple's remarks: Existence of an algebraic gömböc is not obvious. I should have asked first for existence. -Two other questions concerning these objects are: -Is the set of all (algebraic) gömböcs connected? -Are there any gömböcs in dimension $>3$? (There are none in dimension $2$, every -$2-$dimensional convex set has at least $4$ equilibrium points.) - -REPLY [7 votes]: Aside from the last question, about high-dimensional gömböcs, these questions are certain to all be open, and are also totally inaccessible via the methods of Domokos and Varkonyi (who invented the gömböc and proved that it satisfied the constraints of Arnold's conjecture). For reference, here is their paper. To address Stopple's comment the paper uses throughout the stability of the desired properties (being a gömböc) under small perturbations, so there are semi-algebraic gömböcs (albeit likely very complicated ones, whose geometry is unlikely to be in any way revealing). -I went to a talk a couple years ago by Domokos and Varkonyi, where they described their methods and listed some of their conjectures. As I understood it at the time, they found the gömböc (and certain other shapes with the same equilibrium properties, albeit with much less deviation from a sphere) essentially by guessing a form for the solution, and then doing a search by computer within the limited parameter space this guess gave them. Certainly the first two questions you ask (about algebraicity of gömböcs and connectivity of the space of algebraic gömböcs) are inaccessible via this method. Indeed, Domokos and Varkonyi do not even have reasonable conjectures as to the minimal number of faces in a polyhedral gömböc, as I recall; their best candidate has many thousands of faces. -I will sketch what I believe to be an answer to your last question, which I'd say is the most interesting of the three you ask (why are algebraic gömböcs better than, say, piecewise smooth gömböcs?). -High-Dimensional Gömböcs Exist -The idea is to take proceed by induction; take an $n$-dimensional gömböc and revolve it around the axis connecting its two equilibria. It's not hard to convince yourself that this is a gömböc; I'll be a bit more explicit, albeit still a bit sketchy. -Let $G$ be a gömböc and $L$ the line passing through the two equilibria of $G$---say $G$ is good if for $x\in L$, the cross-section of $G$ perpendicular to $L$ at $x$ contains $x$, or is empty; note that Domokos and Varkonyi's gömböc is good (in fact it is easy to see that any gömböc is good, but I don't feel like proving this). I claim that if a good gömböc exists in dimension $n$, a good gömböc exists in dimension $n+1$, which suffices by induction. Indeed, place an $n$-dimensional gömböc $G^n$ in $\mathbb{R}^{n+1}$, and let $L$ be the line passing through the two equilibria (called $S$ and $U$, for stable and unstable). Let $G^{n+1}$ be the body formed by revolving $G^n$ about $L$. Then $G^{n+1}$ clearly is convex (by goodness of $G^n$) and has a stable equilibrium at $S$ and an unstable equilibrium at $U$. We must check that it has no other equilibria. But this is clear; indeed, any point on the surface of $G^{n+1}$ other than $S$ or $U$ belongs to some rotated copy of $G^n$, in which it is not an equilibrium point; thus it is not an equilibrium point of $G^{n+1}$. Goodness of $G^{n+1}$ is clear. -Note that there is an interesting question asked in the Domokos/Varkonyi paper, which they answer in dimension 3; namely which configurations of stable and unstable equilibria are permissible among convex homogenous solids? They show that given some convex homogenous solid, they can make small local modifications to add points of equilibria, without affecting other equilibria. It seems to me that their methods generalize to $k$ dimensions straightforwardly, so by producing a high-dimensional gömböc we have also shown that we may have as many stable and unstable equilibrium points as we desire, as long as there is one of each. -Unfortunately Domokos and Varkonyi have not addressed a slightly more subtle question; namely, there are various types of unstable equilibria (e.g. saddles) which may be classified as follows: consider the vector field on a (smooth) gömböc whose value at a point is the force exerted on the gömböc when it is resting on that point; equilibria are points where the vector field vanishes. Then the signature of the derivative of this vector field gives a reasonable classification of equilibria, assuming they are isolated. It would be interesting to know what are the permissible configurations of these slightly more refined equilibria.<|endoftext|> -TITLE: Is modern computability theory "really" about algorithms? -QUESTION [14 upvotes]: Apologies if my question seems overly naive, but I haven't seen/heard/read any good answers. -What is modern computability theory "really" about? The study of feasible(even remotely feasible) algorithms falls under the domain of theoretical and non-theoretical computer science. There is, of course, the a posteriori fact that computability theory tells us a lot about the structure of the natural numbers(I'm thinking of Turing degrees, etc). But, from a certain perspective, this can be seen as a historical coincidence. (I'm not saying that this is necessarily a "correct" perspective). -So what is the motivation for the subject of modern computability theory? - -REPLY [17 votes]: I think it's important to take a historical perspective. There was a time not so long ago when computers as we know them now did not exist. At that stage, coming up with a precise definition of an algorithm or of a Turing machine was a major advance, allowing one to build the earliest modern computers and begin the revolution that we take for granted today. -As actual computers became more powerful, interest shifted from the computable/uncomputable boundary to the feasible/infeasible boundary, where initially the definition of "feasible" was (roughly speaking) "polynomial time." So then we get the P = NP question and the birth of computational complexity theory as we know it today. -As computers became more powerful and more diverse, interest again shifted. People today are increasingly interested in parallel/distributed algorithms, cloud computing, SIMD architectures, etc. Datasets are so large that polynomial time doesn't cut it any more; people want linear time or even sublinear time algorithms. -So at the time of its invention, computability theory was about practical algorithms. The same goes for computational complexity theory and other subjects in computer science. But as technology advances, the definition of "practical" changes, so that the classical subjects no longer line up so nicely with the interests of current practitioners. That doesn't mean that the classical subjects are no longer of interest, because fundamentally important mathematical concepts never go away. But they become more abstract, and it takes a broad perspective to see their motivation and to be able to tell which problems are still of importance today. For example, in my opinion, some of the most exciting developments in computability theory today are its unexpected connections with differential geometry, as for example described in this paper by Soare. This work is very far removed from "practical algorithms" but illustrates how the study of fundamental mathematical concepts can reap unexpected dividends and is therefore worth pursuing even if immediate applications are not visible.<|endoftext|> -TITLE: Symplectic structures on a homotopy complex projective space -QUESTION [7 upvotes]: For $n>2$, there are infinitely many differentiable structures on the homotopy type of $\mathbb{C}P^n$. I want to know which differentiable structures support a symplectic form. -My question is as follows. Let $M$ be a closed symplectic manifold which is homotopy equivalent to a complex projective space. Can we say that $M$ is homeomorphic or diffeomorphic to the standard one? - -REPLY [9 votes]: As far as I know your question is completely open. At the present moment no one knows if for $n>2$ there is a symplectic structure on any manifold homotopic to $\mathbb CP^n$ but not diffeomorphic to $\mathbb CP^n$. In fact our knowledge in these type of questions equals to zero. Namely, the following is open: -Question. Let $M^{2n}$ be any closed manifold ($n>2$) admitting an almost complex structure $J$ and a class $h\in H^2(M,\mathbb R)$ with $h^n\ne 0\in H^{2n}(M^{2n})$. Is it true that there is a symplectic form $w$ on $M^{2n}$ in the class $h$? -Your question about $\mathbb CP^{n}$ is open as well for $n=2$. We don't know yet if there exist manifolds homeomorphic to $\mathbb CP^2$ but not diffeomorphic to it. The only thing that can be said is that if such a symplectic $4$-fold exists it would be of general type. -But if I would bet, I would say that the answer to your question is no...<|endoftext|> -TITLE: conditional equality symbol -QUESTION [13 upvotes]: Is there a standard notation (perhaps $A \stackrel{\leftarrow}{=} B$) meaning "in all situations where $B$ is defined, $A$ is defined and equals $B$"? -The kind of situation in which such a notation would be useful is the teaching of formulas like $$\lim_{x \rightarrow a} (f(x)-g(x)) = \lim_{x \rightarrow a} f(x) - \lim_{x \rightarrow a} g(x).$$ When I teach such formulas I take pains to teach them as theorems, with hypotheses that must be satisfied (in this case, the existence of $\lim_{x \rightarrow a} f(x)$ and $\lim_{x \rightarrow a} g(x)$) before the truth of the formula can be concluded, and I call to the students' attention the asymmetry of the situation (whenever the RHS is defined the LHS is defined and must be equal to it, but it is emphatically NOT always the case that when the LHS is defined the RHS must be defined and must be equal to it). I feel that one way to help students remember what the theorem says would be to use a variant of the equals sign when summarizing the theorem by a formula. -Has anyone introduced such a symbol? I think it would be at least as useful as the ":=" ("is defined as") symbol. - -REPLY [6 votes]: Freyd and Scedrov, in their book Categories, Allegories, use for this 'directed equality' a peculiar symbol that they call a Venturi tube and that looks a bit like $\mathrel{>=}$, so that $x \mathrel{>=} y$ means if $x$ is defined then so is $y$ and $x=y$. You can find some discussion at the nLab page on Kleene equality (the symmetric version of this) and in this nForum thread.<|endoftext|> -TITLE: Geometric meaning of torsion in homotopy groups -QUESTION [6 upvotes]: It is not too hard to understand the geometric meaning of torsion in homology groups of CW complexes. However, I thought it would be interesting to hear how people describe/think of the geometric meaning of torsion in the homotopy groups of a CW-complex. - -REPLY [12 votes]: Well, the silly answer is that $f:\mathbb{S}^k\to X$ represents a torsion element of order $p$ if $p \cdot f:\mathbb{S}^k\to X$ extends along $\mathbb{S}^k \hookrightarrow D^{k+1}$ to a map $\varphi: D^{k+1}\to X$. -The slightly less silly answer --- the slightly deeper answer, that is --- is that, equivalently, $f$ itself extends to a map $\tilde f: P^{k+1}(p)\to X$. Here $P^{k+1}(p)$ is the CW complex built of a point, a $k$-cell and a $(k+1)$-cell, where the $(k+1)$-cell is attached by a degree-$p$ map ${"p}$; and it's called the $(k+1)$th (cyclic) Moore space of order $p$. -For any $p$ the Moore spaces form a suspension spectrum $P^{k+1}(p) \simeq \mathbb{S}^1\wedge P^k(p)$, and there is in fact a long cofibration sequence -$$ \mathbb{S}^1 \overset{"p}{\to} \mathbb{S}^1 \to P^2(p) \to \mathbb{S}^2 \overset{"p}{\to} \dots $$ -which gives rise to a long exact sequence of generalized homotopy groups -$$ \pi_1(X) \leftarrow \pi_2(X,p) \leftarrow \pi_2(X) \overset{p}{\leftarrow} \pi_2(X) \leftarrow \pi_3(X,p)\leftarrow \cdots $$ -which in turn, for $n$ large enough, breaks up into the homotopy Universal Coefficient Theorem -$$ 0\leftarrow \mathrm{Tor}(\pi_n(X),\mathbb{Z}/(p)) \leftarrow \pi_{n+1}(X,p) \leftarrow \pi_{n+1}(X)\otimes \mathbb{Z}/(p) \leftarrow 0.$$ -The letter $P$ is used here because, at least because they have also been called Peterson spaces, and maybe because $P^2(2)\simeq \mathbb{RP}^2$ is the real projective plane. - -Note that I'm using the letter $p$ because when I worry about these things $p$ is usually prime, but that's not necessary in the above. - -We give up continuing the sequence at the first $\pi_1(X)$ because, without more information, it's not clear that $({"p})_\sharp$ should be a group homomorphism. Someone else can remind me what we can still say about the underlying sets.<|endoftext|> -TITLE: Can we relate Cech cohomology and derived functor cohomology even when the cover we choose isn't nice? -QUESTION [7 upvotes]: In my algebraic geometry class this semester, we've learned about Leray's Theorem, which states that for a sheaf $\mathcal{F}$ on a topological space $X$, and $\mathcal{U}$ a countable cover of $X$, if $\mathcal{F}$ is acyclic on every finite intersection of elements of $\mathcal{U}$ then the Cech cohomology $\check{H}^p(\mathcal{U},\mathcal{F})$ and derived functor cohomology $H^p(X,\mathcal{F})$ agree. -The potential for disagreement between them is covered well in these two MO questions. However, what neither of them seem to address is whether we can salvage any information about $H^p(X,\mathcal{F})$ from $\check{H}^p(\mathcal{U},\mathcal{F})$ even when $\mathcal{U}$ does not have the property that $\mathcal{F}$ is acyclic on all finite intersections, which is what I'd like to find out about here. I'm aware of Hartshorne Lemma 3.4.4, which says that there is a natural map $\check{H}^p(\mathcal{U},\mathcal{F})\rightarrow H^p(X,\mathcal{F})$ which is functorial in $\mathcal{F}$, but this is gotten by abstract nonsense - my feeling is that the existence of this map is not conveying much useful information. For all we know (?), all these maps could be the trivial homomorphism. -What I'm imagining is that perhaps the higher cohomology of $\mathcal{F}$ on the finite intersections of $\mathcal{U}$ can be related to the "difference" between $\check{H}^p(\mathcal{U},\mathcal{F})$ and $H^p(X,\mathcal{F})$, and that when the higher cohomology vanishes (i.e. $\mathcal{F}$ is acyclic), we get back the original theorem (that Cech and derived functor agree). -So, is there a useful relationship betwen Cech and derived functor cohomology even when $\mathcal{U}$ is not a nice open cover with respect to $\mathcal{F}$? Am I mistaken in assuming that the map $\check{H}^p(\mathcal{U},\mathcal{F})\rightarrow H^p(X,\mathcal{F})$ is not (particularly) useful? -Also I would like to avoid if possible the operation of taking the limit over all covers of $X$. I want to relate the specific Cech cohomology with respect to the cover $\mathcal{U}$, whatever its failings may be, with the derived functor cohomology. - -REPLY [8 votes]: There is a Mayer-Vietoris spectral sequence relating the two. This is a "direct" generalization of the Mayer-Vietoris long exact sequence, which is the special case in which your covering has just two open sets. -This is explained, if I recall correctly, in Bott-Tu. - -REPLY [2 votes]: Ken Brown (of Factorization Lemma fame) published a paper relating Cech hypercohomology and and sheaf cohomology, extending a result of Verdier (that is relegated to an appendix in, I believe, SGA 4. Brown calls it "Verdier's Hypercovering theorem"). There is an evident way to compare Cech cohomology and Cech hypercohomology (as discussed in the paper), and so I believe that it should answer your question: -Click! -Edit: If the link does not work, the paper is Abstract Homotopy Theory and Generalized Sheaf Cohomology by K.S. Brown<|endoftext|> -TITLE: Weakest condition for an integrable, almost-symplectic manifold? -QUESTION [11 upvotes]: I was recently speaking with someone who works in Computational Chemistry and they mentioned that in a lot of the computational simulations they do, they have systems that are not symplectic but still integrable. I believe that this means symplectic in the formal sense, but for the record the computation was described to me in terms of Jacobians associated to Hamiltonian flows (Something along the line of the treatment in the Wikipedia article on Symplectic Integrators). The system still has a well-defined non-degenerate bilinear form, but it may not be closed (i.e. an Almost Symplectic Structure). From his examples, I believe that Liouville's Theorem holds for a slightly bigger subcategory of the category of manifolds than the category of symplectic manifolds. I was just wondering what the most general conditions are for having a $2n$-dimensional smooth manifold with global, non-degenerate $2$-form $\omega$ satisfy Liouville's Theorem. I found this paper and I was just wondering if these are the most general conditions for such an integrable system. -Thanks! - -REPLY [4 votes]: Actually, there is a much weaker condition than 'strongly Hamiltonian' if all you want to do is preserve the associated volume form: For a given 'Hamiltonian' $H$ and a non-degenerate $2$-form $\omega$, if you define $X$ so that $\iota_X\omega = -\mathrm{d}H$, then the condition that $L_X(\omega^n) = 0$ simply becomes -$$ -0 = \mathrm{d}\bigl(\iota_X(\omega^n)\bigr) = \mathrm{d}\bigl(-n\,\mathrm{d}H\wedge\omega^{n-1}\bigr) = n\,\mathrm{d}H\wedge \mathrm{d}(\omega^{n-1}),\tag1 -$$ -and this is a single linear first-order partial differential equation for $H$. In fact, if one (uniquely) defines the vector field $Y$ by the relation -$$ -n\,\mathrm{d}(\omega^{n-1}) = \iota_Y\bigl(\omega^n\bigr),\tag2 -$$ -then the partial differential equation is just $\mathrm{d}(H)(Y) = 0$, i.e., $H$ must be constant along the flow lines of $Y$. To see this, just note that, for dimension reasons, we have $\mathrm{d}H\wedge\omega^n = 0$, so -$$ -0 = \iota_Y(\mathrm{d}H\wedge\omega^n) -= \mathrm{d}H(Y)\,\omega^n - \mathrm{d}H\wedge\iota_Y\bigl(\omega^n\bigr) -= \mathrm{d}H(Y)\,\omega^n - \mathrm{d}H\wedge (n\,\mathrm{d}(\omega^{n-1})), -$$ -which, since $\omega^n$ is non-vanishing, shows that (1) is equivalent to -$\mathrm{d}(H)(Y) = 0$. -Of course, if $Y$ vanishes identically (which happens in the symplectic case, of course, but much more generally), then this PDE is trivial and all 'Hamiltonian' flows preserve the 'symplectic' volume. On the other hand, if the flow of $Y$ is ergodic, then only the trivial 'Hamiltonian' flow preserves the volume, and, in particular, only the constants are 'strongly Hamiltonian'.<|endoftext|> -TITLE: Dense sphere packings which are not lattice packings -QUESTION [8 upvotes]: This question is about dense sphere packings in euclidean space $\mathbb R^n$. By a sphere packing I understand any arrangement of mutually disjoint solid open spheres in $\mathbb R^n$, all of the same radius. The density of a packing is -$$\mathrm{lim}_{R \to \infty}\frac{\mathrm{vol }(B(0,R) \cap \mathrm{spheres})}{\mathrm{vol } B(0,R)} $$ -if it exists. Here, $B(0,R)$ is the open ball of radius $R$ centered at $0 \in \mathbb R^n$. -In low dimensions, the highest possible densities of sphere packings are known to be attained by lattice packings, that is, packings such that the centers of the spheres form a discrete subgroup of $\mathbb R^n$ of rank $n$. One could speculate that this is so in all dimensions, but I doubt it very much... - -Is it true that for some (possibly very lagre) integer $n$, there is a sphere packing in $\mathbb R^n$ which has a higher density than any lattice packing? - -Edit -- Note: I didn't mean to ask about an explicit $n$, let alone about explicit packings. So i'm completely satisfied if somebody tells me that there is asymptotically such and such upper bound for lattice packing densities and this and that lower bound for general densest sphere packing densities. - -REPLY [23 votes]: In ten dimensions the best packing known is the Best packing, which is not a lattice packing. Marc Best found a nonlinear $40$-element binary code of block length $10$ and minimal Hamming distance $4$, and one can turn it into a sphere packing in $\mathbb{R}^{10}$ by centering spheres at all the points in $\mathbb{Z}^{10}$ that reduce to it modulo $2$. This packing seems to be better than any lattice packing, but no proof is known. The best lattice packings up through $\mathbb{R}^8$ were determined by the 1930's, but even $\mathbb{R}^9$ isn't known, let alone $\mathbb{R}^{10}$, and there aren't even good enough bounds to prove that nothing is as good as the Best packing. -For some reason, good non-lattice packings are more likely to be known in even dimensions than odd dimensions, at least for dimensions a little less than $24$. For example, (hypothetical) answers are known in $\mathbb{R}^{18}$, $\mathbb{R}^{20}$, and $\mathbb{R}^{22}$, but not in between. I imagine this is an artifact of coding-theory-based constructions. -Probably lattices are suboptimal in all sufficiently high dimensions, but nobody really understands how to think about this problem asymptotically. The best existence results in high dimensions all produce lattices, but that's presumably just because lattices are more tractable than non-lattice packings.<|endoftext|> -TITLE: Is any $(n-1)\times (n-1)$ submatrix of an $n \times n$ Vandermonde matrix invertible? -QUESTION [11 upvotes]: Given an $n \times n$ vandermonde matrix $V$ which is invertible, is any $(n-1) \times (n-1)$ submatrix of $V$ invertible also? -I think the answer is yes, but I don't know how to prove. - -REPLY [5 votes]: Recognizing the functions $e_k$ in David's excellent elaboration of Thierry's answer leads to a simple (equivalent) description and a proof for the problem considered here (but probably not the more general one of $n-1$ arbitrary powers.) -Consider the Vandermonde matrix $$V= -\begin{bmatrix} -1 & 1 & 1 & \ldots & 1 & 1 \\ -x_1 & x_2 & x_3 & \ldots & x_{n-1} & t \\ -x_1^2 & x_2^2 & x_3^2 & \ldots & x_{n-1}^2 &t^2 \\ -\vdots & \vdots & \vdots & \ddots & \vdots \\ -x_1^{n-1} & x_2^{n-1} & x_3^{n-1} & \ldots & x_{n-1}^{n-1} & t^{n-1} -\end{bmatrix}$$ -Claim: The minor obtained from removing the row and column of $t^{n-1-k}$ is non-invertible exactly when the $x_i$ are the $n-1$ (distinct) roots of a polynomial $$a_0t^{n-1}+a_1t^{n-2}+\cdots+a_{n-2}t+a_{n-1}$$ with $a_k=0.$ - -Sketch: Think of $t$ as a variable and the various $x_j$ as parameters. The determinant is a sum of $n!$ terms each with total degree $\frac{n^2-n}{2}$ in $t$ and the $x_j$. It is also a polynomial $f(t)$ with coefficients polynomals in the $x_i$. This determinant is zero if any two of the columns are equal. Hence $f(t)$ is identically zero if any two $x_i$ are equal so $$f(t)=g(t)\prod_{00$ as $k=0$ corresponds to the minor which eliminates the last row. Then what remains is also a Vandermonde matrix and hence non-singular. In the polynomial form of the claim we see that there can't be enough distinct roots. -The fact that $e_{n-1} = x_1x_2x_3\cdots x_{n-1}$ corresponds to the generalization of Barts answer: If some $x_i=0$ and the minor eliminates the first row, we end up with an all zero column and a singular minor. Otherwise we can factor out $x_i \ne 0$ from each column and what remains is again a Vandermonde matrix and nonsingular. -If the minor eliminates neither the first nor the last row, then we can set the parameters to be $x_j=\exp(\frac{2\pi i j}{n-1}),$ the $n-1^{\mbox{st}}$ roots of unity, and have a singular minor since the first and last rows are identical (all 1's). For the problem considered here, this is the only way (up to scaling) to have a repeated row. Also, this is not a different case than $e_k=0$ since we just have the roots of $t^{n-1}-1.$ - -That easy example is unavailable if we restrict to entries from $\mathbb{R}$ and $n>3.$ Then it is especially fortunate to have the description in terms of the $e_k.$<|endoftext|> -TITLE: Does combining Abhyankar's Lemma and embedded resolution give horizontal normal crossings -QUESTION [8 upvotes]: Let $\pi:Y\longrightarrow \mathbf{P}^1_{\mathbf{Z}}$ be a finite surjective flat morphism of schemes, where $Y$ is a normal integral flat projective 2-dimensional $\mathbf{Z}$-scheme, with branch locus $D$. Let us suppose that $\pi$ is tamely ramified. -Question 1. Does this mean that for every prime number $p$ such that some vertical component of $D$ maps to $p$ in $\textrm{Spec} \mathbf{Z}$ does not divide the degree of $\pi$? -By Abhyankar's Lemma, there exists a number field $K$ with ring of integers $O_K$ such that the branch locus $D\subset (B_{O_K})_{\textrm{hor}}$ of the morphism $\pi_{O_K}:Y_{O_K}\longrightarrow \mathbf{P}^1_{O_K}$ is horizontal. -By embedded resolution, there exists a projective birational morphism $f:X\longrightarrow \mathbf{P}^1_{O_K}$ such that $f^\ast D$ is a divisor with normal crossings. -Question 2. Since $D$ is horizontal, does it follow that $f^\ast D$ is horizontal. I thought so but wasn't completely sure. -Question 3. Does the (edit: normalization of the) fibre product $Y^\prime$ of $Y_{O_K}$ and $X$ over $\mathbf{P}^1_{O_K}$ give a morphism $Y^\prime\longrightarrow X$ with branch locus a horizontal normal crossings divisor? I would be surprised if the branch locus picked up vertical components... - -REPLY [4 votes]: Question 1: it is not exactly the meaning of tameness (ramification index prime to $p$ and separable residue extension). But if $p>\deg \pi$, then $\pi$ is tame at $p$. -Question 2: The strict transform of $D$ is horizontal because it is finite birational to $D$, but the preimage of $D$ by $f$ is not horizontal in general. -Question 3: I don't think so. Consider the following example. Let $x$ be a coordinate of $P^1$ and consider the cover induced by the equation $y^3=x^3+p^2$. Then $D$ is the divisor $V(x^3+p^2)$. To solve the singularity of $D$, you have to blowup $P^1$ along the point $(x,p)$. Then you get a vertical ramification at $p$. (In general you need to normalize the fiber product of $X$ with $Y$).<|endoftext|> -TITLE: Galois action on Betti cohomology? -QUESTION [10 upvotes]: Hi all, -Given a variety $X$ over the real numbers, we can consider the singular cohomology of the space $X(\mathbb{C})$, with coefficients in $\mathbb{Q}$, say. The action of complex conjugation on $X(\mathbb{C})$ induces an action on these cohomology groups. How do you compute/describe this action in concrete examples? For example, what happens in the case of $H^1$ of a curve of genus $g$, or even in the supposedly trivial case where $X = Spec(\mathbb{R})$? Other examples would also be interesting, or general hints on how to understand the action. - -REPLY [19 votes]: Suppose that $X$ is smooth and projective. Then Hodge theory gives a decomposition -of $H^i(X(\mathbb C),\mathbb C)$ into the direct sum of $H^{p,q}$'s (with $p + q = i$). -Now there are two complex conjugations acting on $H^i$ with $\mathbb C$ coefficients: the -"trivial action" just coming from conjugating the coefficients, which I'll denote by $c$ (and which is conjugate-linear as an automorphism of $H^i(X(\mathbb C),\mathbb C)$ by its very definition), -and the non-trivial one coming from the action of complex conjugation on $X(\mathbb C)$ itself (which is $\mathbb C$-linear), -which I'll denote by $Fr_{\infty}$ (the Frobenius at $\infty$). -It is a familiar fact from beginning Hodge theory that $c$ interchanges $H^{p,q}$ and $H^{q,p}$. What you can check is that $Fr_{\infty}$ also interchanges $H^{p,q}$ and -$H^{q,p}$. -Another fact is that if $X$ is geometrically connected, then $Fr_{\infty}$ acts on the top dimensional cohomology as multiplication -by $(-1)^d$, if $X$ is $d$-dimensional. (This is a special case of the fact that the top-dimensional etale cohomology of a geometrically connected smooth projective $d$-dimensional variety is always the $-d$th Tate twist.) -These two facts taken together serve to establish most of the claims in David Speyer's answer, for example. -What can be hard to work out in general is what $Fr_{\infty}$ does on $H^{p,p}$. Just as an example, if $X$ is Spec $\mathbb R$, a single point, then the cohomology is just a one-dimensonal $H^{0,0}$, and $Fr_{\infty}$ acts trivially. If we then take a pair of such points, we get a two-dimensional $H^{0,0}$, again with $Fr_{\infty}$ acting trivially. -On the other hand, if $X$ is Spec $\mathbb R[x]/(x^2 + 1)$, so that $X$ is a pair of complex conjugate points, then the cohomology of the complex points is again a two-dimensional $H^{0,0}$, but with $Fr_{\infty}$ acting with one $+1$ and one $-1$ eigenspace (because it switches the two points). -One can make this example a bit more interesting by blowing up $\mathbb P^2$ at (a) a pair of points each defined over $\mathbb R$, and (b) a pair of complex conjugate points. -In case (a) one gets $X$ whose complex points have a three-dimensional $H^{1,1}$ with $Fr_{\infty}$ acting by $-1$, while in case (b) one again gets a three-dimensional $H^{1,1}$, but now the eigenvalues of $Fr_{\infty}$ are $(1,-1,-1)$. (In each case the $H^{1,1}$ is spanned by the fundamental class of a line in $\mathbb P^2$ together with the fundamental classes of the two exceptional divisors. In case (a) each of -these fundamental classes is defined over $\mathbb R$, and so each contributes an eigenvalue -of $-1 = (-1)^1$ --- here the exponent $1$ is because these are fundamental classes of curves --- while in case (b) we see that $Fr_{\infty}$ is switching the two exceptional divisors.)<|endoftext|> -TITLE: Inverse function in singular points -QUESTION [5 upvotes]: Hi there! -First of all i'm not a matematician, i'm just mechanical engineer who interested in some math. -Probably trivial question. Suppose I have a mapping $F: \mathbb{R}^n \to \mathbb{R}^n$ with Jacobian $J(x)$. And at some point $x_0$ Jacobian degenerate $\det J(x_0) = 0$. Is there general condition, for the existence of continuous inverse function near $x_0$, which is weaker than inverse function theorem. For example $y=x^3$ has continuous inverse $x = \sqrt[3]{y}$ near $x = 0$, but this fact is not provided by IFT. -Some usefull thinks I found in book "The implicit function theorem: history, theory, and applications" by Steven George Krantz, Harold R. Parks, but there are only special cases. -Thanks for all inputs. - -REPLY [7 votes]: One result is that small Lipschitz perturbations of the identity on a Banach space are homeomorphisms onto an open set: see here for a more precise statement. Note that the classic Inverse Mapping Theorem is usually proved as a consequence of this. -Another one deals with a strongly monotone map on a real Hilbert space, $F:H\to H$, that is, a map satisfying $(F(x)-F(y),x-y)\ge c |x-y|^2$ for all $x$ and $y$. A continuous strongly monotone map on $H$ is a homeomorphism onto $H$ (see e.g.Thm 11.2 in Klaus Deimling's book Nonlinear functional analysis, where you can also find more advanced inversion theorems). For one variable, real-valued functions this reduces to the well-known criterium. -In the case of $\mathbb{R}^n$ let's recall the theorem of Invariance of Domain. Injectivity is usually simpler to check than surjectivity, and in the case of a continuous map $f:U\to \mathbb{R}^n$ on an open set $U$ of $\mathbb{R}^n$ guarantees that $f$ is a homeomorphism onto an open set $f(U)$.<|endoftext|> -TITLE: A question concerning separate and joint continuity of bilinear maps -QUESTION [10 upvotes]: Suppose that $V$ is a locally convex topological vector space and $f:V^2 \to V$ is a bilinear map. Suppose that $C \subseteq V$ is compact and convex, $f$ maps $C^2$ into $C$ and -$f \restriction C^2$ is separately continuous. -Must $f \restriction C^2$ be jointly continuous? -In the particular application I have in mind, $V = \ell_\infty^*$ with the weak* topology. -Moreover the function $f$ is injective. -I suspect even in this setting that this is false. -I am also interested in a good reference for the optimal results of concerning separate and joint continuity of bilinear maps. -Ideally this would be written for someone who is not a functional analyst. - -REPLY [9 votes]: A personal obsession is (weakly) almost period functions. Let $G$ be a discrete group (you can work more generally) and for $f\in \ell^\infty(G)$ let $O(f)$ be the set of translate of $f$ by the group action. Set -$$ AP(G) = \{ f\in \ell^\infty(G) : O(f)\text{ is relatively compact}\} \\$$ -$$WAP(G) = \{ f\in \ell^\infty(G) : O(f)\text{ is weakly relatively compact}\}$$ -Then these are unital sub-$C^*$-algebras of $\ell^\infty(G)$ and so have character spaces $G^{AP}$ and $G^{WAP}$. You can extend the product from $G$ to these: then the product on $G^{AP}$ is jointly continuous, but that on $G^{WAP}$ only separately jointly continuous. Translated back to algebra, this means that the measure spaces $M(G^{AP})$ and $M(G^{WAP})$ become Banach algebras for the convolution product. The continuity translates to say that the product on $M(G^{WAP})$ is separately weak$^*$-continuous, and that on $M(G^{AP})$ is even jointly weak$^*$-continuous on bounded sets. -So that gives an example: if $C$ is the closed unit ball of $M(G^{WAP})$ and $f$ is the product map, then $f$ satisfies your requirements, but $f$ is not jointly continuous unless $WAP(G) = AP(G)$ (I think-- this last claim needs a little chasing of definitions). If $G$ is abelian then $G^{AP}$ is the classical Bohr compacitification, but $G^{WAP}$ is much larger. There are books by Berglund (and coauthors) on this topic; they do, IMHO, need a functional analysis background. -Surely there are easier examples for what you want though. (And $f$ is not injective in my example).<|endoftext|> -TITLE: Blocks of the category of representations of $GL_n({\mathbb F}_q)$ -QUESTION [8 upvotes]: Let k be an algebraically closed field of characteristic $\ell$, and let $q = p^r$ be -a prime power with $p \neq \ell$. Suppose I have a cuspidal representation $\pi$ of -$GL_n({\mathbb F}_q)$, for some $n < \ell$. -The supercuspidal support of $\pi$ consists of $m$ copies of a supercuspidal representation -$\sigma$ of $GL_{d}({\mathbb F}_q)$ for some integers $m$ and $d$ with $md = n$. -In this setting, there also exists a cuspidal representation $\pi'$ of -$GL_m({\mathbb F}_{q^d})$ that has supercuspidal support equal to $m$ copies of the trivial character. -Let $A$ be the block of the category of $W(k)[GL_n({\mathbb F}_q)]$-modules containing $\pi$, and let $A'$ be the block of the category of $W(k)[GL_m({\mathbb F}_{q^d})]$-modules -containing $\pi'$. Are $A$ and $A'$ equivalent as categories? (More precisely, is there an equivalence of $A$ and $A'$ that takes $\pi$ to $\pi'$?) I can prove this when $\pi$ -is supercuspidal; i.e. $m=1$, but computations from character theory and considerations from the Langlands program make me suspect that it's true in general. -David Helm - -REPLY [3 votes]: I may come a little late, but I think Theorem B' on page 52 of Bonnafé and Rouquiers' paper at IHES does what you want.<|endoftext|> -TITLE: how to determine whether an ideal is prime or not by an algorithm -QUESTION [12 upvotes]: Given polynomials $f_{1},\cdots,f_{n}\in \mathbb{C}[x_{1},\cdots,x_{m}]$, do we have an algorithm to determine whether the ideal $I=(f_{1},\cdots,f_{n})$ is prime ideal or not? Of course, we assume the polynomials are irreducible. - -REPLY [2 votes]: Buchberger's algorithm should do, Faugere's F4 is also one. However, this is generally for any ideal and not necessarily for irreducible. Is it something specific to irreducible polynomials that you are looking for?<|endoftext|> -TITLE: Are problems in complexity theory dependent on set theory? -QUESTION [8 upvotes]: I was pondering the fact that maybe the classical hard complexity-theoretic questions are undecidable, not because they are so themselves, but because some set-theoretic foundations makes the complexity-theoretic foundations shaky. -My thoughts was that perhaps something like the Continuum hypothesis makes P vs NP undecidable. -So my question is, is there a "finitary" or otherwise obviously sane environment for complexity theory that would discount this theory immediately? -I'm aware of simpler structures where P vs NP has been decided, but I don't know how that would fit in. -I apologize in advance if this doesn't make sense. - -REPLY [6 votes]: The statement that P=NP can be expressed in first-order arithmetic, and that part of mathematics is unaffected by the known methods of proving set-theoretic independence results (forcing, inner models).<|endoftext|> -TITLE: Fibers of fibrations of a 3-manifold over $S^1$ -QUESTION [18 upvotes]: Given a fiber bundle $S\hookrightarrow M \rightarrow S^1$ with $M$ (suppose compact closed connected and oriented) 3-manifold and $S$ a compact connected surface, it follows form the exact homotopy sequence that $\pi_1(S)\hookrightarrow \pi_1(M)$. Does this imply that the "fiber" of a 3-manifold which fibers over $S^1$ is well defined? -The answer should be NO, so I am asking: are there simple examples of 3-manifolds which are the total space of two fiber bundles over $S^1$ with fibers two non homeomorphic surfaces? -EDIT: the answer is NO (see Autumn Kent's answer). I'm just seeking for a "practical" example to visualize how this phoenomenon can happen. - -REPLY [27 votes]: There are simple examples with $M = F \times S^1$ for $F$ a closed surface of genus $2$ or more. Choose a nonseparating simple closed curve $C$ in $F$, then take $n$ fibers $F_1,\cdots,F_n$ of $F\times S^1$, cut these fibers along the torus $T=C\times S^1$, and reglue the resulting cut surfaces so that $F_i$ connects to $F_{i+1}$ when it crosses $T$, with subscripts taken mod $n$. The resulting connected surface is an $n$-sheeted cover of $F$ and is a fiber of a new fibering of $M$ over $S^1$. The monodromy of this fibering is a periodic homeomorphism of the new fiber, of period $n$.<|endoftext|> -TITLE: Pullbacks of canonical divisors along branched maps -QUESTION [9 upvotes]: Let $f\colon X \to Y$ be a finite map of smooth surfaces. Let the divisor $D$ of $Y$ be the branch locus of $f$. We assume that $D$ is a union of nonsingular curves intersecting transversally with no three components meeting at one point. -Claim: -$2K_X = f^*(2K_Y + D),$ -where $K_X$ is the canonical divisor of $X$ and $K_Y$ is the same for $Y$. -How do you prove this claim? - -REPLY [7 votes]: As Felipe and Karl correctly pointed out, the formula you write is not true in general. However, if you read Vakil's paper, you see that he's considering a very particular situation, namely Galois coverings with Galois group $G=(\mathbb{Z}_p)^3$, where $p=2,3$. -Let me explain how the formula works in a simpler case, namely $G=(\mathbb{Z}_2)^2$, the so-called bidouble covers (see Catanese' paper [Ca1] in Vakil bibliography). Then you can try to prove it in Vakil's cases (maybe after reading Pardini's paper on abelian covers). -Let us call $\chi_i$, $i=1,2,3$ the three non-trivial characters of $G$. Then the branch locus $D$ of $f$ can be written as -$D=D_1 + D_2 +D_3$, -where $D_i$ correspond to $\chi_i$. The divisors $D_i$ are smooth and intersect transversally. -Now we can factor $f$ as -$X \stackrel{g}{\longrightarrow} Z \stackrel{h}{\longrightarrow} Y$, -where $h$ and $g$ are double covers branched over $D_1+D_2$ and $h^*D_3$, respectively. Note that in general the intermediate cover $Z$ is singular! -By using the formulae for double covers, we can write -$2K_Z=h^*(2K_Y+D_1+D_2), \quad 2K_X=g^*(2K_Z+h^*D_3)$ -that is, putting things together, -$2K_X=g^*h^*(2K_Y+D_1+D_2+D_3)=f^*(2K_Y+D)$.<|endoftext|> -TITLE: Explicit description of boundary map in algebraic K-theory -QUESTION [10 upvotes]: Recall that for a DVR A with fraction field F and residue field k, there is a "localization" fiber sequence in algebraic K-theory, -$$K(k) \rightarrow K(A) \rightarrow K(F).$$ -In Remark 5.17 of his "Higher Algebraic K-theory: I" paper, Quillen gives an explicit description of the corresponding boundary map $\partial:\Omega K(F) \rightarrow K(k)$, saying the proof will be in a later paper. My question is, has a proof appeared in the literature? I'd also be happy with proofs in the literature of any similar descriptions, e.g. involving the S-dot construction. -Thank you for reading! - -REPLY [21 votes]: Apparently there is no proof in the literature. Let me provide a proof here, in case it's helpful. Actually, first let's give the set-up, then a precise statement, then the proof. To tell the truth, the set-up takes the most time. So this fact is harder to state than to prove, which may be why it's not so useful in practice. But I think it does give valuable intuition for the boundary map. -The set-up: for a noetherian commutative ring R, let $mod_R$ denote its abelian category of finitely-generated modules. Then for a DVR $A$ with residue field $k=A/\mathfrak{m}$ and fraction field F, the composite -$$mod_k \hookrightarrow mod_A \rightarrow mod_F$$ -has a canonical trivialization ($0\simeq M\otimes_A F$ whenever $\mathfrak{m}\cdot M=0$), and Quillen's devissage and localization theorems combine to show that the induced map on classifying space of Q-constructions -$$|Q(mod_k)|\rightarrow |Q(mod_A)| \rightarrow |Q(mod_F)|$$ -is a fiber sequence (of pointed spaces -- or of spectra after group-completing w.r.t. $\oplus$). The goal is to describe the "boundary", or "monodromy" map -$$\partial: \Omega |Q(mod_F)|\rightarrow |Q(mod_k)|$$ -associated to this fibration. -The description will use a different model for the left-hand side $\Omega |Q(mod_F)|$. Namely, let $\mathcal{C}$ denote the following symmetric monoidal category: - -Objects of $\mathcal{C}$ are tuples $(L_0,L_1,V)$ where $V$ is a finite-dimensional $F$-vector space and $L_0,L_1$ are spanning f.g. $A$-submodules of $V$ such that $ L_0\subset L_1$ and $\mathfrak{m}\cdot L_1 \subset L_0$. -Morphisms in $\mathcal{C}$ from $(L_0,L_1,V)$ to $(L_0',L_1',V')$ are isomorphisms $\alpha:V\simeq V'$ such that $\alpha(L_0)\supset L_0'$ and $\alpha(L_1)\subset L_1'$ (so $\alpha$ realizes $L_1/L_0$ as a subquotient of $L_1'/L_0'$). -The symmetric monoidal structure is $\oplus$. - -Now consider the composition -$$|\mathcal{C}|\rightarrow |(mod_F)^{\simeq}|\rightarrow \Omega |Q(mod_F)|.$$ -(By $(-)^{\simeq}$ we mean underlying groupoid.) Here the first map is induced by the forgetful functor $\mathcal{C}\rightarrow (mod_F)^{\simeq}$, and the second map is the usual inclusion of objects of an exact category as loops in the Q-construction. Then on the one hand it is easy to check that the first map is an equivalence (fixing $V$, the space of choices for $L_0$ is contractible because it filters down to zero; and fixing $V$ and $L_0$, the space of choices for $L_1$ is contractible because it has a maximal element $\mathfrak{m}\cdot L_0$); but on the other hand by the comparison of the group-completion and Q-construction approaches to K-theory (see Algebraic K-theory: II) the second map is a group-completion. Thus the composition is a group-completion, so one way to produce maps $\Omega |Q(mod_F)|\rightarrow |Q(mod_k)|$ (such as our desired $\partial$) is to produce a symmetric monoidal functor $\mathcal{C}\rightarrow Q(mod_k)$, then apply geometric realization and group-complete. -This leads us to the precise statement: let $f:\mathcal{C}\rightarrow Q(mod_k)$ denote the symmetric monoidal functor -$$f(L_0,L_1,V) = L_1/L_0.$$ -Then the group-completion of $|f|$, identified via the above with a map $\Omega |Q(mod_F)|\rightarrow |Q(mod_k)|$, gives the boundary map in the localization sequence. -Now for the proof. We will use the following characterization of boundary maps in general. Given a fiber sequence $F\rightarrow E\rightarrow B$ and a map $f:\Omega B\rightarrow F$, to give a homotopy from $f$ to the boundary map is equivalent to giving a nullhomotopy of the composite $\Omega B\overset{f}{\longrightarrow} F\rightarrow E$, plus a homotopy between $id_{\Omega B}$ and the map $\Omega B\rightarrow \Omega B$ gotten by combining the two different nullhomotopies of the composition $\Omega B\overset{f}{\longrightarrow} F\rightarrow E\rightarrow B$. (I'm being careless with signs.) -So let us provide this data in our case. The whole time we will be considering maps of spectra out of the group completion of $|\mathcal{C}|$, so we can and will specify all our data just at the level of $\mathcal{C}$. There are two pieces of data: - -We need a nullhomotopy of the composite $\mathcal{C}\overset{f}{\longrightarrow} Q(mod_k)\rightarrow Q(mod_A)$ after geometric realization. This is provided by a composite of two natural transformations, one of which goes the wrong way. The first is the obvious identification of $L_1/L_0$ with a subquotient of $L_1$; the second is the obvious identification of $0$ with a subquotient of $L_1$. -We need a homotopy of the resulting map $|\mathcal{C}|\rightarrow \Omega |Q(mod_F)|$ (gotten by combining the nullhomotopy of 1. with the nullhomotopy provided by our fiber sequence) with the canonical map considered above. This is clear: when we tensor the null-homotopy of 1. with $F$ over $A$ and use $0\simeq (L_1/L_0)\otimes_A F$ and $L_1\otimes_A F\simeq V$, we see exactly the usual way of mapping $|(mod_F)^\simeq|$ to $\Omega |Q(mod_F)|$, the one used to compare group-completion to Q-construction. - -That gives the proof.<|endoftext|> -TITLE: equivariant cohomology with respect to a loop group -QUESTION [5 upvotes]: Let $G$ be a compact connected simply connected Lie group. Let $LG$ be the corresponding -loop group. What is the cohomology of its classifying space (i.e. what is the equivariant -cohomology of a point with respect to $LG$?) I would like to express it in terms of -the Lie algebra of $G$. -Is the corresponding $dg$-algebra formal? - -REPLY [3 votes]: You can compute $H^\ast(LBG)$ as the Hochschild cohomology -$$HH^\ast(C_\ast(G), C^\ast(G)),$$ -where $C_\ast(G)$ is the singular chain complex of $G$, equipped with the Pontrjagin product, and $C^\ast(G)$ are the cochains, with the $C_\ast(G)$-module structure dual to the obvious one on $C_\ast(G)$. If you like, you can then replace $C^\ast(G)$ with the Chevalley-Eilenberg complex $K$ for the Lie algebra, and $C_\ast(G)$ with its dual $K^\ast$. At this point, though, it is perhaps not so obvious how to recover the ring structure on $K^*$ which lets you define the Hochschild cohomology.<|endoftext|> -TITLE: Optimal Countdown -QUESTION [14 upvotes]: Many know the TV game Countdown, whose French version Des chiffres et des lettres has lasted since 1965. -The rules of the count are as follows: you are given natural integers $n_1,\ldots,n_6$ and a target $N$. You are free to employ the four operations $+,\times,-,\div$. You may employ each $n_j$ at most once. You must end with the result $N$. -For mathematicians, a colleague of mine suggests to modify the rule that way: you are given $k\ge1$. You are free to choose $n_1,\ldots,n_k$. Then you must realize the targets $1,2\ldots,N$. How do you choose $n_1,\ldots,n_k$. What is the largest possible $N_k$ ? -Examples: - -$k=1$, nothing much interesting, $N_1=1$ -$k=2$, then $(1,3)$ yields $N_2=4$ -$k=3$, then $(2,3,10)$ yields $N_3=17$. Optimal ? - -Edit about the rules. Parentheses are allowed (and useful). Division $a/b$ is possible only when $b$ divides $a$ in the usual sense of integers. You may have negative integers, but it does not help. - -REPLY [5 votes]: The last results I got using my program CEB are: - - For k=5, the best solution $(n_1,n_2,n_3,n_4,n_5)$ ≤ (200,200,200,200,200) is (2,3,4,63,152) and we can get all numbers up to $N_5$=450. - - For k=6, the best solution $(n_1,n_2,n_3,n_4,n_5,n_6)$ ≤ (10,20,30,40,50,80) is (2,3,24,37,47,66) and we can get all numbers up to $N_6$ = 3398.<|endoftext|> -TITLE: Maximum principle for weak solutions -QUESTION [6 upvotes]: maximum principles for parabolic PDEs seem to be well-known, if the solution is a priori $C^2$ (cf. Protter, Weinberger: Maximum principles in differential equations). However, what about weak solutions? To be specific, are there any maximum principles on the nonnegativity of solutions $u\in W^{1,p}(0,T;L^p(\Omega))\cap L^p(0,T;W^{2,p}(\Omega))$, $p\in(1,\infty)$, where $\Omega\subset R^n$ is a bounded domain? For given nonnegative initial data, does the solution remain positive, as long as it exists? -I assume yes, since there are numerous authors that use the results from Weinberger/Protter just for weak solutions. -I would appreciate any hints on this topic. -Best regards, Marc - -REPLY [2 votes]: You might want to distinguish between maximum principles (which assert typically things like "the max of the solution is attained on the boundary / parabolic boundary of the set") and positivity, which assert things like "if the data are non-negative on the (parabolic) boundary, then so is the solution in the entire domain". The latter often can be shown with functional analytic techniques (see previous post). -As to maximum principles for generalized solutions, there is work by Jensen on viscosity solutions of fully nonlinear elliptic problems. And there is work in the 70s that extends maximum principles for elliptic equations to solutions in $W^{2,n}$ where $n$ is the spatial dimension (if I remember correctly).<|endoftext|> -TITLE: Elliptic Curves over Global Function Fields -QUESTION [9 upvotes]: I am currently thinking about a problem, and I feel that by knowing more about elliptic curves over extensions of $\mathbb{F}_q(T)$, for $q$ a power of $p$ say, might lead to insight. I am also inspired by the following sentence of Silverman in his Advanced Topics (introduction to Chapter 3): - -"Thus conjectures about elliptic curves defined over $\mathbb{Q}$ are often first tested and proven in the easier setting of elliptic curves over $\mathbb{F}_q(T)$." - -Unfortunately I can't seem to find a source that deals with this theory systematically, something that studies properties of these curves particular to these global function fields. I put this into Google, and I learned a few nice things, such as, if the corresponding elliptic surface is $K3$, then SHA can be identified with the Brauer group of the surface, but I'm looking for a more text-book or expository type approach that starts at a lower level, say with torsion subgroups, isogenies, Heights and Mordell-Weil Theorem... (and it would be nice if that cohomological fact were also proved therein). Does anybody have any good suggestions? -May I also ask for other ways such curves are 'easier', or better understood, than the -characteristic 0 case? -I apologise in advance if this question is unfit for the site; I've not been a member for very long. - -REPLY [4 votes]: (Warning: another piece of self advertising) Perhaps you might wish to take a look at this paper of mine (joint with A. Bandini and I. Longhi). With the final aim of giving an alternative proof of a classical result of Igusa describing the image of Galois representations attached to non-isotrivial elliptic curves over a global function field $F$ (of characteristic $p>3$), we explain various auxiliary results in the arithmetic theory of such elliptic curves. In particular, using basic properties of Faltings heights of elliptic curves, we give a detailed proof of the function field analogue of a well-known theorem of Shafarevich about F-isomorphism classes of "admissible" elliptic curves over F. -Since we tried to be as elementary as possible, this article might also be viewed as an introduction to various basic topics in the arithmetic of elliptic curves over global function fields (isogenies, Frobenius morphisms, Faltings heights, Shafarevich's theorem, ...) for which we could not find a convenient reference in the literature.<|endoftext|> -TITLE: What is the size of the smallest rigid extension field of the complex numbers? -QUESTION [6 upvotes]: Suppose we consider a rigid extension field $F$, i.e., $\text{Aut}(F) = 1$ over the complex numbers $\mathbb C$. What is the minimal cardinality of $F$? In particular it should hold that in this case $|F| > |\mathbb C|$. -Moreover, if we replace $\mathbb C$ with any other algebraically closed field, what can one say in this case? -Any comment, reference, or pointer is highly appreciated. -All the best, -Sebastian - -REPLY [9 votes]: Firstly, it may be worth linking to this related question on MO. -Pröhle proved that all fields of characteristic 0 can be embedded in a rigid field - see "Does a given subfield of characteristic zero imply any restriction to the endomorphism monoids of fields?" for his particular construction. -Later, Dugas and Göbel showed in - -"All infinite groups are Galois groups over any field" - that for ${\mathbb C}$ it can be done in a field of cardinality the successor cardinal of $2^{\aleph_0}$ which is as good as you could expect. -In follow-up papers -"Automorphism groups of fields I" and "Automorphism groups of fields II", they show that for any group $G$ and any field $K$, there exists an extension with automorphism group isomorphic to $G$ and cardinality $\aleph_0|K||G|$.<|endoftext|> -TITLE: What are the lengths that can be constructed with straightedge but without compass? -QUESTION [5 upvotes]: Most field theory textbooks will describe the field of constructible numbers, i.e. complex numbers corresponding to points in the Euclidean plane that can be constructed via straightedge and compass. This field is the smallest field of characteristic 0 that is closed under square root (i.e. is Pythagorean) and is closed under conjugation. -I'm interested in know: What is the field of numbers that can be constructed if we disallow compass and use only straightedge? -I have not checked this up, but it seems that this question led Hilbert to formulate his 17th problem, particularly the version involving polynomials with rational coefficients (rather than the real coefficients which Artin proved). I'm also interested in knowing more about this history too. - -REPLY [5 votes]: The short answer is that nothing is constructible. As is standard, we begin with the two points $(0,0)$ and $(1,0)$. Then we can draw a line between them, and that's it. We can't draw any more lines, and hence we can't construct any new points. The Euclidean rules say that we are only allowed to draw a new line if we are joining two already-constructed points, and a point can only be constructed if it is the intersection of two lines (or, irrelevant to this discussion, two circles or a line and a circle). -However, suppose you begin with a finite collection of points $(x_1,y_1),\ldots, (x_n,y_n)$. Let $C$ be the set of points constructible from this set using only a straightedge (unmarked). If a point $(x,y)$ is in $C$, then $x$ and $y$ can be formed from $x_1,\ldots,x_n,y_1,\ldots,y_n$ using the operations $+$, $-$, $\cdot$, and $\div$ (since new points are created as intersections between lines). However the converse of this is not true (as the two-point example shows). I suppose you could say more about what $C$ looks like, but it would probably be messy.<|endoftext|> -TITLE: About representation theory of Heisenberg group -QUESTION [8 upvotes]: Actually I am an undergraduate student, but I want to study Heisenberg groups over arbitrary field. -Firstly, why is this group important? I know that the Heisenberg group is important in the field of quantum physics, but in mathematics how is it? And secondly I want to know the current studies of the classifications of the irreducible representations of Heisenberg groups over an arbitrary field(like over $\mathbb{Q}_p,\mathbb{R}, \mathbb{C}$ or finite field). Would you recommend any references in this area? (e.g. survey papers or influential papers) -Thanks. - -REPLY [8 votes]: Here is one important way in which the Heisenberg group is important : -Let $F$ be a local field of characteristic not equal to 2 (so for example, one of your fields $\mathbb{Q}_p$ or $\mathbb{R}$ or $\mathbb{C}$ above), and let $\psi$ be a nontrivial unitary additive character of $F$. Let $V$ be a symplectic space over $F$ with symplectic form $\langle , \rangle$, and form the Heisenberg group $H(V)$. This is the group which is set theoretically $V \times F$, with group operation $(v_1, x_1)(v_2,x_2) = (v_1+v_2, t_1+t_2 + \frac{1}{2} \langle v_1,v_2 \rangle)$. -By the Stone-Von-Neumann theorem, $H(V)$ has a unique irreducible smooth (or unitary) representation $(\rho_{\psi}, W)$ over $\mathbb{C}$ with central character $\psi$. Note that $Sp(V)$ acts on $H(V)$ in a natural way through its action on $V$. Then, by Schur's Lemma, for every $g \in Sp(V)$, there exists an intertwining operator $\phi_g$ between $(\rho_{\psi}, W)$ and $(\rho_{\psi} \circ g, W)$, which is unique up to multiplication by $\mathbb{C}^*$. -We therefore get a projective representation -$$Sp(V) \rightarrow PGL(W)$$ -$$g \mapsto \phi_g$$ -It can be shown that this projective representation lifts to a representation of $\widetilde{Sp(V)}$, where $\widetilde{Sp(V)}$ is a naturally defined double cover of $Sp(V)$, called the Metaplectic group. This representation, denoted $\omega_{\psi}$, is called the Weil representation. Both the Weil representation and the metaplectic group are important in number theory and representation theory. If you're curious, Wee Teck Gan has a nice survey article on how the metaplectic group arises in the Langlands program : http://www.math.ucsd.edu/~wgan/ICCM.pdf -Best, -Moshe - -REPLY [6 votes]: This seems like a reasonable survey article: -http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.bams/1183547543 -Also there is a book by Deitmar and Echterhoff - "Principles of Harmonic Analysis", which I found quite readable, but that treats coefficients in $\mathbb{R}$ only. But this book is probably more likely to be useful for graduate students. - -REPLY [2 votes]: You should really study the representation theory in the wider context of a semidirect product. George Mackey wrote the basic theory on this, and you'll almost certainly understand more by working out what the theory says for these special cases (whether discrete groups or more general locally compact groups).<|endoftext|> -TITLE: Normal measures on $P_{\kappa }(\lambda )$ extend the club filter -QUESTION [5 upvotes]: This is (a variation on) exercise 20.4 in Jech's "Set Theory." Let $j : V \to M$ witness $\lambda$-supercompactness of $\kappa$, and consider the normal measure $U$ on $P_{\kappa}(\lambda)$ consisting of those $x$ such that $j[\lambda] \in j(x)$. (How do you make the left quotation mark symbol to denote 'j-image-of-lambda'?) -We want to show that this measure extends the club filter. -This hint is as follows: Suppose $C$ is club. Then define $D = j[C]$. Then: - -$D$ is a directed subset of $j(C)$. -$D$ has size $|C| \leq \lambda ^{< \kappa} < j(\kappa )$. -Therefore $\bigcup D \in j(C)$. -$\bigcup D = j[\lambda ]$ - -I'm fine with 1. I'm not sure about 2 - where is the argument taking place, in $V$ or in $M$, or both? For 3, it appears the underlying argument is this: -$V \vDash \forall E \subset C\ (E$ directed and $|E| < \kappa \Rightarrow \bigcup E \in C)$ -and so -$M \vDash \forall E \subset j(C)\ (E$ directed and $|E| < j(\kappa) \Rightarrow \bigcup \in j(C))$ -I can accept this assuming that 2 means "$D \in M$ and $M \vDash |D| < j(\kappa )$." I'm having trouble with 4 as well - I believe that $j[\lambda] \subseteq \bigcup D$, but why does the reverse inclusion hold, i.e. why is it that $x \in C, \beta \in j(x) \Rightarrow \beta \in j[\lambda]$? - -REPLY [3 votes]: Suppose $\kappa$ is $\lambda$-supercompact for some $\lambda \geq \kappa$, and let $j: V \rightarrow M$ be an elementary embedding with critical point $\kappa$ such that $j(\kappa) > \lambda$ and $M^{\lambda} \subseteq M$ for some inner model $M$. First, observe that $V$ and $M$ agree on $P_{\kappa}\lambda$ because $M$ is closed under ${<}\kappa$ sequences. In particular, this means that $\lambda^{{<}\kappa} \leq (\lambda^{{<}\kappa})^M$ since $M \subseteq V$. But this then means that $j(\kappa) > (\lambda^{{<}\kappa})^M \geq \lambda^{{<}\kappa}$ because $j(\kappa)$ is inaccessible in $M$ and $j(\kappa)$ is greater than both $\lambda$ and $\kappa$. Next, note that any $x \in P_{\kappa}\lambda$ will be a subset of $\lambda$ having size less than the critical point $\kappa$ so that $j(x) = j''x \subseteq j''\lambda$. -[Specifically, if for some $\alpha < \kappa$, we have a bijection $f: \alpha \rightarrow x$, then $j(f)$ will be a bijection between $j(\alpha) = \alpha$ and $j(x)$. So every element of $j(x)$ is of the form $j(f)(\beta)$ for some $\beta < \alpha$, but $j(f)(\beta) = j(f(\beta)) \in j''x$ since $\beta$ is also below the critical point.] -Also, $M$ will contain $h = j \upharpoonright \lambda$ by its closure under $\lambda$ sequences. Therefore, $M$ will have $j''P_{\kappa}\lambda = \{j(x)| x \in P_{\kappa}\lambda\} = \{j''x| x \in P_{\kappa}\lambda\} = \{h''x| x \in P_{\kappa}\lambda\}$. Now letting $g: P_{\kappa}\lambda \rightarrow \lambda^{{<}\kappa}$ be a bijection in $V$, we will have a bijection $j(g) \upharpoonright j''P_{\kappa}\lambda: j''P_{\kappa}\lambda \rightarrow j''\lambda^{{<}\kappa}$ in $M$. Therefore, $M$ will have the range of $j(g)$, which is exactly $j''\lambda^{{<}\kappa}$. Now, since $C$ has size at most $\lambda^{{<}\kappa}$ (in $V$), we may let $e: \lambda^{{<}\kappa} \rightarrow C$ be a surjection. Then $j(e) \upharpoonright j''\lambda^{{<}\kappa}: j''\lambda^{{<}\kappa} \rightarrow j''C$ is a surjection in $M$ so similarly, its range, $D = j''C$, will be in $M$. But $M$ will also know that $j''\lambda^{{<}\kappa}$ has size $\lambda^{{<}\kappa} < j(\kappa)$ because $M$ can construct $j \upharpoonright \lambda^{{<}\kappa}$ from $j''\lambda^{{<}\kappa}$ by virtue of $j$ being order-preserving. Therefore $\bigcup D \in j(C)$ by the ${<}j(\kappa)$-directed closure of $j(C)$ in $M$, as you mention. -Also, if $x \in C$, then $|x| < \kappa$ and $x \subseteq \lambda$ so $j(x) = j''x \subseteq j''\lambda$. Therefore, $\bigcup D = \bigcup j''C \subseteq j''\lambda$.<|endoftext|> -TITLE: Does a generic normal measure extend the club filter? -QUESTION [5 upvotes]: This question is related to this one. The setup is as follows: -In $V$, $\kappa$ is supercompact and $\mathbb{P} = \mathrm{Coll}(\kappa, \aleph_2)$ is the Levy collapse. $G$ is $(V,P)$-generic, and in $V[G]$, $C \subset P_{\kappa}(\lambda)$ is club. Let $j : V \to M$ be an embedding witnessing $\lambda$-supercompactness of $\kappa$. Lift this embedding to $j^{\ast} : V[G] \to M[G \times H]$ where we may assume $H$ is generic over $V[G]$ for the poset $\mathrm{Coll}^{V[G]}(j(\kappa), \aleph_2)$ (note $\aleph_2 ^{V[G]} = \kappa$). Now in $V[G \times H]$ we define $U \subset P_{\kappa}^{V[G]}(\lambda)$ by: -$x \in U$ iff $j[\lambda] \in j^{\ast}(x)$. -This $U$ belongs to $V[G \times H]$, but $V[G]$ "would" think its a normal measure on $P_{\kappa}^{V[G]}(\lambda)$, i.e. it's $\kappa$-complete if we restrict to $<\kappa$-sequences in $V[G]$ and it's normal if we restrict to regressive functions $f : P_{\kappa}^{V[G]}(\lambda) \to \lambda$ where $f \in V[G]$. My question: -Does $U$ also extend the club filter, if we restrict to clubs in $V[G]$? - -REPLY [2 votes]: With the help of Jason's answer to the question I linked to, I think I might be able to solve this one. The key is to show that $P_{\kappa}^{V[G]}(\lambda) \in M[G\times H]$. Let's simply denote this set by $X$. An element $x$ of $X$ can be regarded as a function $x : \omega_1 \to \lambda$, which will be a subset of $\omega_1 \times \lambda$. A nice name for such a subset is a map, in $V$, from $\omega_1 \times \lambda$ to the collection of antichains in $\mathbb{P}$. Since $\mathbb{P}$ has the $\kappa$-chain condition and $\mathbb{P} \in M$, $M$ correctly knows the set of antichains of $\mathbb{P}$. Since $M^{\lambda} \subset M$, $M$ correctly knows the set of nice $(V,\mathbb{P})$-names for subsets of $\omega_1 \times \lambda$, let's call this set $Y$. -Now $\mathbb{P}$ names are $j(\mathbb{P})$ names (since $\mathbb{P} \subset j(\mathbb{P})$), so we get that -$X = \{ \dot{x}^{G\times H}\ |\ \dot{x}^{G\times H} : \omega_1 \to \lambda, \dot{x} \in Y \} $ -This is since $\dot{x}^{G \times H} = \dot{x}^G$ for nice $(V,\mathbb{P})$-names. So $X \in M[G \times H]$ as desired. - -Recall, we want to show that $U$ extends the club filter, so take $C \in V[G]$ club in $X$. Following the hint in the previous question, we want to show that - -$D = \{j^{\ast}(x)\ |\ x \in C\}$ belongs to $M[G \times H]$ -$D$ has size less than $j(\kappa)$ in $M[G \times H]$, and -$\bigcup D = j''\lambda$. - -Then, since $D$ is a directed subset of $j^{\ast}(C)$, elementarity will give us that $j''\lambda \in j^{\ast}(C)$, as desired. -Since $M^{\lambda} \subset M$, we know that $g := j\upharpoonright \lambda \in M$. So for $x \in X$ (and in particular for $x \in C$), $j^{\ast}(x) = j''x = g''x$. Using this it's not hard to see that $\bigcup D = j''\lambda$. It also implies that -$j^{\ast}$ $''X$ $= \{j^{\ast}(x)\ |\ x \in X\}$ $= \{j''x\ |\ x \in X\} = \{g''x\ |\ x \in X\}$ -belongs to $M$. Now if $h : X \to C$ is a surjection in $V[G]$, then $j^{\ast}(h)\upharpoonright j^{\ast}$ $''X$ is belongs to $M[G\times H]$ and its range is $D$, so $D \in M[G\times H]$. -It remains to show $M[G\times H] \vDash |D| < j(\kappa)$. Since $j(\kappa)$ is inaccessible in $M$, there's some bijection $i: \alpha \to Y$ in $M$ for some $\alpha < j(\kappa)$. This gives a surjection $k : \alpha \to X$ in $M[G\times H]$. We can obtain a bijection $l : X \to j^{\ast}$ $''X$ via $l(x) = g''x$. And $j^{\ast}(h)\upharpoonright j^{\ast}$ $''X$ is a surjection onto $D$. So: -$M[G\times H] \vDash |D| \leq |j^{\ast}$ $''X| = |X| \leq \alpha < j(\kappa)$.<|endoftext|> -TITLE: Examples of Galois-invariant central simple algebras which aren't base change? -QUESTION [10 upvotes]: Suppose $L/K$ is a Galois extension of number fields, with Galois group $G_{L/K}$. Write $\mathrm{Br}(L)^{G_{L/K}}$ for the subgroup of central simple algebras $A/L$ which are Galois-invariant; equivalently, these are the algebras such that for $v$ a place of $K$ and $w$ a place of $L$ with $w|v$, the local invariants $\mathrm{inv}_w(A)$ only depend on $v$. The Hochschild-Serre spectral sequence yields an exact sequence -$0\to \mathrm{Br}(L/K) \to \mathrm{Br}(K) \to \mathrm{Br}(L)^{G_{L/K}}\to H^3(G_{L/K},L^{\times}),$ -where the third arrow is represented by the base change map $A\mapsto A\otimes_{K} L$, and the image of the fourth arrow is the "obstruction" preventing a Galois invariant algebra from arising via base change. Now, the $H^3$ appearing here is often zero; for example, it vanishes if all the Sylow subgroups of $G_{L/K}$ are cyclic. My questions: - -In the case $K=\mathbf{Q}$ (say), what is the simplest example of a Galois extension $L/\mathbf{Q}$ and a Galois-invariant central simple algebra over $L$ which is not a base change from any proper subfield? Does $G_{L/\mathbf{Q}}=A_4$ work? -Is there a simple way to compute the obstruction map into $H^3$? It seems like the answer to this must naively be "no", since it cannot be as simple as "do such and such with the local invariants". -How much of the above goes through without assuming $L/K$ is Galois? - -REPLY [12 votes]: [big edit] -(1) Let $L/K$ be as above. Take any non-identity element $\sigma \in G_{L/K}$, and let $F=L^{<\sigma>}$ be the fixed field of the cyclic subgroup generated by $\sigma$. By your comment above about the vanishing of $H^3$, every galois invariant CSA of $L$ is a base change of a CSA of $F$. Hence, there are no galois invariant CSA that are not a base change from any proper subfield. -But, if you fix $L$ and $K$, then there can be galois invariant CSA's of $L$ that are not a base change from $K$. Since the smallest non-cyclic group is $C_2\times C_2$, we would like to search for examples with such a galois group. -For $K=\mathbb{Q}$, probably, the simplest example is: -$$L=\mathbb{Q}(\sqrt{-3},\sqrt{13}),\ (43)=P_1\cdot P_2\cdot P_3\cdot P_4,$$ -$$u=\frac{1}{4}P_1+\frac{1}{4}P_2+\frac{1}{4}P_3+\frac{1}{4}P_4 \in \bigoplus_v Br(L_v)$$ -I will prove below that this element of the Brauer group is not a base change, and that in fact a similar construction can be given for any extension. -(2) Near the end of Tate's "Global Class Field Theory" section in Cassels-Frohlich, there is a small paragraph devoted to $H^3(G_{L/K},L^\times)$: - -"$H^3(G,L^\times)$ is cyclic of order $n/n_0$, the global degree divided by the lowest common multiple of local degrees, generated by $\delta u_{L/K}$ ($\delta:H^2(C_L)\rightarrow H^3(L^{\times})$), the "Teichmuller 3-class." ..." - -Therefore, assume $n_0 < n$ for our galois extension $L/K$. Let $v_0$ be an unramified finite place of $K$ that splits completely (exists by Chebotarev's theorem). We can construct an element of the Brauer group similar to the one above: -$$u := \sum_{w|v_0} \frac{1}{n} w$$ -This is in fact an element of $Br(L)$ since the number of $w|v_0$ is $n$, so that $n\frac{1}{n} = 1 \in \mathbb{Z}$. - -Proposition. For any prime $p$ that divides $\frac{n}{n_0}$, $\frac{n}{pn_0}\cdot u$ is not a base change. - -From which we immediately get: - -Corollary. The map $Br(L)^{G_{L/K}}\rightarrow H^3(L^\times)$ is onto. - -Proof of proposition. Assume that $\frac{n}{pn_0}\cdot u$ is the base change of some $u'$, i.e. -$$u' = \sum_v n_v v \mapsto \sum_v \sum_{w|v} [L_w : K_v] n_v w = \frac{n}{pn_0}\cdot u$$ -Since for any $w|v_0$: $[L_w : K_{v_0}] = 1$, we must have $n_{v_0} = \frac{n}{pn_0}\cdot \frac{1}{n} = \frac{1}{pn_0}$. And since $\sum_v n_v \in \mathbb{Z}$, at least one other place $v_1$ has -$$v_p(n_{v_1}) \le v_p(n_{v_0}) = v_p(\frac{1}{pn_0}) < 0$$ -Where $v_p$ is the usual $p$-adic valuation. So, using that $n_0$ is the lcm of local degrees: -$$v_p([L_{v_1}:K_{v_1}] n_{v_1}) \le v_p(n_0) + v_p(\frac{1}{pn_0}) = -1$$ -Contradicting the zero coefficient of any $w|v_1$ in $u$. -A small computation shows that for the extension $\mathbb{Q}(\sqrt{-3},\sqrt{13})/\mathbb{Q}$, $n_0=2$, and that indeed this is the smallest (discriminant-wise) $C_2\times C_2$ such extension.<|endoftext|> -TITLE: Estimating the probability that one Poisson RV is larger than another -QUESTION [7 upvotes]: Let $X$ and $Y$ be Poisson random variables with means $\lambda$ and $1$, respectively. The difference of $X$ and $Y$ is a Skellam random variable, with probability density function -$$\mathbb P(X - Y = k) = \mathrm e^{-\lambda - 1} \lambda^{k/2} I_k(2\sqrt{\lambda}) =: S(\lambda, k),$$ -where $I_k$ denotes the modified Bessel function of the first kind. Let $F(\lambda)$ denote the probability that $X$ is larger than $Y$: $$F(\lambda) := \mathbb P(X > Y) = \sum_{k=1}^{\infty} S(\lambda, k) = \mathrm e^{-\lambda - 1} \sum_{k=1}^\infty \lambda^{k/2} I_k(2\sqrt{\lambda}).$$ According to Mathematica, the graph of the function $F$ looks like -(source: nyu.edu) -My question:Is there a closed-form expression for the function $F$?If not, what are $\lim_{\lambda \to 0} F'(\lambda)$ and $F'(1)$? What is the asymptotic behavior as $\lambda \to \infty$? - -REPLY [2 votes]: By simple computations : The definition of the modified Bessel function of the first kind yields -$$ -I_k(\lambda)=\sum_{n\geq 0}\frac{1}{n!(n+k)!}\left(\frac{\lambda}{2}\right)^{2n+k} -$$ -so that we get (the sums transpositions are clearly allowed) -$$F(\lambda)=e^{-\lambda-1}\sum_{k\geq 1}\sum_{n\geq 0}\frac{\lambda^{k+n}}{n!(k+n)!}=e^{-\lambda-1}\sum_{n\geq 1}a_n\lambda^n \qquad \mbox{where}\qquad a_n=\frac{1}{n!}\sum_{k=0}^{n-1}\frac{1}{k!}.$$ Thus, deriving under the sign sum -$$ F'(\lambda) = e^{-\lambda-1}\Big(1+\sum_{n\geq 1 }[(n+1)a_{n+1}-a_n)]\lambda^n\Big) -= e^{-\lambda-1}\sum_{n\geq 0}\frac{1}{(n!)^2}\lambda^n -$$ -we obtain the closed form -$$ -F'(\lambda)=e^{-\lambda-1}I_0(2\sqrt{\lambda}). -$$ -One finally get $$F'(0)=e^{-1}, \quad F'(1)=e^{-2}I_0(2)=e^{-2}\sum_{n\geq0}\frac{1}{(n!)^2}$$ -and, using the asymptotic formula when $\lambda\rightarrow+\infty$ for all $k$ -$$ -I_k(\lambda)=\frac{e^{\lambda}}{\sqrt{2\pi\lambda}}\Big(1+O(\lambda^{-1})\Big), -$$ -that -$$ -F'(\lambda)=\frac{e^{2\sqrt{\lambda}-\lambda-1}}{2\sqrt{\pi\sqrt{\lambda}}}\Big(1+O(\lambda^{-1/2})\Big) -$$ -when $\lambda\rightarrow+\infty$.<|endoftext|> -TITLE: Is the colimit of a filtered diagram of module categories in AbCat induced by a cofiltered diagram of rings equivalent to the category of modules over the limit? -QUESTION [6 upvotes]: Given a cofiltered diagram of commutative rings $F:D\to \mathrm{CRing}$, we obtain a filtered diagram $\mathrm{Mod}(F):D^{op}\to \mathrm{ExAbCat}$ (where $\mathrm{ExAbCat}$ is the category of Abelian categories with exact additive functors between them) induced by the contravariant functor $CRing^{op}\to \mathrm{ExAbCat}$ sending the maps $f:A\to B$ in $\mathrm{CRing}$ to the restriction functor, $f^*:B\mathrm{-Mod}\to A\mathrm{-Mod}$. -Then here's the question: -Is $\varinjlim \mathrm{Mod}(F)$ equivalent to $(\varprojlim F)\mathrm{-Mod}$? -Thank you very much for any help! - -REPLY [5 votes]: Unless I am severely mistaken, the answer to the question as stated is "no". (There is by the way some slight disagreement in the literature as to what "filtered category" means; for many, a filtered poset means a nonempty poset where any two elements have an upper bound, but in Sheaves in Geometry and Logic, Mac Lane and Moerdijk mean that any two elements have a lower bound. I don't think this affects my answer.) -Let us take for instance the ring of $p$-adic integers as inverse limit of the (directed in either sense) system -$$\ldots \to \mathbb{Z}/(p^{n+1}) \to \mathbb{Z}/(p^n) \to \ldots$$ -Each of these commutative rings $A$ can be viewed as an $Ab$-enriched category, and the abelian category of left modules over $A$ can be viewed as the $Ab$-enriched category of $Ab$-enriched functors $A \to Ab$. So we are homming into $Ab$, and the usual expectation is that a contravariant hom-functor takes colimits to limits (I will tighten up this statement in a moment), but not limits to colimits; the question as stated has to do with the latter situation. -The filtered colimit (and it would be reasonable to interpret "colimit" in a 2-categorical sense) of the system of full subcategory inclusions -$$\ldots \to Ab^{\mathbb{Z}/(p^n)} \to Ab^{\mathbb{Z}/(p^{n+1})} \to \ldots$$ -can be taken to be the union of these (abelian) categories. The union is just the category of those abelian groups where each object is completely annihilated by some $p^n$ (not uniformly of course; the $n$ depends on the group); in particular, they are all torsion. On the other hand, most objects in the category of modules over the $p$-adic integers, for example the $p$-adic integers itself, are not torsion. So there's your counterexample. -But if we turn the directions around, then indeed (filtered) colimits are taken to limits. There are some slightly tricky issues to deal with; first, "morally", we really ought to be working with 2-categorical limits -- weak limits to be precise. To do it right, we should be working with the category of commutative rings as a 2-category (construing ring homomorphisms as enriched functors and taking 2-cells as natural transformations), but if I'm not mistaken, weak colimits in this 2-category of commutative rings coincide with ordinary colimits of commutative rings, and we can relax. Second, there is the issue of whether colimits of diagrams of commutative rings are identified with the corresponding colimit qua $Ab$-enriched categories -- in fact this is not true in full generality (e.g., the coproduct of $\mathbb{Z}/2$ and $\mathbb{Z}/3$ in commutative rings is the terminal ring, whereas the coproduct in $Ab$-enriched categories is gotten by taking a disjoint sum). However, considering the inclusions -$$\text{CRing} \hookrightarrow \text{Ring} \hookrightarrow \text{AbCat}$$ -one finds that both preserve filtered colimits (for the second inclusion, we're saying one-object $Ab$-enriched categories are closed under filtered colimits), so we make that restriction. -With those caveats squared away, a general fact is that homming into an object like $Ab$, giving here a functor -$$Ab^{-}: \text{CRing}^{op} \to \text{AbCat},$$ -takes (weak) filtered colimits to (weak) limits.<|endoftext|> -TITLE: Looking for papers and articles on the Tarskian Möglichkeit -QUESTION [7 upvotes]: Some background: Łukasiewicz many-valued logics were intended as modal logics, and Łukasiewicz gave an extensional definition of the modal operator: $\Diamond A =_{def} \neg A \to A$ (which he attributes to Tarski). -This gives a weird modal logic, with some paradoxical, if not seemingly absurd theorems, notably $(\Diamond A \land \Diamond B) \to \Diamond(A\land B)$. Substitute $\neg A$ for $B$ to see why it's been relegated to a footnote in the history of modal logic. -However, I've realised that it's less absurd when that definition of a possibility operator is applied to Linear Logic and other substructural logics. I have an informal talk about this earlier in the month. A link to the talk is at http://www.cs.st-andrews.ac.uk/~rr/pubs/lablunch-20110308.pdf -Anyhow, the only non-critical work that I found a reference to is a talk by A. Turquette, "A generalization of Tarski's Möglichkeit" at the Australasian Association for Logic 1997 Annual Conference. The abstract is in the BSL 4 (4), http://www.math.ucla.edu/~asl/bsl/0404/0404-006.ps Basically Turquette suggested applications in m-valued logics for m-state systems. (I've not been able to obtain any notes, slides or other content of this talk, so I would appreciate hearing from anyone who has more information.) -I don't have any applications for it, but I find the properties to be interesting enough to merit a paper (in progress) on adding this operator to various substructural logics, and comparing those logics with themselves augmented by Lewsian modal operators. -My question: Is anyone here aware of other articles or papers on Tarski's Möglichkeit or "extensional" modalities? -Note: This is a question from CS Theory @ Stack Overflow https://cstheory.stackexchange.com/questions/5928/looking-for-papers-and-articles-on-the-tarskian-moglichkeit which a commentator suggested I post in Math Overflow. - -REPLY [6 votes]: Rob, I didn't know this was called the Tarskian Möglichkeit, but Martin Escardo and I have been studying this operator (A -> B) -> A, in the more general case when falsity is an arbitrary formula B, for the past few years, mainly in connection with computational interpretations of classical theorems. If we let B be fixed, then we define -J A = (A -> B) -> A -It is easy to show that this is a strong monad. We call it the "selection monad" or the "Peirce monad", as J A -> A is Peirce's law. In fact, the seemingly absurd theorem you mentioned in your post is the cornerstone for our work on interpreting ineffective principles such as the Tychonoff theorem, for instance. Have a look at some of our papers, e.g. -Martín Escardó and Paulo Oliva. Sequential games and optimal strategies. Proceedings of the Royal Society A, 467:1519-1545, 2011. -Martín Escardó Paulo Oliva, The Pierce translation. Annals of Pure and Applied Logic, 163(6):681-692, 2012. -Or others found on our webpages: http://www.eecs.qmul.ac.uk/~pbo/ -Any paper which mentions "selection functions" or "game" is related to the operator you are asking about. -I must warn we have been studying this operator in the setting of intuitonistic (minimal) logic. But I find it very interesting that you are looking at this in the more refined (substructural) settings of linear logic and Lukasiewicz logic. -Best regards, Paulo.<|endoftext|> -TITLE: Generalization of Borsuk-Ulam -QUESTION [23 upvotes]: Let $n$ be a positive intger. Is the following true? For continuous maps $f: \mathbb S^n \rightarrow \mathbb S^n$ and $g: \mathbb S^n \rightarrow \mathbb R^n$, there exists a point $x \in \mathbb S^n$ such that $g(x) = g(f(x))$. - -REPLY [8 votes]: On the other hand, it is known to be true if f is an involution, i.e. f(f(x))=x. This was proved in a paper by C.T. Yang in Annals of Mathematics in 1954.<|endoftext|> -TITLE: When is a ring the ring of adeles of some global field -QUESTION [6 upvotes]: Given a global field $F$, we can construct the ring of adeles. Given a general locally compact ring $R$, when is it isomorphic to the ring of adeles of some global field $F$ and how can I find $F$ in $R$? - -REPLY [13 votes]: Iwasawa gave a characterisation, assuming you are given a subfield F, discrete and such that the quotient is compact. The other conditions are R a semisimple locally compact commutative topological ring with 1 (shared with F). Then R is the adele ring of the global field F. -Edit: I believe it is known that you can't get F from knowledge of R alone. I don't remember details or a reference, but it is something like the fact that the Dedekind zeta function doesn't determine the number field? In other words the ramification degrees e and residue class extension degrees f can be known for each prime, and this will tell you the adele ring R as a restricted product of local fields. But not the field F. Given R, there may be more than one candidate field it contains.<|endoftext|> -TITLE: Property (T) for pseudogroups -QUESTION [9 upvotes]: Let $H$ be a Hilbert space, $S(H)$ be the inverse semigroup (pseudogroup) of linear maps between (closed) subspaces of $H$ preserving the dot product (the operation is composition of partial maps). Now consider an arbitrary inverse semigroup $A$ and all possible homomorphisms $A\to S(H)$. An inverse semigroup, by definition, is a semigroup with unary operation $^{-1}$ such that $(a^{-1})^{-1}=a, aa^{-1}a=a, aa^{-1} bb^{-1}=bb^{-1} aa^{-1}$ (by Wagner's theorem, similar to the Cayley theorem for groups) these are precisely the semigroups of partial bijections between subsets of a set under composition). - -Is it possible to define property (T) for inverse semigroups using these representations? -Let $I_n(\mathbb Z)$ be the set of restrictions of all operators from $SL_n(\mathbb Z)$ onto all subspaces of $\mathbb{R}^n$, $n\ge 3$. It is an inverse semigroup. Does it have property (T)? - -Note that amenability of inverse semigroups (pseudogroups) has been considered, for example, here. -Update Since there is a confusion about what partial maps are, here is a formal definition. A partial map $X\to X$ is a map from a subset $Y$ of $X$, called domain, to another subset $Z\subseteq X$, called range. If $X$ is a metric space (say, a Hilbert space), a partial isometry is an isometry between subspaces $Y$ and $Z$. The composition of a partial map $f: X\to X$ and another partial map $g: X\to X$ is a partial map $fg$ defined of all $x$ such that $x$ is in the domain of $g$ and $g(x)$ is in the domain of $f$, $fg(x)=f(g(x))$. If $f:Y\to Z$ is a (partial) bijection $X\to X$, then the map $f^{-1}f$ is the identity map on the domain of $f$, and it is an idempotent (obviously). If $f$ is an idempotent partial bijection with damain (=range) $Y$, and $g$ is an idempotent with domain (=range) $Z$, then $fg=gf$ is the idempotent with domain (=range) $Y\cap Z$, so idempotents commute. It is obvious that any set of partial bijections closed under products and taking $^{-1}$ is an inverse semigroup. The question is about representations of inverse semigroups into the inverse semigroup of partial (!) bijective unitary (=preserving the dot product) operators of a Hilbert space. - -REPLY [6 votes]: After looking at the paper of Ceccherini-Silberstein, Grigorchuk, and de la Harpe linked to in the question I think I have a better idea of what is being asked, so let me record some further thoughts here in a separate answer. -Given a set $X$ and a bijection $\gamma : S \to T$ between subsets of $X$ let me denote by $s(\gamma) = S$ (the support) and $r(\gamma) = T$ (the range). As in the above paper let me define a (standard) pseudogroup $\mathcal G$ of transformations on $X$ to be a set of bijections $\gamma : S \to T$ between subsets $S, T \subset X$ such that: -(i) ${\rm id}_{|S} \in \mathcal G$, for any subset $S \subset X$. -(ii) If $\gamma \in \mathcal G$ then $\gamma^{-1} \in \mathcal G$. -(iii) If $\gamma, \delta \in \mathcal G$ then $\delta \circ \gamma \in \mathcal G$, where $\delta \circ \gamma$ is understood to have domain $\gamma^{-1}(r(\gamma) \cap s(\delta))$. -(iv) If $\gamma: S \to T$ is a bijection of subsets $S, T \subset X$ and $S = \sqcup_{1 \leq j \leq n} S_j$ is a finite partition with $\gamma_{|S_j} \in \mathcal G$ for all $1 \leq j \leq n$, then $\gamma \in \mathcal G$. -I will only consider the case when $X$ a discrete space with counting measure, although much of this should work in the setting where $X$ is is a standard Borel space with a (possibly infinite) measure $\lambda$ on $X$ such that $\lambda(s(\gamma)) = \lambda(r(\gamma))$ for all $\gamma \in \mathcal G$. -Let me make now a "quantum" definition. A pseudogroup $\mathcal H$ of quantum transformations on a Hilbert space $K$ is a set of partial isometries on $K$ such that: -(i) $1 = {\rm id}_{K} \in \mathcal H$, and the von Neumann algebra generated by the set of projections in $\mathcal H$ contains a maximal abelian self-adjoint subalgebra (MASA) of $\mathcal B(K)$. -(ii) If $v \in \mathcal H$ then $v^* \in \mathcal H$. -(iii) If $v, w \in \mathcal H$ then $w \cdot v \in \mathcal H$ where $w \cdot v$ is the partial isometry $w( w^*w \wedge vv^* ) v$, e.g., if $v$ is a partial isometry from $K_1$ to $K_2$ and $w$ is a partial isometry from $K_3$ to $K_4$ then $w \cdot v$ is a partial isometry from $v^*( K_2 \cap K_3 )$ to $w( K_2 \cap K_3)$. -(iv) If $v \in \mathcal B(K)$ is a partial isometry and $v^*v = \Sigma_{1 \leq j \leq n} p_j$ is a finite partition of projections with $v p_j \in \mathcal H$ for all $1 \leq j \leq n$, then $v \in \mathcal H$. -A nice example is the set of all partial isometries $S(K)$ described by Mark above. -A pseudogroup of transformations on a set $X$ has a natural representation (which I will call the regular representation) as a pseudogroup of quantum transformations on $\ell^2X$, by viewing a bijection of subsets $\gamma: S \to T$ as the partial isometry $v_\gamma : \ell^2S \to \ell^2T$ given by $v_\gamma(\xi) = \xi \circ \gamma^{-1}$. -Conversely, if $\mathcal H$ is a pseudogroup of quantum transformations on $K$ such that the set of projections in $\mathcal H$ generates a purely atomic MASA then by letting $X$ be the set of rank one projections in $\mathcal H$, we have a natural identification $K = \ell^2X$ and we have that each $v \in \mathcal H$ induces a bijection $\gamma : S \to T$ of subsets of $X$ such that $v = v_\gamma$. -A homomorphism between (quantum) pseudogroups $\mathcal G$ and $\mathcal H$ is a map $\pi: \mathcal G \to \mathcal H$ which preserves the structures $(i)-(iv)$. If both $\mathcal G$ and $\mathcal H$ are pseudogroups of transformations on $X$ and $Y$ then homomorphisms are somewhat rigid, if we restrict to characteristic functions on points we obtain a map $f: X \to Y$ and we then can check that $\pi(\gamma) \circ f = f \circ \gamma$, for all $\gamma \in \mathcal G$. -Given a homomorphism between standard pseudogroups $\mathcal G$ and $\mathcal H$ we obtain a representation of $\mathcal G$ by composition with the regular representation for $\mathcal H$. One could hope that by viewing a pseudogroup of transformatins $\mathcal G$ as a pseudogroup of quantum transformations then maybe the homomorphism (and hence also the representation) structure would be richer. This however does not give a larger class. -Proposition. -Let $\mathcal G$ be a pseudogroup of transformations on a set $X$. If $\pi: \mathcal G \to S(K)$ is a representation then $\pi(2^X)$ generates an abelian von neumann subalgebra of $\mathcal B(K)$, hence every representation of $\mathcal G$ is given by composing a homomorphism into a standard pseudogroup with the regular representation. -Proof. Since $\pi$ preserves (ii) we have that $\pi(1_S 1_T) = \pi(1_S) \wedge \pi(1_T)$, for all $S, T \subset X$. Since $\pi(1_X) = 1$ and since $\pi$ preserves (iv) we have that $\pi( 1_X - 1_S ) = 1 - \pi(1_S)$ for all $S \subset X$. Hence by again using the fact that $\pi$ preserves (iv) we have that $\pi( 1_{S \cup T}) = \pi( 1_S ) \vee \pi( 1_T)$ and -$$ -\pi(1_S) - \pi(1_S) \wedge \pi(1_T) = \pi(1_S - 1_{S \cap T}) -$$ -$$ -= \pi(1_{S \cup T} - 1_T) = \pi(1_S) \vee \pi(1_T) - \pi(1_T). -$$ -A standard fact from operator algebras then shows that $\pi(1_T)$ and $\pi(1_S)$ commute. Indeed, the left hand side $\pi(1_S) - \pi(1_S) \wedge \pi(1_T)$ is the projection onto the closure of the range of $\pi(1_S)(1 - \pi(1_T))$ and satisfies -$$ -\pi(1_S) \wedge (1 - \pi(1_T)) \leq \pi(1_S) - \pi(1_S) \wedge \pi(1_T) \leq \pi(1_S), -$$ -while the right hand side $\pi(1_S) \vee \pi(1_T) - \pi(1_T)$ is the projection onto the closure of the range of $(1 - \pi(1_T))\pi(1_S)$ and satisfies -$$ -\pi(1_S) \wedge (1 - \pi(1_T)) \leq \pi(1_S) \vee \pi(1_T) - \pi(1_T) \leq 1 - \pi(1_T). -$$ -Hence both sides equal $P = \pi(1_S) \wedge (1 - \pi(1_T))$ and we have -$$ -\pi(1_S)(1 - \pi(1_T)) = P \pi(1_S)(1 - \pi(1_T)) -$$ -$$ -= P = P (1 - \pi(1_T))\pi(1_S) = (1 - \pi(1_T))\pi(1_S), -$$ -which shows that $\pi(1_T)$ and $\pi(1_S)$ commute. Since $S$ and $T$ were arbitrary this gives the result. $\square$ -One could certainly define property (T) for pseudogroups of transformations in this setting by requiring that any representation which almost contains invariant vectors must contain a non-trivial invariant vectors. But given the above proposition it is equivalent to say that $\mathcal G$ has property (T) if and only if any quotient of $\mathcal G$ which is amenable must be finite. (Recall that a pseudogroup $\mathcal H$ of transformations on $X$ is amenable if there is a state $\varphi$ on $\ell^\infty X$, such that $\varphi(\gamma \gamma^{-1}) = \varphi(\gamma^{-1} \gamma)$, for all $\gamma \in \mathcal H$). -This definition seems like an interesting property, but it lacks a bit in functoriality. For instance, if $\Gamma$ is a countable group and we consider the pseudogroup $\mathcal G(\Gamma)$ of all maps on subsets of $\Gamma$ which arise from the action of $\Gamma$ on itself by left multiplication. Then I believe that the property that $\mathcal G(\Gamma)$ has no quotient onto an infinite amenable pseudogroup, can be restated in terms of group actions as: Every transitive action of $\Gamma$ on an infinite set $Y$ has no invariant mean. -If $\Gamma$ has property (T) then this condition is satisfied, but this condition will also be satisfied for groups which do not have property (T), although it seems to be non-trivial to show this. If $\Gamma$ is an infinite product of infinite simple property (T) groups then this should be an example. This condition is also satisfied for Tarski monsters (are there Tarski monsters which do not have property (T)?). -Coming to question 2, $I_n(\mathbb Z)$ in this setting is not a pseudogroup of transformations on a set, but rather a pseudogroup of quantum transformations in $M_n(\mathbb R) \subset M_n(\mathbb C)$. Finite dimensional algebras correspond to finite sets in the standard setting and so I think that $I_n(\mathbb Z)$ in this setting should be considered as "finite", and hence will have property (T) but for trivial reasons. -For instance, the definition of amenability for a pseudogroup $\mathcal H$ of quantum transformations on $H$ should be that there exists a state $\phi$ on $\mathcal B(K)$ such that $\phi(v^*v) = \phi(vv^*)$ for all $v \in \mathcal H$. This definition is consistent with the definition in the standard setting. -There may be another perspective in which $I_n(\mathbb Z)$ can be seen to be "rigid", but I'm not sure what it would be. -One could also fix a dimension $d$ and consider the action of $SL_n(\mathbb Z)$ on the Grassmannian $Gr(d, \mathbb R^n)$ and ask if the generated pseudogroup of transformations on $Gr(d, \mathbb R^n)$ does not have any non-trivial amenable quotients ($d \not= 0, n$). (This is no longer the discrete setting.) Using the property (T) of $SL_n(\mathbb Z)$ it should be enough to check the following property (perhaps this is well known or is discussed in the paper of Popa and Vaes that I referred to before, I am not sure): If $Y$ is a non-finite compact Hausdorff space which has a continous action of $SL_n(\mathbb Z)$ and such that there is a continuous $SL_n(\mathbb Z)$-invariant surjection $\pi: Gr(d, \mathbb R^n) \to Y$, can $Y$ have a $SL_n(\mathbb Z)$-invariant probability measure?<|endoftext|> -TITLE: Non-normal domain with algebraically closed fraction field -QUESTION [7 upvotes]: I am looking for an integral domain $A$ with the following properties: - -$A$ is not integrally closed -$A$ has a quotient field $K$ that is algebraically closed and that has characteristic 0 -There is an integral element $x\in K$ (over $A$) such that $A[x]$ is integrally closed. - -Can someone help to tell me if the above is even possible? -Edit: Lubin easily gave me an example. Now I want to consider the case when I replace the condition 2. by: -2'. $A$ has a quotient field $K$ that is real closed. - -REPLY [11 votes]: Try this: Let $B_0$ be the ring of real algebraic integers, and let $B=B_0[1/2]$, so the ring of real algebraic numbers integral except possibly at $2$. But $B[i]$ is equal to the ring of algebraic numbers integral except possibly at $2$, and this is integrally closed. And so we take $A=B[3i]$, not integrally closed, and of course the fraction field is algebraically closed.<|endoftext|> -TITLE: Grothendieck group of vector bundles -QUESTION [5 upvotes]: Let $X$ be a smooth hypersurface in $\mathbb{P}^n$ and let $\mathfrak{P}$ be the completion of $\mathbb{P}^n$ along $X$. -Let $\mathcal{R}$ denote the category of reflexive sheaves on $\mathbb{P}^n$ which are bundles outside a finite set of points in $X$ (the finite set is not fixed, it could/will be different for different sheaves). The motivation for this is Grothendieck's effective Lefschetz condition. -Let $\mathcal{V}$ denote the category of (formal) locally free sheaves on $\mathfrak{P}$ and consider the functor -$$F: \mathcal{R} \to \mathcal{V}.$$ -Then by Grothendieck's Lefschetz conditions, this functor is an equivalence of categories. This therefore implies that -$$K_0(\mathbb{P}^n) \cong K_0(\mathfrak{P}).$$ -On the other hand, the subgroup $CH^{n}(\mathbb{P}^n) \subset K_0(\mathbb{P}^n)$ maps to zero in $K_0(\mathfrak{P})$. This can be seen in the following way: -take any point $x \in \mathbb{P}^n \setminus X$ and consider the resolution of the skyscraper sheaf $k_x$ -$$0 \to F_{\bullet} \to k_x.$$ -On restricting this resolution to $\mathfrak{P}$ (restrict to any thickening $X_m$ otherwise), we get an exact sequence -$$ 0 \to F_{\bullet}\otimes O_{\mathfrak{P}} \to 0.$$ -Therefore we see that $\sum(-1)^iF_i \mapsto 0$ under the map $K_0(\mathbb{P}^n) \to K_0(\mathfrak{P})$. -Clearly I am making some (stupid) mistake. I would be glad if someone points out what it is. In particular which of the arguments is wrong. -Thanks! - -REPLY [4 votes]: Dear all, -First, I apologize for writing this as an "answer" -- it is just a comment on ABayer's answer (but I don't have permission to comment on answers). ABayer's point is right on: the kernels and cokernels in the standard resolution of a skyscraper sheaf of a point are not reflexive. So you cannot break things up into short exact sequences. One case where you do get a short exact sequence is for P^2. But Grothendieck rules out the equivalence in this case (because he requires depth at least 3 or 4, can't remember which at the moment). -Best regards, -Jason<|endoftext|> -TITLE: Can the partial sums of a series be uniformly distributed modulo 1? -QUESTION [5 upvotes]: Earlier I asked in Is a sequence of the following type uniformly distributed modulo 1? whether the partial sums of the harmonic series is uniformly distributed modulo 1. Here I ask for necessary and sufficient conditions for the partial sums of a series to be uniformly distributed modulo 1 (look at the original question for the definition of uniformly distributed modulo 1). -Suppose $(a_n)_{n=1}^\infty$ is a sequence of positive numbers such that the sum of $a_n$'s diverges. Define $b_N = \displaystyle \sum_{n=1}^N a_n$. Under what conditions imposed on $a_n$ (such as growth rate) can we say that $(b_N)_{N=1}^\infty$ is uniformly distributed modulo 1? - -REPLY [5 votes]: Every sequence of reals that is uniformly distributed modulo 1 arises as the partial sums of a divergent series of positive terms, so you are really asking for characterizations of sequences that are uniformly distributed modulo 1. Weyl’s criterion seems to fit the bill. - -REPLY [4 votes]: I don't think any kind of growth rate condition will do it for you by itself. (You could just tweak the $b_n$'s without messing up the growth rate to avoid some set). -You might look at a very interesting paper of Boshernitzan ("Uniform distribution and Hardy fields). He gives beautiful necessary and sufficient conditions on the $(a_n)$ for uniform distribution mod 1 provided that the $(a_n)$ look like $f(n)$ for polynomially-growing functions in a reasonable class (a "Hardy field"). -One such Hardy field is the collection of all functions that can be obtained starting from $x$, allowing yourself all real constants, allowing addition, multiplication, exponentiation, logarithms etc. (so that for example it contains the function $(x^{2.76}+15x\log x)/(x+e^{-1.4\sqrt x})$). The theorem in that paper shows that the sequence $f(n)$ is uniformly distributed modulo 1. -ADDED: (in response to Aaron Meyerovitch's question). He asks Can you tweak $b_n=\sqrt n$ to have $b_n′$ increasing with $\sqrt{n−1} < b_n′ < \sqrt{n+1}$ and avoid a uniform distribution? Answer: no. More generally if $(b_n)$ is uniformly distributed and $b_n'=b_n+o(1)$, then $b_n'$ is also uniformly distributed. For a proof, use Weyl's criterion and look at the difference of the expressions you obtain using the $b_n$'s and the $b_n'$s. I don't really see this suggested restriction as being in the spirit of an honest growth condition on the $b_n$'s FWIW. - -REPLY [2 votes]: There are many theorems in Kuipers and Niederreiter to the effect that if $b_n$ has a certain growth rate then it is uniformly distributed modulo one. Presumably you could turn those into results about $a_n$. -EDIT: Let me elaborate on this. Corollary 2.1 is Fejer's Theorem: Let $f(x)$ be defined for $x\ge1$ and differentiable for $x\ge x_0$. If $f'(x)$ tends monotonically to $0$ as $x\to\infty$ and if $\lim_{x\to\infty}x|f'(x)|=\infty$ then the sequence $f(1),f(2),\dots$ is uniformly distributed modulo one. -The authors point out that this implies the following sequences are u.d. mod 1; -$\alpha n^{\sigma}(\log n)^{\tau}$ with $\alpha\ne0$, $0\lt\sigma\lt1$, $\tau$ arbitrary; -$\alpha(\log n)^{\tau}$ with $\alpha\ne0$ and $\tau\gt1$; -$\alpha n(\log n)^{\tau}$ with $\alpha\ne0$ and $\tau<0$. -Later in the book we get other general theorems which imply u.d. mod 1 for -$\alpha n(\log n)^{\tau}$, $\alpha\ne0$, $\tau\gt0$; -$n\log\log n$; -$\alpha n^{\sigma}$, $\alpha\ne0$, $\sigma\gt0$, $\sigma$ not an integer; -$\alpha n^k(\log n)^{\tau}$ with $\alpha\ne0$, $k$ a positive integer, $\tau\lt0$; same is true for $\tau\gt1$; -and there is more. Combining these with the fact that if $u_n$ is u.d. mod 1 and $\lim_{N\to\infty}N^{-1}\sum_{n=1}^Nx_n=0$ then $u_n+x_n$ is u.d. mod 1 should give you the wiggle room to get some results on growth rates of $a_n$ implying u.d. mod 1.<|endoftext|> -TITLE: Non-oriented version of Poincaré duality -QUESTION [6 upvotes]: Hi. Is there a result in the spirit of Poincaré duality but for non-oriented manifolds? Thanks. - -REPLY [8 votes]: A high level purely homotopical answer that does not require explicit consideration -of homology or cohomology is given in Theorem 19.1.5 of Parametrized Homotopy Theory, -by Johann Sigurdsson and myself. That result works equivariantly, where orientation -is not as well understood as one would like. Nonequivariantly, the result gives a -description of the spectrum $k\wedge M_+$ for any spectrum $k$ and smooth closed -manifold $M$ as a function spectrum defined in terms of parametrized spectra. That -may sound daunting, but it is really very natural. -Here $k$ doesn't even have to be a ring spectrum.<|endoftext|> -TITLE: Functoriality of the center of a category -QUESTION [11 upvotes]: The center of a category $C$ is defined to be $\text{Z}(C) := \text{End}(1_C)$; here $1_C$ is the identity functor $C \to C$. See this question for an important application of the center. Now this n-lab entry states that $\text{Z}$ is functorial with respect to equivalences. But I think that we have to be careful here: -If $F: C \to D$ is an equivalence of categories, then we have to choose an inverse equivalence $G : D \to C$ and an isomorphism $e : 1_D \cong FG$ in order to define $\text{Z}(F) : \text{Z}(C) \to \text{Z}(D)$, namely by $s \mapsto e^{-1}(FsG) e$. -Thus in order to get a functor, we have to take the following category: Objects are categories, and a morphism $C \to D$ is a triple $(F,G,e)$, consisting of functors $F : C \to D, G : D \to C$ and an isomorphism $e : 1_D \cong FG$. It is clear how to define the composition and the identity. Then $\text{Z}$ is a functor on this category to the category of commutative monoids. -a) Is this correct so far? Has this already written down somewhere in the literature? -I'm wondering myself why this is not remarked in the context of the Reconstruction Theorem of Rosenberg, since there we have to use $C \cong D \Rightarrow \text{Z}(C) \cong \text{Z}(D)$ in order to reconstruct the structure sheaf. -b) Are my remarks superfluous since category theorists usually regard an equivalence not as a mere functor, but rather as an adjunction, whose functors are equivalences? Namely, this data includes the data required to define the functor $\text{Z}$ above. -By the way, the decategorified version of this is the center of a monoid (especially of a group). If $f : M \to N$ is a monoid isomorphism, then $f^{-1}$ is unique and $f f^{-1}$ is equal to $1_N$, thus we don't have to make any choices. - -REPLY [7 votes]: It should be mentined that any localisation functor $F: C \to D$ induces a morphism $Z(F): Z(C) \to Z(D)$. A reference for this is Gabriel's thesis "Des categories abeliennes" (p. 446).<|endoftext|> -TITLE: Why is the cuspidal spectrum discrete? -QUESTION [6 upvotes]: I have a short question concerning the spectral theory of automorphic forms. What is the main property of the unipotent group $N$, which consist of matrices in the form \begin{pmatrix} 1 & t \\ 0 &1 \end{pmatrix} in $GL(2)$, which provides that the cuspidal spectrum decomposes discretely? -The background: Consider $G = GL_2(\mathbb{A})$, $\Gamma =GL_2(\mathbb{Q})$ and $Z$ the centrum of $G$, then we decompose -$$Ind_{ \Gamma Z}^G 1 = \pi \oplus \pi^\bot,$$ -where $\pi = Ind_{N\Gamma Z}^G$. The projection onto $\pi$ is given in terms of the integral -$$ P : \phi(g) \mapsto \int\limits_{N(\mathbb{A})} \phi(ng) d g.$$ -Now given a bounded function $f$ on $\Gamma \backslash G / \Gamma$, with suitable decay properties, we can define the operator -$$Tf : \phi \mapsto f * \phi.$$ -Why is $(1-P)Tf(1-P)$ a Hilbert Schmidt operator? -My guess is that an answer should include the Iwasawa decomposition and an according decomposition of the integral operator. - -REPLY [6 votes]: This result is due to Gelfand, Graev, Piatetski-Shapiro and has a short proof. I suggest you read Bump: Automorphic forms and representations, Prop. 3.2.3, pp. 285-289. -Let me switch to $G=\mathrm{PGL}_2(\mathbb{R})$ and $\Gamma=\mathrm{PGL}_2(\mathbb{Z})$ for simplicity. -The idea is to consider right convolutions $\rho(\phi)$ by smooth and compactly supported functions $\phi:G\to\mathbb{C}$, which act on the space of automorphic functions $L^2(\Gamma\backslash G)$. This family of operators is sufficiently rich to distinguish automorphic functions: if $f\neq g$ then there is $\phi$ such that $\rho(\phi)f\neq\rho(\phi)g$. Thinking of $\rho(\phi)$ as an operator on $(\Gamma\cap N)\backslash G$, its kernel is given by -$$K(g,h)=\sum_{\gamma\in\Gamma\cap N}\phi(g^{-1}\gamma h).$$ -As $N\cong\mathbb{R}$ and $\Gamma\cap N\cong\mathbb{Z}$, the sum can be analyzed by Poisson summation, using the characters of $N$. It turns out that by subtracting the term corresponding to the trivial character, i.e. -$$K_0(g,h)=\int_{N}\phi(g^{-1} n h) \ dn,$$ -the rest of the kernel decays rapidly at infinity, hence defines a compact operator. However, on the cuspidal space the operator with kernel $K_0(g,h)$ acts by zero, hence the operator $\rho(\phi)$ is compact on the cuspidal space. In particular, $\rho(\phi)$ has an eigenvalue with finite multiplicity on any right $G$-invariant subspace of cuspidal functions, and from here the result follows by a standard linear algebra argument. -EDIT: I fixed some inaccuracies.<|endoftext|> -TITLE: Splitting an evaluated complete graph -QUESTION [7 upvotes]: Suppose we are given a positive integer $k$. Let $K_k$ denote the complete (undirected and simple) graph with vertices $1, 2, \dots, k$. The set of edges of $K_k$ is the set $E_k = \{ \{ x,y \} \mid \ 1 \leq x < y \leq k\}$. -A valuation of $K_k$ is a function $\omega: E_k \rightarrow \mathbb Z$. A splitting of $K_k$ is a partition $\{ 1, 2, \dots, k \} = A \cup B$ of the vertices of $K_k$ into two nonempty sets $A$ and $B$. -Given a valuation $\omega$ of $K_k$ and a positive integer $n$, we call a splitting $A \cup B$ of $K_k$ $n$-valid, if the number $$\sum_{(x,y) \in A \times B} \omega(\{ x,y \})$$ is divisible by $n$. -Using Ramsey's theorem, one can prove that for every positive integer $n$, there exists a positive integer $k$ such that the following condition holds: -For any valuation $\omega$ of $K_k$, there is a splitting of $K_k$ that is $n$-valid. -If there exists at least one, there has to exist a smallest $k$ satisfying the above condition which we denote by $\eta(n)$. -Using the Combinatorial Nullstellensatz from N. Alon, one can show that $\eta(p) = 2p$ for odd primes $p$. -I now want to know if $\eta(n) = 2n$ is true for every integer $n \geq 3$. - -REPLY [2 votes]: The answer is NO, at least for $n=4$. I show here that $\eta(4) \leq 7$. So -let $\omega$ be a valuation on $K_7$ ; we will show that it is $4$-valid. -First, we need some notation : for $A \cup B$ a nontrivial partition -of $V=\lbrace 1,2,3, \ldots ,7\rbrace$, let -$$ -s(A,B)=\\sum_{(x,y) \in A \times B} \omega(\{ x,y \}), \ f(A)=s(A,V\setminus A). -$$ -We can now write some useful relations : -$$ -f( i,j)=f(i)+f(j)-2\omega(i,j), -f(i,j,k)=f(i)+f(j)+f(k)-2(\omega(i,j)+\omega(i,k)+\omega(j,k)) -$$ -(where, for simplicity, we write $f(i)$ instead of $f(\lbrace i \rbrace)$, -$f(i,j)$ instead of $f(\lbrace i,j \rbrace)$, etc). -Assume, by contradiction, that $\omega$ is not $n$-valid. Then the values -of $f$ are all $1,2$ or $3$ modulo $4$. For convencience's sake, we now -take $\omega$ and $f$ to take values in $\frac{\mathbb Z}{4\mathbb Z}$. -By easy double-counting, the sum $f(1)+f(2)+f(3)+ \ldots +f(7)$ is even -(this is twice the sum of all the values of $\omega$), so that at least one -$f(i)$ is even. Then $f(i)=2$. So the set $X=\lbrace i\in [1...7] | f(i)=2 \rbrace$ -is not empty. -For any $i,j$ in $X$, we have $f(i,j)=4-2\omega(i,j)$, hence -$\omega(i,j)=1$ or $3$ and $2\omega(i,j)=2$ in both cases. If $X$ contained -more than two elements, we could find $i \lt j \lt k$ in $X$ and compute -$f(i,j,k)=f(i)+f(j)+f(k)-2(\omega(i,j)+\omega(i,k)+\omega(j,k))=2+2+2-2-2-2=0$, -a contradiction. So $|X| \leq 2$. -Similarly, we show that the set $Y=\lbrace i\in [1...7] | f(i)=1 \rbrace$ -contains at most two elements. Reasoning as before, we have -$2\omega(i,j)=0$ for any $i \lt j$ in $Y$. Recall that some $x$ in -$[1...7]$ satisfies $f(x)=2$. We have -$f(x,i,j)=-2(\omega(x,i)+\omega(x,j))$. By the pigeon-hole principle, -if we had $|Y| \gt 2$ we cound find $i \lt j$ in $Y$ such that -$\omega(x,i)=\omega(x,j)$, hence $f(x,i,j)=0$, a contradiction. So $|Y| \leq 2$. -Using the symmetry $x \mapsto -x$ in $\frac{\mathbb Z}{4\mathbb Z}$, -we see that $|Y'| \leq 2$ where $Y'=\lbrace i\in [1...7] | f(i)=3 \rbrace$. -So, we have shown that on $[1..7]$, $f$ takes each of its three values -($1,2$ or $3$ mod $4$) at most twice. This contradicts the pigeonhole principle -and finishes the proof.<|endoftext|> -TITLE: Example of a finitely generated infinite group with a non-inner automorphism of finite order -QUESTION [17 upvotes]: Does there exist an infinite finitely generated group $G$ together with a finite group $B$ of automorphisms of $G$ such that - -The non-identity elements of $B$ are not inner automorphisms of $G$; -For every element $g \in G, g\neq 1,$ the finite set $g^B$ is a generating set for $G$? - -It looks like a Tarski Monster $p$-group $T_p$ (an infinite group in which every non-identity element has order $p$) might be a contender for $G$. If $T_p$ has an automorphism of finite order coprime to $p$, then this would work perfectly. Unfortunately I'm not sure that it does. - -REPLY [11 votes]: I am pretty sure this example can be constructed. It might be not easy, though, since you need to have finite order automorphisms of the Tarski monsters without non-trivial fixed points. I do not remember if that has been done for Tarski monsters. For other monsters it was done, I think, by Ashot Minasyan in Groups with finitely many conjugacy classes and their automorphisms. Ashot is sometimes on MO, perhaps he can give more details. If not, you can ask him directly. -I talked to Olshanskii and Osin today about this problem. They confirmed the answer. Moreover, they reminded me that there exists a paper Obraztsov, Viatcheslav N. -Embedding into groups with well-described lattices of subgroups. (English summary) -Bull. Austral. Math. Soc. 54 (1996), no. 2, 221–240 where he constructs Tarski monsters with arbitrary prescribed outer automorphism group. It is not difficult to deduce from the construction that if the prescribed outer automorphism group is, say, of order 3, then it does not have non-trivial fixed points which gives the property from the question. -Another, more straightforward, way to construct an example is the following. Fix the free group $F_2$ and its obvious automorphism $\alpha$ of order 2 that switches the generators. Now consider the procedure of creating the Tarski monster quotient of $F$ described in the book by Olshanskii "Geometry of defining relations...". At each step together with a new relation $r_i=1$, impose the relation $\alpha(r_i)=1$. It is not difficult to see that $\alpha$ will be an outer automorphism of the resulting Tarski monster without non-trivial fixed points.<|endoftext|> -TITLE: Background to learn about manifolds -QUESTION [5 upvotes]: Greetings -As a necessity to go forward with physics, I find myself in the need to learn about manifolds. Being an engineering student, I don't have the chance to study topology in all its glory. -So, can any one point me to the right direction, i mean things I must know first and the materials to fill the gaps before approaching manifolds? -Thanks - -REPLY [2 votes]: Arkadiusz Jadczyk very much recommended M.Fecko's textbook ("Differential geometry and Lie groups for physicists").<|endoftext|> -TITLE: Reverse Engineering to find deformation problem (from cohomology groups)? -QUESTION [9 upvotes]: One of my favorite explanation of the cohomology groups of low degree is that they arise as the automorphism group, tangent space and obstruction space (or where the obstruction lives) of a certain deformation problem. -My question is, is it possible to reverse the process of going from deformation problem to cohomology groups? Say I have H^0, H^1 and H^2 of a certain sheaf, is it possible to find a deformation problem such that these groups control, i.e. such that they arises as the Automorphism-Tangent-Obstruction of that deformation problem? -To be more concrete, say we have a smooth scheme X over k, and our sheaf is the tangent sheaf T_X of X, how do I "find" the problem of "smooth deformation of X"? -(I know I have been pretty vague. I know how to interpret cohomology classes as gerbes/torsors, but I'm not quite satisfied (I like these interpretations though). I want a way to find deformation problem that produces gerbes as obstructions...) -Quote from wikipedia: Reverse engineering is the process of discovering the technological principles of a human made device, object or system through analysis of its structure, function and operation. [I hope my usage of this phrase is correct.] - -REPLY [3 votes]: In this generality, I would say that this is not possible: the same cohomology can be responsible for controlling quite different deformations problems. -Just an example: in formal deformation quantization you have the dgla of multivector fields "controlling" both, the classical deformation problem of deforming a Poisson structure, and second, by the (highly non-trivial) formality theorem, the deformation problem of quantising the algebra of smooth functions. Even the moduli spaces of the deformation problems are the same... -So two deformation problem of quite different nature arise from the Schouten algebra. -So little cousing of this is perhaps easier to understand: on a symplectic manifold the (abelian) dgla of differential forms controlls the deformation theory of the symplectic form up to formal diffeos, the (formal series in the) second deRham cohomology yields the parametrization of the moduli space. But also the star products quantizing the symplectic form are controlled by this: the equivalence classes of formal star products are again formal series in the second deRham cohomology. Of course, this is just a special case of the above Kontsevich classification, but there are much simpler proofs in this case.<|endoftext|> -TITLE: Sum of squares of determinants of principal minors -QUESTION [5 upvotes]: I am interested in computing the sum of squares of determinants of principal minors. Let $A$ be an $n\times n$ positive semidefinite matrix and $A_S$ be a principal minor of $A$ indexed by the set $S \subseteq \{1,\ldots,n\}$. The classical result (without squares) is: -$\sum_{S \subseteq \{1,\ldots,n\}} \det(A_S) = \det(A+I)$ -Are there any results on computing -$\sum_{S \subseteq \{1,\ldots,n\}} \det^2(A_S)$ -or any other powers? - -REPLY [7 votes]: The identity you mention does generalize to sums of powers, but I don't know if it can give you anything computationally efficient. Given a set $X\subset \mathbb R$, let $D(X^n)$ denote all $n\times n$ diagonal matrices with diagonal elements from $X$. Then if you take $X_k=\{1,\omega,\cdots,\omega^{k-1}\}$ the $k$-th roots of unity, the following holds -$$\sum_{S\subset \{1,2,\cdots,n\}}\det(A_S)^k=\frac{1}{k^n}\sum_{M\in D(X^n)}\det(A+M)^k$$ -the proof is basically the same as the one for the case $k=1$ with a few more algebraic manipulations.<|endoftext|> -TITLE: How to keep subsets disjoint? -QUESTION [14 upvotes]: Given positive integers $n$ and $k\le 2^n$, how to choose a subset $C\subset\{0,1\}^n$ of size $|C|=k$ to maximize the number of pairs $(c_1,c_2)\in C\times C$ with the supports of $c_1$ and $c_2$ disjoint (in other words, with $c_1$ and $c_2$ orthogonal)? If $k=1+n+...+\binom ns$, should one choose $C$ to be the set of all vectors with at most $s$ coordinates equal to $1$? -Some equivalent restatements: - -How to choose a family of $k$ subsets of a fixed $n$-element set to maximize the number of pairs of disjoint subsets? -How to choose a binary code of length $n$ and size $k$ to maximize the number of pairs of codewords with disjoint supports? -How to choose a simplicial complex on $n$ vertices with $k$ faces to maximize the number of pairs of disjoint faces? - -(For the last restatement observe that the optimal set $C$ is monotonic, aka "downset".) - -UPDATE -As indicated by Sergey Norin (see his answer below), this problem originates from a question of Erdos, and is considered in a 1985 paper by Alon and Frankl. However, establishing a rather strong result for $k$ "small", their paper does not give any information in the case where $k=2^{\gamma n}$ with $\gamma>1/2$. -DISCUSSION There is interesting phenomenon here which seems to be unexplored as yet. For (roughly) $k<2^{n/2+\epsilon}$, a construction from Alon-Frankl shows that the number of pairs can be $\Omega(k^2)$ for a suitable choice of $C$. However, for $k>2^{\gamma n}$ with $\gamma>1/2$, this construction does not work. Indeed, it is not difficult to show that for $k$ "large" the situation breaks in the sense that $\Omega(k^2)$ cannot give the right order of magnitude any longer. It is quite possible that in this case, the set of all vectors with small support is optimal, exhibiting a kind of threshold behavior. - -REPLY [6 votes]: As far as I know, a paper of Alon and Frankl "The Maximum Number of Disjoint Pairs -in a Family of Subsets" (available here) contains state of the art knowledge on the problem. -Briefly outlining some of its conclusions, let me mention that for a wide range of values of $k$ keeping the sets in $C$ supported on disjoint subsets of $[n]=\{1,2,\ldots,n\}$ works much better than keeping them individually small. For example, for even $n$ and $k=2^{n/2 +1}-1$, if we choose $C$ to consist of sets of smallest possible size than typical set in $C$ will have size $\Omega(n/\log{n})$ and two such sets almost surely intersect. On the other hand, if we choose $A$ to be the set of all subsets of $\{1,\ldots,n/2\}$, $B$ to be the set of all subsets $\{n/2+1, \ldots, n\}$ and $C =A \cup B$, then at least half of the pairs of sets in $C$ are disjoint. Solving a problem of Erdős, Alon and Frankl show that this example is essentially the best possible.<|endoftext|> -TITLE: Analysis and finitely generated groups -QUESTION [11 upvotes]: Dear all, this is perhaps a bit a vague question, but some references would already be very helpfull. -So let $G$ be a finitely generated group and choose some finite set of generators. This allows to define the length of a group element as usual, by the minimal number of generators needed to write it as a product of these generators. Using this length you can define all sort of analytic functions by replacing the "n" in various summation formulas by the $\mathrm{Length}(g)$ and the summation is now over the group elements $g$. To have a concrete example in mind, the "exponential series" is now e.g. -\begin{equation} - \mathrm{lexp}(z) = \sum_{g \in G} \frac{z^{\mathrm{Length}(g)}}{\mathrm{Length}(g)!}. -\end{equation} -Since for a given length there are at most exponentially many group elements, the length-exponential function is entire. For $G = \mathbb{Z}$ and $1$ as generator this reproduces the usual exponential series up to a factor $2$ as negative and positive $n \in \mathbb{Z}$ contribute with the same power of $z$. -So my question is: what is known about the analytic features of such functions (depending on the choice of generators, depending on the group itself, etc). I guess there should be some literature on the market, but I'm really not in this buisness... - -REPLY [2 votes]: One special case of groups, where one certainly gets rather quickly explicit and nontrivial expressions should be finite or affine Coxeter groups, that are finite/infinite and defined by involution generators and relations from their Dynkin diagrams - more my topic than graphs ;-) -If the Coxeter-diagram belongs to a finite/affine Lie algebra root system resp. Weyl group (which is the generic case!), these rather strong structures should give you enough informations to control the elements of the group ordered by their length. -I want to be specific in two cases (finite/infinite) I found rather quickly in the relevant literature, but are without factorial in the numerator: -For a finite Weyl-group $W$ acting as reflection group on an $n$-dimensional vectorspace the well-known Chevalley-Solomon theorem (often used "the-other-way-around") asserts, that: -$$W(z):=\sum_{g\in G} z^{Length(g)}=\prod_{\alpha\in \Delta^+}\frac{1-z^{Height(\alpha)+1}}{1-z^{Height(\alpha)}} = \prod_{k=1}^n\frac{1-z^{d_k}}{1-z}$$ -where $d_k$ are the fundamental degrees of the reflection action, i.e. the degrees of a homogenious basis of the invariant part of the polynomial ring over $V$ (having $n$ variables). The middle term is crucial for the proof (and pretty), but the product over the entire root system $\Delta$ is not helpful for our question ;-) -Example For $S_n$ (root system $A_{n-1}$) we have $d_k=k$ (each elementary symmetric polynomial), thus: -$$\sum_{g\in S_n} z^{Length(g)}=\prod_{k=1}^n\frac{1-z^{k}}{1-z}$$ -For an affine Weyl group $\tilde{W}$ i.e. with Dynkin-diagram as some finite $W$ (suppose irreducible) with one node added, Bott's theorem states that (omitting $d_k=1$-Terms): -$$\tilde{W}(z)=W(z)\prod_{k=1}^n\frac{1}{1-z^{d_k-1}}$$ -Example $\tilde{A}_n$ is derived from closing the $n$-chain $A_n$ (i.e. $S_{n+1}$) with an additional $x$. Hence is generated very similar to the symmetric group but infinite (all non-mentioned pairs elements commute!): -$$G=\langle t_1,t_2,...t_n,x\rangle\qquad (t_it_{i+1})^3=1\quad (t_1x)^3=(xt_n)^3=1$$ -Hence the length-generating function now has a pole: -$$\tilde{W}(z)=W(z)\prod_{k=2}^n\frac{1}{1-z^{k-1}}=\frac{1}{(1-z)^n}\prod_{k=2}^n\frac{1-z^k}{1-z^{k-1}}$$ -Finally I must of course mention that the beatifully exotic root systems of Nichols algebras, that would even count the lengths of Coxeter gruppoids ;-) -This was all written down without much further thought, but if there's still interest in the topic (?) I'd be happy about a further discussion! Maybe concerning factorial or something else.... -SOURCES: - -Definitely Humphreys "Reflection Groups and Coxeter groups" -the formulas also directly online e.g. the "survey" www.math.umn.edu/~reiner/Papers/SteinbergNotes.ps). -Some infinite worked-out examples e.g. in dml.ms.u-tokyo.ac.jp/PSRT/PSRT_26/PSRT_26_093-102.pdf and many more)<|endoftext|> -TITLE: Distribution of zeros of a polynomial mod. a prime -QUESTION [6 upvotes]: Let $\mathbb P$ be the set of prime numbers. -Is there a non constant polynomial $f \in \mathbb Z[X]$ such that the set $$I_f := \{ \textstyle\frac{z}{p} : z \in \mathbb Z, p \in \mathbb P, p \mid f(z) \}$$ -is dense in $\mathbb R$? -(I suppose that the following much stronger statement holds: Every polynomial $f \in \mathbb Z[X]$ which has a irreducible factor of degree at least $2$ satisfies the above condition.) - -REPLY [2 votes]: In my paper, Polynomial congruences and density, Math Mag 80 (2007) 299-302, I prove this related result: -Let $f(t)=t^eg(t)$ where $e$ is a nonnegative integer, $g$ is a polynomial of degree at least 2 with integer coefficients, and $g(0)\ne0$. Define $T_f$ by $$T_f=\lbrace r/m:\gcd(r,m)=1,{\rm\ and\ }m{\rm\ divides\ }f(r)\rbrace.$$ Then $T_f$ is dense in the reals. -The proof uses nothing you wouldn't see in an undergrad intro Number Theory course (but of course it's weaker than what you want since I need all denominators, not just primes).<|endoftext|> -TITLE: homotopy type of loop space components -QUESTION [5 upvotes]: Let $X$ be a closed manifold, and let $X^{S^1}$ denote the free loop space of $X$, that is, the set of continuous maps $S^1 \rightarrow X$. Let $Y$ denote a component of $X^{S^1}$. -What conditions ensure that $Y$ is homotopic to a finite CW complex? (by finite, I mean, the total number of cells is finite). -Edit - thanks to Somnath, sufficient conditions are certainly when $X$ is an Eilenberg-Maclane space, as then the based loop space is contractible. Are there any other different sufficient conditions? - -REPLY [8 votes]: There are some negative results about. For instance, a theorem of Sullivan and Vigué-Poirrier states that if $M$ is a closed manifold with $\pi_1(M)$ finite, and if the cohomology algebra $H^*(M;\mathbb{R})$ requires at least two generators, then the Betti numbers of $M^{S^1}$ are unbounded. The finiteness assumption implies that $M^{S^1}$ has only finitely many path components, and so at least one of them is not a finite CW-complex.<|endoftext|> -TITLE: Extensions of torsion modules -QUESTION [7 upvotes]: Given a regular local ring $R$ and an $R$-algebras $S$, which is torsion free and finitely generated (even free if needed) as an $R$-module. -Assume we have a nontrivial surjective map $f: M \rightarrow T$, where $M$ is a projective $S$-module, finitely generated and torsion free, and $T$ is a torsion module over $S$. If $N$ denotes $ker(f)$, we get an exact sequence: $0\rightarrow N\rightarrow M\rightarrow T\rightarrow 0$. -Given another torsion module $Q$, when is the induced map $f^{\*}: Hom_S(M,Q)\rightarrow Hom_S(N,Q)$ non trivial, when is it trivial? -My first idea was to use the long exact $Ext$-sequence: Since $M$ is projective we have $Ext^1_S(M,Q)=0$, thus if $f^{\*}=0$, the sequence gives an isomorphism $Hom_S(N,Q)\cong Ext^1_S(T,Q)$. -So what can be said about the groups $Ext^1_S(T,Q)$? Are they always/sometimes/never trivial? Can we compute them if we assume that one of these moudles is a simple $S$-module? -Are there other approaches to this question? - -REPLY [4 votes]: Actually, it would be easier to look at the other end of the exact sequence. Namely, your map $f^*$ is trivial implies the map $g^*: Hom_S(T,Q) \to Hom_S(M,Q)$ is an isomorphism. -Now, since $M$ is projective, the support of $Hom_S(M,Q)$ is equal to the support of $Q$. Thus we have $Supp(Hom_S(T,Q)) = Supp(Q)$, which implies $$Supp(T) \supseteq Supp(Q) \ \ (1)$$ -When $Q$ is simple, (so $Q=S/m$ where $m$ is a maximal ideal) as you alluded to in the last paragraph, then $(1)$ is also sufficient, provided that the surjection $M \to T$ is minimal when localizing at $m$, as you can easily check for yourself. -Another situation when $(1)$ also suffices is when $T,Q$ are both cyclic $S$ module and $M=S$ (you always need the map $M\to T$ to be minimal, may be that what you meant by "non-trival" surjection?) -Other than what described above, I think what you want will fail most of the times, even with $(1)$.<|endoftext|> -TITLE: an example of a semigroup with solvable word problem but unsolvable power problem -QUESTION [5 upvotes]: We say that a semigroup $S$ has solvable power problem if there is an algorithm that takes as input an element $s \in S$ and decides whether or not there exist $m,n \in \mathbb{N}$ with $m \neq n$ and $s^m=s^n$. - -Does anybody know an "easy" (like finitely presented with relatively few relations) example of a semigroup with solvable word problem but unsolvable power problem? - -I would also be interested in an example of a group with solvable word - problem but unsolvable power problem, if anybody has such an example. - -REPLY [6 votes]: The only known way to construct this example (say, in the case of groups, the case of semigroups is similar) is the following. First consider the free Abelian group $F$ with free generators $a_1,a_2,...$. Pick a recursively enumerable non-recursive set $I$ and impose relations $a_n^{m!}=1$ if $n$ is the $m$th number from $I$ (we assume that there exists a computer that lists numbers in $I$ in some order one by one). That group, call it $A$, has solvable word problem. Indeed, consider any word $w=a_{i_1}^{k_1}\ldots a_{i_s}^{k_s}.$ That word is equal to 1 in $A$ iff each $k_i$ is divisible by $m!$ such that $a_i$ is the $m$-th number in $I$. That gives restriction to $m$. So given $w$ we start the computer that lists $I$ and wait till we have the first (not in the natural order!) $k_1+...+k_s$ numbers from $I$ listed. The power problem in $A$ is not decidable of course. Since $A$ has solvable word problem, by Higman's theorem, it embeds into a finitely presented group $G$. By Clapham's theorem, we can assume that $G$ has decidable word problem. But the power (order) problem in $G$ is not decidable. -There are no examples which are given by very few defining relations. The reason is that the solvability of the order problem is not easy to achieve without Higman embedding (I do not know any such way without getting undecidable word problem also) which is very implicit.<|endoftext|> -TITLE: Lie group examples -QUESTION [38 upvotes]: I'm looking for interesting applications of Lie groups for an introductory Lie groups graduate course. In particular I'd like to hear of non-standard examples that at first sight do not seem to be related to Lie groups (so please don't suggest well-known things like Clifford algebras or triality that appear in standard Lie groups texts such as Fulton and Harris). Here are some examples of the sorts of things I'm looking for: -*The cohomology of a compact Kaehler manifold is a representation of SL2, so the Hopf manifold cannot be Kaehler. -*q-binomial coefficients are unimodal, as they are characters of representations of SL2 -*Hilbert's theorem on the finite generation of rings of invariants can be proved using invariant integration on compact Lie groups. -*Holomorphic modular forms are really highest weight vectors of discrete series representations of certain Lie groups. -*Most closed 3-manifolds are quotients of SL2(C) by discrete subgroups. -*Bessel functions cannot be expressed using elementary functions and indefinite integration. (Differential Galois theory was one of Lie's original motivations, but seems to have been eliminated from texts on Lie theory.) -*Classifying manifolds up to cobordism uses orthogonal groups. - -REPLY [6 votes]: What could be a nicer example than Milnor invariants? At first sight, these "higher linkage" invariants seem to have nothing to do with Lie groups at all; you can even define them using Habiro moves to make any connection to Lie algebras even more mysterious. Or, you can speak of the Baguenaudier puzzle, whose solution involves Habiro moves in disguise, as discussed by Przytycki and Sikora. -And yet, the natural home for Milnor invariants is the group $D(H)$, which is the kernel of the left bracketting map $L(H)\otimes H \to L(H)$, where $L(H)$ denotes the free Lie algebra over $H$, the first homology of the link complement. -The group $D(H)$ also comes up in other contexts which at first sight don't seem to have anything to do with free Lie algebras. It's related to the rational homology of the outer automorphism groups of free groups, as first observed by Kontsevich. Morita, and Conant-Vogtmann, took this idea and ran with it. The group $D(H)$ was also used by Dennis Johnson to study the relative weight filtration of the mapping class group of a surface.<|endoftext|> -TITLE: Finding citations for 'well-known' results -QUESTION [23 upvotes]: I am almost ready to submit my most recent paper, and I find myself in a problem that has already occurred multiple times in my short publishing career. In this paper, I wish to state a result which I consider 'well-known', but a skimming of all the likely textbooks and survey articles doesn't yield a nice statement that I can cite. For reference, the result in question is the following: - Let $X$ be a smooth, affine variety over $\mathbb{C}$, with coordinate ring $\mathcal{O}_X$. Then there is a natural isomorphism of $\mathcal{O}_X$-modules from the Kahler differentials $\Omega(\mathcal{O}_X)$ of $\mathcal{O}_X$ to the global 1-forms on $X$ with regular coefficients. - -This is a result whose proof I know, and is homework-level difficulty, but including the proof in my short paper would require terminology and techniques I'd rather not introduce and consume precious space. It's also not a necessary result for the paper; I am including it to justify the study of Kahler differentials to an audience which might include differential geometers. -So what does one do in this situation? The lazy solution is to include some weasel words to avoid finding a citation ("it is a straight-forward exercise to show that..."), but this seems like a dangerous policy to employ in general. However, finding a citation is proving unreasonably time-consuming, since it's not in the books I know (Hartshorne, Eisenbud, Kunz), and each new book/article I skim has its own notation and assumptions. -Also, while I'd be extremely grateful for a citation for the specific result above, my question is about what to do in this kind of situation. I'm trying not to get the answers mixed up. - -REPLY [5 votes]: I've seen "details are presented in the arXiv version of this paper" several times. The only down side I see to this is that you do need to write up the details.<|endoftext|> -TITLE: Helly theorem + Nerve -QUESTION [11 upvotes]: Consider nerve $\mathcal N$ of a finite set of convex sets in $\mathbb R^n$. -Helly theorem says that $\mathcal N$ is completely determined by its $n$-skeleton, say $\mathcal N_n$. -It seems that not all finite simplicial complexes with dimension $\le n$ can appear as $\mathcal N_n$. -(For example take 2-dimenisonal simplicial complex which is homeomorphic to $\mathbb{R}\mathrm{P}^2$, -it can not appear as $\mathcal N_3$ for finite set of convex sets in $\mathbb R^3$.) - -Is it possible to describe all finite simplicial complexes which can appear as $\mathcal N_n$? - - -The question is inspired by a short discussion here. -I am sure a lot should be known, but a quick search gave me nothing. - -REPLY [3 votes]: I really wish I knew more about general nerve complexes, so I'm afraid I don't have much to bring to your question. But information on intersection graphs is a lot easier to figure out, so I thought I'd add some remarks to what Gil wrote: -Non-planar Graphs -First, I want to explain how his remark about non-planar graphs works: say you have an edge between two vertices $u$ and $v$ of a graph $G$. Split the edge with a new vertex $e$ to get the new graph $G^\star$. Now, suppose that $G^\star$ is the intersection graph of some arrangement of convex sets in the plane, with sets $U$, $V$ and $E$ that correspond to the vertices of the same name. Then since there are edges $(u,e)$ and $(v,e)$ in $G^\star$, you can pick points $a\in U\cap E$ and $b\in V\cap E$, and the segment $[a,b]$ is of course contained in $E$. -Do this for all vertices of $G^\star$ that arose from adges in $G$. This gives you a collection of edges that must be pairwise disjoint since none of the $e$ vertices are adjacent to each other, they're only adjacent to vertices in $V(G)$. Now, contract the convex sets of type $U$ that correspond to vertices in $V(G)$: this keeps the segments disjoint (except for endpoints) and gives a picture of $G$ on the plane. Thus the convex arrangement fr $G^\star$ exists only if $G$ is planar. -Boxes and Graphs -Also, I think it's important to stress how special interval graphs are. To me, one of the best way to see this is to consider the natural generalization: instead of intervals in $\mathbb{R}$, consider boxes: cartersian products of intervals in $\mathbb{R}^d$. Then, F. S. Roberts proved (in his PhD thesis I think) that any graph can be realized as the intersection graph of a collection of $d$-boxes, and we can even take $d\leq |V(G)|/2$. (The smallest dimension you can choose for a graph $G$ is called the boxicity of $G$.) -If you'd rather look at more general convex sets, then it's been known for even longer that you can realize a graph as the intersection pattern of convex sets of $\mathbb{R}^3$ (and thus showing some sort of counterpoint to Gil Kalai's answer in dimension 2). The obvious conclusion: nerve complexes carry a lot more information than graphs; nothing new or surprising there, but these examples can help you really appreciate that. -References -Roberts, F. S. (1969), "On the boxicity and cubicity of a graph", in Tutte, W. T., Recent Progress in Combinatorics, Academic Press, pp. 301–310,<|endoftext|> -TITLE: Removable sets for harmonic functions and hardy spaces of general domains -QUESTION [8 upvotes]: Hi, -Let $\Omega$ be a domain of the complex plane. The Hardy space $H^p(\Omega)$ is defined, for $1 \leq p<\infty$, as the class of functions $f$ that are holomorphic on $\Omega$ such that $|f|^p$ has a harmonic majorant on $\Omega$, i.e. there is a function $u$ harmonic on $\Omega$ such that -$$|f(z)|^p \leq u(z) $$ -for all $z \in \Omega$. -For $p=\infty$, $H^\infty(\Omega)$ is the class of bounded holomorphic functions on $\Omega$. -I came upon the following question : -Let $E$ be a compact subset of the real line, and suppose that $E$ has zero length. Let $\Omega$ be the complement of $E$. Does $H^p(\Omega)$ consist only of the constant functions? -For $p=\infty$, the answer is yes : one can use Cauchy's formula to extend any bounded holomorphic function on $\Omega$ to a bounded holomorphic function on $\mathbb{C}$, and that function is now constant by Liouville's theorem. -What about the case $p < \infty$? Any reference would be quite useful. -Thank you, -Malik - -REPLY [6 votes]: Yes, it is true that $H^p(\Omega)$ consists only of the constant functions for all $1\le p\le\infty$. This is because all nonnegative harmonic functions $f\colon\Omega\to\mathbb{R}$ are constant. -You can prove this using the properties of Brownian motion. If $\Omega$ is a connected open subset of the plane and $B_t$ is a two dimensional Brownian motion with $B_0\in\Omega$ then either (i) $B_t\in\Omega$ for all times $t$, with probability one, or (ii) $B_t\in\mathbb{R}^2\setminus\Omega$ for some time $t$, again with probability one. Also, which case holds depends on the domain $\Omega$ but not on the initial distribution of $B$. If case (ii) holds then $\Omega$ is called a Greenian domain. As I'll show below, the $\Omega$ you are considering is not a Greenian domain, which implies that all nonnegative harmonic functions on $\Omega$ are constant (so $H^p(\Omega)$ consists of the constant functions). For this terminology, I'm going from Foundations of Modern Probability, Second Edition, Olav Kallenberg, Chapter 24. -The idea is that a twice continuously differentiable function $f\colon\Omega\to\mathbb{R}$ is harmonic if and only if $f(B_t)$ is a local martingale when run up until the first time at which $B$ exits $\Omega$. This follows from Ito's formula -$$ -f(B_t)=f(B_0)+\sum_{i=1}^2\int_0^tf_{,i}(B_s)\\,dB^i_s+\frac12\int_0^t\Delta f(B_s)\\,ds -$$ -The first two terms on the right hand side are local martingales, and the third term will be a local martingale if and only if $\Delta f=0$, so that $f$ is harmonic. Now, if $\Omega$ is not a Greenian domain and $f$ is a nonnegative harmonic function then $f(B_t)$ is defined at all times and is a nonnegative local martingale. By martingale convergence (see my blog post on martingale convergence, or any textbook on martingale theory), this implies that $f(B_t)$ converges to a limit as $t\to\infty$ with probability one. However, two dimensional Brownian motion is recurrent, so that it keeps getting arbitrarily close to any given point. Then, $f(B_t)$ can only converge for continuous $f$ if $f$ is constant. -We can prove the following theorem so that, by your observation that bounded holomorphic functions on $\Omega$ are constant (similar reasoning shows that bounded harmonic functions on $\Omega$ are constant), it follows that $\Omega$ is not a Greenian domain and that all nonnegative harmonic functions on $\Omega$ are constant. - -Theorem: If $\Omega$ is a connected open subset of $\mathbb{R}^2$ then the following are equivalent. - -Every bounded harmonic function $f\colon\Omega\to\mathbb{R}$ is constant. -Every nonnegative harmonic function $f\colon\Omega\to\mathbb{R}$ is constant. -Every harmonic function $f\colon\Omega\to\mathbb{R}$ which is bounded below is also constant. -$\Omega$ is not a Greenian domain. - - -As argued above, statement 4 imples 2. Also, statements 2 and 3 are equivalent by adding a constant to $f$ to make it nonnegative. Also, 3 clearly implies 1. So, it just remains to show that 1 implies 4. I'll prove the contrapositive, so supposing that $\Omega$ is Greenian, we just need to construct a non-constant bounded harmonic function on $\Omega$. -Let $\tau$ be the first time at which $B_t$ exits $\Omega$ which, as we are assuming it is Greenian, is finite with probability one. Also, for each $x\in\Omega$, let $\mathbb{P}\_x$ be the probability measure for the Brownian motion started at $B_0=x$, and $\mathbb{E}\_x$ denotes expectation with respect to $\mathbb{P}\_x$. Then, for any bounded measurable function $g\colon\mathbb{R}^2\setminus\Omega\to\mathbb{R}$ we can define -$$ -\begin{align} -&f\colon\Omega\to\mathbb{R},\\\\ -&f(x)=\mathbb{E}\_x\left[g(B_\tau)\right]. -\end{align} -$$ -Then $f$ is a bounded smooth function satisfying $f(B_t)=\mathbb{E}[g(B_\tau)\mid\mathcal{F}_t]$ (whenever $t < \tau$). So, $f(B_t)$ is a martingale run up until time $\tau$. It follows that $f$ is harmonic. Furthermore, by martingale convergence, $f(B_t)\to g(B_\tau)$ as $t\uparrow\uparrow\tau$ (with probability one). To ensure that $f$ is non-constant, we just have to make sure that $g(B_\tau)$ is not equal to a constant with probability one. However, $B_\tau$ cannot be almost surely constant (because two dimensional Brownian motion has zero probability of hitting any given fixed point). So, $B^i_\tau$ is not almost-surely constant for at least one of $i\in\{1,2\}$ and we can take $g(x)=1_{\{\vert x\vert\le K\}}x_i$ for large enough $K$, making $f$ a bounded non-constant harmonic function on $\Omega$ and contradicting property 1 above. - -Update: I just want to add some comments on the use of Brownian motion in this answer. We have a domain $\Omega$ on which we know that all bounded harmonic functions are constant, and want to show that all nonnegative harmonic functions are constant. Or, looking at the contrapositive, given a nonnegative and non-constant harmonic function $f\colon\Omega\to\mathbb{R}$, we want to construct a non-constant bounded harmonic function $g\colon\Omega\to\mathbb{R}$. One way you might think of doing this is to construct $g$ directly from $f$. Say, by capping $f$ at some positive value $K$. You could try looking at $\min(f,K)$ and then choose $g$ to be the maximal harmonic function with $g\le\min(f,K)$. However, this does not work. There can exist positive harmonic functions $f$ on a domain for which every bounded harmonic function $g\le f$ is non-positive everywhere (this is related to the existence of positive martingales which tend to zero). So, this approach would give you $g\equiv0$, which is not very helpful. For example, let $S\subseteq\mathbb{C}$ consist of the negative real axis $S=\{x\in\mathbb{R}\colon x\le0\}$ and $\Omega=\mathbb{C}\setminus S$. Then, $f(z)=\Re[\sqrt{z}]$ defines a nonnegative harmonic function on $\Omega$ (taking the square root with positive real part). However, $f$ vanishes on $S$, so any bounded harmonic function $g\le f$ must be nonnegative on $S$. This implies that $g\le0$ everywhere. So, constructing $g$ directly from $f$ in this way is tricky (maybe there is some way around this). Instead, the approach I took was to use the existence of a nonnegative harmonic function to imply some property of the domain, which then allows us to construct bounded harmonic functions. Taking the expected value of a function of a Brownian motion when it first exits the domain provides one way of doing this, so it is natural to try to show that Brownian motion must exit the domain at some time. In the example just mentioned, there do exist bounded harmonic functions -- e.g., $g(z)=\Re[(1+\sqrt{z})^{-1}]$, which tends to $1/(1-x)$ for $x\in S$, so $g(z)=\mathbb{E}\_z[1/(1-B_\tau)]$. -Also, the result I stated in the answer relating the existence of nonnegative harmonic functions to the existence of bounded harmonic functions is intimately related to the behaviour of 2 dimensional Brownian motion. In particular, it fails in $n > 2$ dimensions, for which Brownian motion is not recurrent. Consider, for example, $\Omega=\mathbb{R}^n\setminus\{0\}$. If $f$ is a nonnegative harmonic function on $\Omega$ and $B$ is a Brownian motion in $\mathbb{R}^n$ then it is true that $f(B_t)$ will converge as $t\to\infty$. However, as $B$ is not recurrent ($\Vert B_t\Vert\to\infty$) in more than 2 dimensions, this does not allow us to conclude that $f$ is constant. In fact, there do exist non-constant and positive harmonic functions on $\Omega$, such as $f(x)=\Vert x\Vert^{2-n}$. However, all bounded harmonic functions are constant. You can prove this by extending any such bounded harmonic $g$ to all of $\mathbb{R}^n$ and applying Liouville's theorem. Or martingale convergence can be used to show that $g(B_t)=\mathbb{E}[X\mid\mathcal{F}\_t]$, where $X=\lim\_{t\to\infty}g(B_t)$ is in the tail $\sigma$-algebra of $B$ which, by Kolmogorov's zero-one law, is almost surely constant. In any case, the result I stated in the answer fails in $n > 2$ dimensions in which Brownian motion is not recurrent.<|endoftext|> -TITLE: Determining the asymptotic behavior of a series -QUESTION [8 upvotes]: I am trying to determine the behavior of the following series as $n\to\infty$. Let $0<\mu<1$ be fixed and for every positive integer $n\geq 1$, consider the function $f_n(t)$ of a real variable $t$ defined by the series $\sum_{k=0}^\infty\mu^k(1-\mu^kt)^n$. I want to determine how $f_n(t)$ behaves as $n\to\infty$ for $0 -TITLE: Cube-free infinite binary words -QUESTION [19 upvotes]: A word $y$ is a subword of $w$ if there exist words $x$ and $z$ (possibly empty) such that $w=xyz$. Thus, $01$ is a subword of $0110$, but $00$ is not a subword of $0110$. I'm interested in right-infinite words over a two-letter alphabet that do not contain subwords of the form $xxx$, where $x$ is a word of one or more letters. (For example, the Thue-Morse word, the Kolakoski word, Stewart's choral sequence, and so on.) In particular, I would like to know if there are any general statements about all such words. For example, are there an infinite number of them? Is there any way to classify them? -It's possible that one way to classify cube-free infinite binary words is to group them according to their subwords. For example, the Kolakoski word has the subword $00100$ whereas the Thue-Morse word does not, so they belong in different classes. The words created in Tony's answer (see below) have the same subwords (the Thue-Morse word is recurrent), so they belong to one class. I suppose there an infinite number of these classes. -Another possible way to classify cfib words is to group them according to their subword complexity. For example, Stewart's choral sequence has a subword complexity of $2n$ (where $n$ is the length of the subword), so we can group it with other cfib words with subword complexity $2n$. Is the subword complexity of the Kolakoski word known? - -REPLY [4 votes]: Lemma: Someone only able to speak an infinite cubefree word can convey information at least $1/24$ of the speed of an ordinary person able to speak an arbitrary binary sequence. -Proof: Consider the following period-$12$ sequence over the alphabet $\{ 0, 1, ?\}$: -$S = (001?010?1011)^{\infty}$ -Now suppose $T$ is a sequence derived from $S$ by replacing each $?$ with a $0$ or $1$. - -$T$ clearly has no words of the form $xxx$ for a single digit $x$. - -If we take alternate digits in $T$, we obtain $(010011)^{\infty}$. The only cubes in this sequence have length a multiple of $6$, hence: - -If $T$ contains a word of the form $WWW$, where $W$ has even length, then $W$ has length divisible by 12. -Also, $T$ contains a word of the form $WWW$ where $W$ has length $3$ if and only if two question marks in the same dodecad $(001?010?1011)$ are both zero. Assert that this will not happen. -If $T$ contains a word of the form $WWW$, where $W$ has length divisible by $3$, then $W$ has length divisible by $12$. - -Hence we are only worried about words of length coprime to or divisible by $12$. - -If the length of $W$ is $\pm 1 \mod 12$, then the existence of $WWW$ is ruled out by the proof of the non-existence of $xxx$. -Similarly, it is easy to observe that there are none of length $\pm 5 \mod 12$. - -So any word of the form $WWW$ in $T$ requires $W$ to have length divisible by $12$. - -Now begin with any tripleless sequence over $\{A, B\}$ (such as the Thue-Morse sequence) and apply the substitution rules: -$A \mapsto (001001011011)$ -$B \mapsto (0011010?1011)$ -Then we obtain an infinite word over $\{0, 1, ?\}$ such that: - -Every replacement of the question marks with digits yields an infinite cubefree word. -Every icositetrad contains one question mark. - -The result follows.<|endoftext|> -TITLE: Which convex bodies roll along closed geodesics? -QUESTION [21 upvotes]: An ellipsoid could be rolled (without slippage) on a horizontal plane so that its point -of contact traces out a closed geodesic on its surface: - -          - -Q1. Which other convex bodies (in $\mathbb{R}^3$) share this property? -A version of this question was raised in an earlier MO question -("Rolling a convex body: Geodesics vs. rolling curves") -in which Andrey Rekalo pointed to the book -Geometry of nonholonomically constrained systems -by Richard H. Cushman, Hans Duistermaat, and Jędrzej Śniatycki. -This book indeed describes the relevant equations of motion, but I am -not -finding it easy to extrapolate from their discussion to answer the -posed question. -Q2. What conditions suffice to guarantee that -a particular closed geodesic is the trace of a rolling curve? -In the case of the ellipsoid, the normal to the point of rolling -contact aims at all times through the center of gravity. -Perhaps it suffices that the center of gravity lies in the plane -determined by that normal and the tangent to the rolling curve. -I was particularly wondering: -Q3. Will a Zoll surface, all of whose geodesics -are closed, roll along (some of its) geodesics? -It would be quite interesting if the answer is 'Yes', although I think -this is unlikely. -(Zoll surfaces were discussed in two earlier MO questions: -"Surfaces all of whose geodesics are both closed and simple" -and -"Riemannian surfaces with an explicit distance function?.") -Thanks for any suggestions or pointers! - -REPLY [10 votes]: I don't have a definitive answer, but surely any convex surface $S$ in 3-space that has reflectional symmetry about a plane $P$ will roll along the geodesic $S\cap P$. If you take such a surface with 120-fold icosohedral symmetry group in $O(3)$, you'll get quite a few such geodesics, more than the (generic) ellipsoid has.<|endoftext|> -TITLE: Topological space associated to a real or complex scheme -QUESTION [5 upvotes]: Hi, consider a scheme $X$ of finite type over $\mathbb{R}$ (or $\mathbb{C}$). In Hartshorne's appendix B on 'transcendental methods' it is shortly mentioned how to assign a reasonable topological space to $X$ and he says that one can use the 'same glueing data'. My question is two-parted: - -Does this construction actually define a functor $R_{\mathbb{R}}:Sch/\mathbb{R}\to Top$ from schemes of finite type over $\mathbb{R}$ to topological spaces? (I suppose that a positive/negative answer would hold for $\mathbb{C}$ also.) - -And: - -Does anyone know where this is written up rigorously, I mean without just saying that one just 'uses the same glueing data' and everything is fine? - -Thank you! - -REPLY [6 votes]: Dear mustafa-kava, Amnon Neeman has written a rather down-to-earth book Algebraic and Analytic Geometry dedicated to a proof of Serre's celebrated GAGA theorem. Serre's article is 42 pages long and Neeman's book exactly ten times as long: 420 pages. This is because the book is extremely detailed and starts from the ground up (It comes from a fourth-year undergraduate course). -Section 4, called "The complex topology", is 20 pages long and seems to be what you want (at least in the complex case), with all details spelled out at an elementary level. -Let me add that Neeman really knows what he is talking about: he has proved a remarkable criterion to decide whether an algebraic variety is affine, given that its analytification is Stein. -As for the real algebraic geometry, the standard reference is Bochnak-Coste-Roy's Real Algebraic Geometry. What you want to study is Chapter 11 Topology of real algebraic varieties. Be warned that this is an advanced monograph and that general knowledge of schemes is insufficient : techniques like positivstellensatz, real spectra, Nash functions,...are de rigueur here. On the positive side you will be delighted by astonishing results like : every $\mathcal C^\infty $ compact manifold is diffeomorphic to an algebraic subvariety of some $\mathbb R ^p$ !<|endoftext|> -TITLE: Universal covers of punctured hyperbolic surfaces -QUESTION [6 upvotes]: Suppose S is a genus g surface with n punctures satisfying the hyperbolicity condition 2g + n - 2 > 0. If n > 0 the fundamental group of the surface is a free group on 2g + n - 1 := m generators. -If we look the universal covers of different punctured surfaces with the same m (e.g., thrice-punctured sphere and once-punctured torus for m = 2) in, say the hyperbolic plane or the Poincare disc model, how do they differ? The "only" apparent difference is in the number of punctures which should give rise to a difference in the lifts of the punctures to the boundary of the disc. The fundamental groups are isomorphic, but they must act differently to produce quotient surfaces of different genera. How? -How does the set of lifts of punctures on the boundary relate to the standard Farey set? -Thanks a lot in advance! - -REPLY [8 votes]: The simplest case, where $g=0, n=3$ and $g = n = 1$ yield isomorphic groups (free of rank 2), can be written explicitly using a little $2 \times 2$ matrix calculation. We choose a fundamental domain in the upper half-plane made out of two vertical lines with real parts $-1$ and $1$, and two semicircles whose diameters are the real intervals $[-1,0]$ and $[0,1]$. Since the boundaries are geodesics, it suffices to find Möbius transformations that transform the endpoints appropriately. -For $g=0, n=3$, we choose generators $\begin{pmatrix}1& 2 \\ 0 & 1 \end{pmatrix}$ to glue the vertical lines together, and $\begin{pmatrix}1& 0 \\ 2 & 1 \end{pmatrix}$ to glue the semi-circles. If you like modular curves, this quotient is called $Y(2)$, and classifies isomorphism classes of elliptic curves equipped with an ordered list of all 2-torsion points. -For $g=1, n=1$, we choose $\begin{pmatrix}1& 1 \\ 1 & 2 \end{pmatrix}$ to glue the left vertical line to the right semicircle, and $\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$ to glue the left semicircle to the right vertical line. This quotient is a leaky torus. -The above results form a special case of a general phenomenon (mentioned by Sam Nead), where the loops around punctures give unipotent (aka parabolic) generators of $\pi_1$, and handles give a pair of hyperbolic generators. -I don't have a good answer concerning the set of lifts of the boundary points - it will always be a disjoint union of $n$ orbits under the transformation group, but it can vary widely, since the transformation group has a continuous family of representations in $PSL_2(\mathbb{R}$.<|endoftext|> -TITLE: What relationship, if any, is there between the diameter of the Cayley graph and the average distance between group elements? -QUESTION [26 upvotes]: It's known that every position of Rubik's cube can be solved in 20 moves or less. That page includes a nice table of the number of positions of Rubik's cube which can be solved in k moves, for $k = 0, 1, \ldots, 20$. (Some of the entries of the table are approximations, but they're good enough for the purposes of this question.) -In particular, the median number of moves needed is 18, and in fact about 70 percent of all positions require eighteen moves. -This seems a bit counterintuitive to me -- I'd expect the median to be around half the maximum number of moves needed. Consider for example generating $S_n$, the symmetric group on $n$ elements, from the adjacent transpositions $(k, k+1)$ for $k = 1, 2, \ldots, n-1$. The number of such transpositions needed to get from the identity permutation to any permutation $\sigma$ using adjacent transpositions is the number of inversions of that permutation -- that is, the number of pairs $(i,j)$ such that $i < j$ and $\sigma(i) > \sigma(j)$. The maximum number of inversions of a permutation in $S_n$ is ${n \choose 2}$. The mean is ${1 \over 2} {n \choose 2}$; this is also the median if it's a whole number; and the distribution is symmetric around ${1 \over 2} {n \choose 2}$. Another similar case is $(Z/2Z)^n$ generated by the generators of the factors -- the diameter is $n$, the typical distance between elements is $n/2$. -My question: which of these situations is, in some sense, "more typical"? More formally, what's known about the relationship between the diameter of the Cayley graph of a group and the typical distance between two vertices? And is there a third case, where the median or mean distance between two random elements is less than half the diameter? Here I'm looking for something like the distribution of Erdos numbers as given by Grossman -- the maximum Erdos number is 15 but the median is only 5 -- although of course there is the complication here that the collaboration graph is very far from being vertex-transitive. - -REPLY [16 votes]: To expand on JSE's answer a bit: for simple finite groups of Lie type, the Gowers "quasirandom groups" argument with the additional trick of Nikolov-Pyber shows that once one has generated $|G|^{1-\delta}$ elements, three times that many steps give the whole group, for some $\delta>0$ which depends only on the Lie rank of the group (I think it is $1/9$ or so for $SL_2$). I've just written this up for my representation theory class -https://people.math.ethz.ch/~kowalski/representation-theory.pdf (Wayback Machine) -(section 4.7.1). -Such ideas are also already present in the work of Helfgott on growth in $SL_2$ -- part (b) of his "Key proposition" -- but the sharp constant $3$ was not; the crucial group-theoretic idea in the argument of Gowers is the use of the large size of a non-trivial representation of the group.<|endoftext|> -TITLE: How to memorise (understand) Nakayama's lemma and its corollaries? -QUESTION [201 upvotes]: Nakayama's lemma is mentioned in the majority of books on algebraic geometry that treat varieties. So I think Ihave read the formulation of this lemma at least 20 times (and read the proof maybe around 10 times) in my life. -But for some reason I just cannot get this lemma, i.e. I have tendency to forget it. Last time this happened just a couple of days ago, in the book of Shafarevich (Basic Algebraic geometry in 1.5.3.) This lemma is used to prove that for finite maps between quasiprojective varieties the image of a closed set is closed, and again this lemma sounded as something foreign to me (so again I went through the proof of the lemma)... -Question. Is there a path to get some stable understanding of Nakayama's lemma and its corollaries? I would be especially happy if there were some geometric intuition underlying this lemma. Or some geometric example. Or maybe there is a nice article of this topic? Some mnemonic rule? (or one just needs to get used to the lemma?) - -REPLY [12 votes]: I just came up with a geometric interpretation for Nakayama's lemma, and I'm surprised that no one here has already mentioned it! -The statement is the following: given a ring $A$, an ideal $I$, and a finitely generated $A$-module $M$, if $IM=M$ then we can find an element $a\in I$ such that $(a-1)M=0$. -In the geometric picture, we see $A$ as functions on a space $X$ (the spectrum $\mathrm{Spec}(A)$), with $I$ corresponding to functions that vanish on a closed subset $Z$ (the subset $V(I)$); $M$ is the module of global sections for a "vector bundle" $\mathcal F$ (the sheaf of modules $\widetilde M$). The elements of $IM$ represent those global sections vanishing along $Z$, so the condition $IM=M$ says that these are all the sections possible. In other words, $\mathcal F$ is identically zero along $Z$, and so its support $\mathrm{Supp}(\mathcal F)$ must be disjoint from $Z$. Now the conclusion of Nakayama's lemma simply affirms the existence of a "bump function" $a$, that is, a function being $0$ along $Z$, and $1$ along the support of $\mathcal F$: i.e. it lies in $I$ and it acts on $M$ like $1$! -If one works out the details of commutative algebra, - -the condition "$\exists\ a\in I$ s.t. $(a-1)M=0$" is equivalent to $V(I)\cap V(\mathrm{Ann}(M))=\emptyset$ (the "partition of unity"); -the support $\mathrm{Supp}(M)$ is contained in $V(\mathrm{Ann}(M))$; -for finitely generated module $M$, the condition $IM=M$ implies $V(I)\cap\mathrm{Supp}(M)=\emptyset$, moreover $\mathrm{Supp}(M)$ can be shown to coincide with the closed subset $V(\mathrm{Ann}(M))$. - - -When $M$ is not finitely generated, $\mathrm{Supp}(M)$ does not behave well, so we have courterexamples - -$IM=M$ does not imply $V(I)\cap\mathrm{Supp}(M)=\emptyset$: the classical counterexample $A=\mathbf Z_{(p)}$ with its maximal ideal, and $M=\mathbf Q$; -$\mathrm{Supp}(M)$ can be non-closed: $A=\mathbf Z$, $I=2\mathbf Z$, and $M=\bigoplus_{p\ge3}\mathbf F_p$; -$\mathrm{Supp}(M)$ can be closed yet different from $V(\mathrm{Ann}(M))$: $A=\mathbf Z$, $I=2\mathbf Z$, and $M=\bigoplus_{n\ge1}\mathbf Z/3^n\mathbf Z$. - - -So, to conclude, Nakayama's lemma (in the above form) says that the support of a finitely generated $A$-module $M$ can be identified with the closed subset $V(\mathrm{Ann}(M))$, and given any closed subset $Z=V(I)$ disjoint from it, we can find a "bump function" separating the two. -In real life, however, we are usually just using another (weaker) property of the closedness: the support is closed under specialization. Since for an $A$-module $M$ to be non-zero, it must at least have some support, and by specialization, it must be supported at least at one closed point (corresponding to a maximal ideal). Therefore, if we can verify that $M$ is not supported at any of the closed points, (which is provided, say, by the condition $IM=M$ for $I$ contained in the Jacobson radical $J(A)$), then $M$ must be zero.<|endoftext|> -TITLE: Is this a proper application of the Lowenheim-Skolem Theorem to a proper class? -QUESTION [7 upvotes]: Please don't get put off by the length, all the questions are quite simple, but given the quasi-mathematical context I tried to be precise with the formulation. The more mathematically interesting title question (and the one that's most important for my purposes) is the last one, so if anything please take a look at the end. -Here goes. In his paper Models and Reality (1980) pg. 468 (http://www.jstor.org/stable/2273415) Hilary Putnam states and proves the following theorem: - -Theorem. $ZF$ plus $V=L$ has an $\omega$-model which contains any given countable set of real numbers. - -I suspect his terminology might be idiosyncratic, so I'll point out that by an $\omega$-model he means "a model of set theory in which the natural numbers are ordered as they are 'supposed to be'; that is, the sequence of 'natural numbers' of the model is an $\omega$-sequence." -My first (not-so-interesting) question is this: - -Is this a proper model-theoretic theorem? - -By "proper" here I mean a theorem that is about the properties of a countable set of (first-order) sentences as they are reflected in the properties of their models. That is to say, in the case of Putnam's theorem (which I think is improper), it seems to me that it doesn't really say anything about $ZF$ - instead it describes the metatheory in which it is interpreted using model-theoretic language (and succeeds in doing so by assuming an outer $\omega$ and $\mathbb{P}(\omega)$ which exist independently of the metatheory or the particular set theory being interpreted.) -And the related question: - -If I am wrong and it is a proper theorem, then how would one put it in more - contemporary model-theoretic - terminology? - -Now to the title question. In his "proof" (quotation marks because I'm not yet sure whether it is a proof or a plausibility argument) of the Theorem, after reducing the statement to something equivalent, he writes: - -Now, consider [the sentence '$\mathcal{M}$ is an $\omega$-model for $ZF$ plus $V=L$ and $s$ is represented in $\mathcal{M}$'] in the inner model $V=L$. For every $s$ in the inner model-that is, for every $s \in L$-there is a model-namely $L$ itself-which satisfies "$V=L$" and contains $s$. By the downward L-S Theorem, there is a countable submodel which is elementary equivalent to $L$ and contains $s$. - -So here's my question: - -Are we allowed to use the downward LST on a proper class-sized inner model such as $L$ in the above case? - -In general it seems to me obviously not - any version of the strong LST uses some notion of cardinality which surely cannot apply to $L$. What am I missing here? Putnam does add that strictly speaking the Skolem hull construction is also needed, but I don't see how that would help. Can it? -I will tag this as a reference request too, in case someone knows whether this theorem has been published elsewhere - by Putnam or otherwise. (His footnote says that he proved it in 1963 but provides no more information.) - -REPLY [10 votes]: The result you state is not provable in ZFC, and cannot be formalized without going to a stronger theory. -We cannot apply Lowenheim-Skolem (LS) to proper classes; the issue is that we do not have a truth predicate, so the construction of hulls cannot be formalized "from within". Note that partial truth predicates are definable, with truth for $\Sigma_n$ statements being itself $\Sigma_n$, but we cannot define a predicate that works simultaneously for all $n$. I suspect Jech's set theory book addresses this issue. (It is more than a weakness of our approach; Tarski's undefinability of truth theorem shows it is an insurmountable obstacle.) Note that the technical problem disappears when working with set models, since the satisfaction relation becomes $\Delta_1$ definable. -There ought to be an obstacle, since otherwise we could conclude in ZFC that there are set models of ZFC, against the incompleteness theorem. -Assuming stronger axioms, we can formalize LS for certain classes. For example, the existence of $0^\sharp$ is essentially giving us a truth predicate for $L$, and so it allows us to form set sized Skolem hulls of $L$. In fact, $0^\sharp$ gives us many $\alpha$ such that $L_\alpha\prec L$. -Many arguments in set theory take advantage of the reflection theorem, which allow us to find set models of any finite subtheory of ZFC, and so in many arguments we can make do with this weak version of LS. But this does not suffice for Putnam's intended application. -Putnam's notation is standard, though. In an $\omega$-model we always identify standard parts with their transitive version, so in particular, we identify the ${\mathbb N}$ of the model with true $\omega$. Of course, we need this, or it wouldn't make sense to say that a real belongs to the model. Assuming a bit more than ZFC, Putnam's theorem is an easy consequence of Shoenfield's absoluteness. -A good reference for Putnam's proof, the philosophical arguments it generated, and the mathematics behind it, is the PhD thesis of my friend and colleague Johannes Hafner, "A Critical Assessment of Putnam's Model-Theoretic Argument", UC Berkeley, 2006, written under Charles Chihara and Paolo Mancosu.<|endoftext|> -TITLE: Units in quaternionic algebras -QUESTION [9 upvotes]: Let $H$ be a quaternionic algebra over ${\bf Q}$, and let $R$ denote a maximal ${\bf Z}$-order in $H$. Is there a theorem on the structure of the units in $R$ analogous to the Dirichlet unit theorem? Is there an analogous theorem for the $S$-units? - -REPLY [12 votes]: If $H$ is definite, then the group of units of $H$ is finite. If $H$ is indefinite, then the group of units is a pretty chunky group; it embeds as a cocompact discrete subgroup of $SL(2, R)$, and the rank of its abelianization (which, if I remember correctly, can be interpreted geometrically as twice the genus of an associated modular curve) can be arbitrarily large. -All of this follows from general theorems on arithmetic subgroups of algebraic groups; this is a classical theory going back to Borel, Harish-Chandra and Ono in the 60s, and there is a nice summary in Gross's paper "Algebraic modular forms". -EDIT: You asked about S-units. Let S be a set of finite primes. If $H$ is definite, then the group of S-units in H will be a discrete cocompact subgroup of -$\prod_{p \in S} (H \otimes \mathbb{Q}_p)^\times$. -If every place in $S$ is ramified in $H$, then the kernels of the reduced norm maps $(H \otimes \mathbb{Q}_p)^\times \to \mathbb{Q}_p^\times$ are compact, so this group above maps with finite kernel to a discrete subgroup of $\prod_{p \in S} \mathbb{Q}_p^\times$ and hence its abelianization has rank equal to the size of S. This argument can be pushed substantially further: you can show that if H is a quaternion algebra over a totally real field F which is totally definite (i.e. definite at every infinite place of F), then the map from S-units of H to S-units of F has finite kernel and cokernel, so the abelianization of the S-units of F has rank $|S| + [F : \mathbb{Q}] - 1$. -My favourite case, though, is when $F = \mathbb{Q}$, $H$ is definite, and $S$ consists of a single finite place $p$ which is not ramified in $H$. Then the S-units of $H$ embed into $GL(2, \mathbb{Q}_p)$ as a discrete co-compact subgroup $\Gamma$, and the space of continuous functions on the quotient $GL(2, \mathbb{Q}_p) / \Gamma$ gives a representation of $GL(2, \mathbb{Q}_p)$ with many fascinating properties (it is an example of one of Matt Emerton's completed cohomology spaces).<|endoftext|> -TITLE: Are the associative grassmannian and the quaternionic projective plane diffeomorphic? -QUESTION [17 upvotes]: I have a doubt which I hope the MO community can quickly resolve. -The associative grassmannian is an eight-dimensional homogeneous space $G_2/SO(4)$. It can be identified with the space of quaternion subalgebras of the octonions. It has a $G_2$-invariant riemannian metric making it a rank-2 riemannian symmetric space. The Poincaré polynomial of its de Rham cohomology is $1 + t^4 + t^8$. This is also the Poincaré polynomial for the quaternionic projective plane $\mathbb{HP}^2$, also an eight-dimensional homogeneous space $Sp(3)/Sp(2)\times Sp(1)$. It is the space of quaternionic lines in $\mathbb{H}^3$. It has an $Sp(3)$-invariant riemannian metric making it into a rank-1 riemannian symmetric space. -The difference in the ranks shows that as riemannian symmetric spaces, the associative grassmannian and the quaternionic projective plane are not isometric, but perhaps one is obtained from the other by squashing (also known as the canonical variation). Anyway, my real question is whether they are diffeomorphic. -Question: Are the associative grassmannian and the quaternionic projective plane diffeomorphic? -Thanks in advance. - -REPLY [9 votes]: To expand on Oscar's answer: the principal $SO(4)$ bundle over -$G_2/SO(4)$ gives us -$$ \qquad \qquad \qquad \ldots \to \pi_2(G_2) \to \pi_2(G_2/SO(4)) $$ -$$ \to \pi_1(SO(4)) \to \pi_1(G_2) \to \pi_1(G_2/SO(4)) $$ -$$ \to \pi_0(SO(4)) \to \ldots \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad$$ -and $\pi_2(G_2) = \pi_1(G_2) = \pi_0(SO(4)) = 1$, -so $\pi_1(G_2/SO(4)) = 1$, $\pi_2(G_2/SO(4)) = Z_2$.<|endoftext|> -TITLE: Critical Radius for Infinite Dimensional Sphere Packing -QUESTION [5 upvotes]: Hello. I'd like to consider the open unit ball in an infinite dimensional Hilbert space and ask when can we fit infinitely many open balls of radius $r<1$ inside. -For example, when $r=1/(1+\sqrt2)$, we can pick an orthonormal basis $(x_1,...)$ for our Hilbert space and put the centers of the balls at $(1-r)x_i = \sqrt2/(1+\sqrt2)x_i$ for each $i$. The distance between any two centers is thus $\sqrt2/(1+\sqrt2)\sqrt2 = 2r$, so indeed the balls just kiss each other. -Can we fit any larger balls? What is the critical radius $r_\infty$ such that for $r>r_\infty$ we may only fit finitely many balls of radius $r$, but for smaller $r$ we may fit infinitely many? - -REPLY [2 votes]: The optimality of your configuration can be shown as a plain consequence of the Kirszbraun theorem. -(I happened to ask myself this problem too, and eventually added this short section in a wiki article, thinking that it could be useful one day --not completely true, since your question has been already answered by Mikael de la Salle).<|endoftext|> -TITLE: Life after Hartshorne (the book, not the person)... -QUESTION [67 upvotes]: I was wondering what material in algebraic geometry is crucial and is a logical step for a serious graduate student in algebraic geometry once they've finished Hartshorne. Good answers could include a list of areas of algebraic geometry or important topics that an algebraic geometer must learn along with good references (i.e. accessible to someone with the background of Hartshorne), preferably in the order he or she should/could learn them. Papers in algebraic geometry tend to draw from so many areas within the field itself that I was wondering what people thought was the best order and way of acquiring that material. - -REPLY [13 votes]: As others have said, after Hartshorne you could branch into too many different things, whose relative importance is subjective. Here is my take on what could constitute a good post-Hartshorne curriculum: -Intersection theory - -Intersection theory, William Fulton. - -Classical foundational book, beautifully written. - -3264 and all that, David Eisenbud and Joe Harris. - -This one actually tells you how to compute Chow rings in a multitude of situations. But it also contains a wealth of useful information apart from how to compute Chow rings, such as geometry of Grassmanians, Chern classes, Hilbert schemes and a toolbox for Riemann-Roch. -Complex Algebraic Geometry - -Complex Geometry, An Introduction, Daniel Huybrechts - -This concise book covers nicely the foundational material of complex-analytic approach (which people used to learn from chapters 0 and 1 of Griffiths-Harris). - -Hodge theory and complex complex algebraic geometry, I and II, Claire Voisin. - -It took me time to develop love for this book. Probably because it strives to reach modern techniques as fast as possible, making it hard for someone just entering the world of Hodge theory. -Derived categories of coherent sheaves - -Fourier-Mukai Transforms in Algebraic Geometry, Daniel Huybrechts - -Just as all other of Huybrechts' books, this one is a true gem. The book teaches you how not to be scared of $D^b (Coh (X))$. -Birational geometry -There is wealth of different literature on this topic, but I think the most useful as a first book is this truly wonderful book: - -Positivity in algebraic geometry I & II, Robert Lazarsfeld - -Very clearly written and contains a wealth of examples and beautiful geometry. -Geometry of special varieties. -For special varieties (curves, various types of surfaces, special three- and four-folds, etc.) people have accumulated a large amount of more ad-hoc methods. At the basic level, contemplating the geometry of simplest such varieties provides an excellent playground to truly absorb the power of abstract machinery from Hartshorne. Further on, while doing research you will probably need a very good knowledge of the geometry of your particular variety at hand, so you cannot go wrong with investing some time to learn this material. I will mention here only the basics: -Curves: - -Chapter 4 of Hartshorne or -Chapter 19 of Ravi Vakil's notes - -Provide the absolute minimum here. -Surfaces: - -Chapter 5 of Hartshorne, -Complex algebraic surfaces, Arnaud Beauville -Algebraic surfaces and holomorphic vector bundles, Robert Friedman - -The first two items are standard sources, while R. Friedman's book is an interesting synthesis of classical geometry of surfaces and analysis of coherent sheaves on them, beautifully written. - -Lectures on K3 surfaces, Daniel Huybrechts - -K3 surfaces is truly an amazing class of varieties whose geometry is studied through a multitude of different techniques, and this book does an excellent job showing those multiple facets of research in the area. - -The geometry of cubic hypersurfaces, Daniel Huybrechts - -These notes are still work in progress, but as the previous item these notes talk about an amazing variety of techniques: moduli spaces, Hodge theoretic methods, etc. There is also a discussion of higher dimensional cases of three- and four-folds. - -The rest is more idiosyncratic since it is the field I am interested in - -Moduli spaces of sheaves. - -The Geometry of Moduli Spaces of Sheaves, D. Huybrechts and M. Lehn - -A systematic treatment of foundations of moduli of semistable sheaves. The book discusses an incredible amount of general techniques used in this area. - -Lectures on Vector Bundles, J.Le Potier. - -Not the best read with respect to foundations (for that see the previous item in the list), but fully explains the best-studied case of moduli of sheaves on $\mathbb{P}^2$, which still serves as a beacon for research in this area. - -Fundamental Algebraic Geometry. Grothendieck's FGA explained, B. Fantechi et al. - -An introduction into some more advanced fundamental techniques useful for moduli problems: descent theory, Hilbert and Quot schemes, elementary deformation theory and Picard scheme. - -Deformation theory, R. Hartshorne - -A nice concise account of algebraic approach to deformation theory, with a lot of examples.<|endoftext|> -TITLE: Bundle Gerbes as Characteristic Classes -QUESTION [8 upvotes]: Perhaps this is a bit naïve, but I was wondering if it possible to (at least formally) represent Bundle Gerbes as Characteristic Classes. Disclaimer: My understanding of Bundle Gerbes is limited to this paper of Hitchin so perhaps I'm not thinking of this correctly. Just for reference, a Bundle Gerbe is defined by specifying an open cover $\{U_i\}$ of a manifold $M$ that has associated to it maps $g_{ijk} : U_i \cap U_j \cap U_k \rightarrow S^1$ that satisfy certain cocycle-like conditions, $g_{jkl} g_{ikl}^{-1} g_{ijl} g^{-1}_{ijk}$. One can define connective structures with $3$-form curvatures $H$ on Bundle Gerbes that define principle circle bundles on the Loop Space of $M$ (See Hitchin, Page 4). These connective structures are classified by their curvatures, $[H / 2\pi] \in H^3(M,\mathbb{Z})$ just like the curvature $2$-form of a line bundle generates the first Chern Class. Explicitly, my question is the following: - -Can we expand the definition of a bundle gerbe on a manifold $M$ to an arbitrary compact, finite-dimensional Lie Group $G$ by considering a Bundle Gerbe to instead be the set of maps $g_{ijk} : U_i \cap U_j \cap U_k \rightarrow G$? If $\dim G = n$, will $H^{n}(M,\mathbb{Z})$ classify the Principle $G$-bundles on $\Omega M$? - -Again, my understand of gerbes is quite insufficient so perhaps this is "obvious" in some other literature. If this is the case, could you please cite a reference? -Thanks! -PS: I'm not sure if the compactness is truly necessary, I just added it with the hope that its more likely in the compact case - -REPLY [8 votes]: The fact that you are dealing with compact and/or finite dimensional Lie groups is completely irrelevant. The fact that these group are Lie is also partially irrelevant (unless you care about putting connections on your bundle gerbes, in which case it becomes very relevant). -More relevant is whether the groups abelian or not. -A priori, the cocycle relation only makes sense for abelian groups. -But there is also a theory of non-abelian (bundle) gerbes, where you allow non-abelian groups. The cocycles have two kinds of data: Maps -$\alpha_{ij}:U_i\cap U_j\to \mathrm{Inn}(G)$ and maps -$g_{ijk}:U_i\cap U_j\cap U_k \to G$, -where $\mathrm{Inn}(G)$ denotes the group of inner automorphisms of $G$. -These non-abelian gerbes are classified by $H^2(-,Z(G))$, the second Cech cohomology group with coefficients in the sheaf of $Z(G)$-valued functions. [that's a non-trivial theorem] -That was the case of a trivial band. - -A band is the same thing as an $\mathrm{Out}(G)$-principal bundle. -Say you are given an $\mathrm{Out}(G)$ principal bundle $P$, described by transition functions -$b_{ij}:U_i\cap U_j\to \mathrm{Out}(G)$. Then you can twist the above definition as follows: -The cocycles now consist of maps -$\alpha_{ij}:U_i\cap U_j\to \mathrm{Aut}(G)$ and maps -$g_{ijk}:U_i\cap U_j\cap U_k \to G$, -where the $\alpha_{ij}$ are lifts of the $b_{ij}$. -The gerbes with band $P$ are classified by a set that is either ♦ empty, or ♦ isomorphic to $H^2(-,Z(G)\times_{\mathrm{Out}(G)} P)$, the second Cech cohomology group with coefficients in the sheaf of sections of -$Z(G)\times_{\mathrm{Out}(G)} P$. -Whether or not that set is empty depends on the value of an obstruction class that lives in $H^3(-,Z(G)\times_{\mathrm{Out}(G)} P)$. -It's non-empty iff that obstruction vanishes. - -Finally, to answer your last question. -If $G$ is a Lie group and you have a bundle gerbe with connection (trivialized over the base point), then you get a $G$-principal bundle, but only on a subspace of the based loop space $\Omega M$. -It's the subspace consisting of those loops over which the band $P$ and its connection trivialize.<|endoftext|> -TITLE: Is the direct image of a constant sheaf a constant sheaf? -QUESTION [7 upvotes]: Is the direct image of a constant sheaf a constant sheaf? I'm not an expert on sheaf theory and can't find this anywhere - -REPLY [7 votes]: When the map is well behaved, say a locally trivial fibration over a base $B$ with fiber $F$, then the direct image of a constant sheaf will be locally constant (all of the fibers of this sheaf will be equal to the cohomology of the constant sheaf on $F$), but won't be constant in general. -The obstruction is the monodromy, the defect of compatibility between all local trivializations -along a loop in the base space.<|endoftext|> -TITLE: Unique preduals up to (nonisometric) isomorphism? -QUESTION [6 upvotes]: It's well known that there are Banach spaces which has a unique isometric predual-- for example, any von Neumann algebra. As other questions on here (for example, Isomorphisms of Banach Spaces ) have shown, there are also Banach spaces $E$ with non-isomorphic (either isometric, or not) $E_1$ and $E_2$ with $E_1^*$ and $E_2^*$ both (isometrically) isomorphic to $E$. -However, I don't know the answer to the following: - -Is there a Banach space $E$ such that, if $F$ is another Banach space with $T:F^*\rightarrow E^*$ a (not necessarily isometric) isomorphism, then necessarily $T$ is weak$^*$-continuous? - -This is clearly true if $E$ is finite-dimensional, or more generally, reflexive. But I cannot think of a non-reflexive example. (I sort of think that the James space might be an example, but I can't see how to show this). - -REPLY [9 votes]: No. Take $E=F$ and perturb the identity on $E^*$ by a rank one operator that is not weak$^*$ continuous; such an operator existing is equivalent to $E$ being non reflexive.<|endoftext|> -TITLE: Low degree cohomology of Eilenberg-MacLane space K(G,2)? -QUESTION [8 upvotes]: Recall that an Eilenberg-Maclane space $K(G, n)$ is characterized by $\pi_i(K(G,n)) = G$ if $i=n$ and is trivial otherwise. (Of course $G$ should be abelian if $n>1$.) -I'm aware that computing $H^j(K(G,n), \mathbb Z)$ for general $j$ and $n$ is not so easy (see, e.g., here), but I'm hoping that for certain small values of $j$ and $n$ it's easier. -My question: Is there a good reference for $H^j(K(G,2), \mathbb Z)$, where $j \le 4$ and $G$ is finite abelian (or just cyclic)? - -REPLY [9 votes]: For a finite cyclic group G, in the range you ask for you get cohomology groups -$$\mathbb{Z}, 0, 0, G \cong Ext(G, \mathbb{Z}), 0.$$ -One sees this by for example computing the Leray--Serre spectral sequence for -$$K(G, 1) \to * \to K(G,2).$$<|endoftext|> -TITLE: Explicit constructions of K(G,2)? -QUESTION [17 upvotes]: Recall that an Eilenberg-Maclane space $K(G, n)$ is characterized by $\pi_i(K(G,n)) = G$ if $i=n$ and is trivial otherwise. (Of course $G$ should be abelian if $n>1$.) -Let $G$ be a finite abelian group. -Below I describe cell complexes $X_1$ and $X_2$ with $\pi_2(X_i) = G$ and $\pi_0(X_i)$ and $\pi_1(X_i)$ both trivial. By standard results it is possible to add 4-cells to $X_i$ to kill off $\pi_3$, then add 5-cells to kill off $\pi_4$, and so on. -My questions: -(1.i) Does there exist in the literature an explicit description of the 4- and 5-cells one would need to add to $X_i$ in order to turn it into a $K(G,2)$? (I'm only interested in dimensions 4 and 5, not higher.) -(2) More generally, are there explicit descriptions of $K(G,2)$ in the literature? (I'm already aware of making $K(G, 1)$ into a group and then applying the bar construction.) -Definition of $X_1$: A single 0-cell. A 2-cell $c_g$ for each element $g\in G$. A 3-cell $d_{g,h}$ for each $(g,h)\in G\times G$, with $\partial d_{g,h} = c_g + c_h - c_{gh}$. -(This starts out similarly to a standard construction of $K(G, 1)$, but the higher dimensional cells will necessarily be more complicated. Obvious candidates for the boundaries of 4-cells would include $d_{g,h} - d_{fg,h} + d_{f,gh} - d_{f,g}$ for all $(f,g,h)\in G\times G\times G$, and also Hopf maps to the 2-cells $c_g$ for each $g$, and also $d_{g,h} - d_{h,g} + x$, where $x$ is a map to $c_h\cup c_h$ which exhibits the commutativity of $\pi_2(c_g\cup c_h)$.) -Definition of $X_2$: Let $G = \mathbb Z/k_1 \times\cdots\times \mathbb Z/k_m$, a prodict of cyclic groups. $X_2$ has $m$ 2-cells $e_1,\ldots, e_m$ and $m$ 3-cells $f_1,\ldots, f_m$, with $\partial f_i = k_i\cdot e_i$. - -REPLY [6 votes]: This is a comment relating the other answers more than anything else. -Following are three isomorphic simplicial abelian groups which are Eilenberg-Maclane spaces $K(G,n)$. - -The result of applying the Dold-Kan correspondence to the chain complex which has a copy of $G$ in degree $n$ and zero in all other degrees. This is an instance of the answer given by Tim Porter. -The result of taking the levelwise tensor product of $G$ with the free simplicial abelian group on the pointed simplicial set $\Delta^n/\partial\Delta^n$. This is an instance of the simplicial variant of the answer by BS. -The simplicial abelian group $\tilde{H}^{n-1}(sk^{n-1}\Delta^\bullet,G)$, where $\tilde{H}^\ast$ is reduced cohomology, $sk^k$ denotes the k-th skeleton, and $\Delta^\bullet$ stands for the standard functor from the simplicial category $\Delta$ into topological spaces. - -You can prove that these are all isomorphic by showing that the normalized chain complex associated to each of them is the chain complex concentrated in degree $n$ with $G$ at that level. This is obvious for the first one, and easy for the second one. It is also not terribly difficult for the last one: it starts with observing that $sk^k \Delta^n$ is homotopy equivalent to a wedge of $\binom{n}{k+1}$ spheres of dimension $k$ (the number $\binom{n}{k+1}$ of spheres is also the number of monomorphisms of $\Delta^k$ into $\Delta^{n-1}$). -This also recovers the number of copies of $G$ appearing in Andre's answer: $\binom{n}{2}$ copies of $G$ for the $n$-simplices. This binomial coefficient is also easily seen to be the number of non-degenerate $n$-simplices in $\Delta^2/\partial\Delta^2$ for $n>0$, so the same answer could be easily obtained from the second description above.<|endoftext|> -TITLE: Relative Lie Algebra cohomology and sheaf cohomology -QUESTION [8 upvotes]: (I apologize in advance for the vagueness of my question). Let $G$ be a reductive algebraic group over $\mathbb C$ with Lie algebra $\frak g$ and Borel $B$. I have seen casual references to the fact that relative Lie algebra cohomology (sometimes also called cohomology of a pair) with coefficients in a $B$-module $V$ is, with the right choice of pair of Lie algebras, related to sheaf cohomology of $G/B$ with coefficients in $V$. Is there a reference for this fact, and is there a precise statement one can make? In particular, which Lie algebras should one use? (My guess would be the pair $(\frak b, \frak h)$, where $\frak b := $ Lie($B$) and $\frak h$ is a Cartan of $\frak b$). - -REPLY [7 votes]: As Ben points out, Kostant's papers are a fundamental reference for transition between Bott's work (Annals of Mathematics 66, 1957) involving the flag variety and a more algebraic formulation involving Lie algebra cohomology for the nilradical $\mathfrak{n}$ of a Borel subalgebra. I think the rough intuition here is that for a suitable assignment of positive or negative roots, $\mathfrak{n}$ approximates the flag variety in the classical finite dimensional representation theory setting. -Yet another viewpoint was offered in the 1970s by Bernstein-Gelfand-Gelfand in the context of category $\mathcal{O}$. (For a fairly short treatment of some of these connections see Chapter 6 of my 2008 AMS book GSM 94 where Delorme's formulation in terms of relative Lie algebra cohomology is outlined.) The relative Lie algebra technology was further developed by Borel and Wallach in their monograph (AMS, 2000). -P.S. To clarify the "relative" aspect of the cohomology here, my understanding (probably incomplete) is that in the narrow setting of finite dimensional representations of a semisimple $\mathfrak{g}$, the essential relative Lie algebra cohomology computations involve the pair $(\mathfrak{g},\mathfrak{h})$; here $\mathfrak{h}$ is a Cartan subalgebra lying in $\mathfrak{b}$. But in more sophisticated study of unitary representations of a corresponding noncompact Lie group, with a maximal compact subgroup $K$, the appropriate relative cohomology arises from Harish-Chandra's $(\mathfrak{g}, K)$-modules.<|endoftext|> -TITLE: Is the reduction of a flat, finite, surjective scheme over an integral base still flat? -QUESTION [7 upvotes]: Is the reduction $X_{red}$ of a flat, finite, surjective scheme $X$ over an integral base $S$ still flat? -I could possibly add that I am already aware we can assume the base $S$ to be local and complete, and we can assume $X$ is local (and henselien). So far this hasn't seemed to help me. -My intuation is that I should be trying to show that the dimension of the residues of the sheaf of nilpotents $\mathcal{N}$ in the structure sheaf $\mathcal{O}_{X}$ of $X$ is constant over $S$. - -REPLY [5 votes]: In analytic geometry, you can see to the Douady's example in Fischer book (p.151) where $X:=\lbrace{(x,s,t)\in {\Bbb C}^{3}: x^{3}+sx +t=0; 27t^{2}+4s^{3}=0}\rbrace$, $S:=\lbrace{(s,t): 27t^{2}+4s^{3}=0\rbrace}$ and $f:X\rightarrow S$ is a finite, surjective and flat map (induced by the canonical projection). Then, it easy to show that the restriction $f: X_{red}\rightarrow S$ is not flat !<|endoftext|> -TITLE: Unitary groups over number fields -QUESTION [9 upvotes]: When defining unitary groups over number fields, one usually takes $F$ to be a totally real number field, $E$ a CM quadratic extension of $F$, and $V$ a hermitian space attached to $E/F$. Then $U(V)$ is simply the isometry group for the hermitian form attached to $V$. -My question is: why does one take $F$ to be totally real and $E$ CM? The definition makes perfect sense without this, but most references (at least in the context of automorphic forms) make these assumptions. - -REPLY [11 votes]: Just to add to Jon Yard's answer: when one defines a unitary group as described in the OP, the group that one gets after extending scalars from $F$ to $F_v = \mathbb R$ for each archimedean place $v$ of $F$ is indeed a unitary group in the usual sense, i.e. the set of complex matrices preserving some non-degenerate Hermitian form. -If one were to apply the defintion in a more general setting, say to a quadratic extension $E$ over $F$ which is not a CM extension of a totally real field, the groups at infinity could end up being $GL_n(\mathbb R)$ or $GL_n(\mathbb C)$, which are not unitary groups.<|endoftext|> -TITLE: Limits of Nilpotent and Quasi-nilpotent Operators in a $\mathrm{II}_1$-factor -QUESTION [8 upvotes]: A bounded operator $A$ in a Hilbert space is called nilpotent if there exists $n$ such that $A^{n}=0$. An operator is called quasi-nilpotent iff -$$ -\limsup_{n\to\infty}{ \|A^{n}\|^{1/n}}=0. -$$ -Every nilpotent operator is clearly quasi-nilpotent. The family of quasi-nilpotent operators is very important for the hyperinvariant subspace problem. For instance, it was proved by Haagerup and Schultz, that if the Brown measure of an operator in a $\mathrm{II}_1$-factor is concentrated in more than one point then it has a non-trivial hyperinvariant subspace. A subspace is called $A$-hyperinvariant if it is invariant for all the operators that commute with $A$. -I seem to recall from Herrero's book that the norm closure of the nilpotent and quasi-nilpotent operators in $B(H)$ is pretty well understood. However, I don't have a copy with me at the moment. My question is: Is it known what is the norm closure of the nilpotent operators or/and quasi-nilpotent operators in the hyperfinite $\mathrm{II}_1$-factor? In any other $\mathrm{II}_1$-factor? -Thanks! - -REPLY [4 votes]: The definition of quasi-nilpotent is that the spectrum of $A$ is zero, which is the same as $\|A^n\|^{1/n} \to 0$. -See Apostol's paper On the norm-closure of nilpotents, Rev. Roumaine Math. Pures Appl. 19 (1974), 277-282 or the later paper Apostol, C.; Foiaş, C.; Pearcy, C. -That quasinilpotent operators are norm-limits of nilpotent operators revisited. -Proc. Amer. Math. Soc. 73 (1979), no. 1, 61–64 for info about the closure of the nilpotent operators. For recent papers giving info about hyperinvariant subspaces for quasi-nilpotent operators, see recent papers by Foias, Pearcy, et al.<|endoftext|> -TITLE: When does the conormal bundle sequence split? -QUESTION [6 upvotes]: Let $X\subset \mathbb{P}^n$ be a smooth projective variety with ideal sheaf $I_X$. The conormal sequence is given by -$$ -0\to I_X/I_X^2\to \Omega_{\mathbb{P}^n}|_X\to \Omega_{X}\to 0. -$$ -For which varieties $X$ is the sequence above split? -If I'm not mistaken, if $X$ a hypersurface, the sequence is split if and only if $X$ has degree 1. - -REPLY [10 votes]: You are right. This is a result due to Van de Ven. -[A. Van de Ven, A property of algebraic varieties in complex projective spaces. In: Colloque Géom. Diff. Globale (Bruxelles, 1958), 151–152, Centre Belge Rech. Math., Louvain 1959. MR0116361 (22 #7149) Zbl 0092.14004] -Even more is true. Recently Ionescu and Repetto proved the following generalization of Van de Ven's Theorem. - -Let $X \subset \mathbb P^n$ be a smooth subvariety. If there exists a curve $C \subset X$ - such that the restriction to $C$ of - the conormal sequence of $X$ splits - then $X$ is linear. - - -Let me sketch a short elementary proof (of Van de Ven's result not its generalization) in the case of hypersurfaces. I will phrase it in the analytic category but once it is translated to the algebraic category, working with infinitesimal neighborhoods, I believe that what will emerge is one of the proofs in the literature. -If the normal sequence splits then we can define a foliation $\mathcal L$ by (germs of) lines everywhere transverse to $X$ at a -neighborhood $U$ of $X$. Since the complement of $X$ is Stein we can extend $\mathcal L$ to the whole $\mathbb P^n$. -Therefore $\mathcal L$ is defined by a global section of $T \mathbb P^n(d-1)$ for some -$d \ge 0$. With the help of Euler's sequence, this section can be presented as a homogeneous vector field $v$ -on $\mathbb C^{n+1}$ with coefficients of degree $d$. -To compute the tangencies between $\mathcal L$ and $X$ we have just to contract -the differential $dF$ of a defining equation $F$ of $X$ with $v$. If $F$ is not linear -then the divisor on $X$ defined by the tangencies between $\mathcal L$ and $X$ (defined by $F=dF(v)=0$) will be non-empty contradicting the transversality between $X$ and $\mathcal L$.<|endoftext|> -TITLE: Integer sequences and integer coefficients in recurrence relations. -QUESTION [10 upvotes]: Background: As a result of teaching recurrence relations in various courses over the years, I am working on a paper on periodic integer sequences generated by second-order, linear, homogeneous recurrences with constant coefficients. -Briefly, if the recurrence generates a sequence of integers, must the coefficients be integers? -More precisely: Consider the recurrence defined by $G_{n+2} = r G_{n+1} + s G_n$, with $G_0$ and $G_1$ given integers. Suppose no lower-order recurrence generates $G$, and further, that $G$ is a sequence of integers. Must $r$ and $s$ be integers? -(1) The condition about lower-order is required. It is not hard to come up with 0th- or 1st-order examples which can also be generated by second-order recurrences with non-integer coefficients. -(2) It is not hard to see that $s$ must be an integer. A simple induction proof shows that for all $n\ge0$, $G_n G_{n+2} - G_{n+1}^2=(-s)^n(G_0 G_2 - G_1^2)$, from which is follows easily that $s$ is an integer. -(3) It seems that it should be easy to show that $r$ is an integer. However, by working backwards, it is not difficult to find a sequence generated by $G_{n+2} = G_{n+1} / 2 + G_n$ which begins with an arbitrary number of integers, the last of which is odd, so that the next term is not an integer. $G_0 = 32$ and $G_1 = 64$ give 311 as the 8th term. If $p$ is a prime factor of the denominator of $r$, it seems that the powers of $p$ in $G_n$ should eventually decrease by at least one (possibly more, if the power of $p$ dividing the denominator of $r$ is greater than 1) each time. -This problem has been nagging me for a while now. I'm not a number theorist (interests lie mainly with polyhedral geometry and education), so I apologize if there is a well-known result which proves this. The extension to sequences of integers generated by higher-order recurrences should be clear, so if there is a general result in such cases, that would be nice, too. - -REPLY [14 votes]: This is Fatou's lemma. One reference is Exercise 4.1(a) of my book Enumerative Combinatorics, vol. 1. This is repeated as Exercise 4.2(a) at http://math.mit.edu/~rstan/ec/ec1.pdf.<|endoftext|> -TITLE: central extensions of Diff(S^1) and of the semigroup of annuli -QUESTION [20 upvotes]: $\mathit{Diff}(S^1)$ refers to the group of orientation preserving diffeomorphisms of the circle. The semigroup of annuli $\mathcal A$ is its "complexification": the elements of $\mathcal A$ are isomorphism classes of annulus-shaped Riemann surfaces, with parametrized boundary. -Both $\mathit{Diff}(S^1)$ and $\mathcal A$ have central extensions by $\mathbb R$, and my question is about their relationship. - -♦ The group $\mathit{Diff}(S^1)$ carries the so-called Bott-Virasoro cocycle, which is given by -$$ -B(f,g) = \int_{S^1}\ln(f'\circ g)\;\; d\;\ln(g'). -$$ -The corresponding centrally extended group is $\widetilde{\mathit{Diff}(S^1)}:=\mathit{Diff}(S^1)\times \mathbb R$, with product given by $(f,a)\cdot(g,b):=(f\circ g,a+b+B(f,g))$. -♦ The elements of the central extension $\widetilde{\mathcal A}$ of $\mathcal A$ have a very different description. -An element $\widetilde{\Sigma}\in\widetilde{\mathcal A}$ sitting above $\Sigma\in\mathcal A$ -is an equivalence class of pairs $(g,a)$, where $g$ is a Riemannian metric on $\Sigma$ compatible with the complex structure, and $a\in\mathbb R$. -There's the extra requirement that the boundary circles of $\Sigma$ be constant speed geodesics for $g$. -The equivalence relation involves the Liouville functional: -one declares $(g_1,a_1)\sim (g_2,a_2)$ if $g_2=e^{2\varphi}g_1$, and -$$ -a_2-a_1=\int_\Sigma {\textstyle\frac 1 2}(d\varphi\wedge \ast d\varphi+4\varphi R), -$$ -where $R$ is the curvature 2-form of the metric $g_1$. - -It is reasonable to believe that the restriction of the central extension $\widetilde{\mathcal A}$ to the "subgroup" $\mathit{Diff}(S^1)\subset \mathcal A$ is $\widetilde{\mathit{Diff}(S^1)}$. -But I really don't see why that's should be the case. - -Any insight? How does one relate the Bott-Virasoro cocycle to the Liouville functional?? - -REPLY [4 votes]: Let me give a method for answering the problem. I haven't yet done the relevant integral to get an actual answer. -In the setting you originally laid out for $\mathcal{A}$, actual diffeomorphisms are not included; however, you can find an annulus close to a diffeomorphism $f$ by taking an annulus of the form $[0,\epsilon] \times S^1$ with $\epsilon$ very small, and parametrizing the two sides differently: parametrize the left boundary by $x \mapsto (0,f(x))$ and parametrize the right boundary by $x \mapsto (1,x)$. Call this parametrized annulus $A_0(f)$. -$A_0(f)$ is not of the type that you apply the Liouville functional to, since the left boundary is not constant-speed geodesic. So let's construct another metrized torus where both boundaries are constant-speed. Let $B_\epsilon:[0,\epsilon] \to [0,1]$ be a bump function which is $1$ in a neighborhood of $0$ and $0$ in a neighborhood of $\epsilon$. Let $A_1(f)$ be the annulus $A_0(f)$ with the metric rescaled as -$$ -g_1(s,t) = \exp\Bigl(-2B_\epsilon(s)\ln \bigl(f'(t)\bigr)\Bigr)g_0(s,t) = -\exp\Bigl(-2B_\epsilon(s)\ln \bigl(f'(t)\bigr)\Bigr)(ds^2 + dt^2). -$$ -Then $A_1(f)$, with the same boundary parametrization as in $A_0(f)$, is now parametrized by constant-speed geodesics. Call $(A_1(f), 0) \in \widetilde{\mathcal{A}}$ the canonical annulus of the diffeomorphism $f$. -Now suppose we have two diffeomorphisms $f,g$. We can glue the associated cannonical annuli $A_1(f), A_1(g)$. The result will represent the diffeomorphism $g \circ f$; but the metric will not be the canonical metric on $A_1(g \circ f)$. We have to rescale the metric by an appropriate scaling factor $\phi$ to get to the canonical metric. (Actually, the conformal width of the annulus is also now $2\epsilon$ rather than $\epsilon$, but that shouldn't matter.) The integral in the problem statement for this rescaling presumably reduces to the Bott-Virasoro integral as $\epsilon$ gets sufficiently small. -I'd still like to understand this geometrically well enough to not have to do the integral.<|endoftext|> -TITLE: surfaces of constant centro-affine curvature -QUESTION [9 upvotes]: It is well-known that every closed surface in $\mathbb R^3$ having constant Gauss curvature is a round sphere. Inspired by this question, I'd like to ask whether a similar rigidity holds for centro-affine curvature. -More precisely, let $M\subset\mathbb R^3$ be a smooth closed convex surface (i.e., the boundary of a convex body) enclosing the origin. Its centro-affine curvature at a point $p\in M$ can be defined as -$$ -K(p)\cdot\langle p,\nu(p)\rangle^{-4} -$$ -where $K(p)$ is the Gauss curvature and $\nu(p)$ is the outer normal vector at $p$. (More generally, for a hypersurface in $\mathbb R^n$ it is $K(p)\cdot\langle p,\nu(p)\rangle^{-(n+1)}$.) -The nice thing about centro-affine curvature is that it is invariant under volume-preserving linear transformations. In particular, it is constant if $M$ is an ellipsoid centered at the origin (because such ellipsoids are equivalent to spheres). Is the converse true? In other words, is it true that every closed convex surface with constant centro-affine curvature is an ellipsoid? -Remark. For curves in $\mathbb R^2$, the condition that the centro-affine curvature is constant boils down to a second-order ODE whose solutions are ellipses and only ellipses. In $\mathbb R^3$, it is a PDE that seems to have many more solutions locally (just like in the case of constant Gauss curvature). So, if there is rigidity, it should be global only. -[EDIT] Another possible definition of centro-affine curvature is what the Legendre transform (the natural bijection between a convex body and its polar) does to the volume locally. I'm adding 'convexity' tag because some extremal properties of ellipsoids might be relevant here. - -REPLY [2 votes]: "The nice thing about centro-affine curvature is that it is invariant under volume-preserving linear transformations. In particular, it is constant if M is an ellipsoid centered at the origin (because such ellipsoids are equivalent to spheres). Is the converse true? In other words, is it true that every closed convex surface with constant centro-affine curvature is an ellipsoid?" -YES, it's true ! An elegant short proof may be found in the following paper: -Leichtweiss, Kurt On a problem of W. J. Firey in connection with the characterization of spheres. Festschrift for Hans Vogler on the occasion of his 60th birthday. Math. Pannon. 6 (1995), no. 1, 67–75.<|endoftext|> -TITLE: Measures on infinite dimensional Banach spaces -QUESTION [6 upvotes]: Does there exist a Borel measure or any valid measure on an infinite dimensional Banach space such that a bounded open set in this space has a positive measure ? - -REPLY [6 votes]: We can give a construction of a standard translation-invariant Borel measure in -$R^N$(here $N$ denotes a set of all naturall numbers), which obtains the value one on the infinite-dimensional cube $[0;1[^N$. Actually, we are -free from the demand of sigma-finiteness, because the space $R^N$ is covered by the uncountable -family of pairwise disjoint shifts of $[0;1[^N$. Measures with above-mentioned properties are adopted as partial analogs of the Lebesgue measure in the infinite-dimensional topological vector space $R^N$. Partial analogs of the Lebesgue measure in general Banach spaces are assumed as translation-invariant Borel measures which obtain the numerical value one on the unit sphere or on the standard infinite-dimensional parallelepiped ( generated by any basis ). -The fundamental works of English mathematicians C. Rogers and D. -Fremlin are devoted to problems of the existence of such measures in non-separable Banach spaces. I have considered -the following problem posed by C. Rogers (1998): -Does there exist a such translation-invariant Borel measure in $\ell^{\infty}$ which obtains the -numerical value one on the closed unite sphere?( here $\ell^{\infty}$ denotes a non-separable Banach space of all bounded real-valued sequences equipted with standard norm) -My result asserts that this question is not solvable within the the theory $ZF+DC$. -On the one hand, we can construct a "consistent" extension of the theory $ZF+DC$ where this question is solvable positivelly( such a theory is the so called "Solovay model") -On the other hand, we can construct a "consistent" extension of the theory $ZF+DC$ where this question is solvable negativelly ( such a theory is "ZF+AC+"there is no a measurable cardinal") -The proof of these facts can be found in -"G.R.Pantsulaia, On ordinary and standard products of infinite family of σ-finite measures and some of their applications -Acta Mathematica Sinica, English Series (2011) 27: 477-496, March 01, 2011" -You have mentioned that in separable Banach spaces there is no a translation-invariant Borel measure which obtain a numerical value one on the unite ball. But if we consider a question asking whether there is a translation-invariant Borel measure in a separable Banach space which obtain a numerical value one on the infinite-dimensional parallelepiped ( generated by any Markushewicz basis, in particular, by Schauder basis ) then the answer to this question is yes.<|endoftext|> -TITLE: Are there uncountably many cube-free infinite binary words? -QUESTION [21 upvotes]: In Cube-free infinite binary words it was established that there are infinitely many cube-free infinite binary words (see the earlier question for definitions of terms). The construction given in answer to that question yields a countable infinity of such words. In a comment on that answer, I raised the question of whether there is an uncountable infinity of such words. My comment has not generated any response; perhaps it will attract more interest as a question. -I should admit that I ask out of idle curiosity, and have no research interest in the answer; it just seems like the logical question to ask once you know some set is infinite. - -REPLY [7 votes]: I just saw and answered the earlier question, but perhaps I should repeat my post: -Here are some deep facts relating to binary cfw's: -1) The set of right infinite binary cube-free words is a perfect set in the topological sense: For any given such sequence, there is a distinct one which agrees with it to the nth letter. In particular, there are uncountably many binary cfw's. -2) Given any finite binary sequence, it is decidable whether it extends to an infinite binary cube-free word. -3) The number of (finite) binary cfw's of length n grows exponentially with n. -These results (and analogous ones for k-power free words over various alphabets) are proved in -http://dl.acm.org/citation.cfm?id=873885 and http://www.sciencedirect.com/science/article/pii/0195669895900519<|endoftext|> -TITLE: Is it OK for a referee to acknowledge identity with a previous referee? -QUESTION [9 upvotes]: As with this older related question, question anonymous for obvious reasons. -If I have been asked to review the same paper twice, is it OK to acknowledge in my review that I am the same person as one of the referees from last time? I'm sure cryptographers would frown on that sort of thing, but it's not clear that there is actually a practical problem with this. -Thanks! - -REPLY [2 votes]: On a couple of occasions I have been sent the same paper more than once to referee by different editors on behalf of different journals. In one occasion I even got the paper a third time. -Since I don't think it's good use of my time, not to mention that it goes against the spirit of the peer review process, to rewrite the referee report so that it may pass for report by a different referee, whenever I get a paper a second time, I disclose it immediately to the editor while sending them the previous referee report. Prior to this, of course, I check that the new version of the paper has not changed: this is some times not as easy to do as one would expect, since some authors try to disguise this by changing fonts, formats,...<|endoftext|> -TITLE: Number of certain positive integer solutions of n=x+y+z -QUESTION [11 upvotes]: Does anyone know how to estimate (as $n$ tends to infinity) the number of solutions of -$$n=x+y+z$$ -where $x,y,z$ are positive integers with $x$ coprime to $y$ and to $z$? -Computer experiments suggest that there are roughly $cn^2$ solutions, where $c>0$ is an absolute constant. - -REPLY [6 votes]: For $n$ prime my heuristics tells me that $c=\frac{1}{2}\prod_p\left(1-\frac{2}{p^2}\right)$, the product being over all primes. Is this supported by computer experiments? If yes, I will share more details.<|endoftext|> -TITLE: Inverse Image as the left adjoint to pushforward -QUESTION [9 upvotes]: This is a repost of a question on Math stackexchange. No one is biting at it there, so I guess it is harder than I thought. -Assume $X$ and $Y$ are topological spaces, $f : X \to Y$ is a continuous map. Let ${\bf Sh}(X)$, ${\bf Sh}(Y)$ be the category of sheaves on $X$ and $Y$ respectively. Modulo existence issues we can define the inverse image functor $f^{-1} : {\bf Sh}(Y) \to {\bf Sh}(X)$ to be the left adjoint to the push forward functor $f_{*} : {\bf Sh}(X) \to {\bf Sh}(Y)$ which is easily described. -My question is this: Using this definition of the inverse image functor, how can I show (without explicitly constructing the functor) that it respects stalks? i.e is there a completely categorical reason why the left adjoint to the push forward functor respects stalks? - -REPLY [5 votes]: Here is a pretty messy proof that is not categorical. I recently solved this problem and decided to share my solution. I apologize if this is not what you are looking for. -Some notation: - -$X,Y$ will be topological spaces, and $f:X\to Y$ a continous map. We will denote $F$ (resp. $G$) to be a sheaf on $X$ (resp. $Y$). -$U$ (resp. $V$) will denote an open set in $X$ (resp. $Y$). -We will denote $\overline{f}G$ to be the presheaf on $X$ given by $\overline{f}G(U) = \lim ~ G(V)$, where $V\supseteq f(U)$. - -If $P$ is a presheaf on $X$, we use the notation $(U,s)$, where $U\ni x$ and $s\in P(U)$, to denote the image of $s\in P(U)$ in the stalk $P_x$ at the point $x$. Thus, $(U,s) = (U',s')$ (where $U'\ni x$) if and only if there is an $U''\ni x$ and $s''\in P(U'')$ such that $U''\subseteq (U\cap U')$ and $s|_{U''} = s'|_{U''} = s''$. \ -Let $P^+$ be the sheafification of $P$, so it comes equipped with a morphism $P\to P^+$. For $s\in P(U)$, we define $\tilde{s}$ to be the image of $s\in P(U)\mapsto \tilde{s}\in P^+(U)$. The important fact for us is that if $a\in P^+(U)$, then for any $x\in U$, there is an $U_x\ni x$, and $s\in P(U_x)$, such that $a|_{U_x} = \tilde{s}$. \ -Therefore, if have a sheaf $H$ on space $X$, and two morphisms $\varphi,\varphi':P^+\to H$, such that $\varphi(U)(\tilde{s}) = \varphi'(U)(\tilde{s})$, for all $s\in P(U)$, and all $U$, then $\varphi = \varphi'$. \ -Now we let $F$ be a sheaf on $X$, with $f:X\to Y$ a continuous map. For an open set $U$ in $X$, $\overline{f}f_*F(U)$ is the direct limit of $F(f^{-1}V)$, where $f^{-1}V\supseteq U$. We denote $(f^{-1}V,s)$ to be the image of $s\in F(f^{-1}V)$ in its direct limit. Hence $\widetilde{(f^{-1}V,s)}\in f^{-1}f_*F(U)$. The sheaf morphism $f^{-1}f_* F\to F$ we denote by $\alpha$. The map $\alpha(U):f^{-1}f_* F(U) \to F(U)$ satisfies the property that $\widetilde{(f^{-1}V,s)} \mapsto s|_U$, where $V$ is an open set in $Y$, and $f^{-1}V\supseteq U$, and $s\in f_*F(V) = F(f^{-1}V)$. \ -For a sheaf $G$ on $Y$, we denote the sheaf morphism $G\to f_*f^{-1}G$ by $\beta$. First, we have a presheaf morphism $\beta':G\to f_*\overline{f}G$ where $\beta'(V)$ is given by $s\mapsto (V,s)$, here we think of $(V,s)$ as a representative in the direct limit $G(W)$ where $W\supseteq f(f^{-1}V)$. And we have a presheaf morphism $\mu: f_*\overline{f}G\to f_*f^{-1}G$ where, $\mu(V)(s) = \widetilde{(f^{-1}V,s)}$. Thus, $\beta(V)$ satisfies the property $s\mapsto \widetilde{(f^{-1}V,s)}$.\ -If we start with a morphism $\psi:G\to f_*F$, then we obtain a morphism $f^{-1}G\to F$ by composition $\alpha\circ f^{-1}\psi$. The map $(\alpha\circ f^{-1}\psi)(U)$ satisfies the property that $\widetilde{ (f^{-1}V,s) }\mapsto \psi(V)(s)|_U$. \ -If we start with a morphism $\varphi:f^{-1}G\to F$, then we obtain a morphism $G\to f_*F$ by composition $f_*\varphi\circ \beta$. The map $(f_*\varphi\circ \beta)(V)$ satisfies the property that $s\mapsto \varphi(f^{-1}V)\widetilde{(f^{-1}V,s)}$. \ -Hence, given a morphism $f^{-1}G\to F$ we can obtain $G\to f_*F$, and given a morphism, $G\to f_*F$ we can obtain $f^{-1}G\to F$. We will show these two transformations are inverses of one another, completing the proof. \ -Let us start with $\varphi: f^{-1}G\to F$. Construct $\psi:G\to f_*F$ and then construct its corresponding morphism $\varphi':f^{-1}G\to F$. We have to show that $\varphi = \varphi'$. Choose an open set $U$ (in $X$). We will show that $\varphi(U) = \varphi'(U)$. The map $\varphi'(U)$ sends $\widetilde{(f^{-1}V,s)} \mapsto \psi(V)(s)|_U$, where $V$ is an open set in $Y$ for which $f^{-1}V \supseteq U$ ($\iff V\supseteq f(U)$). But $\psi(V)(s) = \varphi(f^{-1}V)(\widetilde{f^{-1}V,s})$. As sheaf morphisms are consistent with restrictions and $U\subseteq f^{-1}V$ we have that, $\psi(V)(s)|_U = \varphi(U)(\widetilde{f^{-1}V,s})$. Therefore, $\varphi(U)$ and $\varphi'(U)$ both send $\widetilde{(f^{-1}V,s)}$ to the same section in $F(U)$. This is enough to complete the proof.\ -Now we consider the case $\psi:G\to f_*F$. Construct $\varphi:f^{-1}G\to F$ and then construct its corresponding morphism $\psi':G\to f_*F$. We have to show that $\psi = \psi'$. Choose an open set $V$ (in $X$). We will show that $\psi(V) = \psi'(V)$. The map $\psi'(V)$ sends $s\in G(V)$ to $\varphi(f^{-1}V)\widetilde{(f^{-1}V,s)}$. But $\varphi(f^{-1}V)\widetilde{(f^{-1}V,s)} = \psi(V)(s)|_V = \psi(V)(s)$. Hence, $\psi'(V) = \psi(V)$.<|endoftext|> -TITLE: Local methods in algebraic number theory -QUESTION [5 upvotes]: I'm currently reading about local and global fields in number theory. I have trouble seeing the point or exactly how they help answer questions about e.g. number fields. To be more specific: - -What specifically gets simplified when we move to the local case? -Is there some specific "theme" when using local methods? To be more specific: What type of results can usually be reduced to the local case? What type of results generally can't? -What are the standard methods for lifting local results to global? - -My current problem seems to be that I can't see the forest for the trees. - -REPLY [8 votes]: The completion of a number field at a prime $p$ is a principal ideal domain with a single -maximal ideal. The unit group has a relatively simple structure. Thus two of the main -culprits for making life difficult in number fields disappear locally. In addition, -the difficult part of the Galois group, namely the decomposition group, also disappears -in the local case. -Losely speaking, questions concerning a single prime ideal (inertia group, ramification -groups, \ldots) have a good chance of making sense locally. Global questions, like the -quadratic reciprocity law, don't, at least not in the usual formulation. -The simplest tool is Hensel's Lemma. Other than that I would not speak of a method for -"lifting" results from local to global. What you do is compare the global result with -the collection of all local results, and this is highly specific to the problem you're -looking at. In some cases, like the Kronecker-Weber theorem (see Cassels' book on -local fields), the problems are easily overcome, in others (embedding problems in Galois -theory) they're not.<|endoftext|> -TITLE: Cohomological functor from triangulated category -QUESTION [12 upvotes]: Say we have a cohomological functor F from a triangulated category $C$ to the category $Ab$ of abelian groups, e.g. $F=Hom(x,-)$, where x is an object in $C$. By definition, such a functor transform exact triangles into long exact sequence. And $F(y[i])$ is like the i-th "cohomology group" of the object $y$ w.r.t. the functor F. However, according to the philosophy of derived category, the right thing to look at is the complex, instead of the "cohomology groups". My question is, could there be a complex that "computes" these "cohomology groups"? e.g. Given two objects $x, y$ in $C$, can we find a complex that's like $RHom(x,y)$ whose cohomology computes $Ext_C^i(x,y):=Hom_C(x,y[i])$? -I would guess in full generality the answer is negative. (Otherwise it should have been done by Verdier.) But is there any mild or restrictive assumption that makes this true? -Or just for cohomological functors of the form $Hom_C(x,-)$, can we make the triangulated category $C$ enriched over $D^b(Ab)$ (such that the cohomology of the $Hom$ complex computes the old $Ext^i$'s?) -================ -When I was writing this I feel like triangulated categories might not a good place to play game like this... If the above question isn't too interesting, could anyone just tell me what's the nature playground of questions like this? [Do dg-categories or $A_\infty$-category have an advantage on this?] Thanks. - -REPLY [9 votes]: First, consider the category $\mathcal{S}$ of spectra. Let $S/2$ denote the cofibre of twice the identity map of the sphere spectrum. It is then known that the identity map of $S/2$ has order $4$. If we had a good system of chain complexes $Hom_{\mathcal{S}}(x,y)$ then $Hom_{\mathcal{S}}(S/2,S/2)$ would be the cofibre of twice the identity on $Hom_{\mathcal{S}}(S/2,S)$, but for chain complexes the cofibre of twice the identity always has exponent 2, which gives a contradiction. Thus, for triangulated categories of topological origin, the best you can hope for is to define mapping spectra $F(x,y)$, not chain complexes $Hom(x,y)$. -Next, this paper: - -Fernando Muro, Stefan Schwede, Neil Strickland, Triangulated categories without models, Inventiones Mathematicae, Vol. 170 (2007), No. 2, pp. 231-241, https://doi.org/10.1007/s00222-007-0061-2, https://arxiv.org/abs/0704.1378 - -exhibits a triangulated category $\mathcal{C}$ such there are no nontrivial exact functors from $\mathcal{C}$ to any of the standard examples of triangulated categories, or in the opposite direction. This means in particular that there is no way to define a 'representable' functor $F_{\mathcal{C}}(x,-)$. -However, if $\mathcal{D}$ is a triangulated category arising from a stable model category then one can define mapping spectra $F_{\mathcal{D}}(x,y)$ for objects $x,y\in\mathcal{D}$. I think that is covered by this paper: - -Stefan Schwede, Brooke Shipley, A uniqueness theorem for stable homotopy theory, Mathematische Zeitschrift 239 (2002), 803-828, https://doi.org/10.1007/s002090100347, https://arxiv.org/abs/math/0012021. - -As in David Roberts's comment, I think one can also do the same for triangulated categories that arise from stable $(\infty,1)$-categories.<|endoftext|> -TITLE: Residual Finiteness of Fundamental Group of Compact 3-Manifold -QUESTION [6 upvotes]: Hempel, in his 1987 article "Residual Finiteness for 3-Manifolds", shows that if $M$ is a compact Haken 3-manifold, then $\pi_1(M)$ is residually finite. The outline of the proof is basically: - -Reduce to the case $M$ is closed and irreducible; -Use the JSJ decomposition to split $M$ along tori, getting pieces which are either - -Seifert fibered spaces or -hyperbolic 3-manifolds, via Thurston's hyperbolization theorem. - -Realize $\pi_1(M)$ as the fundamental group of a graph of groups, where the vertex groups come from the Seifert fibered spaces and hyperbolic 3-manifolds, and the edge groups come from the tori we split along. -Show all these vertex groups are essentially linear, and the $\mathbb{Z}\times\mathbb{Z}$ edge groups correspond to maximal unipotent subgroups. -Conclude, using the main theorem of the paper, that $\pi_1(M)$ is residually finite. - -I was wondering if, via the geometrization conjecture, this basic outline now works for any compact 3-manifold $M$. I guess my question is really two (related) questions: - -Does Hempel's argument essentially work in the general case? Are there significant changes necessary? - -and - -If this is correct, are there different arguments? (I actually really like Hempel's proof, but maybe it's not "in" right now.) - -Thanks, Steve - -REPLY [7 votes]: The second question is very similar to this question of Igor Belegradek. -Hempel's argument is still the only known one. It's the 'obvious' way to prove residual finiteness using geometrisation. To see how hard it would be to prove residual finiteness without geometrisation, imagine trying to prove the following much weaker statement with your bare hands: - -Every non-simply connected 3-manifold has a proper finite-sheeted covering space.<|endoftext|> -TITLE: What is the general geometric interpretation of modules in algebraic geometry? -QUESTION [35 upvotes]: Algebraic geometry is quite new for me, so this question may be too naive. therefore, I will also be happy to get answers explaining why this is a bad question. -I understand that the basic philosophy begins with considering an abstract commutative ring as a function space of a certain "geometric" object (the spectrum of the ring). I also understand that at least certain types of modules correspond to well known geometric constructions. For example, projective modules should be thought of as vector bundles over the spectrum (and that there are formal statements such as the Serre-Swan theorem which make this correspondence precise in certain categories). My question is, what is the general geometric counterpart of modules? -This is not a formal mathematical question, and I am not looking for the formal scheme-theoretic concept (of sheaves of certain type and so on), but for the geometric picture that I should keep in mind when working with modules. -I will appreciate any kind of insight or even just a particularly Enlightening example. - -REPLY [6 votes]: These somewhat naive remarks are motivated by the desire to think about the question myself. -Sections of bundles are natural examples of (locally) free modules and the standard examples are the tangent and cotangent bundles of smooth varieties. The need for more general modules is the fact that kernels and cokernels of maps of (locally) free modules need not be so. In addition to ideal sheaves of subvarieties, as noted above the kahler differentials are a standard example of not necessarily locally free sheaves. -These two remarks are connected by the standard exact sequence of a closed subscheme, which exhibits the sheaf of kahler differentials of a local complete intersection, as the cokernel of a map of locally free sheaves. I,.e, the kahler differentials on a closed subvariety of a smooth variety, are the quotient of the kahler differentials on the ambient variety by the conormal sheaf of the subvariety. -Thus, the standard examples of modules are ideals of subvarieties and kahler differentials, and at least for lci subvarieties these arise as either kernels or cokernels of maps of bundles. So geometrically one could say one wants to consider differentials and subschemes, and algebraically one wants to be able to take kernels and cokernels.<|endoftext|> -TITLE: Classification of discrete subgroups of the unitary group -QUESTION [5 upvotes]: Let $U(n)$ be the unitary group. From André Weil's paper "On discrete subgroups of Lie groups" it is well known that discrete cocompact subgroups of $U(n)$ have only a finite number of generators and relations. -Does there exist a full classification of discrete subgroups of the unitary group up to isomorphism? - -REPLY [4 votes]: $U(n)$ is compact. So any discrete subgroup of $U(n)$ is automatically (i) cocompact and (ii) finite. If you are interested in the classification of finite subgroups of $U(n)$, then the main result is Jordan's theorem: There is an integer $J(n)$ such that any finite subgroup of $U(n)$ has a normal abelian subgroup of index $\leq J(n)$. See Terry Tao's blog for a nice exposition. There is a small industry of improving the bounds for $J(n)$; this paper -has some good bounds. -Is this what you wanted to ask?<|endoftext|> -TITLE: Are there any finitely generated artinian modules that are not Noetherian? -QUESTION [10 upvotes]: It is well known that for rings, Artinian implies Noetherian (the famous Hopkins–Levitzki theorem) and it is also well known that there are Artinian modules which are not Noetherian. A simple example can be found in -http://en.wikipedia.org/wiki/Artinian_module#Relation_to_the_Noetherian_condition -Since rings are always finitely generated modules over themselves (all rings considered are unital), it seemed natural to me to ask whether there are finitely generated modules, which are Artinian but not Noetherian (the example given in the reference is clearly not finitely generated). I guess that if the statement "every finitely generated artinian module is noetherian" was true, I would have seen it in any standard text book on algebra, and since I haven't, I guess it's not. But still, I can't find a counter-example for this. Perhaps I'm missing something completely trivial here. I will be happy to see an example of such module or a proof that there are no (just a reference will be much appreciated too of course). - -REPLY [20 votes]: Suppose you have an Artinian but not Noetherian finitely generated $R$ module $M$. Let $0\leq M_1\leq M_2\leq \cdots \leq M_n=M$ be a finite chain of $R$-modules such that each composition factor $M_i/M_{i-1}$ is cyclic for each $i$. -Certainly each composition is Artinian since subquotients of Artinian modules are Artinian. Also one of the composition factors must be non-Noetherian since extensions of Noetherian modules by Noetherian modules are Noetherian. Thus, we may assume that $M$ is a cyclic $R$-module. -Now if $R$ is commutative, $M$ is a quotient ring $R/I$ which is Artinian as such and so Noetherian also, as you say. -If $R$ is non-commutative then I'm not sure what the answer is. - -Added: It seems from the wikipedia article linked from the question that Hartley showed that there are cyclic Artinian and non-Noetherian modules over certain non-commutative rings and Cohn gave another construction nearly twenty years later. See the links I give in the comments on the question for precise references.<|endoftext|> -TITLE: Picard group of Schubert varieties -QUESTION [8 upvotes]: Let $G$ be a semisimple linear algebraic group, $P$ be a parabolic subgroup and $w$ be an element of the Weyl group of $G$. I want to calculate the Picard group of the Schubert variety $X_P(w):=\overline{BwP/P} \subset G/P$. I'm particularly interested in the case of a maximal parabolic subgroup, but I suppose the general case is not much harder. In that case, I read without proof, that $\text{Pic}(X_P(w)) \cong \mathbb{Z}$ (of course if $w$ is non trivial). I would also hope, that there is a very ample generator in that case? -I have no problems to calculate the divisor class group, which is freely generated by the divisorial Schubert subvarieties. So the problem lies mainly in the non-smooth case. -A good reference on this topic would also be nice. - -REPLY [5 votes]: Alex Yong and I work this out in our paper for the case of the Borel subgroup, but I'm pretty sure it's the same for every parabolic. -When is a Schubert variety Gorenstein?, Advances in Math. 207 (2006), 205-220. -Please note our conventions in that paper are backwards from yours in that our Schubert varieties are $\overline{B_-wB/B}$. -We don't say explicitly what happens for groups other than GL_n, but you can do the same thing using the appropriate Monk-Chevalley formula for the group in question. -EDIT: More details upon glancing at my own paper... Mathieu (reference in comments) shows that every line bundle on a Schubert variety is the restriction of a line bundle on the homogeneous space. (Actually, iirc the proof in the finite dimensional case predates Mathieu.) This means the Picard group of the Schubert variety will be the same as that of the homogeneous space unless some nontrivial line bundle restricts to a trivial one. For $G/P$ where $P$ is a maximal parabolic, this only happens if your Schubert variety is a point. -What we do is actually work out the Picard group as a subgroup of the class group.<|endoftext|> -TITLE: equivariant index of Dirac Operator on $S^{2}$ -QUESTION [5 upvotes]: First, I have to admit that I don't have much knowledge of Spin Geometry and Index Theory, the question could be too simple or naive and secondly there may be too many questions. -Let $D$ be the Dirac Operator for standard metric and $S$ be the Spin bundle on $S^{2}$.There is a unique Spin structure on $S^{2}$. How does $D$ look like, Can we write a general form for the harmonic spinors on $S^{2}$ ? -What is the general expression for equivariant index of $D$ ? -If $W \times S$ is the twisted spinor bundle and $D$ is the twisted Dirac operator, we can write the equivariant index as an integral over the fixed point manifold in terms of equivariant Chern Character of $W$ and $\hat{A}$ - genus of the fixed point manifold using the Aiyah-Segal-Singer theorem. What I am interested in is the final expression for $S^{2}$. -What can we say about the product of n $S^{2}$'s. -Thanks - -REPLY [3 votes]: As Sebastian said, the Dirac operator can be identified with the $\bar{\partial}$-operator on the line bundle of degree $-1$ (the inverse of the Hopf bundle) and so its (equivariant) index is trivial. -Now you ask for the equivariant index of the twisted Dirac. The reasonable group to study equivariance is $SU(2)$, because $SO(3)$ does not act on the spinor bundle. First note the the $K$-group of equivariant vector bundles on $S^2$ is $K_{SU(2)} (S^2) = KU_{SU(2)} (SU(2)/U(1)) \cong K_{U(1)} (*) = RU(1)$. Therefore, any $SU(2)$-equivariant vector bundle splits as a sum of line bundles, and the line bundles are precisely all powers of the Hopf bundle. -So this reduces the problem to the study of the kernel of $\bar{\partial}$ on any power of the Hopf bundle, but as a $SU(2)$-representation. -It is a pleasant exercise to prove that the space of holomorphic sections on the $k$th power of the Hopf bundle is canonically isomorphic to the space of degree $k$ homogeneous polynomials in two variables and the $SU(2)$-action on that is given by the action on the variables. -So you get all irreducible representations of $SU(2)$ as the kernel of a twisted equivariant Dirac operator. -This is of course not an accident, and you can read more about this close connection between index theory and representation theory in the papers (I gues that you now some of them): -G. Segal: ''Equivariant $K$-theory'', ''The representation ring of a compact Lie group'' -R. Bott: ''Homogeneous vector bundles'' -M. Atiyah, R. Bott: ''Lefschetz fixed point formula of elliptic complexes''<|endoftext|> -TITLE: Spin TQFT's in dimensions (1+1) -QUESTION [13 upvotes]: I don't seem to be able to find anything written about Spin TQFT's in dimension (1+1). Does anyone know any references? Or is there some reason it is uninteresting? - -REPLY [9 votes]: This is covered in Moore and Segal "D-branes and K-theory in 2D topological field theory". In particular on around page 16 there is a characterization analogous to "1+1 TQFTs = Commutative Frobenius algebras".<|endoftext|> -TITLE: Symmetric functions and regularity -QUESTION [5 upvotes]: Let $f:\mathbb R^2\rightarrow\mathbb R$ be a symmetric function: $f(y,x)=f(x,y)$. It can therefore be written has a function of the elementary symmetric polynomials, here $f(x,y)=F(x+y,xy)$, where $F(\sigma,\mu)$ is defined over $4\mu\le\sigma^2$. -Let me assume that $f$ is of class $\mathcal C^r$. It is clear from the IFT that $F$ is $\mathcal C^r$ too, away from the critical line $4\mu=\sigma^2$. - -What is the regularity of $F$ at the boundary ? My guess is $\mathcal C^{r/2}$ but I have been unable to prove it or to locate such a result. - -For instance, if $f(x,y)=|y-x|^r$ and $r>0$ is not an integer, then $F=|\sigma^2-4\mu|^{r/2}$ is only $\mathcal C^{r/2}$ and not more. - -REPLY [7 votes]: Just note that $F(\sigma,\mu)=f\left(\frac{\sigma+\sqrt{\sigma^2-4\mu}}{2}, \frac{\sigma-\sqrt{\sigma^2-4\mu}}{2}\right)$.<|endoftext|> -TITLE: Non-Standard Prime -QUESTION [9 upvotes]: Hello, -My question is about the non-standard models of the integers. If we add to the Peano's axioms $P$ of arithmetic the following axioms for a fixed constant $c$: -$c \neq 0$, $c \neq 1$, $c \neq 1+1$, $c \neq 1+1+1$, etc... -and $c=ab \implies a=1~~ou~~b=1$. We obtain a system of axioms $S$. -$S$ is consistent, by compacity. If $S$ is not consistent, a finite number of axioms in $S$ (a subset $S'$ of $S$) are not consistent, say axioms of $P$ and $c \neq 0, c \neq 1, c \neq 1+1, ...,c \neq \underbrace{1+1+1+...+1+1}_{k ~~\times}$. So we can consider a prime $p$ greater than $k$. We consider the standard model of $\mathbb{N}$ and we put $c=p$, to obtain a model of $S'$. Hence, $S'$ is consistent. Contradiction. So, $S$ is consistent. -$c$ is prime in a model $M$ of $S$, and $c$ is a non-standard integer. We can consider the field $F=\{x \in M | x< c\}$ obtained by setting -$a~+_F~ b=(a~+_M~b)\mod c$, -$a~\times_F~ b=(a~\times_M~b)\mod c$. We have an inverse for $a$ if $a\mod c \neq 0$. -$F$ is an infinite field. Which field is it isomorphic to ? Is $F$ algebraic over $\mathbb{Q}$ ($\mathbb{Q}$ is included in the field $F$ ) ? -Thanks in advance. - -REPLY [7 votes]: The residue fields $F=M/cM$ of nonstandard primes in models of Peano arithmetic have interesting properties which were investigated by Macintyre. In particular, every such field is pseudofinite (i.e., an infinite model of the first-order theory of finite fields, in other words: a pseudo-algebraically closed field having exactly one extension of each finite degree).<|endoftext|> -TITLE: Semidirect products and PROPs -QUESTION [6 upvotes]: $\newcommand{\p}{\mathcal{P}}$Let G be a group, and let $G_n$ denote the wreath product $G^n \rtimes \Sigma_n$. -There seems to be a notion of a PROP $\p$ where the role of the symmetric groups $\Sigma_n$ is instead played by the groups $G_n$. An example would be $G=SO(N)$ and $\p$ the PROP associated to the framed little N-disc operad. Here $\p(1,n)$ has an action of $G_1^{op} \times G_n$ -- the copy of $G_1^{op}$ rotates the entire disc "counterclockwise" (i.e. an element $g \in SO(N)$ acts via $g^{-1}$), and the action of $G_n$ on $\p(1,n)$ is the evident one. The gluing maps are then suitably equivariant under this simultaneous action of $G$ on the input/output legs, so to speak. -Is there a standard name for this kind of PROP/operad? Cf. how one calls an operad where $\Sigma_n$ has been replaced by $B_n$ a braided operad. I would also be happy to hear of any paper where this kind of gadget has been defined and/or studied. - -Addendum, Jan 23 2013. Sorry if it is poor form to bump an inactive question only to advertise your own work, but I just ran across this old question. When I had thought some more about this I realized eventually that what I described above is really a special case of the notion of a colored PROP/operad, except the collection of colors do not form a set but a category. In this case there is only one color but its automorphism group is $G$, that is, the collection of colors is exactly the one-object category corresponding to the group $G$. -The notion of a PROP/operad which is colored by a category is defined in my paper http://arxiv.org/abs/1205.0420 . This is actually somewhat more general than what Salvatore and Wahl define, even when the category in question is a group. - -REPLY [3 votes]: Dear Dan, -You can find this notion defined in the paper "Framed discs operads and the equivariant recognition principle" by Nathalie Wahl and Paolo Salvatore: http://arxiv.org/abs/math/0106242 . .<|endoftext|> -TITLE: Radical of projection equals projection of radical? -QUESTION [5 upvotes]: Given an Lie Algebra $L$ (of finite dimension and over an algebraically closed field with zero characteristic) and an ideal $I$, is it truth that -$rad\left(\dfrac{L}{I}\right)= \pi(rad(L))$, -where $\pi$ is the projection? -What happens if $L$ isn't finite dimensional? -And if the field has positive characteristic? - -REPLY [2 votes]: A counter-example for the infinite dimensional case: -Let $V$ be a vector space with a basis $\{e_i\mid i\in \mathbb{Z}_+\}$ and $V_p=\langle e_i\mid i=1,\dots,p\rangle$. Consider -$$\mathfrak{t}_\infty:= \{ x \in \mathfrak{gl}(V)\mid x(V_i)\subset V_i \} .$$ -as a lie subalgebra of $\mathfrak{gl}(V)$. -Let $E_{ij}$ be the linear transformations on $V$ such that $E_{ij}(e_j)=e_i$ and $E_{ij}(e_p)=0$ if $p\neq j$. Then, we have the formula -$$[E_{ij},E_{kl}]=\delta_j^kE_{il}-\delta_l^iE_{kj}.$$ -Convince yourself that -$$\mathfrak{h}:=\{x \in t_\infty \mid x(V) \subset \langle e_i\mid i>1\rangle \}$$ -is an ideal of $\mathfrak{t}_\infty$. -So, we have that -$$rad( \mathfrak{t}_\infty / \mathfrak{h})=$$ -$$\langle E_{1j} \mid j>1\rangle / \mathfrak{h}.$$ -Suppose, by absurd, that -$$\pi ( rad(\mathfrak{t}_\infty) ) = $$ -$$rad( \mathfrak{t}_\infty / \mathfrak{h}) =$$ -$$\langle E_{1j}\mid j>1\rangle / \mathfrak{h}.$$ -This implies that -$$\langle E_{1j}\mid j>1\rangle + \mathfrak{h} = $$ -$$rad(\mathfrak{t}_\infty) + \mathfrak{h}.$$ -Therefore, -$$\langle E_{1j}\mid j>1\rangle + \mathfrak{h} \ni E_{12}=$$ -$$x + y \in rad(\mathfrak{t}_\infty) + \mathfrak{h}$$ -for some $x \in rad(\mathfrak{t}_\infty)$ and $y \in \mathfrak{h}$. Then -$$rad(\mathfrak{t}_\infty) \ni $$ -$$[E_{11},x]=$$ -$$[E_{11},E_{12}]-[E_{11},y]=$$ -$$E_{12}+0=E_{12}.$$ -This, again by the formula given above, implies that -$$rad(\mathfrak{t}_\infty)$$ -contains -$$t_\infty ':=\{x \in \mathfrak{t}_\infty\mid x(V_1)=0\textrm{ and } x(V_i)\subset V_{i-1}, i>1\}$$ -that is a non-solvable ideal of $\mathfrak{t}_\infty$. A contradiction.<|endoftext|> -TITLE: Multiplicative Convolution for Binomial Coefficients -QUESTION [9 upvotes]: I know Vandermonde's convolution for binomial coefficients: -$$\sum_{j=0...k} \binom{n}{j} \binom{m}{k-j} = \binom{n+m}{k}$$ -Is there a similar multiplicative convolution? More precisely, is there a simple formula for the coefficients $a(k_j,k'_j,k)$ in the following -identity? -$$\sum_{j} a(k_j,k'_j,k) \binom{n}{k_j} \binom{m}{k'_j} = \binom{n \cdot m}{k}$$ -Thanks. - -REPLY [10 votes]: Riordan and Stein, in "Arrangements on Chessboards" (Journal of Combinatorial Theory, Series A, 12 72-80, 1972) consider the numbers $A(r,s,k)$ defined by -$$\sum_{r,s} \binom{n}{r} \binom{m}{s} A(r,s,k) = \binom{nm}{k},$$ or, as others have pointed out, the number of $r \times s$ $(0,1)$-matrices with $k$ $1$'s and with at least one $1$ in every row and every column. They obtain two formulas for these numbers: -$$A(r,s,k) = \sum_{i,j} (-1)^{i+j+r+s} \binom{r}{i} \binom{s}{j} \binom{ij}{k},$$ -$$A(r,s,k) = \sum_n \frac{r! s!}{k!} S(n,r) S(n,s) s(k,n),$$ -where $S(n,r)$ and $s(k,n)$ are Stirling numbers of the second and first kinds, respectively. -They also obtain the recurrence relation -$$(k+1)A(r,s,k+1) +k A(r,s,k) $$ -$$= rs \left(A(r,s,k) + A(r-1,s,k) + A(r,s-1,k) + A(r-1,s-1,k)\right)$$ -and that $A(r,s,k)$ is the coefficient of $t^k$ in -$$\sum_j (-1)^{s+j} \binom{s}{j} \left( (1+t)^j-1) \right)^r,$$ -as in Richard Stanley's answer.<|endoftext|> -TITLE: If $0^{\sharp}$ exists, then every uncountable cardinal in $V$ is as large as it can be in $L$. -QUESTION [5 upvotes]: According to Wikipedia, if $0^{\sharp}$ exists, then every uncountable cardinal in $V$ satisfies every large cardinal property in $L$ that can be realized in $L$, e.g. weak compactness, total ineffability, etc. It's easy enough to see why every uncountable in $V$ will be inaccessible, or even Mahlo, in $L$. -How can one see that some of the slightly larger large cardinal properties (e.g. weak compactness, total ineffability, etc.) are satisfied in $L$ by the uncountable cardinals in $V$? Is there a good reference for some of these results? - -REPLY [9 votes]: If $0^\sharp$ exists, then every uncountable cardinal -$\kappa$ of $V$ is one of the Silver indiscernibles in $L$, -and this implies that $L_\kappa$ is an elementary substructure of $L$. This implies -that $\kappa$ is a limit cardinal in $L$ and therefore, -since some of the indiscernibles are regular, that $\kappa$ -is inaccessible in $L$. Every order-preserving map on the -indiscernibles induces an elementary embedding $j:L\to L$, -and thus every indiscernible is the critical point of such -a $j$. From this, it follows that every such $\kappa$ has -the tree property in $L$, because if $T$ is any -$\kappa$-tree in $L$, then $j(T)$ has nodes on the -$\kappa$-th level, which gives you a $\kappa$-branch in -$T$. But being inaccessible and having the tree property is -equivalent to being weakly compact, so every such $\kappa$ -is weakly compact in $L$. You can get other properties by -arguing with the embedding like this.<|endoftext|> -TITLE: $p$-adic Langlands correspondence -QUESTION [22 upvotes]: Basic question: Is it correct that the $p$-adic Langlands correspondence is known for $GL_2$ only over $Q_p$ but not other $p$-adic fields? If so, I would like to request some light to be shed on this restriction, i.e., why only $Q_p$ but not its extensions. Any reference for this (and prospects) would be much appreciated. -Thanks! - -REPLY [18 votes]: Regarding prospects for extending the correspondence to $GL_2(F)$ for other $F$, -one could look at Paškūnas's papers "Coefficient systems and supersingular representations", "Towards a modulo $p$ Langlands correspondence for $GL_2(F)$" (joint with C. Breuil), -and "Admissible unitary completions of locally $\mathbb Q_p$-rational representations of -$GL_2(F)$", available on his website and/or the arXiv. -There is also Breuil's ICM talk from last summer, "The emerging $p$-adic Langlands program", available at his website. -This gives a very nice survey of the whole state of the theory (which has remained relatively stable since then). - -Some commentary on Paškūnas's papers: In the $GL_2(\mathbb Q_p)$ case, Breuil found that the numbers of irred. supersingular reps. of $GL_2(\mathbb Q_p)$ mod $p$ matches with the numbers of $2$-dim'l irred. mod $p$ reps. of $G_{\mathbb Q_p}$, and that there is even a natural way to match them (which is e.g. compatible with Serre's conjecture on weights of modular forms giving rise to mod $p$ global Galois reps.). -It was then natural to conjecture that the same was true for $GL_2(F)$. -The first of these papers has the goal of verifying this conjecture. Indeed, -it succeeds in constructing the right number -of supersingular reps. mod $p$ of $GL_2(F)$. However, it was later realized that there was -no way to match these with irred. Galois reps. in any way that is compatible with the Buzzard--Diamond--Jarvis (BDJ) conjecture (the generalization of Serre's conjecture to Hilbert modular forms). -The second paper extends the techniques of the first, and shows in fact that when -$F \neq \mathbb Q_p$ there are many, many more supersingulars than there are $2$-dim'l. -irreps of $G_F$. It attempts to find order among this chaos by identifying certain classes of supersingulars which seem to have something to do with the Galois side (in the sense -that they match with the predictions of the BDJ conjecture). -The third paper shows how to lift mod $p$ representations to $p$-adic Banach spaces -representations in interesting ways, and so can be thought of as (i) giving some evidence -that there will be a $p$-adic local Langlands for $GL_2(F)$, but also (ii) showing that -understanding it will be at least as difficult as understanding the mod $p$ situation.<|endoftext|> -TITLE: What is the homotopy theory of categories? -QUESTION [18 upvotes]: I've heard that Grothendieck, in his letter "Pursuing Stacks," wanted to find alternative models for the classical homotopy category of CW complexes and continuous maps (up to homotopy), and one of his proposed ideas was a "homotopy theory of categories." What does this mean, precisely? -I know that any category corresponds to a simplicial set (its nerve), and an equivalence of categories introduces a homotopy equivalence (in the category of simplicial sets) of the associated nerves. I also know that there is a characterization of (the nerves of) categories among simplicial sets in terms of a unique filler extension condition. If this extension condition is weakened, so that one gets the notion of a quasicategory or $\infty$-category, one can obtain a model structure where the quasicategories are the fibrant objects. -But if we want to just work with ordinary categories, is there a natural model structure on simplicial sets in which they are the fibrant objects? And if so, is this Quillen equivalent to the (Quillen/Serre) model structure on topological spaces? - -REPLY [2 votes]: The first question was already answered David Roberts and Jonathan Chiche. Let me address the second one. It's not reasonable to expect that such a model structure exists. We can ask instead whether there is a model structure on simplicial sets in which nerves of fibrant categories (in Thomason's model structure) are fibrant. And, in fact, such a model structure does exist. We can just transfer the Quillen model structure on simplicial sets along the double subdivision functor. Note that the Thomason's model structure can be now transferred from this new model structure. Also, this model structure on simplicial sets presents the homotopy theory of spaces since weak equivalences in it are just ordinary weak equivalences. -Now, just for completeness, let me answer the question in the post. Of course, there are are a lot of model structures on simplicial sets in which the nerves of categories are fibrant. For example, we can take the left Bousfield localization of the Joyal model structure with respect to the maps $\Delta^n \amalg_{\Lambda^n_k} \Delta^n \to \Delta^n$. The trivial model structure also satsfies the conditions. These model structures do not present the homotopy theory of spaces. If we want to keep the class of weak equivalences the same as in the Quillen model structures, then I can show that there is no model structure in which nerves of categories are fibrant. Actually, I will prove a stronger statement: - -If there is a model structure on simplicial sets in which $\Delta^1$ is fibrant and contractible, then its homotopy category is thin. - -First, note that if $f : X \to Y$ is a weak equivalence between fibrant objects and $A$ is cofibrant, then $f$ has the weak right lifting property with respect to $A$. That is, for every map $g : A \to Y$, there is a map $g' : A \to X$ such that $f \circ g'$ is homotopic to $g$. -Now, let $A$ be a cofibrant simplicial set in the hypothetical model structure. Consider the inclusion of the left endpoint $f : \Delta^0 \to \Delta^1$. It is a weak equivalence between fibrant objects by the assumptions. Let $g : A \to \Delta^1$ be the constant map at the other endpoint. Then the observation in the previous paragraph implies that $g$ is homotopic to $g'$, the constant map at the left endpoint. Let $A \amalg A \to C(A)$ be a cylinder object for $A$. The previous observation implies that there are no 1-simplices in $C(A)$ between two components (if there is, then $g$ and $g'$ cannot be homotopic). -Thus, any cylinder object $A \amalg A \to C(A)$ equals to $s_1 \amalg s_2 : A \amalg A \to A_1 \amalg A_2$ for some maps $s_1 : A \to A_1$ and $s_2 : A \to A_2$. Moreover, there are retractions $r_1 : A_1 \to A$ and $r_2 : A_2 \to A$ of $s_1$ and $s_2$, respectively. Thus, two maps $f,g : A \to B$ are homotopic if and only if $f$ factors through $s_1$ and $g$ factors through $s_2$. But, since $s_1$ and $s_2$ have retractions, all maps factor through them, so any two maps are homotopic. This implies that any two maps in the homotopy category are equal.<|endoftext|> -TITLE: cohomology of BG, G compact Lie group -QUESTION [20 upvotes]: It has been stated in several papers that $H^{odd}(BG,\mathbb{R})=0$ for compact Lie group -$G$. However, I've still not found a proof of this. I believe that the proof is as follows: ---> $G$ compact $\Rightarrow$ it has a maximal toral subgroup, say $T$ ---> the inclusion $T\hookrightarrow G$ induces inclusion $H^k(BG,\mathbb{R})\hookrightarrow -H^k(BT,\mathbb{R})$ ---> $H^*(BT,\mathbb{R})\cong \mathbb{R}[c_1,...,c_n]$ where the $c_i$'s are Chern classes of degree $\deg(c_k)=2k$ ---> Thus, any polys in $\mathbb{R}[c_1,...,c_n]$ are necessarily of even degree. Hence, -$H^{odd}(BG,\mathbb{R})=0$ -Is this the correct reasoning? Could someone fill in the gaps; i.e., give a formal proof of this statement? - -REPLY [8 votes]: Just for completeness, here's another argument without spectral sequences via rational homotopy theory. -Recall a theorem of Hopf, which states that the rational cohomology of a path-connected H-space of finite rational cohomology type (finite dimensional rational cohomology in each degree) is a free graded commutative graded algebra (cga) $(\wedge V,0)$ (The reason is, that the cohomology has a Hopf algebra structure.) This applies to a compact lie group $G$ and implies immediately, that the rational cohomology serves as a minimal model of $G$ and since the cohomology of a compact Lie group is finite dimensional, $V$ has to be concentrated in odd degrees. -The long exact sequence of the universal fibration $G\rightarrow EG\rightarrow BG$ shows, that $\pi_i(BG)\cong\pi_{i-1}(G)$ since $EG$ is contractible. Recall that the minimal model of a space $X$ is a cdga whose underlying cga is the free cga generated by $\pi_*(X)\otimes\mathbb{Q}$ with a differential $d$, i.e. has the shape $(\wedge(\pi_*(X)\otimes\mathbb{Q}),d)$, so the minimal model of $BG$ has the shape $(\wedge(\pi_{*+1}(G)\otimes\mathbb{Q}),d)$. Since the minimal model of $G$ is concentrated in odd degrees, the one of $BG$ is concentrated in even degrees, so the differential must vanish for degree reasons and the minimal model of $BG$ is just $(\wedge(\pi_{*+1}(G)\otimes\mathbb{Q}),0)$. Since the cohomology of the minimal model of a space is the cohomology of the space, we get the claim.<|endoftext|> -TITLE: Is the set of cube-free binary sequences perfect? -QUESTION [13 upvotes]: This question is inspired by this one. In that thread, it's established that there are uncountably many cube-free infinite binary strings (where $x \in 2^{\omega}$ is cube-free iff $\forall \sigma \subset x,\ \sigma \sigma \sigma \not \subset x$). Here $\subset$ denotes "substring" which should be distinguished from "initial segment" which I'll denote by $\sqsubset$ if the need arises. -So let $C \subset 2^{\omega}$ be the subset of Cantor space consisting of all cube-free sequences. It's uncountable, and easily seen to be closed, hence it contains a perfect set. -Question 1. Is $C$ perfect? -Let's define: -$T_C = \{ \sigma \in 2^{<\omega} : (\exists x \in C)(\sigma \sqsubset x)\}$ -$\ \ \ \ \ = \{ \sigma \in 2^{<\omega} : (\forall n > |\sigma|)( \exists \tau \in 2^n )(\sigma \sqsubset \tau,\ \tau$ cube-free$)\}$ -$T_P = \{\sigma \in 2^{<\omega} : (\forall \tau \sqsupset \sigma, \tau \in T_C)(\exists \rho_1, \rho_2 \sqsupset \tau$ both in $T_C)(\rho_1 \perp \rho_2)\}$ -Question 2. $T_C$ is evidently $\Pi ^0 _1$, but is it in fact recursive? -If we let $P$ denote the maximal perfect subset of $C$ obtained by iteratively removing isolated points of $C$ until this procedure stabilizes (taking intersections at limit stages), then $P$ determines a tree which I believe would be the tree $T_P$ defined above. -Question 3. What is the complexity of the tree determined by $P$? -If $T_P$ is indeed the tree determined by $P$, then the tree determined by $P$ is at worst $\Pi ^0 _3$, but can we do better? - -UPDATE: The answers are: - -Yes -Yes -Since $C$ is perfect, $P = C$ and so by the answers to 1 and 2, the tree determined by $P$ is just the tree determined by $C$, which is recursive. - -I've learnt this after discussing it with Robert Shelton, one of the authors of the paper Gjergji linked to in his response below. In fact, what I've gathered is that there's a function $f$, better than being on the order of $n^2$, such that to determine whether an arbitrary finite string $\sigma$ has a cube-free infinite extension, it suffices to check whether it has one of length $f(|\sigma|)$ (where $|\sigma|$ is the length of $\sigma$). I suppose this would mean furthermore that the tree $T_C$ is not merely recursive, i.e. $\Delta_1$, but in fact $\Delta_0$. - -REPLY [12 votes]: I found an article of J.D. Currie and R.O. Shelton that answers the first question. "The set of k-power free words over Σ is empty or perfect". -Added: -Note that Currie has a number of papers on power free words, and many of them are on the arxiv.<|endoftext|> -TITLE: The diophantine eq. $x^4 +y^4 +1=z^2$ -QUESTION [25 upvotes]: This question is an exact duplicate of the question -Does the equation $x^4+y^4+1=z^2$ have a non-trivial solution? -posted by Tito Piezas III on math.stackexchange.com. - -The background of this question is this: Fermat proved that the equation, $$x^4 +y^4=z^2$$ -has no solution in the positive integers. If we consider the near-miss, $$x^4 +y^4-1=z^2$$ -then this has plenty (in fact, an infinity, as it can be solved by a Pell equation). But J. Cullen, by exhaustive search, found that the other near-miss, $$x^4 +y^4 +1=z^2$$ -has none with $0 < x,y < 10^6$ . -Does the third equation really have none at all, or are the solutions just enormous? - -REPLY [4 votes]: This is not an answer but just a probabilistic (i.e. heuristic) argument which is too long for a comment. -I want to argue that the equation $x^4+y^4+1=z^2$ is likely to have non-trivial solutions and that the fact that there are none with $y<10^8$ isn't exactly a good evidence for lack of solutions. First, looking modulo 4 and 5 we notice that $x, y$ must both be divisible by 10. Also note that modulo other primes there doesn't seem to be any unusual obstructions, i.e. roughly half of the values of $\{x^4+y^4+1; x, y \in \mathbb{Z}/p\mathbb{Z}\}$ are quadratic residues modulo $p$. -We then write our equation as $10^4(x^4+y^4)+1=z^2$. A random number of order $n$ has probability roughly $\tfrac{1}{2\sqrt{n}}$ of being a square but here we know that it is a square modulo $10^4$, so the conditional probability is roughly $4/\sqrt{n}$. So for fixed $y$ the expected number of $x$ which would make $10^4(x^4+y^4)+1$ a perfect square is roughly -$$ -\sum_{x=1}^\infty \frac{1}{25\sqrt{x^4+y^4}}\sim \frac{2}{27y}, -$$ -the last comparison is asymptotic for large $y$ (the exact constant is, of course, not $2/27$) but the value of the sum is somewhat smaller for small $y$. -The sum $\sum_{y=1}^\infty \frac{2}{27y}$ diverges (which suggests there are infinitely many solutions) but if you sum only over $y\leq 10^8$ the sum is still only roughly $1.4$<|endoftext|> -TITLE: Reflexive sheaves on singular surfaces -QUESTION [6 upvotes]: Let $S$ be a normal surface over an algebraically closed field $k$ and let -$s$ be a point of $S$. Let ${\mathcal F}$ be a reflexive sheaf on $S$ of generic rank $n$ . Consider the (derived) fiber of ${\mathcal F}$ at $s$. -Are the dimensions of its cohomologies a priori bounded (for fixed $n$)? - -REPLY [2 votes]: $\def\cO{\mathcal{O}}$ I can give a positive answer for $H^0$ in the case that $S = \mathrm{Spec} A$ for a graded ring $A$ and $\mathcal{F}$ is the sheaf associated to a graded $A$-module. -The proof will consist of a number of elementary reductions, followed by a result of Atiyah, ad then some more elementary arguments. -Since $A$ is normal, we know that $\mathrm{Proj} \ A$ is a smooth curve $X$. Let $g$ be the genus of $X$; let $\cO(1)$ be the line bundle on $X$ coming from the graded ring $A$; let $k$ be the degree of $\cO(1)$. These are all constants. -For any vector bundle $E$ on $X$, the section module is -$$\Gamma(E) := \bigoplus_{n=-\infty}^{\infty} H^0(X, E \otimes \cO(n)).$$ -Section modules are reflexive and every graded reflexive $A$-module is the section module of a vector bundle. The rank of $\Gamma(E)$ is the rank of the vector bundle $E$, which we will call $r$. -Step 1: We may assume that $E$ is not a direct sum of lower rank vector bundles. This is because $\Gamma$ and $\otimes_A k$ both commute with direct sum so, if we can bound the irreducible case, we just have to add together our bounds for each partition of $r$. -Let $d$ be the degree of $E$. -Step 2: We may assume that $0 \leq d < kr$. This is because tensoring with $\cO(1)$ shifts $d$ by $kr$ and gives the same section module up to a grading shift, so we get the zero fiber after this tensoring operation. -Step 3: Here is where we pull out the early parts of Atiyah's "Vector Bundles over an Elliptic Curve". As explained in section I.4 of Atiyah's paper, $E$ has a filtration $0=E_0 \subset E_1 \subset E_2 \subset \cdots E_r=E$ such that $E_i/E_{i-1}$ is a line bundle $L_i$ and the degrees of the $L_i$ obey -$$|\deg L_i - \deg L_1| \leq C_1(g,r)$$ -for a constant depending only on $g$ and $r$ (Atiyah's lemmas 4 and 6). Since $\sum \deg L_i = d$, we can rewrite this as -$$|\deg L_i - d/r | \leq C_2(g,r).$$ -(This uses that $E$ is not a nontrivial direct sum, as this is a standing assumption for Atiyah.) Since $0 \leq d \leq kr$, we can rewrite this as -$$|\deg L_i| \leq C_3(g,r,k).$$ -So there are only finitely many sequences $(\deg L_1, \cdots, \deg L_r)$. We will prove a bound for each one. -Step 4: We prove the following result by induction on $r$: Continue to fix the genus $g$ of $X$. Let $(d_1, \ldots, d_r)$ be a given sequence of integers. Then there are constants $P$ and $Q$, and a sequence of integers $(C_P, C_{P+1}, \ldots, C_Q)$, such that, for any vector bundle $E$ on $X$ with a filtration of degree sequence $(d_i)$, the zero fiber of $\Gamma(E)$ is supported in grades between $P$ and $Q$, and the part in grade $i$ has dimension at most $C_i$. -The base case, $r=0$, is vacuous. -So, let $E$ lie in a short exact sequence $0 \to L \to E \to E' \to 0$, where we have bounds of the assumed form for $E'$ and we know the degree of $L$. We will, of course, be using the short exact sequences -$$0 \to H^0(L \otimes \cO(n)) \to H^0(E \otimes \cO(n)) \to H^0(E' \otimes \cO(n)) \to H^1(L \otimes \cO(n))$$ -for various values of $n$. -When $n$ is negative enough, meaning less than $-\deg L$ and less than $P'$, then the first and third term vanish, so the second one does as well and we get no contribution in those grades. -When $n$ is positive enough, meaning larger than $2g-\deg L$, the fourth term vanishes. So, for $n$ slightly larger than that, we have a complex with exact rows: -$$\begin{matrix} -0 & \to & H^0(L \otimes \cO(n-1)) \otimes H^0(\cO(1)) &\to& H^0(E \otimes \cO(n-1)) \otimes H^0(\cO(1)) &\to& H^0(E' \otimes \cO(n-1)) \otimes H^0(\cO(1)) &\to& 0 \\ -& & \downarrow & & \downarrow & & \downarrow & & \\ -0 & \to & H^0(L \otimes \cO(n)) &\to& H^0(E \otimes \cO(n)) &\to& H^0(E' \otimes \cO(n-1)) &\to& 0 \\ -\end{matrix}$$ -For large enough $n$, we inductively know that the third column is surjective; for $n$ larger than $-\deg L + 2g$, the first column is (by the standard Riemman-Roch argument). So, by the snake lemma, we get that the middle column is surjective for large enough $n$, and there is no contribution to the zero fiber in that degree. -We are left bounding the size of the zero fiber in the intermediate degrees. We can just bound the size of $\Gamma(E)$ itself. From Riemman-Roch, $\dim H^0(X, E \otimes \cO(n)) \leq (d+rk) - r(g-1)$, and we are done.<|endoftext|> -TITLE: Outer automorphisms of finite extensions -QUESTION [16 upvotes]: Let $H$ be a finite index subgroup of a finitely generated group $G$. Assume that $Out(H)$ is finite. Can $Out(G)$ be infinite? - -REPLY [9 votes]: I think the answer is yes. Let $A$ be a f.g. group. Assume that $\text{Out}(A)=1$ ($\text{Out}(A)$ finite would probably be enough) and that $A$ contains a central copy of $C^{(\infty)}$, the direct sum of countably many copies of the cyclic group $C$ of prime order $p$. Then $\text{Out}(A\times C)$ is infinite: indeed if $f$ is a homomorphism from $C$ to the center of $A$, then if we set $u_f(a,t)=(af(t),t)$, then $f\mapsto u_f$ injects $\text{Hom}(C,Z(A))$ into $\text{Out}(A\times C)$. Actually all these automorphisms are identity on the finite index subgroup $A$, and this is essentially the reason I don't expect Mark's argument to work. -It remains to find an example of such a group $A$. I expect some complicated nilpotent-by-abelian group over $\mathbf{F}_p[t]$ to work (or Kac-Moody groups??), but I'm not prone to go into computations and maybe somebody else has one example. -Edit: Anton Klyachko's answer to my subsequent question confirms the existence of such a group $A$. Therefore, the answer to the question here is positive: - -for finitely generated groups, having infinite outer automorphism group does not pass to finite index subgroups.<|endoftext|> -TITLE: Do you believe P=NP? -QUESTION [5 upvotes]: Do you believe P=NP? -I've seen some mathematicians say that if P=NP their work would be worthless and restricted to enunciating theorems. They seem to believe that there exist an almost philosophical impediment to P=NP. Do you agree with that? Does the possibility of P=NP bother you? - -REPLY [31 votes]: Contrary to a popular misunderstanding: if P = NP, then the proof of any statement $A$ can be found by an algorithm in time polynomial in the length of the shortest proof of $A$, not in the length of $A$ itself. Moreover, the exponent of the polynomial could easily be so large as to make this algorithm practically worthless. But most importantly: the shortest, machine-generated, proof of some theorem is highly unlikely to be the most elegant, illuminating, or just human-comprehensible, proof. Thus this idea that under P = NP, mathematics would be reduced to “enunciating theorems”, is completely misguided.<|endoftext|> -TITLE: A question on K_1 of an elliptic curve -QUESTION [12 upvotes]: Consider an elliptic curve $E/ \mathbb{Q}$, with a regular model $\mathcal{E} / \mathbb{Z}$. We have (Beilinson regulator) maps -$$ K_1(\mathcal{E})^{(2)} \to K_1(E)^{(2)} \to H_D^3(E_{/ \mathbb{R}} , \mathbb{R}(2) )$$ -from (an Adams eigenspace of) K-theory (with rational coefficients) to Deligne cohomology of $E$. Call the first map $\iota$ and the second map $r$. Note that this map does NOT lie in the index range where the Beilinson conjectures predicts that $r$ is an isomorphism on the image of $\iota$ after tensoring with $\mathbb{R}$. Now, is anything known at all about $r$ or $r \circ \iota$, for elliptic curves in general or for some specific curve/class of curves? Unless I am mistaken, the Deligne cohomology group in question is always a one-dimensional real vector space. My main question is the following: - -After tensoring everything with $\mathbb{R}$, is the the map $r \circ \iota$ zero or surjective??? - -I would also be interested in the following questions: - -Is anything known about the two K-groups here? Finite generation? Rank? Can you write down a nonzero element? -Is the map $\iota$ injective? (This could be asked in much more generality for K-groups of regular models.) - -I'd be grateful for any hints, even those based on unproven conjectures. -EDIT: Maybe one can approach this question from another point of view. I am quite sure that the following is true (have to check though). The cokernel of $r \circ \iota$ can be identified with the Gillet-Soulé arithmetic Chow group $\widehat{CH}^2(\mathcal{E}) \otimes \mathbb{R}$. Furthermore, this group is generated by arithmetic cycles of the form $(Z,g) = (0,\alpha)$, where $\alpha$ is a real harmonic $(1,1)$-form on the complex torus $E(\mathbb{C})$. So the question becomes: Do all arithmetic cycles of this form lie in the group generated by arithmetic cycles of the forms $(div(f), - \log \| f \|^2)$ and $(0, \partial u + \bar{\partial} v)$? - -REPLY [8 votes]: Let me explain why Beilinson's conjecture implies that $\iota$ is the zero map (thus your first question has conditionally a negative answer). -Let $\mathcal{E}$ be a proper regular model of $E$ over $\mathbf{Z}$. The morphism $E \to \mathcal{E}$ induces a $\mathbf{Q}$-linear map $\iota : K_1(\mathcal{E})^{(2)} \to K_1(E)^{(2)}$. The image of $\iota$ is the integral subspace $K_1(E)^{(2)}_{\mathbf{Z}}$, which is also written $H^3_{\mathcal{M}/\mathbf{Z}}(E,\mathbf{Q}(2))$ in cohomological notations. -Now, what does Beilinson's conjecture predict for this group? We are concerned here with the motive $h^2(E)$, whose $L$-function is $L(H^2(E),s)=\zeta(s-1)$, and we are looking at the point $s=2$. Thus we are in the case of the "near central point" (see for example -Schneider, Introduction to the Beilinson conjectures, Section 5, Conjecture -II, or the articles by Beilinson and Nekovar mentioned in my comment). -Since the $L$-function has a pole, we have to introduce the group $N^1(E)=(\operatorname{Pic}(E)/\operatorname{Pic}^0(E)) \otimes -\mathbf{Q}$ which is isomorphic to $\mathbf{Q}$ (more generally, the dimension should be equal to the order of the pole). There is a natural injective map $\psi : N^1(E) \to H^3_{\mathcal{D}}(E_{\mathbf{R}},\mathbf{R}(2))$. Then Beilinson's conjecture asserts that $r \oplus \psi$ induces an isomorphism -$$(H^3_{\mathcal{M}/\mathbf{Z}}(E,\mathbf{Q}(2)) \otimes_{\mathbf{Q}} \mathbf{R}) \oplus (N^1(E) \otimes_{\mathbf{Q}} \mathbf{R}) \xrightarrow{\cong} H^3_{\mathcal{D}}(E_{\mathbf{R}},\mathbf{R}(2)).$$ -Since the target space is $1$-dimensional and $N^1(E) \cong \mathbf{Q}$, this predicts in particular that $H^3_{\mathcal{M}/\mathbf{Z}}(E,\mathbf{Q}(2))=0$. -Moreover, it can be shown that the map $r$ is nonzero. As pointed out by profilesdroxford, the group $H^3_{\mathcal{M}}(E,\mathbf{Q}(2))$ is generated by symbols of the form $(P,\lambda)$ where $P$ is a closed point of $E$ and $\lambda \in \mathbf{Q}(P)^*$. I found a reference for this in Beilinson, Notes on absolute Hodge cohomology (Beilinson attributes this construction to Bloch and Quillen). Furthermore, in the same article the regulator of such elements is computed (in a more general setting). After some computations it turns out that $r([P,\lambda])$ is proportional to $\log | \operatorname{Nm}_{\mathbf{Q}(P)/\mathbf{Q}}(\lambda) |$. Thus $r$ is nonzero. Another useful reference is Dinakar Ramakrishnan's article on regulators. -It would be also interesting to compare the above construction with the construction proposed by profilesdroxford in his answer.<|endoftext|> -TITLE: Abstract Jordan Decomposition different from usual Jordan Decomposition -QUESTION [10 upvotes]: It's known that if $L\subset gl(V)$, with $V$ finite dimensional, is a semisimple Lie algebra, then the abstract and usual Jordan decompositions in $L$ coincide. Is it possible to provide a counter-example if $L$ isn't semisimple? -Remark: The underlying field is algebraically closed of characteristic $0$ . - -REPLY [17 votes]: I'm tempted to amplify Ben's precise short answer by emphasizing how subtle the notion of abstract (or intrinsic) Jordan decomposition really is. You start with a matrix Lie algebra which is (1) required to be isomorphic to its image under the adjoint representation (in other words, centerless). Then it makes sense to interchange the Jordan decompositions in the two settings. But these might not agree for every matrix realization, unless you also require (2) that the given Lie algebra satisfy a complete reducibility theorem for finite dimensional representations. Then the proof I first saw in Bourbaki, which goes back to Richard Brauer's early work influenced by Weyl, takes over. (It needs characteristic 0 and at first an algebraically closed field to apply Weyl's complete reducibility theorem directly.) -Ben's example already fails (1), as does a general linear Lie algebra, whereas a centerless solvable Lie algebra only fails (2). -The somewhat elaborate-looking Bourbaki argument tempts people to take shortcuts (even in one published textbook), but as far as I can see the more sophisticated proof is really needed.<|endoftext|> -TITLE: Parametrization of 2-dimensional torus -QUESTION [7 upvotes]: The units with norm $+1$ in a pure cubic number field $K$ generated -by a cube root of $m = ab^2$, where $a$ and $b$ are coprime and -squarefree integers, correspond to integral points on the torus -$$ R_{K/\mathbb Q}^{(1)}: - X_1^3 + ab^2X_2^3 + a^2bX_3^2 - 3abX_1X_2X_3 = 1. $$ -According to Voskresenskii (Algebraic Groups and their birational -Invariants), all tori of dimension $2$ such as the one above -are rational. -I am having problems with finding such a rational parametrization. -The surface has three singular points at infinity, all of them -defined over the normal closure of $K$; the line through the pair -of conjugate singular points is necessarily contained in the surface -and defined over $K$. - -Is there a way of finding a parametrization -from the singular points or the three lines connecting them? - -A different idea is looking at the tangent plane in $(1,0,0)$. -It intersects the surface in a singular cubic, which can be -parametrized via sweeping lines and produces the parametrization -$$ X_1 = 1, \quad X_2 = \frac{3t}{b+at^3}, \quad X_2 = \frac{3t^2}{b+at^3}. $$ -By looking at the tangent plane at these rational points I would get a -2-parameter family of rational points; the calculations are, however, -quite involved. So: - -Is there a slick way of obtaining this family? -Once I have written down the 2-parameter family of rational points, how -can I show that the parametrization includes all rational points? - -An additional question in this connection is the following: -conics such as $x^2 - my^2 = 1$ can be parametrized by trigonometric -or hyperbolic functions. - -Are there (periodic) analytic functions that parametrize the -cubic surface above? - -I have always wondered why there are so many books on diophantine equation, but few if any explaining some simple geometric techniques useful for finding rational points on algebraic surfaces. Is there such a book out there? - -REPLY [6 votes]: I decided to have a go at finding a parametrisation by following the instructions of Coray and Tsfasman (reference in my comment above), using Magma. Amazingly enough, it works, even working generically with $a,b$ variables. -Here's what I did. Working over the field $K(\omega, \sqrt[3]{a^2b})$, find the three singular points of $X$. Writing $c = \sqrt[3]{a^2b}$, they are $(0:c:a/c:1)$, $(0:\omega^2 c:\omega a/c:1)$, $(0:\omega c:\omega^2 a/c:1)$. There's an obvious rational point $(1:1:0:0)$. The Cremona transformation $f \colon \mathbb{P}^3 \to \mathbb{P}^3$ associated to these four points can be found by linearly mapping them to the standard basis points, applying the standard Cremona transformation $(1/X_0: 1/X_1: 1/X_2: 1/X_3)$, and reversing the linear map. It turns out that $f^{-1}(X)$ is (modulo some rubbish supported on the four planes where $f$ is not defined) a quadric surface $Y$, defined by -$$ -X_0^2 + \frac{1}{3a} X_0 X_2 + \frac{1}{27a^2} X_2^2 + \frac{1}{27a^2b}(X_0-X_1)X_3. -$$ -The surface $Y$ has an obvious rational point $(0:1:0:0)$, so projecting away from that gives an isomorphism $Y \to \mathbb{P}^2$. The inverse of that isomorphism, composed with $f$, gives a rational map from $\mathbb{P}^2$ to $X$. The equations are (cut & pasted from Magma): -729*c^6*x^6 + 729*c^6/a*x^5*y + 324*c^6/a^2*x^4*y^2 + 81*c^6/a^3*x^3*y^3 + - 12*c^6/a^4*x^2*y^4 + c^6/a^5*x*y^5 + 1/27*c^6/a^6*y^6 + -3*c^3/a*x^2*y*z^3 + - -c^3/a^2*x*y^2*z^3 + -2/27*c^3/a^3*y^3*z^3 + 1/27*z^6 -729*c^6*x^6 + 729*c^6/a*x^5*y + 324*c^6/a^2*x^4*y^2 + 81*c^6/a^3*x^3*y^3 + - 12*c^6/a^4*x^2*y^4 + c^6/a^5*x*y^5 + 1/27*c^6/a^6*y^6 + -9*c^3*x^3*z^3 + - -6*c^3/a*x^2*y*z^3 + -c^3/a^2*x*y^2*z^3 + -2/27*c^3/a^3*y^3*z^3 + 1/27*z^6 -243*a*c^3*x^5*z + 162*c^3*x^4*y*z + 45*c^3/a*x^3*y^2*z + 6*c^3/a^2*x^2*y^3*z + - 1/3*c^3/a^3*x*y^4*z + -1/3*x*y*z^4 --9*c^3/a*x^3*y*z^2 + -3*c^3/a^2*x^2*y^2*z^2 + -1/3*c^3/a^3*x*y^3*z^2 + 1/3*x*z^5 - -I'm sure these can be tidied up a lot, but notice that they are at least defined over the original base field, since $c$ only ever appears as $c^3$. -I'll post the Magma code (only 20 lines) if anybody's interested.<|endoftext|> -TITLE: Is it possible to check a graph for acyclicity in $O(|V|)$ time? -QUESTION [6 upvotes]: I wonder if I can make an algorithm to check if a given graph $G=(V,E)$ is acyclic or not with the complexity of $O(|V|)$. -I modified the BFS algorithm to do this, but the complexity seems to be $O(|V|+|E|)$. - -REPLY [9 votes]: The answer for directed acyclic graphs is no: in any standard adjacency-list based representation you have to look at $\Omega(m)$ of the edges; it is not enough to look at $O(n)$ vertices and edges, even if the vertices are already labeled by indegree and outdegree and the adjacency lists are partitioned into incoming and outgoing edges. -For, suppose that you have a directed acyclic graph in the form of a directed path (with $n-1$ edges), partitioned into four equal length subpaths $a_1\dots a_{n/4}$, $b_1\dots b_{n/4}$, $c_1\dots c_{n/4}$, and $d_1\dots d_{n/4}$. For some number of pairs $(i,j)$, add to this path the additional edges in the three-edge path $a_ib_jc_id_j$. This allows the addition of numbers of additional edges ranging from none to $\Omega(n^2)$, and the graph formed in this way is acyclic. But, if your algorithm fails to examine at least one of the edges in any one of these three-edge paths, it could instead be replaced by a path $a_ic_ib_jd_j$, producing a graph with the same indegrees and outdegrees that is not acyclic and that your algorithm cannot distinguish from the given acyclic graph. -This construction doesn't work directly for adjacency matrix representations but in that representation, too, you need more than a linear number of queries; see the Aanderaa–Karp–Rosenberg conjecture and note that the property of containing a cycle is monotone.<|endoftext|> -TITLE: Pullback along the Torelli map is an isomorphism -QUESTION [8 upvotes]: I've been told many times that the Torelli map $J:\mathcal{M}_g\to \mathcal{A}_g$ for ($g\geq 2$, and at least on the level of coarse moduli spaces, over $\mathbb{C}$) gives an isomorphism of Picard groups. On the level of rational picard groups, they're both generated by the determinant of the Hodge bundle, but why is this true on the nose? I'll be more than happy with a proof or with a reference to a detailed proof (actually slightly prefer the latter). - -REPLY [9 votes]: I see two ways interpreting "on the nose": -1) At the level of the orbifold Picard group, or that is at stack level, and in my view this is the correct framework, both Pic_{orb}(M_g) and Pic_{orb}(A_g) are generated by the Hodge class. (The Hodge class on M_g is by definition the pull-back of the Hodge class -on A_g.) The reference for this result is Arbarello and Cornalba "Picard groups of the moduli space of curves" (1987). -2) At the level of integral Picard groups, the answer is still positive but messier. -Pic(M_g) is the sugroup of Pic_{orb}(M_g) generated by the multiple m\lambda of the Hodge class which actually descends to the coarse moduli space. As far as I know, the actual value of this multiple is not known. The bundle L with class m\lambda descends to M_g if every automorphism of a curve C, acts trivially on the fibre of L over C. It is certain to be an even number, for \lambda does not even descend to the smooth part of M_g but 2\lambda does. -The references are Mumford's paper "Stability of proj vareties" (1977) and "Abelian quotients of Teichmüller groups" (1967).<|endoftext|> -TITLE: Geometric interpretation of Cartan's structure equations -QUESTION [22 upvotes]: Given a linear connection on a Riemmanian manifold $M$ and $\phi^1,...,\phi^n$ a local frame for $T^*M$ we can define the connection 1-forms $\omega^j_i$. We define the curvature 2-forms by $\Omega_i^j=\frac{1}{2}R_{klij}\phi^k \wedge \phi^l$. -We have the following identities also known as Cartan's first and second structure equations: -i) $d\phi^j=\phi^i \wedge \omega_i^j + \tau^j$ where $\tau^1,...,\tau^n$ are the torsion 2 forms. -ii) $\Omega_i^j=d\omega_i^j-\omega_i^k \wedge \omega_k^j$ -I have two questions: -1)Is there a geometric meaning attached to these equations? -2) Why are these equations important and what are they useful for? - -REPLY [15 votes]: The $1$-forms $\omega^i_j$ define an affine connection on the tangent bundle, and the first structure equation gives the formula for the torsion tensor. -It is equivalent to the equation -$$ -\nabla_X Y - \nabla_Y X - [X,Y] = \tau(X,Y), -$$ -where the connection $\nabla$ is defined using the $1$-forms $\omega^i_j$. I think of these equations as describing what happens when you parallel transport a vector along a 1-parameter family of curves with respect to the connection. -The second structure equation is equivalent to -$$ -[\nabla_X, \nabla_Y]Z - \nabla_{[X,Y]}Z = R(X,Y)Z -$$ -and defines the curvature tensor. I personally find these equations to be inscrutable, but if you study how families of geodesics vary, then the curvature tensor arises naturally as part of the Jacobi equation. This gives it a nice geometric interpretation.<|endoftext|> -TITLE: polarization formula for homogeneous polynomials -QUESTION [8 upvotes]: given a homogeneous polynomial p of dgree n on $R^d$, there is a unique symmetric n-linear functional $B$ on $(R^d)^n$ such that $p(x)=B(x,..,x)$. The question is: Can we get $B$ by means of a polarization formula as in the case $n=2$ for quadratic forms ? -Thanks in advance. - -REPLY [6 votes]: I'm assuming that $R$ are the reals. In any case, you need to be able -to divide by $n!$. -Given that, the answer is yes. I can't locate a reference, but here's -the formula for $n=3$, say: -$$ 6 B(x,y,z) = p(x+y+z) - p(x+y) - p(y+z) - p(z+x) + p(x) + p(y) + -p(z) -$$ -which should give you a hint as to the general case. -Edit by Denis Serre. This suggests the general formula -$$n!B(x_1,\ldots,x_n)=\sum_I(-1)^{n-|I|}p(x_I),\qquad x_I:=\sum_{i\in I}x_i.$$ -Further edit by JMF. The formula is proved in this preprint by Erik G.F. Thomas A polarization identity for multilinear maps<|endoftext|> -TITLE: commutativity constraint in Grothendieck's motives -QUESTION [8 upvotes]: This is a basic question about Grothendieck's conjectural category $M_k$ of pure motives (over a field $k$). This construction first produces a category (the "false category of motives") which need not be Tannakian; the category undergoes a modification of the commutativity constraints which produces a category (the "true" category) which could be Tannakian (this is spelled out on page 451 Jannsen's great paper). -The modification requires that the Kunneth components of the diagonal be algebraic (part of the standard motivic conjectures). -Question: Is there an alternate (geometric) construction (modulo standard conjectures) which produces the "true" category bypassing this artificial commutativity modification? -Does this happen in Voevodsky's setup? -Thanks! - -REPLY [18 votes]: I don't know any alternate geometric construction of the commutativity contraint, but there is a natural way to explain this, as a simple fact of homological algebra (and assuming very optimistic conjectures about mixed motives). -Let $DM_{gm}(k)$ be Voevodsky's triangulated category of geometric (=constructible) motives over $k$, with rationnal coefficients. For any (smooth) $k$-scheme, we have its homological motive $M(X)$ in $DM_{gm}(k)$ (this construction being functorial in $X$). Voevodsky proved that the full pseudo-abelian subcategory of $DM_{gm}(k)$, closed under arbitrary Tate twists, spanned by objects of shape $M(X)^\wedge=\underline{Hom}(M(X),\mathbf{Q})$ for $X$ smooth and projective is equivalent to the category $Chow(k)$ of Chow motives (the cohomological version, constructed from rational equivalence on cycles). -Now comes the conjectural part. Assume that there exists a motivic $t$-structure on $DM_{gm}(k)$; see this paper of Beilinson for what this means and for the link with the standard conjectures. Let us denote by $MM(k)$ the heart of this $t$-structure. The objects of the category $MM(k)$ are, by definition, mixed motives. This category is tannakian: for any prime number $\ell$, the $\ell$-adic realization functor defines a conservative $t$-exact and symmetric monoidal functor from $DM_{gm}(k)$ to the bounded derived category of finite dimensional $\mathbf{Q}_\ell$-vector spaces (this is part of what it means to be a motivic $t$-structure). -Call a mixed motive $M$ pure of weight $n$ if $M[-n]$ is pure of weight zero in the sense of Bondarko (see this paper for the link between $t$-structures and weight structures): this simply means that $M[-n]$ belongs to $Chow(k)$, seen as a subcategory of $DM_{gm}(k)$. For $X$ smooth and projective over $k$, as $M(X)^\wedge$ is pure of weight zero, for any integer $n$, the cohomology object $H^n(X)=\tau^{\leq n}\tau^{\geq n}(M(X)^\wedge)[n]$ is pure of weight $n$. This implies that we have a (non canonical) isomorphism $M(X)^\wedge\simeq \oplus_n H^n(X)[-n]$ (the so called `Chow-Künneth decomposition'. -Denote by $M(k)$ the full subcategory of $MM(k)$ which consists of finite sums of objects $M$ of $MM(k)$ which are pure of weight $n$ for some integer $n$. We then have a natural functor -$$X\mapsto \oplus_n H^n(X)$$ -from the category of smooth and projective $k$-schemes to $M(k)$. This functor factors (by construction) through the category $Chow(k)$: -$$Chow(k)\to M(k)$$ -and factors through numerical equivalence (just because the motivic $t$-structure is compatible with the monoidal structure, using Poincaré duality in $DM_{gm}(k)$), inducing an equivalence of categories -$$Chow(k)/\text{numerical equivalence} \simeq M(k)$$ -Now, the point is that $M(k)$ is a tannakian subcategory of $MM(k)$, and the modified commutativity constraint of the tensor product on the category of pure motives up to numerical equivalence is the one inherited from the natural monoidal structure of $M(k)$ through the equivalence above.<|endoftext|> -TITLE: (A very limited instance of) Lagrange's Theorem's converse and A_5 -QUESTION [11 upvotes]: Suppose $G$ is a finite simple group and $|G|$ is a multiple of $60$. Does it follow that $G$ has a subgroup isomorphic to $A_{5}$? If so, can this be proven without using the Classification? - -REPLY [10 votes]: I think the answer to the question is yes, but it is very unlikely that it can be proved without using the classification of finite simple groups. -Note that $A_5$ of order 60 is the only simple group order for which this statement is true, because for all higher order simple groups $G$, there will be groups $L_2(p)$ with order divisible by $|G|$ that do not contain $G$ as subgroup. -Let's quickly look at all families if finite simple groups. I hope someone will correct any mistakes I make! -The Suzuki groups (Lie type $^2B_2$) are not divisible by 3, so we can forget them. All other finite simple groups have order divisible by 12, so their order is divisible by 60 if and only if it's divisible by 5. -The claim is clearly true for $A_n$, $n \ge 5$. -It is well-known that $L_2(q)$ contains $A_5$ if and only if $q \equiv \pm{1} \bmod 5$, which is equivalent to $|G|$ divisible by 5. -$U_3(q)$, $L_3(q)$, $G_2(q)$, $^3D_4(q)$ all contain $L_2(q)$ and also have order divisible by 5 if and only $q \equiv \pm{1} \bmod 5$. -$^2F_4(2^{2e+1})$ contains $^2F_4(2)$, which contains $A_5$. -$^2G_2(3^{2e+1})$ never has order divisible by 5. -$S_4(q)$ contains $L_2(q^2)$, which always contains $A_5$ for all $q$. -All remaining groups of Lie type contain $S_4(q)$ and hence contain $A_5$. -It is easily checked, for example by looking at their lists of maximal subgroups in the ATLAS or on -http://brauer.maths.qmul.ac.uk/Atlas/v3/ -that the sporadic groups contain $A_5$.<|endoftext|> -TITLE: How to find nearest lattice point to given point in R^n ? Is it NP ? -QUESTION [8 upvotes]: Consider some lattice in R^n. -Take some point "P" in R^n (which does not belong to this lattice in general). -What are the algorithms to find some nearest lattice point to "P" ? -"Nearest" - means in the sense of the standard Eucleadian distance. -Lattice - means the standard thing - takes some vectors h_1 ... h_n and consider all their combinatioons with INTEGER coefficients. (Let us assume that h_i are independent and their number is exactly the dimension of the ambient R^n). -I think this should pretty well-studied question... -Can someone recommend some good literature for it ? -What are the general results known for it ? -What are the algorithm complexity ? -Is it NP problem ? - -REPLY [12 votes]: This problem is often called the "closest vector problem" for lattices (especially by people in theoretical computer science). The real issue is what sort of lattice you have. For example, for the integer lattice $\mathbb{Z}^n$ the problem is easy, and it's not too hard to do it for other famous lattices, for example root lattices. So if you have a specific, not very complicated lattice in mind, you shouldn't necessarily despair. -However, for an arbitrary lattice it's hard. The strongest result I know of offhand is that Dinur, Kindler, Raz, and Safra proved that it is NP-hard to approximate the distance to the closest lattice point by a factor of $n^{c/\log \log n}$ in $n$ dimensions (Combinatorica 23 (2003), 205–243); however, it may have been slightly improved since then. Using LLL lattice basis reduction, you can get an approximation to within a slightly subexponential factor in polynomial time. -For a good overview, see the book Complexity of lattice problems: a cryptographic perspective by Micciancio and Goldwasser.<|endoftext|> -TITLE: On closed totally disconnected subgroups of connected real Lie groups -QUESTION [5 upvotes]: So the following statement seems to be obvious but I don't see how to prove it: -Q: How does one prove that a closed totally disconnected subgroup of a connected real Lie group is discrete? -Note that it is essential to take into account the group structure since the Cantor set is a closed totally disconnected subset of $\mathbf{R}$ which is not discrete. - -REPLY [2 votes]: To complement what has been said earlier: a Lie group $G$ has no small subgroup, i.e. there exists a neighborhood $U$ of the identity $e\in G$ such that the only subgroup of $G$ contained in $U$, is $\{e\}$. [To see it, let $B$ be a small ball around $0$ in the Lie algebra of $G$, such that the exponential map $\exp$ induces a diffeomorphism between $B$ and its image. Set $U=\exp(B/2)$. Then for every $x\in U\backslash\{e\}$, there exists $p>1$ such that $x^p\notin U$.] -Let then $H$ be a closed subgroup of $G$. If $H$ is totally disconnected, then there is a basis of neighborhoods of $e$ in $H$, consisting of open subgroups of $H$. So $H\cap U=\{e\}$, meaning that $H$ is discrete.<|endoftext|> -TITLE: Naturally occurring orderings -QUESTION [9 upvotes]: The are many orderings that naturally occur in interesting but seemingly unrelated circumstances. Here are some examples: - -The volume spectrum of orientable hyperbolic 3-manifolds has order type $\omega^\omega$. -Ordinals that play important roles in Conway's $\mathbf {On_2}$, most notably $\omega^{\omega^\omega}$, the algebraic closure of $2$. See Lenstra's papers 1 2, Conway's ONAG, and Lieven's blog posts. -The set of fusible numbers has order type $\epsilon_0$ (quite likely but not proven, see my note). -The Sharkovsky ordering of natural numbers, which does not have order type of an ordinal. -There are proof theoretical ordinals, which I know little about. - -Do you know any other examples or see any connection among aforementioned examples? Most of the examples above are ordinals, but other interesting examples are welcome. - -REPLY [9 votes]: Alright, I'll put my comment as an answer and hopefully get this off the no-upvoted-answers queue. :) -Here's another nice-but-surprising way to get $\omega^\omega$: Let $\|n\|$ denote the smallest number of 1's needed to write n using any combination of addition and multiplication, e.g., $\|7\|=6$ as shortest way for 7 is $7=(1+1+1)(1+1)+1$. (This is known as the "integer complexity" of n; it's sequence A005245.) -Now, for any n, we have the lower bound $\|n\|\ge 3log_3 n$. So subtract this off and consider $\delta(n):=\|n\|-3log_3 n$. Then the set of all values of $\delta$ is a well-ordered subset of $\mathbb{R}$, with order type $\omega^\omega$. -For a proof, I refer you to my preprint: http://arxiv.org/abs/1310.2894<|endoftext|> -TITLE: Lattice reduction in R^3 (R^4) or what is fundamental domain for SL(3,Z) , (SL(4,Z)) ? -QUESTION [5 upvotes]: Consider a lattice in R^3. -Is the some "canonical" way or ways to choose basis in it ? -I mean in R^2 we can choose a basis |h_1| < |h_2| and |(h_2, h_1)| < 1/2 |h_1|. -Considering lattices with fixed determinant and up to unitary transformations we get standard picture of the PSL(2,Z) acting on the upper half plane, which has a fundamental domain Im (tau)>1 Re(tau) <1/2. -What are the similar results for other small dimensions R^3, R^4, C^4, C^8 ? -What are the algorithms to find such a lattice reductions ? - -REPLY [7 votes]: The book by Terras "Harmonic analysis on symmetric spaces and applications" volume 2 has some stereoscopic pictures of the fundamental domains for some similar groups.<|endoftext|> -TITLE: Quasi-compact maps in Number Theory -QUESTION [9 upvotes]: Can someone give me an example of a non-quasi-compact morphism of schemes which arises naturally in the field of Algebraic Number Theory? - -REPLY [7 votes]: Rex, about your scheme-theoretic image problem, I don't know how "contrived" the following example is (I am afraid it is not particularly related to number theory). -Notations: $R$ is a discrete valuation ring, $t$ a uniformizer, $R_n:=R/(t^{n+1})$ ($n\in\mathbb{N}$), $X_n=\mathrm{Spec}\,R_n$, $A=\prod_n R_n$. -Take $X:=\coprod_n X_n$ and $Y:=\mathrm{Spec}\,A$. There is a natural open immersion $f:X\to Y$ since each $X_n$ embeds in $Y$ as an open and closed subscheme. -The scheme-theoretic image of $f$ is $Y$: since $Y$ is affine, it just means that each $x\in A$ vanishing on each $X_n$ is zero, which is obvious. (In fact, $A=\Gamma(X,\mathcal{O}_X)$). -But $X$ is not topologically dense in $Y$: indeed, consider $x=(t,t,\dots)\in A$. Then $x$ is locally nilpotent on $X$ but not nilpotent on $Y$, hence the open set $D(x)\subset Y$ is nonempty and disjoint from $X$. - -REPLY [4 votes]: You might see non-quasi-compact maps in the context of universal covers of maximally degenerate pointed curves, and depending on who you ask, this might be called algebraic number theory. Specifically, a maximally degenerate positive genus (proper) curve $X$ has the form of a connected graph made out of finitely many projective lines intersecting transversely, where each line has exactly 3 special points (namely intersections and markings). If $x$ is a marked point, then $\pi_1^{geom}(X,x)$ is a finitely generated free group. One then has a universal cover $(\tilde{X},\tilde{x}) \to (X,x)$, where $\tilde{X}$ is a tree of projective lines. The covering map is étale but not quasi-compact. -Gerritzen and van der Put wrote a book (Schottky Groups and Mumford Curves, Springer LNM 817) describing some number-theoretic data like theta functions on these objects. Some brief web searching suggests that there seem to be some more modern treatments using rigid analytic techniques (that I don't really understand). -I'm afraid this doesn't answer the more focused question about scheme-theoretic image that you posed in the comments.<|endoftext|> -TITLE: Conformal structure determined by principal curvatures -QUESTION [5 upvotes]: On a smooth surface of positive Gauss curvature that is embedded in Euclidean 3 space the principal curvatures seem to define a smooth field of ellipses whose major and minor axes are lined up with the directions of curvature and whose ratio of lengths is k1/k2, the ratio of the principal curvatures. -My first question is whether there is true or whether there is a flaw in this observation. -If true, the embedding determines a conformal structure and I assume that the associated Riemannian metric is determined by the usual equations -1 + |u|/ 1 - |u| = k1/k2 or |u| = k1 - k2/ k1 + k2 with arg(u) = 0. So ds^2 = (dz + udz_)^2 -My second question is whether this is correct. -Finally, if this construction works what can be said about the possible conformal structures and Riemannian geometries that can be obtained in this way from embeddings of surfaces of positive curvature in 3 space? Can the new metric be realized in 3 space? -Can this be generalized? - -REPLY [3 votes]: When defined as a tensor taking values in the normal bundle the second fundamental form of a hypersurface in $\mathbb{R}^{n+1}$ depends only on the affine structure on the ambient space, not on its metric structure. On the locus where it is non-degenerate it determines a conformal structure which has been used in studying the affine geometry of the hypersurface at least since the 1930's (see the second volume of Blaschke's book Vorlesungen über Differentialgeometrie entitled Affine Differentialgeometrie). If one restricts attention to those aspects of the geometry of the hypersurface preserved by unimodular affine transformations, then a particular representative, called the Blaschke or equiaffine metric, is determined. With respect to a Euclidean unit normal, it is the second fundamental form multiplied by the $-1/(n+2)$ power of the Gaussian curvature. That is, if the hypersurface is written locally as the graph of $f$ then the metric is given in coordinates $|\det f_{ij}|^{-1/(n+2)}f_{ij}$, where $f_{ij} = \tfrac{\partial^{2}f}{\partial x^{i} \partial x^{j}}$. When the second fundamental form is locally strictly convex, this metric is Riemannian, and much use of it has been made in the study of such hypersurfaces. To my taste the best introduction to these matters is provided by two papers of Calabi: Complete affine hyperspheres. I (there is no sequel) and Hypersurfaces with maximal affinely invariant area. -When the hypersurface is non-degenerate there can be defined the affine normal, which is a vector field transverse to the hypersurface that transforms equivariantly under the action of unimodular affine transformations on the ambient space. Via the splitting induced by this normal the flat affine connection on the ambient space induces an affine connection on the hypersurface. The difference of this affine connection with the Levi-Civita connection of the equiaffine metric is a tensor, which if its contravariant index is lowered using the metric, is completely symmetric. This completely symmetric covariant tensor three tensor (the Pick form) and the equiaffine metric encode the geometry of the hypersurface. Such a pair can be considered in the abstract, and the local obstructions to realizing it as that induced on a non-degenerate hypersurface are given in terms of the curvature of the affine connection and the Pick form, and, if I remember right, can be found in the textbook Affine differential geometry of Nomizu and Sasaki. Where such obstructions come from can be seen as follows. The map assigning to a point of the hypersurface the annihilator of its tangent space takes values in the projectivization of the vector space dual to the ambient space; this map is an immersion if and only if the hypersurface is non-degenerate. In this case the pullback via this map of the flat projective structure induces a flat projective structure on the hypersurface. A particular connection representative of this projective structure is distinguished by the requirement that its difference with the Levi-Civita connection of the equiaffine metric be the negative of the Pick form. Then for an arbitrary pair comprising a metric and a completely symmetric covariant three tensor to be that induced on an immersed hypersurface, a necessary condition is that the connection formed by adding to the Levi-Civita connection the negative of the three tensor (with an index raised) be projectively flat. -As far as I am aware the question of when such a pair can be realized globally via a proper embedding has not been studied other than in some particular cases, determined by conditions on the affine shape operator of the putative embedding, conditions which can be stated solely in terms of the Ricci curvature of the metric and the Pick form. Much of it boils down to controlling the size of the Pick form. For instance, the heart of Calabi's paper on complete affine hyperspheres is a differential inequality for the Laplacian (in the equiaffine metric) of the norm of the Pick form. In this regard, in addition to the aforementioned papers of Calabi, N. Trudinger and X.-J. Wang's survey The Monge-Ampère equation and its geometric applications and J. Loftin's Survey on affine spheres are good starting points. -In the two-dimensional case the conformal structure determines a complex structure, and it helps to use it. For this two basic references are Changping Wang's Some examples of complete hyperbolic affine $2$-spheres in $\mathbb{R}^{3}$ and Calabi's Convex affine maximal surfaces.<|endoftext|> -TITLE: PBW theorem over a Q-algebra, without freeness or flatness -QUESTION [11 upvotes]: Let $k$ be a commutative ring with $1$. Let $L$ be a $k$-Lie algebra, which is not necessarily free as a $k$-module. Let $S\left(L\right)$ denote the symmetric algebra of $L$ (over $k$), constructed as a quotient of the tensor algebra $T\left(L\right)$ of $L$. Let $U\left(L\right)$ denote the universal enveloping algebra of $L$ (over $k$), constructed as the quotient of the tensor algebra $T\left(L\right)$ modulo the two-sided ideal generated by all elements of the form $x\otimes y-y\otimes x-\left[x,y\right]$ with $x\in L$ and $y\in L$. -The canonical filtration of the tensor algebra $T\left(L\right)$ descends to a filtration of the universal enveloping algebra $U\left(L\right)$. The associated graded algebra of $U\left(L\right)$ - let us call it $GU\left(L\right)$ - is commutative (as is easily seen) and generated by the elements $\overline{\sigma\left(v\right)}\in U_1\left(L\right) / U_0\left(L\right)$ for $v\in L$ (where $\sigma$ denotes the map $L\to T\left(L\right)\to U\left(L\right)$). Thus, there exists a surjective $k$-algebra homomorphism $S\left(L\right)\to GU\left(L\right)$ which maps $v$ to $\overline{\sigma\left(v\right)}$ for every $v\in L$ (according to the universal property of the symmetric algebra). -One version of the Poincaré-Birkhoff-Witt theorem (abbreviated PBW theorem) says that under certain conditions, this homomorphism is actually an isomorphism. The Wikipedia page says that it is so if any of the following four cases holds (here I am quoting Wikipedia): -(1) $L$ is a flat $k$-module, -(2) $L$ is torsion-free as an abelian group, -(3) $L$ is a direct sum of cyclic modules (or all its localizations at prime ideals of $k$ have this property), or -(4) $k$ is a Dedekind domain. -A reference is given to a paper which I have no access to: -P.J. Higgins, Baer Invariants and the Birkhoff-Witt theorem, J. of Alg. 11, 469-482, (1969) -Most internet sources which prove PBW only prove it under the condition that $L$ is a free $k$-module. (Out of these proofs I consider Garrett's version most readable.) I am interested in a proof in case (2). I know that it is enough to consider the case when $k$ is a $\mathbb Q$-algebra. -The following two sources might give such a proof, if only I could understand them: -Source 1: -T. Ton-That, T.-D. Tran, Poincaré's proof of the so-called Birkhoff-Witt theorem Rev. Histoire Math., 5 (1999), pp. 249-284. As this is only formulated for $k$ a field, this needs some modifications, but that's not my main worry. I fail to understand this paragraph on pages 277-278: -"The first four chains are of the form -$U_1 = XH_1,\ U'_1 = H'_1Z,\ U_2 = YH_2,\ U'_2 = H'_2T$, -where each chain $H_1$, $H'_1$, $H_2$, $H'_2$ is a closed chain of degree $p - 1$; therefore by induction, each is the head of an identically zero regular sum. It follows that $U_1$, $U'_1$, $U_2$, $U'_2$ are identically zero, and therefore each of them can be considered as the head of an identically zero regular sum of degree $p$." -I don't understand the "It follows that $U_1$, $U'_1$, $U_2$, $U'_2$ are identically zero" part. This seems to be equivalent to $H_1 = H'_1 = H_2 = H'_2 = 0$, which I don't believe (the head of an identically zero regular sum isn't necessarily zero), but the authors are only using the weaker assertion that each of $U_1, U'_1, U_2, U'_2$ is the head of an -identically zero regular sum of degree $p$ - which, however, is still far from being obvious to me. -Source 2: -P.M. Cohn, A remark on the Birkhoff-Witt theorem, J. London Math. Soc. 38, 197-203, (1963). This gives a rather strange argument, which doesn't really rhyme for me. Probably I don't understand it though. If anybody could write it up in modern terms I would be very thankful. (If you want to know where exactly I am stuck, it's "$1w=w\in M_n$" on page 202, but I fear that there are also some things I have not really grasped before that point.) -Is there any accessible (I'm not at a university campus right now, and I need this rather soon) source for a proof of PBW in case (2)? -UPDATE (5 JUNE 2011): -(a) I do have Higgins's paper now. It is a beautiful and well-written piece of mathematics; I can hardly believe my eyes that something that well-written has been published in a journal! -This paper does not explicitly prove PBW in the cases (1) and (2), but the things it proves (combined with Lazard's theorem that flat modules are direct limits of free modules) are enough to conclude that PBW holds in the cases (1) and (2). -(b) Emanuela Petracci has a proof of PBW in case (2), and she even claims that it is Cohn's proof. This is probably just modesty, as she shows substantially more. I am going to read the proof when I have more time. -(c) My question about Source 1 still stands, although I don't really need that proof now that I know better ones. -(d) I have read the Deligne-Morgan proof; it is beautiful (although I would hardly call it straightforward, Theo; it is algebraic acrobatics in its purest form). - -REPLY [9 votes]: I liked the Cohn paper, although it's been a while since I read it, and I read it more to understand the counterexample in the second half than the proof in the first half. A proof in the case when your $k$ is a commutative algebra over $\mathbb Q$ is available in: - -Deligne, Pierre; Morgan, John W. Notes on supersymmetry (following Joseph Bernstein). Quantum fields and strings: a course for mathematicians, Vol. 1, 2 (Princeton, NJ, 1996/1997), 41--97, Amer. Math. Soc., Providence, RI, 1999. MR1701597. - -The proof is pretty much straightforward: when $k \supseteq \mathbb Q$, then the "symmetrization" map $S(L) \to T(L)$ that sends $l_1\cdot \dots \cdot l_n \mapsto \sum_{\sigma \in S_n} l_{\sigma(1)} \otimes \dots \otimes l_{\sigma(n)}$ exists, and you can check directly that the composition composition $S(L) \to T(L) \to U(L)$ is an isomorphism, is my memory is correct. Unfortunately, I don't have an electronic copy available. -(Deligne and Morgan also cite - -Corwin, L.; Ne'eman, Y.; Sternberg, S. Graded Lie algebras in mathematics and physics (Bose-Fermi symmetry). Rev. Modern Phys. 47 (1975), 573--603. MR0438925. - -to claim that actually only you need $2,3$ to be invertible. This cann't be right, given the counterexample in the paper by Cohn.) -Actually, looking again at the Cohn paper, he argues in two steps: - -Reduce to the case when $k \supseteq \mathbb Q$. -Prove it there. - -The first step is simple enough: all you need to prove is that $L \to U(L)$ is injective, and since $L \to L\otimes_{\mathbb Z}\mathbb Q$ is injective (since $L$ is torsion-free), it's enough to check that $L\otimes_{\mathbb Z}\mathbb Q \to U(L) \otimes_{\mathbb Z}\mathbb Q$ is injective, and for this it's enough to work over $k \otimes_{\mathbb Z}\mathbb Q$. -Ok, so the second step is the one that Deligne and Morgan do. And in Cohn's paper he actually spells out the "you can check directly" in my glib write-up above. Because, of course, it takes some work. One way to do this, and Cohn does roughly this, is by describing the inverse to the symmetrization map $S(L) \to U(L)$ in explicit enough detail that he can show that the inverse map makes sense over any $k$ --- Cohn does this by studying the case when $L$ is free. -Actually, Cohn does the following. He builds the map $S(L) \to U(L)$, which is a filtered map, and hence descends to a map $GS(L) = S(L) \to GU(L)$. He supposes that on associated graded the map is not an isomorphism, so that means that there is $w \in S(L)$ homogeneous of degree $n$ so that its image in $U(L)$ is in the $(n-1)$st filtration. Then if $M_{\leq \bullet}$ is a filtered $L$ module for which $L$ acts by moving up at most one degree, $w : M_{\leq \bullet} \to M_{\leq \bullet + n - 1}$, because we wrote the image of $w$ in $U(L)$ as a product of $n-1$-and-fewer things. But on the other hand, $S(L) \to U(L)$ is an $L$-module homomorphism, and Cohn proves that the image of $w$ in $U(L)$ sends $1\in S(L)$ to $w\in S(L)$, by showing this when $L$ is free.<|endoftext|> -TITLE: Taniyama's original conjecture -QUESTION [10 upvotes]: I've just read on Wikipedia that the original Taniyama conjecture about L-functions of elliptic curves over an arbitrary number field was still unproven. -This made me want to know more about this conjecture, but after a quick glance at the first results displayed by Google, I couldn't find out anything else than the modularity theorem. -So what did Yutaka Taniyama have on his mind? - -REPLY [13 votes]: There is an article by S. Lang on this subject that appeared in the Notices of the AMS (11/1995); it contains the exact statements of Taniyama's problem(s) on this subject. I reproduce two below (this is taken from a longer list of problems, not all are relevant to the question, and not all are in the article). However, this is mainly a historical article, and part of a controversy regarding proper attribution/naming of the problem over the rationals; thus, while having no personal opinion on this, I should perhaps point out that the views expressed in the article I link to might not be universally accepted. As said by François Brunault (at least this is how I interprete his statement) what precisely was asked in 1955, today seems mainly of historical (not so much mathematical) relevance. - - -12. Let C be an elliptic curve defined over an algebraic number field $k$, and $L_C(s)$ denote the $L$-function of $C$ over $k$. Namely, - $$\zeta_C(s) = \frac{\zeta_k(s) \zeta_k(s − 1)}{L_C(s)}$$ - is the zeta function of $C$ over $k$. If a conjecture of Hasse is true for $\zeta_C(s)$, then the Fourier series obtained from $L_C(s)$ by the inverse Mellin transformation must be an automorphic form of dimension $−2$, of some special type (cf. Hecke). If so, it is very plausible that this form is an elliptic differential of the field of that automorphic - functions. The problem is to ask if it is possible to prove Hasse’s conjecture for $C$; - by going back this considerations, and by finding a suitable automorphic form from which - $L_C(s)$ may be obtained. -13. Concerning the above problem, our new problem is to characterize the field of elliptic - modular functions of "Stufe" $N$, and especially, to decompose the Jacobian variety $J$ of this function field into simple factors, in the sense of isogeneity. - It is well known, that, in case N = q is a prime number, satisfying $q = 3 \pmod{4}$, $J$ contains elliptic curves with complex multiplication. Is this true for general $N$? - - -After reproducing these two problems the article continues "As Shimura has pointed out, there were some questionable aspects to the Taniyama formulation in problem 12. First, the simple Mellin transform procedure would make sense only for elliptic curves defined over the rationals; the situation over number fields is much more complicated and is not properly understood today, even conjecturally." -(ADDED: It occured to me that the combination of my answer and my comment, could lead to a misinterpretation. The content I reproduced, in particular the final assertion, are not those aspects on which different opinions exist.)<|endoftext|> -TITLE: Rational points on algebraic curves over $\mathbb Q^\text{ab}$ -QUESTION [17 upvotes]: Motivation: -Let $\mathbb{Q}_{\infty,p}$ be the field obtained by adjoining to $\mathbb{Q}$ all $p$-power roots of unity for a prime $p$. The union of these fields for all primes is the maximal cyclotomic extension $\mathbb{Q}^\text{cycl}$ of $\mathbb{Q}$. By Kronecker–Weber, $\mathbb{Q}^\text{cycl}$ is also the maximal abelian extension $\mathbb{Q}^\text{ab}$ of $\mathbb{Q}$. -A well known conjecture due to Mazur (with known examples) asserts, for an elliptic curve $E$ with certain conditions, that $E(\mathbb{Q}_{\infty,p})$ is finitely generated. This is the group of rational points of $E$ over $\mathbb{Q}_{\infty,p}$ (not a number field!). -A theorem due to Ribet asserts the finiteness of the torsion subgroup $E(\mathbb{Q}^\text{ab})$ for certain elliptic curves. -Questions: -(a) Can one expect to find elliptic curves (or abelian varieties) $A$ with $A(\mathbb{Q}^\text{ab})$ finitely generated? -(c) Can one expect to find curves $C$ of genus $g >1$ with $C(\mathbb{Q}^\text{ab})$ finite? - -REPLY [14 votes]: Actually Ken Ribet proved that if $K$ is a number field and $K(\mu_{\infty})$ is its infinite cyclotomic extension generated by all roots of unity then for every abelian variety $A$ over $K$ the torsion subgroup of $A(K(\mu_{\infty}))$ is finite: -http://math.berkeley.edu/~ribet/Articles/kl.pdf . -On the other hand, Alosha Parshin conjectured that if $K_{p}$ is the extension of $K$ generated by all $p$-power roots of unity (for a given prime $p$) then the set $C(K_{p})$ is finite for every $K$-curve $C$ of genus $>1$: -http://arxiv.org/abs/0912.4325 (see also http://arxiv.org/abs/1001.3424 ).<|endoftext|> -TITLE: How transitive are the actions of symplectomorphism groups ? -QUESTION [15 upvotes]: This question is motivated by the classical fact from differential geometry : -Let $M$ be a smooth manifold of dimension at least $2$. Then for any $n$ the diffeomorphism group $\textrm{Diff}(M)$ acts transtively on the configuration space of $n$-points in $M$ or equivalently it acts $n$-transitively on $M$. -As I recall, it is known that the symplectomorphism group $(M,\omega)$ acts transitively on $M$, which is assumed to be symplectic. My question is then the following : -Let $(M,\omega)$ be a symplectic manifold. -(i) When does $\textrm{Symp}(M,\omega)$ act $n$-transtively for $n\geq 2$ ? -(ii) If the answer above is NOT ALWAYS then what is known ? -As some background, the usual way one proves (rather the only way I know how to prove this) the first fact is by showing the following : -(i) for two sets of distinct $n$ points in $M$ given by $\{p_1,\ldots,p_n\}$ and $\{q_1,\ldots,q_n\}$ which are close, we find disjoint disks $D_i$'s containing $p_i,q_i$. This requires dimension at least $2$. Use some diffeomorphism of $D_i$ that is smoothly identity at the boundary and looks like a rotation inside $D_i$ that swaps $p_i$ and $q_i$. -(ii) Define the natural equivalence relation on $n$-tuples and observe that the configuration space of $n$-points in $M$. By (ii) each equivalence class is open. It is alsoclosed being the complement of open sets. Since the configuration space is connected (this requires dimension at least $2$) this means there is only one equivalence class. -Does this idea work in the symplectic setting - perhaps by taking paths $\gamma_i$ from $p_i$ to $q_i$ and getting Hamiltonian vector fields via $\omega(\gamma_i',\cdot)$ ? - -REPLY [5 votes]: The main question has already been answered, but I wanted to add the following remark. When you have a group acting transitively on a certain space then you know that the space in question will be a homogenous space that is isomorphic to the quotient of the group by the stabilizer of a point. -Every possible homogenous space is thus characterized by a closed subgroup of the group. -I read many years ago a paper by Banyaga; I think this one -A. Banyaga, Isomorphisms between classical diffeomorphism groups., Geometry, topology, and dynamics (Montreal, PQ, 1995), 1–15, -CRM Proc. Lecture Notes, 15, Amer. Math. Soc., Providence, RI, 1998. -Here he characterized several types of geometric structures by the isomorphism groups. The question was for example: Suppose that you know that the group of symplectomorphisms of two symplectic manifolds are isomorphic, can you then reconstruct or identify the underlying symplectic manifolds. As far as I remember, he characterized several structures like smooth manifold, symplectic structures, contact structures, volume forms etc. where knowledge of the isomorphism group (forgetting topology and only remembering the algebraic structure) allows you to identify the underlying geometric structure (in the symplectic case up to rescaling). -For a certain isomorphism group to have such a property, he needed the group to act n-transitively (realized by an isotopy with arbitrarily small support) for any n. I do not remember the details anymore, but the result was somehow that the stabilizer of any point in such a group is a maximal subgroup and if we have a group isomorphism between two groups mapping one maximal subgroup to another one, then the isomorphism induces an identification between the underlying geometric space. -I apologize for not remembering more details, but I found the result quite interesting at time I read this.<|endoftext|> -TITLE: Crystalline Characters -QUESTION [8 upvotes]: Let $K$, $L$ be finite extensions of the $p$-adic numbers. Suppose $\chi:G_K\rightarrow L^{\times}$ is crystalline. It is my understanding that if either $K$ or $L=\mathbb{Q}_p$, then $\chi$ must be a Tate twist of an unramified character. Is there a classification of crystalline characters in general? - -REPLY [7 votes]: [EDIT: I misunderstood the question -- I thought the poster wanted to know why all crystalline characters are unramified twists of powers of cyclotomic when $L$ or $K$ is $\mathbb{Q}_p$, and wrote out a detailed proof. I realise now that the poster already knew this but wanted to understand the classification in the general case. I thought I'd leave this post here anyway in case anyone finds it useful.] -Step 1: Consider the space $\mathbb{D}_{\mathrm{dR}}(\chi)$. This is a free rank 1 module over $K \otimes_{\mathbb{Q}_p} L$, and the steps in the Hodge filtration are $K \otimes_{\mathbb{Q}_p} L$-submodules. By assumption, $K \otimes_{\mathbb{Q}_p} L$ is a field, there is only one Hodge-Tate weight. By twisting by a power of the cyclotomic character, we can assume without loss of generality that the Hodge-Tate weight is 0. -Step 2: Now consider the space $\mathbb{D}_{\mathrm{cris}}(\chi)$. This is a free rank 1 module over $L K_0$, with a Frobenius that is L-linear and $K_0$-semilinear. (Here $K_0$ is the maximal unramified subfield of $K$). If $[K_0 : \mathbb{Q}_p] = d$, then $\varphi^d$ is linear, so acts as multiplication by a scalar $\mu \in L$. -Step 3: Consider the unramified character $G_{K_0} \to L$ mapping geometric Frobenius to $\mu^{-1}$. Tensoring with this character, we may assume that $\mu = 1$. -Step 4: Now choose a basis of $\mathbb{D}_{\mathrm{cris}}(\chi)$ over $LK_0$, and suppose that (with respect to this basis) $\varphi$ acts as multiplication by $\alpha$. The condition that $\varphi^d = 1$ translates to $\alpha \varphi(\alpha) \dots \varphi^{d-1}(\alpha) = 1$. If $K = \mathbb{Q}_p$, then this is just $\alpha = 1$. If $K \ne \mathbb{Q}_p$, then $L = \mathbb{Q}_p$, and $\alpha \in K_0^\times$ satisfies $N_{K_0 / \mathbb{Q}_p} \alpha = 1$. Hence $\alpha = \varphi(x) / x$ for some $x \in K_0^\times$, by Hilbert 90; and after changing basis by $x$ we again have $\alpha = 1$. -So we've shown that $\mathbb{D}_{\mathrm{cris}}(\chi)$ is isomorphic (as a filtered $\varphi$-module) to $\mathbb{D}_{\mathrm{cris}}$ of the trivial character, i.e. $\chi$ is trivial (and hence crystalline!). QED.<|endoftext|> -TITLE: "Secondary operations" for a group acting on a chain complex -QUESTION [12 upvotes]: Suppose a group G acts on a chain complex K and induced action on H(K) is trivial. What "secondary operations" on H(K) can be defined in this situation? - -Example. If $G=\langle\sigma\rangle/\sigma^n$ acts trivially on H(K) then $x-\sigma x=dl(x)$ (for some function $l$) and a secondary operation $x\mapsto l(x)+\sigma l(x)+\dots+\sigma^{n-1}l(x)$ is well-defined mod n. And this operation is non-trivial (consider a complex $Z[G]\to Z[G]$, $x\mapsto (1-\sigma)x$). -So looks like these operations has something to do with group homology, but details elude me. - -Update. Two nice answers explain what is the meaning of the operation from the example above (and how it can be defined for an arbitrary group). -But does this construction give all operations? I.e. what structure on H(K) one needs to recover K (up to q/iso)? Like, - -associative multiplication on K $\Leftrightarrow$ $A_\infty$-structure on H(K); -G-action on K $\Leftrightarrow$ ??? on H(K). - -(Perhaps, there is a very general answer: not just for k[G] but for an arbitrary algebra — or even arbitrary operad, maybe. Probably, Tyler Lawson's comment is relevant — if somebody could elaborate on that...) - -REPLY [3 votes]: (Nothing really new here, but just to link $A_\infty$-description and the explicit construction.) -Action of G on K induces $A_\infty$-action of Z[G] on H(K)=:H. And this is all operations one can define on H(K), since this structure defines K up to (equivariant) q/iso (ref). -Now, let's rewrite this in more concrete terms. $A_\infty$-action of Z[G] on H is just a collection of maps $m_i\colon G^{i-1}\times H\to H$ aka $m_i\colon G^{i-1}\to\operatorname{Hom}(H,H)$ ($i=2,3,\dots$). This maps are subject to some relations. In particular, if $m_{i+1}$ is first non-zero higher action, then $m_{i+1}$ satisfies cocycle condition and gives an element in $H^i(G,\operatorname{Hom}(H,H))$. For i=2 it is exactly the element from Tom Goodwillie's answer (and, probably, it also coincides with $d_i$ from Tyler Lawson's answer, although I don't have a proof).<|endoftext|> -TITLE: Left-bracketed Ackermann function also not primitive recursive? -QUESTION [11 upvotes]: The original Ackermann function $\varphi\colon \mathbb{N}\times\mathbb{N}\times\mathbb{N}_0\to \mathbb{N}$ as defined in [1] was invented to prove that there is a function that is recursive but not primitive recursive. -It can be given by the following recursion: - -$\varphi(a,b,0) = a+b$ -$\varphi(a,b,n+1) = (x\mapsto \varphi(a,x,n))^b(\alpha(a,n))$ - -Where $\alpha(a,0)=0$, $\alpha(a,1)=1$ and $\alpha(a,n)=a$ for $n\ge 2$ are initial values and $(x\mapsto f(x))^k$ is the k-times composition of the function $x\mapsto f(x)$. -The function $n\mapsto \varphi(n,n,n)$ is not primitive recursive because - informally speaking - it grows too quickly. -These operations are right-bracketed, this does not matter for $\varphi(a,b,1)=ab$ and $\varphi(a,b,2)=a^b$, but for the next higher rank it is important -$\varphi(a,b,3)=\underbrace{a^\land (a^\land (...(a^\land a)))}_{b+1\; \text{occurences of}\; a}$ where $a^\land b:=a^b$. -If we would choose left-bracketing the functions would not grow so quickly. The left-bracketed operations would be defined as: - -$\psi(a,b,0) = a+b$ -$\psi(a,b,n+1) = (x\mapsto \psi(x,a,n))^b(\alpha(a,n))$ - -Again $\psi(a,b,1)=ab$ and $\psi(a,b,2)=a^b$, but here the forth operation would be $\psi(a,b,3)=a^{a^b}$ -My question is now whether the left-bracketed operations still grow fast enough for not being primitive recursive, i.e. is $n\mapsto \psi(n,n,n)$ still not primitive recursive? -[1] Ackermann, W. (1928 ). Zum Hilbertschen Aufbau der reellen Zahlen. Math. Ann., 99, 118–133. - -REPLY [8 votes]: Sorry for reviving this ancient post, but it seemed to me that this question shouldn’t be left unanswered. -It is easy to establish by induction that for $a,b,n\ge2$, the function $\psi(a,b,n)$ is strictly increasing in all three coordinates, and $$\psi(a,b,n)>a,b,n.$$ -Then we can show -$$\tag{$*$}\psi(a,b,n+1)\ge\psi(b,a,n)$$ -for $a,b,n\ge2$ by induction on $b$: clearly, $\psi(a,2,n+1)=\psi(\psi(a,a,n),a,n)\ge\psi(2,a,n)$, and going from $b$ to $b+1$, we have -$$\psi(a,b+1,n+1)=\psi(\psi(a,b,n+1),a,n)\ge\psi(\psi(b,a,n),a,n)\ge\psi(b+1,a,n)$$ -using monotonicity and $\psi(b,a,n)\ge b+1$. -Thus, by $(*)$ and monotonicity, we get -$$\psi(a,b,n+2)\ge\bigl(x\mapsto\psi(a,x,n)\bigr)^b(a),$$ -which implies -$$\psi(a,b,2n-2)\ge\varphi(a,b,n)$$ -for all $a,b,n\ge2$ by induction on $n$. Consequently, $\psi$ (or its diagonal, for that matter) is not bounded by any primitive recursive function.<|endoftext|> -TITLE: Entropy of first return map and suspension flows -QUESTION [11 upvotes]: There are some well know formulas of Abramov about derived systems. -Firstly let $(X,\mu,f)$ be a probability preserving system and $A\subset X$ is measurable such that $\bigcup_{n\ge0}f^nA=X$. Let $\mu_A$ be conditional probability of $(\mu,A)$ and $f_A:A\to A$ be the first return map with respect to $A$, that is, $f_A(x)=f^{k(x)}x$ where $k(x)=\inf[k\ge1: f^kx\in A]$ (well defined up to a null set). Then Abramov proved that -$h(f_A,\mu_A)\cdot\mu(A)=h(f,\mu)$. -Secondly let $(X,\mu,f)$ be a probability preserving system and $r:X\to (c,\infty)$ be a roof function with $c>0$ and $\int rd\mu<\infty$. Then consider the suspension space $X_r=[(x,y):x\in X,0\le y\le r(x)]/\sim$ where $(x,r(x))\sim(fx,0)$, the suspension measure $\tilde{\mu}$ given by -$\tilde{\mu}(A)=\int_X |A_x|d\mu(x)/\int_X rd\mu$, and the suspension flow -$\tilde{f}_t:X_r\to X_r,[x,y]\mapsto[x,y+t]$. Abramov also proved that -$h(\tilde{f},\tilde{\mu})\cdot\int_X rd\mu=h(f,\mu)$. -I think there are direct/intuitive proofs of these two entropy formulas. Any explanation will be great. -Thanks~ - -For example the first can be derived from the discrete version of the second one: -The first return time $k:A\to\mathbb{N}$ can be viewed as a discrete roof function on $A$. Then suspension is $A_k=[(x,k):x\in A,k=0,1,\cdots, k(x)]/\sim$, which is isomorphic to $X$. And the map $A_k\to A_k,[x,k]\to[x,k+1]$ is isomorphic to $T$ on $X$ (note that $\int_A k(x)d\mu(x)=1$, or equally $\int k(x)d\mu_A(x)=1/\mu(A)$). By identifying $A$ with $A\times[0]\subset A_k$, we see that -$h(f,\mu)/\mu(A)=h(f_A,\mu_A)$. -Still I have no idea about the proof about the entropy of suspension flows~ - -Finally I understood Abramov's proof (indeed his proof is very clear). -Pick $t\in(0,c)$ (fixed from now on) and consider the subset $A=[(x,s):0\le s < t]\subset X_r$. Then the first return map $\tilde{f}_A$ of $\tilde{f}_t$ with respect to $A$ is given by $(x,s)\mapsto(fx,s-r(x))$, where $s-r(x)\in\mathbb{T}_t$ (Note that we can view $A=X\times\mathbb{T}_t$ and $\tilde{f}_A$ as a fiber extension of $f$). - -He showed that $h(\tilde{f}_A,\tilde{\mu}_A)=h(f,\mu)$ (since the extension is isometric on the fiber). -As the first return map of $(X_r,\tilde{f}_t)$, $h(\tilde{f}_A,\tilde{\mu}_A)\cdot\tilde{\mu}(A)=h(\tilde{f}_t,\tilde{\mu})$. - -Plugging in $\tilde{\mu}(A)=\frac{t}{\int rd\mu}$, he got the desire formula $h(\tilde{f}_t,\tilde{\mu})=h(f,\mu)\cdot\frac{t}{\int rd\mu}$. - -REPLY [13 votes]: What may seems intuitive to you depends on your mental model[s] for entropy. This is not static, since we change our mental models and learn new models with time and experience. -One good intuitive way to think about entropy is in terms of information theory, as developed by Claude Shannon, where he showed how entropy can be interepreted as the lower bound number of the number of bits required to encode a process, as the encoding becomes more efficient with larger and larger batches. -These days it is commonplace to talk about units such as Mbps to measure digital data connections, or channel capacity, without all the complications of noise that were the big concern in Shannon's day. The entropy of a measure-preserving dynamical system is the number of bits per unit time needed to describe an orbit of an average point using a codec (coding-decoding algorithm) of limiting high efficiency. More precisely, it is the infimum of the mean number of bits per unit time per point to describe the orbits of a large batch of uniformly-distributed randomly-chosen points for a long time. -The theorems of Abramov, thought about in this way become obvious. They're generalizations of the idea that the number of megabits per minute is 60 times the number of megabits-per-second. The orbits contain the same information. [It's important here that the invariant measure is preserved, up to a constant, by the modification of the measurable dynamical system]. The only thing that has changed is the measurement of space$\times$time; the formula gives ratio of volumes of space-time. -By the way, the corresponding formulas fail for topological entropy in place of measure-theoretic entropy. (One characterization of topological entropy is the supremum of measure-theoretic entropy over all possible invariant measures). - For example, the shift S on two symbols (i.e., arbitrary two-sided-infinite sequences of 0's and 1's) has topological and measure-theoretic -entropy 1 bit per unit of time. Let $X$ be the union of $S$ with a cycle of length $n$, -i.e., addition mod n acting on integers mod n. The topological entropy of X is still -1 bit per unit time, but the measure-theoretic entropy is the ratio of the volume of S to the volume of $X$ per unit time. If you slow the two parts down by taking a section $A$, the measure-theoretic entropy changes as per the formulas you've given, but the topological entropy depends only on the intersection of $A$ with the shift portion. In general, the change in topological entropy coming from a change in the time parameter will be quite tricky.<|endoftext|> -TITLE: An elementary question about the cut locus -QUESTION [12 upvotes]: Let $M$ be a Riemannian manifold, $x$ and $y$ are two points in $M$. -Assume that $x$ is not in the cut locus of $y$. Does there exist a neighborhood $U$ of $x$ and a neighborhood $V$ of $y$ such that for every point $u$ in $U$ and for every point $v$ in $V$ we have that $u$ is not in the cut locus of $v$? - -REPLY [8 votes]: For a unit tangent vector $u$ with footpoint $p$ let $t(u)$ be the supremum of positive numbers such that the geodesic $t\to \exp_p(tu)$ is minimizing on $[0,t(u)]$. The cut locus at $p$ is the set of points $\exp_{p}(t(u) u)$ of $M$ for which $t_u$ is finite. -A basic result is that $u\to t(u)$ defines a continuous map from the unit tangent bundle to $(0,+\infty]$ where continuity at $+\infty$ is understood in the obvious way. See e.g. Sakai's "Riemannian geometry", Proposition III.4.1. -Now coming to your question fix $x\in M$ and $y=\exp_x(su)$ for some $u=u(x,y)$ and positive number $s$. Suppose $x^\prime$, $y^\prime$ are near $x$, $y$ respectively, -and write $y^\prime=\exp_{x^\prime}(s^\prime u^\prime)$. -If $y$ is not in the cut locus of $x$, then $t(u)=+\infty$. So $t(v)> s$ on some neighborhood of $u$ in the unit tangent bundle. Since $s^\prime$ and $s$ are almost the same, we conclude that $t(u^\prime)> s^\prime$, i.e. $x^\prime$, $y^\prime$ are not cut points of each other. -In my view Sakai's book is the best source of information about cut and cunjugate loci.<|endoftext|> -TITLE: How to prove that $e^{zw}$ can not be written as a rational expression in functions in $z$ and in $w$? -QUESTION [18 upvotes]: Let $K$ be the field of fractions of -$\mathbb{C}[[z]]\otimes_{\mathbb{C}}\mathbb{C}[[w]]\subset \mathbb{C}[[z, w]].$ Given -a formal power series in $t, f\in \mathbb{C}[[t]],$ is there any simple criterion which -will conclude that $f(zw)$ does not belong to $K?$ I suspect that -$f(zw)\in K$ if -and only if $f(t)$ is a rational function, i.e. belongs to $\mathbb{C}(t).$ I am -especially interested in proving that $e^{zw}\notin K.$ - -REPLY [15 votes]: Theorem 1. Let $f\in\mathbb{C}[[t]]$. Then $f(zw)\in \mathbb{C}[[z]]\otimes_{\mathbb{C}}\mathbb{C}[[w]]$ if and only if $f\in\mathbb{C}[t].$ -Proof. The "if" part is obvious. For the "only if" part assume that $f\notin\mathbb{C}[t]$ but -$$f(zw) = \sum_{i=1}^n g_i(z)h_i(w),$$ -where $g_i\in\mathbb{C}[[z]]$ and $h_i\in\mathbb{C}[[w]]$. We can assume that $n>0$ is minimal. If $g_i(0)=h_i(0)=0$ for all $i$, then $f(0)=0$ and we can divide both sides by $zw$ in the obvious way. Therefore $g_i(0)\neq 0$ or $h_i(0)\neq 0$ for some $i$. Without loss of generality, $g_1(0)=1$, then -$$h_1(w)=f(0)-\sum_{i=2}^n g_i(0)h_i(w),$$ -so that -$$f(zw)=g_1(z)f(0)+\sum_{i=2}^n\bigl(g_i(z)-g_i(0)g_1(z)\bigr)h_i(w).$$ -Taking derivative with respect to $w$ and then dividing by $z$ yields -$$f'(zw)=\sum_{i=2}^n\frac{g_i(z)-g_i(0)g_1(z)}{z}h_i'(w).$$ -Here $f'\notin\mathbb{C}[t]$, but the fractions are elements of $\mathbb{C}[[z]]$, contradicting the minimality of $n$. -Theorem 2. $e^{zw}\notin K$. -Proof. Assume there are $u_i,f_j\in\mathbb{C}[[z]]$ and $v_i,g_j\in\mathbb{C}[[w]]$ such that -$$e^{zw}\sum_{i=1}^m u_i(z)v_i(w)=\sum_{j=1}^n f_j(z)g_j(w),$$ -where the sums are nonzero. We can assume that $n>0$ is minimal, and also that $m>0$ is minimal for this $n$. In particular, the $u_i$'s are independent over $\mathbb{C}$. -We can assume that some $v_i(0)$ or $g_j(0)$ is nonzero, for otherwise we can divide both sides by $w$ in the obvious way. Then -$$ \sum_{i=1}^m u_i(z)v_i(0)=\sum_{j=1}^n f_j(z)g_j(0) $$ -is a nonzero element of $\mathbb{C}[[z]]$, call it $h(z)$. In particular, $g_j(0)\neq 0$ for some $j$. Without loss of generality $g_1(0)=1$, then with the notation -$$\tilde u_i(z):=u_i(z)/h(z)\in\mathbb{C}((z))$$ -$$\tilde f_j(z):=f_j(z)/h(z)\in\mathbb{C}((z))$$ -$$e^{zw}\sum_{i=1}^m \tilde u_i(z)v_i(w)=g_1(w)+\sum_{j=2}^n \tilde f_j(z)\bigl(g_j(w)-g_j(0)g_1(w)\bigr).$$ -The sum on the right hand side is independent of $z$ for otherwise we could contradict the minimality of $n$ by taking the $z$-derivative on both sides and then multiplying by a large power of $z$ to turn elements of $\mathbb{C}((z))$ into elements of $\mathbb{C}[[z]]$. At any rate, the right hand side is a nonzero element $k\in\mathbb{C}[[w]]$, because $k(0)=g_1(0)=1$. With the notation -$$ \tilde v_i(w):=v_i(w)/k(w)\in\mathbb{C}((w))$$ -we conclude -$$e^{-zw}=\sum_{i=1}^m \tilde u_i(z)\tilde v_i(w).$$ -In particular, there is some integer $r>0$ such that -$$(zw)^r e^{-zw}\in \mathbb{C}[[z]]\otimes_{\mathbb{C}}\mathbb{C}[[w]],$$ -which contradicts Theorem 1. The proof of Theorem 2 is complete.<|endoftext|> -TITLE: Niemeier lattices and theta functions -QUESTION [12 upvotes]: I have an extremely elementary question. Let's say someone randomly hands you a theta function associated to a Niemeier lattice (unimodular even, n=24). What can you say about which Niemeier lattice gave you this theta function in the first place? - -REPLY [21 votes]: The theta series for a Niemeier lattice determines the lattice in most cases, but there are five ambiguous pairs. -The theta series of an even unimodular lattice must be a polynomial in the theta series of $E_8$ and $\Lambda_{24}$ (this is a modular forms calculation). Thus, for Niemeier lattices, it must be a linear combination of those for $E_8^3$ and $\Lambda_{24}$. The constant term must be $1$, so there is one remaining degree of freedom. This means the theta series for a Niemeier lattice is determined by how many roots (i.e., vectors of norm $2$) it has. There are five pairs of Niemeier lattices with the same number of roots, but no triples.<|endoftext|> -TITLE: Margulis normal subgroup theorem -QUESTION [5 upvotes]: Margulis' normal subgroup theorem states that any normal subgroup of a higher rank lattice is either finite or of finite index. The obvious question is: can one classify finite normal subgroups of such lattices? (even $SL(n, \mathbb{Z})$ and $Sp(2n, \mathbb{Z})$ would be a good start). - -REPLY [14 votes]: These are the central subgroups, see http://www.mathematik.uni-regensburg.de/loeh/seminars/normal_subgroup_thm.pdf . It is proved that every non-central normal subgroup has finite index (page 7).<|endoftext|> -TITLE: When is the Levi subalgebra an ideal? -QUESTION [7 upvotes]: Let $L$ be a Lie algebra. It is known that $L$ admits a Levi decomposition (possibly non unique): -$L=S\oplus rad(L) $, -where $rad(L)$ is the solvable radical and $S$ is a semisimple subalgebra. -If $L$ is reductive, this is, if -$rad(L)=Z(L)$, -where $Z(L)$ is the center, then $S$ is an ideal. -Is the converse true? Or is it possible to exhibit a non reductive Lie algebra that admits a Levi decomposition such that $S$ is an ideal? -(edited) Is there any method to determine whether is possible to have $S$ as an ideal or not? -Remarks: $L$ is finite dimensional and the underlying field is algebraically closed of characteristic $0$ . The direct sum sign means direct sum of vector subspaces. - -REPLY [6 votes]: To answer the original question: Certainly it's possible for $L$ to be non-reductive (and non-solvable) while a Levi subalgebra $S$ is an ideal. Just form the direct sum as Lie algebras of your favorite semisimple Lie algebra and a nonabelian solvable Lie algebra. I think the additional edited question about when $S$ can be an ideal is more interesting. Throughout the discussion it's important to work just with finite dimensional Lie algebras over a field of characteristic 0, since otherwise the classical results break down. -What's involved here is usually called the Levi-Mal'cev Theorem, given thorough treatments in older books such as Jacobson Lie Algebras (III.9) and Bourbaki -Groupes et algebres de Lie (Chapter I, Section 6.8). (The brief online treatment in the Springer math encyclopedia isn't quite correct, but there may be other online sources.) -Both parts of the theorem are relevant here: the existence in $L$ of a semisimple subalgebra $S$ complementary (in the vector space sense) to the solvable radical; the conjugacy of all such subalgebras $S$ under the subgroup of automorphisms of $L$ generated by all exp(ad x) with x in the nilradical. Bourbaki states as Corollary 4 of the theorem that every ideal in $L$ is the direct sum of its intersections with $S$ and with $rad(L)$ (the former being a Levi subalgebra in the given ideal). In particular, when $S$ itself is an ideal of $L$ it must be the unique Levi subalgebra. All of this follows easily from the Levi-Mal'cev Theorem, though the proof of the theorem itself is nontrivial. The conjugacy of Levi factors is really essential here, not just the existence. -To return to the header of the original question, it should be clear by now that $S$ is an ideal precisely when $L$ is a Lie algebra direct sum of a semisimple Lie algebra and a solvable Lie algebra.<|endoftext|> -TITLE: Why are connective spectra called "connective"? -QUESTION [5 upvotes]: Recall that a spectrum is called connective if it is $(-1)$-connected (that is, its homotopy is concentrated in nonnegative degrees). -However, this left me scratching my head a bit. Why "connective"? Is there some geometric intuition behind it that I'm missing? - -REPLY [9 votes]: Maybe just because (-1)-connected is a mouthful? And connected would not be the right term, since that would imply trivial $\pi_0$. Other than that, I don't know.<|endoftext|> -TITLE: Encrypting a message for multiple recipients -QUESTION [13 upvotes]: Let $m$ be a secret message that needs to be sent to $n >1$ recipients. Let each recipient $r_i$ have a public key $p_i$ and private key $s_i$. Is there a scheme such that we can encrypt the message $m$ using the $n$ public keys and produce a encrypted message $E(m)$ such that only the $n$ intended recipients can decipher the message $m$? -One method could be to encrypt the message $n$ times, using each recipients public key, and append them all. That is, if $e_i$ is the encrypted message $m$ using the public key $p_i$, then the encrypted message sent to all the recipients could be $E(m) = p_1|e_1|p_2|e_2 \dots p_n|e_n$, where $|$ is the concatenation operation. But this will increase the size of the encrypted message by $O(n)$. So my question really is – can we keep the message length manageable and simultaneously allow multiple recipients to securely decrypt the message. -Is there a special name given to this kind of cryptography problem? The problem seems natural and perhaps has been investigated. I will be grateful for pointers to literature in this area. - -REPLY [20 votes]: This problem is often called broadcast encryption: how can you set up a system that will enable transmission of an encrypted message to an arbitrarily chosen subset of the people involved? There's a trade-off between two difficulties. If you just give each user an individual key in a generic public key cryptosystem, without setting up some sort of special system, then there's nothing you can do except to append all the encryptions, which is a problem if the number of users is huge. (Imagine broadcasting an event to a billion subscribers. Adrien's observation is an important point, but it doesn't change the scaling.) In the other extreme, you could assign a different key to each subset you might ever want to broadcast to. This works fine if you care only about a few subsets, but if you want a lot of flexibility then you get too many keys for the users to keep track of. It turns out that there are nontrivial solutions to this problem. The original paper is by Fiat and Naor (in Crypto '93), and searching for "broadcast encryption" online will give tons of follow-up papers offering extensions and improvements. -You might also be interested in "attribute-based encryption", in which each user is associated with a list of different attributes and you can encrypt messages so they can be decrypted by just the users with desired combinations of attributes. This is a somewhat different approach, but it has some beautiful consequences. The original paper is by Sahai and Waters (Eurocrypt 2005), and again there have been many further papers.<|endoftext|> -TITLE: Closest vector problem (=nearest lattice point) is trivial for "reduced lattice" ? -QUESTION [6 upvotes]: Consider some lattice in R^n. Take some point "P" in R^n (which does not belong to this lattice in general). The problem is to find "nearest" lattice point. The problem is known NP-hard in general it is named "closest vector problem". -Is it correct that "hardness" comes from the "lattice reduction" step ? -I mean if in our lattice we can easily get the shortest and any other vector or basis we want then problem becomes not NP-hard ? -Some related posts: -How to find nearest lattice point to given point in R^n ? Is it NP ? -Lattice reduction in R^3 (R^4) or what is fundamental domain for SL(3,Z) , (SL(4,Z)) ? -How "often" does LLL-reduction produce "optimal" result ? Is there condition (or informal understanding) on lattice that it LLL is optimal ? - -REPLY [12 votes]: It actually remains hard, although not as hard. See the paper Hardness of approximating the closest vector problem with pre-processing by Alekhnovich, Khot, Kindler, and Vishnoi (FOCS 2005, http://www.math.ias.edu/~misha/papers/cvpp.pdf). They show that even if you are allowed to do arbitrary pre-processing after having been given the lattice basis but before seeing the target vector, you still can't get a better approximation factor than $(\log n)^{1/2-\varepsilon}$, assuming your calculations after seeing the target vector must run in polynomial time, and assuming that there are no quasi-polynomial-time algorithms for NP. -However, you do get real benefit from being able to do lattice basis reduction in advance (not just LLL reduction, but something more computationally intensive). For example, Lagarias, Lenstra, and Schnorr showed that using Korkine-Zolotarev reduction, you could get an $n^{1.5}$ approximation factor (Combinatorica 10 (1990), 333–348). In FOCS 2004, Aharonov and Regev got a $\sqrt{n/\log n}$ approximation factor with pre-processing. However, in all these algorithms the pre-processing requires more than polynomial time.<|endoftext|> -TITLE: Invertible matrices of natural numbers are permutations... why? -QUESTION [37 upvotes]: I have heard the following statement several times and I suspect that there is an easy and elegant proof of this fact which I am just not seeing. - -Question: Why is it true that an invertible nxn matrix with non-negative integer entries, whose inverse also has non-negative integer entries, is necessarily a permutation matrix? - -The reason I am interested in this has to do with categorification. There is an important 2-category, the 2-category of Kapranov–Voevodsky 2-vector spaces, which in one incarnation has objects given by the natural numbers and 1-morphisms from n to m are mxn matrices of vector spaces. Composition is like the usual matrix composition, but using the direct sum and tensor product of vector spaces. The 2-morphisms are matrices of linear maps. -The above fact implies that the only equivalences in this 2-category are "permutation matrices" i.e. those matrices of vector spaces which look like permutation matrices, but where each "1" is replaced by a 1-dimensional vector space. -It is easy to see why the above fact implies this. -Given a matrix of vector spaces, you can apply "dim" to get a matrix of non-negative integers. Dimension respects tensor product and direct sum and so this association is compatible with the composition in 2-Vect. Thus if a matrix of vector spaces is weakly invertible, then its matrix of dimensions is also invertible, and moreover both this matrix and its inverse have positive interger entries. Thus, by the above fact, the matrix of dimesnions must be a permutation matrix. -But why is the above fact true? - -REPLY [6 votes]: Here is a particularly simpleminded proof (which is probably the same as some of the others). Suppose that $A$ and $B$ are $n \times n$ matrices with non-negative entries and that $AB=D$ is a diagonal matrix. For each non-zero entry $a_{ij} \gt 0$ of $A$ we have $b_{j \ell}=0$ for all $\ell \ne i$ since $d_{i \ell} = 0$. Likewise, for each $b_{ji} \gt 0$ we have $a_{kj}=0$ for all $k \ne i.$ -Now suppose that $D$ is invertible, then each row of A has a non-zero entry. We will now see that it is the only non-zero entry of its row and of its column. For each $i$ there is some $j$ with $a_{ij} \gt 0.$ Then $b_{ji} \gt 0$ since nothing else in its row is. Since $d_{mi}=0$ for $m \ne i$, $a_{ij}$ is the only non-zero entry in its column. And since no two rows of $B$ are dependent, for every $k \ne j$ there is an $m \ne i$ with $b_{km} \gt 0.$ Since $d_{im}=0$ it follows that $a_{ik}=0$ so indeed $a_{ij}$ is the only non-zero entry of its row and column. Similarly, $b_{ji} \gt 0$ is the only non-zero entry of its row and column. -Hence the non-zero entries of $A$ are the same as those of a permutation matrix $P$ and those of $B$ are the same as those of $P^t=P^{-1}$. If these non-negative matrices are integer matrices with $AB=I$ then the non-zero entries are all $1$ so $A=P.$<|endoftext|> -TITLE: Are the Millennium Prize Problems all decidable? -QUESTION [11 upvotes]: I am an inexperienced logician, so I may be completely missing something major in this question. I also may be misconstruing the idea of decidability. However, I was wondering if all 6 of the remaining Millennium Prize Problems are decidable in the sense of Gödel. -If any of the associated theories were not decidable, wouldn't that have far-reaching applications in the world of mathematics? -Thanks in advance, and I hope that my question makes sense. - -REPLY [42 votes]: There are very few results which allow us to know that a mathematical claim will be provable or disprovable within ZFC without actually proving or disproving it. To the best of my knowledge, the only exceptions are theories which have quantifier elimination. Few1 open mathematical problems which people are interested in are of this sort, and none of the Millenium problems are. So any of the Millenium problems could be independent of ZFC (except for the Poincare conjecture, because it has been proved!) -You might be particularly interested in Scott Aaronson's survey on whether or not it is likely that $P \neq NP$ is independent of ZFC. -1 Here is an example of a question which I know is decidable in ZFC, yet whether the answer is "yes" or "no" is open. - -Do there exist $44$ vectors $(u_i, -> v_i, w_i, x_i, y_i)$ in - $\mathbb{R}^5$, each with length $1$, - and with the dot product between each - pair $\leq 1/2$? See Wikipedia for background. - -This is the a first order question about real numbers, so it is decidable by Tarski's theorem. The analogous result for four dimensional vectors was only obtained in 2003; if you can get the answer for $5$ dimensions, it should be publishable in a good journal. I think this about as interesting a question as one can find which is definitely settleable in ZFC, yet still open. Most questions mathematicians care about are not of this form (and, in my opinion, are much more interesting).<|endoftext|> -TITLE: how to find the varieties whose cohomology realizes certain representations? -QUESTION [29 upvotes]: The cohomology of Shimura varieties and Drinfeld shtukas is conjectured to realize the representations sought for in the Langlands programme/conjectures, the cohomology of Deligne-Lusztig varieties realizes representations of the classical groups over finite fields: How did people find those varieties? - -REPLY [14 votes]: The construction of Deligne-Lusztig representations is a very natural analog of the construction of discrete series representations of real groups, and follows a very general philosophy (associated I think with Gelfand and Kazhdan and most probably known to Drinfeld at the time) that all representations of reductive groups over any field are "forms of principal series". -Namely we learn from Harish Chandra and Gelfand--Graev--Piatetskii-Shapiro the idea that representations of a reductive group are "always" labeled by conjugacy classes of tori and characters of the latter (up to Weyl group symmetry). -How do we construct these representations? The untwisted case is principal series, ie the case of a split torus. Then we are supposed to perform parabolic induction: look at the big G-representation of functions (or some form of cohomology) on the total space of the natural torus bundle $G/N\to G/B$ over the flag variety and decompose according to the torus action along the fibers (which commutes with G). The theory of (standard/principal series) intertwining operators realizes a Weyl group symmetry on these induced representations. -The philosophy (explained beautifully by Kazhdan in various places, such as his ICM address) is that all other "series" are given as "forms" of the principal series -- ie over the algebraic closure one only has principal series. Put more poetically, the dual of G should have an algebraic structure, so that representations over various fields can be simply specified by descent data from representations over field extensions, so at the end of the day they come from principal series. Over your given field you prescribe a conjugacy class of tori by descent data, which is a conjugacy class of Weyl group representations of your Galois group. So eg over R we just need a conjugacy class of involutions in the Weyl group, over a finite field we can take any conjugacy class in the Weyl group, etc. Now the idea (not yet fully realized in its optimal strength) is that we should twist principal series over the algebraic closure using the "Weyl group action by intertwiners" to define the desired series of representations over the given field. -In any case this idea has a very concrete geometric realization. Recall principal series are functions or cohomologies of a torus bundle over the flag variety. Now given a field k (for simplicity assume the Galois group is cyclic, eg real or finite fields) we can decompose the flag variety over the algebraic closure of k into a collection of G(k) invariant subsets, by looking at flags which are in a given relative position (an element of $W$) with their Galois translates. In the case of $SL_2(R)$ this gives us the decomposition of $CP^1$ into $RP^1$ and the upper+lower half planes. In the finite field case these are Deligne-Lusztig varieties. This is a twisted k-version of G/B, and has a natural version of G/N over it, which is a torsor for precisely the k-torus we specified before by the same Weyl group element. So we can look at functions/cohomologies twisted by characters of this torus, finding a series of representations just as predicted by Harish Chandra. The principal series corresponds to the flag variety over k. For $SL2R$ (an example of a real group with compact torus) we recover the discrete series (though note the representation constructed this way is not irreducible, since we're considering upper and lower half planes together).<|endoftext|> -TITLE: Maneuvering with limited moves on $S^2$ -QUESTION [29 upvotes]: This question comes to me via a friend, and apparently has something to do with quantum physics. However, stripped of all physics, it seems interesting enough on its own. I assume someone has asked this question before, but I have no idea what to search for: -Suppose we have points $P$ and $Q$ on $S^2$, and two available rotations: specifically, I am interested in rotations by $\pi/4$ radians about the $x$ and $z$ axes. Given $\varepsilon > 0$, is there an effective algorithm for applying these rotations to $P$ so that it is within Euclidean distance $\varepsilon$ of $Q$? -Edit: Update retracted. I'm curious about the general situation as well, where the two rotations are arbitrary (and, obviously, send $P$ to a dense subset of $S^2$). - -REPLY [9 votes]: Using properties of commutators in a neighborhood of identity (beautifully described in Bill Thurston's answer, on which the following is a comment), Solovay-Kitaev algorithm in quantum computation produces words of length $O(\log^{2+}(1/\epsilon))$ for any dense subgroup; see, for example, [1] for a very nice exposition. -For dense subgroups generated by elements with algebraic entries, the spectral gap result proved in [2] implies existence of words of length $O(\log(1/\epsilon))$. -[1] C.M. Dawson and M.A. Nielsen, The Solovay-Kitaev algorithm, Quantum Inf. Comput., 6, 81-95, 2006. -[2]J. Bourgain and A. Gamburd, On the spectral gap for finitely generated subgroups of SU(2), Inv. Math. 171, 83-121, 2008. -(It appears to be of interest (and challenge) to extend the result in [2] to arbitrary dense subgroups, and to find an effective algorithm producing words of length $O(\log(1/\epsilon))$).<|endoftext|> -TITLE: Relationship between Hilbert schemes and deformation spaces -QUESTION [8 upvotes]: Hi, I'm just starting to learn about deformation theory (via Hartshorne's Deformation theory, as well as Fantechi's section of FGA explained), and I feel like I'm confused about fundamental concepts. So please indulge me even if the question is well-known, or trivial. -So suppose we have a projective variety $Y \subseteq \mathbb{P}_k^n$. Then we can consider two natural objects associated to it, namely, the Hilbert scheme, which parametrizes all the objects with the same Hilbert polynomial as $Y$, and the (versal?) deformation space, which represents the deformation functor $F: (Art)_k \to (Sets)$. -My question is, is there a relationship between these two spaces? My hunch is that the Hilbert scheme is included in some way (perhaps via immersion, although that sounds kind of strong) to the deformation space - I guess this means that any deformation over an Artin ring doesn't change the Hilbert polynomial - but I can't formulate any coherent and believable conjecture right now. (If the question doesn't make sense, could you answer the right question that most closely approximates it?) -Thank you for reading. - -REPLY [10 votes]: Here's one interpretation if you're content with thinking about the tangent spaces of these functors: -Suppose that $Y$ is a smooth (you can get away with less here) subvariety of $\mathbb{P}^n$. You have the short exact sequence -$0 \rightarrow T_Y \rightarrow i^*T_{\mathbb{P}^n} \rightarrow N_{Y/\mathbb{P}^n} \rightarrow 0.$ -Infinitesimal embedded deformations of $Y$ inside $\mathbb{P}^n$ are given by global sections $H^0(Y, N_{Y/\mathbb{P}^n})$. Infinitesimal deformations of $Y$ abstractly are parameterized by $H^1(Y, T_Y)$. There is a forgetful map of functors sending an embedded deformation to the abstract deformation (that is, forget the embedding). Once you know that those functors are (pro)representable (or even if you don't), then the differential map for the corresponding morphism of spaces/functors is given by the connecting homomorphism in the long exact sequence of cohomology (from the above short exact sequence): -$\delta: H^0(Y, N_{Y/\mathbb{P}^n}) \rightarrow H^1(Y, T_Y)$<|endoftext|> -TITLE: Math and Wormholes -QUESTION [35 upvotes]: Hopefully, MathOverflow is the correct place for this. -I had a student approach me and ask me what kinds of mathematics goes into the study of wormholes. She specifically asked whether there is any topology involved, but I'd appreciate it if I could give her a more full answer. So, topology answers are the best but other areas of math would be nice. -I am a topology Ph.D. student but know almost zero physics, just in case my background helps with your answer. Thanks. - -REPLY [102 votes]: A bit of General Relativity and Causality theory -One feature of general relativity is that the space-time is modelled as a Lorentzian manifold. The Lorentzian metric on the manifold has signature (-+...+), and thus distinguish between time-like, space-like, and null directions. A $C^1$ curve in your manifold is said to be time-like if its tangent vector is always time-like (similarly space-like/null). -For physical reasons we expect our manifold to be time-orientable: the set of time-like vectors in the tangent space has two connected components. We assume there is a continuous (non-vanishing) time-like vector field, which denotes the direction of "future". -A point $p$ is said to be in the chronological future of another point $q$ if there is a time-like, futurely directed curve originating from $q$ to $p$. -Asymptotic regions -For (possibly outdated by modern cosmological observation) physical reasons, one may assume that the local geometry of space-time far away from (large) gravitating bodies is nearly flat. Let $M$ be a Lorentzian manifold which for now we assume to be diffeomorphic to $\mathbb{R}\times \Sigma$, with the restriction of the metric on the $\Sigma$ leaves Riemannian. An asymptotic end of $\Sigma$ is a connected component of $\Sigma \setminus K$ for some compact set $K$ that is diffeomorphic to $\mathbb{R}^d\setminus B_1$. This induces a nice coordinate system on $M\setminus (\mathbb{R}\times K)$ (which allows us to talk about the radius function $r$). We make the flatness assumption (which can be justified by possibly enlarging $K$) that in each asymptotic region, future (or null) directed null rays $\gamma$ that remain in the asymptotic region for all parameter $s\in [s_0,\infty)$ will have $\lim_{s\to\infty} r(\gamma) = \infty$. -Fix now one asymptotic region, call it $M_\infty$. Its corresponding domain of outer communications $D_\infty$ consists of all points of $M$ which can be reached from $M_\infty$ by both a future time-like curve and a past time-like curve. -Intuitively speaking, each domain of outer communications is a connected region which can "communicate" with the asymptotic region through exchange of light-signals. In particular, it lies outside any black holes or white holes. -Worm holes -A simple description of a worm-hole space-time is just a connected Lorentzian manifold with multiple domains of outer communication. -For them to be of physical interest, we require there to exist a time-like or null curve originating from one domain of outer communication and ending in another. This represents the ability to send signals or even travelers between two "almost disjoint" regions of the universe. - -So far it should be clear that you need some knowledge of differential/pseudo-Riemannian geometry at the very least to study worm holes. - -The mathematical study of wormholes essentially boils down to writing down - -Sufficient conditions to rule out their existence -Explicit examples of wormhole solutions - -Let me start with the latter to motivate the former. The most classical solution to Einstein's equation in general relativity that admits a wormhole is the Reissner-Nordstrom charged black hole solution. This solution models a charged black hole, and the explicit analytic extension of the solution to the interior of the black hole reveal some possible structure. In particular, observers who fall into the black hole may not necessarily encounter a singularity in finite time (as opposed to the situation in Schwarzschild black holes). And based on the analytic extension, it is possible that such an observer will emerge after some time into another domain of outer communications by being spat out of a white whole. -Two key things to notice: - -This worm hole is "one-way only" -I kept using the phrase analytic extension - -As it turns out, under fairly generic conditions on the properties of the universe and the matters residing within it (in particular the dominant energy condition which also played a strong role in the Positive Mass Theorem of Schoen and Yau), worm holes can "only be one way". (Actually, with certain additional assumptions worm holes cannot even exist.) Results of this type are captured (as easy corollaries) under the name "Topological Censorship". The original version of the result states that the domain of outer communication must be (under certain reasonable physical assumptions) homeomorphic to $\mathbb{R} \times (\mathbb{R}^d\setminus B_1)$. A recent improvement by Galloway and collaborators states that (under reasonable physical assumptions including global hyperbolicity [more on which later]) any future directed time-like or null curve that starts from one asymptotic region and ends in another (a priori possibly different) asymptotic region must be entirely contained in one domain of outer communication. (Which basically states that wormholes don't exist unless you violate one of the physical assumptions; and if you remove global hyperbolicity, you can possibly obtain one-way worm holes.) -(There is also a much more classical result in the reverse direction: if a space-time admits a Cauchy hypersurface [a special type of a spatial slice] that has non-trivial topology, it must be geodesically incomplete [which suggests that there should be a black hole].) -The second issue at play is this analytic extension business. If you know a bit about the wave equation, please recall that it has the finite speed of propagation property. That is, when solving the initial value problem, the value of the solution at any given point in space-time depends can only on a compact sub-set of the initial space-like slice. (Or, in physical description, information does not travel faster than the speed of light.) Applying this to Einstein's equation of general relativity, one obtains the notion of "Maximal Globally Hyperbolic Extension" corresponding to a set of given initial data: the largest set in the space-time which is completely determined by the given initial data. -A priori, it is not clear whether the maximal globally hyperbolic extension can be further extended as a solution to the Einstein's equations. (Of course, such an extension will be non-unique, but it is interesting to ask whether they exist at all.) And in fact, in certain situations, more extension can be constructed by (real) analytic continuation of the solution. The case of Reissner-Nordstrom is one particular one, and it is due to this possible extension that we have the wormhole behaviour. -An open problem in mathematical general relativity is whether one can rule-out such extensions "generically". The "strong cosmic censorship conjecture" posits that the answer is positive: that the existence of such extensions requires the space-time to be conspiratorially nice. -Going back to the issue of constructing solutions/examples: as any example, to be physically meaningful, will be required to solve Einstein's equation, topological methods tend to be less useful in this context (unless you mean Gromov's h-principle), as geometry and analysis tend to be more "rigid" structures. - -In classical mathematical GR, as you can see, topology has not come into play strongly. (The "problems" introduced by topology tend to be considered as pathological; lots of work were done in the 60s and 70s to discuss sufficient conditions to rule out such pathologies.) Differential/pseudo-Riemannian geometry and mathematical analysis (esp of PDEs) are generally used more heavily. -But that is not to say that topology has no role. It is more that the role of topology is much less clear, beyond the more simple applications of point-set topology, and some slightly less trivial applications based on connections between Riemannian geometry and topology. -In conclusion, let me give you some possible references if your student can be satisfied by black holes instead of wormholes. Some classical results to look into that in some way involves topology: - -The Singularity Theorem of Hawking and Penrose (see, Wald, General Relativity; O'Neill, Semi-Riemannian Geometry will applications to relativity) -The Black Hole Topology Theorem (see this paper for the geometric aspects; the topological aspects requires knowing about the Yamabe problem)<|endoftext|> -TITLE: Freiheitssatz implies a finitely generated one relator group embeds in a two-generator one relator group? -QUESTION [10 upvotes]: I've read that every finitely generated one relator group embeds in a two generator one relator group, and that this fact follows from the Freiheitssatz. -Unfortunately, the only proof I can find of this fact applies B.H. Neumann's proof for denumerable n-relator groups, and it doesn't seem to use the Freiheitssatz. Further, I haven't found any mention of this in Lyndon and Schupp, but it's possible I overlooked a more general theorem from which this follows. -My question is: does this fact truly follow from the Freiheitssatz? Is the proof trivial? I apologize if it is; unfortunately I am new to one relator groups. - -REPLY [12 votes]: Yes, this follows from the Freiheitssatz. Assume that the 1-relator group is defined by $G=\langle g_1,\ldots, g_n | R(g_1,\ldots,g_n)\rangle$, such that the relator $R(g_1,\ldots, g_n)$ is cyclically reduced, and involves the generators $g_1$ and $g_n$ non-trivially. By the Freiheitssatz, the subgroups $\langle g_1,\ldots, g_{n-1}\rangle$ and $\langle g_2,\ldots g_n\rangle$ are free groups of rank $n-1$ freely generated by these elements. Then embed $G$ in an HNN extension $\langle g_1,\ldots, g_n, t | R(g_1,\ldots,g_n), tg_it^{-1} = g_{i+1}, i=1,\ldots ,n-1 \rangle = \langle g_1,t | R(g_1, tg_1t^{-1},\ldots, t^{n-1}g_1t^{1-n}\rangle$ by eliminating generators and relators. -By permuting the labels, one may guarantee that the relator involves $g_1, g_n$ unless the relator involves only one generator. In that case, if the relator is of the form $g_1^k$, then do a Nielsen transformation $h_1=g_1g_n^{-1},h_2=g_2,\ldots, h_n=g_n$. The group with this set of generators has presentation $\langle h_1,\ldots, h_n | (h_1h_n)^k=1\rangle$. One may apply the previous construction to this presentation.<|endoftext|> -TITLE: Representation theory of $S_n$ -QUESTION [9 upvotes]: I need to understand the representation theory of $S_n$ (symmetric group on $n$ letters) and so could someone suggest a good reference for this. There is a similar question on mathoverflow here -A learning roadmap for Representation Theory -Most of the responses to the above question give references for representation theory of Lie groups. Also the usual reference Fulton and Harris has too many exercises (on which I don't want to spend too much time ) and I find it difficult to read. -Another reference which was suggested was Flag varieties by Lakshmibai and Brown. This seems to be a good reference, but are there any other references. -EDIT: By mistake I did not notice something in the above mentioned book and so some of my remarks are being edited. Sorry. - -REPLY [4 votes]: If you like combinatorics, you may enjoy learning about the representations of $S_n$ by reading Chapter 7 of Stanley's Enumerative Combinatorics, Volume 2.<|endoftext|> -TITLE: The relationship between low dimensional topology and dynamics -QUESTION [7 upvotes]: I am just curious how dynamics get connected with low dimensional topology. Or it is just that we have now powerful computing machines therefore it is natural to use them on topological problems. What kind of problems researchers are interested in working at the intersection of dynamics and low dimensional topology. -Any suggestion for reading is greatly appreciated. The phrase sound fascinating, but have almost no understanding. Thanks. - -REPLY [6 votes]: The machinery of Markov partitions and stable/unstable foliations for Anosov and Axiom A diffeomorphisms was adapted to several different bits of low dimensional topology. In the Nielsen-Thurston classification of surface mapping classes alluded to by others, Markov partitions are the machinery behind the construction and study of pseudo-Anosov homeomorphisms, and the stable/unstable foliations are very explicit in that study. In the Bestvina-Feighn-Handel classification of outer automorphisms of free groups, the concepts of Markov partitions lie behind the theory of train tracks and relative train tracks, and the lamination part of the theory comes through quite explicitly in the theory of attracting laminations for outer automorphisms.<|endoftext|> -TITLE: Moduli of pointed Curves -QUESTION [11 upvotes]: Every curve of genus $g\leq 2$ has non trivial automorphisms. So the fiber of the forgetful morphism $\pi:\bar M_{g,1}\rightarrow\bar M_{g}$ over $[C]$ is not isomorphic to $C$ but to $C/Aut(C)$ (I am considering the coarse moduli space instead of the stack). -The general fiber of $\pi:\bar M_{1,2}\rightarrow\bar M_{1,1}$ is isomorphic to $\mathbb{P}^{1}$ and the same is true for $\pi:\bar M_{2,1}\rightarrow\bar M_{2}$. Now $\bar M_{1,2}$ is a rational surface and $\bar M_{2,1}$ is rational $4$-fold. -The question is the following: -Does anyone know an explicit description of $\bar M_{1,2}$ and $\bar M_{2,1} ?$ -My feeling is that $\bar M_{1,2}$ should be a blow-up of a ruled surface. - -REPLY [6 votes]: The moduli space $\overline{M}_{1,2}$ can not be the blow up of $\mathbb{P}^{2}$ in $2$ points because its rational Picard group has rank $2$, indeed it is generated by the divisor parametrizing genus $0$ irreducible nodal curves with $2$ marked points and the divisor parametrizing reducible curves whose components are a smooth genus $1$ curve and a smooth genus $0$ curve with $2$ marked points. -The moduli space of genus $1$ stable curves with $2$ marked points is a rational surface with four singular points. Two singular points lie in $M_{1,2}$, and are: -a singularity of type $\frac{1}{4}(2,3)$ representing an elliptic curve of Weierstrass representation $C_{4}$ with marked points $[0:1:0]$ and $[0:0:1]$; -a singularity of type $\frac{1}{3}(2,4)$ representing an elliptic curve of Weierstrass representation $C_{6}$ with marked points $[0:1:0]$ and $[0:1:1]$. -The remaining two singular points lie on the boundary divisor $\Delta_{0,2}$, and are: -a singularity of type $\frac{1}{6}(2,4)$ representing a reducible curve whose irreducible components are an elliptic curve of type $C_{6}$ and a smooth rational curve connected by a node; -a singularity of type $\frac{1}{4}(2,6)$ representing a reducible curve whose irreducible components are an elliptic curve of type $C_{4}$ and a smooth rational curve connected by a node. -From this one can prove that: -The moduli space of genus $1$ stable curves with $2$ marked points is isomorphic to a weighted blow up of the weighted projective plane $\mathbb{P}(1,2,3)$ in its smooth point $[1:0:0]$. In particular $\overline{M}_{1,2}$ is a toric variety.<|endoftext|> -TITLE: Proof of an 'easy' exercise in a book of Tits -QUESTION [5 upvotes]: In 'Buildings and Finite $BN$-Pairs', Jacques Tits gives the following statement which is left as an easy exercise. -Let $G_1,G_2,G_3$ be three subgroups of a group $G$. Then the following conditions are equivalent. - -$G_2G_1 \cap G_3G_1 = (G_2 \cap G_3) G_1$ -$(G_1 \cap G_2) \cdot (G_1 \cap G_3) = (G_2G_3) \cap G_1$ -If the three cosets $xG_1$, $yG_2$ and $zG_3$ have pairwise nonempty intersection, then $xG_1 \cap yG_2 \cap z G_3 \neq \emptyset$. - -I know it is not usual to ask for the solution of an exercise from a book on MO. However after hours of trying hard with friends, I dedided to post it anyway. Indeed, the problem seems not easy at all to me! Therefore I think it is worth to ask it here. -Does anyone know a reference for a proof or know a solution? -As usual, thanks in advance. - -REPLY [7 votes]: (1) implies (2): -Indeed, the left side of (2) is obviously contained in the right side. So pick an arbitrary $h=g_2g_3=g_1 \in (G_2G_3)\cap G_1$. Now, $g_3=g_2^{-1}g_1$ is contained in $G_2G_1$ but also in $G_3G_1$ and hence by (1) in $(G_2\cap G_3)G_1$. So there is $g_{23}\in G_2\cap G_3$ and $\tilde{g}_1\in G_1$ such that $g_3=g_2^{-1}g_1=g_{23}\tilde{g}_1$. Thus $g_2g_{23}=g_1\tilde{g}_1^{-1}\in G_1\cap G_2$ and $g_{23}^{-1}g_3=\tilde{g}_1\in G_1\cap G_3$, therefore -$h=g_2g_3 = (g_2g_{23})(g_{23}^{-1}g_3)\in (G_1\cap G_2)\cdot(G_1\cap G_3)$. -(2) implies (3): -Wlog assume x=1. From $G_1\cap yG_2\neq\emptyset$ follows $y\in G_1G_2$; similarly we find $z\in G_1G_3$. -Thus $y=g_1g_2$, $z=\tilde{g_1}g_3$ for suitable elements $\tilde{g_1},g_1\in G_1$, $g_2\in G_2$, $g_3\in G_3$. -From $yG_2\cap zG_3\neq\emptyset$ we get $y^{-1}z\in G_2G_3$, hence -$$ -g_2^{-1} g_1^{-1} \tilde{g_1}g_3 \in G_2 G_3 -\implies -g_1^{-1} \tilde{g_1} \in (G_2 G_3) \cap G_1. -$$ -But using (2), wlog we may assume $g_1\in G_1\cap G_2$ and $\tilde{g}_1\in G_1\cap G_3$. Thus $y\in G_2$ and $z\in G_3$, and so $xG_1\cap yG_2 \cap zG_3 = G_1\cap G_2 \cap G_3 \neq \emptyset$. -(3) implies (1): -The right side of (1) is clearly contained in the left side, so we just need to prove the reverse inclusion. So assume $h\in G_2 G_1 \cap G_3 G_1$. Then $h \in G_2 y \cap G_3 z$ for suitable $y,z\in G_1$. Thus $y\in G_1\cap yG_2$ and $z\in G_1\cap zG_3$. Thus by (3) there exists $g_1 \in G_1 \cap G_2 y \cap G_3 z$, and we have $1 \in G_2 y g_1^{-1} \cap G_3 z g_1^{-1}$. This implies $yg_1^{-1} \in G_2$ and $zg_1^{-1}\in G_3$. Therefore $hg_1^{-1} \in G_2 yg_1^{-1} \cap G_3 zg_1^{-1} = G_2 \cap G_3$, and thus $h\in (G_2 \cap G_3) G_1$ as claimed.<|endoftext|> -TITLE: Rational approximations on the circle -QUESTION [14 upvotes]: The well-known Liouville theorem asserts that an irrational algebraic number $\alpha$ cannot have too good rational approximations, namely $|\alpha-p/q|\ge C(\alpha)/q^k$ where $k$ is the degree of $\alpha$. I wonder whether a similar result holds for the argument of an algebraic number that happens to lie on the unit circle. -For example, consider $\theta=\arccos(1/3)$. It is the argument of a root $\alpha$ of the polynomial $x^2-\frac23x+1$. Is there a similar lower bound for $|\theta/\pi-p/q|$, or equivalently, for $|\alpha^q-1|$? (Note that $|\alpha^q-1|\approx q\cdot |\theta/2\pi-p/q|$). -More generally, let $\alpha\in\mathbb C$ be an algebraic number, $|\alpha|=1$ and $\alpha$ is not a root of unity. Is it always true that $|\alpha^q-1|\ge C(\alpha)/q^k$ where $k$ depends only on the degree of $\alpha$ (or perhaps equals the degree minus one)? - -REPLY [5 votes]: The answer is affirmative by a result of Fel'dman: An improvement of the estimate of a linear form in the logarithms of algebraic numbers (Russian), Mat. Sb. (N.S.) 77 (119) 1968, 423–436, MR0232736. -See also Theorem 3.1 in Baker's Transcendental Number Theory.<|endoftext|> -TITLE: Local normal forms of covariantly constant selfadjoint (1,1)-tensors -QUESTION [7 upvotes]: Consider the pair $(g, L)$, where $g$ is a pseudo-Riemannian (i.e., -nondegenerate and of arbitrary signature) metric and $L$ is an (1,1) $g$-selfadjoint tensor field. -Does there exist a local description of such pairs; for example in the form -"in a certain coordinate system $g$ and $L$ are given by formulas ...". -For me, the state of art is as follows: for Riemannian metrics, the answer was known to classics: The existence of such $L$ implies the local -decomposition of $g$ into a direct product: $g= g_0+ ... + g_k$, where -$g_0$ is flat (we can think therefore that $g_0= dx_1^2 +...+dx_m^2$ in a certain coordinate system) and each $g_i$ has irreducible holonomy group. For such metrics $g= g_0+ ... + g_k$, the tensor $L$ also can be decomposed into the product $L= L_0+...+L_k$; moreover, $L_0$ is given by arbitrary symmetric matrix with constant entries and other $L_i$ are proportional to identity with constant coefficients (the coefficients depend on the component). -For the pseudo-Riemannian metrics one can do the same splitting if $L$ has different eigenvalues; so the interesting case is when $L$ has one real eigenvalue or two conjugated complex eigenvalues. And this case looks quite open for me; only the special case when -$L^2= 0$ or $L^2= -1$ are known. -Does anybody know more? - -REPLY [5 votes]: Well, I don't have the complete answer, but then I don't think that a 'complete' answer is going to be simple. For example, you haven't ruled out the case $L=0$, which amounts to giving a `normal form' for all pseudo-Riemannian metrics. What is true, of course, is that the algebraic type of $(g,L)$ is the same at all points (assuming that $M$ is connected) and, as Willie Wong points out, you will at least have to classify those algebraic types as a starting point. -Of course, you can get complete answers in some algebraic types. For example, when $L$ has a single real eigenvalue $r$ (which must be constant), you can assume that this eigenvalue is $0$ by replacing $L$ by $L-rI$, and the `least degenerate' case would presumably then be when $L^n=0$ but $L^{n-1}\not=0$. (I'm assuming that the dimension of the manifold is $n$.) Then it is not difficult to show that there is a (local) frame field $e_1,\ldots,e_n$ such that $e_i = L^{i-1}e_1$ for $i = 2,\ldots,n$ and, moreover, such that $g(e_i,e_{n+1-i}) = 1$ with all other $g(e_i,e_j) = 0$. (You may have to replace $g$ by $-g$ in order to arrange this, but that's a trivial matter.) Moreover, this frame field is unique up to replacing $e_i$ by $-e_i$ for all $i$. Since $L$ is $g$-parallel, it follows that this frame field is also $g$-parallel. In particular, the connection is flat and one can (locally) choose coordinates $x^i$ such that $dx^i(e_j) = \delta^i_j$. In these coordinates, $g$ and $L$ have constant coefficients. -Added after Vladimir's comment: Yes, it appears that the case of more than one Jordan block is more interesting. I did a back-of-the-envelope calculation for the case of a $5$-dimensional manifold $M$ with a pair $(g,L)$ where $L$ is nilpotent with $2$ Jordan blocks, one of size $3$ and one of size $2$. The result is that there are $4$ algebraic types possible ($2$ if you allow the replacement of $g$ by $-g$), and each of them exists as a $1$-parameter family of inequivalent types. One member in each family is 'flat', i.e., $g$ and $L$ have constant coefficients in the appropriate coordinate system, but the others aren't flat, even though they are homogeneous (with a $7$-parameter family of $L$-preserving isometries). Hmm.<|endoftext|> -TITLE: What does progressively measurable actually entail? -QUESTION [6 upvotes]: There is a definition that has always left nagging questions in my mind. To set it up, let $(\Omega, \cal{A}, (\cal{F}_t)_{t\geq 0}, P)$ be a filtered probability space. From Comets & Meyre's Calcul stochastique et modèles de diffusions, -Definition 3.2. A real-valued function $\phi$ on $\mathbb{R}^+\times \Omega$ is progressively measurable if, for all $t\in \mathbb{R}^+$, the mapping $(s,\omega) \mapsto \phi(s, \omega)$ on $[0, t]\times \Omega$ is $\cal{B}[0,t] \otimes \cal{F}_t$-measurable. -Clear enough. The authors go on to define $M^2(\mathbb{R}^+)$ to define the set (space) of all progressively measurable stochastic processes $\phi$ such that -$\mathbf{E}\int_{\mathbb{R}^+} \phi^2(t, \omega) dt < \infty$. -(This formula is verbatim from the book.) -My question is: does one simply interpret the expression on the left as a double integral à la the Fubini-Tonelli theorem, $\int_\Omega\int_{R^+} \ldots dt dP$, and if so, does progressive measurability ensure that $\phi$ is $\cal{B}(\mathbb{R}^+)\otimes \cal{A}$ measurable, so that this interpretation makes sense? -My attempts to realize $\phi$ as the limit of a sequence of measurable functions (like $\phi|_{[0, n]\times\Omega}$) have yielded nothing to convince me yet. - -REPLY [6 votes]: The formula -$$\mathbb{E} \int_{R+} \phi^{2}(t,\omega)dt$$ -is a double integral a la Fubini-Tonelli. And if you did back there is probably a condition on the filtration saying that $\mathcal{F}_t \subset \mathcal{A}$ for all $t \ge 0$. So that progressive measurability does imply that $\phi^{2}(t,\omega)$ is $\mathcal{B}(\mathbb{R}^{+}) \otimes \mathcal{A}$ measurable. But making this integrand measurable isn't the main purpose of the progressive measurability condition. The main point is so that something like $f(t,X_t)$ where $X_t$ is an adapted process is again an adapted process. The integrability condition is to give $\mathcal{M}^{2}$ a Hilbert space structure.<|endoftext|> -TITLE: Geodesic Triangles in Finite Element Method -QUESTION [7 upvotes]: I've been working on a new method for 2-dimensional FEM on Riemannian Manifolds that involves using geodesic Triangles instead of approximating them in an embedded form using "traditional" triangles. -I'm quite far along, and up until now I wasn't even sure if my Method would only work on an abstract level or if it can actually be applied. So far, it works for trivial cases ($\mathbb{R}^2, S^2$,...), which is "good enough" at this point. -Now my problem is this: I've been searching high and low (using Zentralblatt MATH) for any research related to mine in any way. So far, all I could find is a single paper (http://vs24.kobv.de/documents-matheon/580/6086_sander_geodesic_fe.pdf). -I find it hard to believe that nothing else exists on this matter. -My questions are: - -Does anyone know any research done in this direction? -Or is the Idea "useless" (as in, it wouldn't improve the result)? -Or have I simply been searching wrong? - -Any feedback is appreciated - -REPLY [3 votes]: Interesting question! There's recent work on isogeometric analysis (eg. Hughes et al in CMAME 2005), who seek to combine FEM with NURBS. This may be a useful literature to look at, for potential applications of your own ideas. -In essence, if one is to study PDE on surfaces, then an intrinsic (geodesic) mesh may confer advantages in terms of accuracy.<|endoftext|> -TITLE: Characterization of the Poisson law -QUESTION [9 upvotes]: This semester, I teach an introduction to probability course tailored for students with no science background and so with very very little prerequisites. We started with the basics of analytic combinatorics then moved on to random variables and the study of common laws (binomial, hypergeometric, geometric, Poisson). The audience being what it is, I try to avoid as much as possible calculus derivations of probability facts. -For some aspects of the course, it worked out well (for instance the derivation of the expectation of the binomial law) but because I am barely more knowledgeable than my students when it comes to probability, I have been unable to answer this question: - -Is there a set of natural probability properties which characterizes the discrete Poisson law? - -If yes, then I could use this as a definition of the Poisson law, which would suit my students better than saying "it's the law such that $P(X=k)=e^{-\lambda}\frac{\lambda^{k}}{k!}$". By natural above, I want to convey the meaning that I hope they can be formulated using natural language (like, say, memorylessness) rather than using analytic objects. -More precisely, what I have in mind is the following: - -Is there such a set of properties which would make it at least a little intuitively plausible that the sum of two variables following Poisson law also follows Poisson law? - -Of course, the proof of the above fact is completely elementary, but it would still be above the level of everyone in the audience except perhaps the 3 top students. -Note that I would be happy even if proving that this set of properties characterize Poisson law turned out to be much harder than anything I will do in this course (or even much harder than anything I know myself about probability), because what I am looking for is not logical rigour but rather psychological efficiency: in 10 years, my students will have completely forgotten what a derivative is, but I would like them to be able to recollect something if confronted with an epidemiological survey using random variables (at least my most successful students use this course to strengthen their math knowledge before studying medicine). -I realize this question is very elementary, and would understand if it is deemed inappropriate, but the standard references I might consult on the subject will invariably (and with good reasons) develop much more calculus that my students will ever know before dealing with such questions (typically, they will characterize the Poisson law as the limit of the binomial law via Stirling's formula). - -REPLY [3 votes]: The following heuristic argument is finitary apart from the sum of the exponential series. -Consider a birth clinic observing on average $\lambda$ male births and $\lambda$ female births per day. Let $p_k$ be the probability that on a given day we have exactly $k$ boys; then this same $p_k$ is also the probability that we have exactly $k$ girls. Consider the days where together $n$ kids are born. On such days the boys and girls are binomially distributed among them, i.e., we have -$${\bf P}[B=k\ \wedge\ G=n-k]={1\over 2^n}{n\choose k}\ .$$ -As boys and girls are born independently of each other when their time has come the left side of this equation has the value $p_k p_{n-k}/P_n$ where $P_n={\bf P}[B+G=n]$. It follows that -$$p_k p_{n-k}={P_n\over 2^n}{n\choose k}$$ -and in particular -$${p_0 p_n\over p_1 p_{n-1}}={1\over n}\ .$$ -Since this holds for any $n\geq1$ we conclude that -$$p_n={\mu^n\over n!}p_0\quad (n\geq1)\ ,\qquad \mu:={p_1\over p_0}\ .$$ -The condition $\sum_{n\geq 0} p_n=1$ gives $p_0=e^{-\mu}$, so -$$p_n={\mu^n\over n!}e^{-\mu}\qquad(n\geq0)\ .$$ -Now we observe -$$\lambda={\bf E}(B)=\sum_{k=0}^\infty k p_k=e^{-\mu}\sum_{k=1}^\infty {k\over k!}\mu^k =\mu\ ,$$ -so that definitively $p_k={\lambda^k\over k!}e^{-\lambda}$.<|endoftext|> -TITLE: $\mathbb{P}^n$ is simply connected -QUESTION [17 upvotes]: In his chapter about Hurwitz' theorem for curves, Hartshorne shows that $\mathbb{P}^1$ is simply connected, i.e. every finite étale morphism $X \to \mathbb{P}^1$ is a finite disjoint union of $\mathbb{P}^1$s. In an exercise the reader is invited to show that $\mathbb{P}^n$ is simply connected, using the result for $\mathbb{P}^1$. -I have no idea how to do this. Perhaps someone can give a hint? There are closed immersions $\mathbb{P}^1 \to \mathbb{P}^n$, along which we may pull back a finite étale morphism, but the trivializations don't have to coindice ... perhaps we can resolve this using cohomology theory? I'm a bit confused since $\mathbb{P}^n$ is $n$-dimensional, but this is in Hartshorne's chapter about curves. I don't want to use the more advanced material of SGA. - -REPLY [2 votes]: Let me give another answer, even though it does not fit into Hartshorne's context: -Show that $\pi_1(\mathbb{P}^n)$ has to be abelian. -Use Kummer-Theory to relate coverings to torsion in $Pic (\mathbb{P}^n)=\mathbb{Z}$, see e.g. Milne's Etale Cohomology, Prop 4.11. This implies that there are no nontrivial étale coverings of degree prime to the base characteristic. -Then use Artin-Schreier theory to relate the rest of the coverings to $\Gamma(\mathbb{P}^n,\mathcal{O}_{\mathbb{P}^n})/(F-1)\Gamma(\mathbb{P}^n,\mathcal{O}_{\mathbb{P}^n})=0$, and $H^1(\mathbb{P}^n,\mathcal{O}_{\mathbb{P}^n})^F=0$, where $F$ is the Frobenius, see e.g. Milne's, Prop 4.12.<|endoftext|> -TITLE: Is every p-point ultrafilter Ramsey? -QUESTION [16 upvotes]: A non-principal ultrafilter $\mathcal{U}$ on $\omega$ is a p-point (or weakly selective) iff for every partition $\omega = \bigsqcup _{n < \omega} Z_n$ into null sets, i.e each $Z_n \not \in \mathcal{U}$, there exists a measure one set $S \in \mathcal{U}$ such that $S \cap Z_n$ is finite for each $n$. -A non-principal ultrafilter $\mathcal{U}$ on $\omega$ is Ramsey (or selective) iff for every partition as above, there exists a measure one set $S$ such that $|S \cap Z_n| = 1$ for each $n$. -Clearly, every Ramsey ultrafilter is a p-point. What is known about the converse? -I couldn't find anything, not even a consistency result, in any searches I've done or sources I've checked. Is very little known/published about the converse? - -REPLY [9 votes]: Another small and slightly trivial addendum: -If there are no p-points, then every p-point is a Ramsey ultrafilter. (Duh!) -As Andreas Blass remarked above, this situation is consistent, which is easier to prove than the consistency of a -unique p-point. ("It is usually significantly harder to prove there is a unique object than to prove there is none". See Shelah's Proper and improper forcing VI.5)<|endoftext|> -TITLE: learning $\mathbf{A}^1$-homotopy theory -QUESTION [9 upvotes]: If you wanted to learn $\mathbf{A}^1$-homotopy theory, which sources in which order would you use? - -REPLY [5 votes]: Aravind Asok has an entire website devoted to pointing out resources for learning $\mathbf{A}^1$-homotopy theory. It is organized quite well. The concept list section of the page has lots of wikipedia-like entries on topics related to $\mathbf{A}^1$-homotopy theory. -http://a1homotopy.tiddlyspot.com/ - -REPLY [2 votes]: A book that might be helpful, that is probably mentioned on website above is http://www.amazon.com/Motivic-Homotopy-Theory-Nordfjordeid-Universitext/dp/3540458956/ref=sr_1_1?ie=UTF8&qid=1303257360&sr=8-1 -Also, people now call it Motivic instead of $\mathbb{A}^1$ sometimes.<|endoftext|> -TITLE: Is there a formula to count how many different topological regular maps can be created with n faces (on a sphere)? -QUESTION [5 upvotes]: Notes: - -For "regular" I intend maps in which the boundaries form a 3-regular planar graph -For "different" I intend maps that cannot be topologically transformed one into another (faces have to be considered unnamed) - - -I've been looking for a formula, but it is too difficult for me. Maybe it has a simple solution but I don't see it. -This was my best guess, but I already know that it is not correct because full of symmetries, as it can be verified manually. -General formula: - -$$2\sum_{s_{(f-3)}=2f-5}^{2f-5+2} \text{...}\sum_{s_2=5}^{s_3} \sum_{s_1=3}^{s_2} s_1\left(s_1-1\right)\left(s_2-3\right)\text{...}\left(s_{(f-3)}-(2f-5-2)\right) $$ - -Examples: - -4 faces = $$2\sum_{s_1=3}^5 s_1\left(s_1-1\right) $$ -5 faces = $$2\sum_{s_2=5}^7 \sum_{s_1=3}^{s_2} s_1\left(s_1-1\right)\left(s_2-3\right)$$ - -Here are the first results that can be found manually (excluding symmetries): - -2 faces = 0 possible regular map (an island and the ocean) (not to be counted, because not regular) -3 faces = 1 possible regular map (an island with two regions and the ocean) (two islands and the ocean wouldn't be regular) -4 faces = 3 possible regular maps (can be verified adding a face from the previous map) -5 faces = 20 possible regular maps (ERROR: There were duplicates) -6 faces = 329 possible regular maps (ERROR: There were duplicates) -... - -These are all maps up to 5 faces (ERROR: contains duplicates): - -Image for the comment on "triangulations of the sphere" - -And without duplicates: - -MODIFIED: 20/Apr/2011 - Removed "3-connected" from the question. See comment below. -MODIFIED: 21/Apr/2011 - Added a picture with all regular maps up to 5 faces -MODIFIED: 21/Apr/2011 - Added a picture for the comment on "triangulations of the sphere" and multiple edges, related to the dual graph of the original 3-regular planar graph -MODIFIED: 27/Apr/2011 - Manually computed number of different maps of 6 faces = 329, added numerical IDs to the maps -MODIFIED: 29/Apr/2011 - The manual computation of the number of regular maps contains some duplicated (Homeomorphic pairs) -MODIFIED: 29/Apr/2011 - Just to leave things a little more clean (I removed the duplicates ... I hope) - -REPLY [3 votes]: Take a look at the following two papers which will give you a feel for how these sorts of enumerations are tackled (typically by methods of generating functions), and will also serve as a bit of a guide to the early literature: -Here's a fairly accessible and readable article by Tutte which doesn't cover your case but serves as an introduction to how map enumeration problems have been treated mathematically. -Here's a review by Bender and Richmond which also describes how to solve some more general versions of map enumeration problems. I'm not sure if it covers your case or not, since I haven't quite managed to formalize what you're looking for, but if you read these papers you should have a good shot at figuring out for yourself where you need to look next. -The duals of the maps you are studying are definitely planar triangulations, and I think that since you don't allow loops in your maps that the duals must be 2-(vertex-)connected triangulations. Since you are considering the "ocean" to be distinct from any of the regions in the islands, I believe this gives your triangulations a root vertex. So my gut feeling is that you want to look at 2-connected rooted triangulations (which if I'm not mistaken were treated by Tutte), but I'll leave it to you to check this guess. I'm still confused about the rooting, but I think looking at triangulations is better than in terms of cubic maps - if you think of your maps without taking duals, then some pairs of vertices have multiple edges (the results derived in the paper mentioned by Igor Rivin are precisely where maps with such multiple edges are excluded). -Lastly, (just for fun, perhaps) there is also a large literature in the high-energy physics / field theory community on planar map / planar graph enumeration due to a connection with diagrammatic expansions in integrals over matrix ensembles. For this angle, you might start with this survey by Di Francesco.<|endoftext|> -TITLE: Probability of having a bounded ratio of two types of balls in each of 'S' bins after random partitioning of a fixed number of balls -QUESTION [5 upvotes]: Let's say I have a bag with $A$ red balls, $B$ blue balls, and a total number of balls $N = A + B$. With uniform probability, and sampling without replacement from the $N$ balls, I fill an integer number of bins, $S$, with exactly $L$ balls each, where $N = S*L$. As such, the bag of all $N$ balls should be empty by the end of the procedure. -What is the probability of having at most $k \leq B$ blue balls in every one of the $S$ bins? -Edit - I would be very much interested in approximate solutions! This problem should essentially come down to something akin to making sure a fixed number of points randomly placed on a matrix are at least some fixed distance apart (causing them to end up in different bins), so surely there have been problems like this tackled in the literature? -Edit 2 - As per Peter Shor's recommendation, I'm interested in examples where $S$ and $L$ are of fixed size, but where $S$ is large, at least $10^3$ to $10^4$ or so, and $L$ is typically significantly smaller than $S$ (around $10$ to $10^2$ or so). - -Another way of asking this question might be as follows - -Let's say that I have a matrix composed of $S$ blocks of $L$ cells (pick whatever geometry you'd like), where $S > 10^3$ and $L$ is approximately $10^1$ ~ $10^2$ or so. Some fixed number of cells, $A$, store the value '0', while the remaining $B$ cells store the value '1'. To be clear, there are a total of $A + B = S*L = N$ cells in the matrix. -What is the probability that all $S$ blocks contain at most $k$ cells with the storage value '1' (i.e. at most $k$ of the $B$ cells)? - -REPLY [4 votes]: For some parameters, you will not see much of a difference between the exact value and assuming that the balls are independent, or that the bins are independent. The independent approximations are much easier to calculate exactly. -You can estimate the probability using inclusion-exclusion. Whether the computable estimates are useful depends on the parameters. -The probability that there is at least one bin with more than $k$ blue balls equals the sum over the subsets $\beta$ of the bins -$$\sum_\beta (-1)^{|\beta|-1} P(\text{all bins in $\beta$ have more than $k$ blue balls})$$ -$$=\sum_i (-1)^{i-1} {S \choose i} P(\text{the first $i$ bins have more than $k$ blue balls})$$ -Each partial sum up to an odd $i$ is an upper bound, while each partial sum up to an even $i$ is a lower bound. The terms increase and then decrease roughly like a Poisson distribution, or the terms in the Taylor series for $\exp(x)$. -You can evaluate the probability that the first $i$ bins each have more than than $k$ blue balls as a sum over the possible number of blue balls in each. I don't know whether this can be simplified significantly, but when $(L-k)$ and $i$ are not too large you can calculate these explicitly: -$${S L \choose B}^{-1}\sum_{k \lt b_1,...,b_i \le L} {L \choose b_1} {L \choose b_2} ... {L\choose b_i} {L(S-i) \choose B-(b_1 + ... + b_i)}.$$ -The terms become $0$ when $(k+1)i \gt B$, but that might not be useful. -Better is to collect like terms based on the $L-k-1+i \choose i$ possible multisets $\{b_1,...b_i\}:$ -$${S L \choose B}^{-1}\sum_\lambda {i \choose c_{k+1}(\lambda)~ c_{k+2}(\lambda) ... c_L(\lambda)} {L \choose k+1}^{c_{k+1}(\lambda)} ... {L \choose L}^{c_{L}(\lambda)} {L(S-i) \choose B-|\lambda|}.$$ -Here $c_n(\lambda)$ is the count of parts of size $n$ in $\lambda$, and $\lambda$ has $i$ parts, all of size from $k+1$ to $L$. -For example, suppose $S=1000$, $L=10$, $k=4$, and $B=1000$. -If you approximate the colors of the balls as independent Bernoulli random variables with parameter $1/10$, then the probability none of the $1000$ bins will have $5$ or more blue balls would be $(1-16349374/10^{10})^{1000} =0.19470389363...$. -If you approximate the bins as independent, then the probability that none of the $1000$ bins will have $5$ or more blue balls would be $(1-311442378665580894806964843/191843012418970806869358022430)^{1000}$ $= (1-0.00162342)^{1000} = 0.196962429... $ -If I have calculated correctly, inclusion-exclusion gives the following sum: -$1$ $- 1.6234231038105990908$ $+ 1.2919877435018249272$ $- 0.67199565557524499763$ $+ 0.25695414151721206859$ $-0.077037323936817040460$ $+ 0.018861575339778788209$ $-0.0038785183878317956045$ $+0.00068369841392455375101$ $-0.00010494404713497539259$ $+0.000014199866369591258663$ $-1.7106130082560862260*10^{-6}$ $+ 1.8497238308747411999*10^{-7}$ $-...$ $=0.19206027...$ -The chances that there are at most $k$ blue balls in each bin is the complement. -Here is some Mathematica code for calculating this: -subsetToMultiset[sub_] := Table[sub[[i]] - i, {i, Length[sub]}] -Clear[lambdas]; -lambdas[i_, k_, l_] := lambdas[i, k, l] = - k + 1 + Map[subsetToMultiset, Subsets[Table[n, {n, l + i - k - 1}], {i}]] -counts[lambda_, i_, k_, l_, s_, b_] := - Product[Binomial[l, j]^Count[lambda, j], {j, k + 1, l}] - Multinomial @@ Table[Count[lambda, j], {j, k + 1, l}] - Binomial[l (s - i), b - Total[lambda]] -ieTerm[i_, k_, l_, s_, b_] := - (-1)^i Binomial[s, i] - Sum[counts[lambdas[i, k, l][[ind]], i, k, l, s, b], - {ind, Length[lambdas[i, k, l]]}] - /Binomial[s l, b] -Table[{i, N[ieTerm[i, 4, 10, 1000, 1000], 20]}, {i, 0, 12}] -Total[%][[2]] - -This takes $7$ seconds on my computer.<|endoftext|> -TITLE: computing abelianizations -QUESTION [6 upvotes]: Suppose I have a finitely presented group $G,$ and a subgroup $H$ of $G$ given by its finite generating set (given as words in the generators of $G.$ I want to know whether $H/[H, H]$ is finite. Is this question tractable (for your favorite definition of "tractable" -- decidable would be a good start...) - -REPLY [10 votes]: It is undecidable even when $G$ is a direct product of two free groups. Look at Corollary C on page 2 in this paper.<|endoftext|> -TITLE: Propositions equivalent to the completeness of the real numbers -QUESTION [18 upvotes]: Can anyone point me to a reasonably comprehensive article (or book chapter) explaining which basic theorems of calculus are equivalent to the completeness axiom of the reals and which ones aren't? -Here "equivalent" means equivalent relative to a base system that includes all the ordered field axioms, plus naïve set theory, plus (optionally) the Peano axioms (which one probably needs if one wants to use the natural numbers as an index-set, e.g. in the Nested Intervals Property). -At first I thought reverse mathematics would be the place to look, but a little bit of poking around now leads me to think that reverse mathematics in the usual sense deals with more arcane issues, with base systems that are at once weaker and stronger than what I have in mind: Konig's infinity lemma isn't provable in all of them, but the Intermediate Value Theorem is. -(Stephen Simpson, in his Wikipedia article http://en.m.wikipedia.org/wiki/Reverse_mathematics, writes: -"... RCA0 is sufficient to prove a number of classical theorems which, therefore, require only minimal logical strength. These theorems are, in a sense, below the reach of the reverse mathematics enterprise because they are already provable in the base system. The classical theorems provable in RCA0 include: ... Basic properties of the real numbers (the real numbers are an Archimedean ordered field; any nested sequence of closed intervals whose lengths tend to zero has a single point in its intersection; the real numbers are not countable). ... The intermediate value theorem on continuous real functions.".) -So, reverse mathematics may not be the place to turn for answers to questions like "Is the completeness of the reals equivalent to the Mean Value Theorem?" (answer: yes); but I'm sure someone has considered such questions systematically. Perhaps somebody wrote a beautiful Monthly article a few decades ago that explained things so clearly as to make the whole matter seem trivial, with the result that the article was forgotten? :-) - -REPLY [11 votes]: Since the article I was looking for doesn't seem to exist, I decided to write one myself; the current draft can be found at http://jamespropp.org/reverse.pdf . -Comments are welcome!<|endoftext|> -TITLE: Matrix with max sum on all diagonals -QUESTION [7 upvotes]: I knew this problem a long time ago but could not recall the solution. I know the solution is interesting though. Anybody want to help? Thanks. -There is a $n \times n$ matrix. A collection of $n$ entries of the matrix is called good if no two of them have a same row or column. For each good collection we consider the sum of its elements, and let $M$ be the maximum among all these sums. Now any entry of the matrix belongs to at least one good collection with sum $M$. Prove that every good collection has sum $M$. - -REPLY [7 votes]: Let $\circ$ denote the Hadamard product, $A\circ B=(a _{ij} b _{ij}) _{1\le i,j\le n}$, and let $f(A)$ denote the sum of the entries of $A$. Each good set corresponds to a permutation in $S _n$. Let our matrix be $C$, then we are given that there are $k$ permutation matrices $P _{\sigma_i}$, $\sigma _i\in S _n$, such that $f(P _{\sigma _i}\circ C)=M$, and $\sum P _{\sigma _i}$ has all entries $\geq 1$. -Let $\pi\in S_n$ be so that, $M _{0}=f(P _{\pi}\circ C) < M$, then we look at $\sum P _{\sigma _i}-P _{\pi}$, it has nonnegative entries and equal sums of rows and columns. Such a matrix can be written as a sum $\sum P _{\tau _i}$, $\tau _i\in S_n$, this is a simple exercise in combinatorics (maybe you have heard it in the form, every bipartite regular multigraph can be written as a union of perfect matchings). So we have -$$kM-M _{0}=f\left((\sum P _{\sigma _i}-P _{\pi})\circ C\right)=f\left((\sum P _{\tau _i})\circ C\right)=\sum f(P _{\tau _i}\circ C)\le (k-1)M$$ -which gives us a contradiction. So $f(P _{\pi}\circ C)=M$, but $\pi$ was arbitrary, so we are done.<|endoftext|> -TITLE: Hardy spaces: analysis <---> martingales -QUESTION [11 upvotes]: Let $H^p$ be the Hardy space of analytic functions on the open unit disk $\mathbb{D}$: $f \in H^p$ if $f$ is analytic on $\mathbb{D}$ and $\sup_{r < 1} \int_0^{2\pi} |f(re^{i\theta})|^p d\theta < \infty$. -Consider a filtration generated by a 2-d (complex) Brownian Motion $B$. The martingale hardy space $\mathcal{H}^p$ defined on some time interval $[0,T]$, say, is the set of martingales $M$ such that $M^* = \sup_{t \in [0,T]} |M_t| \in L^p$. This definition is mostly interesting for $p=1$, as for $p>1$, $\mathcal{H}^p$ can be associated with a regular $L^p$ space of martingales. -If $B$ starts at zero, let $\tau$ be the hitting time of the boundary of $\mathbb{D}$. Then a connection between these two spaces is the following: for $f$ analytic on the unit disk, $f(B_{t \wedge \tau}) \in \mathcal{H}^p$ if and only if $f \in H^p$, and this mapping is continuous. -This allows you to associate $H^p$ to a subspace of $\mathcal{H}^p$. For studying $\mathcal{H}^p$, it would be useful to have a more complete representation of part of $\mathcal{H}^p$ in terms of functions evaluated on $B$. Specifically, for martingales that run on the whole time interval. Can this be obtained by using another hardy space, such as the Hardy space $h^p$ on $\mathbb{R}^2$? Can anything else be said relating hardy spaces of martingales and hardy spaces of functions? - -REPLY [5 votes]: Charles Fefferman proved that the space of functions of bounded mean oscillation $BMO$ is the dual of the Hardy space $H^1$. The key was the observation that the classical Hardy space $H^1$ is a natural substitution for $L^1$, and $BMO$ is a natural substitution for $L^{\infty}$. -One result of comparing classical Hardy and martingale Hardy spaces is that we can establish that the dual of the martingale Hardy space $\mathcal{H}^1$ is martingale $BMO$. -Check out "Martingale Inequalities: Seminar Notes on Recent Progress (Mathematics Lecture Note Series)" by Adriano Garsia<|endoftext|> -TITLE: Functoriality of the cotangent bundle -QUESTION [19 upvotes]: Recall that to any manifold $X$, I can assign in a canonical way a manifold $\mathrm T ^* X$, the total space of the cotangent bundle over $X$. Recall also that, unlike the tangent bundle construction, the map $X \mapsto \mathrm T^* X$ is not an endofunctor on the category of manifolds: whereas tangent vectors push forward along smooth maps, cotangent (co)vectors do not (they also do not pull back). -Nevertheless, $X \mapsto \mathrm T^* X$ is functorial for some restricted classes of maps. For example, there is a category whose objects are manifolds and whose morphisms are étale maps, and the cotangent construction is (covariantly) functorial for this category. -My question is: - -Do the étale maps comprise the largest class of morphisms of manifolds for which $\mathrm T^*$ is functorial? - -In my particular situation, I have a (surjective) submersion $Y \to X$, and I can construct by hand a (Poisson) map $\mathrm T^* Y \to \mathrm T^*X$ covering it, because I know of some extra structure for $Y,X$. But I would like to know if there is a more canonical reason that I have this map. - -REPLY [4 votes]: I do not know a larger class of smooth mappings; and I considered this question intensively when co-writing the book "Natural operations in differential geometry, Springer-Verlag, 1993"(pdf). -See also 26.11 -- 26.16 in this book for a determination of all natural transformations $T T^* \to T^*T$, viewed as functors on the category of $m$-dimensional manifolds and local diffeomorphisms -(etale mappings), and similar questions.<|endoftext|> -TITLE: Is independence meaningful for commutative $C^*$-algebras? -QUESTION [5 upvotes]: I don't know very much about spectral theory so probably the answer to my question has a basic reference which I would appreciate. -Let's say I have two self-adjoint operators on a Hilbert space and that they commute. I know that, since they generate a commutative $C^*$ algebra, on the one hand they can be viewed as continuous functions on some compact Hausdorff space. On the other hand, individually, they can be viewed as multiplication operators $L^2$ on some measure space. -Now, one (extremely basic) question I have is whether it can be arranged for them to be viewed as multiplication operators on the same measure space. I am guessing the answer is yes, but I would like a reference for it or a counterexample since I don't really know. -Now I am sort of assuming the answer to question 1 is yes, but here's question 2. Let's say I already have two multiplication operators in hand, $\phi_1 : X \to {\mathbb R}$ and $\phi_2 : X \to {\mathbb R}$ where $X$ has measure $\mu$. I would say that $\phi_1$ and $\phi_2$ are independent if the pushforward of $\mu$ by $\phi_1 \times \phi_2 : X \to {\mathbb R}^2$ is the product measure of the pushforwards of $\mu$ by the maps $\phi_1$ and $\phi_2$ individually. Notice that if I compose $\phi_1 \circ T$ and $\phi_2 \circ T$ with any measure-preserving map, the resulting maps are still independent since the pushed-forward measures do not change at all. -My second question is whether the notion of independence makes sense for such operators (or more generally for members of a commutative $C^*$ algebra) without having a measure at all. For example, is being independent an operator-theoretic property of $\phi_1$ and $\phi_2$? - -REPLY [6 votes]: For your first question, yes, you can view two commuting normal operators as functions on the same measure space. Let us for simplicity assume that $T_i : i=1,2$ are self-adjoint acting on a Hilbert space $H$. Then there exists a measure $\mu$ defined on $\mathbb{R}^2$ with values in projections oh $H$, so that: $\mu(A)$ and $\mu(B)$ commute for any Borel subsets of $\mathbb{R}$; $\mu(A)$ and $\mu(B)$ are perpendicular whenever $A$ and $B$ are disjoint; and $\mu$ is $\sigma$-additive (where you use the strong operator topology to define limits of sums of perpendicular projections). Moreover, you have: -Given $c_1, c_2$ and $C_1, C_2$ with $c_i < C_i$, $\mu( [c_1, C_1] \times [c_2,C_2])$ is the orthogonal projection onto the set $\{\xi \in H: c_i \Vert \xi \Vert \leq \Vert T_i \xi \Vert \leq C_i \Vert \xi \Vert\}$ and $T_i = \int x_i d\mu(x_1,x_2)$ (here $x_1,x_2$ are the two coordinates on $\mathbb{R}^2$ and the convergence of the integral is in the strong sense, i.e., at every vector). This means in particular that for any $\xi$, $$\langle T_i \xi ,\xi \rangle = \int x_i \langle d\mu (x_1,x_2)\xi ,\xi \rangle $$ (note that $\langle \xi ,\mu (\cdot)\xi\rangle $ is a positive scalar-valued measure). -Now you would prefer to have scalar-valued measures. To do so, denote by $M$ the von Neumann algebra generated by $T_1,T_2$, and find a sequence of vectors $\xi_i$ in your Hilbert space so that $M\xi_i \perp M\xi_j$ for all $i\neq j$ and $\textrm{span}(M\xi_i : i=1,2,\dots)$ is dense in $H$. Let us arrange that $\sum \Vert\xi_i\Vert^2 =1$. Let $H_i$ be the closure of $M\xi_i$. Then if you denote $\mu_i(B) = \langle \xi_i \mu (B),\xi_i\rangle$, you find that $M\xi_i \cong L^2(\mathbb{R}^2,\mu_i$ in a way that takes $T_i$ to the multiplication operator by $x_i$. The original action of $T$ on $H$ can then be viewed as multiplication by $x_i$ on the disjoint union of measure spaces $\sqcup_i (\mathbb{R}^2,\mu_i)$ with the probability measure given by the sum of the $\mu_i$'s. -Any book that does the spectral theorem will have an equivalent statement. Probably you can find it in Conway's functional analysis book. -Regarding your second question: it does make sense to talk about independence of operators, provided that you have chosen a measure (i.e., a state on your $C^*$-algebra). The definition for multiplication operators is exactly the one for random variables: $\phi_i: (X,\mu)\to \mathbb{R}$ are independent if the push-forward of $\mu$ is the product measure. But this of course depends on the measure $\mu$ in a crucial way. For example, if you take $x_1,x_2: (\mathbb{R^2},\mu)\to \mathbb{R}$ and you take $\mu$ to be the Lebesgue measure on $[0,1]^2$, then the $x_i$'s will be independent. But take instead the $\delta$-measure concentrated on the diagonal $\{(x_1,x_2):x_1=x_2,0\leq x_j\leq 1/\sqrt{2}\}$ (i.e., $\mu(B)$ is the Lebesgue measure on the diagonal of the intersection of $B$ with a diagonal line segment from the origin to $(1/\sqrt{2},1/\sqrt{2})$). In that case the push-forwards of $\mu$ by $x_1$ and $x_2$ are equal and are not the product of the individual push-forwards. Thus it is absolutely essentially that you talk about independence of functions on measure spaces (=self-adjoint operators in an abelian $C^*$-algebra, =random variables) with respect to the chosen measure (=state, =expectation). - -REPLY [4 votes]: If $X$ is a separable compact Hausdorff space then for any Radon measure $\mu$ with dense support we have $C(X) \subset L^\infty(X, \mu)$, so the answer to your first question is yes. -Independence will depend on the measure you take, consider for instance the difference between taking Lebesgue measure on $[0, 1] \times [0,1]$ versus taking a measure which gives most of its weight to the diagonal. In the former case functions of $x$ will be independent from functions of $y$, while this is not the case in the latter. -These types of results would be covered in any introductory book on operator algebras. For instance R. Douglas "Banach algebra techniques in operator theory" (http://www.ams.org/mathscinet-getitem?mr=1634900).<|endoftext|> -TITLE: How does all of the bundles over a certain manifold characterize the homotopy class of the base manifold? -QUESTION [8 upvotes]: It is known that if $f:M\rightarrow N$ is a homotopy equivalent, then the the process of pullback gives a one-one correspondence between bundles over $N$ and $M$ up to isomorphism. Is the converse( that if $f$ gives a 1-1 correspondence between them, then $f$ is a homotopy equivalence)true? Or any counterexample? -By the way, are bundles of a fixed rank over a compact manifold finite up to isomorphism? And is there any characterization of them like the holomorphic line bundles as a first cohomology group of a certain sheaf? If so, is there any way to calculate? - -$M,N$ are smooth manifolds, and I originally thought of real vector bundles. But since this is a wild guess, it might be better to think the most general bundles you could imagine that homotopy invariance holds, even though I only know the real case. -It seems algori's answer apply to all kind of vector bundles I can imagine, though a bit beyond me. Thanx very much. And any reference for your arguments? -And I'd like to know whether homotopy invariance holds in fiber bundles. - -Thanks all of you. But formally whose answer should I accept? - -REPLY [2 votes]: All the above answers are great, but as you are just learning this stuff, the following might be helpful. Vector/$G$-bundles contain actual geometric information! The tangent bundle, for example, tells you a lot about the geometric structure of the underlying manifold (some definitions of curvature etc.). Now a homotopy theorist can measure these things using cohomology and homotopy as described above by Algori and Mark Grant. So cohomology restricts what kind of bundles you can have (are they trivial? how many linearly independent sections might they have?). Chern-Weil theory might be something to look up if this sounds cool. -Something to note though, all vector bundles are homotopy equivalences. -Also, Peter May wrote a little memoir called Classifying spaces and fibrations where he writes down classifying spaces for fibrations with fiber $F$. These have significantly less structure than a plain old fiber bundle. Granted, there are restrictions on when we have classification, but usually they are pretty harmless, and any compact smooth manifold should satisfy them.<|endoftext|> -TITLE: When do 0-preserving isometries have to be linear? -QUESTION [6 upvotes]: Let $\langle \mathbf{V},+,\cdot,||.|| \rangle$ be a normed vector space over $\mathbb{R}$. - -Let $f : \mathbf{V} \to \mathbf{V}$ be an isometry that satisfies $f(\mathbf{0}) = \mathbf{0}$ . - -What conditions on the vector space would or would not force $f$ to be linear? - -examples: finite dimensional, complete, norm induced by an inner product, strictly convex - -REPLY [14 votes]: If you assume $f$ to be surjective then $f$ has to be linear without any assumptions on $V$ by the Mazur-Ulam theorem. Wikipedia doesn't offer much more information than a link to the beautiful recent proof by J. Väisälä.<|endoftext|> -TITLE: Can a Lie group as an abstract group be given more than one topology making it a Lie group? -QUESTION [10 upvotes]: I am an optical engineer, so please forgive any ignorance my questions betoken. I am interested in whether one can tear down the manifold of a finite dimensional Lie group, -leaving an abstract group, and then give the group another manifold structure that makes it again into a Lie group and get anything essentially different -in the process. I know the manifold replacement can be done in some special cases. Here are the examples I have been thinking about and could come up -with answers to. In all examples I can make progress on, the Lie groups arising from the different manifolds built on the same set are isomorphic -(as Lie groups that is: i.e. they have the same Lie algebra, fundamental group and connected components) - they are of course isomorphic as abstract groups!!!. -I have a hunch that this is generally true, otherwise one might get weird examples where an abstract group might have several different Lie algebras, and -I'm sure I'd have read about that somewhere by now! If someone knows anything about the general case, I'd be most interested to hear about it. -Example : Take the additive group $\left(\mathbb{R},\cdot +\right)$, it is its own Lie algebra. As the lie algebra I denote it $\mathbf{g}$ and the Lie group $G$ - -the exponential map is the identity map. Now take one of the everywhere discontinuous, bijective solutions $\phi$ of the Cauchy functional equation -$f\left(x\right) + f\left(y\right) = f\left(x + y\right)$ as detailed in the end of Chapter 2 of Hewitt and Stromberg "Real and Abstract Analysis"; $\phi$ is -now an exponential map from the $g$ onto a new Lie group $G^\prime$, to wit the same group $\left(\mathbb{R},\cdot +\right)$ but now given a new topology -generated by images $\phi\left(U\right)$ of open intervals in $\mathbb{R}$. This topology is of course totally disconnected in the group topology for $G$; -however $\phi$ is continuous, indeed $C^\infty$ when thought of as a map from $g$ to $G^\prime$, the latter with the new topology. Indeed, the Lie groups -arising from the two topologies are isomorphic, their topologies are homoemorphic. -I have a University of Pittsburgh 2007 PhD. Thesis "On the Uniqueness of Polish Group Topologies" by Bojana Pejic that I believe may be relevant, but so far have not made -great headway in understanding it. If someone could point me to other material on this question, I'd be grateful. - -REPLY [19 votes]: To answer the question of the title: given a group, there may be infinitely many different topologies that render it a Lie group: $\mathbb{R}^n$ and $\mathbb{R}^m$ (where $m$ and $n$ are positive integers) are isomorphic as groups since they are $\mathbb{Q}$-vector spaces of the same dimension, hence they are isomorphic to the direct sum of continuously many copies of $\mathbb{Q}$. But they are not isomorphic as topological spaces, let alone Lie groups. -However, given two semi-simple Lie groups, I think they are isomorphic as Lie groups as soon as they are isomorphic just as groups.<|endoftext|> -TITLE: Examples of Brown (co)fibration categories that are not Quillen model categories? -QUESTION [21 upvotes]: K.S. Brown has shown that much of abstract homotopy theory can be carried out in the setting of Brown (co)fibration categories [MR0341469]. The decisive property, immediate from the axioms, is that any morphism can be factored into a cofibration, followed by a weak equivalence. -H.-J. Baues [MR0985099], Cisinski [MR2729017] and Radulescu-Banu [arxiv.org/abs/math/0610009] then followed a similar path. Now someone wants to convince me that this is the proper setting for abstract homotopy theory. To begin with, I do like the simplicity of the axioms. Still, I'd like to be convinced of the practical necessity of this approach. Therefore my question: -Are there examples of Brown (co)fibration categories that are not already Quillen model categories? -More precisely, does there exist a pair (C,W) consisting of a category C and a subset W in Mor C (weak equivalences) such that (C,W) can be equipped with the structure of a Brown fibration (or cofibration) category, but not with the structure of a Quillen model category? -I would particularly be interested in examples in which the lifting axioms of Quillen are the obstacle. If they fail to be a Quillen model category just because they lack limits or colimits, I would be less enthusiastic. -On the other hand, I would also welcome examples in which it is relatively easy to show that they are Brown, but relatively hard that they are Quillen. -I would also welcome good general stability properties. For instance, in the Brown case there are no big problems if you want to enlarge the set of weak equivalences, as long as the larger set satisfies (2 of 3) and as long as the resulting larger set of acyclic cofibrations is stable under pushouts (incision). -To summarise, I want to be able to exclaim: "Good that we have the Brown apparatus!" - -REPLY [11 votes]: Consider an abelian category $A$ (or, more generally, an exact category in the sense of Quillen), then the category of complexes of $A$ is a category of cofibrant objects with the quasi-isomorphisms as weak equivalences and the degreewise (admissible) monomorphisms as cofibrations. This is not a Quillen model category in general (if you restrict your attention to bounded complexes, Quillen lifting axioms correspond exactly to the fact that there are enough injectives in $A$, which might fail in general, as can be seen by contemplating the opposite category of the category of sheaves over a sufficiently general topological space). -Also, even in the case when we have enough injectives, it is also possible to consider degreewise split monomorphisms as cofibrations (keeping the same weak equivalences), and then, Quillen axioms fail unless quasi-isomorphisms are all chain homotopy equivalences (in the case of an abelian category, this means that $A$ is semi-simple). -These examples are instances of a more general situation: consider a category of cofibrant objects $C$ with class of weak equivalences $W$. Then, for any class $S$ of maps of $C$, one can define a new class of maps $W(S)$ as the smallest one which contains $W\cup S$ and which satisfies the following properties: it has the two out of three property, and the class of cofibrations which are in $W(S)$ is closed under pushouts and finite sums. The good news are that $C$ is still a category of cofibrant objects with the same cofibrations but with $W(S)$ as class of weak equivalences. -This process is exactly what you need to define the notion of quasi-isomorphism of complexes of an exact category: starting from the category of bounded complexes with degreewise split monomorphisms and chain homotopy equivalences as weak equivalences (which is then a Quillen model category modulo the existence of finite (co) limits), one gets quasi-isomorphisms as the class $W(S)$ where $S$ consists of maps $X\to 0$, where $X$ runs over the family of complexes associated to admissible short exact sequences -$$0\to A\to B\to C\to 0$$ -There is a non-abelian version of this construction: consider a (small) category $C$. We may then consider the category $s(C)$ of simplicial objects in the free completion of $C$ by finite sums. Then, considering the termwise split monomorphisms, there is a smallest class of maps $W$ such that $s(C)$ is a category of cofibrant objects with $W$ as weak equivalences and such that any simplicial homotopy equivalence is in $W$. The $(\infty,1)$-category (obtained by considering the Dwyer-Kan localization of $C$ by $W$) corresponds to the free completion of $C$ by finite homotopy colimits; for instance, for $C$ the terminal category, $s(C)$ is simply the homotopy theory of finite simplicial sets. If you do the same construction by replacing $C$ by its completion under small sums and -by replacing $W(S)$ by its closure under small sums and realizations, then you obtain the homotopy theory of cofibrant simplicial presheaves over $C$ (for the projective model structure), except that you didn't use any complicated tool to define it (no small object argument, no lifting theorem, in fact, you don't need to know the model category of simplicial sets at all). Of course, this apparent simplification comes at a price: you don't know how to construct homotopy limits in this language. The advantage is that you already are able to speak of homotopy colimits, so that you may use this to understand how to construct homotopy theoretic structures (e.g. Quillen model categories). -Another nice example of a category of cofibrant objects which is not a model category is the category of finite CW-complexes. One may argue that this is a subcategory of the category of cofibrant objects of a Quillen model category. But this is in fact always the case: any category of (co)-fibrant objects can be embedded very nicely in a proper simplicial model category; see Theorems 3.2, 3.10 and 3.25 and Remark 3.13 of my paper -Invariance de la K-théorie par équivalences dérivées, J. K-theory 6 (2010).<|endoftext|> -TITLE: Questions on smoothness of Riemann metrics -QUESTION [9 upvotes]: I've heard assertions of the sort: - -Let there be a Riemann metric (not very smooth, say of class $C^1$ or $C^2$ or maybe $C$?) in a neighbourhood of a point on a manifold. Then it is possible to choose coordinates so that the metric is $C^\infty$ or even analytic in them. -In case of 3-dimensional manifolds it is possible to choose such coordinates globally, so the manifold becomes a smooth one. In the case of higher dimensions $n\ge4$ it is not true. - -Are those assertions true? I've heard them some time ago and not sure I remember all the details. Is it a well-known thing? Are there some detailed references? - -REPLY [8 votes]: I confirm the Anton's answer (No, and the phenomenon is essentially local), but I suggest another explanation which works for C^1 2-dimensional metrics. -We will look for a counterexample in the class of metrics such that they are C^2 everywhere except for some line, where they are C^1. Then, it is possible and relatively easy to cook an example such that the curvature of the metric is discontinuous at this special line; you can do it in the class of confomally flat metrics such that the conformal coefficient depends on one variable only and the line is where this variable is a constant. -Since in order to determine the curvature of a metric you only need the distance function corresponding to this metric, and distance function does not depend on how smooth is your atlas, you can not make this metric smooth by the change of the atlas. - -REPLY [7 votes]: If you combine the work of Jost-Karcher on almost linear co-ordinates with the work of DeTurck-Kazdan and Shefel on harmonic co-ordinates (I recommend a paper of Stefan Peters on a proof of the Gromov convergence theorem), you get the following: -If there exist local co-ordinates in which a Riemannian metric $g$ is $C^1$ and has bounded sectional curvature, then there exist local (harmonic) co-ordinates in which the metric is $C^{1,\alpha}$ for every $\alpha > 0$. If, in addition to this, the covariant derivatives of the Ricci tensor up to order $k$ are locally bounded, then there exist local harmonic co-ordinates in which the metric is $C^{k+1,\alpha}$ for any $\alpha > 0$. If, in particular, the covariant derivatives of Ricci of all orders are bounded, then there exist local harmonic co-ordinates in which the metric is $C^\infty$.<|endoftext|> -TITLE: Asymptotic behaviour of a sequence -QUESTION [6 upvotes]: Hello, -I am interested in some kind of sequence that are "not finitely recurrent". -Let $a_i$ be a sequence taking values in $\{0,1\}$. -Consider the sequence $(u_i)$ such that $u_0=1$, and for any positive integer $n$, -$u_n =\sum_{i=1}^n u_{n-i}a_i$ -It is easy to prove that there exists some positive term $\lambda$ such that $(u_n)$ growth faster than $(\lambda-\epsilon)^n$ and slower than $(\lambda+\epsilon)^n$ for any $\epsilon>0$. -I would like to know if it is possible to get better bounds, even by adding if necessary some hypothesis on the coefficients $(a_i)$ (but of course this should not be a periodic sequence, otherwise this is well known and too easy). -Thanks by advance for your comments ! - -REPLY [4 votes]: Let me slightly change the notation. You have some sequence $\mathbf{a}=a_0,a_1,\cdots$ with each $a_i$ either $0$ or $1$. Define a sequence $u_0,u_1,u_2,\cdots$ by setting $u_m=0$ for $m \lt 0$, $u_0=1$ and for any $n \ge 0$, $u_{n+1}=\sum_{i=1}^{\infty}a_iu_{n-i}.$ As you note in a comment, if $a_1=1$ then $u_i$ is a non-decreasing and eventually increasing sequence (excluding a trivial case). The sequence $u_i$ will eventually be increasing as long as the set of $j$ with $a_j=1$ has a $\gcd$ of $1$. -Consider the power series $f(r)=r-(a_0+\frac{a_1}{r}+\frac{a_2}{r^2}+\cdots).$ Then $f(r)$ is defined and increasing for $r \gt 1.$ There is a unique $\lambda=\lambda_{\mathbf{a}}>1$ with $f(\lambda)=0.$ I think that there should be a constant $c \le 1$ with $\lim \frac{a_n}{c\lambda^n}=1$. We can also say that $\lambda \le 2$ with equality only in the degenerate case that $a_i=1$ for all $i$. For any $1 \lt r \lt 2$ we can create an $\mathbf{a}$ with $\lambda_{\mathbf{a}}=r$ by simply choosing the $a_i$ one at a time to keep the (non-decreasing) partial sums of $f(r)$ non-negative. -We can partially order the possible sequences $\mathbf{a}=(a_i)$ by saying $\mathbf{a}<\mathbf{b}$ when $a_i=b_i$ for $0 \le i \lt j$ and $0=a_j -TITLE: When is a blow-up Cohen-Macaulay? -QUESTION [10 upvotes]: Let $X$ be a smooth projective variety and $Z$ a closed subscheme. Let $X'$ be the blow-up of $X$ with center $Z$. - -Under what conditions on $Z$ is $X'$ - Cohen-Macaulay? - -In the case $Z$ is non-singular, the blow-up $X'$ will also be non-singular, so in particular CM. At the other extreme, any birational morphism is the blow-up of some ideal, so if $Z$ is horrible, there is no hope of having Cohen-Macaulayness. -I'm sure this question has been studied in the literature before and I'd be interested in references for sufficient conditions when $X'$ is CM. The case I find most interesting is when $Z$ is a locally complete intersection. - -REPLY [7 votes]: The local complete intersection case is actually straightforward: - -Claim - Let $X$ be CM and $Z$ an lci subscheme. Then $Bl_ZX$, the blow-up of $X$ along $Z$ is also CM. - -Proof: -Let $\pi:Bl_ZX\to X$ denote the blow-up. Clearly, $X\setminus \pi^{-1}Z\simeq X\setminus Z$ is CM. -Since $X$ is CM, so is $Z$. -Since $\mathscr I$, the ideal sheaf of $Z$ is locally generated by a regular sequence, the sheaf of rings $\oplus \mathscr I^n/\mathscr I^{n+1}$ is locally a polynomial ring and hence $E=\pi^{-1}Z\to Z$ is a $\mathbb P^r$-bundle where $r={\rm codim}_X Z-1$. -Since $Z$ is CM, so is $E$. Finally, since $E$ is a Cartier divisor which is CM, so is $Bl_ZX$ along $E$. $\square$ -(For more on this see $\S$5 of this paper.)<|endoftext|> -TITLE: Logic in mathematics and philosophy -QUESTION [69 upvotes]: What are the relations between logic as an area of (modern) philosophy and mathematical logic. -The world "modern" refers to 20th century and later, and I am curious mainly about the second half of the 20th century. -Background and motivation -Logic is an ancient area of philosophy which, while extensively beein studied in Universities for centuries, not much happened (unlike other areas of philosophy) from ancient times until the end of the 19th century. The development of logic in the first part of the 20th century since Frege, Russell and others is a turning point both in logic as an area of philosophy and in mathematical logic. In the first half of the 20th century there were close connections between the development of logic as an area of philosophy and the development of mathematical logic. Later, in addition to its interest for mathematicians and philosophers logic became a central applied field in computer science. -My question is about relations between logic as part of philosophy and mathematical logic from the second half of the 20th century when its seems that connections between these two areas have weakened. So I am asking about formal models developed in philosophy that had become important in mathematical logic and about works in philosophical logic that were motivated or influenced by developments in mathematical logic. -I am quite curious also about the reasons for the much weaker connections between mathematical logic and formal models developed by philosophers at the later part of the 20th century. (This makes this question much less board than it seems. Another thing I am curious about is to what extent for the applications to computer science formal models described by philosophers turned out to be useful. -Update: To complement the excellent answers already given I will try to ask some additional researchers in relevant fields to contribute directly or through me. -Update While it was clear in some of the answers let me make it explicit that I refer also to relations between philosophy and set theory. -Related MO question: In what ways did Leibniz's philosophy foresee modern mathematics? -Has philosophy ever clarified mathematics? - -REPLY [5 votes]: Here is an answer via a picture (Joel Davis Hamkins) -See also this and this blog posts.<|endoftext|> -TITLE: What are the differences between the terms prevariety, variety, abstract variety, and algebraic variety? -QUESTION [9 upvotes]: What are the differences between the terms prevariety, variety, abstract variety, and algebraic variety? Sometimes I see these terms used in the literature without explicit definitions or references, and it is not clear that all of them have standardized definitions. - -REPLY [10 votes]: There is no standardized definition. Using one of these terms always implies there is a ground field $k$, and (pre)varieties are $k$-schemes of finite type, possibly with certain restrictions (separated, reduced, irreducible...). -As far as I know, "prevariety" and "abstract variety" are no longer in use. "Prevariety" used to mean "possibly nonseparated" (at least in Mumford's Red Book, where prevarieties are in addition irreducible and reduced). "Abstract" stands for "not embedded in projective space". -Regrettably, some authors use the term "(algebraic) variety" without bothering to define it. In general, such carelessness is an indication that the objects in question are assumed separated and reduced (other cases not being worthy of consideration), or even that $k=\mathbb{C}$ (for the same reason).<|endoftext|> -TITLE: pigeonhole principle -QUESTION [5 upvotes]: Hi everyone -I saw a question on Mathoverflow asking for some applications of pigeonhole principle, among the answers I saw a problem set which was proposed by Prof. Richard Stanley and in this problem set there was a question which I am interested on it, here it is: -Consider these two sequences $a_1< a_2 < \cdots < a_n$ and $b_1 >b_2 > \cdots >b_n$ such that $$\{a_1,\cdots a_n,b_1\cdots b_n\}=\{1,2,\cdots 2n\}$$ show that -$$\sum_i|a_i-b_i|=n^2$$ -I have no idea how to do this. Perhaps someone can give a hint? I try to consider some cases but the answer was long and boring, I think there is a nice trick. - -REPLY [12 votes]: Consider the sum $S = \sum (a_i + b_i ) + \sum |a_i - b_i|$. The first term evaluates to $n(2n+1)$. The second term is unknown. Claim: $S$ is equivalent to $2\sum_{n+1}^{2n} i = n(3n+1)$ which implies the result you want. -It is easy to see that $S$ is equivalent to twice the sum of $\sum \max(a_i,b_i)$. Using the pigeonhole principle you can show that $\max(a_i,b_i) > n$. - -REPLY [4 votes]: I'll expand slightly on Dan's suggestion. -Any such pair of sequences $(\{a_{i}\}, \{b_{j}\})$ can be obtained from the pair of sequences $(\{i\}, \{2n+1-j\})$ by finitely many iterations of the following operation. -For some $i$ and some $j$ with $a_{i} = k, b_{j} = k+1$, set $a_{i} = k+1, b_{j} = k$. -Now just check that the claim holds for the pair of sequences $(\{i\}, \{2n+1-j\})$, and that the operation I described leaves the sum invariant.<|endoftext|> -TITLE: Vector bundles on affine scheme -QUESTION [11 upvotes]: I have already asked similar questions before, but now I realized that there a nice -general way to ask what I want. Namely let $X$ be a normal affine variety over -a field $k$. Assume first that $k$ is finite. Then is it true that -1) IF $X$ is smooth, then the set of isomorphism classes of vector bundles of given rank on $X$ is finite? -2) More generally, is it true that for any $X$ the set of isomorphism classes of Cohen-Macaulay torsion free sheaves of fixed generic rank is finite? -When the field $k$ is arbitrary then I would expect that there exists a finite-dimensional -(over $k$) family of vector bundles or Cohen-Macaulay sheaves which contains -every isomorphism class. Is this true? - -REPLY [5 votes]: As shown by David, the answer is no for line bundles. -More generally, the answer in no even for higher dimensional vector bundles. In fact, in his paper "Vector bundle over affine surfaces birationally equivalent to ruled surfaces" Murthy proves the following results: -(1) Let $V$ be an irreducible affine non-singular surface defined over an algebraically closed field $k$ and such that $V$ is birational to $C \times \mathbb{P}^1$, vhere $C$ is a curve. Then any vector bundle over $V$ is a direct sum of a trivial bundle and a line bundle. In general, there are infinitely many equivalence classes of line bundles (as shown by David for $V= \mathbb{P}^1 \times \mathbb{P}^1 \setminus \Delta$). -(2) There exist an affine, nonsingular rational variety of dimension $3$ over which there are infinitely many non-isomorphic indecomposable vector bundles of rank $>1$.<|endoftext|> -TITLE: Are there ever three perfect powers between consecutive squares? -QUESTION [16 upvotes]: I have not seen this question treated in the literature. Does anyone have more information? There are several OEIS sequences (A097056, A117896, A117934) dealing with this question, but no answers. - -REPLY [24 votes]: If you let $a_1,a_2,\dots$ represent the sequence $1,4,8,9,\dots$ of perfect powers, it is a conjecture of Erdos that $$a_{n+1}-a_n >c'n^c.$$ -The weaker conjecture $$\liminf a_{n+1}-a_n =\infty$$ is known as Pillai's conjecture. -This seems to be much beyond the reach of the strongest methods available today. Note, however, that your conjecture in the OP implies $$a_{n+2}-a_n>cn^{1/2}$$ which is a very strong, related result. To give an idea of how hard it is to prove that consecutive powers tend to space out more and more, consider how hard it was to prove Catalan's conjecture (consecutive powers can not be 1 apart, except for 8 and 9) or even the still open Hall's conjecture which asserts that if $k=x^3-y^2\neq 0$ then -$$k>\max\{x^3,y^2\}^{1/6}-o(1)$$ which is related to some abc type conjectures. -Now on the specific problem you mention, there is some work done on powers in relatively short intervals. Start with - -J. H. Loxton, Some problems involving powers of integers, Acta Arith., 46 (1986), - pp. 113–123 - -where it is proved that the interval $[N,N+\sqrt{N}]$ contains at most -$$\exp(40(\log\log N \log\log\log N)^{1/2})$$ integer powers. This is far from the bound of $3$ but it is a nice result. There was a gap in the paper above, but it was corrected by two different authors, see - -D. J. Bernstein, Detecting perfect powers in essentially linear time, Math. Comput., - 67 (1998), pp. 1253–1283 - -and - -C. L. Stewart, On heights of multiplicatively dependent algebraic numbers, Acta - Arith., 133 (2008), pp. 97–108 -C. L. Stewart, On sets of integers whose shifted products are powers, J. Comb. Theory, Ser. A, - 115 (2008), pp. 662–673 - -In the first paper by Stewart above, it is conjectured that there are infinitely many integers $N$ for which the interval $[N,N +\sqrt{N}]$ contains three integers one of which is a square, one a cube and one a fifth power. He also conjectures that for sufficiently large $N$ this interval does not contain four distinct powers and if it contains three distinct powers then one of is a square, one is a cube and the third is a fifth power, which is a refined version of your conjecture. -For more references and a longer survey see Waldschmidt's "Perfect Powers: Pillai’s works and their developments". - -REPLY [20 votes]: It follows from the ABC conjecture that there are only finitely many. First, show (unconditionally) that for all but finitely many such triples of perfect powers, at most one of them is a cube. So, the others are fifth or higher powers, say $n^2n^2, C = O(n)$ and the radicals of $A$ and $B$ are $O(n^{2/5})$, so the ABC conjecture bounds $n$.<|endoftext|> -TITLE: Principal congruence subgroups of $SL(n, \mathbb{Z})$ -QUESTION [7 upvotes]: I want to know whether the principal congruence subgroups of $SL(n, \mathbb{Z})$ are characteristic? please suggest me a reference. - -REPLY [8 votes]: Yes, they are caracteristic. -Let $\rho: GL(n, \mathbb{Z}) \to GL(n, \mathbb{Z}/m\mathbb{Z})$ be reduction mod $m$ and $\Gamma_n(m) := ker(\rho) \cap SL(n, \mathbb{Z})$ be a congruence sugroup. -According to the discussion in Automorphisms of $SL_n(\mathbb{Z})$ the automorphisms of $SL(n, \mathbb{Z})$ for $n > 2$ are generated by - -$X \mapsto (X^T)^{-1}$ -$X \mapsto AXA^{-1}, A \in GL(n, \mathbb{Z}).$ - -Obviously, $\Gamma_n(m)$ is invariant under these automorphisms and thus is characteristic. -In case $n = 2$ there is one more automorphisms of $SL(2, \mathbb{Z})$ that also leaves $\Gamma_n(m)$ invariant (for a description of this automorphism see the paper of Hua and Reiner mentioned in the link above).<|endoftext|> -TITLE: least condition for the Fourier transform to be integrable -QUESTION [5 upvotes]: I want to prove that if $f \in C^{1}(\mathbb{R})$ is compactly supported then its Fourier transform is integrable. -I was able to prove the result for $f \in C^{2}(\mathbb{R})$ and compactly supported. I used the fact that if $f \in C^{2}(\mathbb{R})$, then $\hat{f}$ is bounded by $\frac{c}{1+{|x|}^{2}}$. So it is integrable. -I failed to prove it if $f \in C^{1}(\mathbb{R})$ - -REPLY [3 votes]: Here is a link to the paper about necessary conditions for the integrability of the Fourier transform: -http://www.heldermann-verlag.de/gmj/gmj16/gmj16043.pdf -It is stated in that paper that sufficient conditions for the integrability of the Fourier transform are given in the book -R. M. Trigub and E. S. Bellinsky, Fourier analysis and approximation of functions. -Kluwer Academic Publishers, Dordrecht, 2004.<|endoftext|> -TITLE: Fontaine's rings of periods -QUESTION [20 upvotes]: I've been trying lately to understand Fontaine's rings of periods, $B_{\mathrm{dR}}$, $B_{\mathrm{cris}}$, etc. However, I have a really hard time understanding and appreciating how to think about and use these. These rings seems so incredibly complicated and unintuitive it boggles my mind; I never seem to be able to remember their construction. -So, how do I think about, learn and use these obscure objects? (Do I really need to know all the details in their construction to appreciate and use them?) -In addition, I know the context and Fontaine's original motivation for considering these rings, but have they found any unexpected uses outside their intended domain? - -REPLY [7 votes]: I would say that Bloch-Kato conjecture (giving the exact value of complex $L$ functions of motives) was a rather unexpected application.<|endoftext|> -TITLE: In a compact lie group, can two closed connected subgroups generate a non-closed subgroup? -QUESTION [15 upvotes]: Let $H$,$K$ be closed connected subgroups of a compact Lie group $G$. Let $L:=\langle H,K \rangle$ be the subgroup they generate, ie, the smallest subgroup of $G$ containing them both. Must $L$ be closed? -Notes: - -False if $G$ is not compact. -False if $H$, $K$ not connected: consider two $\mathbb{Z}/2\mathbb{Z}$ subgroups in $O(2)$ generated by reflections in irrationally related axes. - -Is there a way to jack up (2) to a connected counterexample, for instance? - -REPLY [12 votes]: Maybe it's helpful to formulate a somewhat different version of Mikhail's answer, from the viewpoint of the Borel-Tits development of reductive algebraic groups over an arbitrary field. When the field is $\mathbb{R}$, much of the basic structure of Lie groups is recaptured this way including the structure of compact connected Lie groups: such a group is always the group of real points of a connected reductive algebraic group over $\mathbb{C}$. (The book by Onishchik and Vinberg captures much of the information, but at the cost of leaving most proofs as structured exercises and working always over a base field of characteristic 0. The main ideas for algebraic groups work in much more generality.) -In the early part of Borel's book Linear Algebraic Groups you find in I.2.2 a general formulation which implies in particular that the answer to the question asked here is yes. (Indeed, $L$ is connected as well as closed.) The treatment in 7.5 of my book with the same title is a little more explicit, but doesn't refer to fields of definition since those are deferred to the end of the book. As far as I know there is no completely explicit formulation for compact Lie groups in the textbook literature, though I hope I'm wrong. -Unfortunately there seems to be no self-contained textbook treatment of the structure of compact Lie groups, though the representation theory is treated for example in the Springer GTM by Brocker and tom Dieck. Instead, compact Lie groups are typically discussed as special cases of more general Lie groups in larger books.<|endoftext|> -TITLE: Rings with all modules projective ? -QUESTION [9 upvotes]: Is there a classification of the commutative rings (with unit) such that each module over the ring is projective ? - -REPLY [17 votes]: They're called "semisimple artinian" rings. Prove that a ring $R$ (no commutativity is required) is semisimple artinian iff (equivalently) -0) (definition is most books in Ring Theory) $R$ is right artinian and has no nonzero nilpotent right ideals. -1) Any right R-module is projective. -2) Any right R-module is injective. -3) Any simple right R-module is projective. -4.1) Any right R-module is semisimple -4.2) R is a semisimple right module over itself (if you want, $R_R$ equals its socle). -5) $R$ consists of the sum of (finitely many) right ideals.<|endoftext|> -TITLE: About Frobenius of Witt vectors -QUESTION [6 upvotes]: Let $k$ be a characteristic $p$ alg. closed field, Let $W(k)$ be the Witt vectors， Let $\sigma$ be the Frobenius, then we also have $\sigma: W(k)^{\times} \to W(k)^{\times}$, where $W(k)^{\times}$ are the units in $W(k)$. Thus we can define a map $f: W(k)^{\times} \to W(k)^{\times}$, $f(x) = \frac{\sigma(x)}{x}$. My question is, is $f$ surjective? -Here is what I think is a proof. -Suppose $a \in W(k)^{\times}$, write $a$ as $（a_0, a_1, \ldots）$, suppose $x =(x_0, x_1, \ldots)$, then we are looking for $x$ such that $\sigma(x)=x\cdot a$, which means $x_0^p =x_0a_0$ and $x_1^p =x_1 a_0^p + x_0^pa_1$, etc, and clearly, we can solve $x_0$ in the first equation, then solve $x_1$, etc since $k$ is alg. closed. -Is the proof correct? And is there any other proof? -Also, is the alg. closedness necessary? Of course, if $k= \mathbb{F}_p$, $f$ is identity map,but what about $k$ other than $\mathbb{F}_p$$? -Thank you! - -REPLY [14 votes]: You may find the following more transparent, since it uses only the fact that the Witt vectors are a complete DVR with residue field $k$. Call the Witt vectors $R$, and let $y$ be a unit for which you want to find $z$ with $z^\sigma=yz$. First do it mod $p$, by solving $\zeta^p=\eta\zeta$ for $\zeta$ in $k$, where $\eta$ is the image of $y$ in $k$. Now you can assume that you have $z\in R$ satisfying $z^\sigma\equiv yz \mod{(p^m)}$, in other words $z^\sigma \equiv yz + p^m\delta \mod{(p^{m+1})}$. Now you want to adjust $z$ to $z'=z+p^m x$ so that $z'$ satisfies your congruence modulo $(p^{m+1})$. This boils down to solving $\xi^p - \xi \eta + \delta = 0$ in $k$, which you can do. So you see that you don't need $k$ to be algebraically closed, just separably closed.<|endoftext|> -TITLE: Associated graded of filtered module-algebra over a Hopf algebra -QUESTION [5 upvotes]: I ran across the following statement in a paper, and it seems fishy to me: -Lemma: If $A$ is any Hopf algebra, and if $U$ is an $\mathbb{N}_0$-filtered $A$-module algebra, then $U$ and $\mathrm{gr} (U)$ are isomorphic as $A$-modules. -There is no proof of the lemma, it just states that it is a well-known fact. -Such an isomorphism cannot be canonical: consider just the case that $A = k$ is a field and $U$ is any filtered $k$-algebra. In this situation there are plenty of vector space isomorphisms of $U$ with $\mathrm{gr}(U)$, just by pulling back a basis of each $U_{n} / U_{n-1}$, but these are hardly canonical. -So if the lemma is true, it is saying that there is some way to choose one of these maps so that it is an isomorphism of $A$-modules. -Question -Is the lemma true? If no, what is a counterexample? If yes, could you please provide a proof or a reference? - -REPLY [2 votes]: The statement is definitely false. For example, let $A = \mathbb k[x]$ be the group algebra of $\mathbb Z$. Let $U$ be the two-dimensional module in which $x$ acts by $\bigl( \begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \bigr)$. The span of $\bigl( \begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\bigr)$ is a submodule on which $x$ acts by $1$, and the quotient is also one-dimensional with $x=1$. So we give $U$ a filtration $0 \subset \mathbb k \bigl( \begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\bigr) \subset U$. Then the associated graded $\operatorname{gr}(U)$ is the two-dimensional $A$-module on which $x$ acts by $\bigl( \begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix} \bigr)$. This is not isomorphic to $U$ as and $A$-module, since in $U$ the $x$-action is not diagonalizable. -In general, you should only expect an isomorphism $\operatorname{gr}(U) \cong U$ if you have some semisimplicity. For example, when $A$ is the universal enveloping algebra of a semisimple Lie algebra, and $U$ is locally finite-dimensional (each filtered piece is finite-dimensional), or the group algebra of a finite group in characteristic $0$ (or prime to the order of the group) and $U$ is arbitrary, then you can build a noncanonical $A$-module isomorphism $\operatorname{gr}(U) \cong U$. -You should write to the author of the paper.<|endoftext|> -TITLE: Techniques for computing fundamental units in cubic extensions -QUESTION [5 upvotes]: Are there any slick ways of computing the fundamental unit for the cubic polynomial of the form $X^3+aX+b$ over $\mathbb{Q}$? The simplest example would be $X^3+X-1$, where a root $\alpha$ is a unit with inverse $\alpha^2+1$. What about the general cubic of this form? -There's one example in Milne's Algebraic Number Theory notes, but the method only works for a some cubics of this form. Some people I talked to suggested finding the regulator, but that doesn't seem much easier either as I don't see how this would be accessible through anything but the zeta-function. - -REPLY [9 votes]: Use Artin's inequality: if $K$ is a cubic field with one real embedding and $v > 1$ is a unit in $O_K$ then $|\text{disc}(K)| < 4v^3 + 24$. (I use inequalities on elements of $K$ via the one real embedding of $K$.) The proof of Artin's inequality is a pain. As a corollary, if $u > 1$ is a unit in $O_K$ and $4u^{3/2} + 24 < |\text{disc}(K)|$ then $u$ must be the fundamental unit of $O_K$. (Proof: Let $\varepsilon > 1$ be a fundamental unit, so $u = \varepsilon^n$ for some $n \geq 1$. If $n \geq 2$ then Artin's inequality with $v = \varepsilon$ implies $|\text{disc}(K)| < 4\varepsilon^3 + 24 = 4u^{3/n} + 24 \leq 4u^{3/2} + 24$, which contradicts the hypothesis of the corollary and also explains where the hypothesis comes from.) -Example: $K = {\mathbf Q}(\sqrt[3]{2})$, so $|\text{disc}(K)| = 108$. A unit in $O_K$ is $\sqrt[3]{2}-1$, which is between 0 and 1. Its reciprocal is $u := 1 + \sqrt[3]{2} + \sqrt[3]{4}$. Numerically, $4u^{3/2} + 24$ is around 54.1, which is less than 108, so $u$ is a fundamental unit of $O_K$. -Example: $K = {\mathbf Q}(\alpha)$ where $\alpha^3 +\alpha - 1 = 0$. Then $|\text{disc}(K)| = 31$. -The unique real root of $X^3 + X - 1$ is around .68, so less than 1. Its reciprocal -will be a unit greater than 1. If we identify the real root of $X^3 + X - 1$ with $\alpha$ then $u := 1/\alpha = \alpha^2 + 1$ is a unit greater than 1 in $O_K$. From a computer, -$4u^{3/2} + 24$ is around 31.096, which is greater than $|\text{disc}(K)| = 31$, so it looks like we can't use Artin's inequality to deduce that $u$ is the fundamental unit. However, you could modify the corollary to show $u$ is no worse than the square of a fundamental unit and then check $u$ is not a square in $O_K$ by showing it isn't a square mod $\mathfrak p$ for some prime $\mathfrak p$ in $O_K$. That would prove $u$ is a fundamental unit of $O_K$.<|endoftext|> -TITLE: Non-Kahler manifolds and the dd^c-lemma -QUESTION [6 upvotes]: Is there an explicit example of non-Kahler manifold $M$ such that $M$ satisfies the dd^c lemma ? - -REPLY [12 votes]: Here is an example of a Moishezon manifold which is easy to visualize. Take a high degree (e.g. a quintic) hypersurface $Z$ in $\mathbb{P}^{4}$ which has a single ordinary double point. Let $X$ be a small resolution of $Z$. Explicitly, a small analytic neighborhood of the singularity can be identified with the vertex of a cone over a two dimensional quadric and you just need to blow-up the Weil divisor which is the preimage of one ruling. The threefold $X$ is compact complex manifold and does not admit any Kaehler structure. The last statement follows for instance from a theorem of Smith-Thomas-Yau which states that a threefold with a single node will admit a symplectic small resolution only if the three sphere that vanishes at the node is homologous to zero. The high degree condition on $Z$ ensures that the vanishing cycle is not homologous to zero, hence the statement.<|endoftext|> -TITLE: Which finite nonabelian groups have long chains of subgroups as intervals in their subgroup lattice? -QUESTION [9 upvotes]: Given N, what is a finite non-abelian (and preferably non-solvable) group G in whose subgroup lattice Sub[G] there is an interval that is a chain of length at least N? -Since N can be arbitrarily large (but fixed), perhaps there is no easy answer. In that case, can someone suggest which sorts of groups to look at to find intervals that are chains (say, on the order of 10 subgroups in the chain)? -Thanks in advance! -Edit: Thanks to Carnahan's answer, I see that I should have ruled out direct products of cyclic groups with nonsolvable groups. What I'm interested in are intervals in the subgroup lattice of the form: -$\{ K : H \leq K \leq G \}$ -where $H$ is a corefree subgroup of $G$. - -REPLY [9 votes]: I think you can get arbitrarily long chains of this type in the simple groups ${\rm PSL}(2,p)$ for $p$ prime. -We make use of the maximal dihedral subgroups of order $p-1$. For given $N$, choose $p$ such that $(p-1)/2 = q^Nr$ with $q$ an odd prime and $r>2$. Then there is a chain of subgroups -$H_0 < H_1 < \cdots H_N < G$ -where $H_i$ is dihedral of order $2q^ir$.<|endoftext|> -TITLE: Integral identity for Legendre polynomials -QUESTION [5 upvotes]: How does one prove the following integral identity, where $P_n(x)$ is the $n$th Legendre polynomial? -$$ -\int_0^1 P_n(2t^2-1) dt = \frac{(-1)^n}{2n+1} -$$ -Notes & Background - -A variant of this appears in, for instance, Erdelyi et al "Higher transcendental functions" 10.10(49), but with nothing in the way of explanation. -This comes up in harmonic analysis on $U(3)$, when comparing Gelfand-Tseltin bases associated to different choices of nested sequences $U(3) \supset U(2) \supset U(1)$. -Eventually, I'll be looking for a $q$-analogue, related to harmonic analysis on $U_q(3)$, so a proof that will transport well would be my true desire. - -REPLY [4 votes]: Nice idea. As far as I'm concerned, the above comments are "answers", since they check out. I might as well record the details: -$$ -\int_0^1 \frac{dx}{\sqrt{1-2(2x^2-1)t + t^2}} -= \frac{1}{2\sqrt{t}} \int_0^1 \frac{dx}{\sqrt{\frac{(1+t)^2}{4t} -x^2}} = \frac{1}{2\sqrt{t}}\arcsin\left(\frac{\scriptstyle 2\sqrt{t}}{\scriptstyle1+t}\right). -$$ -The half-angle formula for $\sin$ reduces this to -$$ -\frac{1}{\sqrt{t}} \arcsin \sqrt{\frac{t}{1+t}} = \frac{1}{\sqrt{t}} \arctan \sqrt{t}, -$$ -which has the desired power series expansion. -Thanks.<|endoftext|> -TITLE: Names of noncompact riemannian symmetric spaces? -QUESTION [5 upvotes]: Irreducible riemannian symmetric spaces come in pairs: one compact and one not compact, usally called the noncompact dual. -The compact symmetric spaces include spheres, complex and quaternionic projective spaces, compact simple Lie groups, Grassmannians of different denominations,... They all have established names and to a large extent an established notation. I'm presently writing a paper where many low-dimensional riemannian symmetric spaces make their appearance, both compact and noncompact, and I have found that I do not know the names to some (indeed, most) of them. -Of course, the noncompact duals of the spheres are the hyperbolic spaces, and I am guessing (someone will surely correct me if I'm wrong) that the noncompact duals of the complex and quaternionionic projective spaces are called the complex and quaternionic hyperbolic spaces, respectively. I am also guessing that these are denoted $\mathbb{C}H^n$ and $\mathbb{H}H^n$; although I do not remember where I have actually seen this notation before. But how about the noncompact duals of the other symmetric spaces? -Questions: -Is there an accepted terminology and/or notation for the noncompact dual of the grassmannians of real, complex, quaternionic, associative, special lagrangian,... subspaces? How about for the noncompact dual of a simple Lie group (other than $SU(2)$)? -Thank you. - -REPLY [3 votes]: The Riemannian noncompact duals of of Grassmannians and other Hermitian symmetric spaces possess names according to their realizations: - -As bounded models: Cartan domains (4 classical and two exceptional types), or, -In the unbounded realization: Hermitian upper half space,Quaternionic upper half space and Siegel upper half space for the first three classical types. -(Some of them ) As tube domains. - -In addition, the symmetric cones are also named after their realizations. - -REPLY [2 votes]: I cannot remember having seen the family $\mathrm{SL}(n;\mathbb{R})/\mathrm{SO}(n)$ named or denoted otherwise. More generally, people usually say that they consider a symmetric space of non-compact type $X=G/K$, but do not name them (except for rank one ones, for which your denomination is the only I know, while the notation varies: $\mathbb{C}\mathrm{H}^n$, -$\mathrm{H}_{\mathbb{C}}^n$ probably are the most common).<|endoftext|> -TITLE: Holomorphic line bundles on a punctured disc -QUESTION [12 upvotes]: Is every holomorphic line bundle on the - say - punctured unit disc $\dot{\Delta} \subseteq \mathbf{C}$ trivial? Griffiths-Harris (p. 39) prove that $H^{p,q}_{\overline{\partial}}(\Delta) = 0$ (for $q \geq 1$), and mention that by replacing discs by annuli that one could prove also $H^{p,q}_{\overline{\partial}}(\dot{\Delta}^k \times \Delta^\ell) = 0$. They seem to imply that in particular $H^{p,q}_{\overline{\partial}}(\dot{\Delta}) = 0$. If this were true, using Dolbeault's theorem and the Kummer sequence one could conclude $H^1(\dot{\Delta},\mathcal{O}^\times_{\dot{\Delta}}) = 0$, hence that every holomorphic line bundle on $\dot{\Delta}$ is trivial. - -REPLY [13 votes]: Yes, every holomorphic vector bundle of any rank is trivial on the punctured disk $\dot{\Delta}$ . Indeed, since $\dot{\Delta}$ is a Stein manifold ( like any non-compact Riemann surface ! ) the Oka meta-principle (here a Theorem of Grauert ) says that the classification of holomorphic vector bundles on that manifold is the same as that of topological vector bundles. But since the punctured disk is homotopically equivalent to a circle , all topological complex vector bundles are trivial, hence the triviality of all holomorphic vector bundles. -(Do not confuse with the Möbius vector bundle, which is a non-trivial real vector bundle!) -Bibliography If you want to read complete proofs, you can consult Otto Forster's Lectures on Riemann Surfaces. -On page 229, Theorem 30.3 states that every holomorphic line bundle on a non-compact Riemann surface $X$ is trivial, and immediately below (on the same page) Theorem 30.4 proves that holomorphic vector bundles of any rank on $X$ are also trivial.<|endoftext|> -TITLE: Symmetric group irreps in tensor products of exterior products of the standard representation -QUESTION [13 upvotes]: Let $\mathbb{C}^n = V + \mathbb{C}$ be the defining representation of the symmetric group. Is there a nice formula for how $\Lambda^i V \otimes \Lambda^j V$ splits into irreps? - -REPLY [21 votes]: Congratulations, you have asked one of the few questions of this type for which there is a positive answer. Such a formula is the main result (Theorem 2.1) of Remmel's paper "A formula for the Kronecker products of Schur functions of hook shapes". -A few points of vocabulary to get you oriented: The irreducible representations of $S_n$ are indexed by the partitions of $n$. The partition $\bigwedge^k V$ corresponds to the partition $(n-k, 1,1,\ldots, 1)$ where there are $k$ $1$'s. These partitions are called "hooks". (Note the degenerate cases: $k=0$ is $(n)$ and gives the trivial representation; $k=n-1$ is $(1,1,\ldots,1)$ and gives the sign representation.) The tensor product of $S_n$ representations corresponds to the combinatorial operation known as Kronecker product. So the title of Remmel's paper tells you that it is answering your question, and googling on "Kronecker hook" will turn up more references. -Remmel defines a double hook to be a partition of the form $(q,p, 2, \ldots, 2, 1,\ldots, 1)$ where there are $k$ occurrences of $2$ and $\ell$ occurrences of $1$. His result is that the only irreps that appear in $\bigwedge^i V \otimes \bigwedge^j V$ are hooks and double hooks, that the multiplicities with which they appear are $\leq 2$, and he gives an exact formula for which ones appear with which multiplicities.<|endoftext|> -TITLE: Homotopy type of Hilbert schemes of points of $\mathbb C^2$ -QUESTION [14 upvotes]: Let $X=\mathbb C^2$, let $X^{[n]}$ be the Hilbert scheme of length $n$ 0-cycles in $X$, and let $X^{[n]}_0$ be the closed subscheme formed by the 0-cycles supported at 0. As far as I know $X^{[n]}_0$ and $X^{[n]}$ have the same homotopy type. Can anybody suggest a proof? (according to Nakajima this can be proved by adapting an argument of Slodowy (Four lectures on simple groups and singularities, Section 4.3) but I am unable to do it...) - -REPLY [6 votes]: Here's how I understand the argument in Slodowy's book. Sticking with Tamas' notation, let $X$ denote the Hilbert scheme and $D$ the punctual Hilbert scheme. You'd like to define a deformation retraction from $X$ to $D$ by sending every point to its limit under the dilation action. Unfortunately, as Ben points out, $D$ is not fixed by the dilation action, and this function isn't even continuous! So that doesn't work. -Instead, you choose a closed tubular neighborhood $A$ of $D$ such that $D$ is a deformation retract of $A$; this is possible by a basic result in algebraic topology (Slodowy cites Spanier's book). You then use the dilation action to define a deformation retraction from $X$ to $A$. More precisely, send every element $x\in X$ to the first element of $A$ that it hits when you dilate it inward. This is a perfectly well-behaved map, and it does the trick.<|endoftext|> -TITLE: irreducibility of generic linear combination of polynomials? -QUESTION [9 upvotes]: I would be shocked if the following were not true, but I can't seem to see a proof. -Claim: -Let $R$ be an integral domain containing an AC (uncountable if you wish) field $k$. Let $a, b \in R$, and suppose that the ideal $(a, b)$ is height 2. -Then for general $\alpha, \beta \in k$, the element $\alpha a + \beta b$ is irreducible. -Thanks! -Sue - -REPLY [11 votes]: This is not true even over $\mathbb C$. Take $\mathbb C[x,y]$ and $x^2, y^2$. You need general combination of a regular sequence of length at least $3$. Search for "local Bertini theorem" and "Flenner". -ADDED: the relevant reference is Satz 4.9 and 4.10 (Die Sätze von Bertini für lokale Ringe by H. Flenner, Mathematische Annalen, (299), 1977). This works for $n$ elements such that the ideal generated by them has height at least $3$ (over a infinite field of char. $0$). There is no hope in char. $p>0$ no matters how many elements you pick, since one can expand Karl's example. -The height at least $3$ condition can't be weakened (think about $x^2, xy, y^2$ over $\mathbb C$). However, in the case of $2$ elements, if you assume $a,b$ are irreducible to begin with, then I would guess what you want has a much better chance.<|endoftext|> -TITLE: Embedding groups into groups with some vanishing homology groups -QUESTION [11 upvotes]: Which finite subsets $S \subset \mathbb{N}$ have the following property : every countable group $G$ embeds into a finitely generated group $\Gamma$ such that $H_i(\Gamma;\mathbb{Z})=0$ for all $i \in S$. -The only positive answer I know here is that $S=\{1\}$ works since every countable group can be embedded into a simple group. I don't know any negative answers. -I'm especially interested in singleton sets $S$ (in particular, $S=\{2\}$ and $S=\{3\}$). -Also, is the question easier if I restrict myself to finitely generated or finitely presentable groups? - -REPLY [10 votes]: To add to Mark Sapir's post, the answer is precisely given as Corollary 5.6 of $\textit{The Topology of Discrete Groups}$ by Baumslag, Dyer, Heller (JPAA 16, 1980): -"Every countable group can be embedded in a 7-generator acyclic group." -Thus all possible $S$ work.<|endoftext|> -TITLE: How is etale cohomology of integer rings related to Galois cohomology? -QUESTION [31 upvotes]: In the paper of Bloch and Kato in the Grothendieck Festschrift, and some other papers relating to the Bloch-Kato conjecture and the ETNC, the cohomology groups -$H^i_{\mathrm{et}}(\operatorname{Spec} O_{K, S}, M),$ -seem to come up often, where $K$ is a number field, $S$ is a finite set of places of $K$, and $M$ is a finite or profinite $G_K = \operatorname{Gal}(\overline{K} / K)$-module unramified at primes outside $S$. -How should one think about these cohomology groups? How are they related to the much more familiar (to me at least) continuous Galois cohomology groups $H^i(G_K, M)$ (or the restricted ramification analogues $H^i(G_{K, S}, M)$)? Why are they the more natural things to work with in this context? - -REPLY [14 votes]: By usual (sometimes not so trivial) homological arguments, one can reduce to the case where $M$ is a finite discrete module over an artinian ring of residual characteristic $p$. In that case, I think you want $S$ to contain places above $p$ as well, even if your $M$ is unramified at $p$, so let me assume this. -The module $M$ induces an étale sheaf $M_{et}$ on $\operatorname{Spec}\mathcal O_{L,S}$ for all finite extension $L/K$. The spectral sequence UPDATE (converging to $H^{i+j}(\operatorname{Spec}\mathcal O_{K,S},M_{et})$) -$$E_{2}^{i,j}=\underset{\longrightarrow}{\operatorname{\lim}}\ H^{i}(\operatorname{Gal}(L/K),H^{j}(\operatorname{Spec}\mathcal O_{L,S},M_{et}))$$ -then induces isomorphisms between $E_{2}^{i,0}$ and $H^{i}(\operatorname{Spec}\mathcal O_{L,S},M_{et})$ or in other words $H^{i}(G_{K,S},M)$ is isomorphic to $H^{i}(\operatorname{Spec}\mathcal O_{K,S},M_{et})$. So you can assume that you are working with Galois cohomology throughout $provided$ you use Galois cohomology with restricted ramification. -Because the Tamagawa Number Conjectures are formulated only in the setting above, Bloch and Kato could have used Galois cohomology instead of étale cohomology everywhere without changing anything. To touch upon your last question, I think there are two reasons why they chose étale cohomology. -First, at least at the time they wrote, Galois cohomology was not the most familiar object of the two. In fact, many classical well-known results were given correct complete proofs only very late (in the late 90s in some cases). On the other hand, SGA (and works of Bloch and Kato themselves) existed as references for étale cohomology. -Second, using étale cohmology, one can formulate the TNC over more general bases than $\operatorname{Spec}\mathcal O_{K,S}$ (for instance any scheme of finite type of $\mathbb Z[1/p]$). This kind of generalization had been the key idea of previous works of Kato and Bloch-Kato on higher class field theory so it is not surprising that they decided to at least allow the same kind of generality in their subsequent works.<|endoftext|> -TITLE: Galois representation associated to a modular form is crystalline iff... -QUESTION [5 upvotes]: I am looking for the reference for the following fact (used, for example, in the proof of theorem 4.4. in Breuil's expose about local-global compatibility at Bourbaki): -For $f$ a modular cuspidal form of weight $k \geq 2$, let $\rho _f$ be the associated Galos representation and let $\pi _p (\rho _{f, |Gal(\overline{\mathbb{Q}} _p/\mathbb{Q}_p)})$ be the smooth representation associated to $\rho _f$ by the Local Langlands correspondance. Then $\rho _{f, |Gal(\overline{\mathbb{Q}} _p/\mathbb{Q} _p)}$ is crystalline iff -$ \pi _p (\rho _{f, | Gal(\overline{\mathbb{Q}} _p / \mathbb{Q} _p)}) ^{GL_2(\mathbb{Z} _p)}$ is non-zero. -Additional question: does there exist a generalisation of this fact? For instance, to the totally real setting? - -REPLY [7 votes]: As Keerthi writes in his comment above, this follows from T. Saito's results. The statement in one direction, that non-trivial $GL_2(\mathbb Z_p)$-invariants implies that the associated $p$-adic Galois rep'n is crystalline at $p$, is in fact easier, and goes back to Scholl, and in principle goes back further than that. The point is that modular curves, and the universal elliptic curves over them, have good reduction at $p$ when $p$ doesn't divide the level, and the Galois representations appearing in the $p$-adic etale cohomology of a good reduction variety are crystalline. (As Arno Kret observes, there are technical issues related to the non-compactness of modular curves, but these were essentially resolved by Deligne already in his Bourbaki seminar on the construction of Galois rep's for modular forms, and were treated carefully by Scholl in his paper on motives attached to modular forms.) -The statement in the other direction is more difficult, because one has to show that ramified-at-$p$ modular forms give rise to Galois representations which are non-crystalline; Saito proves this (in fact he proves a much more precise local-global compatibility result). -These results all extend to the Hilbert modular case. In most cases, the Galois representations can be constructed directly from $p$-adic etale cohomology, and then the same geometric arguments apply (on Shimura curves now, rather than modular curves). (I think that Saito has a paper specificallly dealing with this.) In those situations for which the Galois representations are not constructed geometrically, but instead by a $p$-adic limiting process, one can still deduce these kinds of local-global compatibility-at-$p$ arguments in various ways. There is an eigenvariety argument due to Chenevier, which is written up in a paper of his, and in a joint paper of his and Harris. I think there are also arguments by Kisin, by Skinner, and by T. Liu. (I hope that commmenters will clarify or correct this list if it is incomplete or otherwise in error.) -One caveat: without chasing references, I'm not sure whether one can show that the monodromy operator is non-trivial in these limiting cases when it should be (the monodromy operator is hard to keep track of in arguments via $p$-adic families), so it may be that in these cases, when $\pi_v$ is Steinberg for a prime $v$ above $p$, one doesn't know whether the Galois representation is semi-stable but non-crystalline (as it should be) rather than crystalline. -(Perhaps someone can comment on this.) -Also, Ana Caraiani has introduced a geometric method which pins down the monodromy operator in these limiting cases (and with $GL_2$ replaced by $GL_n$). Her results at the moment only cover primes $v$ not dividing $p$, but it seems likely that she will be able to extend them to the case of $v$ dividing $p$. -In general, to see the state of the art on such local-global compatibility results at primes dividing $p$ (for $GL_n$, not just $GL_2$), you should look at the recent preprints on this topic by Barnet-Lamb--Gee--Geraghty--Taylor, available on Taylor's web-site. The general statement is given by Theorem A of the second preprint. (The assumption about Shin regular weight appearing in that result is what would be removed by the generalization of Caraiani's work to the case $v$ divides $p$ mentioned above.)<|endoftext|> -TITLE: Generalized Vieta-product -QUESTION [9 upvotes]: It's known that -$$S_2={2\over\pi} = {\sqrt{2}\over 2}{\sqrt{2+\sqrt{2}}\over 2}{\sqrt{2+\sqrt{2+\sqrt{2}}}\over 2}\dots$$ -The terms in the product approaches 1, the same holds for the following convergent series, with $\phi$ the golden ratio -$$S_1 = {\sqrt{1}\over\phi}{\sqrt{1+\sqrt{1}}\over\phi}{\sqrt{1+\sqrt{1+\sqrt{1}}}\over\phi}\dots$$ -Let $$S_n = {\sqrt{n}\over c_n}{\sqrt{n+\sqrt{n}}\over c_n}\dots$$ -Where $c_n$ is the solution to the equation $x=\sqrt{n+x}$ -Is there a simpler formula for $S_n$? -What is the asymptotic behavior (Big-O) of $S_n$ as $n->\infty$? - -REPLY [6 votes]: I doubt there exists a closed formula for $n\ne 2$. In the case $n=2$ such formula exists only thanks to the double-angle formula for cosine. -Let $n$ be fixed and $c=c_n$. Notice that $n=c^2-c$ and $c\to\infty$ as soon as $n\to\infty$. -Denote by $p_k$ the $k$-th multiplier in the product $S_n$. It can be easily seen that -$$c\cdot (p_k^2-1) = p_{k-1} - 1$$ -Consider the functional equation: -$$c\cdot(f(x)^2-1)=f(2cx)-1$$ -with $f(0)=1$ and $f'(0)=1$. Its solution can be expressed as a series: -$$f(x) = 1 + x + \frac{x^2}{2(2c-1)} + \frac{x^3}{2(2c-1)^2(2c+1)} + \frac{x^4(2c+5)}{8(2c-1)^3(2c+1)(4c^2+2c+1)} + \dots.$$ -Then -$$p_k = f\left(\frac{x_0}{(2c)^k}\right)$$ -where $x_0$ is a solution to $f(x_0)=0$. -Now -$$S_n = \prod_{k=1}^{\infty} p_k = \exp \sum_{k=1}^{\infty} \ln\left(1 + \Theta\left(\frac{x_0}{(2c)^k}\right) \right) = \exp \sum_{k=1}^{\infty} \Theta\left(\frac{x_0}{(2c)^k}\right) = \exp \Theta\left(\frac{x_0}{2c-1}\right)$$ -which tends to $1$ as $c\to\infty$. -Therefore, $S_n\to 1$ as $n\to\infty$. -Example. For $n=2$, the functional equation admits the analytic solution $f(x)=\cosh(\sqrt{2x})$ for which $x_0=\frac{-\pi^2}{8}$.<|endoftext|> -TITLE: Complex Lie group without faithful real representations? -QUESTION [7 upvotes]: Does there exist a complex analytic Lie group which doesn't have faithful representations in $GL(N,\mathbb R)$, viewed as a real Lie group? -There are examples of complex Lie groups which do not allow faithful complex representations, like tori $\mathbb C^n/\mathbb Z^{2n}$, but such tori have many faithful real representations. -Also there are examples of real Lie groups without faithful linear representations, like the universal cover of $SL(2,\mathbb R)$ (but they are not complex analytic Lie groups). -How about complex Lie groups without faithful real representations? - -REPLY [16 votes]: Take the complex Heisenberg group of 3 by 3 upper triangular unipotent complex matrices, and mod out by a subgroup $Z\times Z$ in the center.<|endoftext|> -TITLE: Textbook recommendations for undergraduate proof-writing class -QUESTION [31 upvotes]: I am teaching the proof-writing class (for the 3rd time) in the Fall and plan to buck the party line and use a different text than the default Bond and Keane. My parameters are as follows: - -Logic, Sets, Equivalence Relations and Induction should be covered. -Price should be reasonable (say around $30 or less). -Distractions like "historical comments" and "mathematical perspectives" should be kept to a minimum. - -I plan to supplement such a book with "What is Mathematics" by Courant and Robbins. -I would be pleased to hear some recommendations! - -REPLY [4 votes]: My father taught heat transfer in a mechanical engineering department and he had a great trick. Take a good text such as one suggested in another solution: -http://www.pearsonhighered.com/educator/product/Mathematical-Proofs-A-Transition-to-Advanced-Mathematics/9780321390530.page -This book costs $100+ -And then use an older edition for your class. You can buy the 1st edition of that textbook for $9 here: -http://a.co/1Qlm9sT<|endoftext|> -TITLE: What theorem of Liouville's is Gian-Carlo Rota referring to here? -QUESTION [20 upvotes]: I am very curious about this remark in Lesson Four of Rota's talk, Ten Lessons I Wish I Had Learned Before I Started Teaching Differential Equations: - -"For second order linear differential equations, formulas for changes of dependent and independent variables are known, but such formulas are not to be found in any book written in this century, even though they are of the utmost usefulness. -"Liouville discovered a differential polynomial in the coefficients of a second order linear differential equation which he called the invariant. He proved that two linear second order differential equations can be transformed into each other by changes of variables if and only if they have the same invariant. This theorem is not to be found in any text. It was stated as an exercise in the first edition of my book, but my coauthor insisted that it be omitted from later editions." - -Does anyone know where to find this theorem? - -REPLY [18 votes]: This question had been bothering me for a while since I teach the intro differential equations courses occasionally, so I finally looked up the reference Anatoly gave and figured out the details. I'll drop them here in case anyone else can get use out of them. -Starting with the second-order equation - -$y'' + Py' + Qy = 0$ - -we can make the change of variables $w = y \cdot e^{\int P/2}$. This change of variables is pretty clever; if you work it out, it happens to eliminate the first derivative term and gives us a new second-order equation of the form - -$w'' + Q_0w = 0$. - -If you calculate it out, you can find that $Q_0 = Q - \frac14P^2 - \frac12P'$. Nothing fancy here, just what happens when you do the change of variables. This $Q_0$ is the invariant of the second-order equation that is mentioned in the question. -Any two second-order equations that have the same invariant can easily be transformed into one another by a change of variables; simply change variables once to get to the standard form $w'' + Q_0w = 0$ and then change variables back into the other one. -More difficult is that all changes of variables preserve this invariant; proving it for changes of variables y = G x is an easy computation with a bunch of cancellation, but I'm not sure if we need to do more than that to finish the proof. - -From the point of view of someone teaching introductory differential equations, you are normally dealing with second-order equations $y'' + Py' + Qy = 0$ where $P$ and $Q$ are real numbers. In this case, the invariant $Q_0$ is just $-\frac14$ times the discriminant of the auxiliary equation. So the theorem says that any two equations with the same auxiliary equation discriminant can be transformed into each other. - -For example, - -$y'' + 6y' + 10y = 0$ -has $Q_0 \equiv 1$, so - -it must be able to turn into - -$w'' + 1w = 0$ - -via a change of variables. Indeed, if you let - -$w = y e^{3x}$ - -then you get the equation $w'' + w = 0$, and the solution is - -$y \cdot e^{3x} = c_1 \cos x + c_2 \sin x$ -$y = c_1 e^{-3x} \cos x + c_2 e^{-3x}\sin x$. - - -Another example: - -$y'' + 6y' + 9y = 0$ -has $Q_0 \equiv 0$, - -so it must be able to turn into simply - -$w'' = 0$ - -via a change of variables. The change of variables only depends on $P$, and yes, $w = y e^{3x}$ is a pretty good change of variables. Via this route we end up with - -$y \cdot e^{3x} = c_1 + c_2 x$ -$y = c_1 e^{-3x} + c_2 x e^{-3x}$ - -which is of course correct. - -So, in our intro differential equations classes, the invariant is just the familiar fact that if we complete the square of the auxiliary equation, we can see the correct change of variables that will leave us with a bunch of $\cos$, $\sin$, $\cosh$, and $\sinh$ in addition to our exponentials from $P/2$.<|endoftext|> -TITLE: Are there any mathematical objects that exist but have no concrete examples? -QUESTION [14 upvotes]: I am curious as to whether there exists a mathematical object in any field that can be proven to exist but has no concrete examples? I.e., something completely non-constructive. The closest example I know of are ultrafilters, which only have one example that can be written down. MathOverflow user Harrison Brown mentioned to me that there are examples in Ramsey theory of objects that are proven to exist but have no known deterministic construction (but there might be), which is close to what I'm looking for. He also mentioned that the absolute Galois group of the rationals has only two elements that you can write down - the identity element and complex conjugation. -I am worried that this might be a terribly silly question, since typically there is a trivial example of an object, and a definition that specifically did not include the trivial case would be 'cheating' as far as I'm concerned. My motivation for this question is purely out of curiosity. Also, this is my first question on MO, so I probably need help with tags and such (I'm not terribly sure what this would belong to). I think that this should be a community wiki, but I do not have the reputation to make it so as far as I can tell. - -REPLY [14 votes]: You should look at Handbook of Analysis and its Foundations by Eric Schecter. Here is an excerpt from the preface: -Students and researchers need examples; it is a basic precept of pedagogy that every abstract idea should be accompanied by one or more concrete examples. Therefore, when I began writing this book (originally a conventional analysis book), I resolved to give examples of everything. However, as I searched through the literature, I was unable to find explicit examples of several important pathological objects, which I now call intangibles: -finitely additive probabilities that are not countably additive, -elements of $(l_\infty)^*- l_1$(a customary corollary of the Hahn- Banach Theorem), -universal nets that are not eventually constant, -free ultrafilters (used very freely in nonstandard analysis!), -well orderings for R, -inequivalent complete norms on a vector space, -etc. In analysis books it has been customary to prove the existence of these and other pathological objects without constructing any explicit examples, without explaining the omission of examples, and without even mentioning that anything has been omitted. Typically, the student does not consciously notice the omission, but is left with a vague uneasiness about these unillustrated objects that are so difficult to visualize. -I could not understand the dearth of examples until I accidentally ventured beyond the traditional confines of analysis. I was surprised to learn that the examples of these mysterious objects are omitted from the literature because they must be omitted: Although the objects exist, it can also be proved that explicit constructions do not exist. That may sound paradoxical, but it merely reflects a peculiarity in our language: The customary requirements for an "explicit construction" are more stringent than the customary requirements for an "existence proof." In an existence proof we are permitted to postulate arbitrary choices, but in an explicit construction we are expected to make choices in an algorithmic fashion. (To make this observation more precise requires some definitions, which are given in 14.76 and 14.77.) -Though existence without examples has puzzled some analysts, the relevant concepts have been a part of logic for many years. The nonconstructive nature of the Axiom of Choice was controversial when set theory was born about a century ago, but our understanding and acceptance of it has gradually grown. An account of its history is given by Moore [1982]. It is now easy to observe that nonconstructive techniques are used in many of the classical existence proofs for pathological objects of analysis. It can also be shown, though less easily, that many of those existence theorems cannot be proved by other, constructive techniques. Thus, the pathological objects in question are inherently unconstructible. -The paradox of existence without examples has become a part of the logicians' folklore, which is not easily accessible to nonlogicians. Most modern books and papers on logic are written in a specialized, technical language that is unfamiliar and nonintuitive to outsiders: Symbols are used where other mathematicians are accustomed to seeing words, and distinctions are made which other mathematicians are accustomed to blurring -- e.g., the distinction between first-order and higher-order languages. Moreover, those books and papers of logic generally do not focus on the intangibles of analysis. -On the other hand, analysis books and papers invoke nonconstructive principles like magical incantations, without much accompanying explanation and -- in some cases -- without much understanding. One recent analysis book asserts that analysts would gain little from questioning the Axiom of Choice. I disagree. The present work was motivated in part by my feeling that students deserve a more "honest" explanation of some of the non-examples of analysis -- especially of some of the consequences of the Hahn- Banach Theorem. When we cannot construct an explicit example, we should say so. The student who cannot visualize some object should be reassured that no one else can visualize it either. Because examples are so important in the learning process, the lack of examples should be discussed at least briefly when that lack is first encountered; it should not be postponed until some more advanced course or ignored altogether.<|endoftext|> -TITLE: Third differential in Atiyah Hirzebruch spectral sequence -QUESTION [26 upvotes]: Does any one know why $d_3: H^* (X, K^0(point))\rightarrow H^{*+3}(X,K^0(point))$ is actually extended $Sq^3$ to $\mathbb{Z} $ coefficient. - -REPLY [21 votes]: A pretty direct argument was given by Frank Adams in the proof of 16.6 (page 336) in part III of his 1974 Chicago lectures (MR0402720). Thinking of the Atiyah--Hirzebruch spectral sequence for $K^*(X)$ as arising from the Postnikov tower $\{P^n ku\}$ of ku (Adams calls this spectrum $bu$), row 0 and row 2 come from the layers $HZ$ and $\Sigma^2 HZ$, sitting in a cofiber sequence $\Sigma^2 HZ \to P^2 ku \to HZ \to \Sigma^3 HZ$. The $d_3$-differential is induced by the third map, i.e., the first $k$-invariant of $ku$. To see that the $k$-invariant is the integral lift $\beta Sq^2$ of $Sq^3$, i.e., not zero, Adams looks at the third space in the $ku$-spectrum, namely $SU$, and notices that $\beta Sq^2 \ne 0$ in $H^6(K(Z, 3); Z)$ but $H^6(SU; Z) = 0$. This implies that the $k$-invariant is nonzero.<|endoftext|> -TITLE: Random Trigonometric Polynomial -QUESTION [10 upvotes]: Let $t_{1},t_{2},\ldots, t_{n}$ be i.i.d. real Gaussian random variables of zero mean and variance one. Let $a_{1},a_{2},\ldots, a_{n}$ be positive and fixed real numbers and define the random polynomial -$$ -p(z):=\sum_{k=1}^{n}{a_{k}t_{k}z^{k}}. -$$ -Define the random variable -$$ -m=\max_{|z|=1}\Big\{\mathrm{Re}(p(z))\Big\}=\max_{\theta\in (0,2\pi]}\Bigg\{\sum_{k=1}^{n}{a_{k}t_{k}\cos(k\theta)}\Bigg\}. -$$ - -How can one compute the probability - distribution of $m$? Can we compute at - least the first few moments - $\mathbb{E}(m)$ and $\mathbb{E}(m^2)$? - -REPLY [2 votes]: The case of $ a_k=1/k$ and Gaussian coefficents is easily seen that the series is a.s. convergent to a continuous function. (Corollary to Salem-Zygmund, already mentioned) So, in the case you are considering the distribution $m$ will have subgaussian tails. -Under weaker conditions on the iid random coefficents, and $ a_k=1/k$, this is discussed in an article of Michel Talagrand, A borderline Fourier Series" Ann Prob 1995. http://www.jstor.org/pss/2245006<|endoftext|> -TITLE: Applications of commutative algebra -QUESTION [13 upvotes]: Hi. I'm preparing a thesis in commutative algebra, and when I say this to my friends they always ask me what are the applications to "real-world", and I don't know what to answer. This let me think that I'm studying something useless. I'm studying on the Matsumura and on the Herzog-Bruns. Any of you know some applications of this abstract algebra to the real-world? - -REPLY [6 votes]: The answer that I am going to give is implicitly contained in a few answers already given, but it is a bit too implicit, to my taste, so let me give it out and loud: Gröbner bases. When you solve a system of linear equations, you use Gaussian elimination, when you solve a system of polynomial equations of higher degrees, you use Gröbner bases, and it is very clear that solving systems of polynomial equations is something that people have to do for all sorts of applications. -That "very clear" is not just a belief held by a pure mathematician: on a few occasions that I talked about something mathematical to people doing research in some real world questions of statistics, biology, engineering, Gröbner bases would be the only aspect of somewhat advanced algebra, not just commutative algebra, that they would have ever heard of. You can see some relevant bits of software solving applied problems in various areas here: http://www.risc-software.at/en/. -I can't resist from also saying that in some areas of pure maths, for a long time, saying the words "Gröbner bases" was a bit of faux pas, something that a true pure mathematician should rather leave as a discussion topic to people concerned with applications, something as silly and naive and so not worth mentioning as using a calculator to multiply two numbers. However, besides being a useful tool for computations, Gröbner bases and their generalisations also give methods to construct resolutions (starting from work of Anick in 1980s), and in particular to prove that a certain algebra (or an operad) is Koszul etc. So it certainly is something worth being aware of, really. - -REPLY [2 votes]: For applications in physics (string theory) see http://link.springer.com/chapter/10.1007%2F978-1-4614-5292-8_2 (Some Applications of Commutative Algebra to String Theory, by P.S. Aspinwall) and http://arxiv.org/abs/hep-th/0703279 (Topological D-Branes and Commutative Algebra, by the same author).<|endoftext|> -TITLE: $fgf = f$, $gfg = g$, $fg$ not necessarily identity, what is this called? -QUESTION [13 upvotes]: A very simple question, I just totally forgot how it was called, and Google is not helping. -There's a pair of functions $f:X\to Y$, $g:Y\to X$. -$fgf = f$, $gfg = g$, but $fg$ and $gf$ don't need to be identities (and usually are not in interesting cases). -A simple example would be $f(a,b,c)=(a,b)$, $g(a,b)=(a,b,0)$ -What were $f$ and $g$ called? - -REPLY [4 votes]: This is also naturally related to adjointness: -if you consider both $X$ and $Y$ to be poset categories, and add the conditions - -both $f$ and $g$ order-preserving -$x \leq_X g(f(x))$ and $f(g(y)\leq_Y y$ for all $(x,y)\in X\times Y$ - -to your conditions (this is admittedly a little less general, -but satisfied in many natural situations), then: -the conditions are met -$\Longleftrightarrow$ -$f:X\leftrightarrow Y:g$ -is an adjunction.<|endoftext|> -TITLE: "A sea-side town where every house can see the sea" -QUESTION [10 upvotes]: This is a reference request. -The phrase in the title is, if I remember correctly, how Eli Stein described the following set (the definition may be faulty, but I think it is right): - -There exists a set $S$, which is a subset of the unit square $[0,1]^2\subset\mathbb{R}^2$ with full Lebesgue measure, with the property that for any point $p\in S$ there exists a line $\ell \subset \mathbb{R}^2$ such that $\ell \cap S = \{ p\}$. - -I vaguely remember the theorem being attributed to one of the Rieszes (not sure M. or F.; I might also be completely wrong about that). -Can someone tell me whether my rough statement of the theorem above is correct, and better yet, can someone give a reference to the paper which proved it? (I know a paper exists; I remember digging it up in the bowels of the library when I was a beginning graduate student. I am sad to say that two moves later I can no longer find my photocopy of that article.) - -REPLY [12 votes]: This is a Nikodymn set. I haven't seen a citation to Nikodymn's original paper, but the history is breifly discussed (with references) in Stein's survey article Singular Integrals: The Roles of Calderón and Zygmund. -Edit: Stein's Harmonic Analysis cites this to: "O. Nikodym, Sur les ensembles accessibles, Fund. Math. 1927 10:116-168"<|endoftext|> -TITLE: Is there a fast way to compute matrix multiplication mod $p$? -QUESTION [7 upvotes]: I think people have some general strategy to do matrix multiplication fast. But what about for the finite field of $p$ elements? (e.g. when $p=2$, one should have some faster way.) -So my question is, given two integer entry matrix, $A$ and $B$. Is there a fast way to compute $AB \mod p$?. And is there a fast way of computing $A^n \mod p$ for $n$ not too big. (Note that for $n=|\mathrm{GL}_t(\mathbb{Z}/p)|$, we have $A^n=I \mod p$, where $t$ is the size of $A$.) - -REPLY [19 votes]: For any field, we can define the exponent of matrix multiplication over that field to be the smallest number $\omega$ such that $n \times n$ matrix multiplication can be done in $n^{\omega + o(1)}$ field operations as $n \to \infty$. Schönhage proved that it is invariant under taking field extensions, so it depends only on the characteristic of the field. Probably it equals $2$ in every characteristic, but that isn't known. (Certainly $\omega \ge 2$, since it takes at least one operation to get each entry.) -Let $\omega_p$ be the exponent in characteristic $p$. Then one can show that $\limsup_{p \to \infty} \omega_p \le \omega_0$, basically because every low-rank tensor decomposition in characteristic $0$ will work for all but finitely many primes. Over the rationals, you just need to avoid primes that occur in denominators. -However, for small primes the exponent could (as far as we know) be better or worse than in characteristic $0$, and it's even possible that it could be substantially better for all primes, although in that case you can show that as $p \to \infty$ the asymptotics would have to take longer and longer to kick in (i.e., the size of the matrices needed to see the improvement would grow as a function of $p$). -Strassen's exponent $2.8$ algorithm works in every characteristic, and it's the only practical method that achieves exponent less than $3$. However, if you want to do this in practice, it's important to think about issues like cache and memory access. (These issues are often more important than counting arithmetic operations.) Unless you really know what you are doing, it's not worth trying to write your own code, except as a learning exercise. For example, in characteristic $2$ the M4RI code mentioned by unknown (google) seems like it could be a good bet.<|endoftext|> -TITLE: Height of minimal model of ZFC -QUESTION [10 upvotes]: What do we know about the height of the minimal (transitive) model of ZFC, that is, about the least α such that Lα is a model of ZFC? Call this ordinal μ. It is countable, since otherwise we could build L inside the collapse of a countable elementary submodel of Lμ to obtain a lesser such α. Moreover, as I learned here on MO, μ is Δ12 definable in V. - -REPLY [11 votes]: I claim that the ordinal $\mu$, if it exists, is actually -$\Pi^1_1$ definable, which improves on your $\Delta^1_2$ -claim. What I mean by this is that the set of reals coding -a relation on $\omega$ having order type $\mu$ is a -$\Pi^1_1$ set of reals. Even more, I claim that it is a -$\Delta^1_1$ property about reals, among those that do code -well-orders. That is, the set of codes for $\mu$ is the -intersection of WO with a hyperarithmetic set. -First, let's show that the set of reals $x$ coding a -relation $\lhd$ of order type $\mu$ has complexity -$\Pi^1_1$. It is well-known to be a $\Pi^1_1$ property of -$x$ to assert that the relation $\lhd$ it codes on $\omega$ -is a well-order, of some order type $\alpha$. I claim that -$x$ is coding $\mu$ if and only if, in addition, for every -countable structure $M$, if $M$ satisfies V=L and the -ordinals of $M$ have an initial segment isomorphic to -$\alpha+1$, then $M$ thinks that the $\alpha$-th ordinal is -the height of the minimal model of ZFC. -This additional property is indeed $\Pi^1_1$. To see this, -note first that the assertion that $M$ satisfies a certain -theory is an arithmetic property about $M$, and the -existence of the order-isomorphism from $\lhd$ to an -initial segment of $M$ is a $\Sigma^1_1$ assertion, but it -appears in the hypothesis of an implication here, so -altogether the assertion is $\Pi^1_1$. The critical fact we -are using here is that we don't assert that $M$ is fully -well-founded, but only that it is well-founded beyond the -height of the order coded by $x$, which is sufficient to -determine whether $x$ codes $\mu$ or not. -Let us now argue that this property does indeed define the -codes of $\mu$. If $x$ does code $\mu$, then any model $M$ -satisfying V=L and well-founded up to $\mu+1$ will agree -that $\mu$ is the height of the minimal model of ZFC, so -$x$ will pass this definition. Conversely, if $x$ codes a -well-order of order type $\alpha$ and any model $M$ thinks -that $\alpha$ is the height of the minimal model, then it -will be right. -What the argument shows is that the set of codes is a -$\Delta^1_1$ property, that is, a hyperarithmetic property, -on the set WO of codes for well orders. Given that $x$ -codes a well-order, then we can say that $x$ codes $\mu$ if -and only if, every countable $M$ satisfying V=L and -well-founded beyond the height of the ordinal $\alpha$ -coded by $x$ agrees that $\alpha$ is $\mu$; and also, if -and only if there is such an $M$. So we've got a $\Pi^1_1$ -and a $\Sigma^1_1$ characterization, provided that we -already know that $x$ is coding a well-order. -We can get a unique code, rather than a set of codes, by -observing that if $\mu$ exists, then not only is it -countable, but it is countable in $L$, and so there is an -$L$-least code for $\mu$. This code is also -$\Pi^1_1$-definable, since $z$ is this $L$-least code if -and only if it does code a well-order of type $\alpha$, and -all countable models of V=L that are well-founded at least -that high and that think that that ordinal $\alpha$ is -countable, think that $z$ is the $L$-least code of -$\alpha$. The point again is that such models, even if -ill-founded up high, will be right about this information.<|endoftext|> -TITLE: What math institutes offer research in pairs/research in teams? -QUESTION [47 upvotes]: Some math institutes offer programs in which a small number of researchers are enabled to meet at the institute for a week or more. A list seemed as if it could be useful. - -REPLY [5 votes]: The Centro di Ricerca Matematica Ennio De Giorgi at the Scuola Normale Superiore of Pisa usually has a program of research in pairs: -http://crm.sns.it/visit/pairs.html<|endoftext|> -TITLE: Global Sections of the Identity Component of Neron model -QUESTION [6 upvotes]: Let $A$ be an abelian variety over a number field $K$ and consider the Neron model $\mathcal{A}$ of $A$ over $X=Spec{\mathcal{O}_K}$. If $\mathcal{A}^0$ is the identity component of $\mathcal{A}$, then $\mathcal{A}^0$ is an open subgroup scheme of $\mathcal{A}$ that fits into a short exact sequence -$$0 \rightarrow \mathcal{A}^0 \rightarrow \mathcal{A} \rightarrow \Phi_A \rightarrow 0$$ -over $X$. Considering these smooth group schemes as sheaves for the flat (fpqf) topology over $X$, the associated long exact sequence of flat cohomology groups begins -$$0\rightarrow\mathcal{A}^0(X)\rightarrow\mathcal{A}(X)\cong{A(K)}\rightarrow\Phi_A(X)\rightarrow\ldots$$ -where the indicated isomorphism follows from the Neron mapping property. Now, we know that $A(K)$ is finitely generated i.e. it has a free part (copies of $\mathbb{Z}$) and a torsion part (which is finite). Since $A(K)/{\mathcal{A}^0(X)}$ is contained within $\Phi_A(X)$ (which is finite), it follows that the group $\mathcal{A}^0(X)$ of global sections of $\mathcal{A}^0$ must contain all of the free part of $A(K)$. My question is - is it possible in this general scenario to determine how much of the finite part of $A(K)$ is captured by $\mathcal{A}^0(X)$ or do we need additional information on $A$? - -REPLY [7 votes]: First, a remark: the free part of $A(K)$ is a quotient, not a sub, and so it is possible that a point of infinite order in $A(K)$ could have non-trivial image -in $\Phi_A(X)$. Probably what you mean is that $\mathcal A^0(X)$ and $A(K)$ have the same -free rank. -Regarding torsion, my interpretation of your question is that you are asking -about the map $A(K)_{tors} \to \Phi_A(X)$, and are curious is to whether or not it -can have a kernel (so that some part of $A(K)_{tors}$ is contained -in $\mathcal A^0(X)$). -As far as I know, this varies a lot depending on the particular abelian variety, -but in particular cases it has been quite intensively studied. For example, if $p$ -is prime and $A =J_0(p)$ is the Jacobian of the modular curve $X_0(p)$, then Mazur showed -in his Eisenstein ideal paper showed that the map $A(\mathbb Q)_{tors} \to \Phi_A( -\mathrm{Spec}\ \mathbb Z)$ is an isomorphism. I generalized this to subabelian varities -$A$ of $J_0(p)$ in my paper here. -For an example in some sense opposite to this, see this paper of Conrad, Edixhoven, and Stein, -in which they show that if $A = J_1(p)$ (again $p$ is prime), then -$\Phi_A(\mathrm{Spec}\ \mathbb Z)$ is trivial, so that $\mathcal A^0 = \mathcal A$.<|endoftext|> -TITLE: Is every reductive group scheme etale locally trivial? -QUESTION [9 upvotes]: Let $S$ be a scheme over a field $k$, and let $G$ be a reductive group scheme over $S$. Let us call it trivial, if it is a pull-back of a group scheme over $k$ via the structure morphism $S\to k$. Is it always true that $G$ becomes trivial after a certain etale base change $S'\to S$? I am willing to assume that $S$ is smooth if needed. - -REPLY [6 votes]: Reductive groups schemes over $S$ are classified by $H^1_{fpqc}(S,Aut_G)$, see SGA 3 Exp. XXIV.<|endoftext|> -TITLE: Distinct, non-homeomorphic, profinite topologies on a given abstract group ? -QUESTION [6 upvotes]: Just a silly little question which arose in connection with infinite Galois groups and their Krull topology:- can a given abstract group be endowed with distinct, non-homeomorphic, profinite topologies ? (I asked this question several years ago on the Topology Q+A and was told the question is undecidable and has something to do with supercompact cardinals). As I'm not that well-versed as concerns large cardinals etc., could someone verify/elucidate this please ? -Thank you in advance ! Stephan. - -REPLY [4 votes]: Yes. -I have classified some abelian examples: there are uncountably many pairwise non-homeomorphic pro-$p$ topologies that can be placed on the (unrestricted) product of any countable collection of cyclic $p$-groups of unbounded exponent. -The results are presented here, but I am in the process of redrafting http://arxiv.org/abs/1101.3005<|endoftext|> -TITLE: A natural refinement of the $A_n$ arrangement is to consider all $2^n-1$ hyperplanes given by the sums of the coordinate functions. Have you seen this arrangement? Is it completely intractable? -QUESTION [16 upvotes]: The short version -Here is an extremely natural hyperplane arrangement in $\mathbb{R}^n$, which I will call $R_n$ for resonance arrangement. -Let $x_i$ be the standard coordinates on $\mathbb{R}^n$. For each nonempty $I\subseteq [n]=\{1,\dots,n\}$, define the hyperplane $H_I$ to be the hyperplane given by -$$\sum_{i\in I} x_i=0.$$ -The resonance arrangement is given by all $2^n-1$ hyperplanes $H_I$. The arrangement $R_n$ is natural enough that it arises in many contexts -- to first order, my question is simply: have you come across it yourself? -This feels rather vague to be a good question, and after giving some background on where I've seen this I'll try to be a bit more specific about what I'm looking for, but my point is this arrangement has a rather simple and natural definition, and so crops up in multiple places, and I'd be curious to hear about more of them even if you can't specifically connect it to what follows. -What I knew until this week -I came across this arrangement in my study of double Hurwitz numbers -- they are piecewise polynomial, and the chambers of the resonance arrangement are the chambers of polynomiality. I don't want to go into this much more, as it's unimportant to most of what follows Though I will say that conjecturally double Hurwitz numbers could be related to compactified Picard varieties in a way which would connect this arrangement up with birational transformations of those. Also, the name "resonance arrangement" was essentially introduce in this context, by Shadrin, Shapiro and Vainshtein. -It apparently comes up in physics -- I only know this because the number of regions of $R_n$, starting at $n=2$, is 2, 6, 32, 370, 11292, 1066044, 347326352 ... Sloane sequence A034997, which you will see was entered as "Number of Generalized Retarded Functions in Quantum Field Theory" by a physicist. -You might expect by that rate of growth that this hyperplane arrangement is completely intractable, and more specifically, what I would love from an answer is some kind qualitative statement about how ugly the $R_n$ get. Which brings us to: -Connection to the GGMS decomposition -I got to thinking about this again now and decided to post on MO of it because of Noah's question about the vertices of a variation of GIT problem, where this arrangement is lurking around -- see there for more detail. Allen's brief comment there prompted me to skim some of his and related papers to that general area, and I found the introduction to Positroid varieties I: juggling and geometry most enlightening, together with the discussion at Noah's question could give another suggestion why $R_n$ is perhaps intractable. Briefly: -The arrangement $R_n$ is a natural extension of the $A_n$ arrangement. One common description of the $A_n$ arrangement is as the $\binom{n+1}{2}$ hyperplanes $y_i-y_j=0, i,j\in [n]$ and $y_i=0, i\in [n]$. However, one can consider the triangular change of variables -$$y_k\mapsto \sum_{j\leq k} y_j$$. -This changes the hyperplanes to -$$\sum_{i\leq k \leq j} y_k=0.$$ -These hyperplanes are no longer invariant under permutation of the coordinates, and if we proceed to add the entire $S_n$ orbit of them, we get the resonance arrangement $R_n$. -From the discussion on Noah's question and the introduction to the Positroid paper, we see that this manipulation is a shadow of the GGMS decomposition of the Grassmannian, and describing this decomposition in general seems to be intractable in that it requires identifying whether matroids are representable or not. So, what I'd really like to know how is much of the "GGMS abyss", as they refer to it, is reflected in the resonance arrangement? Is it hopeless to describe and count its chambers? - -REPLY [8 votes]: Here's what I know about this arrangement. -Regarding the number of chambers in this arrangment, Zuev obtained the lower bound $2^{(1-o(1))n^2}$. The proof uses Zaslavsky's theorem and a difficult estimate due to Odlyzko. -http://www.doiserbia.nb.rs/img/doi/0350-1302/2007/0350-13020796129K.pdf describes an improvement to that lower bound (including references to Zuev's and Odlyzko's articles). -http://arxiv.org/pdf/1209.2309v1.pdf studies a closely related arrangement and gives a related (but weaker) lower bound using a very elegant method. It turns out the arrangment becomes much easier mod $2$ (even than one might expect). -Whenever one has a central hyperplane arrangement one has a zonotope dual to it. Klivans and Reiner [ http://arxiv.org/pdf/math/0610787v2.pdf ] fix $k$ and look at the zonotope (Minkowski sum of the line segments) generated by all 0/1-vectors of length $n$ with exactly $k$ ones. In particular they are interested in the zonotope considered as symmetric polynomial (the sum $\sum x^m$ over all lattice points $m$ in the zonotope). -Thus the Minkowski sum of their zonotopes over $1\leq k \leq n$ is the zonotope dual to your arrangement. Degree sequences of hypergraphs correspond to integer points in this latter zonotope. Surprisingly the converse is false! This was showed by Liu [http://arxiv.org/abs/1201.5989 ]. -The vertices in the zonotope, and thus the regions in the arrangement, correspond almost to linear threshold hypergraphs (better known as linear threshold (Boolean) functions). A linear threshold hypergraph is determined by $n+1$ real potentially negative numbers $q,w_1,\dots,w_n$; any set $S\subseteq [n]$ with $\sum_{i\in S} x_i \leq q$ is declared an edge. "Almost" above means that the vertices in this zonotope correspond exactly to the linear threshold hypergraphs that can be given with $q = 0$. -What if $q \neq 0$? Given a hypergraph $\mathcal{H}$ with $s$ simplices and degree sequence $v_1,\dots,v_n$, one can look at the values $s-2v_i$, $1\leq i\leq n$. If $\mathcal{H}$ happens to be down-closed, this is what game theorists call the Banzhaf value of the game represented by $\mathcal{H}$ and what computer scientists call the Boolean influence of the Boolean function represented by $\mathcal{H}$. Anyway it is more natural to look at the vector $(s,v_1,\dots,v_n)$ rather than just $(v_1,\dots,v_n)$. This corresponds to looking instead at the zonotope generated by $(1,0,0),(1,0,1),(1,1,0),(1,1,1)$ et.c. (all binary vectors of length $n$, prepended by $1$). The vertices (obviously more than when we didn't have the leading 1's in the generators) in this zonotope correspond exactly to linear threshold hypergraphs. Since each $v_i$ is an integer $0$ and $2^{n-1}$, and $s$ is between $2^n$ there are at most $2^{n + n(n-1)} = 2^{n^2}$ of them, giving an almost matching upper bound to the lower bound above. -The latter zonotope is briefly mentioned in the end of the article http://arxiv.org/pdf/0908.4425.pdf . -I remember having seen a reference to a paper of Terao computing the characteristic polynomial of the arrangemnt for some small values, but cannot locate that reference right now.<|endoftext|> -TITLE: What mathematical treatment is there on the renormalization group flow in a space of Lagrangians? -QUESTION [31 upvotes]: What mathematical treatment is there on the renormalization group flow in a space of Lagrangians? - -REPLY [11 votes]: A small complement to Abdelmalek Abdesselam's answer: on the rigorous, non-perturbative side, there is also a recent (originally two-part, now turned into three-part) exposition by Jonathan Dimock, available in the arXiv's. He uses the $\phi^4$ scalar field theory in 3 dimensions in finite volume as a model for his discussion - the three parts are listed below: - -J. Dimock, "The Renormalization Group According to Balaban I. Small Fields". arXiv:1108.1335 [math-ph] -J. Dimock, "The Renormalization Group According to Balaban II. Large Fields". arXiv:1212.5562 [math-ph] -J. Dimock, "The Renormalization Group According to Balaban III. Convergence". arXiv:1304.0705 [math-ph] - -Tadeusz Balaban refined the method of block-spin renormalization group employed by Gallavotti, Kupiainen and many others for lattice field models in order to analyse "large field" regions, aiming at the treatment of the continuum limit of pure Yang-Mills models in finite volume and 4 dimensions. His long series of papers on the subject from the 80's remain essentially the state of the art towards the rigorous construction of realistic models in quantum field theory in 4 dimensions (see, for instance, the latest of the series), together with the paper of Magnen, Rivasseau and Sénéor, which was motivated by Balaban's work. The third part of Dimock's exposé is meant to establish the convergence of the expansion scheme laid down in Parts I and II.<|endoftext|> -TITLE: Is $PSL(n, Z)$ isomorphic to a subgroup of $GL(n,C)$ or even $GL(n+1,C)$? -QUESTION [10 upvotes]: Is $PSL(n, \mathbb Z)$ isomorphic to a subgroup of $GL(n,\mathbb C)$ or even $GL(n+1,\mathbb C)$? - -REPLY [17 votes]: You can look for a finite subgroup of $PSL(n,\mathbb Z)$ such that every faithful representation of it has dimension $>n+1$. The following works for even $n\ge 6$. -In $GL(n,\mathbb Z)$ take the group of monomial matrices, i.e. products of permutation matrices and diagonal matrices. Intersect with $SL(n,\mathbb Z)$ and let $G$ be the image in $PSL(n,\mathbb Z)$. -$G$ maps onto the symmetric group $S_n$ with kernel $D\cong (\mathbb Z/2)^{n-2}$. In any faithful action of $G$ on $\mathbb C^d$ there is a one-dimensional subspace $L$ acted on nontrivially by $D$. In the action of $S_n$ on the set of nontrivial $1$-dimensional characters of $D$ every orbit has at least $n(n-1)/2$ elements. It follows that there at least that many independent choices of $L$ and therefore $d\ge n(n-1)/2>n$. -One step breaks down when $n=4$, so you'd need a different subgroup in that case.<|endoftext|> -TITLE: Is the pushforward of a line bundle on the smooth locus of a terminal singularity again a line bundle? -QUESTION [10 upvotes]: In algebraic geometry, it is a sad fact of life that pushforward doesn't preserve being a coherent sheaf; for example, the pushforward by the complement of a divisor of the structure sheaf (or more generally a line bundle) has essentially no hope of being again coherent. On the other hand, on a smooth variety, if I pull out a closed subset of codimension $\geq 2$, then the pushforward of the structure sheaf will be be the structure sheaf of the whole thing, essentially by Hartog's theorem. -Now, this is not true for singular varieties; you can have an affine inclusion of varieties where the complement has codimension greater than 1, like removing the singular point from singular plane quadric. -I'm generally curious about when "not too bad" singularities can avoid this problem; I'm particularly interested in whether this works for removing the singular locus of a terminal variety, but would be interested to hear other results along the same lines. - -REPLY [12 votes]: On a normal variety, pushing forward a line bundle from the non-singular locus gives you a reflexive sheaf which is essentially the same as taking a Weil divisor representative of your original line bundle and take the Zariski closure of all the components keeping the same coefficients. So, what you get is a Weil divisor. -In other words, only assuming that your variety is normal, you get that the push forward of a line bundle (i.e., the sheaf version of a Cartier divisor) is a reflexive sheaf of rank 1 (i.e., the sheaf version of a Weil divisor (on a normal variety at least)). -The condition you want/need is that every Weil divisor be Cartier. As Karl said, this is exactly the condition of being factorial.<|endoftext|> -TITLE: Random walk origin return monotinicity -QUESTION [11 upvotes]: Consider a Markov chain on $\mathbb{Z}^d$ with transition kernel $P$ for adjacent vertices (non-diagonal). Essentially this is a $d$ dimensional random walk with the probability of a transition dependent on it's location in the grid. This comes from a random conductance model. The theorem that concerns me is a general result for Markov chains, but I leave this motivation to assist in its proof (see below). -Let $P^{2k}(0,0)$ be the probability of going from the origin and back in $2k$ steps. Moreover, suppose $P$ is reversible. The theorem that concerns me is:  - -$P^{2n}(0,0)$ is decreasing in $n$.  - -I am interested in a probabilistic proof of this. The proof that I know is of a spectral nature: -Define $\langle f,g\rangle:= \sum_{X\in\mathbb{Z}^d} \pi(x)f(x)g(x)$, -where $\pi(x)$ is the stationary measure. This gives an inner product on $L^2(\mathbb{Z}^d)$. In the case of a random conductance model, $\pi(x)$ would be the sum of random edge weights at $x$.  -Then -$P^{2k}(0,0)=\langle \delta_0,P^{2k}\delta_0\rangle$, -and since $P$ is self adjoint with $\|P\|_2\leq 1$, the desired result follows.  -I have tried various approaches such as conditioning on the hitting times of the origin and as well trying to prove the result by induction. I would like to see a proof that showcases a probabilistic argument. For example, is it possible to show the result from the machinery of evolving sets of Morris and Peres? - -REPLY [2 votes]: I am not exactly sure about what is meant by "probabilistic", but there is a simple argument based on the Cauchy inequality (no spectral theory involved) which provides a so-called "ratio limit theorem" for return probabilities. It is valid for any reversible chain with an infinite stationary measure, i.e., any reversible chain which is not positive recurrent. -Let $\phi_n(x)=p_n(o,x)/\pi(x)$ be the density of the $n$-step transition probability from the reference point $o$ with respect to the stationary measure $\pi$ (I use a slightly different notation). I will normalize the measure $\pi$ in such a way that $\pi(o)=1$. Then for any $n,m\ge 0$ -$$ -p_{n+m}(o,o) = \sum_x p_n(o,x) p_m(x,o) = \sum_x p_n(o,x) p_m(o,x) \frac{1}{\pi(x)} = \langle \phi_n,\phi_m\rangle \;, -$$ -where $\langle\cdot,\cdot\rangle$ denotes the scalar product with respect to the stationary measure $\pi$. In particular, for $n=k-1$ and $m=k+1$ -$$ -p_{2k}(o,o) = \langle\phi_{k-1},\phi_{k+1}\rangle_\pi \le \|\phi_{k-1}\| \|\phi_{k+1}\| \;, -$$ -whence -$$ -\frac{p_{2k}(o,o)}{p_{2k-2}(o,o)} \le \frac{p_{2k+2}(o,o)}{p_{2k}(o,o)} \;. -$$ -The above inequalities are strict unless $\phi_{k-1}=\phi_{k+1}$, in which case also $\phi_{k+3}=\phi_{k+5}=\dots=\phi_{k-1}$, which implies that all the return probabilities $p_{2k}(o,o), p_{2k+2}(o,o),\dots$ are the same, which is only possible for a positive recurrent chain. Thus, discarding the positive recurrence case, the sequence of ratios $p_{2k+2}(o,o)/p_{2k}(o,o)$ is strictly increasing. Obviously, its limit can not be greater than 1, whence the claim. -Actually, in the situation considered by the OP the value of the above limit, in other words, the spectral radius of the transition operator, is precisely 1 provided the operator has bounded geometry (i.e., the weights of edges are uniformly bounded from below and above).<|endoftext|> -TITLE: Kuratowski's definition of ordered pairs -QUESTION [18 upvotes]: Among several possible definitions of ordered pairs - see below - I find Kuratowski's the least compelling: its membership graph (2) has one node more than necessary (compared to (1)), it is not as "symmetric" as possible (compared to (3) and (4)), and it is not as "intuitive" as (4) - which captures the intuition, that an ordered pair has a first and a second element. -(source) -Membership graphs for possible definitions of ordered pairs (≙ top node, arrow heads omitted) -1: (x,y) := { x , { x , y } } -2: (x,y) := { { x } , { x , y } } (Kuratowski's definition) -3: (x,y) := { { x } , { { x } , y } } -4: (x,y) := { { x , 0 } , { 1 , y } } (Hausdorff's definition) - -So my question is: - -Are there good reasons to choose - Kuratowski's definition (or did - Kuratowski himself give any) instead of - one of the more "elegant" - sparing, - symmetric, or intuitive - - alternatives? - -REPLY [22 votes]: Of course there are many pairing functions, and they all -have the crucial property that from the pair $(x,y)$, one -can reconstruct both $x$ and $y$. And although your -question has been answered, let me point out that all four -of the ordered pair definitions that you consider have the -property that the von Neumann rank of the pair $(x,y)$ is -strictly greater than the ranks of $x$ and $y$. Thus, for -your functions, if $x$ and $y$ are in $V_\alpha$, then the -pair $(x,y)$ can only be guaranteed to appear by -$V_{\alpha+2}$. -But actually, this rank-increasing feature is sometimes -annoying, and there occasionally arises in set-theoretic -argument the need or desire for a flat pairing function, -a pairing function that does not increase rank in this way. -Specifically, what is desired is a pairing function -$\langle x,y\rangle$ such that whenever $x,y\in V_\alpha$ -for infinite $\alpha$, then also $\langle x,y\rangle\in -V_\alpha$, for the same rank $\alpha$. (Note that one -cannot achieve this for finite $\alpha\gt 1$, since there -are too many pairs to fit.) With such a flat pairing -function, every infinite $V_\alpha$ is closed under -pairing, and this is sometimes important or at least -convenient in inductive arguments, or in arguments about -$\alpha$-strong cardinals and in similar situations, where -one wants to consider only sets of a given rank, but one -also wants to use pairs. -It is a fun exercise to prove that flat pairing functions -exist, and I encourage you to try it on your own, before -reading what I write below. But the definitions are all -somewhat more involved than the comparatively simple -definitions you provide, since they achieve the flatness -property. As Hurkyl says, we ultimately care only about the -existence of the function with the desired properties, -rather than its exact nature. -Here is one way to construct a flat pairing function. -Define $\langle x,y\rangle=x^0\cup y^1$, where $x^0$ is -obtained by replacing every natural number $n$ in any -element of $x$ by $n+1$ and adding the object $0$, whereas -$y^1$ just replaces $n$ inside elements of $y$ with $n+1$. -Thus, we can tell from any element of $x^0\cup y^1$ whether -it came from $x$ or $y$, by looking to see if it contains -$0$ or not, and we can reconstruct the unmodified set by -removing $0$ and replacing all $n+1$ with $n$ again, and so -it is a pairing function. And one can check that $\langle -x,y\rangle$ has the same rank as the maximum rank of $x$ -and $y$, if this max is infinite, and so this is a flat -pairing function, as desired.<|endoftext|> -TITLE: Tree Version of Hechler Forcing -QUESTION [7 upvotes]: In their recent (2009) paper Eventually Different Functions and Inaccessible Cardinals, Brendle and Löwe consider a 'tree version' of the Hechler forcing. This forcing $\mathbb{D}$ consists of nonempty trees $T\subseteq\omega^{<\omega}$ with the property that there is a unique stem $s\in T$ so that for every $t\in T$ extending $s$, $t\frown n\in T$ for all but finitely many $n\in\omega$. The forcing is ordered by inclusion. This forcing allows for a very elegant rank analysis, and many properties about Hechler forcing that are proved to hold using rank arguments can also be proved to hold for $\mathbb{D}$ in a conceptually simpler way. I do not know if the two notions of forcing are equivalent, and based on what is said in the paper I suspect neither do the authors. This is not my real question, though I would be happy to see an answer. -My question is simply whether this specific tree forcing has appeared anywhere in the literature previously. No reference is given in the paper but I ask because I do think I remember seeing it somewhere and now I can't seem to find mention of it in likely places. - -REPLY [9 votes]: This forcing is a special case of forcing with trees that branch into a filter, the filter in this case being the co-finite sets. (This, in turn, can be viewed as a special case of Shelah's creature forcing.) So the example has certainly implicitly appeared in the literature. An early reference to forcing with trees branching into filters is "Combinatorics on ideals and forcing with trees" by Marcia Groszek in J. Symbolic Logic 52 (1987), no. 3, 582–593.<|endoftext|> -TITLE: proportion of positive integers which number of prime divisors has a special remainder. -QUESTION [6 upvotes]: Let $s$ be an integer greater than 1. For each natural number $n$, let $\omega(n)$ be the number of prime divisors (counted with multiplicities) of $n$ modulo $s$. For $i \in \{0,1,\dots, s-1\}$ and a positive integer $k$, set -$$c_i(k) = |\{x \in \{1, 2, \dots, k\} : \omega(x) = i\}|.$$ -Is it true that $\frac{c_i(k)}{k} \rightarrow \frac{1}{s}$ as $k \rightarrow \infty$ for all $i \in \{0,1, \dots, s-1\}$? - -REPLY [4 votes]: Yes, this is true. -More precisely, it is known that, for fixed $s$ and $k \to \infty$, one has -$$ \frac{c_i(k)}{k} = \frac{1}{s} + O(\frac{1}{ (\log k)^A}) $$ -for some $A>0$ . -See 'On the residue class distribution of the number of prime divisors of an integer' by Coons and Dahmen; preprint at http://arxiv.org/abs/0906.1029 -They also discuss the nature of the error-term in more detail, and in particular show that for $s>2$ it cannot be $ O(1/ k^B) $ for $B>0$. -Unrelated side remark: since you define $\omega$ precisely this is not a big deal, but using the standard notation for the number of distinct prime divisors for something else is somewhat 'dangerous' (at least, it almost lead me to give a wrong answer).<|endoftext|> -TITLE: The Guinand-Weil explicit formula without entire function theory -QUESTION [8 upvotes]: I'll admit from the outset that this question is slightly vague. The actual question appears at the end of the post. -The explicit formula of Guinand and Weil can be written in the following way: -For 'nice' g (i.e. in $C_c^\infty(\mathbb{R})$) -$$ -\sum_\gamma \hat{g}(\gamma/2\pi) - \int_\mathbb{R} \frac{\Omega(\xi)}{2\pi}\hat{g}(\xi/2\pi) d\xi -= \int_\mathbb{R} [g(x)+g(-x)] e^{-x/2}d(e^x-\psi(e^x)), -$$ (1) -where the sum is over those $\gamma$ such that $1/2+i\gamma$ a non-trivial zero of the Riemann Zeta function, $\psi(x) = \sum_{n\leq x} \Lambda(n)$ is the Chebyshev prime counting function, and -$$\Omega(\xi) = \tfrac{1}{2}\tfrac{\Gamma'}{\Gamma}(1/4+i\xi/2) + \tfrac{1}{2}\tfrac{\Gamma'}{\Gamma}(1/4-i\xi/2) - \log \pi. -$$ -Here $\gamma$ can possibly be complex. -It is usually proven using a contour integral to capture the zeroes of the Zeta function, then evaluating the integral a different way, making use of the reflection formula along with the arithmetical meaning of $\zeta(s)$ for $\Re s > 1$. (See for instance Montgomery and Vaughan, Multiplicative Number Theory.) -$\Omega(\xi)/2\pi \sim \log \xi /2\pi$, and is the mean density for the number of zeroes to occur in the critical strip with real part $\xi$. On the assumption of the Riemann hypothesis, the left hand side takes the nice form: -$$ -\int_\mathbb{R}\hat{g}(\xi/2\pi)\bigg(\sum_\gamma \delta(\xi-\gamma) - \frac{\Omega(\xi)}{2\pi}\bigg) d\xi. -$$ -The explicit formula therefore expresses a Fourier duality between the error term in the Chebyshev prime counting function and the error term in the zero counting function. The structural reason why this duality arises is not really apparent to me from the contour integral proof above, and is what I'm really getting at with this question. -That said, left at this the question is a little imprecise, and there is something of a lie here because the form of the explicit formula where this becomes apparent involves assuming the Riemann hypothesis. Therefore: -Question: Is there a way not making use of entire function theory proper to show that there exist numbers $\gamma$ with $|\Im \gamma| \leq 1/2$, so that (1) is true? -A proof using harmonic analysis over the adeles would get bonus points. -One reason to be interested in a question like this beyond what I've elaborated above is to ask to what extent explicit formulas like (1) can be replicated for the 'Beurling primes.' - -REPLY [10 votes]: The Riemann zeta function is given for $Re s>1$ -$$\zeta(s) = \prod\limits_{p \; prime} ( 1-p^{-s}) $$ -This product converges absolutely in $Re s >1$, hence it does not vanish in $Re s>1$. Actually the product also converges locally uniformly, which implies that $\zeta(s)$ is holomorphic for $Re s>1$. -The functional equation follows from the Poisson summation formula, which is a Fourier theoretic argument. The functional equation is given by -$$ \Lambda(s) = \pi^{-s/2} \Gamma(s/2) \zeta(s) = \Lambda(1-s).$$ -This implies that all zeros lie inside $0 \leq Re s \leq 1$. We will need to evaluate a certain integral -$$ * = \int\limits_{Re\; s = 1 + \epsilon/2 \cup Re \; s = -\epsilon/2} G(s) \frac{\Lambda'}{\Lambda} (s) d s$$ -We have now for a meromorphic function $F$ and a holomorphic function $G$ in some region $G$ containing the closure of a simply connected region $O$, that the contour integral -$$\frac{1}{2 \pi i} \int\limits_{\partial O} G(z) \frac{F'}{F}(z) d z = \sum\limits_{\rho \; zero \; of \; F \; in \; O} G( \rho) - \sum\limits_{\nu \; pol \; of \; F \; in \; O} G( \nu),$$ -where $\partial O$ denotes the boundary of $O$ and is a Jordan curve by assumption. For this identity to be valid $F$ must have no zero and no pol on the boundary of $O$. -This is what I call the weighted argument principle and can be derived in the same lines as the argument principle. It follows for entire functions easily from the Hadamard factorization theorem, but is a purely local property. -Apply this to the function $F = \Lambda$ and $G$ being holomorphic in $ - \epsilon \leq Re\; s \leq 1$ with $\epsilon>0$ and certain restrictions of the growth. We choose the contour $C=C(T)$ being the boundary of $ - \epsilon/2 < Re\;s < 1 + \epsilon/2$ and $| Im \;s | \leq T$, where \zeta does not vanish on $Im \; s = \pm T$. - This give an expression for $*$ for $T \rightarrow \infty$ involving the nontrivial zeros of $\zeta$. -Using the Euler product, we can also derive a nice explicit expression -$$ \frac{\Lambda'}{\Lambda} (s) = (log \Lambda(s))'= -1/2 \log \pi +\frac{1}{2} \frac{\Gamma'}{\Gamma}(s/2) - \sum\limits_{p \; prime} \frac{p^{-s} \log p }{1-p^{-s}}.$$ -This gives an expression for $*$ for $T \rightarrow \infty$ involving the primes. -Assuming certain boundedness conditions on $\Lambda(s)$ and $G(s)$ in $-\epsilon \leq Re \; s \leq 1+\epsilon$, we are actually allowed to choose $C(T)$ with $T \rightarrow \infty$ and derive the formula as the limit. The boundedness conditions for $\Lambda$ follow from the Hadamard three lines principle or the Phragmen Lindeloeff principle (this is not entire function theory, but this only a complex analysis argument), then the explicit formula follows by choosing $G$ appropiately. -Remark: Inserting a Gaussian function into the explicit formula allows to derive the functional equation for $\zeta$, hence the functional equation is equivalent to Weil's explicit formula. Actually Samuel Patterson states in his famous book on $\zeta$ that they are actually both equivalent to the Poisson summation formula, but I do not know how? Of course the Poisson summation implies Functional equation of $\zeta$... How to go back? -So the philosophy is: Functional equation + Euler product = explicit formula. Another example for this is the relation of the Selberg Zeta function and Selberg trace formula. -You are right that the entire function theory implies that $\Lambda$ has necessary many zeros, but not too many, since it is of exponential type $1$ because of the factor $\Gamma$. If you want to derive this without using the merophorphicity of $\Lambda$, you might want to try to deduce this without knowledge over the primes and by inserting an approppiate chosen test function in the explicit formula. I have never seen this been worked out, but choosing an appropiate function $g$ being supported in $ - \log 2 < t < \log 2$ (so no contribution by finite primes) etc. should lead to a rough asymptotic of the zeros without any information used about the primes, but possibly a weaker error term than in the classical van Mangoldt estimate, which I expect to be a square root of the actual main term. Look at similiar techniques used in Werner Mueller and Erez Lapid's article Chapter 2 of http://www.math.uni-bonn.de/people/mueller/papers/orbint09.11.pdf for the Weyl law. The Selberg trace formula has many analogies with the explicit formula. In Iwaniec - Spectral methods in automorphic forms, you can find an argument using Tauberian theorems, which is weaker, but you get only the main term and no error term. -One interesting, but technical derivation of the explicit formula using only the languages of the adeles, harmonic analysis and no entire function theory at all was given by Ralf Meyer: -http://arxiv.org/pdf/math/0311468v3 -However, the Fourier transform of a function with certain growth properties has some holomorphicity conditions, so you basically will just hide the complex analysis arguments, but be able to deduce the results from Fourier analysis only. If you want to derive the prime number theorem from Fourier analysis, you might want to consider the treatement in Rudin - Functional analysis, which is based on real analysis only. Also there are elementary proofs of the prime number theorem, which I have no idea if they apply to Beurling primes.<|endoftext|> -TITLE: Canonical form for a pair of quadratic forms -QUESTION [5 upvotes]: Could anyone recommend a reference to a canonical form for a pair of quadratic forms over R (not necessarily positively definite)? This is probably related to the Weierstrass elementary divisors (but it is not so easy to find information about them on the web). To be more precise, the case when A=diag(1,1,1,1,-1) and B is a symmetric 5x5 matrix is of interest. - -REPLY [4 votes]: Hi, I just read your question. I hope it's not too late to answer, but I think the following paper contains exactly what you are looking for: -R. C. Thompson: Pencils of complex and real symmetric and skew matrices, Linear Algebra and its Applications, Volume 147, March 1991, 323-371. -The author classifies - up to congruence - any pairs of real symmetric matrices, which amounts to give the classification for any pair of real quadratic forms. -You'll probably find of some interest also this paper of mine: -http://arxiv.org/abs/1106.4678<|endoftext|> -TITLE: Why is the simple trace formula a weaker tool than the Arthur trace formula? -QUESTION [9 upvotes]: What are some concrete examples of theorems which can be deduced from the Arthur trace formula, which do not follow from the simple trace of Kazdhan and Flicker? -(So I do not mean weaker in the sense that the Arthur trace formula implies the simple trace formula.) - -REPLY [3 votes]: This is an addendum to Jason's answer. -Mueller [1] and Lapid-Mueller [2] have exploited Arthur's trace formula to come up with a Weyl law for GL(n) with an excellent error term. (Their method has been adapted by others, Matz, Matz-Templier etc.). Since they are counting all automorphic representations (with a fixed infinitesimal character), the simple trace formula simply cannot detect them. -Much more importantly, Arthur's trace formula has been an indispensible tool in proving important (endoscopic) cases of Langlands' functoriality which hasn't been done (cannot?) by other trace formulas. We now have functoriality so far for classical groups, unitary groups, their inner twists, spin groups. -Nevertheless, Labesse-Mueller proved a weak Weyl law and also Kottwitz could prove the Tamagawa number conjecture (see Jason's answer) using weak test functions in Arthur's trace formula. -[1]: Mueller, Weyl's law for the cuspidal spectrum of $SL_n$ -[2]: Lapid-Mueller, Spectral asymptotics for arithmetic quotients of ${\rm SL}(n,{\mathbb R})/\rm{SO}(n)$ -[3]: Labesse-Mueller, Weak Weyl's law for congruence subgroups<|endoftext|> -TITLE: Path connectedness of varieties -QUESTION [17 upvotes]: Let $X$ be a variety. Then, is $X$ path connected? And by path connected, I mean any two closed points $P, Q$ on the variety can be connected by the image of a finite number of non-singular curves. - -REPLY [7 votes]: In the affine setting over $\mathbb{C}$, an algebraic set is path-connected in the analytic topology if it is irreducible (in fact, its smooth locus is path-connected too). Conversely, it is irreducible if and only if it contains a dense open path-connected subset of smooth points. -See the appendix here.<|endoftext|> -TITLE: primitive roots and primes -QUESTION [10 upvotes]: Given a positive integer $n > 1$, is it true that there exists infinitely many primes $p$ such that $n$ is a primitive root modulo $p$. - -REPLY [11 votes]: It showed up in a recent question so one might wonder. The article Artin's conjecture for primitive roots, Math. Intelligencer, 10 (4) (1988) 59-67 by Ram Murty seems like a good survey. The link is to a dvi copy. It informs one that the result follows from a Generalized Riemann Hypothesis and is unconditionally true for at least one of $2,3,5.$ -The first $12000$ primes for which $7$ is a primitive root run from $11$ to $378011$. This proportion $\frac{12000}{32141} \approx 0.3734$ agrees well with the theoretical expected proportion of $\prod(1-\frac{1}{p(p-1)}) \approx 0.3739$ where the product is over the primes. The distribution according to congruence class $\mod 7$ is $[1, 1748], [2, 2074], [3, 2032], [4, 2058], [5, 2065], [6, 2023].$ This slight deficit in congruence class 1 seems to hold through this range.<|endoftext|> -TITLE: The complexity of the leading fractional bit of a power of a rational number -QUESTION [11 upvotes]: On a mailing list (math-fun) that I subscribe to Dan Asimov asked what's the most efficient way to calculate the leading decimal digits (say 10 of them) of $(p/q)^n \bmod 1$ where $p$ and $q$ are fixed (think of $p/q = 3/2$) and $n$ varies. There were a number of suggestions, but all of them clearly had complexity proportional to $n$. So my question is, for concreteness, let $b(n) = \lfloor 2(3/2)^n \rfloor \bmod 2$ (the leading fractional bit of $(3/2)^n$). Suppose that $n$ is specified in binary. What is the complexity (both time and space) of calculating the function $b(n)$? After thinking about it for a while I wouldn't be surprised if it's exp-space hard. Does anyone know anything about this? - -REPLY [4 votes]: From the review of -Mika Hirvensalo, Juhani Karhumäki, and Alexander Rabinovich, Computing partial information out of intractable: powers of algebraic numbers as an example, J. Number Theory 130 (2010), no. 2, 232–253, MR2564895 (2010j:11117), it looks like there may be something of interest there.<|endoftext|> -TITLE: Orthogonal foliations -QUESTION [8 upvotes]: Consider the manifold $\mathbb{R^2}\setminus \{0\}$, on which the group of rotation acts. The orbits of the group are the circles centered in the origin, and form a foliation of $\mathbb{R^2}\setminus \{0\}$. This foliation will be denoted by $F_1$. The foliation $F_1$ defines univocally another foliation $F_2$, with the following property: the tangent spaces of two leafs of $F_1$ and $F_2$ are orthogonal at the intersection point. In this case $F_2$ is composed by the radial lines from the origin. -My question is the following: to what extent this situation can be generalized, i.e.: assume to have a riemanian manifold $M$, possibly flat, with a foliation $F_1$ defined by the orbits of a group acting on $M$. To what extent does this foliation define univocally an orthogonal foliation $F_2$ with the property that the tangent spaces of any pair of leafs of $F_1$ and $F_2$ are orthogonal at the intersection point(s)? - -REPLY [7 votes]: I cannot resist but mention a related concept, which in a sense generalizes the example you quote. -Let a Lie group $G$ act properly and isometrically -on the complete Riemannian manifold $M$. The action is called -polar if there exists a complete connected submanifold $\Sigma$ that meets all the orbits, -and meets them always orthogonally. Such a submanifold $\Sigma$ is called a section. -It is easy to see that a section must be totally geodesic. If an action admits a section -which is flat in the induced metric, then this action is called hyperpolar. -In the case of linear orthogonal actions (or representations), there is no distinction between polar and hyperpolar actions since the complete totally geodesic submanifolds of Euclidean space are its affine subspaces. -One example of polar representation which is very familiar from basic courses in linear algebra is the $SO(n)$-conjugation of $n\times n$ real symmetric matrices. It is well -known that every symmetric matrix is orthogonally conjugate to a diagonal matrix, so here -the section is given by the subspace of diagonal matrices. More generally, -the standard examples of polar representations are the isotropy representations of symmetric spaces. Conversely, Dadok has shown that these essentially exhaust all the examples. -The orbital foliation of a polar action has many remarkable geometric and topological properties. The story starts with Bott and Samelson in the 1950's and goes on. To mention -a recent result, A. Lytchak and G. Thorbergsson have proven that the orbifold points of the orbit space of a proper and isometric action correspond precisely to the points of the manifold where the slice representation is polar (J. Differential Geom.85 (2010), 117-140). -Polar actions have also been generalized in many directions, e.g. polar foliations, -complex actions, Hilbert space. I leave you with two book references: http://vmm.math.uci.edu/CriticalPointTheory.pdf (link to free book) and J. Berndt, S. Console and C. Olmos, Submanifolds and Holonomy, CRC/Chapman and Hall Research Notes Series in Mathematics 434 (2003), Boca Raton (more recent book). -*Edit:*Ah, I forgot to mention one detail. In your example, you removed the origin to get a second foliation by radial lines. For a polar action, the orbits form a (singular) foliation of course, but the sections form a foliation only if you remove the singular orbits. In fact, a point in the manifold lies in an orbit of lower dimension precisely if it is contained in more than one section.<|endoftext|> -TITLE: Recent Applications of Mathematics -QUESTION [41 upvotes]: What are the recent and new applications of Mathematics in other Sciences ? -Let me try to be more precise about the question: - -By "recent" I mean the last 15 years. -By "new" I want to exclude the standard answers like cryptography or finance -By "applications" I mean a mathematical concept (or even a trick) successfully used in another field (preferably not Theoretical Physics) to solve a problem or to shed a new light on a phenomenon. -I would prefer to see recent applications of modern Mathematics, but new applications of classical results should be considered as a valid answer too. -the answer should not simply be "XXX was successfully applied to YYY". it should contain a short explanation of the mathematical concept involve, and a description of the problem/phenomenon it solved/enlightened. - -The typical example I have in mind does not strictly answer the question (first it is an application of Theoretical Physics to condensed matter, and then I am not giving the required details): it seems that some 2d quantum field theories which had a priori no physical meaning were successfully used to understand properties of graphene (I think this was really unexpected: 2d conformal field theory was considered as a toy model to approach the understanding of more relevant field theories in higher dimension). - -v2: I hope this version is better than the previous one. - -REPLY [2 votes]: Pseudodifferential (PDO) and Fourier integral operators (FIO) and also Wick and antiWick operators have applications in signal processing, especially in analysis of non stationnary signals like speech signals. Examples of such time frequency representation are spectrograms, scalograms which use wavelet theory and similar tools like Gaborets, chirplets, ridglets,...etc. This tools are also used to solve PDEquations as they serve to almost diagonalization of a large class of PDO and FIO.<|endoftext|> -TITLE: Geometric meaning of trigonometric relations -QUESTION [7 upvotes]: According to a paper by Zhiqin Lu in the Mathematical Gazette (the British publication, not the Boston-area newsletter, if that still exists (or even if it doesn't)) in 2007(?), if $u+v+w=\pi$ and $a,b,c \ge 0$, then $a + b + c \ge 2 \sqrt{bc} \cos u + 2 \sqrt{ca} \cos v + 2 \sqrt{ab} \cos w$. It wasn't hard to show that equality holds iff $a:b:c = \sin^2 u : \sin^2 v : \sin^2 w$, but if I recall correctly, that wasn't in the paper. I would think that there must be some natural geometric interpretation of this proposition. What is it? -The case where equality holds says that if $u+v+w = \pi$, then $\sin^2 u + \sin^2 v + \sin^2 w = 2 \sin u \sin v \cos w + 2 \sin u \cos v \sin w + 2 \cos u \sin v \sin w$. That one has a simple geometric interpretation as a sort of mash-up of the law of sines and the law of cosines. But now suppose we say that if -$$\sum_i u_i =\pi$$ -then -$$ -\sum_{i=1}^\infty \sin^2 u_i = \sum_{\text{even }n\ge 2} (-1)^{(n-2)/2}n\sum_{|A|=n} \prod_{i\in A}\sin u_i \prod_{i\not\in A} \cos u_i -$$ -The case of a sum of three variables that add up to $\pi$ is the special case in which all but three of these are 0. Does it have a geometric meaning? -Later edit: So far we have an answer about the inequality, but not yet about the equality. I will probably comment soon on the former. - -REPLY [15 votes]: In an ancient MathLinks topic (post #6; but see below for a copy) I have given a proof of the inequality by reducing it to $\left(\sqrt{a}\vec{p}+\sqrt{b}\vec{q}+\sqrt{c}\vec{r}\right)^2\geq 0$, where multiplication means scalar product of vectors and $\vec{p}$, $\vec{q}$, $\vec{r}$ are unit length vectors chosen in such a way that the angles between them are $\pi-u$, $\pi-v$, $\pi-w$, respectively. This rewrites geometrically as follows: Pick a point $P$ in the plane, and take three points $A$, $B$, $C$ such that $PA=\sqrt{a}$, $PB=\sqrt{b}$, $PC=\sqrt{c}$, $\measuredangle BPC=\pi-u$, $\measuredangle CPA=\pi-v$ and $\measuredangle APB=\pi-w$. Then, the difference between the left hand side and the right hand side of your inequality is $9$ times the square of the distance between the point $P$ and the centroid of triangle $ABC$. Equality thus holds if and only if $P$ is the centroid of triangle $ABC$; this is equivalent to the assertion that the triangles $BPC$, $CPA$, $APB$ have equal areas; this, in turn, is equivalent to the assertion that $\sqrt{a}:\sqrt{b}:\sqrt{c}=\sin u:\sin v:\sin w$ (because the area of triangle $BPC$ is $\frac{1}{2}\cdot PB\cdot PC\cdot \sin\measuredangle BPC=\frac{1}{2}\sqrt{b}\sqrt{c}\sin u$ etc.). - -For better searchability, let me copy my MathLinks posts over here (finding some old post on MathLinks is almost impossible as for now). Note that I do not claim originality for the theorems. - -Theorem 1. Let $x$, $y$, $z$ be three real numbers and $A$, $B$, $C$ three real angles such that $A + B + C = 180^{\circ}$. Then, -$x^2+y^2+z^2\geq 2yz\cos A+2zx\cos B+2xy\cos C$. -Proof of Theorem 1. We will denote by $\measuredangle\left(\overrightarrow{p};\;\overrightarrow{q}\right)$ the directed angle between two vectors $\overrightarrow{p}$ and $\overrightarrow{q}$ (note that this is a directed angle modulo $360^{\circ}$). -For any two vectors $\overrightarrow{p}$ and $\overrightarrow{q}$, we are going to denote by $\overrightarrow{p}\cdot\overrightarrow{q}$ the scalar product of the vectors $\overrightarrow{p}$ and $\overrightarrow{q}$. -For any vector $\overrightarrow{p}$, we are going to denote by $\overrightarrow{p}^2$ the scalar product $\overrightarrow{p}\cdot\overrightarrow{p}$. Every vector $\overrightarrow{p}$ satisfies $\overrightarrow{p}^2=\left|\left|\overrightarrow{p}\right|\right|^2\geq 0$. -Let $\overrightarrow{a}$ be a vector of unit length. Let $\overrightarrow{b}$ be a vector of unit length such that $\measuredangle\left(\overrightarrow{a};\;\overrightarrow{b}\right)=180^{\circ}-C$. Let $\overrightarrow{c}$ be a vector of unit length such that $\measuredangle\left(\overrightarrow{b};\;\overrightarrow{c}\right)=180^{\circ}-A$. Then, -$\measuredangle\left(\overrightarrow{c};\;\overrightarrow{a}\right)=360^{\circ}-\measuredangle\left(\overrightarrow{a};\;\overrightarrow{b}\right)-\measuredangle\left(\overrightarrow{b};\;\overrightarrow{c}\right)$ -$=360^{\circ}-\left(180^{\circ}-C\right)-\left(180^{\circ}-A\right)=C+A=180^{\circ}-B$ -(since $A + B + C = 180^\circ$). -Now, all the vectors $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ have unit length: $\left|\overrightarrow{a}\right|=\left|\overrightarrow{b}\right|=\left|\overrightarrow{c}\right|=1$. Thus, $\measuredangle\left(\overrightarrow{b};\;\overrightarrow{c}\right)=180^{\circ}-A$ yields -$\overrightarrow{b}\cdot\overrightarrow{c}=\left|\overrightarrow{b}\right|\cdot\left|\overrightarrow{c}\right|\cdot\cos\measuredangle\left(\overrightarrow{b};\;\overrightarrow{c}\right)=1\cdot 1\cdot\cos\left(180^{\circ}-A\right)=\cos\left(180^{\circ}-A\right)$ -$=-\cos A$. -Similarly, we obtain $\overrightarrow{c}\cdot\overrightarrow{a}=-\cos B$ and $\overrightarrow{a}\cdot\overrightarrow{b}=-\cos C$. Thus, -$\left(x\cdot\overrightarrow{a}+y\cdot\overrightarrow{b}+z\cdot\overrightarrow{c}\right)^2$ -$=\left(x\cdot\overrightarrow{a}\right)^2+\left(y\cdot\overrightarrow{b}\right)^2+\left(z\cdot\overrightarrow{c}\right)^2$ -${}+2\cdot y\cdot\overrightarrow{b}\cdot z\cdot\overrightarrow{c}+2\cdot z\cdot\overrightarrow{c}\cdot x\cdot\overrightarrow{a}+2\cdot x\cdot\overrightarrow{a}\cdot y\cdot\overrightarrow{b}$ -$=x^2\underbrace{\cdot\left|\overrightarrow{a}\right|^2}_{=1^2}+y^2\cdot\underbrace{\left|\overrightarrow{b}\right|^2}_{=1^2}+z^2\cdot\underbrace{\left|\overrightarrow{c}\right|^2}_{=1^2}+2yz\cdot\underbrace{\overrightarrow{b}\cdot\overrightarrow{c}}_{=-\cos A}+2zx\cdot\underbrace{\overrightarrow{c}\cdot\overrightarrow{a}}_{=-\cos B}+2xy\cdot\underbrace{\overrightarrow{a}\cdot\overrightarrow{b}}_{=-\cos C}$ -$=x^2\cdot 1^2+y^2\cdot 1^2+z^2\cdot 1^2+2yz\cdot\left(-\cos A\right)+2zx\cdot\left(-\cos B\right)+2xy\cdot\left(-\cos C\right)$ -$=x^2+y^2+z^2-2yz\cos A-2zx\cos B-2xy\cos C$. -Since we, obviously, have $\left(x\cdot\overrightarrow{a}+y\cdot\overrightarrow{b}+z\cdot\overrightarrow{c}\right)^2\geq 0$, we thus get $x^2+y^2+z^2-2yz\cos A-2zx\cos B-2xy\cos C\geq 0$, so that $x^2+y^2+z^2\geq 2yz\cos A+2zx\cos B+2xy\cos C$, and Theorem 1 is proven. -Other proofs of Theorem 1 can be found at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5243 and http://www.artofproblemsolving.com/Forum/viewtopic.php?t=42509 . -Theorem 1 is equivalent to the following, also quite useful (for olympiad mathematics and magazine problem sections, that is, although I would not be surprised to see more applications) inequality: -Theorem 2. Let $x$, $y$, $z$ be three real numbers and $A$, $B$, $C$ three real angles such that $A + B + C$ is a multiple of $180^{\circ}$. Then, -$ \left(x + y + z\right)^{2}\geq 4\left(yz\sin^{2}A + zx\sin^{2}B + xy\sin^{2}C\right)$. -We will only show a proof of Theorem 2 using Theorem 1: First, we can WLOG assume that $A + B + C = 180^{\circ}$. This is because the inequality -$ \left(x + y + z\right)^{2}\geq 4\left(yz\sin^{2}A + zx\sin^{2}B + xy\sin^{2}C\right)$ -will not change if we add a multiple of 180° to one of the angles A, B and C (because $ \sin^{2}\left(180^{\circ} + u\right) = \sin^{2}u$ for every u), and consequently, since A + B + C is a multiple of 180°, we can add a multiple of 180° to the angle A such that, after this, we will have A + B + C = 180°. -Now, for A + B + C = 180°, we have -$ \left(180^{\circ} - 2A\right) + \left(180^{\circ} - 2B\right) + \left(180^{\circ} - 2C\right) = 540^{\circ} - 2\cdot\left(A + B + C\right)$ -$ = 540^{\circ} - 2\cdot 180^{\circ} = 180^{\circ}$. -Hence, Theorem 1 (applied to $ 180^{\circ}-2A$, $ 180^{\circ}-2B$, $ 180^{\circ}-2C$ instead of $ A$, $ B$, $ C$) yields -$ x^{2} + y^{2} + z^{2}\geq 2yz\cos\left(180^{\circ} - 2A\right) + 2zx\cos\left(180^{\circ} - 2B\right) + 2xy\cos\left(180^{\circ} - 2C\right)$. -Since $ \cos\left(180^{\circ} - 2A\right) = - \cos\left(2A\right) = - \left(1 - 2\sin^{2}A\right) = 2\sin^{2}A - 1$ and similarly $ \cos\left(180^{\circ} - 2B\right) = 2\sin^{2}B - 1$ and $ \cos\left(180^{\circ} - 2C\right) = 2\sin^{2}C - 1$, this becomes -$ x^{2} + y^{2} + z^{2}\geq 2yz\left(2\sin^{2}A - 1\right) + 2zx\left(2\sin^{2}B - 1\right) + 2xy\left(2\sin^{2}C - 1\right)$ -$ \Longleftrightarrow\ \ \ \ \ x^{2} + y^{2} + z^{2}\geq 4\left(yz\sin^{2}A + zx\sin^{2}B + xy\sin^{2}C\right) - \left(2yz + 2zx + 2xy\right)$ -$ \Longleftrightarrow\ \ \ \ \ x^{2} + y^{2} + z^{2} + \left(2yz + 2zx + 2xy\right)\geq 4\left(yz\sin^{2}A + zx\sin^{2}B + xy\sin^{2}C\right)$ -$ \Longleftrightarrow\ \ \ \ \ \left(x + y + z\right)^{2}\geq 4\left(yz\sin^{2}A + zx\sin^{2}B + xy\sin^{2}C\right)$, -and Theorem 2 is proven. -Theorem 2 also trivially follows from http://www.mathlinks.ro/Forum/viewtopic.php?t=15558 and was also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=3849 ...<|endoftext|> -TITLE: Lie group operation and tangent vectors -QUESTION [6 upvotes]: Suppose we have two differentiable paths $\alpha$ and $\beta$ thru the identity of a Lie group $G$, $\alpha(0)=\beta(0)=e$ the identity element. Denote $\alpha\beta$ the path given by $\alpha\beta(t)=\alpha(t).\beta(t)$ where the dot denotes the group operation. We also have $\alpha\beta(0)=e$. -The paths are differentiable, we can take the derivative, giving tangent vectors $\alpha'(0), \beta'(0), (\alpha\beta)'(0),$ which are all elements of $T_e G =\mathfrak{g},$ the lie algebra. Question: Is is true that $(\alpha\beta)'(0)=\alpha'(0)+\beta'(0)$? If not in general, then under what additional condition is it true? -I know that it's true if $\alpha,\beta$ are 1-parameter subgroups commuting with each other, since in that case $\exp(u+v)=\exp(u)\exp(v)$. -(This question arises when I try to study the tangent space of the space of representations from a fundamental group of a surface into a lie group.) - -REPLY [11 votes]: Here's another way to look at the problem. The derivative of a differentiable map at any point is a linear map of tangent spaces. We have five differentiable maps in play: - -The "pair of paths" map $(\alpha, \beta): (-\epsilon, \epsilon) \times (-\epsilon, \epsilon) \to G \times G$. -The multiplication map $m: G \times G \to G$. -The diagonal $\Delta: (-\epsilon, \epsilon) \to (-\epsilon, \epsilon) \times (-\epsilon, \epsilon)$. -(and 5.) The coordinate inclusions $i_1, i_2: (-\epsilon, \epsilon) \to (-\epsilon, \epsilon) \times (-\epsilon, \epsilon)$ - -We want to say that the derivative of $m \circ (\alpha, \beta) \circ \Delta$ is equal to the derivative of $m \circ (\alpha, \beta) \circ i_1$ plus the derivative of $m \circ (\alpha, \beta) \circ i_2$ (where the derivatives are evaluated at $0 \in (-\epsilon, \epsilon)$). By the chain rule, this follows from the fact that the derivative of $\Delta$ is the sum of derivatives of $i_1$ and $i_2$. - -REPLY [6 votes]: Heres a pretty clean proof. Let $m:G \times G \to G$ denote the multiplication map. Then we have (identifying $T_{e,e} G\times G$ with $T_e G \oplus T_e G$) -$$ -m_*(\alpha'(0), 0) = \frac{d}{dt}\vert_{t=0} m(\alpha, e) = \alpha'(0). -$$ -The same thing shows that $m*(0,\beta'(0)) = \beta'(0)$. By linearity we get $m_*(\alpha'(0),\beta'(0)) = \alpha'(0) + \beta'(0)$.<|endoftext|> -TITLE: complexity of eigenvalue decomposition -QUESTION [20 upvotes]: what is the computational complexity of eigenvalue decomposition for a unitary matrix? -is O(n^3) a correct answer? - -REPLY [2 votes]: I think the other answers are wrong. I periodically look up this problem and I believe it to be open. I will summarize my opinion: - -The symmetric eigenvalue problems is "solved". Wilkinson was able to prove that the QR iteration, with his own special shift strategy, converges cubically. See this for example. -The nonsymmetric eigenvalue problem is still open. The chapter on that subject in Golub and Van Loan says has a discussion on how the Wilkinson shift fails on some nonsymmetric matrices. They also mention that Wilkinson's ad-hoc shift should not be taken "too seriously" and that really it only gives the QR iteration a fresh start and a chance at better convergence. - -As far as I can tell, nobody knows the computational complexity of the approximate eigenvalue problem. -Edit: I stand corrected. Thanks Suvrit.<|endoftext|> -TITLE: reference for p-local and p-complete integers -QUESTION [6 upvotes]: Can anyone suggest a good thorough reference for $p$-localization and $p$-completion of the integers? I'm an algebraic topologist who's found himself washed up without any intuition. -In particular, here are the two questions I'm trying to answer right now: -Question 1: what is $\mathbb{Q}/\mathbb{Z}_p$ tensored with itself? -I mean $\mathbb{Z}_p$ to be the $p$-local integers. The tensor is over $\mathbb{Z}_p$ or $\mathbb{Z}$ - the answer should be the same. -I know $\mathbb{Q}/\mathbb{Z}_p$ is an injective $\mathbb{Z}_p$-module, and $\mathbb{Z}_p$ is a nice local ring, so maybe something can be said about a flat resolution of $\mathbb{Q}/\mathbb{Z}_p$? -Question 2: What is known about the cokernel of the map $\mathbb{Z}_p \rightarrow \mathbb{\hat{Z}}_p$? -I mean the map that $p$-completes the $p$-local integers. I think the cokernel is a rational vector space, but of finite or infinite dimension? Does it have any nice description? - -REPLY [7 votes]: I'm going to switch to a more standard notation, where the localization is $\mathbb{Z}_{(p)}$ and the completion is $\mathbb{Z}_p$. -For your first question, the general rule that concerns us is that the tensor product of a $p$-divisible group with a $p$-torsion group is zero. The reason is that if $a$ is a $p^n$th power and $p^n b = 0$, then $a \otimes b = p^{-n}a \otimes p^n b = 0$. Since $\mathbb{Q}/\mathbb{Z}_{(p)}$ is both $p$-divisible and $p$-torsion, its tensor square is zero. -The cokernel of completion is a $\mathbb{Q}$-vector space, because it admits a $\mathbb{Z}_{(p)}$-action by multiplication, and it is uniquely $p$-divisible (Proof: lift any element to $\mathbb{Z}_p$, subtract the constant term of its $p$-adic expansion, divide by $p$, and check that different lifts yield the same answer). Its dimension over $\mathbb{Q}$ is the cardinality of the continuum, since that is its cardinality as a set. -I think most number theory texts have some discussion of $p$-adic numbers, and Gouvea even wrote a whole book about them.<|endoftext|> -TITLE: Efficient visibility blockers in Polya's orchard problem -QUESTION [13 upvotes]: Polya's orchard problem asks for which radius $\rho$ of trees at each lattice point within a distance $R$ of the origin block all lines of sight to the exterior of the orchard. -         - - -It has been established that rays to infinity are completely blocked iff -$\rho \ge 1/\sqrt{R^2 + 1}$, when $R$ is an integer. -(T.T. Allen, "Polya's orchard problem," -The American Mathematical Monthly -93(2): 98-104 (1986).) -The above shows a quarter of an orchard with $R=6$, $\rho=1/\sqrt{37}=0.164$, -and some random rays. -I am wondering if disks are the most efficient blockers -in terms of area. More precisely: - -For a given $R$, is there a centrally symmetric convex body $K$ - of area less than $\pi \rho^2$ which when translated to all - lattice points within distance $R$ of the origin, block all rays from - the origin to the outside? - -My guess is that the answer is Yes, in which case it would -be interesting to know the shape of the area-optimal blockers. -In particular, are they polygons? -The same question may be posed in $\mathbb{R}^d$: are they polytopes? -Edit. Here is the chord construction for $R=2$ from the first paragraph of Douglas's construction, as I understand it: - -REPLY [5 votes]: Polya's result can't be improved asymptotically. If we take $\rho\sim\frac cR$ with $c<1$ then positive ratio of rays will not be blocked (as $R\to \infty$). This ratio as a function of $c$ can be calculated explicitly, see for example The Statistics of Particle Trajectories in the Homogeneous -Sinai Problem for a Two-Dimensional Lattice. The lattice $\mathbb{Z}^2$ asymptotically isotropic: this ratio does not depend on angles. In partiqular it means that circles asymptotically are best possible blockers: if in some direction the blocker has smaller width (at least from one side from the center; in this direction from one side the blocker looks like a circle of radius $\rho=cR$ with $c<1$) then the ratio of unblocked rays in this direction will be bigger. -For a fixed $R$ best possible blocker can be founf explicitly. As in original proof we can take all integer points inside given circle and consider all rays passing through this points. If two adjacent rays pass through $A_1=(x_1,y_1)$ and $A_2=(x_2,y_2)$ then blockers at this points must block the ray through $A_3=(x_1+x_2,y_1+y_2)$ (which is outside given circle). Bases of the altitudes from $A_1$ and $A_2$ on $OA_3$ must belong to blockers. It means that optimal blocker is convex hull of all such bases.<|endoftext|> -TITLE: Philosophical Question related to Largest Known Primes -QUESTION [23 upvotes]: The other day while discussing math, and primes specifically, the following question came to mind, and I figured I'd ask it here to see what people's opinions on it might be. - - - Main Question: Suppose that tomorrow someone proves that some function always generates (concrete) primes for any input. How should this affect lists such as the Largest Known Primes? - - -Let me give a little more detail to demonstrate why I feel this question is not entirely trivial or fanciful. -Firstly, the requirement that the function be able to concretely generate primes is meant to avoid 'stupid' examples such as Nextprime(n) which, given a 'largest known prime' P, yields a larger prime Nextprime(P). Note however that the definition of Nextprime does not actually explicitly state what this prime is, any implementation of it (in Maple or Mathematica for example) simply loops through the integers bigger than the input, testing each for primality in some fashion. -On the other hand, one candidate for such a function might be the Catalan sequence defined by: -$C(0) = 2$, $C(n+1) = 2^{C(n)}-1$ -Although $C(5) = 2^{170141183460469231731687303715884105727}-1$ is far too large to test by current methods (with rougly $10^{30}$ times as many digits as the current largest known prime), and although the current consensus is that $C(5)$ is likely composite, it does not seem entirely out of the realm of possibility that someone might eventually find some very clever way of showing $C(5)$ is prime, or even that $C(n)$ is always prime, or perhaps some other concretely defined sequence. -The point is this: once one knows that every element of a sequence is prime, does this entirely negate things like the list of largest known primes? Or does the fact that $C(n)$ for $n\geq 5$ has too many digits to ever calculate all of them (instead only being able to calculate the first few or last few digits) mean that even if they were somehow proven prime it would not technically be 'known'? -Note also that in the realm of finite simple groups the analogous question is already tough to decide since there are infinite families of such groups known, but concrete descriptions (such as generators and relations or character tables) are not always available or even computable within reasonable time constraints. Likewise one could pose analogous questions in other branches (largest volume manifolds with certain constraints, etc.) -Anyhow, it seems like a reasonable question for serious mathematicians to consider, so I just want to hear what other's opinions are on the subject (and if anyone can think of a better title, feel free to suggest). - -REPLY [10 votes]: I think you are mis-understanding the purpose of the "largest prime" list. There are areas of mathematics in which we genuinely do not know how to generate or catalogue the objects of certain types. Prime numbers do not fall into this category. -Rather, these lists basically exist as a benchmark for number-theory computer algorithms. For example, if you develop a new supercomputer, you can prove its prowess by testing primality of some bigger numbers not on the list.<|endoftext|> -TITLE: Formal Schemes Mittag Leffler -QUESTION [17 upvotes]: Here is a question that I'm just copying from Math Stack Exchange that I asked awhile ago. It has just been sitting there unanswered, and although I haven't really thought about it since I posted it, I'm still very interested in a nice example if it exists. -Suppose $(\mathfrak{X}, \mathcal{O}_\mathfrak{X})$ is a Noetherian formal scheme and let $\mathcal{I}$ be an ideal of definition. Then we have a system of schemes $X_n=(|\mathfrak{X}|, \mathcal{O}_\mathfrak{X}/\mathcal{I}^n)$. -If the inverse system $\Gamma (X_n, \mathcal{O}_{X_n})\to \Gamma (X_{n-1},\mathcal{O}_{X_{n-1}})$ satisfies the Mittag-Leffler condition (the images eventually stabilize), then we get some particularly nice properties such as $Pic(\mathfrak{X})=\lim Pic(X_n)$. -More generally, we don't have to be worried about converting between thinking about coherent sheaves on the formal scheme and thinking about them as compatible systems of coherent sheaves on actual schemes. -My question: -Is there a known example of a formal scheme for which that system of global sections does not satisfy the Mittag-Leffler condition? -One thing to note is that it can't be affine (the maps are all surjective) or projective (finite dimensionality forces the images to stabilize). -A subquestion is whether or not there is a general reason to believe such an example exists. People I talk to usually say things along the lines of: you definitely have to be careful here because in principle this could happen. But no one seems to have ever thought up an example. -Lastly (still related...I think), is there a known example where you can't think of coherent (or maybe invertible) sheaves as systems because the two aren't the same? - -REPLY [3 votes]: It would take me too much effort to understand and write up the things below in algebraic geometry language because I am from analysis -- but this is where the Mittag-Leffler condition comes from and, originally, it is a broader condition than the one I have seen in books on homological algebra where it is required that the images of an inverse (or projective) spectrum stabilize in the sense that $f_m^n(X_m)=f_k^n(X_k)$ for all $k\ge m$ and $m=m(n)$ sufficiently large (here $f_m^n: X_m \to X_n$ are the linking maps of the inverse spectrum). -This condition is sufficient but not necessary for the vanishing of the derived inverse limit functor (and I suspect that this is very close to what you are interested in). Palamodov constructed this derivative by using injective resolutions but for spectra of abelian groups you can define it ad hoc as the cokernel of the map $\prod X_n \to \prod X_n$, $(x_n)_n \mapsto (x_n- f_{n+1}^n(x_{n+1}))_n$. -In Analysis, the $X_n$ are often complete metric groups and the sufficient condition we use is that $f_m^n(X_m)$ is contained in the closure of $f_k^n(X_k)$ (only for the discrete topologies this coincides with the "algebraic ML-condition). If $X_n$ are even locally convex Frechet spaces than we even have a characacterization for vanishing of the derived inverse limit functor: For every $n$ and every continuous seminorm $p$ on $X_n$ we have that $f_m^n(X_m)$ is contained in the closure of $f_k^n(X_k)$ with respect to $p$. -The situation where Mittag-Leffler originally used such ideas is for an exhaustion of an open set $\Omega\subseteq \mathbb C$ by open and relatively compact sets $\Omega_n$ - and $X_n=H(\Omega_n)$, the space of holomorphic functions on $\Omega$ with the topology of uniform convergence on compact sets. Runge's approximation theorem then implies that the "analytic ML-condition" is satisfied whereas the algebraic is not.<|endoftext|> -TITLE: Positive-Definite Functions and Fourier Transforms -QUESTION [14 upvotes]: Bochner's theorem states that a positive definite function is the Fourier transform of a finite Borel measure. As well, an easy converse of this is that a Fourier transform must be positive definite. -My question is: is there a high-brow explanation for why positive definiteness and Fourier transforms go hand-in-hand? -As I understand it, positive definiteness imposes wonderfully strong regularity conditions on the function. We immediately deduce that the function is bounded above at its value at 0, that it is non-negative at 0 and that continuity at 0 implies continuity everywhere. -A leading example I have in mind comes from probability. One can show (Levy's Theorem) that a sum of iid rv converges weakly to some probability distribution by considering the product of characteristic functions and showing that its tail converges to 1 around an interval containing 0, so by positive definiteness and by the identity $1-\mbox{Re} \phi(2t) \leq 4(1-\mbox{Re} \phi(t))$ this implies convergence to a degenerate distribution. It just seems rather mysterious to me how this kind of local regularity becomes global. -Edit: -To be a little more specific, I understand that the Radon Nikodym derivative is positive and $e^{ix}$ is positive definite. I am more interested in consequences of positive-definiteness on the regularity of the function. For example, if one takes the 2x2 positive definite matrix associated with the function and considers its determinant, it follows that $|f(x)|\leq |f(0)|$. If I take the 3x3 positive definite matrix, I can conclude that if $f$ is continuous at 0, it is then continuous everywhere. My issue is that these types of arguments give me no intuition at all as to what positive definiteness is. -Let me thus add an additional question: what is it about positive definiteness that adds such regularity conditions? - -REPLY [23 votes]: Perhaps the phenomenon you are asking about is: why is the definition of a positive-definite function natural? -One answer is that positive-definite functions are exactly coefficients of group representations, in the following sense. If $\pi : \mathbb{R}\to U(H)$ is a unitary representation of $\mathbb{R}$ on some Hilbert space $H$, and $h\in H$ is a vector, then the function $$t\mapsto \langle \pi (t) h, h\rangle$$ is positive-definite. Conversely, given a positive-definite function $\phi$, there exists a Hilbert space $H$, a vector $h\in H$ and a unitary representation $\pi$ of $\mathbb{R}$ on $H$, for which $\phi(t)=\langle \pi(t)h,h\rangle$. -Indeed, the $n\times n$ matrix occurring in the definition of a positive definite function is nothing more than the Gramm matrix of inner products $\langle \pi (t_i) h, \pi (t_j) h\rangle$; and positivity of this matrix is just a reflection of the fact that the inner product of $H$, restricted to the linear span of $\pi(t_i)h: i=1,\dots,n$ is positive-definite. -The Fourier transform goes from the functions on the group to functions on the space of irreducible unitary representations of the group, and thus switches positivity and complete positivity.<|endoftext|> -TITLE: Finite outer automorphism groups -QUESTION [11 upvotes]: It is a theorem of Gopal Prasad (which I hope I am not misquoting...) that lattices in higher rank linear semi-simple Lie groups have finite outer automorphism groups. Is there some other reasonable class of groups with this finiteness property? - -REPLY [4 votes]: The theorem Igor Belegradek mentions is a more general form of the theorem (due collectively to Bestvina, Feighn, Paulin, and Rips) that a one-ended hyperbolic group has infinite outer automorphism group only if it splits over a two-ended subgroup. See -M. Bestvina, M. Feighn, -Stable actions of groups on real trees. -Invent. Math. 121 (1995), no. 2, 287-321.<|endoftext|> -TITLE: Why is the Arthur trace formula so powerful? -QUESTION [6 upvotes]: Considering the Arthur trace formula, why are the sort of convolution operators, whose "normalized traces" are given in geometric terms and spectral terms, actually able to distinguish all automorphic representations? Can they seperate the cuspidal representation? What about the rest? Both the local and global picture are interesting for me. -Isn't this question not the starting point, which justifies to study functioriality via trace formulas? - -REPLY [5 votes]: I am not sure if I understood your question well. I try to give an answer to the following, -"Why is it expected that the trace formula implies cases of Langlands functoriality?" -hoping that it is the right question. -Very roughly I think the idea is as follows. -Let $G, H$ be two connected reductive groups, and suppose given an L-morphism $LG$ to $LH$, where $LG$ (resp. $LH$) is the Langlands dual of $G$ (resp. $H$), with certain properties. Then one expects (conjecture) to be able to relate automorphic representations of G to automorphic representations of $H$. -With the trace formula one attempts to prove (cases of) this conjecture. -To simplify, assume that the groups are anisotropic. -Let $\pi$ be a cuspdial automorphic representation of G. Assume now, that we are in the following very fortunate situation. Suppose that we have a function g on G(adeles) and h on H(adeles) which have the following properties. -(0) For convenience only: assume that the functions are elementary tensors $g = \bigotimes_v g_v$, and $h = \bigotimes_v h_v$ ($v$ ranges over the places of $Q$). -(1) The trace of the function g acting on an automorphic representation of G is zero unless it is isomorphic to $\pi$. -(2) The trace of the function h acting on an automorphic representation of H is zero unless it is isomorphic to a certain fixed representation $\pi'$. -Unfortunately I do not know how to formulate the third, and most important condition in a precise manner in the above generality. But I will try to give an idea anyway. -(3) The pair of functions (g, h) is associated. This means the following. From the map of L-groups one should be able to transfer conjugacy classes in $H(Q_v)$ to conjugacy classes in $G(Q_v)$. One then asks that the orbital integral of $h_v$ at a conjugacy class $c$ is equal to the orbital integral of $g_v$ at the $G(Q_V)$-conjugacy class obtained from $c$ by transfer. Moreover, the orbital integral at any $G(Q_v)$-conjugacy class which is not a transfer is demanded to be $0$. -Then, using (3) the geometric side of the trace formula of $G$ can be compared with the geometric side of the trace formula for $H$. So applying the trace formula for $G$ and $H$ at the same time, we see that the spectral side of the trace formula for $G$ is equal to the spectral side of the trace formula for $H$. But, by (1) and (2), on these spectral sides only the representations $\pi$, $\pi'$ remain. -The point is then that $\pi \rightarrow \pi'$ is the ''candidate'' for the automorphic representation predicted by Langlands functoriality. -There are many problems with the above. We already saw that it is not clear how to define (3) in general. Moreover, (1) and (2) are often not possible, only for finite sets called packets. Next, there is a problem with non-stable conjugacy; in an algebraic group there are several notions of conjugacy, for example rational conjugacy (two elements are conjugate by an element of G(Q)) and conjugacy over the algebraic closure (two elements are conjugate by an element of G(Q)). There is also a third notion, "stable conjugacy", and one has to work with this one if one wants to have a chance to give a meaning to (3). This also implies that you have to work with a different trace formula instead, "stable trace formula" ... complicated! -An example of (proved) Langlands functoriality is Jacquet-Langlands. One then knows how to give a precise definition of (3), because in this case G and H are inner forms. In my opinion this is explained very well in section 2 of -www.institut.math.jussieu.fr/projets/fa/bpFiles/Intro_Harris.pdf -Another case where (3) is defined is when $H$ is an endoscopic group. This is also explained in the Paris book project by Harris, -www.institut.math.jussieu.fr/projets<|endoftext|> -TITLE: Why is the functional equation of the Riemann zeta function equivalent to the Poisson summation formula? -QUESTION [15 upvotes]: We can derive from the Poisson summation formula the modularity of the Theta function, which results in the functional equation. In his book on the Riemann Zeta function, Patterson mentions also that there is a way back. How can this be done? -Equivalently, how can the modularity of the Theta function (which is Poisson summation for a one parameter family of functions only) be used to derive the Poisson summation formula for general test functions? - -REPLY [9 votes]: The theorem you are looking for is Theorem 10.2.17 in Henri Cohen's book Number Theory: Analytic and Modern Tools (Google Books Link).<|endoftext|> -TITLE: Resources for mathematics advising. -QUESTION [157 upvotes]: This question is possibly ill-advised. (If it is not right for this site I will delete it.) -I, suddenly, have students. -It is very clear to me that there is nothing in my education that has prepared me for the task of training graduate students. -Yes, I know that graduate school is the place where one finally assumes full responsibility for one's own mathematical progress. It is also equally clear to me that there are innumerable things that an advisor might do, unwittingly, to irrevocably damage the career of their own student. This is keeping me up at night. And unlike searching for advice on, say, parenting, it appears that most people keep their opinions on the process to themselves, especially with respect to issues specific to training mathematicians. The more senior people I have approached have generally told me that "things work themselves out". -I see people I know, not so much younger than me, for whom the job market is not working itself out. -I was very lucky, and as a result I have many questions about things I didn't deal with myself. I don't know how to strike the balance between a doable research project and a significant one. I don't know how to help students move from reading background into exploring on their own. I don't know when and how much to help when they are struggling, or what to say when they become unhappy about their progress. -And I don't know where to find resources to do so. As I've said, sometimes I don't know that people take my concerns seriously... my own mentors deal with students at an n'th rate university, rather than an 8n'th. -Any direction would be appreciated. -(This question is anonymous, but not for my own sake.) - -REPLY [3 votes]: Section 2B of Indiana University's How to be a Good Grad Student is Advice for Advisors. The first half of this is very generic, but there are good specific bits of advice in the second part. Also, I think giving your student booklets like How to be a Good Grad Student would probably also help them. -Since the OP mentions that he's new to advising, perhaps The Assistant Professor's Guide to the Galaxy would be helpful. Despite being a guide to the galaxy, it's only 6 pages long, and only section 7 has to do with advising. This section focuses more on what grad students can do for you, but does have one concrete suggestion: - -In my personal work with students, I set goals for them and insist that they document their progress with draft manuscripts. My work with them on these drafts often leads to conference papers. My students always publish before they finish, sometimes jointly with me and sometimes on their own, depending on the degree of my own involvement.<|endoftext|> -TITLE: How "much" does (Grigorieff) forcing destroy an ultrafilter? -QUESTION [20 upvotes]: Introduction. I recently revisited Shelah's model without P-points and I was wondering how "badly" Grigorieff forcing destroys ultrafilters, i.e., what kind of properties can survive the destruction of the "ultra"ness. -An example. Given a free (ultra)filter $F$ on $\omega$, Grigorieff forcing is defined as -$$ G(F) := \{ f:X \rightarrow 2: \omega \setminus X \in F \},$$ -partially ordered by reverse inclusion. A simple density argument shows that "$G(F)$ destroys $F$", i.e., the filter generated by $F$ in a generic extension is not an ultrafilter (the generic real being the culprit). -Of course, there are many forcing notions that specifically destroy ultrafilters (also, Bartoszynski, Judah and Shelah showed that whenever there's a new real in the extension, some ground model ultrafilter was destroyed). -My question is: -If $F$ is destroyed, how far away is $F$ from being the ultrafilter it once was? -Maybe a more positive version: Which properties of $F$ can we destroy while preserving others? -This might seem awfully vague, so before you vote to close let me explain what kind of answers I'm hoping for. - -Positive answers. - -If the forcing is $\omega^\omega$-bounding and $F$ is rapid, then $F$ will still be rapid. That's a very clean and simple preservation. -In Shelah's model without P-points, all ground model Ramsey ultrafilters stop being P-points but "remain" Q-points. - -"Minimal" answers. Is it possible that $F$ together with the generic real generates an ultrafilter, i.e., there are only two ultrafilters extending $F$? For Grigorieff forcing, I'd expect this needs at least a Ramsey ultrafilter. But maybe other forcings have this property? -Negative answers. Say $F$ is a P-point; can $F$ still be extended to a P-point? Shelah tells us that forcing with the full product $G(F)^\omega$ denies this. Is it known whether $G(F)$ already denies this? Do other forcing notions allow this? - -I know there is a lot of literature on preserving ultrafilters (mostly P-points, I think) but I'm more interested in the case where the ultrafilter is actually destroyed. But I'd welcome anything that sheds light on this. -PS: community wiki, of course. - -REPLY [3 votes]: I wanted to add two comments that I received in 'meatspace'. I hope this isn't too inappropriate. - -If $F$ is a P-filter, and the forcing is proper, than $F$ generates a P-filter in the extension. -If $F$ is a Q-filter, i.e., every finite-to-one map becomes injective on a set in $F$, and the forcing is $\omega^\omega$-bounding, then $F$ generates a Q-filter in the extension. - -Proofs of these facts can be found, e.g., in Shelah, Proper and Improper Forcing, Chapter VI, Section 4 and 5 resp. -One more example from myself. - -If $F$ is an idempotent filter then $F$ will remain an idempotent filter in any forcing extension. In particular, if $F$ is an idempotent ultrafilter, it will still extend to an idempotent ultrafilter.<|endoftext|> -TITLE: maximal tori cover compact Lie group -QUESTION [5 upvotes]: Let $G$ be a compact connected Lie group, $T$ be some maximal torus in $G$ (that is, inclusion-maximal connected abelian subgroup). Then the union of tori $gTg^{-1}$, $g\in G$, is the whole $G$. This is well-known (4.21 in Adams book). My question is rather methodological: is there any proof without use of algebraic topology? Adams presents A. Weil's proof, which uses some kind of Lefschetz fixed points theorem. -(Yes, I am sorry but my motivation is mostly that I teach second year students this stuff.) - -REPLY [3 votes]: There are many ways of proving this, using all sorts of different methods. In the second edition my book "Lie groups, Lie algebras, and representations" (following Brocker and tom Dieck) I use the mapping degree theorem. If we fix a single maximal torus $T$ and consider the conjugation map $\Phi:T \times (K/T) \rightarrow K$ be given by $\Phi(t,[x])=xtx^{-1}$. If we can show that this map has nonzero mapping degree, we can conclude it is surjective, which is just what we are trying to show. In Section 11.5, I show that $\Phi$ has mapping degree equal to the order of the Weyl group. This approach also gives a proof of the Weyl integral formula by the same computation.<|endoftext|> -TITLE: Is there a useful generalization of the Schmidt decomposition to the tensoring together of 3 or more vector spaces? -QUESTION [9 upvotes]: I've rewritten the question in math notation, and I've left the old version in physics bra-ket notation here. -Background -A simple consequence of the singular value decomposition is that any vector $v$ in a vector space $V$ formed by the tensor product of two smaller spaces ("subsystems") $U$ and $W$ of dimension $d_U$ and $d_W$, -$v \in V = U \otimes W$, -has a special decomposition in terms of rank-one tensors (aka product states) -$v = \sum_{i=1}^d \lambda_i u_i \otimes w_i $, $\qquad u_i \in U$, $\qquad w_i \in W$, $\qquad d = \mathrm{min}(d_U,d_W)$ -built from the fixed orthonormal bases $\{ u_i \}$ and $\{ w_i \}$. -This decomposition has many nice properties, and it's simple to see that there's no way to generalize it to the case of 3 or more subsystems while keeping all of them. For instance, a generic state in $V = \bigotimes_{n=1}^N V_n$ cannot be expressed as a sum of fewer than $\tilde{d}$ simple tensors, with $\tilde{d} \gg d_n = \mathrm{dim}V_n$ for all $n=1,\ldots,N$. -Vague Question -Is there a useful/natural/canonical decomposition of a vector into a "small" number of orthogonal simple tensors for the case of 3 or more subsystems? At the very least, the GHZ state should take it's canonical form in such a decompostion: -$v_{GHZ} = a_1 \otimes \cdots \otimes a_N + b_1 \otimes \cdots \otimes b_N$, $\qquad a_n, b_n \in V_n$, $\qquad \langle a_n ; b_n \rangle = 0$ for all $n$. -Specific Question -If we guess that the entropy function -$H[\{ p_i \} ] = -\sum_i p_i \ln p_i$ -is appropriate, we can define the "minimum-entropy product-state decomposition" to be the decomposition -$v = \sum_{i=1}^{\tilde{d}} \lambda_i v_i$, $\qquad v_i = \bigotimes_{n=1}^N \psi_i^n$, $\qquad \psi_i^n \in V_n$ -with the minimum value for the entropy $H[\{ p_i = \lambda_i^2 \} ]$, under the condition that the $\{ v_i \}$ are orthonormal. Note that we allow $\langle \psi_i^n ; \psi_j^n \rangle \neq 0$ for $i \neq j$. -The natural questions to ask are: Is this decomposition generically unique (i.e. unique except for some set of measure zero in the global vector space)? Is it continuous? (What other properties should this decomposition have to satisfy to be useful?) -I am 99% sure that in the case of $N=2$, this reduces to the Schmidt decomposition and that the answers to both questions is "yes". -Is any of this sensitive to our choice of the entropy function $H$, as opposed to some other permutation-invariant and majoritization-preserving function of the spectrum? - -REPLY [2 votes]: There is some recent work on tensors (and also the special cases of symmetric and alternating tensors) that admit orthogonal or unitary decompositions (resp., symmetric or alternating decompositions). I will point out, particularly, the following three papers by Elina Robeva and coauthors: - -https://arxiv.org/abs/1409.6685 -https://arxiv.org/abs/1512.08031 -https://arxiv.org/abs/1603.09004<|endoftext|> -TITLE: Non-cohomological proof that the pullback of an ample bundle by a finite morphism is ample -QUESTION [12 upvotes]: It is a standard fact that for any finite morphism of proper Noetherian $A$-schemes ($A$ being Noetherian), the pullback of an ample line bundle is ample. The usual proof of this fact is via Serre's cohomological criterion for ampleness. However, since the statement seems, on its face, to have nothing to do with cohomology, I thought the following question worth asking: -Does anyone know a reasonable proof of this fact that does not go through cohomology? - -REPLY [4 votes]: This has nothinging to do with finiteness. Let $f:X\to Y$ be affine and $L$ a line bundle on $Y$. Then the subschemes $Y_s$ for $s \in \Gamma(Y, L^n)$ pull back to subschemes $X_{f^*s}$. Hence $L$ is ample $\implies$ $Y$ is covered by finitely many affine $Y_s$ $\implies$ -$X$ is covered by finitely many affine $X_{f^*s}$ $\implies$ $f^*L$ is ample.<|endoftext|> -TITLE: Cartan 3-form on a Lie group G -QUESTION [11 upvotes]: Does anyone have a reference to learn more about the Cartan $3$-form on a group manifold $G$? I have read that the WZW term is nothing more than the integral of the pullback of the Cartan $3$-form via $g:W\rightarrow G$ -$WZW = -\frac{1}{6}\int_W \langle \phi_g\wedge[\phi_g\wedge\phi_g]\rangle$, -where $\phi_g=g^\ast(\phi)$ is the pullback of the Maurer-Cartan form, and would like to learn more about the math behind WZW terms. For eg., why is it the generator of $H^3(G,\mathbb{R})$ when $G$ is a connected, simply connected, compact Lie group? - -REPLY [2 votes]: You should check out Theodore Frankel's Geometry of Physics. There is a section on the gauge group . The cartan three form is the three form from the Chern Simons term at a flat connection. A flat connection is when the potential A is equal to the $g^{-1}dg$. In the Chern Simons 3 form you have A^A^A. At pure gauge this becomes $Tr(g^{-1}dg\wedge g^{-1}dg\wedge g^{-1}dg)$. This term is proportional to the cartan 3 form. I hope this helped.<|endoftext|> -TITLE: Matrices that are Hadamard products of $X$ and $X^{-T}$ -QUESTION [6 upvotes]: What are the matrices that you can write in the form $X \odot X^{-T}$, for a complex square matrix $X$, where $X^{-T}$ is the inverse of the complex transpose (not conjugate) and $\odot$ is the Hadamard (component-by-component) product? -In the $2\times 2$ case, you get the group of matrices in the form $$\begin{bmatrix}a & b\\\\ b & a \end{bmatrix},$$ with $a+b=1$, which are closed under matrix multiplication and would form a group were it not for the matrix $a=b=\frac12$ which admits no inverse [EDIT: corrected this assertion, thanks to Denis Serre]. In larger dimension, one sees that all the obtained matrices have the vector of all ones as both a right and left eigenvector. Is this the only restriction? Is the resulting set of matrices closed under multiplication? Is this problem known and studied? -Origin: motivated from this MO question. - -REPLY [3 votes]: There are some properties of this product in Horn and Johnson, "Topics in Matrix Analysis", Cambridge Univ Press 1991.<|endoftext|> -TITLE: Finitely generated Galois groups -QUESTION [16 upvotes]: It is well-known that for a given natural number $n$ there is only finite number of extensions of $\mathbb Q_p$ of degree $n$. This result appears in many introductory books on algebraic number theory. However, these books do not mention that this follows from the fact that the the absolute Galois group $G_{\mathbb Q_p}$ of $\mathbb Q_p$ is finitely generated. In fact, the structure of $G_{\mathbb Q_p}$ implies that there exists a constant $c$ such that the number of extensions of $\mathbb Q_p$ of degree $n$ is at most $c^n$. This is because $G_{\mathbb Q_p}$ is of exponential subgroup growth. So my first question is the following. -Question 1: Is there an alternative way to show that there exists a constant $c$ such that the number of extensions of $\mathbb Q_p$ of degree $n$ is at most $c^n$? -There is a method to show that if a profinite group $F$ does not have many subgroups, then it is finitely generated. For this consider $m_n(F)$ the number of maximal open subgroups of $F$ of index $n$. We say that $F$ is of polynomial maximal subgroup growth (PMSG) if there exists a constant $c$ such that $m_n(F)\le n^c$. We say that $F$ is positively finitely generated (PFG) if there exists $k$ such that $k$ random elements of $F$ generate $F$ with positive probability. In particular, if $F$ is PFG, then $F$ is finitely generated. A theorem of A. Mann shows that PMSG and PFG are equivalent. The group $G_{\mathbb Q_p}$ is prosoluble and finitely generated and so, by another result of A. Mann, $G_{\mathbb Q_p}$ is PMSG. So my second question is the following. -Question 2: Is there an alternative way to show that there exists a constant $c$ such that the number of minimal extensions of $\mathbb Q_p$ of degree $n$ is at most $ n^c$?(An extension $L/K$ is minimal if it does not contain proper subextensions) -The absolute Galois group of $\mathbb Q$ is not finitely generated. However we can look at the restricted ramification case. Let $S$ be a finite number of primes of $\mathbb Q$. Denote by $G_{\mathbb Q, S}$ the Galois group of the maximal extension of $\mathbb Q$ which is unramified outside the primes $S$. I believe that It is unknown whether this group is finitely generated. But what about the number of open subgroups of finite index? -Question 3: Is it true that for a given natural number $n$ there is only finite number of extensions of $\mathbb Q$ of degree $n$ unramified outside the primes $S$? -if the answer on the previous question is yes. -Question 4: Do you know any upper bounds for the number of extensions of $\mathbb Q$ of degree $n$ unramified outside the primes $S$ and for the number of minimal extensions of $\mathbb Q$ of degree $n$ unramified outside the primes $S$? -Class field theory implies that for a given natural number $n$ there are only finite number of soluble Galois extensions of $\mathbb Q$ of degree $n$ unramified outside the primes $S$. This suggest the following question. -Question 5: Is the maximal pro-soluble quotient of $G_{\mathbb Q, S}$ finitely generated? - -REPLY [4 votes]: Regarding the local question, you might be interested in John Jones' website http://hobbes.la.asu.edu/research.html. He has a database where he gives all sorts of data for all the extensions of $\mathbb{Q}_p$ of degree at least through $8$ (polynomials that give the extensions, etc.). His interactive table may go up to degree 12 now (this was the problem his most recent thesis student was working on). Anyway, he has a number of papers on the methods used to compute all the extensions of a given degree, and the references therein may be useful for answering your question (for example I remember finding the mass formula that Pete mentions as well as the Pauli and Roblot paper among them). The main paper about the database is here.<|endoftext|> -TITLE: What would you want on a Lie theory cheat poster? -QUESTION [78 upvotes]: For some long time now I've thought about making a poster-sized "cheat sheet" with all the data about Lie groups and their representations that I occasionally need to reference. It's a moving target, of course -- the more I learn, the more stuff I'd want to see on such a poster! -There are many obvious things I think everybody would agree should be there. The finite and affine Dynkin diagrams, with the Bourbaki numbering and also the coefficients of the simple roots in the highest root. The dimensions of the fundamental representations. Coordinate descriptions of each of the root systems, and of their Weyl groups. The exponents of each group, the Coxeter number, and the structure of the center. The exceptional isomorphisms of low-rank groups. -Only slightly less obvious: Satake diagrams. Dynkin's characterization of the nilpotents. Geometric descriptions of the partial flag manifolds $G/P$. The classification of real symmetric spaces. -One suggestion per answer, please, but otherwise, go wild! I'm not promising to actually make this thing in any timely manner, but would look to the votes on answers to prioritize what actually makes it onto the poster. - -REPLY [2 votes]: For each Dynkin diagram, the involution taking a fundamental weight to its dual. (The main issue being that it's trivial for $D_{even}$, and nontrivial for $D_{odd}$. Mnemonic: what could happen with $D_4$?)<|endoftext|> -TITLE: An elementary problem in Euclidean geometry -QUESTION [22 upvotes]: This problem was first put to me by Luke Pebody (who did not know the answer at the time) and after some work I am yet to find a proof or counterexample. I would be grateful of any insights. -Call a vector $v$ in $\mathbb{R}^2$ 'short' if it has modulus less than 1. Let $v_1,\dots,v_6$ be short vectors such that $\sum_{i=1}^6 v_i = 0$. Prove that some three of the $v_i$ have a short sum. - -REPLY [7 votes]: One word: AoPS - -REPLY [5 votes]: Lemma 1: -If the angle, $\theta$, between two 'short' vectors, $v_1$ and $v_2$ (placed head to tail) satisfies $\theta \leq \pi/3$, then their sum is a short vector. -Proof: -Place $v_1$ at the origin, then the terminal point of $v_1$ lies within the unit ball. Placing the $v_2$ at the tail of $v_1$ creates an angle $\theta \leq \pi/3$. Any arc of radius $r < 1$ traced between $\pi/3$ and $-\pi/3$ from the terminal point of $v_1$ lies within the unit ball as well. Thus $v_1+v_2$ is a 'short' vector. $\blacksquare$ -Since $\Sigma v_i = 0$ we can order the vectors in such a way as to create a convex polygon. If any of the interior angles satisfy the conditions for lemma 1, then we can reduce the problem to a polygon with fewer sides. -Lemma 2: -In a convex n-gon (n=4,5 or 6), with interior angles $\theta_i$, either $\theta_i \leq \pi/3$ for some $i$ or at least one pair of adjacent sides (as vectors) may be interchanged so that an angle less than $\pi/3$ is created. -Proof: -If $v_1$ and $v_2$ create interior angle $\alpha$ and $v_2$ and $v_3$ create angle $\beta$, then interchanging $v_2$ and $v_3$ creates an angle $\alpha+\beta-\pi$ between $v_1$ and $v_3$ (which are now adjacent by exchanging $v_2$ and $v_3$). Aiming for a contradiction, assume that this new angle, $\alpha+\beta-\pi > \pi/3$. Thus, $\alpha+\beta > 4\pi/3$. If this were true for every pair of adjacent angles in a quadrilateral, then $16\pi/3 < 2(\Sigma^4_{i=1} \theta_i)$. $\sharp$ For a 5-gon, $20\pi/3 < 2(\Sigma^5_{i=1} \theta_i)$. $\sharp$ And for a 6-gon $8\pi < 2(\Sigma^6_{i=1} \theta_i)$. $\sharp$ Thus, there exists at least one pair of adjacent vectors which may be exchanged to get an interior angle less than or equal to $\pi/3$. $\blacksquare$ -Thus, in a 6-gon, we can always exchange two vectors to get the sum of two adjacent vectors to be 'short'. Replacing $v_1, ... ,v_5, v_6$ with $v_1, ... , v_4, v_5+v_6$ to get 5 'short' vectors whose sum is zero. If $v_5+v_6$ is pair-able with another vector in such a way that their sum is 'short', then we are done. Otherwise, we get two other vectors whose sum is a 'short' vector, and we reduce to the four vector case. Because we can again reduce the quadrilateral case, we are done unless we must pair the two non-"sum" vectors to get a 'short' vector. If this is the case, I claim we still have the sum of three of the original six vectors is 'short'. -Proof of claim: -Looks like fedja just beat me to it, but this was just too much to write to delete it. :P<|endoftext|> -TITLE: Elementary short exact sequence of sheaves -QUESTION [19 upvotes]: This question arised when I was trying to use this answer to understand Reid's "Young Person's guide to Canonical Singularities". In particular page 352 when computing the blow-up $Y\rightarrow A^2/\mu_3$, the affine plane quotient the cyclic group of order 3, arises to the conclusion that the exceptional divisor is $E\sim P^1$, (no problems there) and $\mathcal{O}_E(-E)\sim \mathcal{O}(3)$ (problems here). -Given a variety $Y$ and an effective Cartier divisor $D$ on it, there seems to be a pretty standard exact sequence: -$$0 \longrightarrow \mathcal{O}_Y \longrightarrow \mathcal{O}_Y(D) \longrightarrow\mathcal{O}_D(D)\longrightarrow 0 $$ -As far as I understand, if $U$ is an open set in $S$ and $D\cap U = div(g)_U$ (for $D$ a hypersurface, if you want, and extend by linearity), then -$$ \mathcal{O}_Y(D)(U)= \{g \in \mathcal{O}_Y(U) \vert div(g)\geq D \}$$ -or equivalently $g/f$ is regular. The first map must be something like $g\rightarrow gf$ maybe with some order. A good answer to my question would include: - -Is this correct? -What is $\mathcal{O}_D(D)$? -What is the second map? -What does $\mathcal{O}_D(-D)$ mean -Why $\mathcal{O}_E(-E) \sim \mathcal{O}(3)$? I understood the RHS is generated by polynomials of degree 3? - -I am aware this is a simple question and probably everyone knows why, but I could not find a proper answer for it. - -REPLY [7 votes]: Diverietti's answer is exhaustive with respect to the first four questions, so let me say something just about the last one. -Miles Reid's example is actually a particular case of the following -more general situation, which is explained for instance in Barth-Peters-Van de Ven book on compact complex surfaces. -Let $n$ and $q$ be natural numbers with $0 < q < n$, $(n,q)=1$ - and let $\xi_n$ be a primitive $n-$th root of unity. -Let us consider the action of the cyclic group -$\mu_n=\langle \xi_n \rangle$ on -$\mathbb{C}^2$ defined by -$\xi_n \cdot (x,y)=(\xi_nx, \xi_n^qy)$. -Then the analytic -space $X_{n,q}=\mathbb{C}^2 / \mu_n$ has a cyclic quotient singularity of type -$\frac{1}{n}(1,q)$, and $X_{n,q} \cong X_{n', q'}$ if and only if $n=n'$ and either -$q=q'$ or $qq' \equiv 1$ (mod $n$). The exceptional divisor on the minimal -resolution $\tilde{X}_{n,q}$ of $X_{n,q}$ is a H-J string - (abbreviation of Hirzebruch-Jung string), that is to say, a - connected union $E=\bigcup_{i=1}^k Z_i$ of smooth rational curves $Z_1, \ldots, Z_k$ with - self-intersection $\leq -2$, and ordered linearly so that $Z_i - Z_{i+1}=1$ for all $i$, and $Z_iZ_j=0$ if $|i-j| \geq 2$. -More precisely, given the continued fraction -$\frac{n}{q}=[b_1,\ldots,b_k]:=b_1- - \cfrac{1}{b_2 -\cfrac{1}{\dotsb - - \cfrac{1}{\,b_k}}}$, -we have -$(Z_i)^2=-b_i, \quad i=1, \ldots, k.$ -In the case considered by Miles Reid, we have $n=3$, $q=1$, i.e. the action is -$\xi_3 \cdot (x,y)=(\xi_3x, \xi_3y)$, -hence the resolution is a unique smooth curve $E:=Z_1$ with self-intersection $(-3)$; this explain why $\mathcal{O}_E(-E)=\mathcal{O}(3)$. -Notice that there is another possible action of $\mu_3$ on $\mathbb{C}^2$, namely -$\xi_3 \cdot (x,y)=(\xi_3x, \xi_3^2y)$. -The corresponding continued fraction is -$\frac{3}{2}=[2,2]=2- - \frac{1}{2}$, -so the resolution is this case is given by two smooth rational curves of self-intersection $(-2)$ intersecting in a single point (this is a Rational Double Point of type $A_2$).<|endoftext|> -TITLE: Which cluster algebras are coordinate rings of double Bruhat cells? -QUESTION [10 upvotes]: Background -A uselessly vague paragraph follows. A cluster algebra is a commutative algebra $A$ with a distinguished set of generators called cluster variables. These cluster variables are grouped into clusters: subsets of cardinality $n$ (for some $n$). Each cluster is equipped with some combinatorial data which allows the complete set of cluster variables (and hence $A$) to be recovered from any given cluster. There is a related extension algebra $U$, called the upper cluster algebra, which coincides with the cluster algebra $A$ in small examples. -A fundamental example of this structure is a double Bruhat cell. Given a complex, reductive group $G$ with Borel $B$ and Weyl group $W$, there is a Bruhat decomposition -$$ G= \coprod_{w\in W} BwB$$ -Choosing the opposite Borel $B^-$ instead gives the opposite Bruhat decomposition -$$ G=\coprod_{v\in W} B^-vB^-$$ -and so -$$G = \coprod_{(w,v)\in W^2} BwB\cap B^-vB^-$$ -The pieces $G_{w,v}$ of this disjoint union are called double Bruhat cells. Then Berenstein, Fomin and Zelevinsky have shown the following. - -Theorem (BFZ05) The coordinate ring of a double Bruhat cell is naturally an upper cluster algebra. - -The Question -Roughly, my question is this. If I have an upper cluster algebra, how can I tell if it is the coordinate ring of a double Bruhat cell? -Note that there may be many different choices of $G,v,w$ which give isomorphic double Bruhat cells $G_{w,v}$, so it seems unlikely there is a constructive way of producing an example. -A more technical aspect of the question is the presence of frozen variables. These are cluster variables which cannot be mutated, and so they are in every cluster. Frozen variables may always be eliminated by setting them equal to 1, without affecting the combinatorics of the clusters. -The coordinate rings of double Bruhat cells will have many frozen variables, and I want to be able to eliminate or add frozen variables without changing the 'class' of cluster algebra. Though this is not a standard terminology, let us call two cluster algebras equivalent if they are related by the equivalence relation generated by 'setting some frozen variables' equal to 1. -Then I ask, If I have an upper cluster algebra, how can I tell if it is equivalent to the coordinate ring of a double Bruhat cell? -This second question is harder, but it does a better job of getting to the combinatorial type of the cluster algebra. - -REPLY [6 votes]: I think this is a hard problem. As evidence, it came up briefly in a conversation I was having with Lauren Williams and Sergey Fomin the other day, and neither of them knew a quick answer. I'll repeat here the one observation which I made at the time: - -The cluster algebra of any double Bruhat cell of type $A$ which looks like $G^{e,w}$ is rigid, in the sense of Quivers with Potential I. - -I suspect that it might be true that all double Bruhat cells are rigid, but I am too lazy to check. -Here is how to read this fact off from the Quivers with Potential paper. First, look at Section 6, where they define rigidity and prove that it is a mutation invariant. Next, look at Example 8.7. The authors check that the quiver in that example is acyclic -- you can check that this is the quiver corresponding to the reduced word $(s_1 s_2 \ldots s_{n-1})(s_1 s_2 \ldots s_{n-2}) \cdots (s_1 s_2 ) s_1$ for the longest element in $S_n$. So this shows that $G^{e, w_0}$ is rigid. Prop. 8.9 shows that, if a cluster algebra is rigid, then its restriction to an induced subquiver is rigid. I think (check this!) that, for any $w \in S_n$, if you choose a reduced word for $w_0$ which has $w$ as an initial segment, then you can get the quiver for $G^{e,w}$ by taking the quiver for $G^{e, w_0}$ corresponding to that reduced word and restricting to the subquiver coming from the word for $w_0$. -I'm looking forward to seeing other answers to this. Greg, I hope you'll report back on anything you work out as well!<|endoftext|> -TITLE: Criteria for topologically finitely generated profinite groups -QUESTION [6 upvotes]: Q1: Do we have a criterion which allows us to say when is a profinite group $G$ topologically finitely generated? -For example, if $G$ is topologically finitely generated then, for a fixed integer $N$, -there are only finitely many open subgroups $H\leq G$ such that $[G:H]=N$. In general, I guess that this condition is not sufficient. -Q2: Where can I find a proof that the absolute galois of $ \mathbf{Q}_p $ is topologically finitely generated? - -REPLY [5 votes]: The following profinite group is not finitely generated but has only finitely many open subgroups of each degree. -$$G=\prod_{n=5}^\infty (Alt_n)^{(n+1)!}.$$ -Note that you need at least $n+1$ elements in order to generate $(Alt_n)^{(n+1)!}$.<|endoftext|> -TITLE: how does one understand GRR? (Grothendieck Riemann Roch) -QUESTION [78 upvotes]: I tried to answer an earlier question as to uses of GRR, just from my reading, although i do not understand GRR. Today i tried to understand the possible idea behind GRR. After editing my answer accordingly, it occurred to me i was asking a question instead of giving an answer. My question is roughly whether the following speculation is in the ball park as to the purpose of GRR. -I've been thinking about Riemann Roch today, and reading Riemann. After dealing with a fixed divisor D, Riemann observes that his result proves every divisor of degree g+1 dominates the pole divisor of a non constant meromorphic function. Then he says that it may be possible to find a special divisor of even lower degree that dominates the poles of a non constant function. I.e. he begins to vary the divisor. By a rank calculation he shows one cannot expect a non constant function unless the pole divisor has degree at least (g/2)+1. -Now following his lead, we are led to vary the curve instead of the divisor. E.g. we might consider the family of curves over the moduli space. Then a good Riemann Roch theorem should let us relate the riemann roch theorem for the curve fibers, to a conclusion for a related sheaf on the base space., like a kunneth type formula, relating cohomology of base space total space and fiber. -I.e. a nice divisor like the canonical divisor on a curve, should be cut out on each curve fiber by a divisor on the total space, by intersecting it with each curve. (e.g. we could restrict the sheaf O(1) on the plane, to every curve of degree 4.) then we can push this sheaf from the total space down to the base space, i.e. the moduli space of curves. A good relative riemann roch theorem would then relate the universal canonical sheaf on the total space, to the canonical sheaves on the curve fibers, and the cohomology of the push down of the universal sheaf to the base space, the moduli space of curves. -Ideally such a relation would let one compute invariants of sheaves on the moduli space that do arise by pushing down sheaves on the total space of curves. Hopeful applications might include finding ample sheaves on Mg, hence proving projectivity, and computing invariants of the canonical sheaf on Mg, hence potentially estimating the kodaira dimension. -Now this is all speculation since i do not understand even the statement of the GRR, and have not read the paper of Harris-Mumford in which the application i cited above is made. Moreover I have never seen any proof of kodaira dimension of Mg using this method. Perhaps someone more knowledgable will comment on these speculative applications? -Is this roughly the idea behind GRR and Mumford's applications of it? I.e. is the idea of GRR to understand the cohomology of a sheaf on a base space which arises as a push down, by restricting it to the fibers of the map? and how helpful is this in practice? -specific question: if chi(O) is constant on fibers, does GRR allow one to determine chi(O) of either total space or base space from the other? - -REPLY [27 votes]: Introduction: GRR gives relations in the tautological ring -I can't speak directly to the potential applications you had in mind as far as Kodaira dimension, but I can say something about Mumford's application of GRR to the moduli space of curves. It seems that it's quite close to what you imagine, and in fact it's very important in the study of (a certain part of) the cohomology ring of the moduli spaces $\mathcal{M_g}$, and its relatives such as Deligne-Mumford space $\overline{\mathcal{M}}_{g,n}$ (now the curve has $n$ marked points, and we compactify by adding certain nodal curves should the points try to collide or the complex structure of the curve degenerate), and even further into Gromov-Witten theory. I'm going to give an overview of this story, building out of your question (I hope). -The part of the cohomology ring of $\mathcal{M_g}$ I'm talking about is called the tautological ring, and a gentle survey-introduction is Ravi Vakil's The moduli space of curves and Gromov-Witten theory. Those notes do not explicitly mention GRR, but they do quote a result that comes directly from Mumford's GRR calculation, and what I'm going to try to do is explain this a little bit. -I'd also like to mention that, as far as I understand it, this direction of application is essentially what Mumford had in mind. The paper he does this calculation is, after all, entitled "Toward an Enumerative Geometry of the Moduli Space of Curves". -Warm-up: Grassmannian -You can see in the first paragraph of Mumford's paper that he is explicitly modeling what he's doing after the cohomology of the Grassmannian, so I'm going to spend a paragraph on them, to motivate what's coming in the moduli space of curves. -On the one hand, we have the schubert cycles, given by the loci of planes that intersect the a fixed flag with given dimensions. On the other hand, since each point represents a vector space, these vector spaces fit together to give a tautological vector bundle, and we can take cycles representing the chern classes of this bundle, and get different classes -- it's not necessarily clear at all that these different tautological cycles should be related, but they are. -Mumford, and many after him, are trying to find similar relations between different tautological classes in $\mathcal{M}_g$. Mumford ends his first paragraph with "Moreover, it appears that many geometrically natural classes are expressible in terms of a small number of basic classes" -- this is akin to that description of the Grassmannian, and it's what GRR will give us. -We start with your basic idea -Rather than get into all the technical details of it, I just want to point out that he proceeds essentially exactly as you imagined here: - -A good relative riemann roch theorem would then relate the universal canonical sheaf on the total space, to the canonical sheaves on the curve fibers, and the cohomology of the push down of the universal sheaf to the base space, the moduli space of curves. - -Rephrased slightly differently: let $\pi:\mathcal{M_{g,1}}\to\mathcal{M_g}$ be the map from a moduli space of curves with one marked point to the moduli space of curves with no marked points -- it turns out this is exactly the universal family. Then we have the universal canonnical sheaf $\omega_\pi$ on $\mathcal{M_{g,1}}$ that you were discussing. We can use this to get cohomology classes in $H^*(\mathcal{M_g})$ in two different ways: first take its chern character, and then push down to $\mathcal{M_g}$, or first push down to $\mathcal{M_g}$, and then take the chern character. -These two alternatives give rise to a priori very different looking cohomology classes on $\mathcal{M_g}$, but GRR says that, after fiddling with Todd classes, they are the same. -Chern then pushforward -Let's see what happens when we take the first path. Since $\omega_\pi$ is one dimensional, taking the chern character is simply exponentiating $c_1(\omega_\pi)$. The class $c_1(\omega_\pi)$ is called the psi class $\psi$. Note that usually this is defined as the first chern class of the tangent bundle to the curve at the marked point, but through the identification of the universal curve with $\mathcal{M_{g,1}}$, these are equivalent. But I should warn you that this is no longer quite true if we start adding more marked points or nodes to our curves. So taking the Chern character of $\omega_\pi$ gives powers of $\psi$ on $\mathcal{M_{g,1}}$, and now if we push these forward we get the Morita-Mumford-Miller kappa classes $\kappa_i=\pi_*(\psi^{i+1})\in H^{2i}(\mathcal{M_g})$ -- indeed, this is the definition of $\kappa_i$. -Pushforward then Chern -Now, what happens if we go in the other direction? To pushforward $\omega_\pi$, we take the cohomology of $\mathcal{M_g}$ -- since $h^0(C, \omega_C)=g$, independent of the curve $C$, we have $\pi_*(\omega_\pi)=\mathbb{E}$, where $\mathbb{E}$ is a dimension $g$ vector bundle on $\mathcal{M_g}$ known as the hodge bundle. More simply, the fiber of $\mathbb{E}$ over a curve $C$ are the sections of the canonical bundle of $C$. The chern classes of the $\mathbb{E}$ are known as the $\lambda$ classes: $\lambda_i=c_i(\mathbb{E})$. Taking the Chern character of the $\mathbb{E}$ then would then give us a bit of a mess of polynomials in the $\lambda$ classes. -Comparing them -So taking the two different paths from $K(\mathcal{M_{g,1}})$ to $H^*(\mathcal{M_g})$ gives two different looking types of tautological classes, the $\kappa$ classes and the $\lambda$ classes. Since we set this up with GRR in mind, we should now see a relation between them. -In this case it turns out that this relationship cleans up rather nicely if we package it in a generating function, and a better answer might explain how, but I'll just note that -since we were working with the relative cotangent bundle to begin with, the relative Todd class can be manipulated into just giving us more $\kappa$ classes, and then with more mucking around with characteristic classes, it turns out we can express this relationship very beautifully in terms of generating functions: -$$\sum_{i=0}^\infty \lambda_i t^i=\exp\left(\sum_{j=i}^\infty \frac{B_{2j}\kappa_{2j-1}}{2j(2j-1)}t^{2j-1}\right).$$ -Here $B_{2j}$ are the Bernoulli numbers, coming from the Todd class. This is the formula in Ravi's notes I alluded to earlier: he cites Faber for this particular expression. -Extensions -I've done this just for $\mathcal{M_g}$ for simplicity, but I want to indicate here that you can get a lot more gas out of the same basic idea. -First, you can add marked points and boundary points and it essentially goes through the same. The universal curve is still just adding another marked, and then forgetting it. What gets a little complicated is that the relative dualizing sheaf $\omega_\pi$ stops being equal to just the cotangent line at the extra point when our curve becomes singular, but we can understand how they differ, and so we get some additional contributions involving the boundary strata -- I think Mumford already started to deal with this, and Faber and Pandharipande certainly dealt with it. -Also, you can extend this to Gromov-Witten theory, and consider moduli spaces of stable maps, and consider a curve with marked points together with a map $f:C\to X$, and play the same game there. Or better, we can first pull back a bundle from $E\to X$, and play the game described above with $f^*(E)$. In case $E$ is a line bundle, $s:X\to E$ a line bundle, and $Y=s^{-1}(0)$ the vanishing set, this can give relationships between the Gromov-Witten invariants of $X$ and $Y$ in terms of the chern classes of $E$. And in case $E$ is a negative, this can express the Gromov-Witten invariants of the total space of $E$ in terms of the Gromov-Witten invariants of $X$ and the chern classes of $E$. This is worked out in Tom Coates's thesis, and in his joint Annals paper with advisor, Givental: Quantum Riemann–Roch, Lefschetz and Serre, and is very important to GW theory, as its the method by which we can understand $X$ that is a complete intersection in a toric variety $Y$ -- this method relates the GW theory of $X$ to that of $Y$, and since $Y$ is toric we can localize with respect to the torus action to compute its Gromov-Witten invariants.<|endoftext|> -TITLE: Extending maps on Riemann surfaces -QUESTION [6 upvotes]: Suppose you have a map $g:\Sigma \rightarrow G$ from a Riemann surface $\Sigma$ to a compact Lie group $G$. What is the obstruction to finding a $3$-manifold $W$, such that $\partial W = \Sigma$, and an extension of $g$ to a map $\tilde{g}:W\rightarrow G$? In the paper I'm reading they say it lies in $H_2(G,\mathbb{Z})$. Why is this true? I mean, the obstruction class to extending $g$ to $\tilde{g}$ is an element in $H^3(W,\pi_2(G))$, which vanishes since $\pi_2(G)=0$ for compact $G$, right? So does this mean that the obstruction to finding a $3$-manifold $W$ with boundary $\Sigma$ lies in $H_2(G,\mathbb{Z})$? By the way, the paper I'm referring is here http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.cmp/1104180750 , see section 4.1 (page 405). Thanks. - -REPLY [10 votes]: This has nothing to do with Lie groups. Let $X$ be any space, let $S$ be a closed, orientable surface and let $f : S \rightarrow X$ be a map. We then get a canonical homology class $f_{\ast}([S]) \in H_2(X;\mathbb{Z})$. If there exists a closed orientable $3$-manifold $M$ with boundary $S$ such that $f$ extends over $S$, then we are done : the manifold $M$ maps into $X$ to provide a homology between $f_{\ast}([S])$ and $0$. Conversely, assume that $f_{\ast}([S]) = 0$. This implies that $f_{\ast}([S])$ is the boundary of a singular chain mapping into $X$. This singular chain can be thought of as a collection of tetrahedra glued together. Aside from the faces of the tetrahedra which lie in $S$, the faces of these tetrahedra are glued together in pairs. Let $M'$ be the result of gluing these tetrahedra together. We thus have $\partial M' = S$ and a map $F' : M' \rightarrow X$ extending $f$. It is a fun exercise to show that $M'$ is a 3-manifold except possibly at finitely many points $x_1,\ldots,x_n$. A small neighborhood of $x_i$ is homeomorphic to the cone on an oriented surface. Cut out these neighborhoods and glue in handlebodies. We get a $3$-manifold $M$ with $\partial M = S$, and from the construction it is clear that we can modify $F'$ to give a function $F : M \rightarrow X$ which still extends $f$. -The point here is that in low degree, bordism agrees with homology.<|endoftext|> -TITLE: How locally ubiquitous are totally real fields? -QUESTION [6 upvotes]: Let $p$ be a fixed prime number. -Question 1: Given a finite extension $K$ of $\mathbb{Q}_p$ is there a totally real extension $F$ of $\mathbb{Q}$ and a place $v$ of $F$ over $p$ such that $F_v = K$? -This is used in the proof of the local Langlands conjecture (thus I am quite sure that the answer is Yes) but I have never seen a reference. My state of knowledge is similar for the next question (and again I would be very grateful for a reference): -Question 2: Given an integer $g \geq 1$ is there a totally real extension $F$ of $\mathbb{Q}$ such that $F \otimes_{\mathbb{Q}} \mathbb{Q}_p = \mathbb{Q}_p \times \dots \times \mathbb{Q}_p$ ($g$ copies). -A common naive generalization of both questions is the following question (which is now a real question): -Question 3: Let $K_1, \dots, K_g$ be finite extensions of $\mathbb{Q}_p$. Is there a totally real extension $F$ of $\mathbb{Q}$ such that $F \otimes_{\mathbb{Q}} \mathbb{Q}_p = K_1 \times \dots \times K_g$? -I would not be surprised if the answer is No due to trivial reasons which I am just not seeing. - -REPLY [6 votes]: The answer to the first question is "yes". See this paper of the Dokchitser brothers, Lemma 3.1 for the case where $K/\mathbb{Q}_p$ is Galois. In the general case, apply the result to the Galois closure $K'$ of $K$ to get $F'$, identify the Galois group of the local fields with a decomposition group $D$ at $p$ inside the global Galois group and take the fixed subfield of the subgroup of $D$ corresponding to $K$. -As Kevin says, unless I am missing something, the Dokchitsers' proof works with minor modifications for all three of your questions. Note that the result for question 3 follows from the previous two (using the slightly more general version of qn 1 in the above link): first take an extension in which $p$ is totally split, then work with each of the places above $p$ separately, using the answer to qn 1.<|endoftext|> -TITLE: What is the difference between a primary representation and a irreducible representation? -QUESTION [6 upvotes]: I am currently reading some of Mackey's work on unitary representation. -Given a locally compact group $G$ and a unitary representation $\pi : G\rightarrow U(H)$. As far as I understood it, the representation $\pi$ is primary, if the von Neumann algebra generated by $\pi(g)$ for all $g \in G$ is a factor, see http://mathworld.wolfram.com/PrimaryRepresentation.html. The representation $\pi$ is irreducible, if there does not exist a nontrivial $G$ invariant closed subspace $H' \subset H$, i.e. $\pi(g) h \in H'$ for all $g \in G$ and $h \in H'$. -See the comments: The countable sum of the same irreducible representation is primary. -When is a primary representations quasi equivalent to an irreducible one? Are they the same if the group is of type 1? Does the decomposition of the group von Neumann algbera $L(G)$ into factors correspond to the decomposition (as a direct integral) in isotypic components? Are the some nice examples, which illustrate that this is to much to hope for. -Motivating example is the Peter Weyl theorem, which states that every irreducible is finite dimensional and -$$L(G) = \bigoplus_{\pi \; \in irr(G)} M_{dim(\pi)}( \mathbb{C}),$$ -where the components $M_{dim(\pi)}( \mathbb{C})$ are the factors. Hence here the factors are quasi equivalent to an irreducible one. -Aside to the original question: Do all unitary representation appear in $L(G)$? - -REPLY [3 votes]: I would recommend reading parts of Jacques Dixmier's book: "$C^\ast$-algebras" (North Holland, 1977 - translated from the french version of 1969), especially Chapters 5 (irreducible and factor representations of $C^\ast$-algebras) and Chapter 13 (the analogue for locally compact groups). -The regular representation of a free group is not type I (as the commutant is not type I), while every irreducible representation is type I (since the commutant is $\mathbb{C}$). Hence a factor representation is not necessarily quasi-equivalent to an irreducible. It is true that, for a type I group, every factor representation is quasi-equivalent to an irreducible one (Proposition 5.4.11 in Dixmier). -For the decomposition of the regular representation and the corresponding decomposition of $L(G)$, see Proposition 18.7.7 in Dixmier.<|endoftext|> -TITLE: Non-integrable ergodic theory -QUESTION [13 upvotes]: Can anyone help me out with proofs/counterexamples? I'm working on an operator-valued multiplicative ergodic theorem and need what may(?) be a well-known fact. This fact (if true) would help me get rid of an annoying asymmetry in the conclusion of a theorem. -I'm assuming that $T$ is an invertible ergodic transformation of a probability space $(X,\mathcal B,\mu)$ and that $f$ is a measurable (but not necessarily integrable) function on $X$. - -Is it true that $f(T^nx)/n \to 0$ a.e. if and only if $f(T^{-n}x)/n\to 0$ a.e.? - -Comments: -(1) In probability language if you define $X_n=f(T^nx)$ this is a stationary sequence of random variables. Borel-Cantelli 1 shows that if $\mathbb E |X_0|<\infty$ (i.e. $f\in L^1$) then $X_n/n\to 0$ as $n\to\pm\infty$: The probability that $|X_n|/n > 1/k$ is $\mathbb P(|X_0| > n/k)$. The sum of this series is over-estimated by $k\mathbb E|X_0| < \infty$. Hence almost surely $|X_n|/n < 1/k$ for all large $n$. Since this is true for all $k$ you get $X_n/n\to 0$. -If the $X_n$ are i.i.d. random variables, then the converse holds by Borel-Cantelli 2. This shows that for i.i.d. random variables, the boxed question has an affirmative answer. -(2) In the case where the $X_n$ are not i.i.d. I believe there are examples where $X_n/n\to 0$ almost surely even though $\mathbb E|X_0|=\infty$. -In the case that $f\in L^1$, both sides of the implication in the main question are true. The unresolved case is $f\not\in L^1$. - -REPLY [10 votes]: $\newcommand{\R}{\mathbb R}$ -$\newcommand{\P}{\mathbf P}$ -$\newcommand{\Z}{\mathbb Z}$ -I found this question very interesting and gave it much thought this week. I believe I have a proof now. I think it would be interesting to see what generalizations one can get from this argument. Below I use probabilistic notation, which I'm more used to. -Let $\{X_n\}_{n\in \Z}$ be a stationary stochastic process, taking values in $\R$. Let $L=\limsup_{n\to -\infty} \frac{X_n}{|n|}$ and $R=\limsup_{n\to \infty} \frac{X_n}{|n|}$. -Theorem: $L=R$, almost surely. -Proof: -It is enough to prove the theorem for ergodic processes. We may also assume, WLOG, that all the values are nonnegative integers. -Let $A$ be the event that for some $n<0$ we have $X_n\ge |n|$ and let $B$ be the event that for some $n>0$ we have $X_n\ge |n|$. -Lemma: $\P(A) \le 2 \P(B) .$ -Proof of lemma: -For a given realization of $X_n$, let $I$ be all indices $i$ for which $T^i X$ is in $A$ and let $J$ be all the indices for which $T^i X$ is in $B$. We claim that the density of $J$ is no more than twice the density of $I$, which then implies the conclusion. -$I$ can be written as $\cup_{n \in \Z} \{n,\ldots,n+X_n\}$, while $J=\cup_{n \in \Z} \{n-X_n,\ldots,n\}$. In particular, $J$ is contained in $\bar{J}=\cup_{n \in \Z} \{n-X_n,\ldots,n+X_n\}$. But if we write $I$ as a union of disjoint intervals, then in $\bar{J}$ each of these intervals is extended to the left by at most the length of the interval. Hence, the density of $\bar{J}\supset J$ is at most twice that of $I$. $\blacksquare$ -Of course, by symmetry, we also have $\P(B) \le 2 \P(A)$. Let $A_K$ be the event that for some $n<0$ we have $X_n \ge \max(|n|,K)$ and define $B_K$ analogously. By applying the lemma to the process $\{Y_n\}_{n\in \Z}$ defined by $Y_n=X_n$ if $X_n\ge K$ and $Y_n=0$ otherwise, we get that $\P(A_K) \le 2 \P(B_K)$ and vice verse. -In particular, $\lim_{K\to \infty} \P(A_K) =0$ if and only if $\lim_{K\to \infty} \P(B_K) = 0$. These limits necessary exists, since these are monotone decreasing events. -For ergodic processes, we have that $L$ and $R$ are a.s. constant. Now, if $L<1$ a.s. then we have $\P(A_K)\to 0$. In the other direction, if $\P(A_K)\to 0$ then a.s. $L\le 1$. Similar implications hold for $R$ and $B_K$. Hence, we get that if $L<1$ then $R\le 1$ and vice verse. By applying these to a rescaled process $X_n / \alpha$, we get that for any $\alpha$ we have $L<\alpha$ implies $R\le \alpha$ (and vice verse), so we must have $L=R$. $\blacksquare$<|endoftext|> -TITLE: A sequence of order statistics from an iid sequence -QUESTION [5 upvotes]: Note: This question was asked in stats.stackexchange.com and math.stackexchange.com, with expired bounties on both sites. -Given a sequence of iid random variables $X_i$ (without loss of generality from $U(0,1)$), an integer $k \ge 1$ and some $p \in (0,1)$, construct the sequence of random vectors $Z^{(j)}$, $j=0,1,...$ in the following way. Let -$$Z^{(0)}=(X_{(1)},...,X_{(k)}),$$ -where $X_{(l)}$ is the $l$-order statistic of sample $\{X_1,...,X_k\}$. Introduce notations -\begin{align} -Z^{(j)}&=(Z_{j,1},...,Z_{j,k}),\\\\ -m_j&=\min(Z_{j-1,1},...,Z_{j-1,k},X_{k+j}),\\\\ -M_j&=\max(Z_{j-1,1},...,Z_{j-1,k},X_{k+j}) -\end{align} -Then -$$Z^{(j)}=(Y_{(1)},...,Y_{(k)})$$ -where $Y_{(l)}$ is the $l$-order statistic of the following set which is - -The set $\{Z_{j-1,1},...,Z_{j-1,k},X_{k+j}\}\backslash m_j$ with probability $p$ -The set $\{Z_{j-1,1},...,Z_{j-1,k},X_{k+j}\}\backslash M_j$ with probability $1-p$ - -The decision between cases 1. and 2. is made independently from the $X_i$ (and hence from the $Z^{(i)}$). -The $Z^{(j)}$ are supported on the $k$-dimensional simplex $S_k = \{(x_1, \dots, x_k) \in \mathbb{R}^k \, | \, 0 \le x_1 \le x_2 \le \dots \le x_k \le 1 \}$. -It appears that the $Z^{(j)}$ converge in distribution. Is this known? Is anything known about the limiting distribution? -For the case $k=1$, the answer is the following. Denote the cdf of $Z^{(j)}$ by $F_j$. -The cdf of $\min(X_{n+1},Z^{(n)})$ (for $U(0,1)$ case) is -$$x+F_n(x)−xF_n(x)$$ -and the cdf of $\max(X_{n+1},Z^{(n)})$ is -$$xF_n(x)$$. -Hence -\begin{align} -F_{n+1}(x)&=p(x+F_n(x)−xF_n(x))+(1−p)xF_n(x)\\\\ -&=px+(p(1-x)+(1-p)x)F_n(x) -\end{align} -Since $p(1-x)+(1-p)x\in(0,1)$ we have that -$$\lim F_{n}(x)=\frac{px}{1-p(1-x)-(1-p)x}$$ -I am looking for general results (case $k>1$) either for the limiting distribution of the whole vector $Z^{(j)}$ or of some of its components (marginal distributions). - -REPLY [5 votes]: Another way to describe this sequence of random vectors is that you have an unordered set of $k$ points, initially sampled independently, and at each step you add a point and then with probability $p$ remove the minimum, and with probability $1-p$ remove the maximum. -Given $x \in (0,1)$, the number of points in the $n$th set lower than $x$ follows a random walk on the set $\{0,1,...,k\}$ whose initial distribution is binomial (and unimportant) and whose transitions occur with the following probabilities except at the edges: - -$-1$ with probability $(1-x)p$: Add a point greater than $x$ and delete the minimum. -$+1$ with probability $x(1-p)$: Add a point smaller than $x$ and delete the maximum. -$0$ with the complementary probability $(1-x)(1-p) + xp$: Add a point greater than $x$ and delete the maximum or add a point smaller than $x$ and delete the minimum. - -The boundary cases are that when there are no points smaller than $x$, this will still increase to $1$ with probability $x(1-p)$, but the chance to stay $0$ is $1-x(1-p)$, and when all $k$ points are smaller than $x$, this will decrease to $k-1$ with probability $(1-x)p$ and stay $k$ with probability $1-(1-x)p$. -The limiting distribution is the same as if you eliminate the chance to pick a new point lower than $x$ and then delete the minimum or to pick a point greater than $x$ and then delete the maximum. -The initial distribution doesn't matter. The limiting distribution of this random walk gives the limiting distributions for the coordinates, since the probability that the $\ell$th point is at most $x$ is the sum of the probabilities that there are exactly $\ell$, $\ell+1$, ... or $k$ points less than $x$. It does not say the joint distribution.<|endoftext|> -TITLE: In knot theory: Benefits of working in $S^3$ instead of $\mathbb{R}^3$? -QUESTION [37 upvotes]: In several textbooks on knot theory (e.g. Lickorish's, Rolfsen's) knots are considered in $\mathbb{R}^3$ or $S^3$. The reason for working in $S^3$ is sometimes given at the beginning of a text as that $S^3=\mathbb{R}^3\cup \{\infty\}$ is compact, so working there is "more convenient". - -What are some specific benefits or - conveniences for working in $S^3$ - instead of $\mathbb{R}^3$? - -I could think of one example: In knot theory we work in piecewise-linear category, so working in $S^3$ allows for considering finite triangulations of knot complements. -Another example is in studying achirality, where Smith theory for fixed points of periodic maps on the sphere is used. -Furthermore, some results in these textbooks are stated for knots in either $\mathbb{R}^3$ or $S^3$, but some other results are stated only for knots in $S^3$. - -Are there results that are correct for - knots in $S^3$ but not for - knots in $\mathbb{R}^3$? - -REPLY [2 votes]: I think it is much more convenient to work in $S^3$ when you want to compute the fundamental group of the complement of a toric knot , than in $R^3.$ -In $S^3$ you use the torus containing the knot to divide $S^3$ into two $S^1 \times D^2$, and then apply Van Kampen theorem. -In $R^3$ the corresponding divison of the complement of the knot is less natural, it is harder to describe and visualize. -And ofcourse the fundamental group is the same if you take the knot-complement in $R^3$ or in $S^3.$<|endoftext|> -TITLE: Euler class of S^1-orbibundle -QUESTION [7 upvotes]: Given a topological space $M$ with a locally free $S^1$ action on it, assume the slice representation holds,(this is often the case, e.g. M is a smooth manifold) then this will make $M$ a principal $S^1$-orbibundle over $M/S^1$. What's the proper way to define the Euler class of this $S^1$-orbibundle? How should the naturality of Euler class be stated in the category of $S^1$-orbibundles? - -REPLY [6 votes]: Here is a topological construction of such a class, in singular cohomology with rational coefficients. -Let $M$ be an $S^1$-space. Then there is the Borel construction $M // S^1 := ES^1 \times_{S^1} M$. It comes with a map $f$ to $BS^1= CP^{\infty}$ which is a fibre bundle with fibre $M$. -Moreover, there is a map $q: M//S^1 \to M/S^1=Q$. -As a warm-up, let me give an alternative construction of the Euler class in the free case, i.e. for principal bundles. In that case, $q$ is a homotopy equivalence. Now put $e:= (q^{\ast})^{-1} f^{\ast} (z) \in H^2 (Q)$, where $z \in H^2 (BS^1)$ is the Euler class of the universal bundle. Then $e$ is the Euler class of the bundle $M \to Q$. -To prove this, use universality to reduce to the case of $S^1$ acting on $ES^1$. In that case, both $q$ and $f$ are homotopy equivalences. There is a sign issue here, but allow me ignore it. -Your question was of course about nonfree actions. The step that fails is that $q$ is an isomorphism in integral cohomology, because it is a homotopy equivalence. However, if the action is locally free (and everything else is nice), then $q$ is still a rational cohomology isomorphism. -This can be seen by several methods; one method is similar to the argument I sketched in my answer to this question Euler characteristic of orbifolds. The idea is that the fibre of $q$ over a point of $M/S^1$ with stabilizer group $G$ is $BG$, which has the rational cohomology of a point. The niceness of the action is needed to promote this observation to small open sets, i.e. to find a cover of $Q$ by open sets $U$ such that $q^{-1}(U) \to U$ is a rational isomorphism. Then apply Mayer-Vietoris to globalize. -Once $q^{\ast}$ is shown to be an isomorphism with rational coefficients, it can be inverted and you can apply the construction from the first part of the answer. Naturality is clear.<|endoftext|> -TITLE: $\Sigma_1$ elementary substructure -QUESTION [5 upvotes]: This is a question I was given in the exam: show that if $\lambda, \kappa$ are uncountable cardinals then -$$(H_\lambda,\in) \prec_1 (H_\kappa,\in)$$ -where $H_\lambda$ is the class of all sets $x$ such that $|TC(x)| < \lambda$. ($TC(x)$ refers to the transitive closure of $x$.) -Equivalently, given any $\Sigma_1$ formula $\varphi(x_1,x_2,...,x_n)$ with free variables shown and $a_1,a_2,...,a_n \in H_\lambda$ then $(H_\lambda,\in) \models \varphi(a_1,...,a_n)$ if and only if $(H_\kappa,\in) \models \varphi(a_1,...,a_n)$. -I have tried hard without success. Can anyone give me a hint? - -REPLY [5 votes]: I assume that you mean to assume $\lambda\leq\kappa$. -For any $\vec x\in H_\lambda$, we may by the Lowenheim-Skolem theorem find an elementary substructure $X\subset H_\kappa$ with $TC(\vec x)\subset X$ and $X$ having size less than $\lambda$. In particular, $X$ has a witness $z$ for your $\Sigma_1$ statement $\exists z\,\psi(\vec x,z)$, so that $\psi(\vec x,z)$ for this particular $z$. Let $X_0$ be the Mostowski collapse of $X$. This is a transitive set of size less than $\lambda$ and hence $X_0\subset H_\lambda$. Furthermore, $\vec x$ is not changed by the collapse. Thus, if $z_0$ is the set $z$ collapses to, then we have $\psi(\vec x,z_0)$ in $X_0$ and hence also in $H_\lambda$. So we've found a witness in $H_\lambda$, and so $H_\lambda$ is $\Sigma_1$ elementary in $H_\kappa$, as desired.<|endoftext|> -TITLE: Known Mirror Calabi-Yau pairs -QUESTION [11 upvotes]: There is a well known class of Calabi-Yau (3 dimensional) pairs constructed by Batyrev. These are resolutions of Calabi-Yau hypersurfaces in reflexive polytops of dimension 4. -Question: Does any body know any other mirror pair, or a family of them, beside this kind of pairs? -For example, how about Calabi-Yau complete intersections in higher dimensional weighted projective spaces or Fano toric varieties? -Warning: My question is only about closed Calabi-Yau 3-folds. - -REPLY [6 votes]: An answer to this question depends on what you mean by "mirror". As to topological mirror symmetry, the Batyrev-Borisov toric construction is the easiest and the most standard. Other examples would be Borcea-Voisin CY3s. If you want to do more, for example instanton computation etc, you need to work on CY3s with small $h^{1,1}$ (or mirrorly $h^{2,1}$) so that you can compute the mirror map, period integral etc. -CY3s with $h^{1,1}=1$ are typically complete intersections of Grassmannians. Batyrev, Ciocan-Fontanine, Kim and van Straten construct mirror manifolds (with mild singularity) of such CY3s via toric degeneration in this paper. The basic idea is easy to understand; they degenerate the ambient Grassmannians to toric varieties and apply the Batyrev-Borisov toric construction. So the essential part is toric in this case, too. -Apart from toric examples, the best known example would be Rodland's pfaffian-7 CY3. In this paper, he constructs a smooth mirror manifold of the pfaffian-7 CY3 by an orbifold method, i.e. taking a special 1-parameter family and take quotient by a finite group. The interesting observation is that the mirror manifold is also mirror of the complete intersection CY3 of Gr(2,7). Later Borisov and Caldararu prove that Rodland's CY3 and the complete intersection CY3 of Gr(2,7) are derived equivalent as HMS indicates. -Recently Kanazawa constructs several new pfaffian CY3 with $h^{1,1}=1$ in this paper, partially solving van Enckevort and van Straten's conjecture. Mirror manifolds (with mild singularity) are also constructed by an orbifold method. Kanazawa's pfaffian CY3s are interesting in the sense that the mirror manifolds have two large complex structure limits (similar to Rodland's case). This kind of phenomenon is also found by Hosono and Takagi in this paper. Their mirror symmetry is based on the standard Batyrev-Borisov toric method (with mild modification), but a non-trivial Fourier-Mukai partner comes into play. -Apart from toric examples, there seems no standard way to construct mirror manifolds for a given smooth compact CY3. Rodland's and Kanazawa's construction is some what an art. People have been working hard to find an intrinsic and systematic mirror construction. SYZ conjecture is one of such attempts.<|endoftext|> -TITLE: Is there an integration free proof (or heuristic) that once differentiable implies twice differentiable for complex functions? -QUESTION [42 upvotes]: The title pretty much says it all. I am revisiting complex analysis for the first time since I "learned" some as an undergraduate. I am trying to wrap my head around why it should be the case that a function which is differentiable once should be differentiable twice. I know a proof (Use Cauchy integral formula and differentiate under the integral sign), but that proof doesn't do a whole lot to explain the magic. -Say you had never heard of complex numbers before, and someone told you that you had a function $f: \mathbb{R}^2 \to \mathbb{R}^2$ given by $f(x,y) = (u(x,y), v(x,y))$, which locally looks like a rotation/expansion (i.e. it satisfies the Cauchy-Riemann equations everywhere). Then why on Earth should this function be $C^\infty$? I would love to see a picture, or just a proof that makes this feel less like a magic trick. I hoped that "Visual Complex Analysis" would help me out here, but this seems to be the one theorem in the book which is not given a geometric motivation. - -REPLY [2 votes]: This appears to be the earliest "topological" proof: -A TOPOLOGICAL PROOF OF THE CONTINUITY OF THE -DERIVATIVE OF A FUNCTION OF A COMPLEX -VARIABLE -BY ROBERT L. PLUNKETT -1958 -http://www.ams.org/journals/bull/1959-65-01/S0002-9904-1959-10251-2/S0002-9904-1959-10251-2.pdf -G.T. Whyburn himself (whose research was foundational to the result) gives his own proof, allegedly inspired by Connell and Porcelli: -Developments in topological analysis, 1961 -http://matwbn.icm.edu.pl/ksiazki/fm/fm50/fm50125.pdf<|endoftext|> -TITLE: Foliating R^3 with straight lines -QUESTION [11 upvotes]: Is there a complete characterization of ways one can foliation the 3-dimensional Euclidean space with straight lines? -For example, one can partition R^3 into parallel planes and fill up each plane with parallel lines. Slightly more involved examples should be possible, like filling space with hyperboloids. -I would like to know what conditions one can say about an arbitrary foliation of R^3 by straight lines. - -REPLY [4 votes]: The leaves of the foliation are regular. So, the space of leaves $X$ induces a submersion $\pi: R^3\rightarrow X$ with leaves of $R.$ Using the standard homotopy sequence, we see that $X$ is simply-connected. Furthermore, $X$ is path-connected since any two leaves can be connected by a path of leaves in the foliation. -We can actually define an embedding $\phi: X\rightarrow R^3$ by letting $\phi(l)$ be the point on the leaf $l\in X$ that is closest to the origin in $R^3.$ This gives us two things: a vector space structure on each leaf by using $\phi(l)$ as the origin and an orientation of $X$ (since only orientable simply-connected surfaces are embeddable in $R^3$). We use the orientation of $X$ to give an orientation on each leaf. -Thus, we have a 1-dimensional orientable vector bundle over a simply-connected space $X$, which means that the vector bundle is trivial. Hence, all smooth foliations with leaves of lines on $R^3$ are diffeomorphic. -I rather wanted to say all continuous foliations of such sort are homeomorphic, but I'm nervous about the embedding of $X.$ - -REPLY [3 votes]: Probably you don't, but if you mean to include arbitrary partitions of space into lines, then a tidy classification will be impossible, since with the axiom of choice one can construct extremely bizarre partitions. Just enumerate the points in space in order type continuum, and iteratively pick a line through the next point missing the previous lines. There are continuum many such lines, since only fewer than continuum many lines were chosen so far. So one can arrange, for example, that no two of the lines in the partition are parallel, that no two of them have the same angle with the axis planes, or each other, and so on. This is because at each step of the recursion, such requirements only rule out fewer than continuum many of the continuum many lines through the desired point.<|endoftext|> -TITLE: Seemingly emergent structures in mathematics -QUESTION [8 upvotes]: I rather suspect that this must have come up here on MO already, but my handful of searches didn't turn up the thread, so... -I'm curious about examples of mathematical structure that seems to arise "from nothing." The example that motivates this is one that I was teaching today, namely, the central limit theorem. -I was trying to convey to my (business math) students how astounding it is that the sampling distributions of the mean of a uniformly distributed random variable approach a normal distribution as the sample size increases. -Out of complete randomness, very specific and rather subtle structure arises (if in the limit). -I'd be amused to see other examples of this perceived phenomenon in different areas of mathematics. Not just structure where it wasn't expected (which is quite cool, but ubiquitous), but structure that seems to "arise from a vacuum." - -REPLY [9 votes]: A nice example of seemingly trivial structure that hides highly nontrivial structure -is that of a projective space. Such a space consists of "points', "lines", and "planes" -with the obvious properties: there is a unique line through any two points, any two -planes meet in a unique line, three points not on a line lie on a unique plane, -and so on. -Surprisingly, any such space has an underlying skew field which coordinatizes -the space so that lines and planes have linear equations. This due to the fact -that the Desargues theorem holds in any projective space. Hilbert (1899) showed (in a highly roundabout way) that one can then define sum and product of -points, and use the Desargues theorem to prove their skew field properties. - -REPLY [6 votes]: I'm not exactly sure what you're after, but one could think of Ramsey theory as saying that any large enough structure will necessarily contain an orderly substructure. Or, even more loosely, that order is unavoidable in a large enough chaos. So I suppose the following would be examples of answers to the question: - -Ramsey's theorem; -van der Waerden's theorem; -the Hales-Jewett theorem; -Szemeredi's theorem; -the Green-Tao theorem. - -And there are many more in this vein. Especially infinite versions of such theorems seem to match nicely with your example of the central limit theorem. - -REPLY [3 votes]: The ADE classification. - -REPLY [2 votes]: Thurston's geometrisation conjecture might be a good example here. If you start with a few simple 3-manifolds, you can combine them to create any 3-manifold you want.<|endoftext|> -TITLE: What are "maps" between proper classes? -QUESTION [12 upvotes]: When defining a functor (between categories), I am usually told that it assigns to each object of the source category an object of the target category. I do not find this very satisfactory since we are dealing with proper classes here. Judging by the definition, it must be possible to have the concept of a "map" between proper classes. I would like to know what exactly that is and how it is defined. -I have attempted to read some books on set theory in search for an answer, but they all treat classes very briefly and never mention the possibility of having anything like a map between two of them. I would be just as happy if you could point me to a book where this is explained. - -REPLY [2 votes]: Instead of MK set theory + Morse ordered pair definition, is ARC set theory (F.A.Muller, "Sets, Classes, and Categories", 2001, Bibliography PDF) with the usual Kuratowski ordered pair an option? -ARC supposedly proves the existence of the $n$-th power-class of the set universe V for any $n \in \mathbb{N}$, and all so-called "good" classes provably exist, good classes being the class of all sets and "the powerclass and the union-class of a good class, and the union-class, the intersectionclass, the complement-class, the pair-class, the ordered pair-class, and the Cartesian product-class of any finite number of members of one good class". -According to the cited paper, ARC is consistent relative to ZFC plus a strongly inaccessible cardinal axiom. -I think that this set theory looks quite nice, so I'm wondering why it didn't take off at all. -The formal proofs should be in Muller's PhD thesis, which I don't have access to.<|endoftext|> -TITLE: On positive matrices and their eigenvectors -QUESTION [7 upvotes]: Let $A$ be an $n\times n$ positive integer-valued matrix, that is every entry of $A$ is a a positive integer. Let $\lambda$ be the Perron-Frobenius eigenvalue and $x = (x_1,...,x_n)^T$ the corresponding positive probability eigenvector: $\sum_i x_i =1, \ x_i > 0$. Denote by $H(x)$ the additive subgroup of $\mathbb R$ whose generators are the coordinates of $x, \ -H(x) = < x_1,...,x_n >$. -Fix any integer $k \geq 1$ and consider the set of positive integer-valued matrices $\mathcal B_k$ formed by all $n\times n$ matrices $B$ satisfying the following conditions: $\lambda^k$ is the Perron-Frobenius eigenvalue for $B$, and if $By = \lambda^k y,\ \sum_i y_i = 1, y_i >0,$ then $H(y) = H(x)$. -My questions are as follows. -(1) Is there an algorithm describing all matrices from $\mathcal B_k$? -(2) How can one find at least one matrix $B$ in the set $\mathcal B_k$ different from $PA^kP^{-1}$ where $P$ is a permutation matrix? -Comments: (i) The case when $\lambda $ is an integer is not interesting, so that one can assume that $\lambda$ is an algebraic number. (ii) I asked a similar question before but these ones seems formulated in more precise form. (iii) Of course, (2) is simpler than (1), and I actually need a constructive answer to (2). -I'll be glad to see any comments, suggestions, references. - -REPLY [2 votes]: This question is very closely related to problems in dimension groups (partially ordered abelian groups with various other properties), specifically stationary ones, for which there is a history going back to the early 80s [e.g., by me, Positive matrices and dimension groups affiliated to C*-algebras and topological Markov chains, J of Operator Theory 6 (1981) 55--74]. -For example, if $A$ has determinant $\pm 1$, then $H(x)$ is an invariant of shift equivalence (related to conjugacy, but somewhat coarser). More generally (without the determinant condition), one considers $\cup_{n=0}^{\infty} H(x)\lambda^{-n}$, the direct limit group, an invariant of shift equivalence. In particular, examples size two examples (with irreducible characteristic polynomial) exist with the same $H$ values, but aren't shift equivalent. Of course, we can also arrange non-shift equivalence by making sure the spectra are different. -However, it is also true that if $\det A =\pm 1$ and $A$ and $B$ are primitive, and the characteristic polynomials of $A$ and $B$ are equal to each other and irreducible over the integers, then $H_A = H_B$ entails that $A$ is conjugate to $B$. This however, is rather special. Drop the irreducibility of the characteristic polynomial and the result fails, as the example above shows. Drop the determinant condition (keeping irreducibility of the char poly), and the union determines the shift equivalence class, almost the same as conjugacy. -Likely what you are looking for is the connection between ideal class structure of orders in number fields and classification of the matrices.<|endoftext|> -TITLE: Complexity of computing matrix rank over integers -QUESTION [11 upvotes]: Does computing the rank of an integer matrix have complexity polynomial in the size of the input? -The Gaussian elimination algorithm is polynomial in the number of elementary operations (addition and multiplication), but the intermediate size of of the matrix entries may go up exponentially. Are there other algorithms with better complexity? Can anyone give a reference? - -REPLY [6 votes]: I asked more-or-less the same question 12 years ago on sci.math.num-analysis, but for the case when the integer entries are small. Thom Mulders replied describing this wonderful modular method. -Edit: the link above seems to be dead, so I include the reply verbatim here -You can use a modular method. When A is your matrix, compute for N different -primes p the rank of (A modulo p). When the product of the primes you use -is sufficiently big, then the rank will be equal to the maximum of the ranks -you have computed. - -To be specific. Let A be an n by m matrix (n<=m) and ||A|| a bound on the -entries in A. When r is the rank of A there is an r by r submatrix B of A -with det(B)<>0. When the product Q of the primes exceeds det(B), there must -be at least one prime p such that the rank of (B modulo p) is r and thus the -rank of (A modulo p) is r. Using Hadamard's bound we have -det(B) <= r^{r/2}||A||^r <= n^{n/2}||A||^n, -so you need at most n((log n)/2 + log ||A||) different primes. -Since the complexity of computing the rank of (A modulo p) is O(mn^2) (assuming -the prime is not too big, e.g. fits into a machine word), the complexity of -this algorithm is (up to some log factors) O(mn^3). This is better than the -complexity (O(mn^4)) of the fraction free algorithm proposed before. - -You can also use a probabilistic version of the above. Take a set S of at least -2n((log n)/2 + log ||A||) different primes and compute the rank of (A modulo p) -for different random primes from S. For such a random prime, the probability -that the rank of (A modulo p) is the rank of A is at least 1/2. When you find -that the rank does not increase for N choices of p, then the probability -that you have found the right rank is at least 1-(1/2)^N. Taking N=20, the -probability of failure is less than one in a million. Increasing the set S -improves the probability of success.<|endoftext|> -TITLE: Is every poset the poset of prime ideals of a ring? -QUESTION [13 upvotes]: The answer to this question, as it is, is trivially false, for one necessary condition is the existence of maximal element(s), i.e., maximal ideals exist and are prime. -My question was inspired from Exercise 1.8 of Atiyah-Macdonald's book in Commutative Algebra: -The set of prime ideals of a nonzero ring has minimal elements with respect to inclusion. -Thus the above is also a necessary condition. My question is if there are more necessary conditions which are also sufficient, i.e., - -Are there necessary and sufficient conditions that a poset must possess so that it is the poset of a ring's spectrum. - -Answers (possibly partial answers) are welcome for both necessary and sufficient conditions as well as imposing restrictions on the ring. For example, the poset corresponding to Noetherian rings must obey the ACC. If the poset is a lattice, then the ring must be local. Do nice conditions occur assuming our ring is Artinian or a Dedekind domain? - -REPLY [15 votes]: Hochster answered this question in his thesis (by finding such conditions). See: Hochster, M. Prime ideal structure in commutative rings. Trans. Amer. Math. Soc. 142 1969 43–60. - -REPLY [10 votes]: Beside considering the article by Hochster you could also take a look at -WILLIAM J. LEWIS, The Spectrum of a Ring as a Partially Ordered Set, -JOURNAL OF ALGEBRA 25, 419-434 (1973). -It is online: -www.maths.manchester.ac.uk/.../Lewis-Thespectrumofaringasapartiallyorderedset.pdf<|endoftext|> -TITLE: When will the pushforward of a structure sheaf still be a structure sheaf? -QUESTION [27 upvotes]: Let $f:X\rightarrow Y$ be a morphism of schemes. - -When $PicY\rightarrow PicX$ is an embedding and $f_{*}\mathscr{O}_{X}$ is invertible, it is the structure sheaf of $Y$. -In the proof of Zariski's Main Theorem, we have: If $f$ is birational, finite, integral, and $Y$ is normal, then $f_{*}\mathscr{O}_{X}$ is the structure sheaf of $Y$. - -My questions are -1) What exactly prevent $f_{*}\mathscr{O}_{X}$ to be a structure sheaf? -2) Is there any necessary and sufficient condition(s) guarantee that $f_{*}\mathscr{O}_{X}$ is a structure sheaf? - -REPLY [24 votes]: Q: Exactly what information is contained in $f_*\mathscr O_X$? Look at the -definition. For any $U\subseteq Y$ open, $f_*\mathscr O_X(U) = \mathscr O_X(f^{-1}(U))$ = -regular functions on $f^{-1}(U)$. So the information in $f_*\mathscr O_X$ is related -to the sets in $X$ of form $f^{-1}(U)$. -Cases where $f_*\mathscr O_X$ contains as little information about $X$ as possible. -If $X$ is irreducible and projective and $f$ is constant, e.g. if $Y$ is affine, then -the only non empty set of form $f^{-1}(U)$ in $X$ is $X$ itself. In this case -$f_*\mathscr O_X$ is a skyscraper sheaf with stalk $k$ supported on the image point -of $f$ in $Y$. There is very little information here about $X$, but perhaps we do -see that $f$ is constant and that $X$ is connected. More generally, if $Z$ is a -projective variety, $Y$ is any variety, and $X = Z\times Y$, and $f:Z\times Y\to Y$ -is the projection, then $f^{-1}(U) = Z\times U$, so an element of $f_*\mathscr -O_X(U)$, i.e. a regular function on $f^{-1}(U)$, is determined by its restriction to -$\{p\}\times U$ for any $p\in X$, i.e., a regular function on $U$ in $Y$. Thus in -this case we have $f_*\mathscr O_X = \mathscr O_Y$. Consequently in this case -$f_*\mathscr O_X$ recovers $Y$, but contains no information at all about $X$. -In general, if $f:X\to Y$ is a projective morphism with every fiber connected, and -$Y$ is any normal variety, then $f_*\mathscr O_X = \mathscr O_Y$, so again -$f_*\mathscr O_X$ contains little information about $X$. Recall that if $X$ is a -projective variety then every morphism out of $X$ is a projective morphism, and more -generally a projective morphism $X\to Y$ is one that factors via an isomorphism of X -with a closed subvariety of $\mathbb P^n\times Y$, followed by the projection -$\mathbb P^n\times Y\to Y$. Suppose that $f:X\to Y$ is any projective morphism. -Then the fibers $f^{-1}(y)$ over points $y \in Y$ are all finite unions of projective -varieties. Therefore for any open set $U\subseteq Y$ containing the point $y$, the -only regular functions in $\mathscr O_X(f^{-1}(U)) = f_*\mathscr O_X(U)$ are constant -on every connected component of the fiber $f^{-1}(y)$. Thus $f_*\mathscr O_X$ can -contain little information about $X$ and $f$, other than at most the connected -components of the fibers. We shall see below that it contains exactly this -information. -Cases where $f_*\mathscr O_X$ contains as much information about $X$ as possible. -If $f:X\to Y$ is a map of affine varieties, then the global sections of $f_*\mathscr -O_X$ determine $X$ completely, since then $H^0(Y,f_*\mathscr O_X) = H^0(X,\mathscr -O_X)$, and then $X = \mathrm{Spec}h^0(X,\mathscr O_X)$, is the unique affine variety -with coordinate ring $H^0(X,\mathscr O_X)$. The generalization of this case is that -of any affine map $f:X\to Y$, since then $X$ can be recovered by patching together -the analogous construction from $H^0(U,f_*\mathscr O_X)$ for affine open sets -$U\subseteq Y$. Thus $X$ is completely determined by $f_*\mathscr O_X$ for any -affine map $f:X\to Y$, and this is essentially the only case. I.e. in general -$f_*\mathscr O_X$ is always a quasi coherent $\mathscr O_Y$ algebra, and if we want -it to determine a variety, as opposed to a "scheme", it is reasonable to assume for -all $U\subseteq Y$ affine open, that $f_*\mathscr O_X(U)$ is a finitely generated k -algebra, as well as an $\mathscr O_Y(U)$ algebra. We may call temporarily such an -$\mathscr O_Y$ algebra "of finite type". Thus if $f:X\to Y$ is any morphism such that -$f_*\mathscr O_X$ is of finite type, then the patching construction above yields not -necessarily $X$, but a variety $Z$ and an affine map $h:Z\to Y$ which factors via a -map $g:X\to Z$, where $f = h\circ g$, and where $g_*(\mathscr O_X) = \mathscr O_Z$. -In particular then, we have $f_*\mathscr O_X = (h\circ g)_*(\mathscr O_X) = -h_*(g_*(\mathscr O_X))= h_*(\mathscr O_Z)$. So since $h$ is affine, $f_*\mathscr O_X = -h_*(\mathscr O_Z)$ determines not $X$, but $Z$. (Kempf, section 6.5.) -The case of an arbitrary projective morphism. -Now when $f:X\to Y$ is any projective morphism, then $f_*\mathscr O_X$ is a coherent -$\mathscr O_Y$-module, hence we get a factorization of $f$ as $h\circ g:X\to Z\to Y$, -where $h:Z\to Y$ is affine, and where also $h_*(\mathscr O_Z) = f_*\mathscr O_X$. -Then $h$ is not only an affine map, but since $h_*(\mathscr O_Z)$ is a coherent $\mathscr -O_Y$-module, $h$ is also a finite map. Moreover $g:X\to Z$ is also projective and since -$g_*(\mathscr O_X) = \mathscr O_Z$, it can be shown that the fibers of $g$ are connected. -Hence an arbitrary projective map $f$ factors through a projective map g with connected -fibers, followed by a finite map $h$. Thus in this case, the algebra $f_*\mathscr O_X$ -determines exactly the finite part $h:Z\to Y$ of $f$, whose points over $y$ are precisely -the connected components of the fiber $f^{-1}(y)$. -One corollary of this is "Zariski's connectedness theorem". If $f:X\to Y$ is projective -and birational, and $Y$ is normal then $f_*\mathscr O_X= \mathscr O_Y$, and all fibers -of $f$ are connected, since in this case $Z = Y$ in the Stein factorization described -above. If we assume in addition that $f$ is quasi finite, i.e. has finite fibers, then -$f$ is an isomorphism. More generally, if $Y$ is normal and $f:X\to Y$ is any birational, -quasi - finite, morphism, then $f$ is an embedding onto an open subset of $Y$ ("Zariski's -'main theorem' "). More generally still, any quasi finite morphism factors through -an open embedding and a finite morphism. - -REPLY [3 votes]: Another issue that has not been addressed is what happens if $f$ is not proper. You may have intended to assume that it is, but it also an interesting question for not necessarily proper morphisms. For that matter, you could ask "if $f:X\hookrightarrow Y$ is an open embedding, when will $f_*\mathscr O_X$ be isomorphic to $\mathscr O_Y$?" You are also writing that "... if $f_*\mathscr O_X$ is a line bundle, then ...". It should be noted that this is actually a strong restriction. For instance if you have a generically finite morphism that satisfies this, then it has to be birational. -For the question of an open embedding the answer is relatively simple. If the complement of $X$ in $Y$ has a non-empty codimension $1$ part, then $f_*\mathscr O_X$ is not even coherent, so little chance there. If the complement is of codimension at least $2$, then this is a condition on the singularities of $Y\setminus X$, and essentially equivalent to $Y$ being $S_2$ along $X\setminus Y$.<|endoftext|> -TITLE: orbifold covering -QUESTION [10 upvotes]: Given two compact surfaces $S_1$ and $S_2$ of genus at least $2,$ it is easy to tell when $S_1$ covers $S_2:$ whenever $\chi(S_2)$ divides $\chi(S_1).$ Now, suppose I have two orbifolds of negative Euler characteristic. There is still the divisibility obstruction, but there must be others. Or must there? - -REPLY [9 votes]: Since $\chi(S_1)$ and $\chi(S_2)$ are non-zero, you know what the degree a possible covering should be: it is -$$d=\chi(S_1)/\chi(S_2).$$ -Let $S_2$ have $k$ cone points of order -$$d_1, \ldots, d_k \geqslant 2.$$ -If a covering $S_1 \to S_2$ exists, then the preimage of the cone point of order $d_i$ is a collection of $k_i \leqslant d$ cone points of order $d_{i1}, \ldots, d_{i_{k_i}}\geqslant 1$ (order 1 = smooth point). Every $d_{ij}$ divides $d_i$ and we have -$$d = d_i/d_{i1} + \ldots + d_i/d_{i_{k_i}}.$$ -A necessary condition for having a covering $S_1 \to S_2$ is therefore the following: - -By adding some auxiliary 1's to the set of all cone orders of $S_1$, we must get a set of natural numbers which can be partitioned into subsets $\{d_{i1}, \ldots, d_{i_{k_i}}\}$ such that every $d_{ij}$ divides $d_i$ and by summing the natural numbers $d_i/d_{ij}$ along $j$ we get $d$, for every $i$. - -The non-trivial problem here is: is this numerical condition enough to guarantee the existence of a covering? The same problem can be rephrased in therms of branched coverings of surfaces, and is called the Hurewitz existence problem. The Hurewticz problem has a positive solution when $S_2$ is not a sphere, i.e. when it is a surface with genus (and cone points), as proved by Husemoller in 1962. I think that this implies that an orbifold covering exists when $S_2$ has positive genus. -When $S_2$ is a sphere there are some cases where the Hurewitcz problem has no solution, i.e. the necessary conditions above do not guarantee the existence of a covering. The general case when tha base hyperbolic orbifold $S_2$ is a sphere with some cone points is open, see some recent papers of Pascali, Pervova and Petronio.<|endoftext|> -TITLE: Where do surreal numbers come from and what do they mean? -QUESTION [31 upvotes]: I know about Conway's original discovery of the surreal numbers by way of games, -as well as Kruskal's way of viewing surreal numbers in terms of asymptotic behavior -of real-valued functions, leading to connections between surreal analysis and the -theory of o-minimal structures (if Kruskal isn't the right attribution here, please feel -free to correct me and educate everyone else). But I feel that, even with those two -viewpoints available, surprisingly few connections between surreal numbers and -the rest of mathematics have emerged over the past four decades. I say "surprising" -because one would expect something so beautiful and natural to have all kinds of -links with other things! -I think one reason the surreal numbers have found so few points of contact with the rest -of contemporary mathematics is that the simplicity relation $a$-is-simpler-than-$b$ does not -have any translational or dilational symmetries. (The simplest number between $-2$ and $+2$ is $0$, but the simplest number between $-2+1=-1$ and $+2+1=3$ is not $0+1=1$ but $0$. Likewise, the simplest number between $1$ and $3$ is $2$, but the simplest number between -$2 \times 1=2$ and $2 \times 3=6$ is not $2 \times 2=4$ but $3$.) In the wake of Bourbaki, mathematicians have -favored structures that have lots of morphisms to and from other already-favored structures, -and/or lots of isomorphisms to themselves (aka symmetries), and the surreal numbers don't -fit in with this esthetic. -Are there any new insights into how the surreal numbers fit in with the rest of math -(or why they don't)? -See also my companion post What are some examples of "chimeras" in mathematics? . -It occurred to me after I posted my question that there is a weak $p$-adic analogue of the 2-adic surreal-numbers set-up, in which one relaxes the constraint that every interval contains a unique simplest number (that's a lot to give up, I admit!). If one defines $p$-adic simplicity in ${\bf Z}[1/p]$ in the obvious way (changing "2" to "$p$" in Conway's definition, so that integers are small if they are near 0 in the usual sense and elements of ${\bf Z}[1/p]$ are small if they have small denominator), then the following is true for $a_L,a_R,b_L,b_R$ in ${\bf Z}[1/p]$: if there is a unique simplest $a$ in ${\bf Z}[1/p]$ that is greater than $a_L$ and less than $a_R$, and there is a unique simplest $b$ in ${\bf Z}[1/p]$ that is greater than $b_L$ and less than $b_R$, then there is a simplest $c$ in ${\bf Z}[1/p]$ that is greater than $a+b_L$ and $a_L+b$ and less than $a+b_R$ and $a_R+b$, and it satisfies $c=a+b$. (Conway's multiplication formula works in this setting as well.) Is this mentioned in the surreal numbers literature, and more importantly, does the observation lead anywhere? - -REPLY [3 votes]: See this paper in regard to the p-adic analogue: -http://arxiv.org/abs/1108.0962<|endoftext|> -TITLE: What are some examples of "chimeras" in mathematics? -QUESTION [25 upvotes]: The best example I can think of at the moment is Conway's surreal number system, which -combines 2-adic behavior in-the-small with $\infty$-adic behavior in the large. The surreally -simplest element of a subset of the positive (or negative) integers is the one closest to 0 -with respect to the Archimedean norm, while the surreally simplest dyadic rational in a -subinterval of (0,1) (or more generally $(n,n+1)$ for any integer $n$) is the one closest to 0 -with respect to the 2-adic norm (that is, the one with the smallest denominator). -This chimericity also comes up very concretely in the theory of Hackenbush strings: -the value of a string is gotten by reading the first part of the string as the unary representation -of an integer and the rest of the string as the binary representation of a number between -0 and 1 and adding the two. -I'm having a hard time saying exactly what I mean by chimericity in general, but -some non-examples may convey a better sense of what I don't mean by the term. -A number system consisting of the positive reals and the negative integers would -be chimeric, but since it doesn't arise naturally (as far as I know), it doesn't qualify. -Likewise the continuous map from $\bf{C}$ to $\bf{C}$ that sends $x+iy$ to $x+i|y|$ is chimeric (one does -not expect to see a holomorphic function and a conjugate-holomorphic function stitched -together in this Frankenstein-like fashion), so this would qualify if it ever arose naturally, but I've never seen anything like it. -Non-Euclidean geometries have different behavior in the large and in the small, but the -two behaviors don't seem really incompatible to me (especially since it's possible to -continuously transition between non-zero curvature and zero curvature geometries). -One source of examples of chimeras could be physics, since any successful Theory Of -Everything would have to look like general relativity in the large and quantum theory in -the small, and this divide is notoriously difficult to bridge. But perhaps there are other -mathematical chimeras with a purely mathematical genesis. -See also my companion post Where do surreal numbers come from and what do they mean? . - -REPLY [4 votes]: Outer automorphisms of free groups have a chimeric nature, somewhat like mapping classes of surfaces but with weirder pieces and stranger stiches. The pieces are ``strata of relative train track maps'', and are somewhat analogous to the subsurfaces of the Thurston decomposition of a mapping class, but the strata can spill over and interact with each other in ways that the subsurfaces cannot. -For instance, you can have two different exponentially growing strata, which as in the surface situation correspond to two different exponenially stretched laminations each having a dense leaf, but one of those laminations contains the other as a sublamination. -You can also have an exponentially growing stratum and a fixed stratum --- the latter analogous to a subsurface on which the mapping class is the identity --- but the lamination corresponding to the exponentially growing stratum scribbles all over the fixed stratum, filling it up with junk. -And then there are the linearly and polynomially growing strata. A linear stratum spills over a fixed stratum, a quadratic stratum spills over a linear stratum, etc. And last but not least, there are the nonexponentially growing strata that spill over exponentially growing strata; I still can't decide whether they grow or they don't grow under iteration of the outer automorphism.<|endoftext|> -TITLE: What is the Picard group of a Schubert variety in the affine Grassmannian? -QUESTION [5 upvotes]: I'm not sure I have a lot more to say than the title. Let $G$ be your favorite simple algebraic group over $\mathbb{C}$, and let $$\overline {\mathrm{Gr}}_\lambda= \overline{G(\mathbb{C}[[t]])\cdot t^\lambda \cdot G(\mathbb{C}[[t]])}/ G(\mathbb{C}[[t]]).$$ It's a commonly cited theorem that $\overline {\mathrm{Gr}}_\lambda$ is a projective variety for every $\lambda$, but the usual tricks for finding the Picard group of a Schubert variety in the finite dimensional case don't work (the group $G(\mathbb{C}[[t]])$ is perfect if $G$ is semi-simple). Is this Picard group computed anywhere in the literature? - -REPLY [6 votes]: [From the comments] The proof of Proposition 13.2.19 in Kumar's "Kac-Moody groups, their flag varieties and representation theory" appears to provide the requested information.<|endoftext|> -TITLE: Completely bounded maps on Mn -QUESTION [5 upvotes]: The aim of this question is to collect nice maps on $M_n(\mathbb{C})$ with the following property: - -$\phi_n:M_n(\mathbb{C})\rightarrow M_n(\mathbb{C})$ with $||\phi||=1$ and $||\phi_n||_{cb}\rightarrow \infty$ when $n\rightarrow \infty$ -Here $||\phi_n||_{cb}$ denotes completely bounded norm of $\phi_n$, i.e. in this special case $||\phi_n||_{cb}=||\phi_{n}^{(n)}||$, where $\phi_{n}^{(n)}:M_n(\mathbb{C})\otimes M_n(\mathbb{C})\rightarrow M_n(\mathbb{C})\otimes M_n(\mathbb{C})$ such that $\phi_{n}^{(n)}([a_{ij}])=[\phi_n(a_{ij})]$ for every $[a_{ij}]\in M_n(M_n(\mathbb{C}))$. - -There are many examples of such maps, the simplest one is the transposition of matrix. -Please contribute yours... - -REPLY [4 votes]: It is well known that each Schur multiplier satisfies $||.||_{cb}=||.||$. These maps are the $D_n$-bimodule maps on $M_n$. For the right module maps, the situation is more complicated. This gives concretes exemples of maps with your property. See the following paper: -http://arxiv.org/PS_cache/arxiv/pdf/1102/1102.4255v1.pdf<|endoftext|> -TITLE: Ring of continuous functions, reference request. -QUESTION [7 upvotes]: I am looking for a reference for the following facts in functional analysis and topology. (If these "facts" are not true, I suppose I'm looking for the closest approximation which is true.) -Let $X$ be a locally compact Hausdorff topological space. Let $C(X)$ denote the ring of continuous complex-valued functions on $X$, endowed with the compact-open topology. Then $C(X)$ is a complete locally convex topological (complex) vector space (this can be found in Kothe, vol. 2, I think). -Now let $Y$ be another locally compact Hausdorff topological space. From Kothe, vol. 2, I know that $C(X \times Y)$ is naturally isomorphic to $C(X) \hat \otimes C(Y)$, where $\hat \otimes$ denotes the completion with respect to the injective tensor product topology. -I believe that pointwise multiplication $C(X) \times C(X) \rightarrow C(X)$ extends (uniquely) to a continuous linear map from $C(X) \hat \otimes C(X)$ to $C(X)$. -If $f: X \rightarrow Y$ is a continuous function, then precomposition with $f$ yields a continuous $C$-algebra homomorphism $f^\ast: C(Y) \rightarrow C(X)$. -I believe the following to be true: -Theorem: For every continuous algebra homomorphism $\phi: C(Y) \rightarrow C(X)$, there exists a unique continuous map $f: X \rightarrow Y$ such that $\phi = f^\ast$. -In other words, I wish that $C( \bullet )$ is a faithful functor from the category of locally compact Hausdorff spaces and continuous maps to the category of rings in the (symmetric monoidal under $\hat \otimes$) category of complete locally convex topological vector spaces and continuous linear maps. -Any references and/or corrections would be very welcome! -But an important note: I am not looking for well-known modifications, like "try the $C^\ast$-algebra instead" or the von Neumann algebra, etc.. I have good reasons for considering the ring $C(X)$ with the compact-open topology, and I don't wish to mess with it. - -REPLY [2 votes]: Let $X=Y_1\sqcup Y_2$, with both $Y_i$ homeomorphic to $Y$. Then $C(X)=C(Y_1)\oplus C(Y_2)$. Given $a\in C(Y)$ let $\phi\colon a\mapsto a\oplus 0$, in the obvious way. This $\phi$ cannot be any $f^*$, since $f^*$ would necessarily map $1\mapsto 1\oplus 1$. I believe, this is a counter example to your putative theorem, which shows that you may want a connectedness hypothesis on your spaces. -For more general information, I second Ramsey's recommendation of "Rings of continuous functions" by Gillman and Jerison. Though, I don't think it has the exact theorem you are looking for. -The strongest relevant result from that book is Theorem 10.8, which states that a homomorphism $\mathfrak{s}\colon C(Y)\to C(X)$ determines a unique continuous map $\tau\colon E\to \upsilon Y$ with the properties like what you want. Here $E$ is a clopen subset of $X$ and $\upsilon Y$ is the (Hewitt) realcompactification of $Y$, which is a bigger space than $Y$. See the book for full details. The hypotheses on $X$ and $Y$ (implicitly) include complete regularity, which is weaker than local compactness. Note that the homomorphism $\mathfrak{s}$ is not assumed to be continuous in any topologies on $C(Y)$ and $C(X)$. Perhaps your continuity requirement is enough to cut $\upsilon Y$ down to $Y$ and give you the desired result.<|endoftext|> -TITLE: "Famous" 2d Riemannian manifolds with non-constant curvature -QUESTION [6 upvotes]: I'm looking for "famous" or otherwise well-known 2d Riemannian manifolds which have non-constant curvatures but have a non-trivial Killing vector field. Of course there are tons of spaces like these, for instance if we parametrize the plane (or a subset of it) by $(r,\phi)$ then any conformal rescaling of the flat metric by a conformal factor which only depends on $r$ will be generally good, i.e. have non-constant curvature but the rotations generated by $\partial/\partial\phi$ will still be a symmetry. -Are there special spaces which are somehow famous or well-studied because of some special property? Ideally, I'm looking for deformations of the Poincare disc. - -REPLY [4 votes]: My favorite 2D metrics of nonconstant curvature admitting a Killing vector field are the so-called Darboux-superintegrable metrics. Their definition is: the space of Killing tensors of degree two (i.e., integrals of the geodesic flow that are quadratic in velocities) is 4-dimensional. These metrics necessary admit Killing vector field. Their local description was known to Koenigs (Note II from Darboux "Sur la theorie generale des surfaces, Vol. IV, 1896) and S. Lie (1882), one can get a list of these metrics from arXiv:math-ph/0307039v1 or, alternatively from the paper arXiv:0705.3592 (the metrics 2a,2b,2c from Theorem 1). The lists are equivalent modulo coordinate transformation and contain also semi-riemannian metrics. -Let me explain why I love these metrics: -(1) They appear in differential geometry and physics. In differential geometry, they appear in studying of projective connections with big group of symmetries and their analysis was one of the main ingredients of the solution of two problems explicitly stated by S. Lie which is one of my bests results. They also appeared in mathoverflow: for example the metrics suggested in the Bryant's answer on -Riemannian surfaces with an explicit distance function? are Darboux-superintegrable. -(2) They appear in physics: many physical phenomena lead to such metrics. There is a big activity about it in the "superintegrability" community (consult, for example, the papers of Winternitz et al in ArXive). -(3) For these metrics, one can answer many natural questions that normally require solving systems of ODE or PDE (such that description of geodesics or eigenvalues of the Laplacian) by algebraic methods, using the Killing tensors. -My personal opinion is that the metrics appeared naturally in differential geometry, because they are "the third best" metrics: the best metric is flat, the second best is of constant curvature, and the Darboux-superintegrable are the next simplest choice. The metrics appeared in physics, since physicists love solvable models, and models described by these metrics are sometimes solvable.<|endoftext|> -TITLE: What universities have laid off tenured math faculty for financial reasons? -QUESTION [59 upvotes]: When a math department lays off tenured staff, people cry out loud. -But, 10 years later, such memories are no longer popular discussion subjects, and so the information doesn't always spread. -Those who lived through it will of course remember. But will the younger get to know of the troubled past of a given university? -I would like to record incidents of universities laying off tenured math faculty for financial reasons. If you know of such an event, please write the name of the university, the year when it happened, and the number of tenured faculty that got laid off. Other relevant information, such whether or not there was a lawsuit, aggravating circumstances, etc. should also be included. -(This is a follow up on this discussion about the VU Amsterdam laying off people.) - -REPLY [3 votes]: Since other answers mention cases which were not financially motivated, here is another one: -Gutkin v. University of Southern California (2002) -The document below concerns the complaint Eugene Gutkin filed in 2001 against University of Southern California, which dismissed him in 2000. He appealed the decision and sought damages, but lost the case. I will quote the paragraph describing the circumstances of his dismissal; the full text is here: -http://law.justia.com/cases/california/court-of-appeal/4th/101/967.html -"Gutkin's complaint alleged that he was a tenured professor of mathematics at the University. In the wake of a dispute over the University's requirement that Gutkin teach extra classes to "make up" for classes he had not been able to teach in the fall 1995 semester, the University initiated dismissal proceedings against Gutkin in October 1996. But Gutkin's dismissal hearing was not scheduled to take place until early December 1998, because, according to Gutkin, "[i]t took [the University] longer than anticipated to effectuate the deceptive alterations to the [faculty] [h]andbook" that would govern the "rigged [dismissal] procedures," and the University "dragged its feet." Gutkin further alleged that by December 1998, "[the University] had not finished its tampering, so the deceptively altered [h]andbook was not ready for posting on the Internet by the time of the scheduled hearing. . . . Consequently, [the University] unilaterally postponed the hearing until February 26, 1999." -The dismissal procedure outlined in the faculty handbook required a hearing before a panel of Gutkin's faculty peers. According to Gutkin, the panel selection process set out in the revised faculty handbook constituted a "charade of impartiality" that resulted in a "sham dismissal procedure." In March 1999, the faculty panel issued a recommendation to the president of the University that found that Gutkin had engaged in "serious neglect of duty," a ground for termination in the faculty handbook. The president of the University terminated Gutkin in March 2000 for "serious neglect of duty," as found by the faculty panel."<|endoftext|> -TITLE: Congruences between Fermat quotients -QUESTION [9 upvotes]: This a repost of a question I asked at Stack Exchange: -https://math.stackexchange.com/questions/35264/congruences-for-fermat-quotients -I didn't get a complete answer to my question, so I'm trying again here. -If $p$ is a prime number and $a$ is relatively prime to $p$, then by Fermat's Little Theorem, the Fermat quotient $q_p(a) = (a^{p-1}-1)/p$ is an integer. To summarize that other post, I stumbled across some congruences relating Fermat quotients and wonder 1) if these and/or similar congruences are known, and 2)where I can find other proofs than my own, either simple enough to post here or written up somewhere. -Here are some examples: -If $p = 2^a-1$ is a Mersenne prime, so $a$ is prime, then $q_p(2) \equiv 2q_a(2) \pmod{p}$. -If $p = 2^n-3$ is prime for some $n$, then $3nq_p(2) + 1 \equiv 3q_p(3) \pmod{p}$ -If $p = 3^n-4$ is prime for some $n$, then $4nq_p(3) + 1 \equiv 8 q_p(2) \pmod{p}$ -I have proved a few others of this nature and seem to have a way to generate more if I wish. -Edit: These and other similar congruences follow from an elementary argument of Edmund Landau. See my answer to my own question belowl. - -REPLY [6 votes]: After further thought, I realized that most of these congruences are all immediate consequences of the well-known property of Fermat quotients -$$ q_p (a^n) \equiv n q_p (a) \pmod{p} $$ -along with the following property: if $p$ is an odd prime and $r$ is an integer and we write $r p = a \pm b$ with $(p,ab)=1$, then -$$ a q_p (b) \pm b q_p(a) \equiv r \pmod{p} $$ -Together, these show that if $rp = a^m \pm b^n$, then -$$ a^m nq_p (b) \pm b^n m q_p (a) \equiv r \pmod{p} $$ -What shocks me is that these latter congruences have a short completely elementary proof but are not mentioned in a lot of the literature on Fermat quotients. I'd add them to the basic properties of Fermat quotients on Wikipedia, except that I doubt one post on a forum constitutes a reliable source. -To prove the second congruence above, note that -$$ b q_p(a) p = b (a^{p-1} - 1) = b \left( (rp \mp b)^{p-1} - 1 \right) \equiv b \left( \mp b^{p-2} r (p-1) +q_p(b) \right) p \pmod{p^2} $$ -and thus -$$ b q_p(a) \equiv \pm b^{p-1} r + b q_p(b) \equiv \pm r + b q_p(b) \pmod{p} $$ -Since $a \equiv \mp b \pmod{p}$ it follows that -$$ \pm b q_p(a) \equiv r - a q_p(b) \pmod{p}$$ -from which the second congruence above follows. -Edit: I finally found a reference! I knew this simple of a result could not be new! The earliest I can find is from a journal called "L'Intermediaire des Mathematiciens", of which I had not previously heard. It seems fascinatingly to be a publication where people could pose short questions hoping that others would answer --- perhaps like a paper version of MO! Does anyone know the history of this journal? Was its function indeed like an early version of MO? -On p.121, volume 20, published 1913, one E. Dubouis poses the question (numbered 3764): How does one prove the theorem of Mirimanoff (C.R. 24 January 1910): `If p is a prime number of the form $2^a 3^b \pm 1$ or of the form $\pm 2^a \pm 3^b$, then at least one of the numbers $2^{p-1} - 1$ and $3^{p-1} -1$ is not divisible by $p^2$?' " -Later in that same volume, p.206, one finds an answer by Endmund Landau: "One can easily prove the more general fact that the following three lines lead to a contradiction: -$$ mp = x+y $$ -$$ p \text{ odd } > 1 \text{ and not dividing } m $$ -$$ x^{p-1} \equiv y^{p-1} \equiv 1 \pmod{p^2} $$ -(These congruences are satisfied for $x=2^a 3^b$, $y = \pm 1$ and for $x = \pm 2^a$, $y = \pm 3^b$, if $2^{p-1} \equiv 3^{p-1} \equiv 1 \pmod{p^2}$; taking $m=1$ and $p$ prime, we recover the question posed by M. Dubouis.) We have: -$$ 1 \equiv x^{p-1} \equiv (mp-y)^{p-1} \equiv -(p-1) mpy^{p-2}+y^{p-1} \equiv -(p-1) mpy^{p-2} + 1 \pmod{p^2}$$ -and $p$ divides $(p-1) m y^{p-2}$, hence $m$, contrary to hypothesis." -So although E. Landau doesn't explicitly state the second congruence above, the proof I gave is almost the same as Landau's.<|endoftext|> -TITLE: Elegant proof that any closed, oriented 3-manifold is the boundary of some oriented 4-manifold? -QUESTION [47 upvotes]: I'm looking for an elegant proof that any closed, oriented $3$-manifold $M$ is the boundary of some oriented $4$-manifold $B$. - -REPLY [23 votes]: Thom wrote two notes in the proceedings of the "Colloque de Topologie de Strasbourg", which was a topology seminar organized by Ehresmann at that time: - -"Quelques propriétés des variétés-bords", Colloque de Topologie de Strasbourg, 1951, no. V. La Bibliothèque Nationale et Universitaire de Strasbourg, 1952. - -"Sur les variétés cobordantes", Colloque de topologie et géométrie différentielle, Strasbourg, 1952, no. 7. La Bibliothèque Nationale et Universitaire de Strasbourg, 1953. - - -These two notes prefigure some of the results and some of the techniques that will appear slightly later in his 1954 paper "Quelques propriétés globales des variétés différentiables" and each of these notes contains a proof of the fact that $\Omega_3=0$. In fact, these two prior proofs are different from the demonstration that follows by specializing the results of his '54 paper, and which was alluded to in the above answers. -His '52 proof (in the 1953 proceedings) is close to Pontryagin's ideas relating framed cobordism groups to stable homotopy groups of spheres. Thom considers there the map $\psi_k:\pi_{n+k}(S^n) \to \Omega_k$ which consists in taking regular preimages of smooth approximations of maps $S^{n+k} \to S^n$. This map is additive and its image consists of the cobordism classes of those $k$-manifolds which can be embedded in $\mathbb{R}^{n+k}$ with trivial normal bundle. Since any closed orientable $3$-manifold is parallelizable, the map $\psi_3$ is surjective (for $n$ large enough) and it suffices to prove that $\psi_3$ vanishes on a generating set of the group $\pi_{n+3}(S^n)$. To conclude his argument, Thom mentions that this stable homotopy group is isomorphic to $\mathbb{Z}_{24}$ (which, I guess, he knew from Cartan or Serre) and, most importantly, that it is generated by the suspension of the quaternionic Hopf fibration with fiber $S^3= \partial D^4$. -In contrast with his '52 and '54 proofs, his '51 proof (in the 1952 proceedings) does not use any kind of Pontryagin-Thom constructions but, as Lickorish and Rourke will do some years later, he uses the fact that any closed oriented $3$-manifold can be presented by a Heegaard splitting. Let $H_g$ be the handlebody of genus $g$, let $\hbox{MCG}(\partial H_g)$ denote the mapping class group of its boundary and for any $f \in \hbox{MCG}(\partial H_g)$, set -$$ -V_f := H_g \cup_f (-H_g). -$$ -Thom starts by observing that, if $V_f$ and $V_{f'}$ bound, then so does $V_{f \circ f'}$ (see Bruno's comment to Daniel's answer). Next, he recalls that Dehn found an explicit and finite system of generators $\{\tau_i\}$ for $\hbox{MCG}(\partial H_g)$ which consists of several Dehn twists: there are 2, 5 and 2g(g-1) such generators for $g=1$, $g=2$ and $g>2$ respectively. Thus, by the previous observation, it suffices to check that $V_{\tau_i}$ is a boundary for any $g\geq 1$ and for any $i$. For any $g>1$, it can be observed that each of the Dehn twists $\tau_i$ is performed along a curve which avoids a full (solid) handle of $H_g$: hence $V_{\tau_i}$ is the connected sum of $S^1 \times S^2$ with a $3$-manifold admitting a Heegaard splitting of genus $g-1$. Since such a connected sum can be realized in dimension $4$ by the attachement of a handle of index $1$, we are allowed to do an induction on the genus $g$. For $g=1$, the $3$-manifolds $V_{\tau_1}$ and $V_{\tau_2}$ are $S^3$ and $S^1 \times S^2$, which are obviously boundaries. -It seems that this early proof that $\Omega_3=0$ has been forgotten since then: this is probably due to the fact that the proceedings is not widely available and is written in French. Personally, I did not know it before I opened this proceedings in the Math' Library of Strasbourg a few months ago! Nonetheless, Haefliger mentions it in his survey paper of Thom's works (Publ. IHES 1988).<|endoftext|> -TITLE: Surreal Numbers as Inductive Type? -QUESTION [23 upvotes]: Prompted by James Propp's recent question about surreal numbers, I was wondering whether anyone had investigated the idea of describing surreal numbers (as ordered class) in terms of a universal property, roughly along the following lines. -In categorical interpretations of type theories, it is common to describe inductive or recursive types as so-called initial algebras of endofunctors. The most famous example is the type of natural numbers, which is universal or initial among all sets $X$ which come equipped with an element $x$ and an function $f: X \to X$. In other words, initial among sets $X$ which come equipped with functions $1 + X \to X$ (the plus is coproduct); we say such sets are algebras of the endofunctor $F$ defined by $F(X) = 1 + X$. Another example is the type of binary trees, which can be described as initial with respect to sets that come equipped with an element and a binary operation, or in other words the initial algebra for the endofunctor $F(X) = 1 + X^2$. -In their book Algebraic Set Theory, Joyal and Moerdijk gave a kind of algebraic description of the cumulative hierarchy $V$. Under some reasonable assumptions on a background category (whose objects may be informally regarded as classes, and equipped with a structure which allows a notion of 'smallness'), they define a ZF-structure as an ordered object which has small sups (in particular, an empty sup with which to get started) and with a 'successor' function. Then, against such a background, they define the cumulative hierarchy $V$ as the initial ZF-structure, and show that it satisfies the axioms of ZF set theory (the possible backgrounds allow intuitionistic logic). By tweaking the assumptions on the successor function, they are able to describe other set-theoretic structures; for example, the initial ZF-structure with a monotone successor gives $O$, the class of ordinals, relative to the background. -Now it is well-known that surreal numbers generalize ordinals, or rather that ordinals are special numbers where player R has no options, or in different terms, where there are no numbers past the 'Dedekind cut' which divides L options from R options. In any case, on account of the highly recursive nature of surreal numbers, it is extremely tempting to believe that they too can be described as a recursive type, or as an initial algebraic structure of some sort (in a background category along the lines given by Joyal-Moerdijk, presumably). But what would it be exactly? -I suppose that if I knocked my head against a wall for a while, I might be able to figure it out or at least make a strong guess, but maybe someone else has already worked through the details? - -REPLY [11 votes]: Heh, I just ran across this old question again and realized that I now know an answer: it's in section 11.6 of the HoTT Book. It's written out there as a higher inductive-inductive type; in other language it says roughly that $\mathbf{No}$ is the initial object among classes $X$ equipped with the following. - -Two binary relations $<$ and $\le$. -For any two sub-sets $L,R\subseteq X$ such that $x^L -TITLE: Do I need to know what an infinity-Gerstenhaber algebra is, and if so, what is it? -QUESTION [10 upvotes]: I am in the following situation. I have two (rather explicit and specific) dg commutative algebras $R,S$ over a field of characteristic $0$. In fact, $S$ is an $R$-algebra, in that I have a map $R \to S$. Because I am interested in computing some derived tensor products $S \otimes_R$, I have worked out a "Koszul" resolution $\tilde S$ of $S$ over $R$. So all seems well. -But! Actually, $S$ and $R$ are both Gerstenhaber algebras (in dg vector spaces; the differential and the Gerstenhaber bracket point in opposite directions), and the map $R \to S$ is a homomorphism of Gerstenhaber algebras. My problem is that I have been so far unsuccessful at giving the resolution $\tilde S$ a Gerstenhaber structure such that the resolved map $R \to \tilde S$ is a homomorphism of Gerstenhaber algebras. -This leads me to two questions. The second question depends on the answer to the first. - -Question 1: Does there necessarily exist a resolution of $S$ that computes the derived $S\otimes_R$ and that is Gerstenhaber in a compatible way? -Question 2 if the answer to 1 is YES: How do I construct it? -Question 2 if the answer to 1 is NO: Certainly my homotopy equivalence $S \leftrightarrow \tilde S$ allows me to move the Gerstenhaber structure on $S$ to something on $\tilde S$. What structure on $\tilde S$ does it move to? - -REPLY [9 votes]: Question 1: Does there necessarily - exist a resolution of S that computes - the derived $S\otimes_R$ and that is - Gerstenhaber in a compatible way? - -Yes. As pointed out in the comments, the category of dg Gerstenhaber algebra admits a model structure in which the weak equivalences are the quasi-isomorphisms, fibrations are degreewise surjections, and cofibrant objects are those dg Gerstenhaber algebras that are free as graded algerbas. -This actually works with dg algebras over any given operad $\mathcal O$ (in place of Gerstenhaber). -This is proved in Hinich's paper (quoted by the nLab: http://ncatlab.org/nlab/show/model+structure+on+dg-algebras+over+an+operad). -Then there is also a natural model structure on the category of dg Gerstenhaber $R$-algebras (there is a more general statement about existence of a model structure on the category of objects under a given one $X$ in a model category $\mathcal C$). -So, the answer to the title of your question is that you don't "need" to know what a $G_\infty$-algebra is. - -Question 2 if the answer to 1 is YES: How do I construct it? - -Shortly, bar-cobar. You can have a look at Homologie et model minimal des algèbres de Gerstenhaber in order to see how it works in details. -Btw, the above paper also tells you what is the definition of a $G_\infty$-algebra. - -Question 2 if the answer to 1 is NO: - Certainly my homotopy equivalence $S\leftrightarrow\widetilde{S}$ - allows me to move the Gerstenhaber - structure on $S$ to something on $\widetilde{S}$. - What structure on $\widetilde{S}$ does it move to? - -Even though the answer to Question 1 is YES, there is still something to say here. -There is on $\widetilde{S}$ a $G_\infty$-structure. This is "just" homotopy transfer formula (and the use of the explicit minimal model for the Gerstenhaber operad). -The homotopy transfer for algebras over operad $\mathcal O$, w.r.t. to a cofibrant resolution $\widetilde{\mathcal O}\to\mathcal O$ is proved in the appendix A.2 of my paper with Van den Bergh (see also Theorem 10.3.6 in Loday-Vallette's Algebraic Operads for the Koszul case).<|endoftext|> -TITLE: Which Fréchet spaces have a dual that is a Fréchet space? -QUESTION [23 upvotes]: I've read the claim that Fréchet spaces that are not Banach spaces never have a dual that is a Fréchet space, but have not been able to find a proof of this statement. Is it trivial or does someone have a reference? - -REPLY [26 votes]: For any locally convex and metrizable space $E$, its strong dual is metrizable if and only if $E$ is normable. - -This and related properties of (F)-spaces are discussed in detail in Topological Vector Spaces I by Köthe (see §29.1, pp. 393-394 in the English edition).<|endoftext|> -TITLE: Why isn't meromorphic continuation enough for converse theorems? -QUESTION [17 upvotes]: This is a very naive question which really does little more than highlight my ignorance of how converse theorems really work. -Take an algebraic gadget which should be conjecturally associated to an automorphic representation. For example, take a finite image continuous complex representation of -the absolute Galois group of a number field. Or take an elliptic curve over a totally real field. There are converse theorems of the form "if the $L$-function of this gadget and the $L$-function of sufficiently many twists of this gadget have analytic continuation, are bounded in vertical strips, and satisfy the expected functional equations, then the gadget is indeed associated to an automorphic representation". For 2-dimensional gadgets (for example 2-dimensional complex Galois representations, or elliptic curves) one only needs to consider abelian twists---this is essentially proved in Jacquet--Langlands. -However it is nowadays pretty standard that these $L$-functions have meromorphic continuation and the expected functional equation. In the finite image case this follows from Brauer's theorem in finite group theory, and in the elliptic curve case this follows from the fact that these curves are now known to be potentially modular, following work of Kisin, Taylor and others. -So one might ask whether meromorphic continuation + functional equation, plus converse theorem techniques, is enough to prove something about the algebraic gadget (other than "it potentially comes from an automorphic representation" -- something which we already know and are using to get the meromorphic continuation). But presumably there is some serious theoretical obstruction to proving anything interesting here, or else it would all have been done for Artin $L$-functions in the 70s. -What is the obstruction? - -REPLY [9 votes]: I think it is also instructive to just look how the entireness condition comes in the proof of the simplest converse theorem: the case of cusp forms for $SL(2,\mathbf{Z})$; there you start with a Dirichlet series -$$L(s)=\sum_{n\geq 1}{a_nn^{-s}}$$ -and form a suitable completed $L$-function -$$\Lambda(s)=(2\pi)^{-s}\Gamma(s)L(s),$$ -and a function -$$ -f(z)=\sum_{n\geq 1}{a_n\exp(2i\pi nz)} -$$ -which is holomorphic and $1$-periodic on the upper half-plane, provided the coefficients of $L$ grow polynomially, which means $L(s)$ converges on some right half-plane. -Now the statement, which is due to Hecke here is (essentially): $f$ is a cusp form of weight $k$ on $SL(2,\mathbf{Z})$ if and only if $L(s)$ is entire with polynomial growth in vertical strips and satisfies $\Lambda(s)=i^k\Lambda(k-s)$. -The proof does not work if you assume merely that $\Lambda(s)$ is meromorphic: one starts by expressing $f(iy)$ as the inverse Mellin transform of $\Lambda(s)$ when the real part of $s$ is large enough, i.e., an integral on a vertical line with large enough real part that the Dirichlet series converges absolutely; then one proceeds to shift the line of integration to the left, so that it becomes possible to apply the functional equation to replace $s$ by $k-s$ and get the modularity relation (of weight $k$) for $z\mapsto -1/z$ (applied to $f(iy)$). The point is that this shift of contour requires that one does not pass through poles in the middle, and that the function which is integrated (here $\Lambda(s)y^s$ roughly) decays fast enough at infinity (to handle the horizontal segments in shifting the contour). -This being said, it is a fact that it is quite hard to find examples of Euler products which (1) converge absolutely in a half-plane, but not everywhere; (2) have analytic continuation (meromorphic) to the whole complex plane; (3) are not related in any way to automorphic forms/Galois representations...<|endoftext|> -TITLE: Martin's "Philosophical Issues about the Hierarchy of Sets" -QUESTION [16 upvotes]: Some months ago (October 2010), in the context of the Workshop on Set Theory and the Philosophy of Mathematics, Professor Donald A. Martin gave a talk entitled "Philosophical issues about the hierarchy of sets". - -Abstract: I will discuss some philosophical questions about the cumulative hierarchy of sets, its levels, and their theories. Some examples: -(1) It is sometimes asserted one cannot quantify over everything. A related assertion is that each of our statements about the universe of sets can from a different perspective be seen as a statement about some Va. Thus the class-set distinction is really a relative one. Does this make sense? Is it right? -(2) Is the first order theory of V determinate? Does every sentence have a truth value? Are there levels of the hierarchy whose first order theories are indeterminate? If so, what is the lowest such level? What about L and the constructibility hierarchy? -(3) There are many examples of proofs of a statement about one level of the hierachy that use principles about a higher level. Under what conditions and in what sense do these count as establishing the lower level statement? -I will discuss these questions mainly from a viewpoint that takes mathematics to be about basic mathematical concepts, e.g., those of natural number, real number, and set. - -I am highly interested in learning how these questions might be answered (as you may problably know from previous questions of mine here in MO), so I would be grateful if anyone could give any information in this respect, especially for those questions of 1 and 3 (I am afraid it is almost impossible to do justice to 2 in a few lines). - -REPLY [12 votes]: Of course there are no universally agreed-upon answers to these philosophical questions, and if you are interested in Martin's views specifically, then I suggest that you read his articles. Meanwhile, allow me simply to explain a few of the issues arising in the specific questions you mention. - -"One cannot quantify over everything." This is a reference to the predicative/impredicative debate in the philosophy of set theory. One of the objections to the replacement and collection axioms is that they are used to describe sets by means of properties of a totality of which they themselves are a member. That is, you define a subset of $B=\{a\in A\mid \varphi(a)\}$, but $\phi(a)$ may be a very complicated property that quantifies over the entire universe, referring to objects and properties of objects, including $B$ itself. But also, it can be a reference to the cumulative view of set theory as building up more and more sets in a process that is never completed, and in this case, it may not be sensible to form sets by means of properties holding in the entire universe, as though it were completed. -"Each of our statements about the universe of sets can from a different perspective be seen as a statement about some $V_\alpha$." The Levy reflection theorem shows that for any assertion $\sigma(x)$, there is an ordinal $\alpha$ such that $\sigma(x)$ is true if and only if it is true in $V_\alpha$, for any $x\in V_\alpha$. That is, $\sigma$ is absolute between $V_\alpha$ and $V$. Going a bit beyond this, consider the theory denoted "$V_\delta\prec V$", which asserts, in the language with a constant for $\delta$, that $\forall x\in V_\delta\, [\varphi(x)\iff \varphi^{V_\delta}(x)]$. This is the scheme asserting that $V_\delta$ is an elementary substructure of the universe. Although some set theorists are surprised to hear it, this scheme is equiconsistent with ZFC, and any model $M$ of set theory can be elementarily embedded into a model of this theory. (This is done by a simple compactness argument; one writes down the theory $V_\delta\prec V$ plus the elementary diagram of $M$, and observes that the reflection theory shows that it is finitely consistent.) Finally, note that in a model of $V_\delta\prec V$, every sentence can be viewed as an assertion about $V_\delta$, rather than about $V$, since they have exactly the same theory. -"Is the first order theory of $V$ determinate?". This question is asking whether there is a fact of the matter in regard to our set-theoretic questions. For example, does it make sense to say that there is ultimately an answer to the question of whether the Continuum Hypothesis is really true? Or whether large cardinals exist? This question is connected in my mind with issues about whether there is a unique structure that we are investigating when we do set theory---the universe $V$ of all sets---or is there instead a multiverse of possibilities? In other words, is there a final truth of the matter in set theory, or is set theory instead something more like geometry, having a plethora of diverse Euclidean and non-Euclidean worlds? In the slides for my talk at the same conference, I explore the multiverse view in detail. -"Are there levels of the hierarchy whose first order theories are indeterminate? What is the lowest such level?" Some set theorists may view questions about the Continuum Hypothesis to be a source of indeterminateness, in the sense that there is no fact of the matter about CH. But CH is a statement expressible in $H_{\omega_2}$, or alternatively in $V_{\omega+2}$. Martin is asking whether we might expect indeterminateness at lower levels. In his talk at the workshop you mention, I recall him saying that he found it unacceptable to think that there would be indeterminateness arising at the level of $V_\omega$, and that arithmetic truth was absolute in some very strong sense. -"There are many examples of proofs of a statement about one level of the hierachy that use principles about a higher level." This is referring to the fact that mathematicians routinely use higher level objects in order to make conclusions about lower level objects. For example, one might use infinite objects (such as automorphisms of field extensions) in order to make conclusions about finite objects, or very large function spaces or ultrafilters in order to make conclusions about a lower level object. Part of Martin's point was the philosophical concern that if there is indeterminism about features of the higher level objects, then they might seem unsuited for this purpose.<|endoftext|> -TITLE: Uses of the Chern--Connes Pairing? -QUESTION [6 upvotes]: The backbone of Connes' approach to noncommutative geometry is the Chern--Connes pairing. By discovering the cyclic homology of an algebra and then pairing it the $K$-theory of that algebra, Connes showed how to find numerical invariants of $K$-theory classes. This is analaogous to the way in which the Chern character map gives numerical invariants of vector bundles. I would like to know what has the Chern--Connes pairing been used for? Besides its intrinsic interest as a noncommutative version of a classical phenomenon, why is it important? - -REPLY [3 votes]: The Selberg principle in representation theory of semisimple $p$-adic groups $G$, says that the orbital integral of a coefficient of supercuspidal representation of $G$, on a hyperbolic conjugacy class in $G$, is $0$. Connes suggested around 1982 that this might be interpreted by means of the Chern-Connes pairing: indeed a coefficient of supercuspidal representation is, up to a scalar factor, an idempotent in the convolution algebra $C_c^\infty(G)$, while an orbital orbital gives a trace over $C_c^\infty(G)$; hence the interpretation of the Selberg principle by saying that certain traces give the zero pairing on the $K$-theory $K_0(C_c^\infty(G))$. This program was implemented for rank 1 groups in this paper: -http://www.theta.ro/jot/archive/1986-016-002/1986-016-002-007.pdf - -REPLY [2 votes]: I guess you named already one if the biggest points yourself: getting invariants! So more specifically, Connes obtained (with Moscovici and others) invariants of pretty ugly foliations, it helps in the classifying certain $C^*$-algebras etc. There are also more algebraic versions in deformation quantization, where people have formulated index theorems using this pairing. In some sense, it is also a new way of understanding index theorems in general, though I must admit that I'm not really an expert here...<|endoftext|> -TITLE: Upper bounds for the sum of primes up to $n$ -QUESTION [15 upvotes]: Let $s(n)$ denote the sum of primes less than or equal to n. Clearly, $s(n)$ is bounded from above by the sum of the first $n/2$ odd integers $+1$. $s(n)$ is also bounded by the sum of the first $n$ primes, which is asymptotically equivalent to $\frac{n^2}{2\log{n}}$. It should thus be possible to find estimates for $s(n)$ using the fact that for an $\epsilon > 0$ and $n$ large enough $s(n) < (1+\epsilon)\frac{n^2}{\log{n}}.$ -I would like to know if there are any known sharp upper bounds for $s(n)$. That is, I am looking for a function $f(n)$ such that for every $n > N_0$ $$ s(n) \leq f(n)$$ -As a way of relaxing the question, $s(n)$ could be regarded as the sum of the primes in the interval $[c,n]$ given a constant $c$. - -REPLY [7 votes]: The following paper gives the asymptotic expansion of the sum of the first $n$ prime numbers. Hence for sufficiently large $n$, the first few positive and negative terms of the asymptotic expansion will give best upper and lower bound on the sum of primes. -http://arxiv.org/pdf/1011.1667.pdf -$$ -\sum_{r \le n}p_r = \frac{n^2}{2}\Bigg[\ln n + \ln\ln n - \frac{3}{2} + \frac{\ln\ln n}{\ln n} - \frac{3}{\ln n}- \frac{\ln^2 \ln n}{2\ln^2 n} -$$ -$$ -+ \frac{7 \ln \ln n}{2\ln^2 n} - \frac{27}{4\ln^2 n} -+ o\Bigg(\frac{1}{\ln^2 n}\Bigg) \Bigg]. -$$<|endoftext|> -TITLE: Checkmate in $\omega$ moves? -QUESTION [103 upvotes]: Is there a chess position with a finite number of pieces on the infinite chess board $\mathbb{Z}^2$ such that White to move has a forced win, but Black can stave off mate for at least $n$ moves for every $n$? - -This question is motivated by a question posed here a few months ago by Richard Stanley. He asked whether chess with finitely many pieces on $\mathbb{Z}^2$ is decidable. -A compactness observation is that if Black has only short-range pieces (no bishops, rooks or queens), then the statement "White can force mate" is equivalent to "There is some $n$ such that White can force mate in at most $n$ moves". -This probably won't lead to an answer to Stanley's question, because even if there are only short-range pieces, there is no general reason the game should be decidable. It is well-known that a finite automaton with a finite number of "counters" can emulate a Turing machine, and there seems to be no obvious reason why such an automaton could not be emulated by a chess problem, even if we allow only knights and the two kings. -But it might still be of interest to have an explicit counterexample to the idea that being able to force a win means being able to do so in some specified number of moves. Such an example must involve a long-range piece for the losing side, and one idea is that Black has to move a rook (or bishop) out of the way to make room for his king, after which White forces Black's king towards the rook with a series of checks, finally mating thanks to the rook blocking a square for the king. -If there are such examples, we can go on and define "mate in $\alpha$" for an arbitrary ordinal $\alpha$. To say that White has a forced mate in $\alpha$ means that White has a move such that after any response by Black, White has a forced mate in $\beta$ for some $\beta<\alpha$. -For instance, mate in $\omega$ means that after Black's first move, White is able to force mate in $n$ for some finite $n$, while mate in $2\omega + 3$ means that after Black's fourth move, White will be able to specify how many more moves it will take until he can specify how long it will take to mate. -With this definition, we can ask exactly how long-winded the solution to a chess problem can be: - -What is the smallest ordinal $\gamma$ such that having a forced mate implies having a forced mate in $\alpha$ for some $\alpha<\gamma$? - -Obviously $\gamma$ is infinite, and since there are only countably many positions, $\gamma$ must be countable. Can anyone give better bounds? - -REPLY [5 votes]: I have an idea for how to get up to $\omega_1^{CK}$. Consider this position: black's king is trapped and white has a mate in one. However, white's king is trapped in a perpetual check. The only way out for the white king is to go along a specific trail, emulating a finite-state machine. Along certain points on the trail, black will need to check white by moving a queen somewhere along an infinitely long line or diagonal. There will be two black queens which are far enough from the rest of the pieces that they will be able to move freely along these lines. As the white king goes along various lines the queens will be forced to emulate a two-register Minsky Machine. Note that with two queens and an infinite chess board there will be enough space to ensure that only one queen will be able to attack the king in at any one time. For the decrement operation black must be able to force white along of two paths, one of which will be obstructed for the queen if she is beyond the '0' square, the other of which will be obstructed only for a queen at the '0' square. -So now we have a Minsky machine. Next we need a source of $\omega$-power. For this, the white king will occasionally be forced to block the line white needs to checkmate black. When white is there, black will not have any checks in one move. However, one of the queens will be in square '0', and it would have been able to force a checkmate were it in any farther square. Now black moves their queen as far back as they please and threaten to checkmate white the next move. White cannot checkmate black because the line is blocked by the white king. The only way for white to avoid checkmate is to move the white king one step further and allow black to continue their perpetual checks, with an arbitrary value on one of the registers. -The way this is currently designed has a problem: If one of the registers is 0, the other must contain the state information. Then, when the first register moves to an arbitrary value, the Minsky machine must fully make use of the values in both registers. However, two-register Minsky machines are too inflexible to do this. Luckily, by moving from lines to diagonals or vice versa, a chess queen can also divide and multiply by 2. That should be enough to adequately manipulate both registers. Another possibility is to add more queens and when multiple queens can check the king along a line, make it unstrategic for all but one of the queens to do so. This approach will also be needed to modify this strategy to work in quarter-infinite and half-infinite boards. -To completely work out this strategy it is necessary to describe in full the chess positions that will make up the trail for the king and to also describe in detail how to maneuver the queens to force them to emulate a Minsky machine. I have some idea for how this will go, but not entirely. Still, I believe this idea can work.<|endoftext|> -TITLE: Avatars of the ring of symmetric polynomials -QUESTION [25 upvotes]: I'm collecting different apparently unrelated ways in which the ring (or rather Hopf algebra with $\langle,\rangle$) of symmetric functions $Z[e_1,e_2,\ldots]$ turns up (for a Lie groups course I will be giving next year). So far I have: -*The ring of symmetric functions -*Irreducible representations of symmetric groups =Schur functions -*Irreducible representations of general linear groups = Schur functions -*The homology of $BU$, the classifying space of the infinite unitary group. (It also turns up in several other related generalized homology rings of spectra.) -*The universal commutative $\lambda$-ring on one generator $e_1$ -*The coordinate ring of the group scheme of power series $1+e_1x+e_2x^2+\cdots$ under multiplication -What other examples have I missed? - -REPLY [11 votes]: Boson-Fermion correspondence in representation theory -of Kac-Moody algebras (this is implicit in Richard's comments). -Closely related to this: coordinates on the "big cell" in -the loop Grassmannian and it's relation to the KP and KdV -equations where Schur polynomials appear as $\tau$-functions -(see Segal-Wilson old paper in Publ. IHES).<|endoftext|> -TITLE: How can we detect the existence of almost-complex structures? -QUESTION [34 upvotes]: Any smooth $k$-manifold $M$ comes with a well-defined map $f:M\rightarrow BGL_{k}(\mathbb{R})$ (up to homotopy) classifying its tangent bundle. Since $GL_{k}(\mathbb{R})$ deformation-retracts onto $O_k$, then $BGL_{k}(\mathbb{R})\simeq BO_k$, which is a cute way (though it's certainly overkill) of proving that every smooth manifold admits a Riemannian metric. An almost-complex structure, on the other hand, is equivalent to a reduction of the structure group from $GL_{2n}(\mathbb{R})$ to $GL_n(\mathbb{C})$, which is the same as asking for a lift of the classifying map through $BU_n\simeq BGL_n(\mathbb{C})\rightarrow BGL_{2n}(\mathbb{R})$. - -Can we detect the nonexistence of a - lift entirely using characteristic - classes? If not, what else goes into the classification? - -I'd imagine these don't suffice themselves. I know that $w_{2n}(TM) \equiv_2 c_n(TM)$, so this holds in the universal case $H^\ast(BO_{2n};\mathbb{Z}/2) \rightarrow H^\ast(BU_n;\mathbb{Z}/2)$. And certainly there are necessary conditions like $w_1(TM)=0$ (which of course just means that $TM$ is an orientable bundle, which is the same as asking that $M$ be an orientable manifold). But I have no idea of what sufficient conditions would look like. I've heard that this problem is indeed solved. Maybe it takes some characteristic class & cohomology operation gymnastics, or maybe it even needs extraordinary characteristic classes. Or maybe there's yet another ingredient in the classification? -Edit: Apparently I misquoted my source, and this is only known stably (which makes sense, in light of Joel's answer and Tom's comments on it). - -REPLY [16 votes]: Edit: Now updated to include reference and slightly more general result. -Edit 2: Includes remark about integrability. -Similar to Francesco Polizzi's answer, there is the following Theorem concerning 6-manifolds. -A closed oriented 6-dimensional manifold $X$ without 2-torsion in $H^3(X,\mathbb{Z})$ admits an almost complex structure. There is a 1-1 correspondence between almost complex structures on $X$ and the integral lifts $W \in H^2(X, \mathbb{Z})$ of $w_2(X)$. The Chern classes of the almost complex structure corresponding to $W$ are given by $c_1 = W$ and $c_2 = (W^2 - p_1(X))/2$. -In fact, a necessary and sufficient condition for the existence of an almost complex structure is that $w_2(X)$ maps to zero under the Bockstein map $H^2(X,\mathbb{Z}_2) \to H^3(X,\mathbb{Z})$. -I think the reason for results such as this and the one mentioned by Francesco is the following. To find an almost complex structure amounts to finding a section of a bundle over $X$ with fibre $F_n=SO(2n)/U(n)$. The obstructions to such a section existing lie in the homology groups $H^{k+1}(X, \pi_k(F_n))$. When $n$ is small I would guess we can compute these homotopy groups and so have a good understanding of the obstructions. For example, in the case mentioned above, n=3, $F_n = \mathbb{CP}^3$ and so the only non-trivial homotopy group which concerns us is $\pi_2 \cong \mathbb{Z}$. This is what leads to the above necessary and sufficient condition concerning 2-torsion. On the other hand when $n$ is large I don't know what $F_n$ looks like, let alone its homotopy groups... -For the proof of the above mentioned result see the article "Cubic forms and complex 3-folds" by Okonek and Van de Ven. (I highly recommend this article, it's full of interesting facts about almost complex and complex 3-folds.) -It is worth pointing out that in real dimension 6 or higher there is no known obstruction to the existence of an integrable complex structure. In other words, there is no known example of a manifold of dimension 6 or higher which has an almost complex structure, but not a genuine complex structure. By the classification of compact complex surfaces, those 4-manifolds admitting integrable complex structures are well understood.<|endoftext|> -TITLE: Degree of image of a polynomial map -QUESTION [6 upvotes]: Let $K$ be a field. Let $V/K$ be an affine variety in $A^m$. Let f be a polynomial map (and hence a "morphism of finite type") $f:V\to A^n$. A theorem of Chevalley's tells us that im(f) is either a variety or "almost" a variety - that is, im(f) is a variety $W$ with perhaps a few varieties of lower dimension cut out from it. -Question: is the degree of $W$ (= Zariski closure of im(f)) bounded solely in terms of m, n, deg(V) and the degree of the polynomials $f_1,\dots ,f_n$ defining f (i.e. $f(\vec{x}) = (f_1(\vec{x}),...,f_n(\vec{x}))$?). -This seems intuitively obvious, but I do not know where to look for a reference. (It's also non-obvious how to adapt the proof of Chevalley's theorem I'm looking at so as to give this.) Does anybody where to look this up and/or how to prove this? - -REPLY [5 votes]: Call $r$ the dimension of $W$, $d$ the degree of $V$, and $e$ the largest degree of one of the $f_i$. Then I claim that $\deg W$ is at most $de^r$. -We may assume that $K$ is algebraically closed. We may also assume that the dimensions of $V$ and $W$ are the same (otherwise, cut $V$ with generic hyperplane sections until they become equal). Then the morphism $V \to W$ generically finite and non-empty fibers. -Now, let $L \subseteq \mathbb A^n$ be a generic linear subspace of codimension $r$ of $W$. The cardinality of the intersection $L \cap W$ is $\deg W$, and this is bounded above by the cardinality of $f^{-1}L$, which is finite. But $f^{-1}L$ is the intersection of $V$ with the zero loci of $r$ generic linear combinations of the $f_i$. Now use the following fact: if $X$ is a (not necessarily equidimensional) closed subset of an affine space, call $\deg X$ its degree (that is, the sum of the degrees of its irreducible components), and $H$ is a hypersurface of degree $e$, then $\deg(X\cap H) \leq e\deg X$. This reduces easily to the case that $X$ is irreducible, which is standard. This implies that the degree of $f^{-1}L$, that is, its cardinality, is at most $de^r$, which gives us what we want.<|endoftext|> -TITLE: Trace theorem for $C^{k,1}$ domains -QUESTION [9 upvotes]: What are the best results on (Sobolev space) trace theorems for $C^{k,1}$ domains? -For $k=0$, e.g., when the domain is Lipschitz, from e.g. the works of Martin Costabel and Zhonghai Ding, it is known that the trace map is bounded (and has a continuous right inverse) from $H^{s}(\Omega)$ to $H^{s-1/2}(\partial\Omega)$ for $s\in(\frac12,\frac32)$. Moreover the endpoint case $s=\frac32$ is claimed by David Jerison and Carlos Kenig but a proof seems to have not appeared. A part of the question is whether or not the claimed proof appeared. The other part is if there is a proof of the obvious extension of this result to $C^{k,1}$ domains. If so, what is the situation of the corresponding endpoint (i.e., $s=k+\frac32$) result? -Update: Essentially complete answer has been found, and recorded below. - -REPLY [2 votes]: Part 2: -As pointed out in the answers, there is a paper by Doyoon Kim which completely answers the second part of my question. An earlier paper by Jürgen Marschall also answers this question, with possible exception of some exponents (interestingly, this paper appeared before Costabel's paper). I say "possible", because the both papers treat the traces for $L^p$-based spaces and Marschall's paper exclude the case $s-\frac1p$ is an integer for the boundedness of the right inverse. -But his methods maybe applicable for $p=2$ (I have not checked) as we know this case is somewhat special. In any case Kim proves the theorem even when $s-\frac1p$ is an integer, for general $p$. -Part 1: -The first part of the question, that is the claim about the endpoint case $s=k+\frac32$ is not true, in the sense that the norm of the trace operator cannot be bounded in terms of the Lipschitz constant alone. Some years after the paper containing the claim appeared, Jerison and Kenig corrected it by publishing a counterexample for domains with $C^1$ boundary, which means there is no hope for the Lipschitz case ($k=0$). Note that the main results in the earlier paper by Jerison and Kenig depend weakly on this claim, and are still true. Their $C^1$ counterexample was based on a counterexample in the Lipschitz case due to Guy David, and I want to record its main idea here. -Consider in $\mathbb{R}^2$ a half disk whose flat part of the boundary is replaced by a sawtooth (or zigzag) with period $2\varepsilon$. -Let us call this domain $\Omega_\varepsilon$. -Note that the Lipschitz constant of the domain does not depend on $\varepsilon$. - -We pick a function $f\in H^{3/2}(\mathbb{R})$ such that $f'(y)$ tends to $\infty$ as $y\to0$. This is possible because $H^{1/2}(\mathbb{R})$ has unbounded functions. Then in the configuration as shown in the picture, we define $g(x,y)=\phi(x,y)f(y)$, where $\phi$ is a suitable cut-off function (say, equal to 1 on a considerable portion of the zigzag boundary). -Obviously, we have $g\in H^{3/2}(\Omega_\varepsilon)$, -but the tangential derivative of $g$ along the zigzag boundary is $\pm\sqrt2f'$ on a large portion of the boundary. This means that the $L^2$-norm of the tangential derivative is at least a constant multiple of $|f'(\varepsilon)|$, which blows up as $\varepsilon\to0$.<|endoftext|> -TITLE: Proof of a concentration compactness lemma -QUESTION [5 upvotes]: Hi I'm stuck with the proof of a concentration-compactness lemma. -We have the following equation in $\mathbb{R}^N, N \ge 3$: -$$ --\Delta u +u=|u|^{p-2}u, -$$ -where $2 < p < 2^{*}$. -The functional associated to that equation is given by -$$ -J(u) = \frac{1}{2}\|u\|^2_{H^1}-\frac{1}{p}\|u\|^p_{L^p}. -$$ -Because the Sobolev imbedding $H^1 (\mathbb{R}^N) \subset L^q (\mathbb{R}^N), 2 < q < 2^{*}$ is not compact, $J$ does not satisfy the Palais-Smale condition. So one considers the family $J_k$ of functionals defined by -$$ -J_k(u) = \frac{1}{2}\|u\|^2_{H^1 (B_k)}-\frac{1}{p}\|u\|^p_{L^p(B_k)}, -$$ -where $(B_k)_k$ is an open cover of $R^N$; $k$ positive integer. -Each $J_k$ now satisfies the PS condition. -The "lemma" says the following: Let $u_k \in H^1_0(B_k)$ be uniformly bounded in $H^1(\mathbb{R}^n)$, i.e. $\|u_k\| \le \Lambda$, with $\Lambda>0$ independent of $k$, and -such that $J'(u_k) \to 0$ as $k \to \infty$. Then, along a subsequence, one of the following holds true: - -either $u_k \to 0$ in $H^1(\mathbb{R}^N)$, -or, there exist $r,\delta>0$, and a sequence $a_k$ in $R^N$ such that -$$ -\liminf_k \int_{B_{r} (a_k)}u^2_k \ge \delta. -$$ -I know how to prove that if 2. does not hold then 1. holds. I need a hint on how to prove that if 1. does not hold then 2. holds. - -Thanks - -REPLY [5 votes]: The concentration-compactness lemma of P.L. Lions is independent of the fact your sequence is Palais-Smale. It holds for general sequences of bounded positive measure, in your case $u_k^2\; dx$. -First normalize your sequence such that $\Vert u_k^2\Vert_2 =1$. -Hence the idea is to consider the concentration function -$$Q_k(r)=\sup_{x\in R^n} \left( \int_{B(x,r)} u_k^2 \;dx\right).$$ -Note that $Q_k$ is non-decreasing and non-negative bounded function on $[0,+\infty[$, with $\displaystyle \lim_{r\rightarrow +\infty }Q_k(r)=1$. -Then there exists a subsequence of $Q_k$, still denote $Q_k$, and a non-decreasing and non-negative bounded function $Q$ on $[0,+\infty[$ such that -$$Q_k(r) \rightarrow Q(r),$$ -for almost all $r$. We can assume that $Q$ is left-continuous and moreover since $Q_k$ is non-decreasing and bounded, we have -$$Q(r)\leq \liminf_{k\rightarrow +\infty} Q_k(r) \hbox{ for all } r.$$ -Let -$$\lambda =\lim_{r\rightarrow +\infty} Q(r).$$ -If $\lambda=0$, this is the vanishing case, else let $r_0$ such that -$$Q(r_0)\geq \frac{\lambda}{2}.$$ -For any $k$ there exists $x_k$ such that -$$Q_k(r_0)\leq \int_{B(x_k,r_0)} u_k^2 \;dx + \frac{1}{k}.$$ -Hence for $k$ big enough, you have -$$\int_{B(x_k,r_0)} u_k^2 \;dx\geq \frac{\lambda}{3}.$$ -In fact, you can refine this case in two -2.1) For all $\epsilon$ there exits $R$ such that -$$\int_{B(x_k,R)} u_k^2 \;dx\geq 1-\epsilon.$$ -This is the compactness-case. -2.2) the measure split on disjoint set arbitrary far, this is the dichotomy case. -You will find all the details in the two papers of Lions.<|endoftext|> -TITLE: Where does the principal ideal theorem (from CFT) go? -QUESTION [19 upvotes]: My impression is that one of the celebrated results of class field theory the principal ideal theorem namely that given a number field $K$ and its maximum unramified abelian extension L, every ideal in the ring of integers $O_K$ of $K$ becomes principal in the ring of integers $O_L$ of $L$. That is, given an ideal $I$ in $O_K$, the ideal $I \dot O_L$ is principal. -This result was originally conjectured by Hilbert in 1900 and reduced to a group theoretic question by Artin which was finally solved by Furtwangler in 1930. -I've never seen any further discussion of the principal ideal theorem - I don't know any generalizations or applications. -As James Milne comments in Remark 3.20 of the fifth chapter of his book on class field theory it's easy to see that there is some finite extension of $K$ for which all ideals of $K$ become principal. He further comments that this extension need not be the Hilbert class field of $K$ (EDIT: see Dror's comment for an example). -Is the principal ideal theorem primarily of historical interest (e.g. because it was a long standing conjecture of Hilbert)? Or does it have some deeper significance? - -REPLY [15 votes]: Among the generalizations that I can recall off the top of my head are: - -the generalization to ray class groups already mentioned by Kevin, proved by Iyanaga pretty much immediately after -Furtwängler's proof; -Furtwängler's own theorem saying that if the class group is an elementary -abelian $2$-group, then its basis can be chosen in such a way that each basis -element capitulates in some quadratic extension; -the theorem of Tannaka and Terada, according to which ambiguous classes in cyclic -extension already capitulate in the genus field (the obvious generalization to central -extensions fails at least group theoretically due to results of Miyake) -the theorem of Suzuki, which claims that in any abelian unramified extension $L/K$, -a subgroup of order $\ge (L:K)$ must capitulate; this was generalized by Gruenberg and -Weiss (Capitulation and transfer kernels). - -Capitulation is also at the center of the Greenberg conjecture in Iwasawa theory. In addition, its analogue in the theory of abelian varieties is the visibility of Tate-Shafarevich groups; for a recent contribution in the other direction see Schoof and Washington's article on´the "Visibility of ideal classes".<|endoftext|> -TITLE: Third bordism group of BG, where G is an arbitrary compact Lie group. -QUESTION [9 upvotes]: Is anything known about $\Omega_3(BG)$, where $G$ is an arbitrary compact Lie group; i.e., is it possible to describe the structure of $\Omega_3(BG)$ for any compact Lie group? I know that $H_3(BG)$ consists completely of torsion when $G$ is compact and I would like (if possible) a similar type of statement for $\Omega_3(BG)$. - -REPLY [11 votes]: If you think about oriented bordism, the answer is that $\Omega_3 (BG) \cong H_3 (BG)$. This is true for any space $X$ instead of $BG$, because of the Atiyah-Hirzebruch spectral sequence and because $\Omega_i (pt)=0$ for $i=1,2,3$.<|endoftext|> -TITLE: What can be said about an infinite linear chain of conjugate prior distributions? -QUESTION [6 upvotes]: We can sample a discrete value from the multinomial distribution. -We can also sample the parameters of the multinomial distribution from its conjugate prior the dirichlet distribution. -Since the dirichlet distribution is part of the exponential family, it too must have a conjugate prior distribution in the exponential family. -I hope you see where I'm going: what happens as this chain of priors is taken to infinity? -For a simpler example, what happens with the self-conjugate Gaussian distribution? - -REPLY [5 votes]: Let's say that you have a distribution $F$ in the exponential family with density -\begin{align} -\newcommand{\mbx}{\mathbf x} -\newcommand{\btheta}{\boldsymbol{\theta}} -f(\mbx \mid \btheta) &= \exp\bigl(\eta(\btheta) \cdot T(\mbx) - g(\btheta) + h(\mbx)\bigr) -\end{align} -Given independent realizations $\{x_1, x_2, \dotsc, x_n\}$ of $F$ (with unknown parameter $\theta$), then the distribution over $\theta$, $F'$, is the conjugate prior of $F$. The density of $F'$ is -\begin{align} -f(\btheta \mid \boldsymbol\phi) = L(\btheta \mid \mbx_1, \dotsc, \mbx_n) -&= f(\mbx_1, \dotsc, \mbx_n \mid \btheta) \\\\ -&\propto \prod_i f(\mbx_i\mid \btheta) \\\\ -&= \textstyle\prod_i\exp\Bigl(\eta(\btheta) \cdot \textstyle T\left(\mbx_i\right) - g(\btheta) + h(\mbx_i)\Bigr) \\\\ -&\propto \textstyle\prod_i\exp\Bigl(\eta(\btheta) \cdot \textstyle T\left(\mbx_i\right) - g(\btheta)\Bigr) \\\\ -&= \textstyle\exp\Bigl(\eta(\btheta) \cdot \bigl(\textstyle\sum_iT\left(\mbx_i\right)\bigr) - ng(\btheta)\Bigr) \\\\ -&= \exp\bigl(\eta'(\boldsymbol \phi) \cdot T'(\btheta)\bigr) -\end{align} -where -\begin{align} -\eta'(\boldsymbol\phi) &= -\begin{bmatrix} -\sum_iT_1(\mbx_i) \\\\ -\vdots \\\\ -\sum_iT_k(\mbx_i) \\\\ -\sum_i1 -\end{bmatrix} -& -T'(\btheta) &= -\begin{bmatrix} -\eta_1(\btheta) \\\\ -\vdots \\\\ -\eta_k(\btheta) \\\\ --g(\btheta) -\end{bmatrix}. -\end{align} -Thus, $F'$ is also in the exponential family ($T'$ replaced $\eta$ and $\eta'$ replaced $T$ since this distribution is over $\theta$ the parameter of the distribution over $x$.) -Interestingly, $\boldsymbol\phi$ has exactly one more parameter than $\btheta$ except in the rare case where natural parameter $\phi_{k+1}$ is redundant, but such a distribution would be very weird (it would mean that the number of observations $\mbx$, that is, $n$, tells you nothing about $\btheta$.) -So, to answer your question, with each conjugate prior you get exactly one more hyperparameter. -There are many conjugate priors of the Gaussian distribution depending on how you look at it. In my opinion, the analogy to the Multinomial-Dirichlet example would set things up as follows: assume that $n$ real-valued numbers are generated by a Gaussian with unknown mean and variance. Then, the distribution of the mean and variance given the data points is a three-parameter conjugate prior distribution whose sufficient statistics are the total of the samples, the total of the squares of the samples, and the number of samples.<|endoftext|> -TITLE: Cohen reals and strong measure zero sets -QUESTION [20 upvotes]: A set of reals $X$ is $\textit{strong measure zero}$ if for any sequence of real numbers $ ( \epsilon_n ) _{n \in \omega }$ there is a sequence of open intervals $ ( a_n ) _{n \in \omega }$ which covers $X$ and such that each $ a_i $ has length less than $ \epsilon _i $. -The Borel Conjecture (BC) is the statement that a set of reals is strong measure zero iff it is countable. (It's easy to see that any countable set is strong measure zero, so BC just says that there are no uncountable strong measure zero sets.) -I heard somewhere that adding a Cohen real necessarily destroys BC, i.e., if $x$ is a Cohen real over $V$, then $V[x] \models \neg BC$. I can see why this is true for adding $\omega_1$ Cohen reals: for any $\epsilon$-sequence in the ground model, and any uncountable set of reals $X$ in the ground model (which is still uncountable after ccc forcing), we can use a single Cohen real to generically choose small intervals which will cover $X$ while staying within the constraints of the $\epsilon$-sequence. So when we add $\omega_1$ Cohen reals successively, every $\epsilon$-sequence shows up at some countable stage, and at the next stage we create a corresponding cover for $X$. Thus $X$ will be strong measure zero in the forcing extension. But here we are using the fact that every $\epsilon$-sequence shows up at some stage, which we then proceed to force over with a Cohen real; and that will not be true if we only add a single Cohen real. -So my question is: does adding a single Cohen real necessarily destroy BC? -And more generally: after adding a single Cohen real, what can we say about "the set of all reals from the ground model"? Is it meager? Measure zero? Does it contain no perfect sets? - -REPLY [12 votes]: (An attempt at an answer, and also my first posting here. Thanks to Andres Caicedo for the reformatting.) -I claim that a single Cohen real makes the set of old reals strong measure zero. -Reals are functions from $\omega$ to 2. -Let ${\mathbb C}$ be Cohen forcing, and let $c$ be the name of the generic real. -Let $(n_k)$ be a sequence of ${\mathbb C}$-names for natural numbers. -I will find a sequence $(s_k)$ of names for finite $01$-sequences -($s_k$ of length $n_k$) -such that ${\mathbb C}$ forces: every old real is in some $[s_k]$. -Let $D_k$ be a dense open set deciding the value of $n_k$ and -containing only conditions of length at least $k$. -Say, each $q$ in $D_k$ decides that the value of $n_k$ is $f_k(q)$, where -$f_k$ is a function in the ground model defined on $D_k$. -Each $f_k$, and also the sequence $(f_k)$, is in $V$. -Now we work in the extension. - (The point is that even though we now know the actual values of - $n_k$, we will play stupid and use the names only, plus the minimal - amount of information that we need from the generic real. - This lets us gauge exactly how much information from the - generic we need.) -In the extension I will define a sequence $(i_k)$ of natural numbers. - Let $i_k$ be the minimal $i$ such that $c \mathord\upharpoonright i$ is in $D_k$, - where $c\mathord\upharpoonright i = c$ restricted to $i$. - (So $i_k$ is at least $k$.) -For each $k$ we now define a $01$-sequence $s_k$ of length $n_k$ as follows: Take $n_k$ successive bits from the Cohen real $c$, starting at position $i_k$. (Formally: $s_k(j) = c(i_k+j)$ for all $j\lt n_k$.) -I claim that "every old real is in some $[s_k]$" is forced. -Assume not, so let $p$ force that $x$ is not covered. -Let $k$ be larger than the length of $p$. So $p$ not in $D_k$. -Extend $p$ to $q$ so that $q$ is in $D_k$, $q$ minimal. -Let $l$ be the length of $q$. -So $q$ forces that $i_k$ is exactly $l$. Also $q$ forces that $n_k = f_k(q)$. -Now extend $q$ to $q'$, using the first $f_k(q)$ bits of $x$. -So $q'$ is stronger than $q$, and $q'$ forces that $s_k$ is an initial -segment of $x$. -mg*<|endoftext|> -TITLE: Coefficients in cohomology -QUESTION [30 upvotes]: (Sorry if this is too elementary for this site) -I’m having some trouble understanding sheaf cohomology. It’s supposed to provide a theory of cohomology “with local coefficient”, and allow easy comparison between different theories like singular, Cech, de Rham and Alexander Spanier. What I don’t understand is: what’s all the fuss with coefficients that vary with each open set? Indeed what’s all the fuss with changing coefficients in an ordinary cohomology theory as in Eilenberg Steenrod? -Homology is trying to measure the “holes” of a space; wouldn’t integer coefficients suffice already? I’m not really sure what cohomology is trying to measure; at least I think the first singular group is trying to measure some kind of “potential difference”, like explained in Hatcher’s book. It gets worse for me when the coefficient group isn’t the integers. But when I get to sheaf cohomology I’m totally dumbstruck as to what it’s trying to measure, and what useful information of the space can be extracted from it. Now if it’s just about comparisons of different theories I can live with that… -Can someone please give me an intuitive explanation of the fuss with all these different coefficients? Please start off with why we even use different coefficients in Eilenberg Steenrod. Sorry if this is too elementary. - -REPLY [33 votes]: This (elementary and perfectly standard) example might help show the power of sheaves with non-constant coefficients: -First, think about the circle $S^1$. Suppose you want to understand (real) line bundles on the circle. You can certainly cover the circle with two open contractible subsets $U_1$ and $U_2$ (which you can take to be the complements of the north and south poles), and we know that any line bundle on a contractible space is trivial. So if you've got a line bundle $L$ over $S^1$, you can restrict it to either $U_i$ and get a trivial bundle $L_i$. $L$ is built from these $L_i$ and the way they they are patched together over $U_1\cap U_2$. -Now what does it mean to patch the $L_i$ together over $U_{12}=U_1\cap U_2$? It means choosing an isomorphism $L_1|U_{12}\rightarrow L_2|U_{12}$. For any $x\in U_{12}$, the restriction of this isomorphism to the fiber $L_x$ over $x$ is an isomorphism between 1-dimensional vector spaces, and so (after choosing bases) can be identified with an element of ${\bf R}^*$ (the non-zero reals). Therefore your patching consists of a continuous map -$$U_{12}\rightarrow {\mathbb R}^*$$ -which is to say, a Cech 1-cocycle for the sheaf of continuous ${\bf R}^{*}$-valued functions. -Now of course you could build a line bundle in some other way, say by starting with two different contractible sets $U_1$ and $U_2$. When do two sets of patching data give isomorphic line bundles? A little thought reveals that the answer is: When and only when the corresponding cocycles give the same class in -$$H^1(S^1,G^{*})$$ -with $ G^{*} $ being the sheaf of continuous ${\bf R}^*$-valued functions. -Therefore line bundles are classified by $H^1(S^1,G^{*})$. Now consider the exact sequence of sheaves -$$0 \rightarrow G \rightarrow G^*\rightarrow {\bf Z}/2{\bf Z}\rightarrow 0$$ -where $G$ is the sheaf of continuous ${\bf R}$ valued functions, and the map on the left is exponentiation. Follow the long exact sequence of cohomology, use the fact that $G$ is acyclic, and conclude that $H^1(S^1,G^*)=H^1(S^1,{\bf Z}/2{\bf Z})={\bf Z}/2{\bf Z}$. In other words, there are exactly two real line bundles over $S^1$ --- and indeed there are: the cylinder and the Mobius strip. -Exercise: Do a similar calculation for ${\bf CP}^1$ (the Riemann sphere). Conclude that the set of (complex) line bundles is in one-one correspondence with $H^2({\bf CP}^1,{\bf Z})={\bf Z}$.<|endoftext|> -TITLE: Curious 'mechanical' immersion of the Klein bottle in R^3 -QUESTION [5 upvotes]: I have a question about a specific immersion of the Klein bottle in -$\mathbb{R}^{3}$ that seems curious in different aspects to me. First, it is -obtained by a mechanical movement of a geometrical circle that is easy to -define analitically. Second, each circle $L$ from this family is meeting -exactly two others and $L$ is linked with the circles that are 'close' to it, -while is unlinked with the 'distant' ones. This model reminds the 'figure 8' -immersion of the Klein bottle (Klein bagel), but is quite different, as the -Klein bagel is obtained by mechanical movement of a 2:1 Lissajous curve -similar to figure 8 (see wiki). -The idea is to take a 'small' circle $l$ and a 'big' circle $L$ so that the -center of $L$ lies on $l$, both in one and the same plane. Then $L$ starts -moving so that its center is describing $l$, and simultaneously $L$ is turning -around $l$ (in fact - around the tangent line to $l$ at the current center of -$L$). It suffices to adjust the speeds so that when the center of $L$ makes a -full round of $l$, circle $L$ achieves a half turn around $l$. Then $L$ -describes an immersed Klein bottle. -Now, my questions about this model are: -1) Is everything OK with this immersion (at least topologically) since it is -not quite easy to visualise the result. -2) How exactly the self-intersection set looks like? I suppose it consists of -one or two closed curves. (By the way, are there any general results saying -what the self-intersection set of an immersed Klein bottle should be?) -3) Are there any references about similar 'mechanical' models of surfaces in -$\mathbb{R}^{3}$? -Note that if everything is alright with the model, we can take a small knot -$l$ instead of a circle and to make $L$ turning $n+1/2$ times around $l$, in -such a way getting many different immersions of the Klein bottle. -Anyway, let me write the analytical description of this model (up to -my fault), although I am not sure it helps understanding the things geometrically. -Here $l:x^{2}+y^{2}=1,\ z=0$, $L:x^{2}+(y-1)^{2}=r^{2},\ z=0;\ r>2$. The -immersion is then given by -$X=r\cos u\cos2t-\sin2t\cos t(1+r\sin u)$ -$Y=r\cos u\sin2t+\cos2t\cos t(1+r\sin u)$ -$Z=\sin t(1+r\sin u)$, -where $t\in\lbrack0,\pi],\ u\in\lbrack0,2\pi]$. -Note that for $r\leq2$ the same example should be valid as well; especially -for $r<1$ it is easier to visualise, but the case $r>2$ seems more enigmatic. -s::l - -REPLY [5 votes]: Not an answer; just an aid to visualization. -If I've interpreted your equations correctly, here are views of the surfaces for three values of $r$:<|endoftext|> -TITLE: universal cover of SL2(R): does it admit central extensions? -QUESTION [17 upvotes]: Is it true that the universal cover of SL2(ℝ) has no non-trivial central extensions... as an abstract group? -(that's certainly true as a Lie group) -Motivation: -I have a projective action of SL2(ℝ) on some Hilbert space H -and I'd like to know that it induces an honest action of its universal cover. -But it's a hassle to show that the action is continuous. -So I'm wondering if there is an alternative argument that uses solely group theory. - -REPLY [10 votes]: As van der Kallen mentions, the answer is: definitely no. Here's argument which is less precise than his, but with a checkable basic reference: thm 11.10 from Milnor's book "Introduction to Algebraic K-theory": if a field F is uncountable then $K_2(F)$ is uncountable as well. Moreover $K_2(F)$ is generated by symbols which already define cocycles on $\text{SL}_2$. So the Schur multiplier of the discrete group $\text{SL}_2(\mathbf{R})$ is uncountable, hence certainly not cyclic. -In the same spirit, the Schur multiplier of the simply connected Lie group, $\text{SL}_2(\mathbf{C})$, viewed as a discrete group, is uncountable.<|endoftext|> -TITLE: an algebraically closed field definable in a real closed field -QUESTION [8 upvotes]: Is it true that an algebraically closed field $k$ definable (in the model theory sense) in a real closed field $\mathcal R$ is the algebraic closure $\mathcal{R}^{alg}=\mathcal{R}(\sqrt{-1})$? -UPDATE: if we prove that the definable set $K \subset \mathcal{R}^n$ that defines $k$ in $\mathcal R$ has the same cardinality as $\mathcal R$ then by categoricity of $ACF_0$ we get that $K$ is isomorphic to $\mathcal{R}(\sqrt{-1})$ (which is definable in $\mathcal R$ in an obvious fashion). Would this isomorphism be definable? - -REPLY [11 votes]: Suppose that $A$ is an infinite definable subset of a real closed field $R$ which is a zero-divisor-free ring under operations whose graphs are definable in $R$. Then $A$ is definably isomorphic to one of $R$, $R(\sqrt{-1})$ or the ring of quaternions over $R$. This is a special case of the main result of: -Otero, Peterzil, and Pillay, -On groups and rings definable in o-minimal expansions of real closed fields, -Bull. London Math. Soc. 28 (1996), no. 1, 7–14.<|endoftext|> -TITLE: Universally measurable sets and weak topology -QUESTION [8 upvotes]: After I posted this question, a couple of months ago, and got from MO-users several -good hints, I think i'm ready, after some study, to ask another related question (or rather, to focus on the main point of my previous question, after I got aware of all the necessary background). -First let me describe the setting: -WEAK TOPOLOGY -Given any Polish space $X$, we denote with $\mathcal{M}(X)$ the set of all Borel-probability measures on $X$. The set $\mathcal{M}(X)$ is endowed with the smallest topology such that the map $\mathcal{M}(f): \mathcal{M}(X)\rightarrow[0,1]$ defined as -$\mu \mapsto \displaystyle \int_{X} f \ d\mu$ -is continuous, for every continuous $f:X\rightarrow[0,1]$, where $[0,1]$ has the usual topology. This topology is called the "weak topology" on $\mathcal{M}(X)$, and is itself a Polish space. -As Gerald Edgar pointed out in my previous question, it turns out that the map $\mathcal{M}(g)$ defined as above, is Borel-measurable for every $g:X\rightarrow [0,1]$ Borel measurable function, i.e. -$\Big(\mathcal{M}(g)\Big)^{-1}\big( (\lambda,1] \big)$ -is a Borel set in $\mathcal{M}(X)$ for every $\lambda \in [0,1)$. -UNIVERSALLY MEASURABLE SETS and FUNCTIONS -Given a Polish space $X$, a subset $A\subseteq X$ is called universally measurable if and only if is $\mu$-measurable, for every (completion of) $\mu\in\mathcal{M}(X)$. The set of universally measurable sets forms a $\sigma$-algebra. A function $f:X\rightarrow[0,1]$ is called universally measurable if the inverse images of Borel sets is universally measurable. -(of course this is equivalent to "inverse image of every $(\lambda,1]$ sub-basic open.." ) -In particular (I look at) the set of universally measurable functions $f$ as the largest set such that $\mathcal{M}(f)$ is well defined. -FACT 1: Universally measurable functions are closed under composition. -FACT 2: Every $\Sigma^{1}_{1}$ set is universally measurable. -FACT 3: ZFC + V=L $\vdash$ "there exists a $\Delta^{1}_{2}$ set which is not universally measurable". -FACT 4: ZFC+ $\Sigma_{n}^{1}$-Determinacy, implies that every $\Sigma^{1}_{n+1}$-set is universally measurable. -FACT 5: ZFC $\Sigma_{n}^{1}$-Determinacy implies that every function $f:X\rightarrow[0,1]$ having graph $X\times[0,1]$ which is a $\Pi_{n}^{1}$-set is universally measurable. This follows from 4, because the set $f^{-1}\big( (\lambda,1] \big)$ is $\Sigma^{1}_{n+1}$. -QUESTIONS -Q1) Prove that if $f:X\rightarrow[0,1]$ is universally measurable, so is $\mathcal{M}(f)$. -I don't have many ideas for proving directly this result. Please let me know if you have any suggestions. -Anyway, I'd be happy enough (and actually equally interested) to prove the following result: -Q2) Prove that if $f:X\rightarrow[0,1]$ is function having $\Pi^{1}_{n}$ graph, so is $\mathcal{M}(f)$. -I'm not sure if the above theorem is right, or perhaps need to be weakened for example by saying that if $f:X\rightarrow[0,1]$ has a $\Pi^{1}_{n}$ graph, then $\mathcal{M}(f)$ has a $\Pi^{1}$graph, for some other $m$, maybe $m=n+1$. -This route would prove that the result of Question 1 holds (under PD), whenever $f$ has a reasonable (i.e. projective) description. -Now I would very much appreciate any suggestion for developing these ideas for Question 2. In particular I don't know precisely how to reason about the graph of $\mathcal{M}(f)$ starting from the graph of $f$. -Unfortunately I'm not a mathematician and I lack proper background to work freely on this problem, and this idea of attacking the problem using PD and by considering the complexity of the graph of $f$ is the only one I had so far. -THank you again for any suggestion! -bye -matteo - -REPLY [6 votes]: After some time I found the solution to Question 1 in the excellent book: -"Stochastic Optimal Control: The Discrete-Time Case", by Dimitri P. Bertsekas and Steven E. Shreve, freely available at link -The answer is YES, see Corollary 7.46.1 page 177, Chapter 7. -For what concerns Question 2 I have not found a solution yet, but I guess that after reading properly this book, I'll be able to solve it by myself. -thank you -matteo<|endoftext|> -TITLE: Fibonacci, compositions, history -QUESTION [22 upvotes]: There are three basic families of restricted compositions (ordered partitions) that are enumerated by the Fibonacci numbers (with offsets): -a) compositions with parts from {1,2} -(e.g., 2+2 = 2+1+1 = 1+2+1 = 1+1+2 = 1+1+1+1) -b) compositions that do not have 1 as a part -(e.g., 6 = 4+2 = 3+3 = 2+4 = 2+2+2) -c) compositions that only have odd parts -(e.g., 5 = 3+1+1 = 1+3+1 = 1+1+3 = 1+1+1+1+1) -The connection between (a) & the Fibonacci numbers traces back to the analysis of Vedic poetry in the first millennium C.E., at least (Singh, Hist. Math. 12, 1985). -Cayley made the connection to (b) in 1876 (Messenger of Mathematics). -Who first established the connection with (c), odd-part compositions? It was known by 1969 (Hoggatt & Lind, Fib. Quart.), but I suspect it was done before that. Thanks for any assistance, especially with citations. -BOUNTY! Not sure how much this incentivizes responses, but it would be nice to figure this out. By the way, I have queried Art Benjamin, Neville Robbins, and Doug Lind about this (Doug modestly mentioned of the 1969 article ``It's even possible this was an original result.''). - -REPLY [7 votes]: Found it! (Sorry, Doug, ha ha.) -Augustus de Morgan added several appendices to his Elements of Arithmetic in the fifth edition, 1846 (available on Google Books). Appendix 10, pages 201-210, is "on combinations." The relevant paragraph is on 202-203. - -Required the number of ways in which a number can be compounded of odd numbers, different orders counting as different ways. If $a$ be the number of ways in which $n$ can be so made, and $b$ the number of ways in which $n+1$ can be made, then $a+b$ must be the number of ways in which $n+2$ can be made; for every way of making $12$ out of odd numbers is either a way of making $10$ with the last number increased by $2$, or a way of making $11$ with a $1$ annexed. Thus, $1+5+3+3$ gives $12$, formed from $1+5+3+1$ giving $10$. But $1+9+1+1$ is formed from $1+9+1$ giving $11$. Consequently, the number of ways of forming $12$ is the sum of the number of ways of forming $10$ and of forming $11$. Now, $1$ can only be formed in $1$ way, and $2$ can only be formed in $1$ way; hence $3$ can only be formed in $1+1$ or $2$ ways, $4$ in only $1+2$ or $3$ ways. If we take the series $1$, $1$, $2$, $3$, $5$, $8$, $13$, $21$, $34$, $55$, $89$, &c. in which each number is the sum of the two preceding, then the $n$th number of this set is the number of ways (orders counting) in which $n$ can be formed of odd numbers. Thus, $10$ can be formed in $55$ ways, $11$ in $89$ ways, &c. - -He established "increasing" and "annexing" in deriving the formula for the number of what we now call compositions. He does not treat either of the other two restrictions mentioned above.<|endoftext|> -TITLE: Subshifts with the same entropy -QUESTION [6 upvotes]: It is known that two Markov subshifts with the same entropy are "almost isomorphic" (up to a subset of measure 0) if the entropy is a logarithm of an integer (see R. L. Adler, L. W. Goodwyn, and B.Weiss. Equivalence of topological markov shifts. -Israel J. Math, 27(1):49--63, 1977). Is it true (known) if the entropy is not a logarithm of an integer? - Update Basically I want to know if the result of AGW is true without the integer assumption. I would appreciate an answer of the form "yes plus reference" or "no plus reference" or "it is still unknown". - -REPLY [2 votes]: The main theorem in Adler, Roy L.; Marcus, Brian -"Topological entropy and equivalence of dynamical systems" -doesn't use the fact that the entropy is log of a natural number. However it states that two ergodically supported topological Markov shifts are almost topologically conjugated if and only if them have the same topological entropy and the same ergodic period. -There is a work of Wenxiang Sun which extends this result for general expansive, ergodically supprted maps that have the shadowing property: -http://iopscience.iop.org/0951-7715/13/3/309<|endoftext|> -TITLE: Rolling a random walk on a sphere -QUESTION [44 upvotes]: A ball rolls down an inclined plane, encountering horizontal obstacles, at which it -rolls left/right with equal probability. There are regularly spaced staggered gaps that let the ball -roll down to the next, lower obstacle. The pattern resembles a binary tree: - -Suppose the vertical and horizontal rolls have equal length $\delta$. -Tracing out the roll contact point on the ball surface we see a random walk, with each step a -geodesic arc of length $\delta$, and $90^\circ$ turns. -I expected that for rational (multiples of $\pi$) $\delta$, the trace would not fill the surface, -but the experiment below for $\delta=\pi/16$ (for 10, $10^2$, $10^3$, $10^4$ downhill steps) -indicates otherwise. - -For which $\delta$ will this trace fill the sphere surface? - -           - -Thanks for any insights! -Answer: The surface will be filled for every $\delta$ except $\pi/2$ and $\pi$. See Scott Carnahan's answer below, -and Dylan Thurston's simplification. I find this answer remarkable! - -REPLY [47 votes]: Let $A = \begin{pmatrix} \cos \delta & -\sin \delta & 0 \\ \sin \delta & \cos \delta & 0 \\ 0 & 0 & 1 \end{pmatrix}$, and let $B = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos \delta & -\sin \delta \\ 0 & \sin \delta & \cos \delta \end{pmatrix}$ be rotation by $\delta$ along the $z$ and $x$ axes, respectively. In suitable coordinates, a progression down one step in the tree is either $AB$ or $AB^{-1}$. -The trace will fill (a.s.) a dense subset of the surface if and only if the closure of the group generated by $AB$ and $AB^{-1}$ is not a subgroup of $SO(3)$ of dimension zero or one. -The dimension zero closed subgroups of $SO(3)$ are either cyclic, dihedral, or symmetries of Platonic solids, and the dimension one closed subgroups are conjugates of $SO(2)$ and $O(2)$. Therefore, it suffices to determine which values of $\delta$ yield a pair of elements in either a conjugate of $O(2)$, or a conjugate of one of the three Platonic groups (isomorphic to $A_4$, $S_4$, and $A_5$). -In order for $AB$ and $AB^{-1}$ to both lie in a conjugate of $O(2)$ it is necessary and sufficient that they have a common eigenvector with eigenvalue $\pm 1$ - this eigenvector is the axis of rotation. Writing this requirement explicitly yields a polynomial identity in $\sin \delta$ and $\cos \delta$ (whose solutions I haven't enumerated yet). Edit: Some straightforward case elimination with the $z$ coordinate of a common eigenvector shows that $\delta$ must be an integer multiple of $\pi/2$. -For the Platonic solutions, we can narrow down the solution set using the criterion that the rotation $(AB^{-1})^{-1}(AB) = B^2$ lies in the group, and Platonic solids have rotational symmetries of order at most 5. This means $\delta$ is a multiple of $\pi/3$, $\pi/4$ or $\pi/5$. -Since the traces of $AB$ and $AB^{-1}$ are both $(\cos \delta)(2 + \cos \delta)$, we can compare with character table entries to see if that number is the trace of an element in a Platonic group. It was pretty easy to eliminate candidates by eyeball in SAGE. -Conclusion: The only values of $\delta$ where the image is not dense are $0$, $\pm \pi/2$, and $\pi$.<|endoftext|> -TITLE: The first eigenvalue of a graph - what does it reflect? -QUESTION [17 upvotes]: A big-picture question: what "physical properties" of a graph, and in particular of a bipartite graph, are encoded by its largest eigenvalue? If $U$ and $V$ are the partite sets of the graph, with the corresponding degree sequences $d_U$ and $d_V$, then it is easy to see that the largest eigenvalue $\lambda_{\max}$ satisfies - $$ \sqrt{\|d_U\|_2\|d_V\|_2} \le \lambda_{\max} \le \sqrt{\|d_U\|_\infty\|d_V\|_\infty}; $$ -in particular, if the graph is $(r_U,r_V)$-regular, then $\lambda_{\max}=\sqrt{r_Ur_V}$. (A reference, particularly for the double inequality above, will be appreciated.) In the general case, the largest eigenvalue also reflects in some way the "average degree" of a vertex - but is anything more specific known about it? To put it simply, - -What properties of a (bipartite) graph can be read from its largest eigenvalue? - - -A brief summary and common reply to all those who have answered so far. - -Thanks for your interest and care! -To make it very clear: I am interested in the usual, not Laplacian eigenvalues. -Although the largest eigenvalue is related to the average degree, for non-regular graphs this does not tell much; hence, I believe, understanding the meaning of the largest eigenvalue in terms of the "standard" properties of the graph is of certain interest. -It is true that different bipartite graphs (as $K_{1,ab}$ and $K_{a,b}$) may have the same largest eigenvalue, but, I believe, this does not mean that the largest eigenvalue cannot be suitably interpreted. -I still could not find a reference to the displayed inequality above. (@kimball: Lovasz does not have it.) - -REPLY [3 votes]: Assume the graph is connected. -Let $\left| 1 \right>$ be the vector of all 1's (in Dirac notation), and let $A$ be the adjacency matrix. Then $\left<1|A|1\right>$ is the number of edges of the graph (well, actually twice the number of edges). Similarly, $\left<1|A^n|1\right>$ is the number of paths of length $n$, where a path is a sequence of vertices with consecutive vertices connected, repetitions allowed in the case of loops. Call this $P_n$. -Since the graph is connected, its adjacency matrix is irreducible and by the Perron-Frobenius theorem the first eigenvalue is simple and the eigenvector $\left| v \right>$ has positive components. Therefore, $\left< v | 1 \right> > 0$, allowing use of the power iteration method. As $n \to \infty$, $A^n\left|1\right>$ approaches the first eigenvector of $A$ (aside from normalization). So, -$$ -\lim_{n \to \infty} \frac{P_{n+1}}{P_n} = -\lim_{n \to \infty} \frac{P_{2n+1}}{P_{2n}} = -\lim_{n \to \infty} \frac{\left<1|A^n A A^n|1\right>}{\left<1|A^n A^n|1\right>} = \lambda_{\textrm{max}}. -$$ -The largest eigenvalue then tells how the number of paths of length $n$ grows, as $n$ grows (keeping in mind that I used a nonstandard definition for "path"). Furthermore, the limit written above approaches $\lambda_\textrm{max}$ from below. -Theorem 4 of Yu, Lu, Tian (2004) is equivalent to $\lambda_\textrm{max} \ge \sqrt{P_4 / P_2}$, but is expressed in terms of the degrees and 2-degrees of the vertices.<|endoftext|> -TITLE: How to decompose an infinite set into two isomorphic ones without choice? -QUESTION [10 upvotes]: Is it provable in ZF that each infinite set $A$ can be partitioned into two sets $B$ and $C$ which are isomorphic to each other, i.e., there is a bijective map from $B$ to $C$? - -REPLY [9 votes]: To answer the question in Todd's answer, the principle $2|A|=|A|$ for all infinite $A$ is strictly weaker than AC. This difficult result was proved by Halpern and Howard for ZFA (allowing atoms) and by Sageev for ZF. - -REPLY [3 votes]: The question seems interesting to me. There's an old result of Tarski that the axiom of choice is a consequence of ZF + the assertion "every infinite set $u$ is in bijection with $u^2$". A proof of this fact is given in Stoll's Set Theory and Logic; see page 126 (in the Dover edition). -However, exercise 10.5 on that page reads, "It is not known whether the formula $2u = u$, which follows from the axiom of choice (Theorem 9.5), implies the axiom of choice. Attempt a proof of this." Stoll's book was written some time ago, but I suspect this implication is still not known.<|endoftext|> -TITLE: Size of stationary sets -QUESTION [7 upvotes]: What can we say about the size of stationary subsets of $P_{\kappa}(\lambda)$ for infinite cardinals $\kappa, \lambda,$ especially when $\kappa=\aleph_1.$ -Please give me some references, if there are any. - -REPLY [6 votes]: Shelah proved using his pcf theory that the least cardinality of a stationary subset of $P_\kappa(\lambda)$ is equal to the least cardinality of a cofinal subset of $P_\kappa(\lambda)$. -See here: M. Shioya: A proof of Shelah's strong covering theorem for $P_\kappa(\lambda)$, Asian J. Math, 12(2008), 83-98.<|endoftext|> -TITLE: What is an example of a ring with two (or more) multiplicative right-identities? -QUESTION [10 upvotes]: I know that if a ring has a multiplicative identity, then the multiplicative identity must be unique. Are there simple-to-describe examples of rings with two (or more) multiplicative right-identities? - -REPLY [3 votes]: I was browsing some "What is...?" threads, and bumped in here: Let $n$ be a positive integer, and consider the subring of the opposite of the ring of $n$-by-$n$ matrices over a commutative unital ring $\mathbb A = (A, +, \cdot)$ consisting of those matrices all of whose rows, except at most for the first, are zero; this has $|A|^{n-1}$ right identities, given by those matrices whose first row is any vector of $A^n$ with first element equal to $1_\mathbb{A}$.<|endoftext|> -TITLE: Action on the highest weight vector of a representation of a semisimple linear algebraic group -QUESTION [9 upvotes]: Let $G$ be a semisimple linear algebraic group, $V$ a $G$-representation and $v \in V$ a vector of highest weight $\lambda$. Is it true, that for any positive root $\alpha \in R^+$ the one dimensional unipotent subgroup $U_{-\alpha}$ acts trivially on $v$ if and only if $\langle \alpha^{\vee}, \lambda \rangle = 0$? -I'm asking, because I want to apply this to the following situation. For any simple root $\alpha$, I denote the corresponding fundamental weight by $\omega$. Let $V=V(\omega)$ be the simple representation of highest weight $\omega$ with highest weight vector $v$. I'm trying to prove, that the stabilizor of the element $[v] \in \mathbb{P}(V)$ is the maximal parabolic that doesn't contain $U_{-\alpha}$. -The only reference I have for representations of semisimple groups is the small chapter in Humphrey's book Linear algebraic groups. I would also be grateful, if someone could give me a bigger reference on that subject. - -REPLY [5 votes]: A couple of added remarks, too long for a comment. For a more comprehensive reference in the algebraic group setting, there is (uniquely) the second edition of Jantzen's book Representations of Algebraic Groups (AMS, 2003). This large book of course contains far more than you need here but also allows the field to be of arbitrary characteristic. While the specific features of the finite dimensional irreducible representations (such as dimensions) usually differ in prime characteristic from the classical case, the "highest weight" technology doesn't really change so much. -On the other hand, if you stay in characteristic 0 the Lie algebra representations provide most of the information you want and are worked out in an example-oriented geometric spirit by Fulton and Harris. But the question you ask is really centered on roots and weights in the context of parabolic subgroups, where Jantzen gives the most extensive treatment. This is dealt with much more concisely in Wilberd's comment.<|endoftext|> -TITLE: element algebraically distinguishable from its inverse -QUESTION [25 upvotes]: (This question came up in a conversation with my professor last week.) -Let $\langle G,\cdot \rangle$ be a group. Let $x$ be an element of $G$. - -Is there always an isomorphism $f : G \to G$ such that $f(x) = x^{-1}$ ? - -What if $G$ is finite? - -REPLY [40 votes]: The Mathieu group $M_{11}$ does not have this property. A quote from Example 2.16 in this paper: "Hence there is no automorphism of $M_{11}$ that maps $x$ to $x^{−1}$." -Background how I found this quote as I am no group theorist: I used Google on "groups with no outer automorphism" which led me to this Wikipedia article, and from there I jumped to this other Wikipedia article. So I learned that $M_{11}$ has no outer automorphism. Then I used Google again on "elements conjugate to their inverse in the mathieu group" which led me to the above mentioned paper. -EDIT: Following Geoff Robinson's comment let me show that any element $x\in M_{11}$ of order 11 has this property, using only basic group theory and the above Wikipedia article. The article tells us that $M_{11}$ has 7920 elements of which 1440 have order 11. So $M_{11}$ has 1440/10=144 Sylow 11-subgroups, each cyclic of order 11. These subgroups are conjugates to each other by one of the Sylow theorems, so each of them has a normalizer subgroup of order 7920/144=55. In particular, if $x$ and $x^{-1}$ were conjugate to each other, then they were so by an element of odd order. This, however, is impossible as any element of odd order acts trivially on a 2-element set. - -REPLY [37 votes]: No, such an isomorphism does not always exist, and the smallest counterexample is $G=C_5\rtimes C_4$ with $C_4$ acting faithfully. It is not hard to see that the only automorphisms of $G$ are inner, and that they cannot map an element of order 4 to its inverse. - -REPLY [18 votes]: Here's a comment which might as well be written down. If $f$ is required to be an inner automorphism, then for $G$ finite this question can be understood using the character table of $G$: - -$x$ is conjugate to its inverse if and only if $\chi(x)$ is real for all characters $\chi$. - -Since $\chi(x^{-1}) = \overline{ \chi(x) }$, one direction is clear. In the other direction, if $\chi(x)$ is real then $\chi(x) = \chi(x^{-1})$ for all characters $\chi$, hence $c(x) = c(x^{-1})$ for all class functions $c$. One also has the following cute result: the number of conjugacy classes which are closed under inversion is equal to the number of irreducible characters all of whose values are real (equivalently, the number of self-dual irreps). Since there exist plenty of groups (even simple groups) whose character tables have complex entries, there are plenty of groups with elements not conjugate to their inverses. -This is one way to address the question for finite groups with no outer automorphisms.<|endoftext|> -TITLE: A Linear Algebra Question -QUESTION [8 upvotes]: Let $M$ be a symmetric square matrix with integer coefficients and $M_k$ the matrix obtained by deleting the k-th line and k-th column. If det(M)=0 does it follow that $\det(M_kM_j)$ is a square? - -REPLY [14 votes]: Abdelmalek's answer was posted while I was finishing up this, but I decided to post it anyway since it gives insight into why the statement is indeed true. -Let $M = (m_{i,j})\in Mat^{n\times n}(\mathbb{Z})$ and define $D_i := Det(M_i)$ -Since det(M) = 0, one has that some row of M is linearly dependent on the others, WLOG say the last row so that $m_{n,j} = \sum_{k=1}^{n-1}a_im_{k,j}$ for some $a_k's$; furthermore since $M$ is integer-valued, upon scaling one can assume the $a_k$'s are integers. -Now $M_n = \begin{pmatrix}m_{1,1}&m_{1,2}&\ldots&m_{1,n-1}\\m_{1,2}&m_{2,2}&\ldots&m_{2,n-1}\\ \vdots&\vdots&\ddots&\vdots\\m_{1,n-1}&m_{2,n-1}&\ldots&m_{n-1,n-1}\end{pmatrix}$ -Lets compare $M_{n-1}$ to $M_n$: -$M_{n-1} = \begin{pmatrix}m_{1,1}&m_{1,2}&\ldots&m_{1,n-2}&m_{1,n}\\m_{1,2}&m_{2,2}&\ldots&m_{2,n-2}&m_{2,n}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\m_{1,n-2}&m_{2,n-2}&\ldots&m_{n-2,n-2}&m_{n,n-2}\\m_{1,n}&m_{2,n}&\ldots&m_{n,n-2}&m_{n,n}\end{pmatrix} =$ -$\small\begin{pmatrix}m_{1,1}&m_{1,2}&\ldots&m_{1,n-2}&a_1m_{1,1}+a_2m_{1,2}+\ldots a_{n-2}m_{1,n-2}+a_{n-1}m_{1,n-1}\\m_{1,2}&m_{2,2}&\ldots&m_{2,n-2}&a_1m_{1,2}+a_2m_{2,2}+\ldots a_{n-2}m_{2,n-2}+a_{n-1}m_{2,n-1}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\m_{1,n-2}&m_{2,n-2}&\ldots&m_{n-2,n-2}&a_1m_{1,n-2}+a_2m_{2,n-2}+\ldots a_{n-2}m_{n-2,n-2}+a_{n-1}m_{n-1,n-2}\\m_{1,n}&m_{2,n}&\ldots&m_{n,n-2}&a_1m_{1,n}+a_2m_{2,n}+\ldots a_{n-2}m_{n-2,n}+a_{n-1}m_{n-1,n}\end{pmatrix}$ -Now since a determinant is unchanged by subtracting a multiple of one column from another, for each $i$ from 1 to $n-2$ subtract $a_i$ times the $i^{th}$ column from the last column, this leaves the following matrix with the same determinant as $M_{n-1}$: -$\begin{pmatrix}m_{1,1}&m_{1,2}&\ldots&m_{1,n-2}&a_{n-1}m_{1,n-1}\\m_{1,2}&m_{2,2}&\ldots&m_{2,n-2}&a_{n-1}m_{2,n-1}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\m_{1,n-2}&m_{2,n-2}&\ldots&m_{n-2,n-2}&a_{n-1}m_{n-1,n-2}\\m_{1,n}&m_{2,n}&\ldots&m_{n,n-2}&a_{n-1}m_{n-1,n}\end{pmatrix}$ -Now do the same thing along the rows of this matrix, giving the following matrix with the same determinant as $M_{n-1}$ -$\begin{pmatrix}m_{1,1}&m_{1,2}&\ldots&m_{1,n-2}&a_{n-1}m_{1,n-1}\\m_{1,2}&m_{2,2}&\ldots&m_{2,n-2}&a_{n-1}m_{2,n-1}\\ \vdots&\vdots&\ddots&\vdots&\vdots\\m_{1,n-2}&m_{2,n-2}&\ldots&m_{n-2,n-2}&a_{n-1}m_{n-1,n-2}\\a_{n-1}m_{1,n-1}&a_{n-1}m_{2,n-1}&\ldots&a_{n-1}m_{n-2,n-1}&a_{n-1}^2m_{n-1,n-1}\end{pmatrix}$. -Note that this matrix is obtained from $M_n$ by multiplying the last row by $a_{n-1}$ and then multiplying the last column by $a_{n-1}$, hence one has $D_{n-1} = a_{n-1}^2D_n$. A similar argument holds for the other $D_k$, thus $det(M_jM_k) = D_jD_k = a_{j}^2a_{k}^2D_n^2$ is indeed always a square.<|endoftext|> -TITLE: Cayley's Theorem and the Yoneda Lemma -QUESTION [10 upvotes]: Hi, from wiki, I know that yoneda lemma is the generalization of Cayley's theorem. But I am not quite understand the intuition behind that. Anyone can help me with that? Cheers! - -REPLY [10 votes]: Let $G$ be a finite group of order $n$ and consider the category $\mathscr G$ with a single object $\{ \bullet \}$ and whose morphisms consist of the elements of $G$, i.e., $\mathrm{Hom}_{\mathscr G}(\bullet, \bullet)=G$. -Let $h^{\bullet}=\mathrm{Hom}_{\mathscr G}(\bullet , {\_} )$ be the Yoneda functor. Then according to Yoneda's lemma -$$\mathrm{Nat}(h^{\bullet},h^\bullet)\simeq \mathrm{Hom}_{\mathscr G}(\bullet,\bullet),$$ -which implies that $h^\bullet$ is a faithful functor. In other words, -$$ h^{\bullet} : \mathscr G \to \mathscr Set$$ -maps $\bullet$ to the set $\mathrm{Hom}_{\mathscr G}(\bullet,\bullet)\simeq_{\mathscr Set} |G|$ (where $|G|$ is the set of elements of $G$) -and also embeds the group (!) $\mathrm{Hom}_{\mathscr G}(\bullet,\bullet)\simeq_{\mathscr Groups} G$ into $\mathrm{Hom}_{\mathscr Set}(|G|,|G|)\simeq S_n$, which gives you Cayley's Theorem.<|endoftext|> -TITLE: What groups have a second maximal subgroup below exactly four maximal subgroups? -QUESTION [7 upvotes]: I am looking for a finite group $G$ with the following property: there is a (core-free) subgroup $H < G$ such that the interval $\{ K : H < K < G\}$ in Sub[$G$] contains exactly four maximal subgroups. (In other words, $[H, G] \cong M_4$.) -I have used GAP to search for such groups and, to my surprise, I could find only three: $S_3$, $C_3 \times C_3$, and $(C_3 \times C_3) : C_2$. So far, all other examples reduce to these after modding out by a normal subgroup (so they are not examples if we require $H$ be core-free). -I've searched through most of the groups of order less than 960. Though, I can't promise my GAP code is free of bugs that may be causing me to miss something. -Question: -Does anyone know of other finite groups, besides $S_3$, $C_3 \times C_3$, and $(C_3 \times C_3) : C_2$, with an upper interval isomorphic to $M_4$? (If not, I would welcome any ideas that could help explain why this should be a rare phenomenon.) - -REPLY [4 votes]: I think that with some work you can characterize the examples where $G$ is solvable as follows. (I did not check carefully, so you can take this with a grain of salt.) -If $H$ has trivial core in $G$ and $[H,G]$ is $M_4$, then one of the following holds: -1) $G$ is the semidirect product $H(V+V)$, where $V$ is an irreducible $F_3[H]$-module such that the only elements of $GL(V)$ commuting with $H$ are the scalar transformations $1$ and $-1$. -(This case occurs when every maximal subgroup containing H has nontrivial core in $G$. Your example $C_2:C_3xC_3$ is of this type. This condition should be sufficient and you can construct tons of examples of this type without too much trouble.) -2) $G$ is the semidirect product $MV$, where $V$ is an irreducible $F_3[M]$-module. There is a $1$-dimensional subspace $W$ of $V$ such that $H=N_M(W)=C_M(W)$, and $C_V(H)=W$. - In particular, no element of $M$ has $W$ as an eigenspace with eigenvalue $-1$. -(This case occurs when some maximal $M$ containing H has trivial core in $G$. Your example $S_3$ is of this type. I don't know if there are lots of examples with $H$ maximal in $M$) -I can recommend that if you are interested in understanding how to find intervals of type M_n, you look at the work of Andrea Lucchini, including his joint paper with Robert Baddeley in Journal of Algebra. There is also recent work of Michael Aschbacher on the interval representation problem in general that is most certainly worth examining.<|endoftext|> -TITLE: Are undecidable consequences of Con recursively enumerable? -QUESTION [6 upvotes]: Let $X\subset\Pi_1^0$ be the set of statements which are -provable in PA$+$Con(PA) but independent of PA. -Is $X$ recursively enumerable? - -REPLY [10 votes]: Here's a proof that doesn't directly diagonalize but instead relies on well-known results, which in turn were proved by diagonalization. So ultimately, it isn't really easier than Emil's, but it may be easier to find and remember. -I claim first that, if a $\Pi^0_1$ sentence $\phi$ is provable in PA plus $\neg\text{Con}(PA)$, then it is already provable in PA. This is probably well known, but here's a proof anyway. The assumption is equivalent to saying that $\text{Con}(PA)$ is provable from PA plus $\neg\phi$. But since $\neg\phi$ is a $\Sigma^0_1$ sentence, one can also prove from PA plus $\neg\phi$ that PA proves $\neg\phi$. Combining the preceding two sentences, we get a proof from PA plus $\neg\phi$ that PA plus $\neg\phi$ is consistent. By Gödel's second incompleteness theorem, it follows that PA plus $\neg\phi$ is inconsistent. This means that PA proves $\phi$, as claimed. -Now consider the transformation $T$ on $\Pi^0_1$ sentences defined by letting $T(\phi)$ be $\text{Con}(PA)\lor\phi$. I claim that $T(\phi)$ is in the set $X$ of the question if and only if PA does not prove $\phi$. To see this, note first that $T(\phi)$ is trivially provable from PA plus $\text{Con}(PA)$. So $T(\phi)\in X$ if and only if PA doesn't prove $\text{Con}(PA)\lor\phi$. That's if and only if PA plus $\neg\text{Con}(PA)$ doesn't prove $\phi$. And, by the claim proved above, that's if and only if PA doesn't prove $\phi$. -So T is a (trivially computable) many-one (in fact one-one) reduction to $X$ of the set of $\Pi^0_1$ sentences not provable in PA. The latter set is known not to be recursively enumerable; therefore neither is $X$.<|endoftext|> -TITLE: Minimizing geodesic on a convex surface -QUESTION [16 upvotes]: Let $\Sigma$ be a smooth convex surface in Euclidean 3-space -and $\gamma$ be a unit speed minimizing geodesic in $\Sigma$. -Assume that for some $a < b < c$, we have -$$\gamma'(a)=\gamma'(b)=\gamma'(c).$$ -Is it true that $\gamma'$ is constant on one of two intervals $[a,b]$ or $[b,c]$? -Comments - -This is a simplification-variation of an other question I heard from Dima Burago. -It is not hard t construct an example of a minimizing geodesic such that $\gamma'(a)=\gamma'(b)$, but $\gamma'$ is not a constant on $[a,b]$. (See the example below.) -I would be also interested in the analog for $n$ points. -This paper: "Total curvature..." by Barany, Kuperberg, Zamfirescu is relevant. -Recently, in "On the total curvature..." by Nina Lebedeva and me we answered the original question of Dima Burago. The problem above remains open, likely the answer is "no". - -Example. I will construct a convex polyhedron, but it is easy to smooth. -Consider polyhedron defined by 5 inequlaities: -$$z\ge 0,\ \ |x+\alpha{\cdot}y|\le \alpha\ \ \text{and}\ \ z\pm\beta{\cdot}y\le \beta$$ -and look at the minimizing geodesic between points $(0,1-\epsilon, \beta{\cdot}\epsilon)$ and $(0,-(1-\epsilon), \beta{\cdot}\epsilon)$. For appropriately chousen $\alpha$, $\beta$ and $\epsilon$ this minimizing geodesic will pass through the faces in this order $$\{z+\beta{\cdot}y= \beta\},\ \ \{x+\alpha{\cdot}y= \alpha\},\ \ \{z=0\},\ \ \{x+\alpha{\cdot}y= -\alpha\},\ \ \{z-\beta{\cdot}y= \beta\}$$ -and it will have the same velocity vector $(1,0,0)$ on both faces $\{z+\beta{\cdot}y= \beta\}$ and $\{z-\beta{\cdot}y= \beta\}$. - -REPLY [7 votes]: Just to help visualize the example, if -I have interpreted the description correctly, -this is one view, for $\alpha=\beta=1$, and $\epsilon=\frac{1}{16}$. - - - -In this view, the $x$-axis is horizontal, the $z$-axis vertical. -Addendum. The shortest path (yellow) from $p_1$ to $p_2$ follows the face sequence (Bk, R, Bt, L, F). - - - I tried to indicate (in the nearly invisible short dashed lines) the initial direction of the path -on the (mauve) Bk and F faces, in their planar unfoldings. I should note that these initial -velocities are not exactly $(1,0,0)$, which presumably is only achieved by "appropriately chosen" -$\alpha, \beta, \epsilon$.<|endoftext|> -TITLE: Is the Euler product formula always divergent for 01$ -(and there it represents the Riemann zeta function). -My question: Is the Euler product always divergent for -$0 < \Re(s) < 1$ ? -I thought that the absolute value of the Euler product formula is positively divergent under the above condition. Is it apparent? - -REPLY [3 votes]: The simplest argument is that the domain of convergence of generalized Dirichlet series are always half planes. Since there is a pole at s=1, it can only converge for $Re(s) >1$. -But the question is still very interesting for two reasons. If you consider instead Dirichlet L-functions based on non-principal characters, then there is no pole at $s=1$ and the above argument does not apply. In recent work with Franca, we argued that the Euler product in fact converges for $Re(s)> 1/2$. -The other reason is that for Riemann zeta and other principal characters, a truncated Euler product can still be made sense of for $Re(s) > 1/2$. -Furthermore, the Euler product formula is valid in the limit $Im(s) \to \infty$ without truncation in a manner that can be made precise. This work is on math.NT<|endoftext|> -TITLE: A question about large cardinal axioms. -QUESTION [5 upvotes]: Can all the large cardinal axioms not presently known to be -inconsistent with ZFC be arranged in a strict linear order based -on the following criterion. Let A(1) and A(2) be any distinct pair -of these axioms and let both be adjoined to ZFC. We will say that -$A(1)\lt A(2)$ if and only if the smallest cardinal number satisfying A(1) -is smaller than the smallest cardinal number satisfying A(2). Should -such a strict linear ordering of these axioms be possible, which of -them ranks the largest? - -REPLY [6 votes]: Just to give a short, explicit example of the local-versus-global issue that Andres mentioned, consider supercompactness (global) and hugeness (local). In terms of consistency strength, hugeness is far stronger than supercompactness. Yet, if there exist both a supercompact cardinal and a huge cardinal, then the first supercompact cardinal is larger than the first huge cardinal.<|endoftext|> -TITLE: Is there a Galois correspondence for ring extensions? -QUESTION [24 upvotes]: Given an ring extension of a (commutative with unit) ring, Is it possible to give a "good" notion of "degree of the extension"?. By "good", I am thinking in a degree which allow us, for instance, to define finite ring extensions and generalize in some way the Galois' correspondence between field extensions and subgroups of Galois' group. -I suppose one can call a ring extension $A\subset B\ $ finite if $B$ is finitely generated as an $A$-module, and the degree would be the minimal number of generators, but is that notion enough to state a correspondence theorem? -Thanks in advance! - -REPLY [4 votes]: see related question -and SGA1 as well as Lenstra's notes -and Manjul Bhargava and Matt Satriano's paper<|endoftext|> -TITLE: Drinfeld't map, centre of quantum group, representation category of quantum group -QUESTION [9 upvotes]: My question is about the Drinfeld't map between $Rep(U_q(\mathfrak{g}))$ and $Z(U_q(\mathfrak{g}))$. I have heard the reference 1989 paper by Drinfeld't "Almost cocommutative Hopf algebras" - but this paper I cannot find in English. -Define the Drinfeld't map as follows. Given a representation $\pi_V: U_q(\mathfrak{g}) \rightarrow End(V)$, $(1 \otimes \pi_V)(R^{21}R) \in U_q(\mathfrak{g}) \otimes End(V)$, so let $c_V = tr((1 \otimes \pi_V)(R^{21}R)) \in U_q(\mathfrak{g})$. It is easy to see that the map $c$ is compatible with the additive structure on $Rep(U_q(\mathfrak{g}))$; is it also compatible with the multiplicative (tensor) structure on that category (my calculations say otherwise, but I may have messed up somewhere)? Then the claim is that $c_V \in Z(U_q(\mathfrak{g}))$ and these elements generate the center under suitable hypothesis. Why is this? -If there is an ambiguity: by $U_q(\mathfrak{g})$, I mean the quantum group defined in Jantzen's "Lectures on Quantum Groups", Chapter 4. -Edit: Cf. the comment below, feel free to make any assumptions that make the problem easier, e.g. characteristic $0$ and $q$ not a root of unity. - -REPLY [2 votes]: The fact that these elements are in the center is quite direct, just using the basic property of $R$ and of the trace, and is true in any quasitriangular Hopf algebra. You really need to know what you want the quantum group for, because it is certainly possible to define the QG so that this is all the center, or so that it is not!<|endoftext|> -TITLE: What is the minimal degree of a smooth projective embedding of a hyperelliptic curve? -QUESTION [14 upvotes]: (partly motivated by this question, but different: Degree of a Variety) -For a hyperelliptic curve $C$ of genus $g$ (over an algebraically closed field of characteristic not two) what is the smallest $d$ for which $C$ can be embedded in some $\mathbb{P}^n$ (I guess $n=3$ wlog) as a smooth curve of degree $d$? Does it depend only on $g$? Can anything be said for an arbitrary curve? - -REPLY [11 votes]: Felipe, I believe the answer here is d=g+3. To see that you can embed your curve in this degree is straightforward - just choose a generic line bundle of degree g+3 and it will work. -In the case of hyperelliptic curves, I don't think you can do better. The key point is that any special linear series on a hyperelliptic curve comes from taking a multiple of the pullback of O(1) on P^1 together with some base points. (You can find this fact in Arbarello-Cornalba-Griffiths-Harris.) Since these cannot give rise to embeddings (either have base points or the associated map factors through the hyperelliptic involution) we conclude the the embedding line bundle has no H^1. The Riemann-Roch gives g+3 as the lower bound for having 4 sections. - -REPLY [2 votes]: I agree with Tom G in the case of hyperelliptic curves. Interestingly, I think the bound for a general curve of degree $g$ should be $d = (3/4)g+3$. More specifically, the Brill-Noether theorem tells us that, for this $d$, there is a line bundle with $\dim H^0(X, L) =4$. I would guess (but don't know a reference) that, for generic $X$, this line bundle gives an embedding $X \to \mathbb{P}^3$.<|endoftext|> -TITLE: Dimensional Analysis in Mathematics -QUESTION [39 upvotes]: Is there a sensible and useful definition of units in mathematics? In other words, is there a theory of dimensional analysis for mathematics? -In physics, an extremely useful tool is the Buckingham Pi theorem. This allows for surprisingly accurate estimates that can predict on what parameters a quantity depends on. Examples are numerous and can be found in this short reference. One such application (pages 6-7 of the last reference) can derive the dispersion relation exactly for short water ripples in deep water, in terms of surface tension, density and wave number. In this case an exact relation is derived, but in general one expects to be off by a constant. The point is that this gives quick insight into an otherwise complex problem. -My question is: can similar techniques be used in mathematics? -I envision that one application could be to derive asymptotic results for say, ode's and pde's under certain asymptotic assumptions for the coefficients involved. For any kind of Fourier analysis, dimensions naturally creep up from physics if we think about say, time and frequency. I find myself constantly using basic dimensional analysis just as a sanity check on whether a quantity makes sense in these areas. -Otherwise, let's say I'm working on a problem involving some estimate on a number theoretic function. If I have a free parameter, can I quickly figure out the order of the quantity i'm interested in in terms of my other fixed parameters? - -REPLY [3 votes]: To add to the replies above, Terry Tao has a blog entry on this -A mathematical formalisation of dimensional analysis | What's new -https://terrytao.wordpress.com/2012/12/29/a-mathematical-formalisation-of-dimensional-analysis/<|endoftext|> -TITLE: Is this ergodic inequality true? -QUESTION [9 upvotes]: Is anything similar to the following inequality true, - -$\displaystyle P\{\max_{n \leq k \leq m} |A_k f - A_n f| > \epsilon\} \leq C \frac{||A_m f - A_n f||_1}{\epsilon}$ - -where $A_n f = \frac{\sum_{i=0}^{n-1} T^i \circ f}{n}$, and $T$ is a measure-preserving transformation? -My motivation for thinking it might be true is that something similar is true for martingales, namely - -$\displaystyle P\{ \max_{n \leq k \leq m} |M_k - M_n| > \epsilon\} \leq \frac{||M_m - M_n||_1}{\epsilon}$, - -by Doob's Submartingale Inequality (Ville's Inequality?), and I know there are many similarities between backward martingales and ergodic averages. However, I can't seem to deduce this from the Maximal Ergodic Theorem in the same way I can for the martingale case. -Any references or counterexamples would be helpful. Ergodic theory is not my specialty. Thank you! - -REPLY [4 votes]: I guess you wanted to define $A_n f =\frac{1}{n}\sum_{k=0}^{n-1}f \circ T^k.$ -Then you can prove that there exists constants $C,B>0$ such that: - -$P\left\{\max_{n\leq k \leq m} |A_k f - A_n f|>\epsilon\right\} \epsilon\}\leq P\{\max_{n\leq k \leq m} |A_k f-\int fdP |>\frac{\epsilon}{2}\}<(m-n)P\{|A_n f-\int fdP |>\frac{\epsilon}{2}\}0$ constants. -On the other hand, I do not see the analog with Martingales. For example, the process -$\{\exp(\lambda \frac{1}{\sqrt{n}}\sum_{k=0}^{[tn]}f\circ T^k-\frac{1}{2}\lambda^2 t),t\geq 0\}$ looks like a Martingale when $n$ is big.<|endoftext|> -TITLE: The rank of a class elliptic curves -QUESTION [5 upvotes]: For elliptic curve $y^{2}=x(x+a^{2})(x+(1-a)^{2})$,($a$ is a rational number and does not equal 0,1,1/2),is its rank always 0? - -REPLY [18 votes]: Your family of elliptic curves are exactly those elliptic curves with torsion subgroup containing $\Bbb Z / 2 \Bbb Z \times \Bbb Z / 4 \Bbb Z$. To see this make the linear change of variables $t = a/(1-a)$ and look at the parameterizations on page 101 of Husemöller's book "Elliptic Curves". -So in particular there are values of $a$ for which your elliptic curve has Mordell-Weil rank 8, the first of which was discovered by Elkies in 2005. See Dujella's website. Moreover, Eroshkin found that there are infinitely many such curves with rank at least 5. -These happen to be my favorite elliptic curves. I'd be very interested to know why you were interested in them.<|endoftext|> -TITLE: On a Theorem of Fontaine -QUESTION [5 upvotes]: Let $S=Spec(R)$ where $R$ is a Henselian local ring with fraction field $K$. Let $G$ and $G'$ be finite, flat group schemes of odd order over $S$ with isomorphic generic fibers (over $Spec(K)$). Does this isomorphism extend to one over $S$? -When $R=\mathbb{Z}_p$, the answer is yes following Fontaine's discussion in his 1975 paper 'Groupes finis commutatifs sur les vecteurs de Witt' (where he works over Witt vectors of a perfect field. Mazur uses this result to prove Theorem I.4 in his Eisenstein Ideal paper). What happens when $R$ is any general local ring (not necessarily Henselian)? Are there rings other than $\mathbb{Z}_p$ for which the answer to the above question is also a yes? - -REPLY [7 votes]: The answer to the edited question is 'yes' if $R$ is a complete DVR in mixed characteristic with absolute ramification index $e< p-1$ (where $p$ is the residue characteristic). This is one of the main results of Raynaud's (p,...,p) paper. He shows in Corollaire 3.3.6 that the functor sending a finite flat group scheme of $p$-power order over R to its generic fiber is fully faithful (when $e< p-1$).<|endoftext|> -TITLE: Is the kernel of a morphism of locally free sheaves still a locally free sheaf? -QUESTION [5 upvotes]: If we assum the underlying scheme is Notherian and reduced. - -REPLY [7 votes]: Let $X$ be a smooth quasi-projective variety over a field of dimension $n$ and $Z\subseteq X$ a subvariety such that the depth of $\mathscr O_Z$ at a fixed (closed) point of $X$ is $d$. Assume that $n-d\geq 3$, i.e., the projective dimension of $\mathscr O_Z$ at that closed point of $X$ is at least $3$. Then consider a locally free sheaf that surjects onto the ideal sheaf of $Z$. That gives a morphism to $\mathscr O_X$ and its kernel cannot be locally free, because then the projective dimension of $\mathscr O_Z$ at that closed point of $X$ would be $2$. -On the other hand, if your morphism is surjective, then what you want is true. Just count the rank of stalks.<|endoftext|> -TITLE: State of resolution in positive characteristic? -QUESTION [21 upvotes]: Heisuke Hironaka's coming talk makes me wonder how the state of the work on that theme is. So far, I noticed (but didn't read) these papers: -Kawanoue, Hiraku, Toward resolution of singularities over a field of positive characteristic. I. Foundation; the language of the idealistic filtration, Publ. Res. Inst. Math. Sci. 43, No. 3, 819-909 (2007). ZBL1170.14012. -Urabe, Tohsuke, New Ideas for Resolution of Singularities in Arbitrary Characteristic. -Edit: Another recent talk by Hironaka (in Vienna). - -REPLY [31 votes]: Also, I'd like to point out this paper of Hironaka... -http://www.math.harvard.edu/~hironaka/pRes.pdf -I haven't read the paper, and also I haven't heard anybody talking about it in the last weeks, which I find a little strange given the problem in question... has anybody here gone through the proof?<|endoftext|> -TITLE: How does the Framed Function Theorem simplify Cerf Theory? -QUESTION [22 upvotes]: A handle decomposition of a manifold $M$ is a useful structure to carry around. It is induced by a Morse function $f\colon\, M\to \mathbb{R}$. -How are two handle decompositions of $M$ related? The space of Morse functions turns out not to be connected. But if one expands the space of Morse functions to include functions $g\colon\, M\to \mathbb{R}$ which are Morse at all but finitely degenerate times, at which birth-death singularities occur, then one obtains a connected space, via Cerf's Theorem. So any two Morse functions $F_{0,1}$ are connected by a 1-parameter family of generalized Morse functions $f_t$ with $t\in[0,1]$ devoid of the worst singularities, but which still have "mild" singularities of codimension 1. We have made progress. However, the global pattern of crossings and birth-death singularities as $t$ runs from $0$ to $1$ might still be quite involved. This pattern can be visualized as a curve in $I\times \mathbb{R}$ called a Cerf graphic (see e.g. Kirby-Gay). Our next goal is to simplify the Cerf graphic, to get rid of as many crossings and birth-death singularities as possible. This entails another expansion of our space of functions, this time allowing elliptic and hyperbolic umbilics, and swallowtails, which are worse sorts of singularities. -An alternative idea to embed the space of Morse functions in a connected space of functions, which does not involve swallowtails and umbilics, is to introduce framed functions. A framed function is a generalized Morse function (Morse at all but finitely many degenerate times, at which birth-death singularities occur), together with an orthonormal framing of the negative eigenspace at the Hessian of the function $f$ at each critical point of $f$. Amazingly, it turns out that the space of framed functions is connected and contractible (see e.g. this MO question). This is (one version of) the Framed Function Theorem. -It's silly, but I'm having difficulty understanding how everything now fits together. I'm interested in the context of TQFT's, in which I would like to slice a manifold with boundary into simple pieces using a height function (in particular, I'm working with compact manifolds rather than with closed manifolds). In order to do this, my Morse functions need to be especially nice, in that they should not have level surfaces crossing the boundary, because then slicing would create corners. -How does the Framed Function Theorem simplify Cerf Theory? Does it completely replace it (in the sense that I can use the Framed Function Theorem instead of Cerf's Theorem to prove the Kirby Theorem (or pretty-much anything else Cerf's Theorem is used for), for example), because I no longer need swallowtails and elliptic and hyperbolic umbilics? -I'm asking this question primarily in order to motivate myself to look more seriously at understanding framed functions. I'd like to understand how exactly they fit into the bigger story first. - -REPLY [10 votes]: The framed function theorem tells you that up to "contractible choice" a compact manifold admits a framed function: i.e., a function as you prescribe. Furthermore, a framed function is supposed to give you a cell structure on the manifold up to contractible choice. -Note: The framing data is there to give you an explicit coordinatization of the cells. -(Actually, Igusa only showed that the space of framed functions is $n$-connected, where $n$ is the dimension of the domain manifold, but ideas of Eliashberg are supposed to give contractibility). -This is supposed to give rise to an alternate proof of Waldhausen's celebrated theorem $$A(X) \simeq Q(X_+)\times \text{Wh}^{\text{diff}}(X).$$ -One idea in the proof is roughly this: suppose for simplicity that $X =\ast$ is a point. -Then there is a model (the "expansion space") for $\text{Wh}^{\text{diff}}(\ast)$ that is a moduli space of finite, contractible, based cell complexes. That is, a point in the expansion space is a point contractible finite cell complex and the topology is defined so that a perturbations are of three kinds: (1) sliding cells over each other, (2) a refinement of the partial ordering of the cells, and (3) an elementary expansion/contraction. -Now suppose that $p : E \to B$ is a smooth fiber bundle whose fibers are contractible manifolds $E_t$ (for $t\in B$; for example, a bundle of $h$-cobordisms of a disk is such a case). Then the framed function theorem implies that $E$ can be equipped with a fiberwise framed function -$f: E \to \Bbb R$. -Now here is the difficult step: the framed function on each $f_t: E_t\to \Bbb R$ is supposed to give rise to a based cell structure on $E_t$ which varies continuously in $t$; these are -supposed to amount to a parametrized family of cell complexes, i.e., a map $B \to \text{Wh}^{\text{diff}}(*)$. -Unfortunately, none of these ideas have appeared (there is a preprint by Igusa and Waldhausen which is supposed to do just this, but it never was released). In my Ph.D. thesis (written in 1989 under the direction of Igusa), I manage to give the details of the geometric construction of the family of cell complexes in the special case of a fiberwise framed Morse function, that is, I assumed there were no birth/death singularities in the family. The associated family of cell complexes had cell slides but no elementary expansions/collapses.<|endoftext|> -TITLE: Consequences of technically proving anything in Coq (on at least Linux) exploiting a bug? -QUESTION [14 upvotes]: Technically, it is possible to prove anything in Coq proof assistant [1] (on at least Linux) due to a programming feature (or bug). This seems tractable when validating large proofs. Human analysis may catch this. -Question: - -What are the consequences of technically proving anything in Coq? - -More or less the question had in mind dishonest human provers presenting seemingly valid long computer readable proofs. -Later Coq developers replied `little hack'. I saw no official plans or replies of breaking the well formed proof (possibly because of backward compatibility being priority). -The basic technical details are: Hostile ocaml plugins (possibly disguised as proofs in FILE.v) can generate their own .vo proofs (of trivial statements), thus subverting coqchk, and don't giving a chance of "coqc" to even see the "anything proof". The plugin does this with exit(2) after writing .vo. This scenario seems interesting when validating large archives. -Here is a sample session, including links to full source code: - joro@j:/tmp/test1$ tar xvf ../proof.tar - fib5.v - bLOB - joro@j:/tmp/test1$ ls -l - total 16 - -rwxr-xr-x 1 joro joro 10301 2011-05-03 12:53 bLOB - -rw-r--r-- 1 joro joro 125 2011-05-03 12:53 fib5.v - joro@j:/tmp/test1$ coqc fib5.v - Trivially true. coqchk may pass - joro@j:/tmp/test1$ ls -l - total 24 - -rwxr-xr-x 1 joro joro 10301 2011-05-03 12:53 bLOB - -rw-r--r-- 1 joro joro 51 2011-05-03 12:55 fib5.glob - -rw-r--r-- 1 joro joro 125 2011-05-03 12:53 fib5.v - -rw------- 1 joro joro 812 2011-05-03 12:55 fib5.vo - joro@j:/tmp/test1$ coqchk fib5 - Welcome to Chicken 8.2pl1 (February 2010) - [intern /tmp/test1/fib5.vo ... done] - ...snip... - Checking library: fib5 - *** vo structure validated *** - checking cst: <>.fib5.thm1 - checking cst: <>.fib5.really - Modules were successfully checked - joro@j:/tmp/test1$cat fib5.v - Theorem thm1: True. - Proof. - auto. - Qed. - Declare ML Module "bLOB" . - Theorem really: True = False. - Proof. - intuition. - Qed. - #note: there is a zero byte at the end of "bLOB". in addition "bLOB" may be "aux.v" - joro@j:/tmp/test1$ coqchk -v - The Coq Proof Checker, version 8.2pl1 (February 2010) - compiled on Feb 27 2010 16:09:50 - -to compile the plugin: -ocamlopt -o bLOB -shared a.ml -the plugin writes valid but unrelated .vo proof and then does exit(2) to prevent coqc from analyzing the "anything proof". -a.ml is here. -tar with the proof is here -Possibly the most likely exploit scenario is: -Here is a proof of X consisting of about $10^5$ files. lemma2817.v is the plugin and it doesn't stop the final proof. -[1] http://coq.inria.fr/ -(crossposted on M.SE) -UPDATE: a.ml (OCAML) here: - open Printf;; - (* #load "unix.cma";; *) - (* compile: ocamlopt -o bLOB -shared a.ml*) - open Unix;; - let x = 1 + 2 ;; - printf "Trivially true. coqchk may pass\n";; - let fd = Unix.openfile "./fib5.vo" [O_RDWR ; O_CREAT] 0o600 in - ftruncate fd 0 ; - write fd "\x00\x00\x20\x08\x84\x95\xa6\xbe\x00\x00\x02\xef\x00\x00\x00\xe6\x00\x00\x02\xbe\x00\x00\x02\x91\xd0\xa0\x24\x66\x69\x62\x35\x40\xc0\x04\x03\xd0\x90\xa2\xb0\x42\x24\x66\x69\x62\x35\x40\xa0\xa0\x24\x74\x68\x6d\x31\x90\xf0\x40\x90\x90\x90\x9c\xa0\xa0\xb0\x90\xa0\x25\x4c\x6f\x67\x69\x63\xa0\x24\x49\x6e\x69\x74\xa0\x23\x43\x6f\x71\x40\x40\x24\x54\x72\x75\x65\x40\x41\x90\x9b\xa0\x04\x0c\x40\x90\x90\x40\x40\x41\x40\xa0\xa0\x26\x72\x65\x61\x6c\x6c\x79\x90\xf0\x40\x90\x90\x90\x9c\xa0\xa0\xb0\x90\xa0\x04\x19\xa0\x04\x18\xa0\x04\x17\x40\x40\x04\x16\x40\x41\x90\x9b\xa0\x04\x08\x40\x90\x90\x40\x40\x41\x40\x40\x40\x40\x40\x40\xa0\xa0\xa0\x2a\x4c\x6f\x67\x69\x63\x5f\x54\x79\x70\x65\xa0\x24\x49\x6e\x69\x74\xa0\x23\x43\x6f\x71\x40\x30\x5f\xb0\x16\xdd\x26\xc4\x94\xf9\x07\x89\x51\x91\x03\xf1\xd3\x06\xa0\xa0\xa0\x27\x50\x72\x65\x6c\x75\x64\x65\xa0\x24\x49\x6e\x69\x74\xa0\x23\x43\x6f\x71\x40\x30\x3f\x7c\xa2\x88\x4a\x72\x6c\x12\x85\x67\x03\x1c\x07\xf8\x24\x86\xa0\xa0\xa0\x27\x54\x61\x63\x74\x69\x63\x73\xa0\x24\x49\x6e\x69\x74\xa0\x23\x43\x6f\x71\x40\x30\x27\xf0\x1f\x3c\xa4\x0a\x05\x89\x0c\xe5\xc4\xb6\x96\xec\x49\x5a\xa0\xa0\xa0\x22\x57\x66\xa0\x24\x49\x6e\x69\x74\xa0\x23\x43\x6f\x71\x40\x30\xb0\xd6\xf6\x26\x6c\xdd\x54\x3f\x23\x97\x33\x9e\xa3\xec\x62\xf1\xa0\xa0\xa0\x25\x50\x65\x61\x6e\x6f\xa0\x24\x49\x6e\x69\x74\xa0\x23\x43\x6f\x71\x40\x30\x07\x94\x7b\x89\xa1\x80\xa3\x50\xc0\x48\x2e\xb7\x0c\x6b\xf3\x1e\xa0\xa0\xa0\x26\x53\x70\x65\x63\x69\x66\xa0\x24\x49\x6e\x69\x74\xa0\x23\x43\x6f\x71\x40\x30\x0d\x59\x83\x8d\x26\x27\x56\x53\x31\x41\x6b\x00\x71\x08\x50\x6b\xa0\xa0\xa0\x29\x44\x61\x74\x61\x74\x79\x70\x65\x73\xa0\x24\x49\x6e\x69\x74\xa0\x23\x43\x6f\x71\x40\x30\xb6\x9e\x13\xbf\x29\x8c\x28\xd9\xe1\x77\xc9\x93\xca\xdf\xae\xfb\xa0\xa0\xa0\x04\x62\xa0\x04\x61\xa0\x04\x60\x40\x30\x10\xcd\x50\xdb\x7b\x31\x4d\xb5\xd1\x18\x3f\x03\x19\xfa\xda\x1d\xa0\xa0\xa0\x29\x4e\x6f\x74\x61\x74\x69\x6f\x6e\x73\xa0\x24\x49\x6e\x69\x74\xa0\x23\x43\x6f\x71\x40\x30\xf0\xb3\xe5\xfb\x02\x4b\x8f\x8c\xdb\x44\x61\x6d\x64\x44\x74\x0c\x40\x40\xb0\x04\x7f\xa0\xa0\x04\x7d\xa0\x28\x43\x4f\x4e\x53\x54\x41\x4e\x54\xb0\x90\x91\xa0\x94\x90\x41\x40\x40\x92\x40\xa0\xa0\x22\x5f\x37\xa0\x29\x49\x4d\x50\x4c\x49\x43\x49\x54\x53\xa0\xa0\xb0\x92\x04\x93\x40\x04\x8f\xe0\x40\x41\x40\x40\x40\x40\xa0\xa0\x91\x04\x06\x40\x40\xa0\xa0\x22\x5f\x38\xa0\x24\x48\x45\x41\x44\xa0\x91\x04\x0d\x90\x90\x04\x0f\xa0\xa0\x22\x5f\x39\xa0\x2f\x41\x52\x47\x55\x4d\x45\x4e\x54\x53\x2d\x53\x43\x4f\x50\x45\xb0\x40\x91\x04\x16\x40\xa0\xa0\x04\x8c\xa0\x04\x28\xb0\x04\x27\x40\x92\x40\xa0\xa0\x23\x5f\x31\x30\xa0\x04\x22\xa0\xa0\xb0\x04\x21\x40\x04\x96\x04\x20\xa0\xa0\x91\x04\x04\x40\x40\xa0\xa0\x23\x5f\x31\x31\xa0\x04\x1f\xa0\x91\x04\x0a\x90\x90\x04\x0c\xa0\xa0\x23\x5f\x31\x32\xa0\x04\x1e\xb0\x40\x91\x04\x12\x40\x40\x40\xa0\xa0\x04\x4f\x04\x49\xa0\xa0\x04\x57\x04\x54\xa0\xa0\x04\x62\x04\x5c\xa0\xa0\x04\x6d\x04\x67\xa0\xa0\x04\x78\x04\x72\xa0\xa0\x04\x83\x04\x7d\xa0\xa0\x04\x8e\x04\x88\xa0\xa0\x04\x99\x04\x93\xa0\xa0\x04\xa4\x04\x9e\x40\xa0\x04\x60\xa0\x04\x70\xa0\x04\x7a\xa0\x04\x84\xa0\x04\x8e\xa0\x04\x98\xa0\x04\xa2\xa0\x04\x6d\xa0\x04\xad\x40\x84\x95\xa6\xbe\x00\x00\x00\x11\x00\x00\x00\x01\x00\x00\x00\x06\x00\x00\x00\x04\x30\x92\x3c\xc2\xa2\xba\xb3\x7b\x2f\x1b\xe7\xea\x03\x44\x0e\x9f\x1b" 0 812 ; - exit 4 ;; - -Update Possibly a more portable solution (Coq): -Theorem really: True = False. -Proof. - external "/bin/sh" "ESCAPE_SEQ; write_vo_proof; nicely_kill_coq ;" True. - (* this invokes /bin/sh *) -Qed. - -tactic external - -REPLY [6 votes]: What are the consequences of technically proving anything in Coq? - -This is a question in the Coq faq.<|endoftext|> -TITLE: Martin's Axiom and Determinacy-axioms: independence results -QUESTION [7 upvotes]: Hello everybody, -I'm searching for references for the following independence assertions: -ZFC + $MA_{\aleph_{1}}$ $\not\vdash$ "Analytic determinacy" -ZFC + $MA_{\aleph_{1}}$ $\not\vdash$ $\neg$ ("Analytic determinacy") -i.e. $MA_{\aleph_{1}}$ does not settle any determinacy question. The question extends also to Projective determinacy. -Also I'd need references for the reversed independence question, i.e. -Analytic determinacy (and Projective det. ) does not settle cardinality issues, so for instance. -ZFC + Analitic-Determinacy $\not\vdash$ CH -ZFC + Analitic-Determinacy $\not\vdash$ $\neg$CH -but also -ZFC + Analitic-Determinacy + "$2^{\aleph_{0}}> \aleph_{1}$" $\not\vdash$ $MA_{\aleph_{1}}$ -and -ZFC + Analitic-Determinacy + "$2^{\aleph_{0}}> \aleph_{1}$" $\not\vdash$ $\neg MA_{\aleph_{1}}$ -where by $MA_{\aleph_{1}}$ I mean the standard instance of Martin's Axiom at $\aleph_{1}$ (which implies $\neg CH$). -Please note that I have at my hands Fremlin's book "Consequences of Martin's Axiom" but it is very hard to read, and in the summary I couldn't find even the work "analytical determinacy" and just a reference to "determinacy". I also have Jech's Set theory. However I need these references for my PhD thesis (just to mention these facts) which i'm writing right now, and I'd rather not invest too much time searching in books at this stage. -So please, if you can, provide precise references. -THank you very much, -bye -matteo - -REPLY [15 votes]: ZFC plus $\text{MA}_{\aleph_1}$ is consistent relative to ZFC, while analytic determinacy has a little bit of large cardinal strength, namely the existence of sharps of reals. So ZFC+MA cannot prove analytic determinacy. On the other hand, analytic determinacy follows from the existence of a measurable cardinal, and the usual way of forcing $\text{MA}_{\aleph_1}$ preserves measurable cardinals (being a small forcing). So ZFC+$\text{MA}_{\aleph_1}$ can't refute analytic determinacy either. Similarly, since you can force either of CH or not-CH with a small forcing, hence preserving measurable cardinals, analytic determinacy cannot decide CH.<|endoftext|> -TITLE: Determinant associated with a group action -QUESTION [11 upvotes]: Let $G$ be a finite group and $S$ be a finite set, with $G$ acting on $S$. I consider indeterminates $x_g$ indexed by $g\in G$ and form the matrix of the group action $A\in M_{S\times S}$. Its entries are indexed by pairs $(s,t)$ of elements of $S$. By definition, $a_{st}$ is the sum of the $x_g$'s over the group elements such that $s^g=t$. - -What are the irreducible factors of $\det A$ ? - -Without loss of generality, we may assume that the action is transitive, because otherwise $A$ is block diagonal. Just sort out the elements of $S$ by orbits. We may also assume that the action is faithful. -This determinant has an obvious factor $\ell:=\sum_{g\in G}x_g$, because $\ell$ is the sum of the entries of every row (or of every column). What does it mean about the action that the quotient $(\det A)/\ell$ be irreducible ? -Notice that if $S=G$ and $G$ acts by multipication, this determinant is that considered by Dedekind and Frobenius, which led the latter to the theory of representations. - -REPLY [5 votes]: I wish to elaborate a bit on Geoffrey Robinson's answer: -Given a representation $D\colon G\to M_{n\times n}(F)$ which is similar to a representation in upper block form -$$ D \sim \begin{pmatrix} - D_1 & * \\ - & D_2 - \end{pmatrix}, -$$ -then the determinant factors into -$$ \Theta_D(x):= \det\left( \sum_{g} x_g D(g) \right) = \det\left( \sum_g x_g D_1(g) \right) - \det\left( \sum_g x_g D_2(g) \right) .$$ -Thus $\Theta_D$ factors accordingly to the composition factors of $D$, even if the characteristic of $F$ divides $|G|$. -If $D$ is absolutely irreducible, then $\Theta_D$ is irreducible (see here), and inequivalent representations yield different irreducible polynomials. In particular, your $\det(A)/l$ is irreducible over $\mathbb{C}$ iff the action is doubly transitive. -The coefficients of $\Theta_D(x)$ as polynomial in the $x_g$'s generate the same field as the values of the character of $D$, and of course $\Theta_D(x)$ remains irreducible over that smaller field (for $D$ absolutely irreducible). If $D$ is irreducible over $F$, but not absolutely irreducible, then $D \cong m(D_1 \oplus \dotsb \oplus D_r)$, where the $D_i$ are algebraically conjugate over $F$, and $\Theta_{D_1}(x)\cdot \dotsm \cdot \Theta_{D_r}(x) $ is irreducible over $F$. Thus in your situation, $\det(A)/l$ is irreducible as a polynomial over $\mathbb{Q}$ iff the permutation module -$$ \mathbb{Q}S \cong \mathbb{1} \oplus( \text{rationally irreducible module}). $$ -This is a slight weakening of double transitivity. There is a paper about this situation: - -Dixon, John D., Permutation representations and rational irreducibility, Bull. Austral. Math. Soc. 71 (2005), p. 493-503 (link). - -There are only few examples where this occurs and the action is not doubly transitive, but there is, to the best of my knowledge, no classification.<|endoftext|> -TITLE: What are the external triumphs of matroid theory? -QUESTION [50 upvotes]: As a relatively new abstraction, matroids clearly enjoy a rich theory unto themselves and also offer a viewpoint that suggests interesting analogies and clarifies aspects of the foundations of venerable subjects. -All that said, a very harsh metric by which to judge such an abstraction might ask what important results in other areas reasonably seem to depend in an essential way upon insights first gleaned from the pursuit of the pure theory. So I'd like to know, please, what specific results a matroid theory partisan would likely cite as the best demonstrations of the power of matroid theory within the larger arena of mathematics. -(I realize that mathematicians in one field will sometimes absorb ideas from another field, then translate back to their preferred language possibly obscuring the debt. So important papers that somehow could not exist without matroid theory should count here even if they never explicitly mention matroids.) - -REPLY [5 votes]: The notion of NBC-basis from matroid theory was used to give an elegant presentation for the cohomology algebra for the complement of a complex hyperplane arrangement, the `Orlk-Solomon Algebra'.<|endoftext|> -TITLE: Sobolev imbedding on Riemannian manifolds -QUESTION [5 upvotes]: Let $(M, g)$ be a non-compact smooth Riemannian manifold of dimension $n \ge 2$, and $G$ a subgroup of the isometry group of $(M,g)$, say with $G$ contained in the component of the identy. -Let $W^{1,2}_{G}(M)=\{f \in W^{1,2}(M)| \quad f\circ \phi =\phi \quad \forall \phi \in G\}$. -Is there any known result concerning the compactness of the Sobolev imbedding $W^{1,2}_{G}(M) \hookrightarrow L^p(M)$ for some subgroup $G$? - -REPLY [2 votes]: Hi, -You need additional geometric condition for the general case but considering the case of $\mathbb{R}^n$ with $G=SO(n)$, i.e. $H^1_ {radial}$, you have compact injection. You will find all the details in chapter 9 of the excellent book of Hebey Nonlinear Analysis on Manifolds: Sobolev Spaces and Inequalities.<|endoftext|> -TITLE: Ideas to face the current publishing issues ? -QUESTION [20 upvotes]: EDIT (Nov. 21, 2011) - follow up -For those who are interested by this question, let me mention that a similar one is discussed quite seriously on Timothy Gowers' blog (many thanks to r0b0t, who pointed this to my attention in a recent comment). -The context -There are many reasons for being unsatisfied with the current situation concerning scientific communication in mathematics. -Let me summarize a few issues that I consider quite important: - -math journal are expensive. -there is an increasing number of papers/preprints and the refereeing process starts to be a bit overtaken. -main preprint archives and review databases do not use the full potentiality of the web. -journals used to be a tool for communicating results... it is no longer the case and they became a kind of "label of quality" for papers (a more pessimistic way of saying this would be that journals became a tool for a blind management of sciences solely based on a quantitative index). - -Some answers -With respect to the first issue John Baez suggested broadly to stop publishing in and collaborating with journals that are highly priced. The main drawback of this is that it is not really a collective answer from the math community; moreover, I guess that only established people can allow themselves to do so. -I remember that Greg Kuperberg had a suggestion about publicity of referee's reports, whenever papers are accepted. -The idea is indeed quite nice but addresses only one issue. My concern here is about a global answer. -A proposal -My question is - -How could we (the math community) answer globally, and collectively? - -I actually have a naive proposal, which emerged after some discussions with Vincent Borrelli. -My dreamed answer is a website on which we can find - -archived preprints and/or links -to all versions of the same document -that exists on the web. -a discussion page associated to each -preprint, so that interested people -can ask/answer questions about the -paper (and even suggest -improvements). -a review page on -which one can find a review of the -paper. It would be like on -Mathscinet and Zentralblatt, except -that the review is a collective work -(nobody asks someone do to it). -There should be rules (the most -obvious one is that authors of the -paper cannot edit the review page). -an editorial board having the -possibility to "publish" some of -preprints. Here "publish" simply -means there is a label on the -document certifying that the -editorial board consider the paper -relevant and (i) either the discussion page converged -to a consensus about the validity of the results, -(ii) or the editorial board have asked experts to write reports on the paper, that -happened to be positive, -and that will be freely available together with the paper. - -One can of course imagine having different types of labels (by fields, by standards of quality, by length, etc... like for usual journals), but starting with one would be already great. -Only point (4) cannot be developed with solely "the help of people who want to help". -Here we would need to start with a board of well-established and famous mathematicians. -My second question is then - -Do you have any suggestion or criticisms to improve this proposal? And would you be ready to help ? - -Please just tell me if this is not the place for such a question, in which case I would simply delete it. -Let me add a more funny question: - -How should we cite labeled papers? I.e., do you have a suggestion for the name of this virtual journal ? - -EDIT : Thierry Zell pointed this MO question about a possibly free alternative to MathSciNet or ZB. It seems to me that all the objections to such a project could have been addressed to wikipedia. Nevertheless, I think it is not the heart of my question. - -REPLY [7 votes]: These are just some idealistic thoughts about a new type of Website -Perhaps a combination of arXiv and a heavily modified version of Stackexchange might be worth a try: -Papers - -Any User can submit papers, like asking questions here on MO -Papers containing Errors, previously published results, etc. can be voted to close/delete -A special System "Edit-Mode" to allow certain User Groups to edit a seperate copy of the paper (correcting mistakes) and allow the submitting User to undo his Mistakes and repost. Furthermore, this "Edit-Mode" will allow other Users to mark passages of a paper as correct. -Papers will be given Points based on: - -The number of times it has been cited -"Upvotes" weighted by the number of "Points" the voting User has - -Papers will be concidered "reviewed" if the entire paper has been marked as correct via "Edit-Mode" by at least X number of Users with >Y number of Points -As a rule, papers are only allowed to be cited when they have reached the "reviewed" rank - -Users - -Various groups of Users (Profs, Postdocs, Students, etc..). Certain groups need to be verified via PostIdent or some other mean. Users in these groups start out with a higher amount of Points -Users gain Points by publishing high ranked papers, writing good reviews, resubmitting corrected Papers... (Basically like MO) -Users can loose Points by plagarism or other wrongful behaviour - -Peer-Review would still be in place. The main questions would be: -Would people bother reviewing other papers? -If there are enough people in this system, and this system has become the main source of Mathematical Papers, the reviewing process should work. Merely the fact that only reviewed papers are allowed to be cited should be enough to "get the ball rolling". I also believe that the reviewing process will be honest, as it is public. -Furthermore, like here on MO, humans love achievements. People love gathering "Points", "Exp", or any other sort of trophies. So yes, I think this alone will motivate alot of people. -How to avoid "double posting" -As on MO, someone is bound to notice that a certain subject has already been researched. I even believe that this system will be even better stopping plagarism than the current peer-review Process. -The biggest problem would be, to get enough people to join this sort of Website. -And this last point is why I think, my proposed website will most likely fail.<|endoftext|> -TITLE: Is every Lie subgroup of GL(V) isomorphic to a (maybe another) closed subgroup of GL(V)? -QUESTION [12 upvotes]: I am gathering material for an exposition and I note that some texts (e.g. Ise and Takeuchi, "Lie Groups I & II", Stillwell, "Naive Lie Theory", Hall, "Lie Groups, Lie Algebras, and Representations") define "Matrix Lie Groups" with the unwonted requirement that the group should be a closed subgroup of $GL\left(V\right)$ (with $V = \mathbb{R}^n, \mathbb{C}^n$). I don't want to make such a restriction in my exposition - it seems a bit clunky to me and is certainly not needed. So the basic point of my question is - is the restriction just a simplification to make a first exposition easier to read (e.g. allows more readily grasped techniques to be used in proofs)? Or is there some deeper justification for it - e.g. the answer to my question if this answer is indeed yes? A simple example is the irrational slope one-parameter subgroup of the 2-torus - it is of course isomorphic to $\left(\mathbb{R},+\right)$. -By "isomorphic" I mean of course isomorphic as Lie groups, not just as abstract groups, everywhere in this question. Furthermore, for the purposes of this question, by "Lie Subgroup" I mean in Rossmann's (Rossmann "Lie Groups, An introduction through linear groups") sense: for a subgroup you use the topology generated by sets of the form $\exp\left(U\right)$ where $U$ is open in the Lie subalgebra of the subgroup you are considering - this is generally not the same as the relative topology gotten from $GL\left(V\right)$ if the subgroup is not closed. If you like, I have seen the term "Virtual Lie Subgroup" for what I mean by "Lie Subgroup" here. Thus, the irrational slope one-parameter subgroup of the 2-torus is not a submanifold of $GL\left(V\right)$, but if Rossmann's group topology is used, you've got a (virtual) Lie Subgroup. -Moreover, I'm not groping here for something like the closed subgroup theorem (Rossmann, section 2.7). One can argue that we study closed $GL\left(V\right)$ subgroups because this theorem guarantees they are Lie groups. I'm interested in whether ALL Lie subgroups of $GL\left(V\right)$ can be thought of as closed matrix groups after a suitable isomorphism. If you like, the isomorphism would be a "change of co-ordinates" to make the problem easier. -There is an MO discussion line that seems related here wherein Greg Kuperberg adapts a proof of Ado's theorem to show that every Lie algebra is the algebra of some closed subgroup of $GL\left(V\right)$. So that means that either my arbitrary group is covered by or covers a closed subgroup of $GL\left(V\right)$ - maybe it's trivial, but I can't see whether this line of reasoning can or can't be furthered to my suspected result. - -REPLY [16 votes]: Any linear Lie group is Lie-isomorphic to a closed subgroup of $GL(V)$: that's a result of Morikuni Goto: Faithful representations of Lie groups. II. Nagoya Math. J. 1, (1950). 91–107. From the review in MR: "A Lie group $G$ is called faithfully representable (f.r.) if there exists a topological isomorphism $\phi$ of $G$ into the general linear group of suitable degree $n$. It is shown ultimately that if $\phi$ exists, then $\phi$ can be chosen so that $\phi(G)$ is closed."<|endoftext|> -TITLE: The top Stiefel-Whitney Class of an Unorientable Manifold -QUESTION [6 upvotes]: The top Stiefel-Whitney class of an orientable manifold is the Euler characteristic mod 2, since it is the mod 2 reduction of the Euler class. Does this still hold for an unorientable manifold? - -REPLY [14 votes]: Every manifold is $Z_2$-orientable, so the Euler class (with $Z_2$-coefficients) is defined and coincides with the top Stiefel-Whitney class. -Back to your question: Yes, the top Stiefel-Whitney class evaluated on the fundamental class of the manifold is equal to the Euler characteristic mod 2 - regardless of the $Z$-orientability of the manifold. -This is Corollary 11.12 in Milnor, Stasheff: Characteristic Classes.<|endoftext|> -TITLE: Conceptualizing Weil Pairing for elliptic curves ( and number fields) -QUESTION [22 upvotes]: There are two explanations in Silverman ( Arithmetic of Elliptic Curves), one in exercises developing the Weil reciprocity law ( for algebraic curves) and then generalizing, and then there is a different, somewhat computational (in my opinion) proof in one of the chapters. -[I should also point out that in case of elliptic curves over complex numbers there is a rather simple description of this pairing in terms of determinant of a matrix!, see Ribet Stein : Hecke Operators ... for example] -While I do understand the proofs, I have heard that there is a conceptual explanation of this pairing? -Is there a uniform construction for the Hilbert Symbols in number field case, which is again a "form" of Weil pairing? - -REPLY [13 votes]: Here is how I like to understand the Weil pairing. -The dual of an abelian variety $A$ is the scheme $\hat{A} = \mathrm{Hom}(A, B\mathbf{G}_m)$. Here $B\mathbf{G}_m$ is the stack of line bundles and $\mathrm{Hom}$ refers to homomorphisms of group stacks. One therefore has a perfect pairing -$A \times \hat{A} \rightarrow B\mathbf{G}_m$. -This actually means something relatively concrete: For each pair of points $(a, a') \in A \times \hat{A}$ one has a one dimensional vector space $L(a,a')$, together with isomorphisms -$L(a_1 + a_2, a') \simeq L(a_1, a') \otimes L(a_2, a')$, -$L(a, a'_1 + a'_2) \simeq L(a, a'_1) \otimes L(a, a'_2)$ -satisfying some compatibility conditions (you can find the details of this definition under the heading of biextensions; there is a nice explanation in SGA7). One of these compatibilities is that the two isomorphisms -$L(a_1 + a_2, a'_1 + a'_2) \simeq L(a_1, a'_1) \otimes L(a_1, a'_2) \otimes L(a_2, a'_1) \otimes L(a_2, a'_2)$ -should coincide. -One consequence of the definition is that $L(0, a') = L(0 + 0, a') \simeq L(0, a') \otimes L(0, a')$ which means that $L(0, a')$ is canonically trivialized, as is $L(a, 0)$ by symmetry. Moreover, the two trivializations of $L(0,0)$ must be the same. -If we choose an $n$-torsion point $a$ of $A$ then $L(a, a')$ will be a line bundle with a trivialization of its $n$-th tensor power. Similarly, if $a'$ is also an $n$-torsion point then $L(a,a')^{\otimes n}$ will come with two trivializations coming from its identification with the canonically trivialized line bundles $L(na, a') = L(0,a')$ and $L(a, na') = L(a,0)$. Comparing these two trivializations, we get an element of $\mathbf{G}_m$ and therefore a pairing -$A[n] \times \hat{A}[n] \rightarrow \mathbf{G}_m$. -However, we notice that the induced trivializations of $L^{\otimes n^2} \simeq L(na, na')$ must coincide. Therefore the image of this map actually lands in $\mathbf{G}_m[n] = \mu_n$. This gives the Weil pairing -$A[n] \times \hat{A}[n] \rightarrow \mu_n$. -I don't know of a way to interpret the Hilbert symbol in quite this way, but I'd be very interested if someone could suggest one!<|endoftext|> -TITLE: Non-standard models of finite set theory -QUESTION [12 upvotes]: It is well known how the intended model and how the (countable) non-standard models of arithmetic look like. -It's also well known how the intended model of set theory with the axiom of infinity replaced by its negation (ZF-Inf) looks like: $\langle V_\omega;\in\rangle$, the hereditarily finite sets with the $\in$-relation. - -But how do (countable) non-standard - models of ZF-Inf look like? - -REPLY [19 votes]: Models of ZF-Infinity that arise from models of PA via binary bits - a method first introduced by Ackermann in 1940 to interpret set theory in arithmetic- end up satisfying the statement TC := "every set has a transitive closure". -It is known that the strengthened theory ZF-Infinity+TC is bi-interpretable with PA, which in particular means that every model of ZF-Infinity+TC is an "Ackermann model" of a model of PA. -However, TC is essential: there are models of ZF-Infinity that do NOT satisfy TC; and therefore such models cannot arise via Ackermann coding on a model of PA. -It is also known that there are "lots of" nonstandard model of ZF-Infinity [i.e., models not isomorphic to the intended model $V_{\omega}$] that are ${\omega}$-models [i.e., they have no nonstandard integer]. -It is possible for a nonstandard ${\omega}$-model of ZF-Infinity to have a computable epsilon relation. Indeed, there is an analogue of Tennenbaum's theorem here: all computable models of ZF-Infinity are ${\omega}$-models. -For more detail on the above, and references on the subject of finite set theory, you can consult the following paper: -http://academic2.american.edu/~enayat/ESV%20%28May19,2009%29.pdf -Ali Enayat -PS. In light of the comments about TC to my posting, it is worth pointing out that even though TC is not provable in ZF-Infinity, the theory ZF-Infinity is "smart enough" to interpret ZF-Infinity + TC via the inner model of sets whose transitive closure exists as a set [as opposed to a definable class; cf. the aforementioned paper for more detail]. -Therefore the relation of TC to ZF-Infinity is analogous to the relation between Foundation (Regularity) to ZF without Foundation since ZF is interpretable in ZF without Foundation via an inner model.<|endoftext|> -TITLE: What does it mean geometrically that an element in a domain is irreducible? -QUESTION [37 upvotes]: Consider a domain $A$ and a non-zero element $f\in A$. That element $f$ is prime if and only if the subscheme $V(f)\subset \operatorname{Spec}(A)$ is integral and this is a completely satisfactory geometric interpretation of primeness. -However I have realized to my annoyance that I can't interpret geometrically what it means that $f$ is irreducible (i.e. not a product of two non-units), a concept that a beginning undergraduate student certainly finds clear and easy! Here is why some naïve guesses turn out to be false : -$f$ irreducible $\nRightarrow V(f)$ irreducible -Take $f=x$ in the ring $\mathbb R[X,Y]/\langle X^2+Y^2-1\rangle =\mathbb R[x,y]$ -$V(f)$ irreducible $\nRightarrow f$ irreducible -Can't be right because $V(f)=V(f^2)$ as topological subspaces of $\operatorname{Spec}(A)$ -The ideal $(f)$ is irreducible $\nRightarrow f$ is irreducible -(An ideal is irreducible if it is not the intersection of two strictly larger ideals) -In the polynomial ring $k[X]$ over the field $k$ , take $f=X^2 \in k[X])$ -$f$ is irreducible $\nRightarrow $ the ideal $(f)$ is irreducible -In $\mathbb Z[\sqrt {-5}]$ notice that $3$ is irreducible but $(3)=(3,1+\sqrt {-5})\cap(3,1-\sqrt {-5})$ -So let me spell out two ( closely related) questions: -What is the geometric meaning of $f$ being irreducible ? -How do you show in/with Algebraic Geometry that an element is irreducible ? -Here I mean, for example, some analogue of the tricks used in Algebraic Number Theory, like saying that in $\mathbb Z[\sqrt {-5}]$, the number $3$ is irreducible, although not prime, because no element of the ring has norm $3$. I find it frustrating that I can't find an elegant argument proving that $x$ is irreducible in the ring $\mathbb R[X,Y]/\langle X^2+Y^2-1\rangle =\mathbb R[x,y]$ mentioned above... -Edit In relation with the comments below, let me add that in the first non-implication above the fact that $x\in \mathbb R[x,y] =\mathbb R[X,Y]/\langle X^2+Y^2-1\rangle $ is irreducible depends crucially on the ground field being $\mathbb R$: the corresponding irreducibility statement is false over $\mathbb C$. -Indeed in $ \mathbb C[x,y] =\mathbb C[X,Y]/\langle X^2+Y^2-1\rangle$ we have -$$ x=1/2(x-iy+i)(-ix+y+1)$$ - -REPLY [7 votes]: Dear Georges, -I don't have an answer for this interesting question but here are some remarks (which raise more questions than answers...). -1) If $A$ is a UFD then $f$ is irreducible iff $(f)$ is a nonzero prime ideal. So in this case we have a geometric characterization. Unfortunately this equivalence already fails for Dedekind domains. One could first investigate the case $A=k[C]$ where $C$ is a nonsingular affine curve over an algebraically closed field $k$ : is there a characterization of the irreducibility of $f$ in terms of $\operatorname{div}(f)$ ? -2) As I mentioned in my comment, for any domain $A$, a nonzero element $f \in A$ is irreducible iff the ideal $(f)$ is maximal among the principal ideals of $A$. This is equivalent to saying that the hypersurface $V(f) \subset \operatorname{Spec}(A)$ is minimal among the hypersurfaces of $\operatorname{Spec}(A)$ (endowed with their scheme structure). However, I find this characterization a little bit artificial, and I don't really know what to do further with it. -3) As you mention in the question, the topological space $V(f) \subset \operatorname{Spec}(A)$ is not enough to tell whether $f$ is irreducible (because $V(f)=V(f^2)$). One can further see that the full scheme structure of $V(f)$ is not sufficient to decide the irreducibility of $f$. An example (already given) is $A=\mathbf{R}[X,Y]/(X^2+Y^2-1)$, $f=\overline{X}$ and $A'=\mathbf{R}[Y]$, $f'=Y^2-1$. We have $A/(f) \cong A'/(f')$ but $f$ is irreducible in $A$ while $f'$ is reducible in $A'$. -So it seems that one needs further information on how $V(f)$ is embedded in $\operatorname{Spec}(A)$ in order to decide the irreducibility of $f$. Again giving just the continuous map $V(f) \to \operatorname{Spec}(A)$ isn't sufficient (think of $V(f^2)$). Of course the embedding of schemes $V(f) \to \operatorname{Spec}(A)$ enables one to recover $(f)$ and thus the nature of $f$, but we would like to use less information. In this direction I think Sándor's idea of looking what happens locally is a very good idea.<|endoftext|> -TITLE: Vanishing locus of a general section of a vector bundle. -QUESTION [9 upvotes]: Let $X$ be a smooth projective variety of dimension $\geq 2$ and $E$ a vector bundle on $X$ of rank $r\geq 2$. Is it true that, if $E$ is globally generated, then the zero locus of a general section of $E$ has codimension $\geq r$ in $X$? - -REPLY [11 votes]: As Donu pointed out, it may happen that the general section has no zeroes. Anyway, the the answer to your question is yes. -This is a special case of the following more general result "of Bertini type" about degeneracy loci of morphism of vector bundles, whose proof can be found in Ottaviani's book Varietà proiettive di codimensione piccola. This book is in italian; if you prefer a treatment in english, look at Lazarsfeld's book Positivity in algebraic geometry II, Section 7.2, and at the references given therein. - -Theorem. Let $\phi \colon F \to E$ be a morphism between vector bundles on $X$, with $\textrm{rank} \, F=s$, $\textrm{rank} \, E=r$, and let $D_k(\phi)$ be the set of points $x \in X$ such that $\phi_x \colon F_x \to E_x$ has rank $\leq k$. -If $F^{*} \otimes E$ is generated by global sections, then for the general $\phi$ either $D_k(\phi)$ is empty or it has the expected codimension $(s-k)(r-k)$. -Moreover, if $F^{*} \otimes E$ is ample and $(s-k)(r-k) < \dim X$, then for the general $\phi$ the locus $D_k(\phi)$ is non-empty, of the expected codimension $(s-k)(r-k)$. - -Now apply this result to the morphism $$\phi \colon \mathcal{O}_X \to E$$ -induced by the section. Since $D_0(\phi)$ is exactly the locus where the section vanishes, you are done.<|endoftext|> -TITLE: Shuffle Hopf algebra: how to prove its properties in a slick way? -QUESTION [29 upvotes]: Let $k$ be a commutative ring with $1$, and let $V$ be a $k$-module. Let $TV$ be the $k$-module $\bigoplus\limits_{n\in\mathbb N}V^{\otimes n}$, where all tensor products are over $k$. -We define a $k$-linear map $\mathrm{shf}:\left(TV\right)\otimes\left(TV\right)\to TV$ by -$\mathrm{shf}\left(\left(a_1\otimes a_2\otimes ...\otimes a_i\right)\otimes\left(a_{i+1}\otimes a_{i+2}\otimes ...\otimes a_n\right)\right)$ -$= \sum\limits_{\sigma\in\mathrm{Sh}\left(i,n-i\right)} a_{\sigma^{-1}\left(1\right)} \otimes a_{\sigma^{-1}\left(2\right)} \otimes ... \otimes a_{\sigma^{-1}\left(n\right)}$ -for every $n\in \mathbb N$ and $a_1,a_2,...,a_n\in V$. Here, $\mathrm{Sh}\left(i,n-i\right)$ denotes the set of all $\left(i,n-i\right)$-shuffles, i. e. of all permutations $\sigma\in S_n$ satisfying $\sigma\left(1\right) < \sigma\left(2\right) < ... < \sigma\left(i\right)$ and $\sigma\left(i+1\right) < \sigma\left(i+2\right) < ... < \sigma\left(n\right)$. -We define a $k$-linear map $\eta:k\to TV$ by $\eta\left(1\right)=1\in k=V^{\otimes 0}\subseteq TV$. -We define a $k$-linear map $\Delta:TV\to \left(TV\right)\otimes\left(TV\right)$ by -$\Delta\left(a_1\otimes a_2\otimes ...\otimes a_n\right) = \sum\limits_{i=0}^n \left(a_1\otimes a_2\otimes ...\otimes a_i\right)\otimes\left(a_{i+1}\otimes a_{i+2}\otimes ...\otimes a_n\right)$ -for every $n\in \mathbb N$ and $a_1,a_2,...,a_n\in V$. -We define a $k$-linear map $\varepsilon:TV\to k$ by -$\varepsilon\left(x\right)=x$ for every $x\in V^{\otimes 0}=k$, and -$\varepsilon\left(x\right)=0$ for every $x\in V^{\otimes n}$ for every $n\geq 1$. -Then the claim is: -1. The $k$-module $TV$ becomes a Hopf algebra with multiplication $\mathrm{shf}$, unit map $\eta$, comultiplication $\Delta$ and counit $\varepsilon$. It even becomes a graded Hopf algebra with $n$-th graded component $V^{\otimes n}$. -2. The antipode $S$ of this Hopf algebra satisfies -$S\left(v_1\otimes v_2\otimes ...\otimes v_n\right) = \left(-1\right)^n v_n\otimes v_{n-1}\otimes ...\otimes v_1$ -for every $n\in \mathbb N$ and any $v_1,v_2,...,v_n\in V$. -I call this Hopf algebra the shuffle Hopf algebra, although I am not sure whether this is the standard notion. What I know is that the algebra part of it is called the shuffle algebra (note that it is commutative), while the coalgebra part of it is called the tensor coalgebra or deconcatenation coalgebra. -Question: Is there a slick, or at least a not-too-long proof (I'm speaking of <10 pages in detail) for the statements 1 and 2? The best I can come up with is this here: -For 1, we WLOG assume that $V$ is a finite free $k$-module (because all we have to prove are some identities involving finitely many elements of $V$; now we can see these elements as images of a map from a finite free $k$-module $W$, and by functoriality it is thus enough to prove these identities in $W$). Then, we have $V^{\ast\ast}\cong V$, and we notice that the graded dual of our above graded Hopf algebra (we don't know that it is a graded Hopf algebra yet, but at least it has the right signature) is the tensor Hopf algebra of $V^{\ast}$, for which Hopf algebraicity is much easier to show. (Note that this only works with the graded dual, not with the standard dual, because $TW$ is free but not finite free.) -For 2, we prove that $v_1\otimes v_2\otimes ...\otimes v_n\mapsto \left(-1\right)^n v_n\otimes v_{n-1}\otimes ...\otimes v_1$ is indeed a $\ast$-inverse of $\mathrm{id}$ by checking the appropriate equalities combinatorially (i. e., showing that positive and negative terms cancel out). -These things are ultimately not really difficult, but extremely annoying to write up. Somehow it seems to me that there are simpler proofs, but I am unable to find any proof of this at all in literature (except of the "obviously" kind of proof). -One reason why I am thinking that there are simpler proofs is that the similar statements for the tensor Hopf algebra (this is another Hopf algebra with underlying $k$-module $TV$; it has the same counit and unit map as the shuffle Hopf algebra, but the multiplication is the standard tensor algebra multiplication, and the comultiplication is the so-called shuffle comultiplication) are significantly easier to prove. In particular, 2 holds verbatim for the tensor Hopf algebra, but the proof is almost trivial (since $v_1\otimes v_2\otimes ...\otimes v_n$ equals $v_1\cdot v_2\cdot ...\cdot v_n$ in the tensor Hopf algebra). -What would Grothendieck do? Is there a good functorial interpretation, i. e., is the algebraic group induced by the shuffle Hopf algebra (since it is commutative) of any significance? - -REPLY [4 votes]: Well, assume again that $V$ is a free $k$-module with base $X=x_i,i∈I$. One has to avoid the fact that $k\langle\langle X\rangle\rangle$ is NOT a Hopf algebra, be it with shuffle or concatenation except when $X=\emptyset$ because you have to take Sweedler's dual and cannot consider complete dual. A striking (but limited to free - finite or infinitely generated - $V$) proof of the first statement goes as follows -a) The Hopf algebra $k\langle X\rangle$ with concatenation as product and co-shuffle as coproduct is graded - in finite dimensions - over $\mathbb{N}^{(I)}$. -b) Then, the shuffle Hopf algebra is exactly the graded dual of it with the pairing given by -$$ -\langle x_{i_1}\otimes\ldots\otimes x_{i_n}\mid x_{j_1}\otimes\ldots\otimes x_{j_n}\rangle=\delta_{i_1,j_1}\ldots \delta_{i_n,j_n} -$$ -and 0 if $n\not=m$. -c) (For statement 2.) the antipode is just $S^*=S$ (the adjoint of $S$).<|endoftext|> -TITLE: Why do sl(2) and so(3) correspond to different points on the Vogel plane? -QUESTION [15 upvotes]: Vogel assigns to every simple metric Lie algebra (and more generally to every simple metric Lie algebra object in a symmetric monoidal category) a point in the orbifold $\mathbb{P}^2/S_3$ (where $S_3$ acts by permuting the 3 projective coordinates) based on the value of the Casimir on the various summands of the symmetric square of the adjoint representation. These three numbers are only defined up to permutation (as there's no natural way to specify which rep is which) and rescaling (because the Casimir itself is only well-defined up to rescaling). -Under this assignment we have $\mathfrak{sl}_2$ and $\mathfrak{so}_3$ going to different points. How is this possible? My best guess is that $\mathfrak{sl}_2$ and $\mathfrak{so}_3$ are different as metric Lie algebras, but that also seems weird. -In the conventions of this paper $\mathfrak{sl}_2$ corresponds to the point $(-1:1:1)$ while $\mathfrak{so}_3$ corresponds to the point $(-1:2:-1)$ and these are different points in $\mathbb{P}^2/S_3$. You can also easily check that the points are different in other convetions. -This question was originally asked in comments by Scott Carnahan, but I wanted to move it up to the main page. - -REPLY [6 votes]: I take it you mean $\mathfrak{sl}_2$ is on the line $\mathfrak{sl}_n$ and $\mathfrak{so}_3$ is on the line $\mathfrak{so}_n$. There is no contradiction because there is a whole line for this metric Lie algebra. The symmetric square of the adjoint decomposes as the trivial representation (which is accounted for by the Killing form) and one other irreducible (of dimension 5). This means you only have one of the three coordinates. Usually the coordinates are the values of the quadratic Casimir on the three non-trivial factors of the symmetric square of the adjoint representation. However looking at the adjoint representation you also know the sum of the three coordinates. This determines a line in the Vogel plane.<|endoftext|> -TITLE: How is the Ising model an example of a lattice model as per Kontsevich? -QUESTION [15 upvotes]: In section 3.2 of Kontsevich's very interesting paper "Notes on motives in finite characteristic,", he gives an axiomatic definition of a "lattice model" attached to a Boltzmann datum $(V_1,V_2,R)$, where $V_1$ and $V_2$ are vector spaces and $R$ is a linear endomorphism of $V_1$ tensor $V_2$. He remarks that the 2-dimensional Ising model is an example. Can someone explain to me what $V_1$, $V_2$, and $R$ are for the 2-dimensional Ising model? - -REPLY [15 votes]: In this section, Kontsevich is describing a pattern of index contraction of a certain tensor which reproduces a statistical mechanics sum over configurations. For his model, you have at each vertex, which I will label by the integer $k$, a copy of the same four index tensor -$$R_{a\mu}^{b\nu}$$ -where $a$ and $b$ are indices for one vector space of dimension $d_1$, $\mu$ and $\nu$ indices are for another vector space of dimension $d_2$. The dimensions of the vector space are going to be the number of discrete states at each site. -You have two different directions, he calls them 1 and 2, but I will call them "up" and "left". The image in your head should be a finite square grid, and at each point there are two incoming arrows, one going up and one going left, and two outgoing arrows going up and left. -He multiplies together all the $R$'s at all the vertices, and contracts the $a$-index of each site with the $b$-index of the site which is "left" of it, and the $\mu$ index of every site with the $\nu$ index of the site which is "up" of it. This contracts all the indices in closed cycles, giving a complex number. -In the spirit of maximum generalization, he defines the construction on the most general graph that allows it: the graph needs up arrows and down arrows, and the up arrows make closed oriented cycles, and the left arrows make closed oriented cycles. Maximally generalizing in another direction, he notes that the same definition works for any objects with a tensor product and a notion of contraction, i.e. for tensor categories. -Sliding the vertices along the two kinds of arrows, you get two independent "flows" which are number conserving on a finite graph, so the trajectories of the flows are cycles, and on each cycle, the flow is a cyclic permutations of the vertices. For the Ising model on an square $N$-by-$M$ lattice with periodic boundaries, one of the two flows moves all the vertices left by one unit, and the other moves all the sites up one unit, and these operations commute, and he likes this condition, so he emphasizes it. -The partition function with this pattern of contraction is then -$$Z = \sum_{a_k,\mu_k} \prod_k R_{a_k\mu_k}^{a_{l(k)}\mu_{u(k)}}$$ -where $l(k)$ is the left-neighbor of $k$, and $u(k)$ is the up-neighbor of $k$. That is, $Z$ is a sum over all possible configurations of values of the indices $a_k, \mu_k$ on the vertices of the graph, of the product of a certain quantity which depends on the value of the indices at the site and the right and up neighbor. In terms of the logarithm of $R$, -$$R_{a\mu}^{b\nu} = \exp(-E(a,\mu;b,\nu)),$$ -where $E$ is the energy function, then -$$Z = \sum_{a_k,\mu_k} \exp(-\sum_k E(a_k,\mu_k;a_{l(k)},\mu_{u(k)}))$$ -where the big sum outside is over all the possible assignments of indices $a_k$ and $\mu_k$ to each of the vertices, and $l(k)$ is the left-map taking vertex number $k$ to the number of the left neighbor, while $u(k)$ is the up-map, taking $k$ to the up neighbor. -To reproduce the Ising model, let $a_k$ and $\mu_k$ take the two "spin" values 0 or 1 (i.e. the two vector spaces $V_1$ and $V_2$ are both 2 dimensional). Then you want to make sure you get a zero contribution unless the index value $a_k$ is equal to $\mu_k$. For this purpose, make the energy function $E(a,\mu;b,\nu)$ infinite unless $a=\mu$ (i.e. make the corresponding $R$ element zero). Then the big sum on the outside collapses to a sum over $a_k$. The following values of $E$ are the finite ones: - -$E(00;00) = E(11;11) = 0$ (the up neighbor and the left neighbor are the same) -$E(00;01) = E(11;10) = J$ (the up neighbor is different) -$E(00;10) = E(11;01) = J$ (the left neighbor is different) -$E(00;11) = E(11;00) = 2J$ (both neighbors are different) - -The nonzero $R$ tensor elements are the exponentials of these. The coupling $J$ is the standard extra energy cost for mismatched neighboring spins. -Reconstructing what he is doing can be a little difficult because he says "correspond" in a way that is vague on the top of page 22, and "an identification" on the bottom of page 21 without specifying the correspondence or the identification. This is why I went into great detail. - -REPLY [3 votes]: For the 2d-Ising model $V_1=V_2$ is a $2$-dimensional vector space. The formula for $R$ may be find in Baxter's book "Exactly Solved Models in Statistical mechanics" (chapter 7 if I remember well). -The point is that the nearest-neighbor lattice Ising model is not a vertex model... but for a square lattice in 2d it is: one has to consider the dual lattice. -EDIT: this paper of Baxter might help you.<|endoftext|> -TITLE: Given a branched cover with branch cycle description $(g_1,...,g_r)$, does $g_i$ generate some decomposition group? -QUESTION [5 upvotes]: Classically: -Let $a_1,...,a_r$ be points in $\mathbb{P}^1_{\mathbb{C}}$, and let $\alpha_1,...,\alpha_r$ be simple loops around the $a_i$, all counterclockwise, and none touching (so $\alpha_1...\alpha_r=1$ in the fundamental group of the projective line minus those points). -An (pointed, to be pedantic) unramified $G$-cover (meaning a normal covering space with deck transformations$=G$) of $\mathbb{P}^1_{\mathbb{C}}-a_1,...,a_r$ is given by a surjection $\pi_1(\mathbb{P}^1_{\mathbb{C}}-a_1,...,a_r) \rightarrow G$. Let $g_i$ be the image of $\alpha_i$. We say that this $G$-Galois branched cover has branch cycle description $(g_1,...,g_r)$ (note that this depends on our choice of the $\alpha_i$'s). This covering map of curves can be extended to a map of (smooth) projective groups. It can then be shown by a simple topological argument that $g_i$ generates the inertia group (=decomposition group in this case) of some point above $a_i$. -My question is whether (and if so, how?) this is also true for the $\overline{\mathbb{F}_p}$ case. -Let me be precise. It is known via Grothendieck that $\pi_1^{(p)}(\mathbb{P}^1_{\overline{\mathbb{F}_p}}-a_1,...,a_r)=\widehat{\langle \alpha_1,...,\alpha_r|\prod \alpha_i =1 \rangle}^{(p)}$ (the $^{(p)}$ indicates that we're taking the inverse limit of all prime-to-$p$ finite quotients). Since these $\alpha_i$'s are given in SGA1 through a rather mysterious method, I wonder if the phenomenon described in the first paragraph is still true. -My question, therefore, is: let $G$ be a prime-to-$p$ group, and let $X\rightarrow \mathbb{P}^1_{\overline{\mathbb{F}_p}}$ be a (pointed, to be pedantic) branched $G$-cover with branch points $a_1,...,a_r$. Let $\alpha_1,...,\alpha_r$ be such that $\pi_1^{(p)}(\mathbb{P}^1_{\overline{\mathbb{F}_p}}-a_1,...,a_r)=\widehat{\langle \alpha_1,...,\alpha_r|\prod \alpha_i =1 \rangle}^{(p)}$ (I'm almost positive that what I'm about to say is false if you're allowed to choose any such $\alpha_i$'s, so let's assume that we're taking the ones from Grothendieck's construction. If you see a better way of saying what the condition should be on the $\alpha_i$'s I would be very interested in that). Let the branch cycle description of this cover be $(g_1,...,g_r)$ (with respect to these $\alpha_i$'s). Is it true that $g_i$ generates the inertia group (=decomposition group in this case) of some point of $X$ above $a_i$? -The topological argument that we were able to use for the $\mathbb{C}$ case seems to no longer apply... - -REPLY [2 votes]: I'm not sure I totally understand your question, but I think the answer is yes almost by definition. The procyclic group topologically generated by you alpha_i is the image of the map -pi_1(Spec O_i) -> pi_1(P^1 - a_1, ... a_r) -where O_i is the completed local ring of P^1 at the puncture a_i (so it's isomorphic to F_q((t)).) I guess the content here is that Grothendieck's comparison is compatible with the passage between local and global. -I was lazy and didn't write in base points; the choice of base point would induce the choice of WHICH decomposition group over a_i you get.<|endoftext|> -TITLE: What is an explicit example of a variety X which is finite over Spec F_p but which does not lift to a scheme Y which is finite and flat over Spec Z_p? -QUESTION [16 upvotes]: What is an explicit example of a variety X which is finite over Spec F_p but which does not lift to a scheme Y which is finite and flat over Spec Z_p? - -REPLY [10 votes]: I received the following very explicit answer via private communication: the algebra -$$ -A = \mathbb{F}_p[x_1,\ldots,x_6]/(x_1^p,\cdots, x_6^p, x_1x_2 + x_3x_4 + x_5x_6) -$$ -does not lift to a finite flat $\mathbb{Z}_p$ algebra. (I am still working out the details of why this does not lift.) This is exampe 3.2(4) of Berthelot-Ogus.<|endoftext|> -TITLE: Galois descent for K-groups (or for étale cohomology groups) -QUESTION [8 upvotes]: Let $F/K$ be a Galois extension of number fields with Galois group $G$. Let $\mathcal{O}_F$ and $\mathcal{O}_K$ be the associated rings of integers, and let $n\geq 1$. - -When is - $$ -K_{2n-1}(\mathcal{O}_F)^G \cong K_{2n-1}(\mathcal{O}_K)? -$$ - -It would be good enough for me to have this on the prime-to-2 parts. So let $p$ be an odd prime. Then -$$ -K_{2n-1}(\mathcal{O}_F)\otimes_{\mathbb{Z}}\mathbb{Z}_p\cong H_{ét}^1(\mathcal{O}_F,\mathbb{Z}_p(n)) -$$ -by the Bloch-Kato conjecture, as proven by Rost, Suslin, Voevodsky and Weibel. So my question becomes: - -when is - $$ -H_{ét}^1(\mathcal{O}_F,\mathbb{Z}_p(n))^G \cong H_{ét}^1(\mathcal{O}_K,\mathbb{Z}_p(n))? -$$ - -A concise reference to a place that summarises everything we know about this would be great! Or maybe it is clear to the experts that this is always true? That would be even better. - -REPLY [3 votes]: Sorry to answer my own question, but according to these notes by Manfred Kolster, Proposition 2.9, the étale cohomology groups in fact always satisfy Galois descent. Should have googled harder before asking.<|endoftext|> -TITLE: Is every homology theory given by a spectrum? -QUESTION [11 upvotes]: Let $E$ be a spectrum. For any CW complex $X$, define $h_*=\pi_i(E\wedge X)$. Then we know that $h_*$ form a homology theory. In other words, there functors satisfy the homotopy invariance, maps a cofiber sequence of spaces to a long exact sequence of abelian groups, also satisfy the wedge axiom in the definition of a homology theory. I want to know the converse case. Is every homology theory given by a spectrum in such way? -Thanks for all your comments. This is not really a problem. Anybody knows how to close it? - -REPLY [19 votes]: For homology theories on CW-complexes or homology theories that map weak equivalences to isomorphisms, that's Brown's representability theorem, which you can find in any textbook on stable homotopy theory. You forgot the important axiom of excision, by the way. The short answer is yes. - -REPLY [16 votes]: The answer is yes, if you replace the wedge axiom with the stronger direct limit axiom -$h_{i}(X) = \mathrm{lim}\ h_{i}(X_{\alpha})$, - where $X$ is the direct limit of subcomplexes $X_{\alpha}$. -As well as Switzer, this is discussed in Chapter 4.F of Hatcher's "Algebraic Topology", Adams' little blue book "Stable homotopy and generalised homology", and Adams' paper "A variant of E. H. Brown's representability theorem".<|endoftext|> -TITLE: lowest weight representation of loop groups -QUESTION [6 upvotes]: I am trying to understand lowest representations of loop groups as developed in Pressley and Segal's book. Specifically I want to be able to compute the weight spaces that appear in a lowest weight representation. I realize there is a formula for this -my question is along the lines of how to apply the formula correctly. -I tried to do a small example with $LSL_3$ (actually $\mathbb{C}^\times_{\theta} \ltimes \tilde LSL_3$) and something fishy happened so I was hoping someone could point out my mistake. The maximal torus is $\mathbb{C}^\times_\theta \times T \times \mathbb{C}^\times$ where $\mathbb{C}^\times_\theta$ is the loop rotations and the other $\mathbb{C}^\times$ is central. -The fundamental weights are $w_0 = (0,0,1)$, $w_1 = (0,-\omega_1,1)$, $w_2 = (0, -\omega_2,1)$. The positive roots are $(0,\alpha_1,0), (0,\alpha_2,0),(1,-\alpha_3,0)$. Where $\omega_i,\alpha_i$ are fundamental weights and positive roots of $SL_3$. Pressley and Segal normalize the Killing form so $\langle H_{\alpha_i},H_{\alpha_i}\rangle = 2$. Choosing coordinates $H_{\alpha_1} = [1\ \ 0]^T$, $H_{\alpha_2} = [0\ \ 1]^T$, $\alpha_1 = [2 \ \ -1]$, $\alpha_2 = [-1 \ \ 2]$, $\omega_1 = [1\ \ 0]$, $\omega_2 = [0\ \ 1]$ and the restriction of the Killing form to the torus is just the Cartan matrix $(B_{11} = B_{22} = 2, B_{12} = B_{21} = -1)$. -I'm interested in the representation $V_{\tilde \lambda}$ of lowest weight $\tilde\lambda = (0, - \alpha_3,3) = w_0 + w_1 + w_2$. Let $\tilde \mu = (m,\mu, 3)$ be a weight of $V_{\tilde \lambda}$. According to Loop Groups (11.1.1) it is the case that $\tilde \mu - \tilde \lambda = (m,\mu +\alpha_3, 0)$ is a sum of positive roots. Viewing $B$ as a map from co-characters to characters and noting that $\alpha_i = BH_{\alpha_i}$ it follows that we can write $\mu = B[a\ \ b]^T$ for some $a,b$. -According to (9.3.7) on pg 180 of Loop Groups the $\tilde\mu =(m, \mu,3)$ which satisfy -$3\langle \mu,\mu\rangle - 6m = 6 = 3\langle -\alpha_3,-\alpha_3\rangle$ -appear among the weight of $V_{\tilde \lambda}$. This says -$m = {1 \over 2}\langle\mu,\mu\rangle-1 = {1\over 2}[a\ \ b]B[a\ \ b]^T - 1 = a^2 + b^2- ab -1$. -Taking $a,b = 0$ produces the weight $\tilde \mu = (-1, 0, 3)$ but then $\tilde \mu - \tilde \lambda = -(1,-\alpha_3,0)$. Which is certainly not a sum of positive roots. So what gives? - -REPLY [4 votes]: The formula for the invariant bilinear form is given in $(4.9.3)$ on -page 64 $$\langle (x_1,\xi_1, y_1),(x_2,\xi_2,y_2) \rangle=\langle -\xi_1, \xi_2 \rangle - x_1 y_2-y_1x_2$$ As I mentioned in the -comments, $(9.3.7)$ becomes then $||\mu||^2-6m=2$. So your last -equation would be $m=\frac{1}{3}(a^2-ab+b^2)-\frac{1}{3}$. As you -said, there's no more worries about negative $m$ and as a consistency -check, for $m=0$, the solutions for $[a \ \ b]$ are $[\pm 1 \ \ 0]$, -$[0 \ \ \pm 1]$, $[1 \ \ 1 ]$, and $[-1 \ \ -1]$. Applying, $B$ -gives you the six weights in the Weyl orbit of $-\alpha_3$ (the -roots).<|endoftext|> -TITLE: Finite-dimensional faithful representations of compact groups -QUESTION [5 upvotes]: Is it true that a compact group always has a faithful, finite-dimensional unitary representation? If not, are there any reasonably simple counter-examples? -I've done some research and know that every group has some faithful representation, all irreducible reps of a compact group are finite, and that the irreducible reps separate the points of the group. However, that doesn't quite answer the question! - -REPLY [5 votes]: Another famous and interesting NASC for a compact (or even locally compact) group to have a faithful finite dimensional representation is that it "not have arbitrarily small subgroups", i.e., that there exist a neighborhood of the identity with no non-trivial subgroup. This was the way that von Neumann solved the Hilbert 5th Problem in the compact case, and is explained (starting on page 1243) in: -http://www.ams.org/notices/200910/rtx091001236p.pdf<|endoftext|> -TITLE: Surfaces containing curves of arbitrarily negative self-intersection -QUESTION [14 upvotes]: Olivier Wittenberg and I are curious about the following : -Let $S$ be a smooth projective complex surface. Are the self-intersection numbers of integral curves on $S$ always bounded below ? Or can $S$ contain integral curves with arbitrarily high negative self-intersection ? - -REPLY [17 votes]: It is a folklore conjecture that surfaces in characteristic zero has bounded negativity. For a nice account of this problem and references, see the two survey articles -Global aspects of the geometry of surfaces by Harbourne, and -Recent developments and open problems in linear series by Bauer et al. -In positive characteristic however, the situation is different and there is a nice counterexample due to Kollar (taken from the 2nd paper above): -Let $C$ be a smooth curve of genus $g\geq 2$ defined over a field - of characteristic $p>0$ and let $X$ be the product surface $X=C\times C$. - The graph $\Gamma_q$ of the Frobenius morphism defined by taking - $q=p^r$--th powers is a smooth - curve of genus $g$ and self-intersection $\Gamma_q^2=q(2-2g)$. - With $r$ going to infinity, we obtain a sequence of smooth curves - of fixed genus with self-intersection going to minus infinity.<|endoftext|> -TITLE: An elementary description of the torsion in $Pic(X)$ for a smooth projective curve $X$ -QUESTION [6 upvotes]: In computing the etale cohomology of curves, one of the key facts one needs is the torsion in $Pic(X) = H^1(X_{et}, \mathcal{O}_X^*)$ for a smooth projective curve $X$. Namely, one shows that the $n$-torsion (for $n$ prime to the characteristic, at least) is given by $(\mathbb{Z}/n\mathbb{Z})^{2g}$ where $g$ is the genus by invoking the theory of the Jacobian. However, I don't really know anything about the Jacobian (and pretty much all that I do know is in characteristic zero, in which case all you need is Abel's theorem for the aforementioned fact), so I am curious: can this fact, describing the torsion in $Pic(X)$, be proved directly in an elementary fashion? - -REPLY [4 votes]: As far as I know, the standard computation of the torsion in $Pic(X$) proceeds as follows: -one considers the exact sequence $0 \to Pic^0(X) \to Pic(X) \to \mathbb Z \to 0,$ -the map to $\mathbb Z$ being the degree map. This shows that the torsion subgroups -of $Pic(X)$ and $Pic^0(X)$ coincide. One then shows that $Pic^0(X)$ is -an abelian variety of dimension $g$ (this is the algebraic theory of the Jacobian), -and one applies the general theory of abelian varieties to compute its torsion. -An (apparently) alternative approach is as follows: one uses an $n$-torsion line bundle on $X$ to construct a degree $n$ cyclic cover of $X$, and thus classifies the $n$-torsion -of the Jacobian in terms of the maps from the etale $\pi_1(X)$ to $\mathbb Z/n\mathbb Z$. -But I don't know how (in positive characteristics, where one can't resort to analytic -arguments) to concretely compute the group of such maps without going back to the -computation in terms of $Pic^0(X)$ and the theory of abelian varieties.<|endoftext|> -TITLE: Birational automorphisms of canonical models -QUESTION [11 upvotes]: Let $X$ be a variety with canonical singularities such that $K_X$ is ample. -Do you have a reference of the fact that every birational map from $X$ to itself is biregular? -Thank you - -REPLY [2 votes]: Let $R$ be the canonical ring of $X$. We know that $R$ is finitely generated. Furthermore -$$X\cong Proj(R) = X_{can}$$ -since $K_{X}$ is ample and $X$ has at most canonical singularities. Now, any birational automorphism $f:X\dashrightarrow X$ induces an automorphism of the canonical ring $R$ which in turns induces a biregular automorphism of $X$. So $f$ itself is biregular. -More generally, if $X,Y$ are projective varieties with $K_{X},K_{Y} $ ample and at most canonical singularities, then any birational map $f:X\dashrightarrow Y$ is indeed biregular.<|endoftext|> -TITLE: If a manifold suspends to a sphere... -QUESTION [21 upvotes]: I have a topological manifold whose suspension is homeomorphic to the sphere $S^{k+1}$. Is it necessarily itself homeomorphic to $S^k$? -I know that this is not true if I replace "suspension" with "double suspension", because I found the helpfully named http://en.wikipedia.org/wiki/Double_suspension_theorem. - -REPLY [34 votes]: Suppose $M$ is a closed $n$-manifold whose suspension is homeomorphic to $S^{n+1}$. -Removing the two "singular" points from the suspension gives $M\times \mathbb R$, while -removing two points from $S^{n+1}$ gives $S^n\times\mathbb R$. Thus $M\times \mathbb R$ and $S^n\times\mathbb R$ are homeomorphic, which easily implies that -$M$ and $S^n$ are h-cobordant, and hence $M$ and $S^n$ are homeomorphic.<|endoftext|> -TITLE: $G$-structures of finite type. -QUESTION [8 upvotes]: A $G$-structure $\pi : B_G \rightarrow M$ is said to be of $finite$ $type$ if $\mathfrak{g}^{(k)} = 0$ for some $k \in \mathbb{N}$, where $\mathfrak{g}^{(k)}$ denotes the $k$th prolongation of the Lie algebra $\mathfrak{g} = T_{e}G$. For finite type $G$-structures, let us call the first $k$ for which $\mathfrak{g}^{(k)} = 0$ the $order$ of the $G$-structure. -For example: -• $O(n)$-structures (Riemannian metrics) are of finite type and order $1$, because $\mathfrak{o} -(n)^{(1)} = 0$. -• But $Sp(n)$-structures (symplectic structures) are not of finite type because the group of symplectomorphisms is infinite dimensional. -• It can also be shown that $CO(n)$-structures (conformal structures) are of finite type and order $2$ (except if the dimension is $2$, in which case it is not of finite type). -Are there any finite type $G$-structures of order greater than $2$? -More generally, are there $G$-structures of any order? -Thanks. - -REPLY [14 votes]: Yes, $G$-structures exist of each finite order. In other words, for every $k\ge1$, there is an $n\ge1$ and a subgroup $G\subset GL(n,\mathbb{R})$ such that its Lie algebra $\frak{g}$ satisfies ${\frak{g}}^{(k-1)}\not=0$ while ${\frak{g}}^{(k)}=0$. -There is no known classification of such algebras, but here is a simple example of an algebra ${\frak{g}}_k\subset {\frak{gl}}(k{+}3,\mathbb{R})$ such that ${\frak{g}}_k$ has order $k$: Let $e_1,\ldots, e_{k+3}$ be the standard basis of $\mathbb{R}^{k+3}$, with dual basis $x^1,\ldots, x^{k+3}$. Let ${\frak{g}}_k$ be the (abelian, nilpotent) subalgebra of ${\frak{gl}}(k{+}3,\mathbb{R})$ with basis $l_1,\ldots,l_k$, where -$$ -l_i = e_{i+3}\otimes x^1 + e_{i+2}\otimes x^2. -$$ -One computes that ${\frak{g}}_1^{(1)}=0$ and that, for $k>1$, the space ${\frak{g}}_k^{(1)}$ has dimension $k{-}1$, with basis $q_2,\ldots,q_k$, where -$$ -q_i = e_{i+3}\otimes (x^1)^2 + 2e_{i+2}\otimes x^1x^2 + e_{i+1}\otimes (x^2)^2. -$$ -Continuing on in this way, one finds that the dimension of ${\frak{g}}_k^{(j)}$ is $k{-}j$ for $0\le j\le k$. -For each $n$, there is an upper bound on the order of the subalgebras of ${\frak{gl}}(n,\mathbb{R})$ of finite type, but I do not know what that is. There are estimates for this upper bound, but I don't think they are very tight. -Meanwhile, a theorem of Cartan (originally proved over $\mathbb{C}$ by a classification (but with some omissions), and later completed by others and worked out over $\mathbb{R}$ as well) says that, if $G\subset GL(n,\mathbb{R})$ acts irreducibly on $\mathbb{R}^n$, then $\frak{g}$ has order $1$, $2$, or $\infty$. The list of the irreducibly acting $G\subset GL(n,\mathbb{R})$ that have order $2$ or $\infty$ is known and can be found in my 1996 survey paper, Classical, exceptional, and exotic holonomies: a status report, in Actes de la Table Ronde de Géométrie Différentielle (Luminy, 1992), Sémin. Congr., vol. 1 (1996), pp. 93–165. See the tables in Appendix A.<|endoftext|> -TITLE: Where is a good place to start learning about the Grothendieck-Teichmuller group? -QUESTION [13 upvotes]: I've had a desire to get some sort of handle on the Grothendeick-Teichmuller group for years now, but I've always felt that I could never find a source that was introductory and readable. -The obvious candidates are Schneps's: Grothendieck's "Long March through Galois theory"' andThe Grothendieck-Teichmuller group $\widehat{GT}$: a survey'. However I find them to be rather technical and off-putting for a person trying to understand this for the first time. -Do you know of an "easy" (by which I mean: well-motivated) introduction to the Grothendieck-Teichmuller group? - -REPLY [12 votes]: V.G. Drinfeld: On quasi-Hopf algebras and on a group that is closely connected with $Gal(\overline{\mathbb{Q}}/\mathbb{Q}$), Algebra i Analiz, 2:4 (1990), 149–181 -I am not saying it is easy to read but it is definitely well-motivated. By the way, the point of view on $\widehat{GT}$ might also be a bit different from what you expect. -In the very same spirit, but less technical, there is an excellent account of Drinfled's theory by Dror Bar-Natan: On Associators and the Grothendieck-Teichmuller Group I, Selecta Mathematica NS 4 (1998) 183-212 ( http://arxiv.org/abs/q-alg/9606021). - -Let me now explain broadly how this works. -Consider the collection $\mathcal B_*=(\mathcal B_n)_n$ consisting of groupoids of parenthesised braids. It is an operad in groupoids (by using the so-called cabling operations). -A more algebroic way to look at it is to consider the free braided monoidal category with one generator $\bullet$, and consider the groupoid of isomorphisms in this category. Objects in this category are actually parenthesisations of $\bullet^{\otimes n}$... so if we fix $n$ we recover $\mathcal B_n$. -Now we would like to define the group $GT$ as the automorphism group of this operad in groupoids $\mathcal B_*$... the main problem is that it is rather small. -The main point is then, for any field $k$, to consider $k$-prounipotent completions $\mathcal B_n(k)$ of $\mathcal B_n$. Now $\widehat{GT}(k)$ is the automorphism group of the operad in $k$-groupoids $\mathcal B_*(k)$. -Using the version of Mac-Lane coherence Theorem for braided monoidal categories, one can prove that such automorphisms are determined by their action on $\mathcal B_3(k)$, and that the relations we have to impose are in $\mathcal B_3(k)$ and $\mathcal{B}_4(k)$. -The relation in $\mathcal B_3(k)$ is the hexagon axiom for braided monoidal categories and comes from the cabling operations from $\mathcal B_2(k)$ to $\mathcal B_3(k)$. The relation in $\mathcal B_4(k)$ is the pentagon axiom for braided monoidal categories and comes from the cabling operations from $\mathcal B_3(k)$ to $\mathcal B_4(k)$.<|endoftext|> -TITLE: What does the term "yoga" mean in mathematics? -QUESTION [37 upvotes]: Just exactly what the title says; often, in mathematics, particularly in the vicinity of Grothendieck, I see reference to "the yoga of...". What exactly does the term "yoga" mean in these contexts? - -REPLY [38 votes]: I've taken "yoga" to mean a part of the body of mathematics which does not consist of many actual theorems or results -- or in fact could not be formalized as just a few theorems -- but rather a collection of principles and techniques that one needs to wrap one's head around completely, after which one will be able to use them almost effortlessly. -As an example, I would say that there is a yoga of generating functions in combinatorics. (Perhaps this is the simplest example of a yoga.) - -REPLY [13 votes]: "Yoga" and "yoke" (as in of oxen) are derived from the same -Indo-European root, meaning a linkage. Of course "linkage", and -"relation", and "connection", and "join", all have mathematical meanings -already, so one must go further afield to talk about two mathematical -concepts being yoked to one another. -When I have seen the word used -by mathematicians (esp. Bott), it is usually in exactly this way -- the yoga of X and Y, -not of X. -(As such I must disagree with the comment -`of course it has nothing to do with any proper meaning of the word "yoga".')<|endoftext|> -TITLE: Which of Quillen's Papers Should I read? -QUESTION [52 upvotes]: I just heard that Daniel Quillen passed on. I am not familiar with his work -and want to celebrate his life by reading some of his papers. Which one(s?) -should I read? -I am an algebraic geometer who is comfortable with cohomological methods in his field, but knows almost nothing about homotopy theory. My goal is to gain a better appreciation of Quillen's work, -not to advance my own research. -Here I tagged this question as "at.algebraic-topology, algebraic-k-theory" because I think these are the main fields in which Quillen worked. Please add or change this if other tags are appropriate. - -REPLY [13 votes]: Quillen's book on Homotopical Algebra is a great pleasure to read, and likely to appeal to a geometer. - -REPLY [11 votes]: There are a number of papers that I have affection for. Those which I don't see listed: -"On the group completion of a simplicial monoid." Before I read this paper I never really understood the appearance of the plus-construction. This essentially proves the +=Q theorem. (A little harder to find: Appendix Q in Friedlander-Mazur's "Filtrations on the homology of algebraic varieties.") -"On the formal group laws of unoriented and complex cobordism theory." A very influential paper, along the likes that Mark Grant mentioned, and one which we've been trying to unravel the consequences of ever since. (Bull. Amer. Math. Soc. 75 1969, 1293–1298.) -"The Adams conjecture." Quillen's proof of the Adams conjecture by making use of a Brauer lift is short but wonderful. (Topology 10 1971 67–80.)<|endoftext|> -TITLE: general point, general line -QUESTION [5 upvotes]: Hello, -could anyone explain the notion of ''general point'' and ''general line'', ''general hyperplane'' in algebraic geometry, What does it mean exactly general line in the 3 dimensional projective space? -Thank you. - -REPLY [8 votes]: In short: The statement "Property $P$ holds for a general hyperplane/line/etc" should mean that property $P$ holds for "almost every" hyperplane, line, etc. Now the cruicial point here is how to formalize the notion of almost every hyperplane - you have to model all hyperplanes: In other words, give them the structure of, say, a variety. Then you can say that some statement holds for almost every hyperplane if the property holds for a Zariski-dense open (Edit 05/08) subspace of the hyperplanes (in this model). -As Francesco already pointed out, this is exactly what the Grassmannian does for the linear subspaces of a vectorspace. It can be modelled as a projective variety via the Plücker embedding and then, it makes sense to say that some property holds for "almost every linear $n$-dimensional subspace". -Now this might help understand S. Carnahan's comment: The Hilbert Scheme is a construction that can serve as such a model for all closed subvarieties of a variety, hence allowing you to speak of a general hypersurface of degree $d$, for instance. - -REPLY [5 votes]: I hope that the following example, in the vein of Daniel's comment, can be useful for you. -Let $H \subset \mathbb{P}^3$ be a fixed plane and let us consider the statement: -"A general line $L \subset \mathbb{P}^3$ intersect $H$ in a single point". -One can think of it in the following way: the Grassmannian of lines $\mathbb{G}(1,3)$, via the Plucker embedding, can be identified with a quadric $X \subset \mathbb{P}^5$. The lines contained in $H$ give a $2$-plane $\Pi_H \subset X$. Therefore the word "general" in the statement precisely means -"a line corresponding to a point in the open dense subset $X \setminus \Pi_H \ $ of $X$". -Analogously, the lines containing a point $p \in \mathbb{P}^3$ form a $2$-plane $\Pi_p \subset X$. Then in the statement -"A general line $L \subset \mathbb{P}^3$ does not contain the point $p$" -the word "general" precisely means -"a line corresponding to a point in the open dense subset $X \setminus \Pi_p \ $ of $X$".<|endoftext|> -TITLE: A compactness property for Borel sets -QUESTION [9 upvotes]: Is the following generalised compactness property of Borel sets in a Polish space consistent with ZFC? -($*$) Let $\mathcal{B}$ be a family of $\aleph_1$-many Borel sets. If $\bigcap \mathcal{B} = \emptyset$, then, for some countable $\mathcal{C} \subseteq \mathcal{B}$, it holds that $\bigcap \mathcal{C} = \emptyset$. -In the special case that $\mathcal{B}$ is a family of open sets, it is not hard to show that property ($*$) is a consequence of $\neg\mathrm{CH} + \mathrm{MA}$ (where $\mathrm{CH}$ is the Continuum Hypothesis and $\mathrm{MA}$ is Martin's Axiom)—more specifically, a consequence of $\mathrm{MA}_{\aleph_1}$. Also, trivially, this special case of ($*$) itself implies $\neg \mathrm{CH}$. -However, I am stuck as to the consistency of ($*$) for general Borel sets. Actually, I have been unable to show consistency even when $\mathcal{B}$ is restricted to families of $F_\sigma$ sets. - -REPLY [5 votes]: Here is a even simpler example. -Let $\{x_{\alpha} \}_{\alpha\in \omega_1}$ be an increasing chain in the Turing degrees. For every $\alpha$, let $B_{\alpha}=\{y\mid y\geq_T x_{\alpha}\}$. Each $B_{\alpha}$ is a boldface $\Sigma^0_3$ set. -Then for any countable ordinal $\beta$, $\bigcap_{\alpha<\beta}B_{\alpha}$ is not empty but $\bigcap_{\alpha<\omega_1}B_{\alpha}=\emptyset$.<|endoftext|> -TITLE: The maximum of a polynomial on the unit circle -QUESTION [45 upvotes]: Encouraged by the progress made in a recently posted MO problem, here is a "conceptually related" problem originating from a 2003 joint paper of Sergei Konyagin and myself. -Suppose we are given $n$ points $z_1,...,z_n$ on the unit circle $U=\{z\colon |z|=1\}$ and $n$ weights $p_1,...,p_n\ge 0$ such that $p_1+....+p_n=n$, and we want to find yet another point $z\in U$ to maximize the product - $$ \prod_{i=1}^n |z-z_i|^{p_i}. $$ -How large can we make this product by the optimal choice of $z$? - -Conjecture. For any given $z_1,...,z_n\in U$ and $p_1,...,p_n\ge 0$ with $p_1+...+p_n=n$, there exists $z\in U$ with - $$ \prod_{i=1}^n |z-z_i|^{p_i} \ge 2. $$ - -Here are some comments. - -If true, the estimate of the conjecture is best possible, as evidenced by the situation where the points are equally spaced on $U$ and all weights are equal to $1$. -We were able to resolve a number of particular cases; say, that where the points $z_i$ are equally spaced on $U$, and also that where all weights are equal to $1$. -The case $n=2$ is almost trivial, but already the case $n=3$ is wide open. -In the general case we have shown that the maximum is larger than some absolute constant exceeding $1$. -Although this is not obvious at first glance, this conjecture is actually about the maxima of polynomials on the unit circle. - -I would be very interested to see any further progress! - -REPLY [17 votes]: Here is a second proof. Let -$$H(z)=\prod_j|z-z_j|^{p_j}.$$ -Let -$$F(z)=\prod_j\left((z-z_j)(1/z-\bar{z}_j)\right)^{p_j/n},\quad |z|\geq 1.$$ -It is easy to see that: -$$|F(z)|=H^{2/n}(z),\quad |z|=1,$$ -and that $F$ maps conformally the exterior of the unit disc on the exterior of a -"star" consisting of $n$ straight segments $[0,a_j]$ with some complex $a_j$. -Moreover, $F(z)\sim z$ as $z$ tends to infinity. -Now we see that the result is equivalent to the following theorem of Dubinin: -Let $K$ be the union of some intervals of the form $[0,a_j]$, and suppose -that capacity of $K$ is $1$. Then $\max_j|a_j|$ takes its minimal value when -the star is symmetric: all |a_j| are equal and the angles between adjacent -intervals are equal. -The reference is Dubinin, Uspekhi Mat. Nauk (=Russian Math. Surveys), 49 (1994). -Fedja's proof is much simpler, so it is a new proof of Dubinin's theorem.<|endoftext|> -TITLE: Crystalline realizations of Artin motives -QUESTION [7 upvotes]: What are the crystalline realizations of Artin motives? -In a paper by Kisin and Wortmann, "A note on Artin motives" (google it and you'll find it immediately), they define a suitable category of motives incorporating crystalline realization in analogy with Deligne's absolute Hodge cycles. However, they don't say explicitly what these realizations are (or at least I couldn't find it). Or can I find it somewhere else? - -REPLY [11 votes]: An Artin motive is just the same thing as a continuous representation of the absolute Galois group $G_K$ (of whatever number field $K$ we are thinking about) with finite image. -For conreteness, let's write it as $G_K \to GL(V)$, where $V$ is a finite dimensional vector space over $\mathbb Q$. (We could incorporate coefficients into the picture, by having $V$ be -an $E$-vector space over some other number field $E$, but this doesn't really change -anything: since the construction of realizations is functorial, the $E$-action will just -come along for the ride.) -Our Artin motive will have a crystalline realization at those place $v$ where the representation is unramified. If $v$ is such a place, lying over the prime $p$, then we -can consider $\mathbb Q_p \otimes_{\mathbb Q} V$; this is now a $p$-adic representation -of $G_K$, unramified at $v$. Restrict it to $G_{K_v}$. Fontaine's theory attaches an $n$-dimensional $K_v$-vector space to $\mathbb Q_p\otimes_{\mathbb Q} V$, its crystalline Dieudonne module, which is equipped with a $\sigma$-linear Frobenius operator; this is -the crystalline realization. (One could define it more geometrically, via crystalline -cohomology, but would get the same answer.) -In this simple context, one can compute the crystalline Dieudonne module in the following -concrete fashion: begin with $\mathbb Q_p\otimes_{\mathbb Q} V$, and give it -a "relative Frobenius" operator by looking at the action of the geometric Frobenius -at $v$ arising from our given $G_K$-action (restricted to $G_{K_v}$). -This is a linear operator, which will (by construction) turn out to be the $d$th -power of the crystalline Frobenius (if $d = [K_v^0:\mathbb Q_p]$, where $K_v^0$ is the -maximal subfield of $K_v$ that is unramified over $\mathbb Q_p$). -Now apply a $\sigma$-linear induction (from $\mathbb Q_p$ to $K_v^0$) to $\mathbb Q_p \otimes_{\mathbb Q} V$, to get an $n$-dimensional vector space over $K_v^0$ with -a $\sigma$-linear "absolute Frobenius", whose $d$th power is (the extension of scalars from to $K_v^0$ of) the "relative Frobenius" introduced above. This is the crystalline Dieudonne module, or, equivalently, the crystalline realization.<|endoftext|> -TITLE: Are measurable automorphism of a locally compact group topological automorphisms? -QUESTION [10 upvotes]: Consider a locally compact group $G$, considered as a measurable space with the completed Borelstructure wrt. the Haarmeasure. Consider a map $f:G \to G$, which is measurable and has an inverse, which is then also measurable. Is $f$ an homeomorphism? -What if $G$ is abelian? If not, what are necessary conditions on $G$, such that this is the case. -For $\mathbb{R}$, it seems to be true: see http://math.uchicago.edu/~henderson/additive.pdf (Wayback Machine). - -REPLY [24 votes]: The basic fact about locally compact groups is that you can recover the topology from the underlying measure space: -This is because, for any measurable subset $X\subset G$ of positive measure, the set -$$ -X^{-1}X:=\{x^{-1}y\mid x\in X, y\in X\} -$$ -is a neighborhood of the neutral element. -Letting $X$ vary along all measurable subsets of positive measure you get a basis of neighborhoods of $e\in G$. By translating by group elements, you get a basis of neighborhoods of any element $g\in G$. And so you recover the topology on $G$. -Corollary: -Since the topology is entirely encoded in the measurable structure, an automorphism that respects the measurable structure, will also respect the topology, i.e., be continuous.<|endoftext|> -TITLE: The number of singular fibres of a semi-stable arithmetic surface over \Z -QUESTION [7 upvotes]: This is an arithmetic follow-up to my previous question Does there exist a non-trivial semi-stable curve of genus >1 with only 4 singular fibres -Let $k$ be an algebraically closed field and let $f:X\longrightarrow \mathbf{P}^1_{k}$ be a semi-stable curve. Let $s$ denote the number of singular fibres. If $X$ is non-isotrivial and of positive genus, we have that $s>2$ (Beauville and Szpiro). As Angelo stated in my previous question, for genus >1 and $k=\mathbf{C}$, Sheng-Li Tan has shown that $s>4$. -Now, let $S=\textrm{Spec} \mathbf{Z}$ and let $X\longrightarrow S$ be a (regular) semi-stable arithmetic surface. Let $s$ be the number of singular fibres. Fontaine has shown that $s>0$ if $X$ is of positive genus. -Question. Let $g>0$ be an integer. Does there exist a semi-stable arithmetic surface $X\longrightarrow S$ of genus $g$ with precisely one singular fibre? -I expect the answer to be yes for $g=1$ but no for $g>1$. -Example. The modular curve $X_1(\ell)$ ($\ell$ big enough) has semi-stable reduction over Spec $\mathbf{Z}[\zeta_{l}]$. This model has precisely one singular fibre. Note that the modular curve $X_1(l)$ does not have semi-stable reduction over $\mathbf{Z}$. - -REPLY [7 votes]: There are some classical examples of such surfaces. For any prime number $p\ge 11$ different from 13, the modular curve $X_0(p)$ has good reduction away from $p$, and semi-stable reduction at $p$ (equal to the union of two projective lines intersecting at supersingular $j$'s). This is proved by Deligne-Rapoport (see also Bouw-Wewers: Stable reduction of modular curves, Prog. In Math. 224 (2004)). -There are however some constraints for such curves over $\mathbb Z$. If $p$ is the unique semi-stable fiber, then Brumer-Kramer (Manuscripta Math.(2001)) showed that $p\ne 2, 3, 5, 7$, and Schoof -(Compos. Math. (2005)) showed that $p\ne 13$.<|endoftext|> -TITLE: Homotopy groups of $S^2$ -QUESTION [39 upvotes]: in the paper -Foundations of the theory of bounded cohomology, -by N.V. Ivanov, the author considers the complex of bounded singular cochains on a simply connected CW-complex $X$, and constructs a chain homotopy between the identity and the null map. The construction of this homotopy involves the description of a Postnikov system for the space considered. In some sense, $S^2$ represents the easiest nontrivial case of interest for this construction, and I was just trying to figure out what is happening in this case. Since the existence of a contracting homotopy obviously implies the vanishing of bounded cohomology, this is somewaht related to understanding why the bounded cohomology of $S^2$ vanishes. -A first step in constructing the needed Postnikiv system is the computation of the homotopy groups of $X$, so the following question came into my mind: -Do there exists integers $n\neq 0,1$ such that $\pi_n(S^2)=0$? -I gave a look around, and I did not find the answer to this question, but I am not an expert of the subject, so I don't even know if this is an open problem. -In -Berrick, A. J., Cohen, F. R., Wong, Y. L., Wu, J., -Configurations, braids, and homotopy groups, - J. Amer. Math. Soc. 19 (2006), no. 2, 265–326 -it is stated that $\pi_n(S^2)$ is known for every $n\leq 64$, and Wikipedia's table -http://en.wikipedia.org/wiki/Homotopy_groups_of_spheres#Table_of_homotopy_groups -shows that $\pi_n (S^2)$ is non-trivial for $n\leq 21$. - -REPLY [46 votes]: SERGEI O. IVANOV, ROMAN MIKHAILOV, AND JIE WU have recently(2nd June 2015) published a paper in arxive giving a proof that for $n\geq2$, $\pi_n(S^2)$ is non-zero. -You can look at it in the following link. -Sergei O. Ivanov, Roman Mikhailov, Jie Wu, On nontriviality of homotopy groups of spheres, arXiv:1506.00952<|endoftext|> -TITLE: Geometric construction of depth zero local Langlands correspondence -QUESTION [20 upvotes]: Dear community, -In light of the recent work of DeBacker/Reeder on the depth zero local Langlands correspondence, I was wondering if there is an attempt to "geometrize" the depth zero local Langlands correspondence. -In particular, in Teruyoshi Yoshida's thesis, one can see a glimpse of this for $GL(n,F)$, where $F$ is a $p$-adic field. Namely : Suppose $k$ is the residue field of $F$. Let $w$ be the cyclic permutation $(1 \ 2 \ 3 \ ... \ n)$ in the Weyl group $S_n$ of $GL(n,F)$. Let $\widetilde{Y_w}$ be the Deligne-Lusztig variety associated to $w$, and denote by $H^*(\widetilde{Y_w})$ the alternating sum of the cohomologies $H_c^i(\widetilde{Y_w}, \overline{\mathbb{Q}_{\ell}})$. Let $T_w(k) = k_n^*$ be the elliptic torus in $GL(n,k)$, where $k_n$ is the degree $n$ extension of $k$. -Since $T_w(k)$ and $GL(n,k)$ act on cohomology, $H^*(\widetilde{Y_w})$ is an element of the Grothendieck group of $GL(n,k) \times T_w(k)$-modules. There is a canonical surjection $I_F \rightarrow k_n^* = T_w(k)$, where $I_F$ is the inertia subgroup of the Weil group of $F$. Therefore, we may pull back the $GL(n,k) \times T_w(k)$ action on -$H^*(\widetilde{Y_w})$ to an action of $GL(n,k) \times I_F$. -By Deligne-Lusztig Theory, as $GL(n,k) \times T_w(k)$-representations, -$$H_c^{n-1}(\widetilde{Y_w}, \overline{\mathbb{Q}_{\ell}})^{cusp} = $$ -$$\displaystyle\sum_{\theta \in C} \pi_{\theta} \otimes \theta$$ -where $C$ denotes the set of all characters of $k_n^*$ that don't factor through the norm map $k_n^* \rightarrow k_m^*$ for any integer $m$ such that $m \neq n$ and $m$ divides $n$, and where cusp denotes "cuspidal part". Here, $\pi_{\theta}$ is the irreducible cuspidal representation of $GL(n,k)$ associated to the torus $T_w(k)$ and the character $\theta$ of $T_w(k)$. -One of Yoshida's main theorems is that in this decomposition $$\displaystyle\sum_{\theta \in C} \pi_{\theta} \otimes \theta,$$ the correspondence $\theta \leftrightarrow \pi_{\theta}$ is indeed the depth zero local Langlands correspondence for $GL(n,k)$, ''up to twisting'' (this twist is unimportant for my question), by comparing with Harris-Taylor. -So my question is : Has anyone tried to generalize this to more general groups, but still working only in depth zero local Langlands? One could try to do this, and then compare to the recent work of DeBacker/Reeder (they write down a fairly general depth zero local Langlands correspondence). In other words, has anyone tried to realize depth zero local Langlands in the cohomology of Deligne-Lusztig varieties outside of the case $GL(n,F)$, which Yoshida did? -A priori the above idea for $GL(n)$ won't work on the nose for other reductive groups since the tori that arise in other reductive groups vary considerably, but something similar might. One would possibly want to try to pull back the action of $T_w(k)$ on cohomology to the inertia group $I_F$ in a more general setting, where now $T_w(k)$ is a more general torus in a more general reductive group. Then, one could compare to DeBacker/Reeder. -I took a look at the case of unramified $U(3)$, and it seems that things will work quite nicely. -My other question is : It might turn out that what I'm proposing is an easy check if one understands DeBacker/Reeder and Deligne-Lusztig enough to write this down in general. If so, then is my original question even interesting? It would basically say that Deligne-Lusztig theory is very naturally compatible with local Langlands correspondence, but the hard work is really in DeBacker/Reeder and Deligne-Lusztig, and putting everything together might not be difficult. Is the original question interesting regardless of whether or not it is difficult to answer? -Sincerely, -Moshe Adrian - -REPLY [2 votes]: Jared and community, -Thank you for your responses! I agree, I think one should be able to do things without explicit equations for the DL varieties. -I was also thinking along a different line than your line of attack. Here is a concrete example of how things might go (and I'm not sure if everything that follows is completely correct) : Consider the unramified p-adic group $U(3,F)$ over the p-adic field $F$. Let $k$ be the residue field of $F$. There are two natural types of depth zero supercuspidal Langlands parameters $$\phi_i : W_F \rightarrow GL(3,\mathbb{C}) \rtimes \mathbb{Z} / 2 \mathbb{Z}$$ -for $U(3,F)$, where $W_F$ is the Weil group. In both cases, the image of $I_F$ lands in $\hat{T}$, the dual maximal torus, where $I_F$ is the inertia group. Moreover, if $\Phi_F$ denotes Frobenius, $\phi_1(\Phi_F) = {}^{(13)} w$ and $\phi_2(\Phi_F) = {}^{(12)} w$, where ${}^{(13)} w$ denotes the Weyl group element of $GL(3,\mathbb{C})$ that switches the first and third entries of the maximal torus, and ${}^{(12)} w$ is the Weyl group element that switches the first and second entries of the maximal torus. -Consider a Langlands parameter of type $\phi_1$ (what follows will also work in an analogous way for parameters of type $\phi_2$). For ease of notation, set $w := {}^{(13)} w$. Let $\widetilde{Y_w}$ denote the DL variety associated to $w$. Associated to $w$ is also a twisted torus $T_w$. Unwinding definitions, one gets that $T_w(k) = k_2^1 \times k_2^1 \times k_2^1$, where $k_2$ denotes the degree 2 extension of $k$, and $k_2^1$ denotes the group of norm $1$ elements from $k_2$ to $k$. This torus splits over $k_2$, and the $k_2$-points are given by $T_w(k_2) = k_2^* \times k_2^* \times k_2^*$. -The twisted torus $T_w(k)$ and $U(3,k)$ both act on $H_c^*(\widetilde{Y_w})$. As $U(3,k) \times T_w(k)$-representations, we have that $H_c^*(\widetilde{Y_w})$ contains -$$\displaystyle\sum_{\theta \in C} \pi_{\theta} \otimes \theta$$ -where $C$ denotes the set of all characters of $T_w(k)$ in general position, and where $\pi_{\theta}$ is the representation of $U(3,k)$ associated to the torus $T_w(k)$ and the character $\theta$ of $T_w(k)$. The question is to get a $U(3,k) \times I_F$-action on $H_c^*(\widetilde{Y_w})$ so that we can get a local Langlands correspondence (and in the case of $U(3,F)$, it turns out that we don't even need the full Weil group). Note that we have the norm map on tori : $$N : T_w(k_2) \rightarrow T_w(k)$$ -Finally, as in the example of $GL(n,F)$ above (which Yoshida considers), one can consider the canonical surjection $$\alpha : I_F \rightarrow k_2^*$$ We may now consider the maps $$\alpha_1 : I_F \rightarrow T_w(k_2) = k_2^* \times k_2^* \times k_2^*$$ -$$x \mapsto (\alpha(x), 1, 1)$$ -$$\alpha_2 : I_F \rightarrow T_w(k_2) = k_2^* \times k_2^* \times k_2^*$$ -$$x \mapsto (1, \alpha(x), 1)$$ -$$\alpha_3 : I_F \rightarrow T_w(k_2) = k_2^* \times k_2^* \times k_2^*$$ -$$x \mapsto (1, 1, \alpha(x))$$ -We may pull back the action of $T_w(k)$ on $H_c^*(\widetilde{Y_w})$ to $I_F$ by the maps $\alpha_i \circ N$. In particular, form the direct sum $$ \displaystyle\bigoplus_{i=1}^3 (\alpha_i \circ N)$$ -This direct sum is an $I_F$-action on $H_c^*(\widetilde{Y_w})$, which is what we sought. If $\theta$ is a character of $T_w(k)$ in general position, denote by $\eta_{\theta}$ the pullback of $\theta$ to $I_F$ by $ \displaystyle\bigoplus_{i=1}^3 (\alpha_i \circ N)$. Via this action, $H_c^*(\widetilde{Y_w})$, as a $U(3,k) \times I_F$-module, contains -$$ \displaystyle\sum_{\theta \in C} \pi_{\theta} \otimes \eta_{\theta}$$ -With a fair amount of work (essentially in progress by Joshua Lansky and myself, and by results of Joshua Lansky and Jeffrey Adler), one can show that $\pi_{\theta} \leftrightarrow \eta_{\theta}$ "is the local Langlands correspondence". To be precise, I would have to say something about L-packets (since the L-packet of this Langlands parameter consists of more than one representation), but I will brush that under the rug for now. With much less work (I think), one can show that $\pi_{\theta} \leftrightarrow \eta_{\theta}$ agrees with the correspondence of DeBacker/Reeder (in particular, DeBacker/Reeder's correspondence is therefore correct for $U(3,F)$). Again, I am ignoring the entire L-packet for now (but essentially, you can conjugate $\theta$ around to various isomorphic tori in the group and get an L-packet of representations, as in DeBacker/Reeder). Moreover, since we are dealing with $U(3,F)$, it turns out that we don't even need to worry about ``twisting'', as we did with $GL(n,F)$. -This above construction is analogous to the $GL(n,F)$ example. In the case of $GL(n,F)$, one could pull the torus action back to $I_F$ using the norm map as above, take the direct sum of the analogous $\alpha_i \circ N$, and get the local Langlands correspondence ``up to twisting''. So really (as far as I can tell) this is what is being done in the case of $GL(n,F)$ as well. -It doesn't seem completely unreasonable that a similar construction might be able to be carried out in more generality (maybe this is wishful thinking) .<|endoftext|> -TITLE: The affine Grassmannian and the Bogomolny equations -QUESTION [11 upvotes]: In "Electric-Magnetic Duality and The Geometric Langlands Program", Sections 9 and 10, Kapustin and Witten describe certain convolution varieties in the affine Grassmannian (and more generally, in the Beilinson-Drinfeld) as moduli spaces of solutions to "the Bogomolny equations with 't Hooft operators added." While I can roughly make sense of what they are doing, it is not such easy reading for a mathematician, and of course, the proofs are pretty loose in nature. My (admittedly very vague) question is - -Have any mathematicians followed up on this description i.e. written things in more mathematical language and done the proofs rigorously, or used it to understand the affine Grassmannian better? - -REPLY [5 votes]: For those who could be interested, I worked out a formal construction of the E3-structure on the derived Satake category here, following the arguments hinted at by Lurie.<|endoftext|> -TITLE: blowing curves down -QUESTION [10 upvotes]: If I blow up $\mathbb{P}^2$ in $2$ points I get $3$ exceptional divisors - two are disjoint and one intersects both of them. What object do I get if I blow down this intersecting curve? It's a minimal surface with Picard number $2$ and no $(-1)$-curves, what is that? $\mathbb{P}^1\times\mathbb{P}^1$? - -REPLY [2 votes]: Up to automorphisms, the map $\mathbb{P}^2\to \mathbb{P}^1\times \mathbb{P}^1$ is in fact given by $(x:y:z)\mapsto ((x:y)(x:z))$, one easily sees that it blows-up $(0:0:1)$ and $(0:1:0)$ and that it contracts the line of equation $x=0$ onto the point $((0:1),(0:1))$. The inverse is $((a:b),(c:d))\to (ac:bc:ad)$, and it blows-up $((0:1),(0:1))$ and contracts the two lines $a=0$ and $c=0$ (passing through the point $((0:1),(0:1))$) onto $(0:0:1)$ and $(0:1:0)$. -One funny other comment is that if you take the real sphere $S$ of dimension $2$ and take the stereographic projection, it gives a birational map $S\to \mathbb{R}^2$ which is not defined at the north pole (the point of the projection). Seeing $S$ in $\mathbb{P}^3$ as a a smooth quadric and putting $\mathbb{R}^2$ into the projective plane, we obtain a birational map $S\to \mathbb{P}^2$ which blows-up the point $q$ (north pole) and contract the two imaginary lines being the intersection of the quadric with the tangent plane at $q$. -Over the reals, $S$ is not isomorphic to $\mathbb{P}^1\times \mathbb{P}^1$ (look at the topology of the real part) but extending to $\mathbb{C}$ it is, and the stereographic projection is exactly the inverse of the map you wanted (up to change of coordinates).<|endoftext|> -TITLE: Is there a classification of open subsets of euclidean space up to homeomorphism? -QUESTION [24 upvotes]: I hope this question is reasonable enough to have a well known answer. i.e either there is a simple invariant (like the homotopy groups) that characterizes the homeomorphism type of such set among other such sets or there is an explanation of why such classification is difficult. partial results, like the well known fact that every two open connected and simply connected subsets of the plane are homeomorphic, would be interesting too. thanks. - -REPLY [4 votes]: Here's a descriptive set-theoretic view of the problem of classifying open subsets of $\mathbb{R}^2$ up to homeomorphism. The eventual answer will be a formal version of "it's pretty darn hard, but maybe not inconceivably hard." Note that I haven't actually done any work here besides placing the work of others in the correct order. -First, a couple of words on classification problems in general (read: propaganda). When somebody has classified some objects $\mathcal{A}$ in terms of some objects $\mathcal{B}$, two natural questions to ask are -(1) How hard is it to compute the particular $B \in \mathcal{B}$ corresponding to a given $A \in \mathcal{A}$?, and -(2) How hard is it to distinguish elements of $\mathcal{B}$ from each other? -If either of these tasks is particularly hard, one could argue that the classification isn't all that useful. For example, in the current problem one could say "invoke the axiom of choice to wellorder the quotient space of open sets modulo homeomorphism and classify an open set by the location of its equivalence class in the order," which makes (1) rather challenging. Alternatively, one could simply beg the question by saying each open set is in a unique homeomorphism-class, so the homeomorphism-classes themselves provide a classification, which makes (1) easy (in a sense) but there's no reason to think (2) can be done simply. -One way to formalize these concepts is through Borel reducibility of equivalence relations. If $E$ and $F$ are equivalence relations on standard Borel spaces $X$ and $Y$ (resp.), we say $E$ is Borel reducible to $F$, written $E \mathrel{\leq_B} F$, iff there is a Borel function $\phi: X \to Y$ such that $x_0 \mathrel{E} x_1 \iff \phi(x_0) \mathrel{F} \phi(x_1)$. Loosely, the relation $E \mathrel{\leq_B} F$ means "any definable classification of $Y/F$ is also a definable classification of $X/E$" (or, put more succinctly, "$F$ is at least as complicated as $E$"). This gives us a meaningful way to compare various classification problems.(*) -The $\leq_B$ relation is well studied (but there's still a lot we don't know!) in the case that $E$ and $F$ are in themselves in some sense definable. Of particular note is when they are orbit equivalence relations induced by a continuous action of a Polish group on a Polish space. Of even more particular note is when this Polish group is $S_\infty$, the group of permutations of the natural numbers (including those of infinite support). This group is especially important from a logical point of view. Given some countable (let's say relational) language $\mathcal{L}$, you can look at the $\mathcal{L}$-structures with universe $\mathbb{N}$. Then the group of automorphisms of $\mathbb{N}$ respecting $\mathcal{L}$ is a closed subgroup of $S_\infty$. In particular, isomorphism of $\mathcal{L}$-structures with universe $\mathbb{N}$ is Borel reducible to to an equivalence relation induced by a nice action of $S_\infty$. -Moreover, there's an $S_\infty$-universal equivalence relation $E_{S_\infty}$ in the sense that every equivalence relation induced by a nice $S_\infty$ action Borel reduces to $E_{S_\infty}$. In a precise sense, $E_{S_\infty}$ is at least as complicated as the isomorphism relation for any sort of countable structure in a countable language (for example, isomorphism of countable graphs, isomorphism of countable groups, etc.). -Now, the punchline of all this is: there's a Borel reduction of $E_{S_\infty}$ to the homeomorphism relation on open subsets of $\mathbb{R}^2$ (equipped with the natural interpretation of the Effros Borel structure). This follows from -Camerlo and Gao, "The completeness of the isomorphism relation for countable Boolean algebras," Trans AMS 353 (2001), no. 2, 491-518 -and the the pitch-perfect response by André Henriques to a rather nebulous question of mine in the comments after his answer. The idea is that there is a Borel reduction of $E_{S_\infty}$ to homeomorphism of closed subsets of Cantor space (again with the corresponding Effros Borel structure), constructed in Camerlo-Gao. Fixing some copy of Cantor space in the plane, this reduction can then be composed with the Borel map $\phi$ that assigns to each closed subset $K$ of Cantor space its complement in $\mathbb{R}^2$. Since it is known from results of Richards (thanks, Agol!) that $\phi(K_0)$ and $\phi(K_1)$ are homeomorphic iff $K_0$ and $K_1$ are, this composition is indeed a Borel reduction. In fact, if you restrict your attention to nowhere dense subsets of Cantor space you get a Borel reduction to a part of the homeomorphism relation induced by a Polish group action (I'm not sure whether the entire homeomorphism relation is so induced). -So, in other words, any classification of open subsets of $\mathbb{R}^2$ up to homeomorphism is also a classification of any countable structure in any countable language up to isomorphism! So, that's what I meant when I said it's "pretty darn hard." -So then what did "maybe not inconceivably hard" mean? Some equivalence relations are so complicated that they are not classifiable by countable structures. That is, they can not be Borel reduced to any isomorphism relation of $\mathcal{L}$-structures as described above. For example, homeomorphism of closed subsets of Cantor space is classifiable by countable structures (Stone duality), while isomorphism of type $II_1$ von Neumann factors is not classifiable by countable structures (Sasyk-Törnquist 2009). I have no idea where this problem sits with respect to this notion (to show that this homeomorphism relation is classifiable by countable structures it would be enough to show that there's a Borel way of associating with each open subset $U$ of the plane a closed subset of Cantor space homeomorphic to $U$'s space of ends). For more examples see: -http://www.math.ucla.edu/~greg/223d.1.09f/countable.pdf -(*)Fine print disclaimer: not everybody thinks this is the "right" hierarchy to use. One alternative is to "effectivize" everything (working lightface rather than boldface: think "recursive" instead of "Borel"), although this is quite challenging and restrictive. But, aside from these, I'm not familiar with any notions with such a rich structure theory. Note that I'm not including the "I'll know a classification when I see it" standpoint (which is fine for SCOTUS, but isn't particularly amenable to formal analysis).<|endoftext|> -TITLE: When is a finite dimensional real or complex Lie Group not a matrix group -QUESTION [27 upvotes]: I have a smattering of knowledge and disconnected facts about this question, so I would like to clarify the following discussion, and I also seek references and citations supporting this knowledge. Please see my specific questions at the end, after "discussion". - -Discussion and Background -In practice, groups that do not have any faithful linear representation seem to be seldom (in the sense that I believe it was not till the late 1930s that anyone found any). -By Ado's theorem, every abstract finite dimensional Lie algebra over $\mathbb{R}, \mathbb{C}$ is the Lie algebra of some matrix Lie group. All Lie groups with a given Lie algebra are covers of one another, so even groups that are not subsets of $GL\left(V\right)$ ($V = \mathbb{R}, \mathbb{C}$) are covers of matrix groups. -I know that the metaplectic groups (double covers of the symplectic groups $Sp_{2 n}$) are not matrix groups. And I daresay it is known (although I don't know) exactly which covers of semisimple groups have faithful linear representations, thanks to the Cartan classification of all semisimple groups. But is there a know general reason (i.e. theorem showing) why particular groups lack linear representations? I believe a group must be noncompact to lack linear representations, because the connected components of all compact ones are the exponentials of the Lie algebra (actually if someone could point me to a reference to a proof of this fact, if indeed I have gotten my facts straight, I would appreciate that too). But conversely, do noncompact groups always have covers which lack faithful linear representations? Therefore, here are my specific questions: - -Specific Questions -Firm answers with citations to any of the following would be highly helpful: -1) Is there a general theorem telling one exactly when a finite dimensional Lie group lacks a faithful linear representation; -2) Alternatively, which of the (Cartan-calssified) semisimple Lie groups have covers lacking faithful linear representations; -3) Who first exhibited a Lie group without a faithful representation and when; -4) Is compactness a key factor here? Am I correct that a complex group is always the exponetial of its Lie algebra (please give a citation for this). Does a noncompact group always have a cover lacking a faithful linear representation? -Many thanks in advance. - -REPLY [3 votes]: My knowledge is piecemeal, and this is really a comment rather than an answer. I believe that the following is a good way of organizing the problem. -The problem is which connected Lie groups have faithful representations. -Let $g$ be a Lie algebra and $G$ its simply connected group. When can a given discrete normal (necessarily central) subgroup of $G$ occur as the kernel of a continuous homomorphism $G\to GL_n(\mathbb R)$ for some $n$, or equivalently as the kernel of a continuous homomorphism $G\to GL_n(\mathbb C)$ for some $n$. -Such homomorphisms correspond to Lie algebra homomorphisms $g\to gl_n(\mathbb C)$, thus to complex-linear Lie algebra homomorphisms $g_{\mathbb C}\to gl_n(\mathbb C)$ where $g_{\mathbb C}=g\otimes \mathbb C$, thus to complex-analytic homomorphisms $G_{\mathbb C}\to gl_n(\mathbb C)$ where $G_{\mathbb C}$ is the relevant simply connected complex group. -The canonical map $G\to G_{\mathbb C}$ has a discrete kernel. It can be nontrivial, for example in the case $g=sl_2(\mathbb R)$. So in order for a given discrete $D$ in the center of $G$ to be the kernel of a representation it is necessary for $D$ to contain this. -If that test is passed, then the question arises, for a simply connected complex group such as $G_{\mathbb C}$, whether a given discrete subgroup can be the kernel of a complex-analytic representation. In the semisimple case maybe the answer is always yes? I don't know. In the nilpotent case the answer is always yes for the trivial subgroup. But it's no for some nontrivial discrete central subgroups of the Heisenberg group, as we recently learned at another MO question.<|endoftext|> -TITLE: Ergodicity of Convoluted White Noise -QUESTION [5 upvotes]: I have a question regarding ergodicity in infinite dimensional spaces. -Let $\mathcal{D}$ be the space of distributions on a Schwartz space, and let $\mu$ be the white noise process which exists by the Bochner-Minlos theorem. -We can define the translation $\tau_x \phi$ of a distribution $\phi$ by: -$$ \langle \tau_x \phi, \varphi \rangle = \langle \phi, \tau_x \varphi \rangle, $$ -for all $\varphi \in C^\infty_c(\mathbb{R}^n)$, where $\tau_x \varphi$ is the usual translation of a function. The set of translations form a group acting on $\mathcal{D}$. -My first question is: -Is the group $\lbrace \tau_x : x\in \mathbb{R}^n \rbrace$ ergodic with respect to the white noise measure? -Now let $\varphi \in C^\infty_c(\mathbb{R}^n)$, then we can define the convolution $\varphi * \zeta(\omega)$, where $\zeta$ is a white-noise distributed variable. This is a $C^\infty$ random variable. My second question is: -Is $\varphi * \zeta(\omega)$ ergodic with respect to translations? -My intuition is that the first answer is true, and the second answer is "true, assuming the support is sufficiently small", but I don't even know what tools to use to tackle this problem. - -REPLY [6 votes]: The answer is yes for both situations because mixing implies ergodic. -To make things precise, let $S=S(\mathbb{R}^n)$ be the Schwartz space of -smooth real valued functions on $\mathbb{R}^n$ which decay faster than any negative power of the -Euclidean norm. Let $S'$ be the dual space of tempered distributions. -For $f\in S$ and $\phi\in S'$ denote by $\phi(f)$ the duality pairing. -Let the cylindrical $\sigma$-algebra $\mathcal{C}$ on $S'$ be -the smallest such that -$\phi\mapsto \phi(f)$ from $(S',\mathcal{C})$ to $(\mathbb{R}, {\rm Borel\ sets})$ -is measurable, for any $f\in S$. -I assume the measure $\mu$ you are talking about is the centered Gaussian probability -measure on $(S',\mathcal{C})$ with covariance -$$ -\int_{S'} \phi(f)\phi(g)\ {\rm d}\mu(\phi)= -\int_{\mathbb{R}^n} f(x) g(x) \ {\rm d}^n x\ . -$$ -Let $x\in\mathbb{R}^n$ we define translation by $x$ by: -$(\tau_x f)(y)=f(x-y)$ for test functions, $(\tau_x \phi)(f)=\phi(\tau_{-x} f)$ -for distributions, and by $(\tau_x F)(\phi)=F(\tau_{-x}\phi)$ -for complex valued $\mathcal{C}$-measurable functionals on $S'$. -To show the mixing property, you need to prove that for any such functionals -$F$, $G$ in $L^2(S',{\rm d}\mu)$ one has -$$ -\int_{S'} F(\tau_x \phi) G(\phi)\ {\rm d}\mu(\phi)\rightarrow \left(\int_{S'} F(\phi)\ {\rm d}\mu(\phi)\right) \times\left(\int_{S'} G(\phi)\ {\rm d}\mu(\phi)\right)\ \ \ (\ast) -$$ -when $x$ goes to infinity. -First note that the span $V$ of monomials $\phi(f_1)\cdots \phi(f_n)$ -is dense in $L^2(S', {\rm d}\mu)$. That's basically the Wiener-Bargmann-Ito-Segal -isomorphism with Fock space. -Also note that $||\tau_x F||_2=||F||_2$. So if you approach $F$ with $F_n$ -in $V$ the norm of the difference is the same for all translates. -So a simple three epsilon argument allows to reduce the proof of $(\ast)$ to the case -of monomials. -If you evaluate -$$ -\int_{S'} -\phi(\tau_x f_1)\cdots\phi(\tau_x f_n) -\phi(g_1)\cdots\phi(g_m) -\ {\rm d}\mu(\phi) -$$ -using the so-called Wick theorem (due in fact to Isserlis) -you will see the factors -$$ -\int_{S'} \phi(\tau_x f_i)\phi(g_j)= -\int_{\mathbb{R}^n} f_i(y-x)g_j(y)\ {\rm d}^n y -$$ -which couple the $f$'s and the $g$'s go to zero. -The same proof goes in the mollified case too. -The essential ingredient here is that the kernel -of the covariance decays fast enough. -Also remark that this generalizes to measures which are not Gaussian. -One just needs to decompose moments into cumulants aka connected correlation -functions. The key input which generalizes the single factor decay -between $f_i$ and $g_j$ above is called the clustering property. It is one of the Osterwalder-Schrader -axioms in constructive quantum field theory which physically corresponds to -the uniqueness of the vacuum.<|endoftext|> -TITLE: Computability and complexity of computing $|Hom(G,H)|$ for finitely presented groups G, H. -QUESTION [5 upvotes]: In the general case, I want to say that determining $|Hom(G,H)|$ is incomputable, arguing that you could use the number to test for simplicity of a presentation, but I am new to this area and I keep finding flaws with my argument. -While I believe the general case is incomputable, there are computable special cases. One in particular that interests me is: compute $|Hom(\pi(S),G)|$ for the fundamental group of a surface $S$, given by a triangulation, and $G$ finite. This arises in Mednykh’s Formula for a 2D TLFT invariant ( $|G|^{\chi(S)-1}|Hom(\pi(S),G)|$), which one can approximate (details in a paper to appear by Gorjan Alagic and myself) efficiently on a quantum computer. However, I have been unable to find any information on the classical complexity of finding $|Hom(\pi(S),G)|$ (with $\pi(S),G$ given in any way) to contrast with the quantum case, or even a discussion of when $|Hom(G,H)|$ is computable and what the complexity of computing it should be. -So, that leaves me with the possibly too broad: - -When is $|Hom(G,H)|$ computable for finitely presented $G,H$ and in these special cases what is the classical complexity of computing it? - -REPLY [4 votes]: This answer is really just intended to add some keywords to the discussion. -If $G=\langle x_1,\ldots,x_m\mid r_1,\ldots,r_n\rangle$ then the set $\mathrm{Hom}(G,H)$ is naturally in bijection with the set of solutions to the system of equations -$r_1(x_1,\ldots,x_m)=1$ -$\ldots$ -$r_n(x_1,\ldots,x_m)=1$ -in $H$. -For this reason, the study of $\mathrm{Hom}(G,H)$ is sometimes called 'algebraic (or Diophantine) geometry over $H$'. See the masses of recent literature on the Tarski Problem and related matters, with the key works by Sela and Kharlampovich--Miasnikov, for the case in which $H$ is free. In this case, $\mathrm{Hom}(G,H)$ is infinite if and only if it's non-trivial, which one can determine using Makanin's Algorithm.<|endoftext|> -TITLE: Roth's theorem and Behrend's lower bound -QUESTION [15 upvotes]: Roth's theorem on 3-term arithmetic progressions (3AP) is concerned with the value of $r_3(N)$, which is defined as the cardinality of the largest subset of the integers between 1 and N with no non-trivial 3AP. The best results as far as I know are that -$CN(\log\log N)^5/\log N \ge r_3(N) \ge N\exp(-D\sqrt{\log N})$ -for some constants $C,D>0$. The upper bound is by Tom Sanders in 2010 and the lower bound is by Felix Behrend dating back to 1946. My question is this: even though the upper and lower bounds are still quite a bit apart, I hear mathematicians hinting that something close to Behrend's lower bound might be giving the correct order, such as in Ben Green's paper "Roth's theorem in the primes", and I am curious as to where this is coming from. Is it because there has been no significant improvement on the lower bound (whereas there has been lot's of work on the upper bound)? Or in analogy to a similar type of problem? Or maybe just a casual remark? Or some other reason? -Thank you. - -REPLY [19 votes]: Dear Yui, -It's only slightly more than a casual remark. Our inability to find a better example is certainly a big reason for believing that Behrend's bound is correct. Julia Wolf and I slightly rehashed the proof of Behrend's bound -http://arxiv.org/abs/0810.0732 -When formulated this way, I think the construction looks both fairly natural and fairly unimprovable. -Also, there are beginning to be hints as to the correct behaviour coming from apparently similar equations such as $x_1 + x_2 + x_3 = 3x_4$. I think that Schoen and others, using work of Sanders, may have improved the bounds for this equation to $N \exp(-\log^c N)$, though I'm not certain about this. -Despite these remarks it is not known whether better bounds for Roth's theorem follow from any other more "natural" conjectures, such as the Polynomial Freiman-Ruzsa conjecture, so any suggestion that Behrend is sharp is somewhat tenuous. Some other people think differently, I believe - that it may not be sharp. - -REPLY [8 votes]: I am looking forward to answers better than mine, but as a start: -My understanding is that, yes, part of the reason is how difficult it is to improve the lower bound. After all, after Behrend it took 60+ years to get an improvement on it. And, it is (as far as I knnow) thus not at all clear where a much larger example should come from; e.g., there is no construction of a much larger set where one suspects the set has the property but one can 'just' not prove it. -In contrast, for the upper-bound the progress was more continous with a variety of improvements over the years. And, also at the moment there is an ongoing effort (with the details of which I am unfamiliar, but there is a Polymath-project, Polymath6, see here) to get further improvements. By exploiting a recent advance on a closeley related problem. -This closely related problem is a so-called finite field analogue; instead of considering the problem for integers one rather considers it for subsets of $r$-dimensional vector-spaces over the field with $3$ elements, so $\mathbb{F}_3^r$.` -And, let $r'_3(r)$ denote the analogue constant. -This constant $r'_3(r)$ is very similar, but in certain aspects easier to handle. -For example, for this constant $r'_3(r) \ll 3^r / r $ is known since well more then a decade; this corresponds to $N/ \log N$ as $3^r$ is the size of the structure. -Very recently, this was improved to -$$ r'_3(r) \ll 3^r / r^{1+\epsilon} $$.` -And, there is work done to carry-over this progress to the other situation, for an upper bound of $N / (\log N)^{1+\epsilon}$ by Katz–Bateman. -So for the upper-bound there is continous progress and further hope for progress, as there are ideas for improvement. While the lower bound somehow seems more stubborn and undpredictable. -There are various blog posts on the finite field analogue and also the actual problem asked about. -One by Tao http://terrytao.wordpress.com/2007/02/23/open-question-best-bounds-for-cap-sets/ -yet note this is four years old, and there was progress since. (It also mentions differing opinions on he finite field analogue; so there is no universal conjecture there.) -And a couple of recent and long ones by Gowers -http://gowers.wordpress.com/2011/01/11/what-is-difficult-about-the-cap-set-problem/ -http://gowers.wordpress.com/2011/01/18/more-on-the-cap-set-problem/ -http://gowers.wordpress.com/2011/02/05/polymath6-a-is-to-b-as-c-is-to/ -So, in summary, I believe that a large part is the unclearness where a much larger bound should possibly come from; while the upper bounds seems more flexible: they are very hard to actually improve, yet there seems more hope and ideas what could work. -However, as there seems to be no clear consensus on the finite field analogue, it might be the case that the opinions are actually not as fixed on the problem at hand either. -And, to end with a purely sociological reason: if a very convincing argument was known, it should be well-known. So none might be known. -I hope to be wrong on the last point, and learn one from this question.<|endoftext|> -TITLE: Indecomposable vector spaces and the axiom of choice -QUESTION [11 upvotes]: It is a known result by A. Blass that the axiom of choice is equivalent to the assertion that every vector space has a basis. (Rubin's Equivalents of the Axiom of Choice: form B) -It is also known that the axiom of choice equivalent to the existence of a field $F$ such that for every vector space $V$ over $F$ has the property that if $S\subseteq V$ is a subspace of $V$ then there is $S'\subseteq V$ such that $S\oplus S'=V$. -In this case, assuming the axiom of choice does not hold, we have a vector space without a basis somewhere in the model, and over every field there is a vector space $V$ that has a subspace without a complement. -My question: Can we derive from the absence of choice that there will be a vector space (over some field? any field?) such that for any nontrivial subspace $S$ of $V$ there is no subspace $S'$ of $V$ such that $S\oplus S'=V$? -My initial instinct was to try and take a vector space without a basis and start decomposing it, somehow deriving if we stop at a successor stage we have found such subspace, and if we happened to hit a limit stage with an empty subspace then we can construct a basis. I think I overlooked something because I couldn't finish my argument. -Edit: Just to be clear, I am not looking whether or not this is consistent with ZF, but rather if the negation of choice implies there exists such vector space. - -REPLY [4 votes]: This is not meant to be a complete answer to the question (which I find very interesting), but rather an explanation for why the particular approach you described is bound to run into difficulties. I'm new here, and I'm not sure this belongs as an "answer;" please let me know if this is inappropriate. -It is consistent with ZF that there is a vector space (over the field of two elements) which does not have a basis and yet every nontrivial subspace admits a decomposition into two complementary subspaces. In fact, one such example is quite familiar. -Set $V = \mathcal{P}(\omega)/\mathrm{FIN}$, viewed as a vector space over $\mathbb{Z}/2\mathbb{Z}$ (with, of course, symmetric difference serving as the addition operation). A fairly straightforward Baire category argument shows that $V$ does not have a basis in a model of ZFDC + all sets of reals have the Baire property. However, any subspace $W \subseteq V$ with more than two elements can be decomposed as $S_0 \oplus S_1$: simply fix some element $A \in W$ which does not almost contain every element of $W$ and set -$S_0 = \{B \in W : B \mbox{ is almost contained in A}\}$ and -$S_1 = \{B \in W : B \mbox{ is almost disjoint from A}\}$. -I'm sure the same argument actually works for $\mathcal{P}(\omega)$ itself, but I hate thinking about finite sets. -Edit: OK, now I'm worried about offending finite sets. I don't hate thinking about you in general, but I'd rather not think about you in the particular Baire category argument I have in mind.<|endoftext|> -TITLE: Limits in functor categories -QUESTION [12 upvotes]: Let $C$ have all limits of diagrams indexed by $J$, and for each $b\in B$ let $E_b:C^B \rightarrow C$ the evaluation functor that evaluates at $b$. -Given a diagram $F: J \rightarrow C^B$, -we can calculate its limit by doing it pointwise, i.e. since for each $b\in B$, -$E_b \circ F : J \rightarrow C$ has a limit $v_b$ we may construct a functor in $V: B \rightarrow C$ that's a limit of $F$ by letting $V(b) = v_b$ on objects $b\in B$ and blah blah for arrows. -My question is: if $F$ has a limit, does each $E_b\circ F$ have a limit? -This is similar to the problem of mono/epimorphisms, answered in -Can epi/mono for natural transformations be checked pointwise? -So basically if some property is pointwise true in the original category $C$ then it's true in any functor category $C^B$, but which properties are true the other way around? - -REPLY [14 votes]: What you're asking is whether every limit in a functor category $[B,C]$ is a pointwise limit. The answer is yes if C is complete, but not always otherwise. Kelly gives an example in Basic Concepts of Enriched Category Theory, section 3.3, of a limit in a functor category that is not a pointwise limit. -I don't know of any striking examples of always-pointwise properties. One obvious example is that of being an absolute limit or colimit (i.e. one preserved by every functor). The latter are characterized by the fact that a category is Cauchy complete if and only if it has all absolute colimits.<|endoftext|> -TITLE: How much larger is the powerset of a transfinite set? -QUESTION [8 upvotes]: This may seem a vague question, but in the process of explaining the hierarchy of transfinite sets to my students, I've had to use the Cardinality(powerset(S)) > Cardinality(S) argument, which appeared to me to be extremely weak in indicating how much larger the powerset is. Is there a better argument? -In contrast, Cantor's diagonalization argument shows that the set of reals is very much larger than the set of natural numbers -- the argument shows that there is a vast number of reals unaccounted for in any attempted bijection between the naturals and the reals. -But then, when we get to the hierarchy of the alephs, I depend upon the argument that the powerset(S) is larger than S. Perhaps it's the version of that argument that I use, but all I'm able to show is that there is a single element of the powerset(S) unaccounted for in any attempted bijection between the powerset(S) and S. While this shows that the powerset is larger, it doesn't give any indication that it must be very, very much larger (and we are talking about infinite sets here). This is a hard way to impress my students about the vastness of each larger transfinite set. -Is there a better argument for showing that the powerset of a transfinite set is vastly larger than the set? I'm, in particular, looking for an argument that I can explain to talented high school math students. - -Added later: -Thanks for all the comments so far. As many have pointed out, I've been a little too vague, so let me be more precise. -The argument I present to students that the set of reals is (vastly) larger than the set of naturals is exactly the one that Jason mentions below in the first sentence of his second paragraph. Namely, in Cantor's diagonalization argument, one simply chooses a different digit in the kth position of the kth real in the supposed ordering of the reals. This will ultimately provide a large number of reals not in the proposed bijection between naturals and reals. -In contrast, the argument that I've used to show my students that the powerset(S) is larger than S is based on the Russell Paradox. Suppose there were a one-to-one correspondence between elements of S and subsets of S. Let Subset A be those elements of S which are each matched to a subset that contains it. Let Subset B be the rest of the elements of S, namely each element which has been matched to a subset not containing that element. Since there is a supposed one-to-one correspondence, subset B should be matched to some element of S, but this results in a contradiction. Therefore there is at least one element in the powerset of S (this subset B) which doesn't have a match. So the powerset(S) is larger than S... by at least one element. -So that is the root of my complaint: that the argument I'm using to show that the reals are larger than the naturals demonstrates a vast number of reals that are not covered by any trial bijection. But the argument I'm using to show that the powerset(S) > S shows only one element larger. -So I'm either misunderstanding the conclusion of the powerset argument I'm using (it really does show that the powerset is vastly larger), or I need a better elementary argument than that... - -REPLY [3 votes]: This is not an answer per se, but it sums up many of the points made in earlier answers. -If the question is: "How many more elements does $\mathcal P(S)$ have than $S$?" then the answer is of course: $|\mathcal P(S)|$ ! -(Why? Because $\mathcal P(S)$ has $|\mathcal P(S)\smallsetminus S)|$ more elements than $S$, and we know that $|\mathcal P(S)\smallsetminus S)|=|\mathcal P(S)|$.) This may seem like a really trivial remark, but the point is this: if you only understand sets of size $|S|$, then $\mathcal P(S)$ has so many elements that you don't understand how many more it has. So sentences like "one more", "two more", ... , "$\aleph_0$ more", and even "$|S|$ more" can't even begin to describe it.<|endoftext|> -TITLE: Automorphism groups and etale topological stacks -QUESTION [6 upvotes]: Recall that an etale topological stack is a stack $\mathscr{X}$ over the category of topological spaces (and open covers) which admits a representable local homeomorphism $X \to \mathscr{X}$ from a topological space. Equivalently, it is a topological stack arising from an etale topological groupoid. It is well known that a differentiable stack is etale if and only if all of its automorphism groups are discrete, but the proof involves foliation theory. It seems this proof cannot be extended to the topological setting. However, clearly every etale topological stack has discrete isotropy groups. This begs the question: -If a topological stack has all of its isotropy groups discrete, is it necessarily etale? -EDIT: By a topological stack, I mean a stack $\mathscr{X}$ over the category of topological spaces (and open covers) which admits a representable epimorphism $X \to \mathscr{X}$ (not necessarily a local homeomorphism). This is equivalent to saying $\mathscr{X}$ is the stack of torsors for a topological groupoid. -Remark: This question is equivalent to asking if a topological groupoid all of whose isotropy groups are discrete must be Morita equivalent to an etale topological groupoid. - -REPLY [5 votes]: Here's a counterexample: the stack associated to the relative pair groupoid of the map -$$ -([0,1]\times\{0\}) -\cup -(\{1\}\times[0,1]) -\cup -([1,2]\times\{1\})\;\;\to\; [0,2] -\qquad\qquad\qquad\qquad\qquad -$$ -$$\qquad\qquad\qquad(x,y)\;\;\mapsto\;\;\; x$$ -Equivalently, this stack can be described as the pushout in the 2-category of stacks of the diagram $[0,1]\leftarrow \{1\} \rightarrow [1,2]$ (where we identify a space with the stack it represents).<|endoftext|> -TITLE: What's the difference between a Riemann theta and a Siegel theta function? -QUESTION [10 upvotes]: One of the things I'm working on has required me to look into the literature of multidimensional theta functions, and I've gotten a bit confused on a few naming details. -A look at the DLMF says that "the" multidimensional theta function is the Riemann theta function, -$$\mathop{\Theta}\left(\mathbf{z}\mid\boldsymbol{{\Omega}}\right)=\sum_{{\mathbf{n}\in{\mathbb Z}^g}}e^{{2\pi i\left(\frac{1}{2}\mathbf{n}\cdot\boldsymbol{{\Omega}}\cdot\mathbf{n}+\mathbf{n}\cdot\mathbf{z}\right)}}$$ -with $\mathbf{z}$ a complex vector and $\mathbf \Omega$ a complex symmetric matrix with symmetric positive definite imaginary part. I find that there is a nice implementation for numerically computing this function, but it is only available in Maple and Java. Since I work in Mathematica, I decided to see if there was already something in Mathematica before going through the trouble of translating code. -Peering at the docs for Mathematica netted the Siegel theta function. However, looking at how the Siegel theta function was defined, this and the Riemann theta function seem to be the same thing! -Not having Maple to check if the results from their implementation of Riemann theta and the results from Mathematica's implementation of Siegel theta agree, and not being able to access the (older) references to these functions, I now wish to ask: is there no difference whatsoever between these two except in name (and if there are, how are these two different)? Did Riemann and Siegel independently study the exact same multidimensional theta function? How come it seems that there is no reference to these two being synonymous? - -REPLY [10 votes]: It's the same function. I checked with Maple 15: -$r := RiemannTheta([0.5+I, 2 I], Matrix(2, 2, [[I, 0], [0, I]]), [])$ -$evalf(r)$ -$-6.586149971*10^6-2.132900065*10^{-8} I$ -Mathematica 8 gives -$N[SiegelTheta[\{\{I, 0\}, \{0, I\}\}, \{0.5 + I, 2 I\}]]$ -$-6.58615*10^6 - 8.06571*10^{-10} I$ -as Deconinck et al. write: "There are as many different conventions for writing the Riemann theta function as there are names for it."<|endoftext|> -TITLE: Towards a metric characterization of Euclidean spaces -QUESTION [16 upvotes]: I want to obtain a metric characterization of the classical finite dimensional spaces of Euclidean geometry. -Motivation: Suppose $A$ and $B$ live in an $n$-dimensional Euclidean space. They are each assigned the task of constructing an equilateral triangle of side length 5. Subject $A$ finds first one point $p_1$. If $n>0$, he can then find another point $p_2$ at distance 5 from $p_1$. If $n>1$, he can then find $p_3$ at distance 5 from $p_1$ and $p_2$. Then $B$ proceeds similarly. He selects a point $q_1$ (probably different from $p_1$), then finds another point $q_2$ and $q_3$. We wouldn't expect him to get stuck before constructing $q_3$, since $n$ is the same for both subjects. I will say a space is all-equal if the possibility of completing a figure is independent of the starting points chosen. Are Euclidean spaces all-equal? Are there non Euclidean all-equal spaces? -Precise statement -Definition: A metric space $X$ is all-equal if every time $S$ and $T$ are isometric subspaces of $X$ (with a selected isometry $S\to T$), the isometry extends to an automorphism of $X$. -Question 0: Is every Euclidean space all-equal? -To state the second question we must first observe some limitations. Being connected our life intervals, we cannot visit more than one connected component of our space, and being imprecise our measurements, we cannot distinguish directly between a space and its completion. Finally, existing no natural unit of measurement, the correct category to state these questions is that of spaces in which the distance is defined up to scale. That is, it takes values in a 1-dimensional module $M$ over $[0,+\infty)$, with no multiplication structure (although lengths can be tensorised to obtain areas). -More definitions: A congruence space is a pair $(X,M,d)$ where $X$ is a set, $M$ is a 1-dimensional $[0,+\infty)$ module and d is a distance in $X$ taking values in $M$. The morphisms from $(X,M,d)$ to $(Y,N,e)$ will be the subsimilarities, that is, the pairs $(f,\alpha)$ where $f$ is a function from $X$ to $Y$ and $\alpha$ is a morphism from $M$ to $N$ such that for each $x$, $x'$ in $X$ we have $e(fx,fx')\leq\alpha(d(x,x'))$. Similarities are similarly defined using an $=$ sign instead of $\leq$. Subspaces are defined by restriction, and the identity of $(X,M,d)$ is the obvious $(id_X,id_M)$. It's easy to prove that every isomorphism is a similarity. A congruence space $X$ is all-equal if every time $S$ and $T$ are similar subspaces, the similarity extends to an automorphism of $X$. -Question 1: Is every connected complete all-equal space a finite dimensional real inner-product space? -Remarks: if the connectedness hypothesis is dropped, some discrete spaces would be all-equal. If we work in the usual category of metric spaces, projective and hyperbolic spaces could be all-equal, and also Euclidean spheres, both with their inner metric and with the subspace metric. -Partial results and lines of thought: -EDIT: As noted by Sergei, there are counterexamples. Generally, if $(X,d)$ is an all-equal metric space, and $p>1$, then $(X,d^{\frac 1p})$ could be an all-equal space. This is similar to the example in which However, I was thinking of spaces in which the distance from $x$ to $y$ is measured using a signal that travels form $x$ to $y$ at constant speed. Hence I require that the space have the following property: -If $x$ is a point and $a$ and $b$ are lengths, then $N_b(N_a(\{x\}))=N_{a+b}({x})$ (where the $N$ stands for open neighborhood of a set). -Related MO questions: -Characterizations of Euclidean space -Easy proof of the fact that isotropic spaces are Euclidean -Characterization of Riemannian metrics - -REPLY [9 votes]: I can complete Anton's plan with an additional assumption that geodesics do not branch. I also assume local compactness (otherwise there are too many technical details to deal with). More precisely, I prove the following: -Let $X$ be a geodesic space. Suppose that - -For every finite subset $Q\subset X$, every similarity $Q\to X$ can be extended to a bijective similarity $X\to X$. A similarity is a map that multiplies all distances by a constant. (It is easy to see that, in the case of a length metric, the more complicated definition from the question reduces to this.) -$X$ is locally compact. -For every two distinct points of $X$ there is a unique line containing them. (A line is a subset isometric to $\mathbb R$. The existence of lines follows from the two other assumptions.) - -Then $X$ is isometric to $\mathbb R^n$ for some $n$. -Proof. First observe that the group of similarities acts transitively on pairs (oriented line, point not on this line). Indeed, given two such pairs $(\ell_1,p_1)$ and $(\ell_2,p_2)$, it suffices to find similar triples $p_1,x_1,y_1$ and $p_2,x_2,y_2$ where $x_i,y_i$ is a positively oriented pair of points on $\ell_i$. And it is easy to see that such triples $p_i,x_i,y_i$ realise all similarity types of, e.g. isosceles triangles. -Next we construct perpendiculars. Lines $\alpha$ and $\beta$ intersecting at a point $q$ are said to be perpendicular if there exists an isometry that fixes $\alpha$ and maps $\beta$ to itself by reflection in $q$. It is easy to see that this relation is symmetric. Further, for every line $\ell$ and every point $p\notin\ell$ there exists a unique perpendicular to $\ell$ containing $p$. To prove existence, pick two points $x,y\in\ell$ such that $|px|=|py|$. There is an isometry that fixes $p$ and exchanges $x$ and $y$. It acts on $\ell$ by reflection in the midpoint $q$ of $[xy]$, hence the line $(pq)$ is perpendicular to $\ell$. Uniqueness follows from the fact that two distinct reflections of $\ell$ cannot fix $p$, otherwise their composition shifts $\ell$ along itself and fixes $p$, and some iteration of this shift would break the triangle inequality. -This argument also shows that the base $q$ of the perpendicular is the nearest point to $p$ on $\ell$, and the distance from $p$ to $x\in\ell$ grows monotonically with $|qx|$. -Clearly all right-angled triangles with given leg lengths $a$ and $b$ are isometric; denote their hypotenuse by $f(a,b)$. Then $f$ is strictly monotone in each argument and positively homogeneous: $f(ta,tb)=tf(a,b)$. -Next, we show that the sum of Busemann function of two opposite rays is zero. Or, equivalently, if $\gamma$ is an arc-length parametrized line, $q=\gamma(0)$ and a line $(pq)$ is perpendicular to $\gamma$, then $B_\gamma(p)=0$ where $B_\gamma$ denotes the Busemann function of $\gamma$. Suppose the contrary. We may assume that $|pq|=1$ and $B_\gamma(p)=c>0$. This means that for every point $x\in\gamma$ we have $|px|-|qx|\ge c$. Let $s\in\ell$ be very far away and let $p_1$ be the perpendicular from $q$ to $(ps)$. The above inequality implies that $|pp_1|\ge c$, hence we have an upper bound $|qp_1|\le\lambda<1$ where $\lambda$ is determined by $f$ and $c$ and does not depend on $s$. Let $q_1$ be the base of the perpendicular from $p_1$ to $(qs)$, then rescaling the above inequality yields that $|p_1q_1|\le\lambda|qp_1|\le\lambda^2$. The next perpendicular (from $q_1$ to $p_2$ on $(ps)$) has length at most $\lambda^3$ and so on. Summing up these perpendiculars, we see that $|ps|\le 1/(1-\lambda)$ which is not that far away, a contradiction. -The fact that opposite rays yield opposite Busemann functions implies that the Busemann function of a line $\gamma$ is the only 1-Lipschitz function $f$ such that $f(\gamma(t))=-t$ for all $t$. And, given local compactness and non-branching of geodesics, this implies that $X$ is split into lines parallel to $\ell$, where a line $\gamma_1$ is said to be parallel to $\gamma$ if the Busemann function of $\ell$ decays with unit rate along $\gamma_1$ (to construct a parallel line, just glue together two opposite asymptotic rays). Further, Busemann functions of parallel lines coincide, due to their above mentioned uniqueness. -Now it is easy to see that the level set of a Busemann function (which is also a union of perpendiculars to a given line at a given point) has the same isometry extension property and is a geodesic space. Then we can carry induction in the parameter $d$ defined as the maximum number of pairwise perpendicular lines that can go through one point. The cases $d=1$ and $d=2$ can be done by hand, then use isometries exchanging perpendicular lines to do the induction step.<|endoftext|> -TITLE: Characterization of Tychonoff spaces in terms of open sets -QUESTION [6 upvotes]: Metrizability and complete regularity are topological properties that are, in a sense, different from the Hausdorff condition because they are not defined purely in the terms of the open sets, but rather using some external object, namely $\mathbb R$. Now a space is metrizable if it has the weak topology induced by a function $d : X \times X \rightarrow \mathbb R$ satisfying the properties of a metric, but metrization theorems tell us that an equivalent condition is that $X$ is regular and has a sigma-discrete basis (and this is purely topological). Is there a similiar characterization of Tychonoff spaces that makes no reference to $\mathbb R$ whatsoever? - -REPLY [3 votes]: Suppose that $(X,\mathcal{T})$ is a topological space. Then define a relation $\prec$ on $\mathcal{T}$ by letting $U\prec V$ if and only if $\overline{U}\subseteq V$. Define a relation $\ll$ on $\mathcal{T}$ by letting $U\ll V$ if and only if there is some $(U_{r})_{r\in[0,1]\cap\mathbb{Q}}$ such that $U_{0}=U,U_{1}=V$ and $\overline{U_{r}}\subseteq U_{s}$ whenever $r -TITLE: For quasi-coherent D-Modules -QUESTION [7 upvotes]: It is well-know that the category of coherent D-modules over a smooth algebraic $k$-scheme is a Tannakian category. So it is equivalent to the category of finite representations of some affine group scheme G/k. My question is do we have the similar statement for quasi-coherent D-modules? I hope that the category of quasi-coherent D-modules is equivalent to the representation (not necessarily finite) category of some affine group schemes. Is that true, is there any reference for that? -I hope that any quasi-coherent D-module is the union of its coherent sub D-modules. If the answer to the above question is true then this is true. If the above is wrong. I still believe this is true. At least I think it is true for char$k$=0, where a quasi-coherent D-module is a quasi-coherent sheaf with a flat connection. If this is true, could you give me any reference? - -REPLY [7 votes]: The proper Tannakian theory that has a chance to encompass categories of coherent or quasicoherent D-modules deals with tensor categories WITHOUT a faithful fiber functor - for example the categories of (quasi)coherent sheaves on quasicompact varieties, or more generally geoemtric stacks (see Lurie's article or this MO question). (Once you have D-modules that are not local systems, measuring them at a single point can't be faithful any more..) -Of course to talk about quasicoherent sheaves we need to drop rigidity since -they are not dualizable objects, but that's not a serious problem, since quasicoherent sheaves (eg D-modules) are unions of coherent ones (eg coherent D-modules, which does NOT mean coherent as O-modules). -In any case in this context one can hope to recover not an affine group scheme but rather the underlying variety or stack again (the usual Tannakian story is the case of the classifying stack pt/G of an affine group scheme G). Here one builds a stack of all fiber functors (not necessarily faithful), and you can hope to reconstruct the original variety/stack by this procedure (the content of Lurie's theorem). -So how might this apply to D-modules? D-modules on X (coherent or quasicoherent) are the same as O-modules (coherent or quasicoherent) on the de Rham space (see e.g. this MO question). So it's reasonable to ask whether -Tannakian reconstruction rebuilds the de Rham space of X from the tensor category of D-modules on X. I believe the answer is yes - certainly there's a natural map -from the de Rham space of X to the Tannakian functor of D-modules (for every point -in X you get a fiber functor and infinitesimally nearby points give isomorphic fiber functors). I would imagine this is an equivalence but haven't thought it through.<|endoftext|> -TITLE: Questions about $\aleph_1-$closed forcing notions -QUESTION [15 upvotes]: "Foreman`s maximality principle" states that every non-trivial forcing notion either adds a real or collapses some cardinals. This principle has many consequences including: -1) $GCH$ fails everywhere, -2) there are no inaccessible cardinals, -3) there are no $\kappa-$Souslin trees, -4) Any non-trivial $c.c.c.$ forcing adds a real, -5) Any non-trivial $\kappa^+-$closed forcing notion collapses some cardinals. -Consistency of (1) is proved by Foreman-Woodin, (2) clearly can be consistent and the consistency of (4) is shown in "Forcing with c.c.c forcing notions, Cohen reals and Random reals". -My interest is in the consistency of (5). Let's consider the case $\kappa=\omega.$ -Question 1. Is it consistent that any non-trivial $\aleph_1-$closed forcing notion collapses some cardinals? -The above question seems very difficult, and it is not difficult to show that for its consistency we need some very large cardinals. But maybe the following is simpler: -Question 2. Is it consistent that any non-trivial $\aleph_1-$closed forcing notion of size continuum collapses some cardinals? Does its consistency imply the existence of large cardinals? - -REPLY [3 votes]: The following results are obtained in a joint work with Yair Hayut and are now presented in our joint paper On Foreman's maximality principle: - -Theorem 1. (Assuming the existence of a strong cardinal) There is a model of $ZFC$ in which for all uncountable (regular) cardinals $\kappa,$ any $\kappa$-closed forcing notion of size $\leq 2^{<\kappa}$ collapses some cardinals. - -Note that in such a model $GCH$ should fail everywhere and hence some very large cardinals are needed. The following theorem is in the opposite direction: - -Theorem 2 (Assuming the existence of a strong cardinal and infinitely many inaccessibles above it) There is a model of $ZFC$ in which $GCH$ fails everywhere and for each uncountable (regular) cardinal $\kappa,$ there exists a $\kappa$-closed forcing notion of size $2^{<\kappa}$ which preserves all cardinals.<|endoftext|> -TITLE: Dual Borel conjecture in Laver's model -QUESTION [18 upvotes]: A set $X\subseteq 2^\omega$ -of reals is of strong measure zero (smz) if $X+M\not=2^\omega$ -for every meager set $M$. (This is a theorem of Galvin-Mycielski-Solovay, -but for the question I am going to ask we may as well take it as a definition.) -A set $Y$ is strongly meager (sm) if $Y+N\not=2^\omega$ for every Lebesgue null -set $N$. -The Borel conjecture (BC) says that every smz set is countable; the dual Borel conjecture (dBC) says that every sm set is countable. -In Laver's model (obtained by a countable support iteration of Laver reals -of length $\aleph_2$) -the BC holds. Same for the Mathias model. -In a paper that I (with Kellner+Shelah+Wohofsky) just sent to arxiv.org, we claim that it is not clear if Laver's model satisfies the dBC. -QUESTION: - Is that correct? Or is it perhaps known that Laver's model has uncountable sm - sets? -Additional remark 1: Bartoszynski and Shelah (MR 2020043) proved in 2003 -that in Laver's model there are no sm sets of size continuum ($\aleph_2$). -(The MR review states that the paper proves that the sm sets are exactly -$[\mathbb R] ^{\le \aleph_0}$. This is obviously a typo in the review.) -Additional remark 2: If many random reals are added to Laver's model (either -during the iteration, or afterwards), then BC still holds, but there will be sm -sets of size continuum, so dBC fails in a strong sense. - -REPLY [5 votes]: In analogy with the fact that Lusin set are strongly null, Pawlikowski has shown that Sierpinksi sets are strongly meagre, so one might ask if there are -are Sierpinski sets in the iterated Laver Model. The answer is no; indeed - any set of positive outer measure contains an uncountable null set after adding a Laver real -Let $\psi:\omega\to\omega$ be a function growing so quickly that -$$\lim_{n\to\infty} \prod_{j=n}^\infty \left(1- 2^{-\psi(j)} \right) =1 $$ -Then let $L:\omega\to \omega$ be the generic real added by $\mathbb L$ and define -$A$ to be the set of those $f\in 2^\omega $ such that -$f$ is constantly $0$ on the interval $[L(n), L(n) + \psi(n)] $ for all but finitely many $n$. - $A$ is Lebesgue null and it will be shown that $|A\cap X|=\aleph_1$. -Suppose not and let $T\in \mathbb L$ and $C\in [X]^{\aleph_0}$ be such that -$T$ forces that $X\cap A\subseteq C$. For $s\in T$ extending the root of $T$ -define $S_s $ to be the set of all $f\in 2^\omega$ for which there exists infinitely many $ n\in \omega$ such that $ s^\frown n\in T$ and $ f(n+j) = 0 $ for all $j\leq \psi(|s|) $. Then each $S_s$ is of measure one. Since $X$ has positive outer measure there is some $x\in \bigcap_{s\supseteq \text{root}(T)}S_s\cap(X\setminus C)$. It follows that there is then -$\tilde{T}\subseteq T$ such that $\tilde{T}\in \mathbb L$ and -$x$ is constantly $0$ on the interval $[m,m+\psi(|s|)]$ -for all $s^\frown m\supseteq \text{root}(\tilde{T})$. -In other words, $\tilde{T}$ forces that $x$ is in $A$ contradicting that $x\in X\setminus C$.<|endoftext|> -TITLE: Gluing along closed subschemes -QUESTION [9 upvotes]: Let $Z \to X$ be a closed immersion of schemes. Is it true that for every morphism $Z \to Y$, the pushout $X \cup_Z Y$ in the category of schemes exists? If yes, a) does it turn out to be simply sthe pushout in the category of locally ringed spaces, b) is the natural morphism $Y \to X \cup_Z Y$ a closed immersion? -In his paper "Gluing Schemes and a Scheme Without Closed Points", Karl Schwede studies such questions. In particular, he gives an affirmative answer if everything is affine (Theorem 3.4), but also for arbitrary schemes if we also assume that $Z \to Y$ is a closed immersion (Corollary 3.9). My intuition says that it should be also true if we drop this condition, but on the other hand the paper shows with some examples that our intuition may be wrong in the context of pushouts. I'm aware that colimits of schemes are not well-behaved in general, but in my research it would be useful to construct pushouts also when just one inclusion is a closed immersion. -As with this MO question, I think it is not sufficient just to say that some pushout does not exist just because you don't see one. I'm interested in rigorous proofs. - -REPLY [2 votes]: To complement Laurent Moret-Bailly and Karl Schwede's answers (10 years ago!), the pushout of a closed immersion $Z\to X$ along an affine morphism $Z\to Y$ always exists in the category of algebraic spaces. This is a special case of Theorem 1.8 of https://arxiv.org/abs/2205.08623.<|endoftext|> -TITLE: Complex analytic space with no (positive dim.) subscheme ? -QUESTION [7 upvotes]: Is there an example of a complex analytic space $X$ that doesn't have any (not necessarily open or closed) positive dimensional subspace $Y$ which is analytically isomorphic to (the complex analytic space associated to) a scheme? - -Edit: after D.Arapura's comment, we require $X$ to have dimension $>1$. -If I remember correctly, there are non "Abelian" complex tori $X=\mathbb{C}^n/\mathrm{Lattice}\;\;$ that do not have any positive dimensional analytic subvariety. Can a counterexample be derived from this? -Also, any complex algebraic space has an open subspace which is a scheme. So the counterexample (if it exists) must be searched outside algebraic spaces. - -What if the question is modified by requiring that $X$ has no $Y$ that is locally closed in the analytic Zariski topology (where opens are complements of analytic subspaces)? - -REPLY [2 votes]: Take a complex 2-torus $X$ without curves, and hence, without non-constant meromorphic functions (see e.g. Shafarevich, Basic algebraic geometry, chapter VIII, \S 1, example 2). The only locally closed subsets of $X$ will be $X$ itself, $\varnothing$, finite subsets and the complements of finite subsets. $X$ minus a finite subset can't be algebraic: if it were, it would have a non-constant meromorphic function, and then so would $X$. -[upd: locally closed here means locally closed in the analytic Zariski topology, so this answers the second question and not the first.]<|endoftext|> -TITLE: Short five lemma in Banach spaces -QUESTION [5 upvotes]: Denote by $\mathbf{Ban}$ the category of Banach spaces and bounded linear maps and by $\mathbf{Banc}$ the subcategory of Banach spaces and linear contractions. The isomorphisms of $\mathbf{Ban}$ are the bounded linear bijections (open mapping theorem) and in $\mathbf{Banc}$ they are the isometric (linear) isomorphisms (easy computation). $\mathbf{Ban}$ is additive and has all finite limits and finite colimits, so we can speak of (short) exact sequences. -proposition: the short five lemma holds in $\mathbf{Ban}$. -proof: courtesy of the open mapping theorem, the diagram-chase proof for abelian groups transfers verbatim. -Now for my question: -Q: does the five short lemma also hold in $\mathbf{Banc}$? -Since $\mathbf{Banc}$ is not additive, a few words about the definitions. The crucial thing to notice is that kernels and cokernels in $\mathbf{Ban}$ are also kernels and cokernels in $\mathbf{Banc}$. For the sake of illustration, take the case of cokernels. In $\mathbf{Ban}$ they are given by the quotient map $\pi:B\to B/M$ with $M$ a closed subspace of $B$. If $T$ is a bounded linear map whose kernel contains $M$, then it factors uniquely through $\pi$ and the norm of the factorization equals the norm of $T$. But this is precisely what is needed for $\pi$ to be a cokernel in $\mathbf{Banc}$. -So now, we have two short exact sequences connected by maps $g$, $f$ and $h$ with $g$ and $h$ isometric isomorphisms and $f$ a linear contraction. The short five lemma holds if $f$ is an isometric isomorphism -- is there a way to draw diagrams? In any case, just look at short five lemma. -Start by noticing that by the above proposition $f$ must be a bijection and thus it has an inverse. What we need to prove is that this inverse is contractive. Alternatively, since we already know that $f$ is injective, the following easy result offers another route. -proposition: a map $f$ is a quotient iff it is surjective on open unit balls iff it is dense on unit balls. -Applying the above and chasing down an element of the open unit ball of the codomain of $f:B\to B'$, we get is that there is a $b$ in the open unit ball of $B$ and an $m$ in the kernel of the quotient in the top short exact sequence such that $f(b + m) = b'$. But of course, this does not get us very far. -At this point, I am beginning to suspect that the lemma does not hold, but I have been unable to rig a counterexample. The hypothesis is very strong and rules out the "usual" gang of suspects. -regards, -G. Rodrigues - -REPLY [10 votes]: How about considering $\mathbb R^2$ with $l^1$ and $l^2$ norms. The restriction to the $x$-axis is an isometry. The quotient onto the $y$-axis, also an isometry. But not an isometry on the whole space.<|endoftext|> -TITLE: Numerical Differentiation. What is the best method? -QUESTION [10 upvotes]: What is the best method for 1D numeric differentiation? Something as glorious as Gaussian quadrature for numeric integration. -Maybe differential quadrature is such a method? What is its accuracy? -I'm well aware that it is really easy to have symbolic differentiation in the program (automatic differentiation or truly symbolic algorithm). However to use such methods it is necessary to rewrite all functions to be differentiated. Thus one can't differentiate functions imported from libraries. -I need differentiation almost with the machine precision. - -REPLY [3 votes]: Complex step differentiation (CSD) is well known as an efficient numerical differentiation method: -$$f^\prime(x)=\mathrm{Im}\frac{f(x+\mathrm{i}h)}{h}+O(h^2),\quad\mathrm{i}:=\sqrt{-1}.$$ -This method requires the function to be analytic (differentiable as a complex function). -We cannot apply this method to higher order derivatives, but since this method does not use differences it is useful to avoid subtractive cancellation (loss of significance caused when we try to subtract similar numbers). For the derivation and implementations, check the article on SIAM News and the post on Mathworks Blog. -A noteworthy alternative to CSD is the approach by dual numbers. Dual numbers are defined by: -$$x=a+b\epsilon,\quad a,b\in\mathbb{R},\quad\epsilon^2=0.$$ -By using Taylor expansion and dual numbers, we obtain -$$f(a+b\epsilon)=f(a)+bf^\prime(a)\epsilon\quad(\because\epsilon^2=0).$$ -Hence we can obtain the derivative by dual numbers.<|endoftext|> -TITLE: Looking for an appealing counterexample in probability -QUESTION [17 upvotes]: There is a commonly-encountered-but-wrong rule of thumb that says something like - -If a probability distribution is positively skewed, its mean is greater than its median. - -(You sometimes also see it phrased in terms of mean and mode). I'm looking for a nice counterexample to this rule. What do I mean by "nice"? Hard to say exactly, of course, but something like: - -It must be continuously differentiable. -It must have only one maximum and at most two points of inflection. -Its graph must be "obviously" positively skewed and the mean must be "obviously" less than the median. - -I'm aware of the Weibull distribution, which can satisfy (1) and (2) - but for me the graph appears too close to being unskewed and the mean appears too close to the median for it to be an intuitively obvious counterexample. So I'm looking for something where the properties are more easily visible. -Alternatively: if such a distribution did exist it seems to me that it would be quite well-known. So it seems plausible that the rule of thumb might be almost true, in the sense that there might exist a true statement of the form - -If a probability distribution is positively skewed, its mean can be at most __ less than its median. - -I'm also very interested to see answers of this form. - -REPLY [12 votes]: You are looking for random variables $X$ of median $m(X)$ such that -$$ -E((X-E(X))^3)>0\quad\mbox{and}\quad E(X)< m(X). -$$ -Assume there exists two nonnegative random variables $Y$ and $Z$ and an independent Bernoulli sign $B=\pm1$ with $E(B)=0$ such that $X=Y$ on $[B=1]$ and $X=-Z$ on $[B=-1]$. -Then $m(X)=0$ and $E(X)=\frac12(E(Y)-E(Z))$. Furthermore, assuming $E(Y)< E(Z)$, one can prove that -$$ -E(Y^3)>E(Z^3)\quad\implies\quad E((X-E(X))^3)>0. -$$ -To prove this implication, expand $(X-E(X))^3$ using $Y$ and $Z$ to see that $E((X-E(X))^3)>0$ if and only if -$$ -2(E(Y^3)-E(Z^3))+2(E(Z)^3-E(Y)^3)+3(E(Z)-E(Y))(\mbox{var}(Y)+\mbox{var}(Z))>0. -$$ -If $E(Z)> E(Y)$, the second and the third terms are positive, hence if the first term is positive as well, the whole LHS is positive. Thus the implication holds. -Finally, every couple $(Y,Z)$ such that -$$ -E(Y)< E(Z),\qquad E(Y^3)>E(Z^3) -$$ -is a solution. Note that every $X$ such that $m(X)=0$ can be written as above. -First example If $Y$ is distributed like $y$ times a reduced exponential and $Z$ is distributed like $z\sqrt{\pi/2}$ times the absolute value of a reduced normal, then $E(Y)=y$, $E(Y^3)=6y^3$, $E(Z)=z$ and $E(Z^3)=2z^3$, hence every couple of parameters $(y,z)$ such that $y< z< \sqrt[3]{3}y$ yields a solution. -Second example If $Y$ is distributed like $y$ times a reduced exponential and $Z$ is distributed like $z$ times the sum of two independent reduced exponentials, then $E(Y)=y$, $E(Y^3)=6y^3$, $E(Z)=2z$ and $E(Z^3)=24z^3$, hence every couple of parameters $(y,z)$ such that $\sqrt[3]{4}z< y< 2z$ yields a solution.<|endoftext|> -TITLE: metaplectic group does not split -QUESTION [13 upvotes]: I'm trying to understand the Weil representation and hope there are some experts around who can set me straight. Let $F$ be a non-Archimedean local field (I don't mind assuming that the characteristic of $F$ is not $2$ if it simplifies things; also the situation over $\mathbb{R}$ is similar but obviously I need a proof which does not rely on any covering space theory.) and fix a nontrivial continuous additive character $\psi : F \to \mathbb{C}^{\times}$. Then by the Stone-von Neumann theorem there is a unique up to isomorphism irreducible smooth representation of the Heisenberg group with central character $\psi$, and one sees the existence of a projective representation of the symplectic group in this space by the uniqueness part of that theorem. -So my question is: why does this not lift to an ordinary representation? All the articles I have read refer this claim to Weil's original paper, which is quite long, and my French is not so good. I think one can see this from the fact that the metaplectic group (I'm talking about the two-sheeted cover) is not a trivial extension of the symplectic group by using the fact that the latter group is perfect. So if one constructs the metaplectic group via Maslov cocycles, one needs to show a certain cocycle is not a coboundary. But how? -Thanks for the help. I hope my question is clear enough. - -REPLY [2 votes]: I would like the credit to go to Peter Woit for suggesting Section I.6 of Stephen Kudla's "Notes on the Local Theta Correspondence," which contains a nice proof, but he posted this only as a comment and the bounty ends today. -The text of Woit's comment is - -There's a relatively straight-forward argument in Section I.6 of Stephen Kudla's "Notes on the Local Theta Correspondence", available at www.math.toronto.edu/~skudla/castle.pdf This may just be a restatement of the Rao argument. As mentioned elsewhere, it comes down to invoking non-triviality of the Hilbert symbol<|endoftext|> -TITLE: On the Existence of Certain Fourier Series -QUESTION [5 upvotes]: Is there an $f\in L^{1}(T)$ whose partial sums of Fourier series $S_{n}(f)$ satisfies $\|S_{n}(f)\|_{L^{1}(T)} \rightarrow \|f\|_{L^{1}(T)}$ but $S_{n}(f)$ fails to converge to $f$ in $L^1$-norm ? - -REPLY [7 votes]: I was hesitating for a while whether to answer or to vote to close and to refer the OP to AoPS, but, since the question has been upvoted, here goes. -Suppose that $f_k\in L^1$ converges to $f\in L^1$ in the sense of distributions and in measure. Suppose also that $\|f_k\|_1\to \|f\|_1$. Then $f_k\to f$ in $L^1$. -Indeed, let $g$ be a bounded by $1$ infinitely smooth function such that $\int fg>\|f\|_1-\delta$. Then $\int f_kg > \|f\|_1-2\delta$ for large $k$. Now, $\int_{\{|f_k-f|<\delta\}} f_kg\ge \int_{\{|f_k-f|<\delta\}} fg-\delta\ge \int fg-2\delta\ge \|f\|_1-3\delta$ for large $k$ because the integral of a fixed $L^1$ function $fg$ over a set of small measure is small. -So, $\int|f_k|\ge \int_{\{|f_k-f|\ge\delta\}} |f_k|+\|f\|_1-3\delta$ whence the first integral is at most $4\delta$ for large $k$. Thus -$$ -\int |f_k-k|\le \int_{\{|f_k-f|\ge \delta\}} (|f_k|+|f|)+\delta\le 6\delta -$$ -for large $k$. -To apply this to $f_k=S_kf$, one only needs to check the convergence in measure. But it immediately follows from the weak type 1-1 bound for $S_k$ (applied to the difference of $f$ and a trigonometric polynomial approximating $f$ in $L^1$, of course).<|endoftext|> -TITLE: PNT for number fields. -QUESTION [5 upvotes]: Hi there, -Here is a part of the book of Murty "an introduction to Sieve methods and applications" -page 36 -"At this point we invoke some algebraic number theory let $K=Q(\theta)$ where $\theta$ is the solution of the polynomial $f(x)$. The ring of integers $O_{k}$ of K is a Dedekind Domain. it's a classical theorem of Dedekind that for all but finitely many primes $\delta_{f}(p)$ is the number of prime ideals $p$ of $O_{k}$ such that the norm $N_{K/Q}(p)=p$" -Note: $\delta_{f}(p)$:= number of solutions of $f(x)$ modulo $p$ -Question: What can be said about the size of the set of primes breaking this rule? -In my opinion, might they be those dividing $disc(f)$? -Yildo - -REPLY [3 votes]: This statement is true for all $p$ not dividing $disc(f)$, as you say. Moreover you can write down exactly what the primes dividing $p$ are: they're the ideals $(p, g(\theta))$ for each irreducible factor $g$ of $f$. The norm of such a prime is the degree of $g$, so the number of degree 1 primes is the number of roots of $f$ mod $p$.<|endoftext|> -TITLE: Natural transformations as categorical homotopies -QUESTION [59 upvotes]: Every text book I've ever read about Category Theory gives the definition of natural transformation as a collection of morphisms which make the well known diagrams commute. -There is another possible definition of natural transformation, which appears to be a categorification of homotopy: - -given two functors $\mathcal F,\mathcal G \colon \mathcal C \to \mathcal D$ a natural transformation is a functor $\varphi \colon \mathcal C \times 2 \to \mathcal D$, where $2$ is the arrow category $0 \to 1$, such that $\varphi(-,0)=\mathcal F$ and $\varphi(-,1)=\mathcal G$. - -My question is: - -why doesn't anybody use this definition of natural transformation which seems to be more "natural" (at least for me)? - -(Edit:) It seems that many people use this definition of natural transformation. This arises the following question: - -Is there any introductory textbook (or lecture) on category theory that introduces natural transformation in this "homotopical" way rather then the classical one? - -(Edit2:) Some days ago I've read a post in nlab about $k$-transfor. In particular I have been interested by the discussion in the said post, because it seems to prove that the homotopical definition of natural transformation should be the right one (or at least a slight modification of it). On the other end this definition have always seemed to be the most natural one, because historically category theory develop in the context of algebraic topology, so now I've a new question: - -Does anyone know the logical process that took Mac Lane and Eilenberg to give their (classical) definition of natural transformation? -Here I'm interested in the topological/algebraic motivation that move those great mathematicians to such definition rather the other one. - -REPLY [4 votes]: Following the previous indication of Professor Brown I want to add another possible way to see natural transformation which is a generalization of the previous definition. - -Given categories $\mathcal C$ and $\mathcal D$ and two functors between them $\mathcal F,\mathcal G \colon \mathcal C \to \mathcal D$ then a natural transformation $\tau$ can be defined as a functor $\tau \colon \mathcal C \to (\mathcal F \downarrow \mathcal G)$ which arrow components are the diagonal functions, sending each arrow $f \in \mathcal C(c,c')$, with $c,c' \in \mathcal C$ to $(f,f) \in (\mathcal F \downarrow \mathcal G)(\tau(c),\tau(c'))$. - -Edit: I think the definition of natural transformation proposed by professor Brown probably can be even a more natural than the one proposed in the question. -I think that more details are worthed. -The key ingredient for that definition is the concept of arrow category of a given category $\mathbf D$: such category have morphism of $\mathbf D$ as objects and commutative square as morphisms. -This category come equipped with two functors $\mathbf {source}, \mathbf{target} \colon \text{Arr}(\mathbf D) \to \mathbf D$ such that for each object (i.e. a morphisms of $\mathbf D$) $f \colon d \to d'$ we have -$$\mathbf{source}(f)=d$$ -$$\mathbf{target}(f)=d'$$ -while for each $f \in \mathbf D(x,x')$, $g \in \mathbf D(y,y')$ and a morphism $\alpha \in \text{Arr}(\mathbf D)(f,g)$ (i.e. a quadruple $\langle f,g, \alpha_0,\alpha_1\rangle$ where $\alpha_0 \in \mathbf D(x,y)$ and $\alpha_1 \in \mathbf D(x',y')$ such that $\alpha_1 \circ f = g \circ \alpha_0$) we have -$$\mathbf{source}(\alpha)=\alpha_0$$ -$$\mathbf{target}(\alpha)=\alpha_1$$ -it's easy to prove that these data give two functors (which gives to $\text{Arr}(\mathbf D)$ the structure of a graph internal to $\mathbf{Cat}$). -Now let's take a look to this new definition of natural transformation: - -A natural transformation $\tau$ between two functors $F,G \colon \mathbf C \to \mathbf D$ is a functor $\tau \colon \mathbf C \to \text{Arr}(\mathbf D)$ such that $\mathbf{source} \circ \tau = F$ and $\mathbf{target}\circ \tau = G$. - -A functor of this kind associate to every object $c \in \mathbf C$ a morphism $\tau_c \colon F(c) \to G(c)$ in $\mathbf D$, while to every $f \in \mathbf C(c,c')$ it gives the commutative triangle expressing the equality -$$\tau_{c'} \circ F(f)=\tau_{c'} \circ \mathbf {source}(\tau_f)=\mathbf {target}(\tau_f) \circ \tau_c = G(f) \circ \tau_c$$ -certifying the naturality (in the ordinary sense) of the $\tau_c$. -This definition reminds the notion of homotopy between maps $f,g \colon X \to Y$ as map of kind $X \to Y^I$ (i.e. an homotopy as a (continuous) family of path of $Y$). -That's not all, indeed we can reiterate the construction of the arrow category obtaining what I think is called a cubical set -$$\mathbf D \leftarrow \text{Arr}(\mathbf D) \leftarrow \text{Arr}^2(\mathbf D)\leftarrow \dots $$ -where each arrow should be thought as the pair of functors $\mathbf{source}_{n+1},\mathbf{target}_{n+1} \colon \text{Arr}^{n+1}(\mathbf D) \to \text{Arr}^n (\mathbf D)$. -In this way we can associate to each category a cubical set. There's also a natural way to associate to every functor a (degree 0) mapping of cubical sets. -If we consider natural transformation as maps from a category to an arrow category then this correspondence associate to each natural transformation a degree 1 map between such cubical sets (by degree one I mean that the induced map send every object of $\text{Arr}^n(\mathbf C)$ in an object of $\text{Arr}^{n+1}(\mathbf D)$). -I've found really beautiful this construction because it shows an analogy between categories-functors-natural transformation and complexes-map of complexes-complexes homotopies.<|endoftext|> -TITLE: Simplest examples of rings that are not isomorphic to their opposites -QUESTION [36 upvotes]: What are the simplest examples of -rings that are not isomorphic to their -opposite rings? Is there a science to constructing them? - -The only simple example known to me: -In Jacobson's Basic Algebra (vol. 1), Section 2.8, there is an exercise that goes as follows: -Let $u=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1\\ 0 & 0 & 0 \end{pmatrix}\in M_3(\mathbf Q)$ and let $x=\begin{pmatrix} u & 0 \\ 0 & u^2 \end{pmatrix}$, -$y=\begin{pmatrix}0&1\\0&0\end{pmatrix}$, where $u$ is as indicated and $0$ and $1$ are zero and unit matrices in $M_3(\mathbf Q)$. Hence $x,y\in M_6(\mathbf Q)$. Jacobson gives hints to prove that the subring of $M_6(\mathbf Q)$ generated by $x$ and $y$ is not isomorphic to its opposite. -Examples seem to be well-known to the operator algebras crowd: -See for example the paper: "A Simple Separable C*-Algebra Not Isomorphic to Its Opposite Algebra" by N. Christopher Phillips, Proceedings of the American Mathematical Society -Vol. 132, No. 10 (Oct., 2004), pp. 2997-3005. - -REPLY [2 votes]: Many examples are already given; here is another one, just for its own interest: -Let $V$ be a vector space of infinite countable dimension over a countable field $K$. -Let $E$ be the $K$-algebra of endomorphisms of $V$. I claim that $E$ is not isomorphic to its opposite (even as a ring, i.e., as $\mathbf{Z}$-algebra). Precisely: - -(1) for every $g\in E-\{0\}$, $gE=\{gf:f\in E\}$ is uncountable [of continuum cardinal] -(2) there exists $f\in E-\{0\}$ such that $Ef=\{gf:g\in E\}$ is countable: [namely this holds iff $f$ has finite rank (otherwise it has continuum cardinal)] - -Let me justify the non-bracketed assertions. In (2) this holds because if $B$ is a finite subset of $E$ such that $f(B)$ spans $f(E)$, then every element of $Ef$ is determined by its restriction to $B$. -In (1), just fix a line $L$ not in the kernel of $g$, and let $f$ range over the space $Y$ linear maps $V\to L$. Since the dual of $V$ has uncountable dimension, $Y$ has uncountable [continuum] dimension. And $f\mapsto gf$ is injective in restriction to $Y$. -Maybe in this case $E$ and $E^{\mathrm{op}}$ are not elementary equivalent, but this would require another argument.<|endoftext|> -TITLE: Integrating the multinomial over a hypercube -QUESTION [6 upvotes]: I have come across an integral of the form -$$\int_{b}^{a}\cdots\int_{b}^{a} \left( \sum_{i=1}^{n}x_i\right)^mdx_1d x_2\dots dx_n.$$ -I have a solution that makes use of the partition function, but I feel there should be a much nicer solution and I'm sure this has been looked at before. Does anybody know a reference? -Motivation: -This integral has appeared whilst trying to compute moments of the Voronoi cell of the lattice $A_n$ (see page 462 of Sphere Packings, Lattices and Groups) - -REPLY [11 votes]: We want the coefficient of $t^m/m!$ in - $$ \int_b^a\cdots \int_b^a e^{(x_1+\cdots+x_n)t}dx_1\dots dx_n = - \left(\frac{e^{at}-e^{bt}}{t}\right)^n. $$ -Expanding by the binomial theorem and then taking the coefficient of $t^m/m!$ from each term will give a formula.<|endoftext|> -TITLE: Characterization of infinite paths in graphs -QUESTION [6 upvotes]: First an introduction. -A directed graph we all know what is, and a graph is serial whenever -every vertex has a successor. I do not consider the empty graph. A -pair $(\mathcal{G},s)$ is called a rooted graph when $s \in -\mathcal{G}$ and $\mathcal{G}$ is a directed serial graph. -Given a rooted graph $(\mathcal{G},s)$, a $(\mathcal{G},s)$-path is a -function $\lambda: \mathbb{N} \rightarrow V(\mathcal{G})$ such that -$\lambda(0) = s$ and for all $i \in \mathbb{N}$, -$(\lambda(i),\lambda(i+1)) \in E(\mathcal{G})$. Given any rooted graph -$(\mathcal{G},s)$, we define the set $N(\mathcal{G},s)$ as follows: -$$N(\mathcal{G},s) = \{ X \subseteq V(\mathcal{G}) \mid \text{ exists -a $(\mathcal{G},s)$-path $\lambda$ s.t. for all $i \in \mathbb{N}$, } -\lambda(i) \in X \}$$ -My question is, for what sets $N$ does there exist a rooted directed -graph $(\mathcal{G},s)$ such that $N = N(\mathcal{G},s)$? -I am looking for a way of describing (possibly infinite) directed -serial graphs by giving a set of sets of vertices, each of which -corresponds to an infinite path starting in a specific start vertex -$s$. I call these sets neighbourhood sets (a slightly unfortunately -name in graphs, I agree, but it comes from modal logic and its -neighbourhood semantics, which I am applying this to). -I would like to make some restrictions on a family of sets so that I -can say when such a family of sets do has a corresponding graph, -i.e. given a set of sets $N$, which properties must $N$ have in order -to have a rooted directed graph $(\mathcal{G},s)$ such that $N = -N(\mathcal{G},s)$. -Define the non-monotonic core of $N$ as follows: -$$N^{nc} = \{ X \in N \mid \not \exists Y \in N \text{ with } Y \subset X \}$$ -I have come up with a few trivial properties that are all necessary, -but they are not sufficient, not even when restricted to finite -graphs. The properties are as follows: - -Safety: The universe itself (i.e. the vertex set $V$) has to be contained in $N$ -Reflexivity: There is an element $s$ such that when $X \in N$, we have that $s \in X$ -Upwards closed: If $X \in N$ and $X \subseteq Y \subseteq V$, $Y \in N$ -Countable case: If $X \in N^{nc}$, then $ |X| \leq \omega $ -If $X \in N^{nc}$ and $Y \in N^{nc}$ and $|X \cap Y| = |X \setminus Y| = |Y \setminus X| = \omega$, then there has to be at least another (or infinitely many) $Y \neq Z \in N^{nc}$ with $|X \cap Z| = |X \setminus Z| = |Z \setminus X| = \omega$ - -Now, it is trivial to check that the $N(\mathcal{G},s)$ satisfies the -above properties for any directed graph $\mathcal{G}$, but they are -not sufficient. Consider the following set: -$$N = \{ \{s,1,2,3\}, \{s,1,2,4\}, \{s,1,3,4\} ,\{s,2,3,4\}, -\{s,1,2,3,4\} \}$$ -I have proven (in a very brute force manner) that the above cannot -have a corresponding graph, so I will not give the proof here. -I am looking for the missing properties, and work that has been done -in this area. I apologize in advance if I have not given enough -explanation, I have been stuck in this problem too long now to see -this clearly. If I should explain more, or give examples, please let -me know, and I will. -Edit: Added some cases I didn't want to explain, but realized later I should put in anyway, but they both deals with infinite graphs, and I'm first and foremost after managing the subproblem that is finite graphs. - -REPLY [2 votes]: (Not an answer, but too long for a comment) -One way to look at what you're asking for is a theory in some appropriate language which axiomatizes that we have a serial directed rooted graph, and a collection of subsets of the graph which behaves like $N(\mathcal{G},s)$, where the part of the axiomatization that dictates the behaviour of $N(\mathcal{G},s)$ doesn't say anything about the graph relation on $\mathcal{G}$. I'll give an axiomatization where the part that talks about $N(\mathcal{G},s)$ does mention the graph relation, and I'll set it up in such a way that it'll seem unlikely that it can be redone without mentioning the graph relation. I'm not sure about this, it's just an idea, but it's too long for a comment. -The Language -Consider the language $(\bar{V}, \bar{N}, \bar{E}, \bar{s}, \bar{\epsilon})$ - two unary relation symbols, one binary relation symbol, one constant symbol, and another binary relation symbol, respectively. I'm going to set up a theory whose finite models will be precisely (sort of) those in which $(\bar{V}, \bar{E})$ is interpreted as a directed serial graph $\mathcal{G}$, $\bar{s}$ gets interpreted as a member $s$ of $\bar{V}$, $\bar{N}$ gets interpreted as $N(\mathcal{G},s)$, and $\bar{\epsilon}$ gets interpreted as the membership relation, a subset of $\bar{V} \times \bar{N}$. Setting up the right theory is easy, the only part that will require a little explanation is how we ensure $\bar{N}$ gets interpreted correctly. -Axiomatizing $N(\mathcal{G},s)$ when we're allowed to mention the edge relation -Consider a formula like: -$$x_1 = \bar{s} \wedge x_1 \neq x_2 \wedge x_1 \neq x_3 \wedge x_2 \neq x_3$$ -$$\wedge \bar{E}(x_1,x_2) \wedge \bar{E}(x_2,x_1) \wedge \neg \bar{E}(x_1,x_3) \wedge \neg \bar{E}(x_3,x_1) \wedge \neg \bar{E}(x_2,x_3) \wedge \neg \bar{E}(x_3,x_2)$$ -It "says" we have three distinct $x_i$, the first one is $s$, and it tells you exactly which pairs stand in edge relation to one another and which don't. You can also tell by looking at it that it defines a set $\{ x_1, x_2, x_3\}$ which contains an infinite path starting at $s$, namely $x_1, x_2, x_1, x_2, \dots$. -Let's define $\mathcal{R}$ to be the set of formulas in the language $(\bar{s}, \bar{E})$ of the form $\phi(x_1, \dots , x_n)$ which say that the $x_i$ are distinct, and which tell you precisely which pairs of the $x_i$ stand in $\bar{E}$ relation to one another, and which don't. Define $\mathcal{R}^+$ to be those formulas $\phi \in \mathcal{R}$ such that for any rooted directed graph $((V,E),s)$, we have that: -$(V,s,E)\vDash \phi(v_1, \dots , v_n) \rightarrow \{v_1, \dots , v_n\} \in N((V,E),s)$. -$\mathcal{R}^-$ will be those $\phi$ such that: -$(V,s,E)\vDash \phi(v_1, \dots , v_n) \rightarrow \{v_1, \dots , v_n\} \not\in N((V,E),s)$. -Alright, now here's our theory: - -(Rooted serial directed graph) $\bar{V}(\bar{s}) \wedge \forall x \bar{V}(x) \rightarrow [\neg \bar{E}(x,x) \wedge \exists y (\bar{V}(y) \wedge \bar{E}(x,y))]$ -(Schema for what goes in $N$) As $\phi(x_1, \dots ,x_n)$ varies over $\mathcal{R}^+$: $\forall \vec{x} \exists Y \forall x \left (\phi(\vec{x}) \rightarrow \left[ \bar{N}(Y) \wedge \bigwedge _i (x_i \bar{\epsilon} Y) \wedge \left (x \bar{\epsilon} Y \rightarrow \bigvee _i (x = x_i)\right )\right ]\right )$ -(Schema for what stays out of $N$) As $\phi(x_1, \dots ,x_n)$ varies over $\mathcal{R}^-$: $\forall \vec{x} \forall Y \exists x \left (\phi(\vec{x}) \rightarrow \neg \left[ \bar{N}(Y) \wedge \bigwedge _i (x_i \bar{\epsilon} Y) \wedge \left (x \bar{\epsilon} Y \rightarrow \bigvee _i (x = x_i)\right )\right ]\right )$ - -Checking this axiomatization works -It's clear that if $\mathcal{G} = (V,E)$ is a finite directed serial graph and $s \in V$, then $(V,N(\mathcal{G},s),E,s,\in)$ is a model of this theory. Conversely, if we have a finite model $(V,N_0,E,s,\epsilon)$ of this theory, then $(V,E)$ forms a directed serial graph with $s \in V$. Now let $N_1$ consist of those $Y \in N_0$ such that $\forall x (x \epsilon Y \rightarrow V(x))$. I claim that: -$N((V,E),s) = \{ \{x : x \epsilon Y\} : Y \in N_1\}$ -But I won't prove this. -A remark about some "unnaturalness" -It should seem like I could have done things differently so that things looked more natural and the above claim could be proved more easily. The way I've written 2 and 3 are not the most natural, but I've done it for a reason. 2 says that if $\vec{x}$ is a tuple which is sure to contain an infinite path starting at $s$, then there's a member of $N$ consisting precisely of the members of $\vec{x}$. 3 says that if $\vec{x}$ is sure to not contain an infinite path starting at $x$, then there's nothing in $N$ consisting precisely of the members of $\vec{x}$. The formulas in 2 and 3 are in prenex normal form, where the matrix is a conditional where the left side only involves the symbols $\bar{E}$ and $\bar{s}$, and the right side only $\bar{N}$ and $\bar{\epsilon}$. Moreover, the antecedents have variables $\vec{x}$ and the consequents have variables $\vec{x}, Y, x$. -The point -You want a theory equivalent to axiom 1 and schemas 2 and 3 above, but you want to replace 2 and 3 with (probably finitely many) axioms which don't mention the symbol $\bar{E}$. I feel like there ought to be some interpolation-type theorem (along the lines of Craig Interpolation or Lyndon Interpolation) which says this can't happen, i.e. something which says that a theory in which each axiom mention either only $\bar{V}, \bar{E}, \bar{s}$ or only $\bar{V}, \bar{N}, \bar{s}, \bar{\epsilon}$ can't prove a theory which has axioms like those in schemas 2 or 3.<|endoftext|> -TITLE: Dynamics of a random "quadratic" directed graph -QUESTION [7 upvotes]: Let G be a directed graph on N vertices chosen at random, conditional on the requirement that the out-degree of each vertex is 1 and the in-degree of each vertex is either 0 or 2. The "periodic" points of G are those contained in a cycle. What do we know about the statistics of G? For instance, what is the mean number of periodic points, and how do the cycle lengths look? -By comparison, a random directed graph with all out-degrees 1 (which is to say, the graph of a random function from vertices to vertices) has on order of sqrt(N) periodic points on average. -(Motivation: the graph of a quadratic rational function f acting on P^1(F_q) looks like this, and I'm wondering what the "expected" dynamics are.) - -REPLY [3 votes]: Let $G$ be a directed graph on $N$ vertices such that the out-degree of each vertex is 1 and the in-degree is either 0 or $n$. Letting $N=nt$, there are $t$ vertices with in-degree $n$ and $(n-1)t$ vertices with in-degree 0. Assuming the vertices are labeled, the number of such graphs is -$$ -\binom{nt}{t}\frac{(nt)!}{(n!)^t}. -$$ -For $1\leq m\leq t$, the number of such graphs with a fixed $m$-cycle is -$$ -\binom{nt-m}{t-m}\frac{(nt-m)!}{((n-1)!)^m(n!)^{t-m}}, -$$ -so the number of $m$-periodic points amongst all such graphs is -$$ -m\cdot\binom{nt-m}{t-m}\frac{(nt-m)!}{((n-1)!)^m(n!)^{t-m}}\cdot\frac{(nt)!}{m(nt-m)!}=\binom{nt-m}{t-m}\frac{(nt)!}{((n-1)!)^m(n!)^{t-m}}. -$$ -Summing over $1\leq m\leq t$, the average we want is -$$ -\binom{nt}{t}^{-1}\sum_{m=1}^{t}{n^m\binom{nt-m}{t-m}}=-1+\binom{nt}{t}^{-1}\sum_{m=0}^{t}{n^m\binom{nt-m}{t-m}}. -$$ -When $n=1$, the average is $t$ (since all points would be periodic). When $n=2$, as in the question, the sum is -$$ --1+\frac{4^t}{\binom{nt}{t}}\sim-1+\sqrt{\frac{\pi}{2}}\sqrt{N}, -$$ -so the number of periodic points does indeed look random. When $n>2$, I haven't found the identity I need yet.<|endoftext|> -TITLE: Blocking visibility with cylinders -QUESTION [16 upvotes]: Suppose you have a supply of infinite-length, opaque, unit-radius cylinders, -and you would like to block all visibility from a point -$p \in \mathbb{R}^3$ to infinity with as few cylinders as possible. -(The cylinders are infinite length in both directions.) -The cylinders may touch but not interpenetrate, and they -should be disjoint from $p$, -leaving a small ball around $p$ empty. -(Another variation would insist that cylinders be pairwise disjoint, -i.e., not touching one another.) -A collection of parallel cylinders arranged to form a "fence" around -$p$ do not suffice, leaving two line-of-sight $\pm$ rays to infinity. -Perhaps a grid of cylinders in the pattern illustrated left below -suffice, but at least if there are not many cylinders, there is -a view from an interior point to infinity (right below). - -I feel like I am missing a simple construction that would -obviously block all rays from $p$. -Perhaps crossing the cylinders like the poles of a tipi (teepee) -could help, but it seems this would at best lead to inefficient -blockage. -Suggestions welcome—Thanks! -Addendum1. -Perhaps if the weaving above is rendered irregular by displacing the cylinders slightly by different -amounts, so that cracks do not align, then a sufficient portion of the weaving will block all visibility. -Here (left below) is the start of Gerhard's first suggested construction (a portion of the weaving above), which I don't see how to complete. But perhaps -seeing this depiction will aid intuition. - - - - -Addendum2. -To the right above I added (three-quarters of) a forest along the lines (but not exactly as) -Yaakov suggested. - -REPLY [5 votes]: Here is one construction. On the horizontal xy plane place a forest of vertical cylinders of radius r<1/2 (or =1/2 if we allow contacts) centered at each point in $(\mathbb{Z} \backslash{\lbrace0\rbrace})\times(\mathbb{Z}\backslash{\lbrace0\rbrace})$; moreover place 2 similar cylinders parallel to x centered at (y=+/-1, z=0), 2 more parallel to y centered at (x=0, z=+/-1) and the last 2 parallel to z and centered at (x=+/-1, y=0). Then (0,0,0) is blocked by the forest in all directions except those in the xz and yz planes, which are blocked by the other 6 cylinders. The forest clearly does not need to be infinite and it should be easy to find an upper limit on its size. -${\bf UPDATE}$ As pointed out by Mark in a comment, the forest should be based on $(\mathbb{Z} \backslash{\lbrace0\rbrace})\times(\mathbb{Z}\backslash{\lbrace-1,0,1\rbrace})$.<|endoftext|> -TITLE: Does completion commute with localization? -QUESTION [8 upvotes]: Suppose $A$ is a Noetherian (not necessarily local) ring and $\mathfrak{m}\subset A$ a maximal ideal. Then is it true that $$\hat{A}_{\hat{\mathfrak{m}}}=\widehat{A _{\mathfrak{m}}},$$ where hats denote completion and subscripts denote localization? If one uses superscripts to denote completion it would be -$$(A^{\mathfrak{m}})_{\mathfrak{m^{\mathfrak{m}}}}=(A _{\mathfrak{m}})^{\mathfrak{m} _{\mathfrak{m}}}.$$ - -REPLY [19 votes]: It is true. $(\widehat{A}, \widehat{m})$ is a Noetherian local ring so your left hand side could be simplified replacing it by $\widehat{A}$. Now let's use the definitions: $\widehat{A} = \varprojlim A/\mathfrak{m}^n$, whereas $\widehat{A_{\mathfrak{m}}} = \varprojlim A_{m}/(\mathfrak{m}A_{\mathfrak{m}})^n$. The desired equality is the result of localization being exact so that $A_{\mathfrak{m}}/(\mathfrak{m}A_{\mathfrak{m}})^n = (A/ \mathfrak{m}^n)_{\mathfrak{m}}$ and the fact that in $A/\mathfrak{m^n}$ everything outside the maximal ideal is already invertible, so that $(A/\mathfrak{m^n})_{\mathfrak{m}} = A/\mathfrak{m}^n$.<|endoftext|> -TITLE: When is this map completely positive? -QUESTION [6 upvotes]: Consider the complex $n$-by-$n$ matrices $M_n$. -Suppose that $A_i$, for $i=1,\ldots,n^2$, satisfy $\mathrm{Tr}(A_i^* -A_j)=\delta_{ij}$, so that together they form an orthonormal basis for -$M_n$. Define a linear map $T \colon M_n \to M_n \otimes M_n$ by $T(A_i) = A_i \otimes A_i$. - -Question: when is $T$ completely positive? - -For example, if $A_i$ are the matrices with a single entry one and the rest zeroes in some fixed basis of $\mathbb{C}^n$, then $T$ is completely positive. In fact, I think these might be the only examples. If $T$ is completely positive, then the following are equivalent to $A_i$ being matrix units as in the above example: - -each $A_i$ has rank one; -each positive semidefinite $A_i$ has trace one; -the set $\{0,A_1,\ldots,A_{n^2}\}$ is closed under multiplication; -$T(1)$ is idempotent; -$T^*(1) \leq 1$; -$T$ preserves trace. - -These are sufficient conditions, but proving they are sufficient doesn't use $\mathrm{Tr}(A_i^* A_j)=\delta_{ij}$ at all. Are they necessary? - -REPLY [2 votes]: I think I finally have the answer to this question (some 7 years later, and 3 years after first hearing of the problem myself, but never mind). TL;DR: the matrices with a single entry one and the rest zeroes in some fixed basis of $\mathbb{C}^n$ are indeed the only examples. A detailed proof (with diagrams!) will appear shortly on the arXiv, and I will link to it once it's ready, but I will summarise the gist here for posterity. -A linear map $\delta:M_n \rightarrow M_n \otimes M_n$ defined by $\delta(A_{ij}) := A_{ij} \otimes A_{ij}$ for some orthonormal basis $(A_{ij})_{i,j=1,...,n}$ of $M_n$ is necessarily a special commutative $\dagger$-Frobenius algebra in the category $\operatorname{fHilb}$ of finite-dimensional Hilbert spaces and complex linear maps, with co-unit $\epsilon: M_n \rightarrow \mathbb{C}$ defined by setting $\epsilon(A_{ij}) := 1$ (together with their respective adjoints $\delta^\dagger$ and $\epsilon^\dagger$). If $\delta$ is a CP map, then so is $\epsilon$, and $(\delta,\epsilon,\delta^\dagger,\epsilon^\dagger)$ is also a special commutative $\dagger$-Frobenius algebra in the category $\operatorname{CPM}(\operatorname{fHilb})$ of finite-dimensional Hilbert spaces and completely positive maps. -The point is that every isometric comonoid in $\operatorname{CPM}(\operatorname{fHilb})$, such as $(\delta,\epsilon)$ would be when $\delta$ is CP, must necessarily involve pure maps $\delta,\epsilon$. This is because of the purity principle: whenever for pure CP maps $\Psi_i$ and $F$ we have $\sum_i \Psi_i = F$, necessarily $\Psi_i = q_i F$ for all $i$ and some coefficients $q_i \in \mathbb{R}^+$ (not necessarily all non-zero). The broad strokes of the proof are as follows. - -The isometry condition $\delta^\dagger \circ \delta = id$ implies that $\delta$ must be $\mathbb{R}^+$-linear combination (not necessarily convex) $\delta = \sum_i q_i V_i$ of isometries $V_i$ with pairwise orthogonal ranges (i.e. $V_j^\dagger \circ V_i$ is the identity $id$ for $i=j$ and the zero map for $i\neq j$). -The unit laws $(id \otimes \epsilon)\circ \delta = id = (\epsilon \otimes id)\circ \delta $ imply that $\epsilon$ must be pure. -By post-composing the associativity law with $\epsilon$ in all three possible ways, an associativity law for the isometries can be derived. -By post-composing the associativity law for isometries with $\epsilon$, all isometries $V_i$ are identified, and orthogonality of ranges proves that $\delta$ must in fact be pure. -Because $\delta$ and $\epsilon$ are pure, i.e. they arise from the comultiplication and counit of a special commutative $\dagger$-Frobenius algebra of $\operatorname{fHilb}$. Those are exactly the examples given by taking matrices $A_{ij} := |i\rangle\langle j|$ with a single entry one and the rest zero in some fixed orthonormal basis $\big(|i\rangle\big)_{i=1,...,n}$ of $\mathbb{C}^n$.<|endoftext|> -TITLE: Explicit formula for the trace of an unramified principal series representation of $GL(n,K)$, $K$ $p$-adic. -QUESTION [18 upvotes]: Let $K$ be a non-arch local field (I'm only interested in the char 0 case), let $\mathbb{G}$ be a connected reductive group over $K$ and let $G=\mathbb{G}(K)$. If $V$ is a smooth irreducible complex representation of $G$ and if $H$ is the Hecke algebra of locally constant complex-valued functions on $G$ with compact support (fix a Haar measure on $G$ to make $H$ an algebra under convolution), then $V$ is naturally an $H$-module, and every $h\in H$ acts on $V$ via a finite rank operator and hence has a trace. -It is my understanding that in this connected reductive situation, a theorem of Harish-Chandra says that this trace function $t:H\to\mathbb{C}$ can actually be expressed as -$$t(h)=\int_G tr(g)h(g) dg$$ -for $tr:G\to\mathbb{C}$ an $L^1$ function, called the trace of $V$. -If $V$ is finite-dimensional then $tr$ is the usual trace. However I realised earlier this week that I do not know one single explicit example of this function if $V$ is infinite-dimensional. I just spent 20 minutes trying to fathom out what I guessed was probably the simplest non-trivial example: if $G=GL(2,K)$ and $V$ is, say, an unramified principal series representation. I failed :-( I could compute the trace of $h$ for various explicit $h$ (typically supported in $GL(2,R)$, $R$ the integers of $K$) but this didn't seem to get me any closer to an actual formula: in particular, although I could figure out $t$ on various functions I couldn't figure out $tr$ on any elements of $G$. On the other hand I imagine that this sort of stuff is completely standard, if you know where to look. -If $\mu_1$ and $\mu_2$ are unramified characters of $K^\times$ and $V$ is the associated principal series representation of $GL(2,K)$, then what is $tr(g)$ for $g$, say, a diagonal matrix? Or $g$ a unipotent matrix? -[EDIT: Alexander Braverman points out that I have over-stated Harish-Chandra's result: $t$ is only locally $L^1$. Furthermore one has to be a little careful---more careful than I was at least---because $t$ is only defined via some integrals so one could change it on a set of measure zero---hence in some sense asking to evaluate $t$ at an explicit point makes no sense. However he, in his answer, shows how to make sense of my question anyway, as well as answering it.] -[EDIT: Loren Spice points out that my paranthetical char 0 comment is actually an assumption in Harish-Chandra's result, and that apparently local integrability is still open in char $p$. I didn't make a very good job of stating H-C's theorem at all!] - -REPLY [2 votes]: I am not sure, if you are still interested in this, but here is the general computation: -Let $\phi \in C_c^\infty (GL_n(F))$, and let $\pi$ be a super-cuspidal representation of a Levi subgroup $M$ of a parabolic $P$ with unipotent radical $N$, and let $\pi_0 = Ind_{P}^{GL_n(F)} \pi$ be the normalized induced representation (assume unitary, irreducible for safety). Let $K$ be compact open subgroup with $GL_n(F) = P K$. - -Define $ \phi^K (x) = \int\limits_K \phi(k^{-1}xk) d k,$ -Define $ A\phi^K(m) = \Delta_P(m)^{1/2} \int\limits_{N} \phi^K(mn)d n$ for $m \in M$ -Then we have that $A\phi^K \in C_c^\infty(M)$ and the formula -$$ tr \pi_0(\phi) = tr \pi( A \phi^K).$$ - -The same formula is also useful, if the representation is not irreducible (unitarizability is not really an issue, and admissibility follows from the Iwasawa decomposition), but one has to normalize and decompose according to the $K$-isotypes. -So in your situation, you get a Fourier transform of $A \phi^K$. This in addition with Moshe Adrian answer computes all the irreducible, unitary principal series representation, at least in prinicple.<|endoftext|> -TITLE: Area of intersection of a family of circles in the plane -QUESTION [5 upvotes]: Suppose you are given a family F of circles in the plane such that each circle has radius 1. Let G be the family of circles with same centers as in family F but now each circle has radius $r$. Let A be area of union of circles in family F and let B be area of union of circles in family G. Then can we upper bound the number $\frac{B}{A}$ by a function $f(r)$ ? -One trivial thing is that if the family F contains all disjoint circles then $\frac{B}{A}$ is at most $r^2$. But in general case the geometry is getting weird and complicated . - -REPLY [10 votes]: The problem is much simpler than the general Kneser-Poulsen case. Here is an elementary proof of the fact that $B/A\le r^2$ for all $r\ge 1$. To simplify technicalities, I assume that the family of circles is finite (the general case follows as a limit). -Let $p_1,\dots,p_N$ be the centers of our circles. Let $V$ denote the union of the unit circles and $U$ the union of the circles of radius $r$. Divide $U$ into Voronoi regions $U_i$ of the points $p_i$. Namely, $U_i$ is the set of points $x\in U$ such that $p_i$ is nearest to $x$ among $p_1,\dots,p_N$. (For convenience, remove the points having more than one nearest center; this is a zero measure set.) Each $U_i$ is star-shaped w.r.t. the respective point $p_i$. Indeed, if $x\in U_i$, then $|p_ix|\le r$, hence the segment $[p_ix]$ is contained in $U$. And obviously $p_i$ is the nearest center for every point of this segment. -Now apply the $(1/r)$-homothety centered at $p_i$ to each set $U_i$. The resulting sets $U_i'$ are disjoint and contained in $V$, hence -$$ - S(V) \ge\sum S(U_i') = \frac1{r^2} S(U_i) = \frac1{r^2} S(U) -$$ -where $S$ denotes the area. Thus $S(U) \le r^2 S(V)$, q.e.d. -The same argument works in $\mathbb R^n$ (with $r^n$ in place of $r^2$), in Alexandrov spaces of nonnegative curvature, and in Riemannian manifolds of nonnegative Ricci curvature (just combine it with the proof of Bishop-Gromov inequality).<|endoftext|> -TITLE: Which norms have rich isometry groups? -QUESTION [19 upvotes]: Let $n \ge 2$ be some positive integer. Given a norm $p : \mathbb{R}^n \to \mathbb{R}$, one can inquire about the structure and properties of its isometry group, i.e. the group of all bijections $F:\mathbb{R}^{n}\to\mathbb{R}^{n}$ such that $p\left(v\right)=p\left(F\left(v\right)\right)$ for all $v \in \mathbb{R}^n$. By the Mazur-Ulam theorem, the isometries of $p$ are affine transformations, so the subgroup $Iso_{0}\left(p\right)$ of all isometries which fix the origin, consists of linear transformations, and is therefore a closed subgroup of $GL_n (\mathbb{R})$. Essentially, $Iso_{0}\left(p\right)$ is what we get after modding out the obvious distance-preserving functions (the translations), which have no relation to $p$ whatsoever. -For some norms, $Iso_{0}\left(p\right)$ is a very small group. For instance, if $p$ is either the $\ell^\infty$ norm or the $\ell^1$ norm, then the group of $Iso_{0}\left(p\right)$ is finite, and this seems to be the case for all $\ell^q$ norms other than $q=2$. On the other hand, for the euclidean norm ($p = \ell^2$), $Iso_{0}\left(p\right)$ is quite a rich group: it is a Lie group of positive dimension (which increases significantly with $n$) - the orthogonal group $O(n)$. -Obviously, a significant difference between this norm and other possible norms is that it arises from an inner product on $\mathbb{R}^n$. Since there is only one inner product structure on $\mathbb{R}^n$ (up to conjugacy) this essentially gives us just one example of a norm with a rich group of linear isometries. So my questions are: - -Are there any other norms (not induced by an inner product) on $\mathbb{R}^n$ which have a rich group of linear isometries? Here let me define "rich" as having positive dimension as a Lie subgroup of $GL_n (\mathbb{R})$ (every closed linear group is canonically a submanifold of $GL_n (\mathbb{R})$ compatible with the group structure). -If not, why is the property of being induced from an inner product the "right" condition for having many symmetries? That is, from a geometric point of view, why does this property make the norm especially "symmetric" or "smooth"? - -REPLY [5 votes]: Sorry to join so late to the party, but I couldn't help noticing there is a missing class of rich matrix norms (which are not operator norms). -These are the p-Schatten norms on $R^{n^2}$, which see only the singular values of a matrix. -Its immediate to see that these norms are invariant under left and right multiplication by orthogonal matrices. -Hence in particular their isometry groups contain copies of $O(n)$. -So $\dim(Iso(\| \cdot \|_{p-Schatten})) \ge \frac{n(n-1)}{2}$ -Note that for $p \neq 2$ these norms are not induced by inner products.<|endoftext|> -TITLE: extensions, abelian varieties, $\mathbb{G}_m$ -QUESTION [7 upvotes]: Question: why are there no non-trivial extensions of $\mathbb{G}_m$ by abelian varieties? -Specifically, let $A$ be an abelian variety over a field $k$. Then it seems to be well known that any extension $$ 0 \to A \to G \to \mathbb{G}_m \to 0$$ in the category of group varieties over $k$ is split. For instance, this is mentioned in Deligne's Theorie de Hodge I page 426 just before Section 3. Chevalley's structure theorem of algebraic groups certainly implies this. -Question: Is there a simple/direct proof? -Note that there are non-trivial extensions of abelian varieties by tori (semiabelian varieties). -Thanks. - -REPLY [20 votes]: Using the Kummer exact sequence -$0\rightarrow\mu_n\rightarrow\mathbb{G}_m\rightarrow\mathbb{G}_m\rightarrow0$ we -get a long exact sequence -$$ -0\rightarrow\mathrm{Hom}(\mathbb{G}_m,A)\rightarrow\mathrm{Hom}(\mathbb{G}_m,A) -\rightarrow\mathrm{Hom}(\mu_n,A)\rightarrow\mathrm{Ext}^1(\mathbb{G}_m,A). -$$ -As $\mathrm{Hom}(\mathbb{G}_m,A)=0$ this gives an embedding -$\mathrm{Hom}(\mu_n,A)\hookrightarrow\mathrm{Ext}^1(\mathbb{G}_m,A)$ and at -least over an algebraically closed field in which $n$ is invertible we always -have that $\mathrm{Hom}(\mu_n,A)$ is non-trivial and consequently so is -$\mathrm{Ext}^1(\mathbb{G}_m,A)$. -This does not contradict Deligne's claim as he seems to be (somewhat implicitly -it must be admitted) speaking of extensions up to isogeny (in Principle 2.1 this -is made more clear). In fact Chevalley's theorem implies that this is true: -Consider a connected affine subgroup $G'\hookrightarrow G$ for which the -quotient is an abelian variety. Then the kernel of the composite -$G'\rightarrow\mathbb{G}_m$ is affine and embeds in $A$ and thus is finite. On -the other hand $G'\rightarrow\mathbb{G}_m$ must be surjective as otherwise $G'$ -would be finite and hence trivial so $G$ would be an abelian variety which is -not possible. The kernel of $G'=\mathbb{G}_m\rightarrow\mathbb{G}_m$ is then -some $\mu_n$ which shows that the extension is in the image of -$\mathrm{Hom}(\mu_n,A)\hookrightarrow\mathrm{Ext}^1(\mathbb{G}_m,A)$ and in -particular is trivial up to isogeny. -The final conclusion is that $\mathrm{Ext}^1(\mathbb{G}_m,A)=\mathrm{Hom}(\hat{\mathbb Z}(1),A)$. -Addendum: I just realised that I didn't answer the actual question about -whether there is a simpler proof (to the isogeny statement would be my -interpretation). I doubt it (depending of course somewhat on your definition of -simpler). A proof (arguably less simple) avoiding the use of Chevalley's theorem -(but assuming to be on the safe side that the base field is algebraically closed -of characteristic $0$) would be to prove that for some $n$ the pull back of the -extension along the $n$'th power on $\mathbb{G}_m$ is trivial as an $A$-torsor -because then the extension would be described by a $2$-cocycly -$\mathbb{G}_m\times\mathbb{G}_m\rightarrow A$ and any map of that form is -constant. That in turn will follow from the fact that for some $m$ the class of -this extension as an $A$-torsor would be in the image of -$H^1(\mathbb{G}_m,A[m])\rightarrow H^1(\mathbb{G}_m,A)$. This follows from the -fact that $H^1(\mathbb{G}_m,A)$ is torsion and then the conclusion follows as -any class of $H^1(\mathbb{G}_m,C)$ for any finite locally constant sheaf $C$ is -killed by pulling back along some $n$'th power map.<|endoftext|> -TITLE: Is there a theory about these kinds of recurrence equations? Is this a known formula? -QUESTION [5 upvotes]: (New information at bottom) -Hi. -For a while, I've been toying around with solving recurrence equations of the form -$$a_1 = r_{1,1}$$ -$$a_n = \sum_{m=1}^{n-1} r_{n,m} a_m$$ -What are these kind of recurrence equations called? Does one have any references for any general theory behind them? -The goal is to try to find a non-recursive formula for the coefficients of what is called the "Schroder function" of $e^{uz} - 1$, which satisfies the functional equation -$$\chi(e^{uz} - 1) = u \chi(z)$$. -This function can be expressed as -$$\chi(z) = \sum_{n=1}^{\infty} \chi_n z^n$$ -with -$$\chi_1 = 1$$ -$$\chi_n = \sum_{m=1}^{n-1} \frac{u^{n-1}}{1 - u^{n-1}} \frac{m!}{n!} S(n, m) \chi_m$$ -a recurrence of the given form, where $S(n, m)$ is a Stirling number of the 2nd kind. -I managed to find the following formula: -$$a_n = r_{1,1} \sum_{\substack{1 = m_1 < m_2 < \cdots < m_k = n\\ 2 \le k \le n}}\ \prod_{j=2}^{k} r_{m_j, m_{j-1}},\ n > 1$$ -which sums over $2^{n-2}$ terms, namely, all subsets of the integer interval from 1 to $n$ that contain 1 and $n$. However, is there a way to simplify this for the case I gave, where $r_{n,m} = \frac{u^{n-1}}{1 - u^{n-1}} \frac{m!}{n!} S(n, m)$? I note that in cases like the Bernoulli numbers, these kind of recurrences have solutions expressible as nested sums or as products over many fewer terms than above (with a simple linear index). Is such a thing also possible here with the $r_{n,m}$ I just gave? If so, how? Also, is the above formula already known? -ADDENDUM (19 Sep 2011): I later found out on a different group (the Usenet newsgroup sci.math) that such a recurrence may be called a "full-history recurrence", though that term seems a little more general than just referring to the specific kind of sum recurrence mentioned above, and googling it did not turn up much (much less the solution formula mentioned above! (and too bad you can't google math formulas!)), and much of what it did turn up seemed to have to do more with the recurrence in the context of algorithmic theory and computer science than with just pure maths. Is there a better name or something more useful I might try looking up? -EDIT (25 Sep 2011): There is another form of this formula, for the indexing -$$a_0 = \alpha$$, -$$a_{n+1} = \sum_{m=0}^{n} r_{n,m} a_m$$. -This version goes as -$$a_n = \alpha \sum_{\substack{-1 = m_0 < m_1 < m_2 < \cdots < m_k = n-1\\ 1 \le k \le n}}\ \prod_{j=1}^{k} r_{m_j,m_{j-1}+1},\ n > 0$$. -Does that ring a bell better? Notice how we can get, e.g. the Bernoulli numbers by setting $\alpha = 1$, $r_{n,m} = -{n \choose m} \frac{1}{n - m + 1}$. -EDIT/ADD #2 (28 Sep 2011): Unfortunately, Helms' answer did not help very much. However, my access to academic resources is somewhat limited. It would be nice to have a direct reference to something containing the mentioned solution formula or a suitable equivalent and the specific kind of linear recurrence I'm asking about (the general form that is, not necessarily the "Schroder" one above), and also discussion of them of course. -EDIT/ADD #3 (29 Sep 2011): Yeah. "Eigensequence" did not seem to yield much on Google (incl. Google Scholar and Google Books). It looks to be much too broad. Certainly didn't find anything with the solution formulas I mentioned. Could this usage be peculiar to the OEIS site? - -REPLY [3 votes]: I don't know what else this may be called. But if the coefficients $r_{n,m}$ are written as lower triangular matrix $R$ and the coefficients $a_n$ as column-vector $A$ then this is an eigenvector-problem at eigenvalue $\lambda =1$ : -$ \qquad R*A=A*\lambda = A*1$ -(I should note, that the formulation of the problem indicates, that the diagonal of $R$ at least from the second row on seems to be zero here because the sum-index goes only to rowindex $n$ minus 1) -Note 2: in the OEIS as well as in its associated electronic journal I've seen the term "eigensequence" and "invariant sequence" with the same meaning. -Example: -$ \qquad \small -\begin{array} {rrrrrr} - & & & & & | & 1 \\\ - & & & & & | & 2 \\\ - & R*&A =&A & & | & 7 \\\ - & & & & & | & 33 \\\ - & & & & & | & 201 \\\ - - & - & - & - & - & - & - \\\ - 1 & . & . & . & . & | & 1 \\\ - 2 & . & . & . & . & | & 2 \\\ - 1 & 3 & . & . & . & | & 7 \\\ - 1 & 2 & 4 & . & . & | & 33 \\\ - 2 & 3 & 4 & 5 & . & | & 201 - \end{array} $ -Here we have, for a rowindex $n$ : $ \qquad \sum_{m=1}^{n-1} R_{n,m}*A_m = A_n$<|endoftext|> -TITLE: In the dictionary between Poisson and Quantum, what corresponds to Coisotropic? -QUESTION [10 upvotes]: I work entirely over a field of characteristic $0$, in case it matters. -Recall that a Poisson algebra is a commutative algebra $A$ with a bracket $\lbrace,\rbrace : A^{\wedge 2} \to A$ which is (1) a Lie bracket, i.e. it satisfies a Jacobi identity, and (2) a derivation in each variable. Or, maybe even better is that $(A,\lbrace,\rbrace: A^{\wedge 2} \to A)$ is a Poisson algebra if $A$ is a commutative algebra and $\forall a\in A$ $\lbrace a,-\rbrace : A \to A$ is a derivation of $(A,\lbrace,\rbrace)$. -A coisotrope in $A$ is a vector subspace $I \subseteq A$ which is (1) an ideal for the multiplication on $A$ and (2) Lie subalgebra for $\lbrace,\rbrace$. In particular, I do not demand that $I$ be an ideal for the bracket. The most important examples of Poisson algebras are $A = \mathcal C^\infty(M)$, where $M$ is a symplectic manifold; then an embedded submanifold $N \hookrightarrow M$ is coisotropic ($\mathrm T^\perp N \subseteq \mathrm T N$) iff the vanishing ideal of $N$ is a coisotrope. For more details and equivalent characterizations, see: - -Alan Weinstein, Coisotropic calculus and Poisson groupoids, J. Math. Soc. Japan Volume 40, Number 4 (1988), 705-727. http://projecteuclid.org/euclid.jmsj/1230129807 - -One reason to invent Poisson algebras is that the arise naturally as "deformation" or "quantization" problems: a Poisson algebra is the linear or infinitesimal data for a noncommutative algebra. In the most studied symplectic case, the Poisson algebra is in an important sense "maximally Poisson-noncommutative": the Poisson-center of $A$ consists only of (locally) constant functions. After quantization, the corresponding algebras are maximally noncommutative, and so should be algebras of (bounded) operators on some Hilbert space. In this sense, the quantization of a symplectic manifold $M$ is some Hilbert space $H$, or maybe its projectivization $\mathbb P H = H / \mathbb C^\times$. -Then the general dictionary says that lagrangian, i.e. minimal coisotropic, submanifolds of a symplectic manifold $M$ should correspond to elements of $H$ (or maybe elements of $\mathbb P H$, i.e. lines in $H$). My question is to understand a generalization of this that relaxes two things: - -I am interested in algebras that are Poisson but not symplectic, and so might have center; then I do not expect to have as good a "Hilbert-space" description of the quanization. Rather, my quantizations of Poisson algebras $(A,\lbrace,\rbrace)$ are nothing more nor less than (flat) deformations of $A$ in the $\lbrace,\rbrace$ direction. -In the non-symplectic setting, one loses a good theory of "lagrangian" submanifold, and the best stand-ins are the coisotropes. - -My question is then something along the following: - -Suppose I have a Poisson algebra $(A,\lbrace,\rbrace)$ with a coisotrope $I \subseteq A$, and I have a reasonably good quantization of $A$ in the $\lbrace,\rbrace$-direction. What should I expect/hope to see at the quantum level that corresponds to $I$? - -Asking the same question in the opposite direction: - -Suppose I have an associative algebra $B$ with a formal parameter $\hbar$ (and satisfying some strong flatness/topological freeness conditions), such that $B/(\hbar B)$ is commutative. Then the associated graded algebra $A = \bigoplus (\hbar B)^n / (\hbar B)^{n+1}$ is Poisson. What structures (e.g. ideals, left-modules, etc.) on $B$ become coisotropes in $A$ upon taking associated-graded? - -REPLY [6 votes]: This question was already positively answered. -I'd like to add a remark on a situation where things go in a very clear way supporting Waldmann's statementes. Say $G$ is a Poisson Lie group. -A (closed) subgroup $H$ is coisotropic iff the annhilator of its Lie algebra $\mathfrak h$ is a Lie subalgebra in $\mathfrak g^*$. -Quantization of such subgroups are exactly two-sided coideal (this is the subgroup part) which are also one sided ideal (this is the coisotropy assumption). -A two-sided coideal and one sided ideal inside a Hopf algebra is exactly what is needed to define a subalgebra of coinvariants which is a one sided coideal, i.e. a quantum homogeneous spaces seen as a subalgebra of the quantum algebra. This fact reflects the property that modding out a Poisson-Lie group by a coisotropic subgroup you get a Poisson homogeneous space (of quotient type, i.e. such that the quotient map is Poisson). -The idea that one sided ideal is the non commutative counterpart of coisotropy was stated in a paper in the beginning of the 90's by Jiang-Hua Lu: the statement was named coisotropic creed.<|endoftext|> -TITLE: Is there an elegant algebraic proof of this formula for quadratic field discriminants? -QUESTION [14 upvotes]: Consider the Dirichlet series counting discriminants of real quadratic fields. Quadratic field discriminants are "basically" squarefree integers, so the associated Dirichlet series $\sum D^{-s}$ is "basically" $\zeta(s)/\zeta(2s)$. However, there is the funny business at 2, and one derives the formula -$\sum D^{-s} = \frac{1}{2} \big( 4^{-s} - 1 \big) \frac{\zeta(s)}{\zeta(2s)} + \frac{1}{2} \big(1 - 4^{-s} \big) \frac{L(s, \chi_4)}{L(2s, \chi_4)},$ -which is a wee bit messy. (This formula, and all the subsequent ones, include 1 as a "quadratic field discriminant" for convenience.) -However, I was reading a fantastic paper by David Wright, where he considers positive and negative discriminants together, in which case you have the much nicer formula -$\sum |D|^{-s} = \big(1 - 2^{-s} + 2 \cdot 4^{-s} \big) \frac{\zeta(s)}{\zeta(2s)}.$ -This formula is easy to derive from scratch, but he derives it as a consequence of the beautiful formula -$\sum |D|^{-s} = \prod_p \Big( \frac{1}{2} \sum_{[K_v : \mathbb{Q}_p] \leq 2} |\text{Disc}(K_v)|^s_p \Big).$ -He uses the parameterization of quadratic fields by $\mathbb{Q}^{\times} / (\mathbb{Q}^{\times})^2$, which may be thought of as $\text{GL}_1$-orbits on a one-dimensional prehomogeneous vector space, where $\text{GL}_1$ acts by $t(x) = t^2 x$ rather than the usual $t(x) = tx$. He then analyzes these orbits by means of an adelic zeta function; note that with the usual action you recover Tate's thesis. -These formulas generalize quite a bit, with some complications, to $n$th-root extensions -of any global field (with some restrictions on the characteristic). They also allow for twisting by characters, allowing (for example) a nice formula for $\sum \text{sgn}(D) |D|^{-s}.$ Note that the first formula considered looks nicer when viewed as a linear combination of $\sum |D|^{-s}$ and $\sum \text{sgn}(D) |D|^{-s}$. -The MathSciNet review says that "similar results, however, can be obtained by class field theory", and this is also hinted at in Wright's paper, but the details aren't worked out. My first question is the following. -Is there an elegant algebraic proof of the above identity for $\sum |D|^{-s}$ and its generalizations? -And my second question, which is essentially a vaguer version of the first one, is: -What is the best way to think of these formulas? -I quite like the prehomogeneous vector space approach, but I imagine there might be a nice algebraic proof as well, and in particular some kind of ``local-to-global'' principle for quadratic discriminants. I am no expert in class field theory, and I am curious if any of these identities look simple and natural when viewed in the correct light. -Thank you! - -REPLY [12 votes]: The discriminant $D$ of a quadratic number field can be written uniquely as a product of prime discriminants, namely $-4$, $\pm 8$, and $p^* = (-1)^{(p-1)/2}p$ for odd primes $p$. -The Dirichlet series for odd discriminants therefore simply is -$$ \sum_{D \text{ odd}} |D|^{-s} = \prod_{p \text{ odd}} (1 + p^{-s}), $$ -and the contribution of the even prime discriminants is taken care of by the factor -$$ 1 + 4^{-s} + 2 \cdot 8^{-s}. $$ -Both the beautiful as well as the much nicer formula now follow immediately. -Factorization of quadratic discriminants into prime discriminants holds over totally real algebraic number fields with class number $1$ in the strict sense (see e.g. L. Goldstein, On prime discriminants, Nagoya Math. J. 45, 119-127 (1972); J. Sunley, Remarks concerning generalized prime discriminants, Boulder 1972; J. Sunley, Prime discriminants in real quadratic fields of narrow class number one, Carbondale 1979); a weaker version good enough for the purpose of counting discriminants works if the class number in the strict sense is odd.<|endoftext|> -TITLE: When is a restricted enveloping algebra a domain? A finitely generated domain? -QUESTION [8 upvotes]: Suppose $\mathfrak{g}$ is a restricted Lie algebra over a field of characteristic $p>0$. Are there conditions on $\mathfrak{g}$ and its restriction which ensure that its restricted enveloping algebra is a domain? That it is a finitely generated domain? I really care about the graded case, so what if we assume further that $\mathfrak{g}$ is graded, concentrated in positive even degrees, and finite-dimensional in each degree? - -Edit: for example, if the restriction is injective, does that mean that $u(\mathfrak{g})$ is a domain? What if you only know that if $x^{[p]}=0$, then $x=0$ – is that good enough? In the graded case, if the restriction is also surjective in all sufficiently large degrees, is $u(\mathfrak{g})$ a finitely generated domain? -Are there conditions on $\mathfrak{g}$ and its restriction which make $u(\mathfrak{g})$ isomorphic to an ordinary enveloping algebra $U(L)$ of some Lie algebra $L$? - -REPLY [3 votes]: Let $g$ be a restricted Lie algebra over a field of positive characteristic $p$. It is not difficult to see that a necessary condition such that the restricted enveloping algebra $u(g)$ of $g$ is a domain is that $g$ has no nonzero $p$-algebraic elements. (An element $x \in g$ is said to be $p$-algebraic if the restricted subalgebra generated by $x$ is finite-dimensional.) The converse of this property is a well-known open question posed by V. Petrogradsky (see "DNIESTER NOTEBOOK: Unsolved Problems in the Theory of Rings and Modules", Problem 3.59). Apparently, this is the Lie theoretical analog of the Kaplansky Problem about zero-divisors of group algebras of torsion-free groups.<|endoftext|> -TITLE: Universal sets in metric spaces -QUESTION [11 upvotes]: (I am cross-posting this from math.SE as it seems to be slightly over the top for that site.) -I saw in the class the theorem: -Suppose $X$ is a separable metric space, and $Y$ is a polish space (metric, separable and complete) then there exists a $G\subseteq X\times Y$ which is open and has the property: -For all $U\subseteq X$ open, there exists $y\in Y$ such that $U = \{x\mid\langle x,y\rangle\in G\}$. -$G$ with this property is called universal. -The proof is relatively simple, however the $y$ we have from it is far from unique, in fact it seems that it is almost immediate that there are countably many $y$'s with this property. -My question is whether or not this $G$ can be modified such that for every $U\subseteq X$ open there is a unique $y\in Y$ such that $U = \{x\mid\langle x,y\rangle\in G\}$? -Perhaps we need to require more, or possibly even less, from $X$ and $Y$? -Some thoughts: -Firstly $X$ cannot be finite, otherwise there are less than continuum many open subsets, and since $G$ is open we have that the projection on $Y$ is open, since $Y$ is Polish we have that this projection is of cardinality continuum, which in turn implies there are continuum many $y$'s with the same cut. -Secondly, as the usual proof goes through a Lusin scheme over $Y$, and using it to define $G$, I thought at first that using the axiom of choice we can select a set of points on which the mapping to open sets of $X$ is 1-1, and somehow remove some of the sets from the scheme. This proved to be a bad idea, as we remove sets that can be used for other open sets. -Thirdly, I thought about enumerating the open sets according to a rational enumeration so $A_i\subseteq A_j$ if and only if $q_i\le q_j$, and then instead of just placing the open sets of $X$ arbitrarily by the Lusin scheme, we use the rationals somehow. - -REPLY [4 votes]: While idly browsing around I stumbled over the follwing paper and remembered this question: -A.W. Miller, Uniquely Universal Sets, Topology and its Applications 159 (2012), pp. 3033–3041. It's available in various formats here. -Let me quote the abstract (to avoid confusion: Miller's terminology reverses the rôles of $X$ and $Y$ in your question): - -We say that $X \times Y$ satisfies the Uniquely Universal property (UU) - iff there exists an open set $U \subseteq X \times Y$ such that for - every open set $W \subseteq Y$ there is a unique cross section of - $U$ with $U_x=W$. Michael Hrušák raised the question of when does - $X \times Y$ satisfy UU and noted that if $Y$ is compact, then $X$ - must have an isolated point. We consider the problem when - the parameter space $X$ is either the Cantor space $2^\omega$ or the Baire - space $\omega^\omega$. - We prove the following: - -If $Y$ is a locally compact zero dimensional - Polish space which is not compact, then $2^\omega\times Y$ has UU. -If $Y$ is Polish, then - $\omega^\omega \times Y$ has UU iff $Y$ is not compact. -If $Y$ is a $\sigma$-compact subset of a Polish space which is - not compact, then $\omega^\omega \times Y$ has UU. - - -His results are mostly positive: “a certain space or family of spaces has UU” and various permanence properties. One nice “negative” result: - -Proposition 30: There exists a partition $X\cup Y=2^\omega$ into Bernstein sets - $X$ and $Y$ such that for every Polish space $Z$ neither - $Z\times X$ nor $Z\times Y$ has UU. - -He also raises a few questions, e.g.: - -Question 4: Does $(2^\omega\oplus 1) \times [0,1]$ have UU? -Question 6: Does either $\mathbb{R} \times \omega$ -or $[0,1]\times \omega$ have UU? -Or more generally, is there any example of UU for a -connected parameter space? -Question 11: Is the converse of Corollary 10 false? -That is: Does there exist $Y$ such that $\omega^\omega \times Y$ has UU but -$2^\omega\times Y$ does not have UU?<|endoftext|> -TITLE: Formally smooth morphisms, the cotangent complex, and an extension of the conormal sequence -QUESTION [9 upvotes]: I'm reading Daniel Quillen's paper "Homology of commutative rings," in which he proves: -A finitely presented morphism of rings $A \to B$ is - -Formally etale iff $L_{B/A}$ (this denotes the cotangent complex) is homotopy-equivalent to zero -Formally smooth iff $\Omega_{B/A}$ is projective and $L_{B/A}$ is homotopy-equivalent to it (i.e. acyclic outside degree zero). -(It's also true, and elementary (not in the paper), that formally unramified iff $\Omega_{B/A} = 0$.) - -I've heard that these results are true even without finitely presented hypotheses. In fact, I understand that the fpqc localness of projectivity is what one uses to show that formal smoothness is, in fact, a local property (cf. 2 above). I also know how to show that if $A \to B$ is formally smooth, then the differentials are a projective $B$-module (take a quotient by some polynomial ring, $B = C/I$, and show that the sequence $I/I^2 \to \Omega_{C/A} \otimes_C B \to \Omega_{B/A} \to 0$ is actually split exact). -In fact, one can show that if $C$ is a formally smooth $A$-algebra and $B = C/I$, then $B$ is smooth iff the conormal sequence above is split exact. -So I am guessing that there is an extension of the conormal sequence to the cotangent complex, and if this works will prove 2 without finitely presented hypotheses (and thus 1 as well). How does this "long exact sequence" work? - -REPLY [6 votes]: We deduce this from the Jacobi-Zariski exact sequence as follows: -Some notation first: Let $H_i(A,B,W)$ where B is an A-algebra and W is a B-module denote $\pi_i(L_{B/A}\otimes_B W)$. -Then given an $A$-algebra $B$, a $B$-algebra $C$, and a $C$-module $W$, we have a long-exact sequence, called the Jacobi-Zariski sequence: -$$\dots \to H_n(A,B,W)\to H_n(A,C,W) \to H_n(B,C,W)\to H_{n-1}(A,B,W)\to \dots \to H_0(B,C,W)\to 0$$. -Now, let $C$ be an $A$-algebra, let $B$ be a polynomial ring over $A$ with an ideal $I$ such that $B/I\cong C$. -Then we have the following long-exact sequence: -$$\dots \to H_n(A,B,C)\to H_n(A,C,C) \to H_n(B,C,C)\to H_{n-1}(A,B,C)\to \dots \to H_0(B,C,C)\to 0$$. -Since $C$ is a quotient of $B$, it is formally unramified over $B$, so $H_0(B,C,C)\cong \Omega_{C/B}=0$. We also see that $I/I^2$ is precisely $H_1(B,C,C)$ by Proposition 1 of Chapter VI.a of the book Homologie des algèbres commuatatives by Michel André. This gives us a long-exact sequence -$$\dots \to H_n(A,B,C)\to H_n(A,C,C) \to H_n(B,C,C)\to H_{n-1}(A,B,C)\to \dots $$ -$$\to H_1(A,B,C) \to H_1(A,C,C)\to I/I^2 \to H_0(A,B,C)\to H_0(A,C,C)\to 0$$. -We see that the conormal exact sequence is the truncation at $I/I^2$, and that this sequence splits if $H_0(A,C,C)$ is projective and $H_1(A,C,C)$ is $0$. Conversely, if the sequence is split-exact, then $H_0(A,C,C)$ is a direct summand of $H_0(A,B,C)$, which is projective (it might be free, but I don't remember), and the kernel of $I/I^2\to H_0(A,B,C)$ is $0$, which combined with the fact that $H_1(A,B,C)=0$ implies that $H_1(A,C,C)=0$.<|endoftext|> -TITLE: Reference request for projective representations of finite groups over a non-problematic field -QUESTION [11 upvotes]: I would like to get a reference where I can learn about the theory of projective representations of finite groups over the complex numbers (or over any field K such that the order of the given group under study is invertible in K). And how one can relate this to character theory for linear representations (with which I am more familiar). Thanks. - -REPLY [2 votes]: In addition: Karpilovsky, Gregory. -The Schur Multiplier. London Math. Soc. Monographs, 1987. -In this book only $\mathbb{C}$ is considered.<|endoftext|> -TITLE: Explicit extension of Lipschitz function (Kirszbraun theorem) -QUESTION [17 upvotes]: Kirszbraun theorem states that if $U$ is a subset of some Hilbert space $H_1$, and $H_2$ is another Hilbert space, and $f : U \to H_2$ is a Lipschitz-continuous map, then $f$ can be extended to a Lipschitz function on the whole space $H_1$ with the same Lipschitz constant. -Now let's take $H_2$ to be the Euclidean space $\mathbb{R}^n$. My question is: Is there way to explicitly construct this extension? Note that the standard proof (e.g. see Federer's geometric measure theory book or Schwartz's nonlinear functional analysis book) is an existence proof, which uses Hausdorff's maximal principle. -Some remarks: -1) For $n = 1$, the extension can be constructed explicitly, which works even if $H_1$ is only a metric space (with metric $d$): $\tilde{f}(x) = \inf_{y \in U} \{ f(y) + {\rm Lip}(f) d(x,y) \}$. See for example Mattila's book p. 100. -2) For $n > 1$, performing the above extension for each component of $f$ results in blowing up the Lipschitz constant by a factor of $\sqrt{n}$. - -REPLY [4 votes]: If I remember well the Kirszbraun's extension of a $L$-Lipschitz map $f:U\subset H_1\to H_2$ has the following canonical construction, analogous to the one-dimensional case you mentioned (so in a sense it is explicit). -Let $\mathcal{Co} (H_2)$ denote the metric space of all non-empty bounded closed convex sets of $H_2$ endowed with the Hausdorff distance. Let $f_*:H_1\to \mathcal{Co}(H_2)$ be defined by $$f_*(x):=\cap_{u\in U}\overline{B}(f(u),L)$$ -(In other words, $f_*$ takes $x\in H_1$ to the set of the admissible values at $x$ for any $L$-Lipschitz extension of $f$ to $U\cup\{x\}$). This map $f_*$ has the same Lipschitz constant of $f$, w.r.to the Hausdorff distance on $\mathcal{Co}(H_2)$. -Any non-empty bounded closed convex $C$ of a Hilbert space $H$ has a well-defined point $\kappa( C), $ the center of the closed ball of minimum radius containing $C$; this point is unique, and the corresponding map $\kappa: \mathcal{Co}(H)\to H $ is $1$-Lipschitz. -One can therefore define a canonical $L$-Lipschitz extension of $f$ as $\tilde f:=\kappa \circ f_*$. In case $n=1$, the set $f_*(x)$ is just an interval, its end-points are the inf-convolution you mentioned, and the sup-convolution, and this $\tilde f$ is their arithmetic mean.<|endoftext|> -TITLE: Products of Ideal Sheaves and Union of irreducible Subvarieties -QUESTION [7 upvotes]: Assume I have a nonsingular, irreducible, algebraic variety $X$ and irreducible, nonsingular subvarieties $Z_1,\ldots,Z_k\subseteq X$. Let $\mathcal{I}_i$ be the ideal sheaf of $Z_i$ and $\mathcal{I}:=\mathcal{I}_1\cdots\mathcal{I}_k$ the product. My question is whether $\mathcal{I}$ is the ideal sheaf of the union $Z_1\cup\ldots\cup Z_k$. You may assume that $Z_1\cap\ldots\cap Z_k=\emptyset$ or, equivalently, $\mathcal{I}_1+\ldots+\mathcal{I}_k=\mathcal{O}_X$. You may also assume that the $Z_i$ intersect transversally, if that helps. - -REPLY [2 votes]: As Sandor and Martin pointed out above, it is ok if the subschemes intersect in the emptyset pairwise. I'm going to provide 3 examples, I think the third one gives a counter-example to even the transversality statements. -Example 1: Here's an example where it's false without that hypotheses, notice that the varieties are smooth and they intersect pairwise with normal crossings. EDIT: as t3suji pointed out in the comments, these varieties don't intersect transversally in the ambient space, just in the ambient $z = 0$ plane EndOfEdit Consider $X = \mathbb{A}^3$ and set $Z_1 = V(y,z)$, $Z_2 = V(x,z)$, $Z_3 = V(x, z-1)$. Notice that $Z_3$ doesn't intersect any of the other subschemes. -Then, $I_1 \cap I_2 = (z, xy)$. -However, $I_1 \cdot I_2 = (xy, yz, xz, z^2)$. -These ideals are not equal clearly. Now, we can immediately see that multiplying/intersecting by $I_3$ won't change the behavior at the origin at all since the ideal doesn't vanish there, so they are not equal. However, just to be sure, I also did the following computation (with Macaulay2): -$$I_1 \cdot I_2 \cdot I_3 = (xz, yz, z^3 - z^2, yz^2 - yz, xz^2 - xz).$$ -$$I_1 \cap I_2 \cap I_3 = (z^2 -z, xz, xy).$$ -Macaulay2 also confirmed that the ideals were not equal. -Example 2: Here's a different example. a 4th variety that doesn't intersect the others at all. -$X = \mathbb{A}^3$. $I_1 = (x,y)$, $I_2 = (x,z)$, $I_3 = (y,z)$ and $I_4 = (x-1,y-1,z-1)$. Certainly again the $I_4$ doesn't matter, it's just included so that the sum of the ideals is equal to $R = k[x,y,z]$. I wonder if it might be reasonable to say that $Z_1$, $Z_2$ and $Z_3$, as a triple, have transverse intersection at the origin. Anyways: -$I_1 \cap I_2 \cap I_3 = (xy, xz, yz)$ but, -$I_1 \cdot I_2 \cdot I_3 = (yz^2, xz^2, y^2z, xyz, x^2z, xy^2, x^2y)$. -Example 3: Ok, now I'm just going to give three subvarieties which intersect at the origin, pairwise transversally, and such that the product of the ideals is not equal to the intersection. You may add the ideal sheaf of some other variety that doesn't intersect them at all to make the sum of ideals equal the whole structure sheaf. -$X = \mathbb{A}^4 = \text{Spec}k[x,y,u,v]$. $I_1 = (x,y)$, $I_2 = (u,v)$, $I_3 = (x+u, y+v)$. I believe these have pairwise transverse intersection at the origin. Now then, it is true that $I_1 \cdot I_2 = I_1 \cap I_2$, and likewise with any pair. However, Macaulay2 can be used to verify that $I_1 \cdot I_2 \cdot I_3 \neq I_1 \cap I_2 \cap I_3$. Roughly speaking the problem is that $Z_1 \cup Z_2$ has funny intersection with $Z_3$. -Ok, let me now give a proof of a correct statement showing that sometimes they are equal. -Lemma: Suppose that subschemes $Z_1, \dots, Z_k$ have pairwise trivial intersection in some ambient Noetherian scheme $X$. Then $I_{Z_1} \dots I_{Z_K} = I_{Z_1} \cap I_{Z_k}$. -Proof: The statement is local so we may assume that $X$ is the spectrum of a local ring $(R, \mathfrak{m})$. Now, since $I_{Z_1} + I_{Z_2} = R$, at least one of those ideals must equal $R$ (if not, both would be in the maximal ideal $\mathfrak{m}$, and so would their sum). Likewise with all pairs. Therefore, at most one of the ideals $I_{Z_i}$ is not equal to $R$. But now the statement is obvious. $R \cdot R \dots I_{Z_i} \dots R = I_{Z_i} = R \cap R \cap \dots I_{Z_i} \cap \dots R$.<|endoftext|> -TITLE: Global Algebraic Proof of the Kahler Identities? -QUESTION [11 upvotes]: I'm looking at Kahler geometry at the moment and admiring how it manages to do so much with clean global algebraic arguments. One of the big exceptions to all this, however, is the proof of the Kahler identities -$$ -[\Lambda,\overline{\partial}]=-i \partial^\ast, ~~~~~~ -[\Lambda,\partial]=-i \overline{\partial}^\ast. -$$ -In the two standard references, Voisin, and Griff + Harris, the identities are proved using arguments that are local and somewhat analaytic. Does there exists anywhere a nice global algebraic proof? - -REPLY [7 votes]: Actually, there is an argument that doesn't use neither Weil identities nor flat up to a second order coordinates, instead, it uses the symplectic Hodge star, invented by Brylinski. When you have a 2n-dimensional manifold $M$ with the non-degenerate 2-form $\omega$, you can define an operator $*_s:\Lambda^k(M) \mapsto \Lambda^{2n-k}(M)$ by the standard identity $\alpha \wedge *_s\beta = \omega(\alpha,\beta)\omega^n$. Then, for a closed $\omega$ a really simple computation (you can do it only for the case of two variables, and then use induction on dimension) in Darboux coordinates shows that, up to some sign ($(-1)^{k+1}$, i guess), $*_sd*_s=[\Lambda, d].$ It is written in the Brylinski's article "A differential complex for Poisson manifolds". And then you need to observe that for a Kaehler $M$ symplectic and Riemannian Hodge stars differ by an action of $I$. The Kaehler identity $d^*=[\Lambda, d^c]$ follows from that.<|endoftext|> -TITLE: Bridge game with only one suit: strategy -QUESTION [6 upvotes]: This game looks like bridge, but 1- there are only two players Alice and Bob, 2- there is only one suit, whose cards are numbered $1, 2,\ldots,2n$. One deals each player $n$ cards. Therefore Alice knows Bob's cards and conversely; once the cards are dealt, there is no randomness. -Alice play a card, then Bob. The highest card wins the trick. The winner of the trick leads a card and so on. At the end Alice has got $p$ tricks and Bob $n-p$ tricks. The goal for each player is to get as much tricks as possible with the cards (s)he was dealt. -My question is about the strategy of play and the number of tricks you expect to win in a given layout. The answer should not be obvious. Let me give an example with $n=3$. If Alice is dealt $6,4,1$, she gets two tricks by leading first the $1$. If she has instead $6,3,2$, she leads the $3$ (equivalently the $2$). I see easily the way to get as many tricks as possible if $n$ is small, say $n\le6$, but I don't see a generalization. -Of course, the number of expected tricks with a given hand differs whether you begin or you opponent does. - -REPLY [19 votes]: I believe the game you describe is two-person single suit whist and was solved by Johan Wastlund in this paper.<|endoftext|> -TITLE: Sums of uncountably many real numbers -QUESTION [5 upvotes]: Suppose $S$ is an uncountable set, and $f$ is a function from $S$ to the positive real numbers. Define the sum of $f$ over $S$ to be the supremum of $\sum_{x \in N} f(x)$ as $N$ ranges over all countable subsets of $S$. Is it possible to choose $S$ and $f$ so that the sum is finite? If so, please exhibit such $S$ and $f$. - -REPLY [2 votes]: This is a standard result in undergraduate analysis, although it is admittedly somewhat hard to find in the standard references. The following is a very non-standard reference: see the last exercise in II.9.4 of these notes on sequences and series (see p. 69...for now; page numbers are subject to change). They occur in the context of a larger discussion on unordered summation, which is what you are looking into above. The general definition of unordered summability is a bit more complicated (it is a nice special case of convergence with respect to a net, although one needn't use the term), but in the case where the values of the "$S$-indexed sequence" are non-negative, it coincides with what you have given: see Proposition 82. -Note that this fact comes up sometimes in practice. In this math.SE question I set as a challenge to give a proof of the following fact -- there is no function $f: \mathbb{R} \rightarrow \mathbb{R}$ with a removable discontinuity at every point -- which does not use the kind of uncountable pigeonhole principle argument that you need to answer the current question. And I got a very nice answer!<|endoftext|> -TITLE: Evaluating the integral $\int_0^\infty \frac{\psi(x)-x}{x^2}dx.$ -QUESTION [8 upvotes]: Let $\psi(x)=\sum_{n\leq x} \Lambda(n)$ be the weighted prime counting function. I am trying to evaluate the integral $$\kappa:=\int_{1}^{\infty}\frac{\psi(x)-x}{x^{2}}dx$$ in several different ways. Originally, this integral came up as a particular part in a particular case for a a formula for a summatory function I was looking at. From now on, let $\gamma$ refer to the Euler-Mascheroni constant. -(Now Corrected:) I found a fun, elementary approach to this integral which gave $\kappa=-1-\gamma$ if we assume the quantitative prime number theorem. (Precisely, we just need to assume that this integral is absolutely convergent. ) Since I am not too confident about this, I naturally wanted to check by complex analytic methods to see if my answer was correct. My question then is: - -What other ways can be used to prove this identity? - -I feel like knowing many approaches to this problem will give a greater understanding of certain properties of these functions. A friend suggested that it must be related to the logarithmic derivative of $\zeta(s)$, and certain special values, but I cannot see how to use this. -Thanks a lot! -Additional Remark: -I attempted to use the explicit formula for $\psi(x)$, and deduced $\kappa=-\gamma-1$. Originally I felt this was wrong, but after reading Julian Rosen's answer I think it is correct. Here is the alternate solution: -Substituting in the explicit formula, and then integrating termwise we have$$\kappa=\int_{1}^{\infty}\left(-\sum_{\rho}\frac{x^{\rho-2}}{\rho}-\frac{\log2\pi}{x^{2}}-\frac{\log\left(1-x^{-2}\right)}{2x^{2}}\right)dx=\sum_{\rho}\frac{1}{\rho(\rho-1)}-\log2\pi+1-\log2$$since -$$\frac{1}{2}\int_{1}^{\infty}\frac{\log\left(1-x^{-2}\right)^{-1}}{x^{2}}dx=\frac{1}{2}\int_{1}^{\infty}\sum_{i=1}^{\infty}\frac{1}{ix^{2i+2}}dx=\sum_{i=1}^{\infty}\frac{1}{2i(2i+1)}=1-\log2.$$As $$\sum_{\rho}\frac{1}{\rho(\rho-1)}=\sum_{\rho}\frac{1}{\rho-1}-\frac{1}{\rho}=-\sum_{\rho}\frac{1}{1-\rho}+\frac{1}{\rho}=2B=-\gamma-2+\log4\pi$$it follows that $\kappa=-\gamma-1$. - -REPLY [3 votes]: I want to prove it with an elementary approach: -a theorem of Landau say that the PNT is equivalent to $\sum_{n\leq x} \frac{\Lambda(n)}{n}=log(x)-\gamma+o(1)$. -Now using partial summation we have: -$\displaystyle \sum_{n\leq x} \frac{\Lambda(n)}{n}=\displaystyle \int_{1}^{x}\frac {d\psi(t)}{t}=\int_{1}^{x}\frac {d(\psi(t)-t)}{t}+\int_{1}^{x}\frac {dt}{t}=log(x)+\int_{1}^{+\infty}\frac {\psi(t)-t}{t^{2}}-\int_{x}^{+\infty}\frac {\psi(t)-t}{t^{2}}+\frac{\psi(x)-x}{x}-\frac{\psi(1)-1}{1}=log(x)+\int_{1}^{+\infty}\frac {\psi(t)-t}{t^{2}}+o(1)+1$ -so that $\displaystyle \int_{1}^{+\infty}\frac {\psi(t)-t}{t^{2}}=-1-\gamma$.<|endoftext|> -TITLE: What is the standard notation for group action -QUESTION [20 upvotes]: Please let me know what is the standard notation for group action. - -I saw the following three notations for group action. -(All the images obtained as G\acts X for different deinitions of \acts.) -(1) -I saw this one most, but only in handwriting and I like it. But I did not find a better way to write it in LaTeX. -\usepackage{mathabx,epsfig} -\def\acts{\mathrel{\reflectbox{$\righttoleftarrow$}}} - -(2) -It is almost as good as 1, but in handwriting this arrow can be taken as $G$. -\usepackage{mathabx} -\def\acts{\lefttorightarrow} - -(3) -I saw this one in print, I guess it is used since there is no better symbol in "amssymb". -\usepackage{amssymb} -\def\acts{\curvearrowright} - -REPLY [6 votes]: As this was the first hit on google for "latex action arrow", but didn't contain what I wanted, let me post what I figured out. But to address other people's issues with the original question: while I agree that in a sentence one should simply say "Let $G$ act on $X$...", I was interested in drawing a (what some people just generically call "commutative") diagram in order to visually display the relationships between several objects (as you do). -Anyway, there is a symbol $\circlearrowright$, which is almost what I wanted. If you do this: - -\usepackage{graphicx} - G\ \rotatebox[origin=c]{-90}{\$\circlearrowright\$}\ X - -you will get an arrow whose beginning and end are right next to the X.<|endoftext|> -TITLE: Evaluating the integral $\int_{1}^{\infty}\frac{\{u\}}{u^{2}}\left(\log u\right)^{k}du.$ -QUESTION [15 upvotes]: I am trying to find a formula for the following integral for non-negative integer $k$: -$$\int_1^{\infty}\frac{\{u\}}{u^{2}}\left(\log u\right)^{k}du.$$ -My first thought was to use the formula $$\zeta(s)-\frac{1}{s-1}=1-s\int_1^\infty u^{-s-1}\{u\}du$$ where $\{u\}$ refers to the fractional part. We can then take derivatives with respect to $s$ and use the Laurent expansion for $\zeta(s)$. It follows that each integral must be a finite linear combination of the Stieltjes Constants. All of the coefficients must be integers, and $\gamma_n$ can only appear if $n\leq k$. (This checks out numerically for $k=0,1,2$) -Unfortunately, I am not sure what the pattern is, but I feel these particular integrals must be very common, and must have been dealt with before. I am hoping someone can give me a reference, or give a solution. -Thanks a lot, - -REPLY [21 votes]: Let $a_k$ be the integral. Then -$$\begin{eqnarray*} - \sum_{k \ge 0} \frac{a_k}{k!} t^k &=& \int_1^{\infty} \frac{ \{ u \} }{u^2} e^{t \log u} \, du \\\ - &=& \int_1^{\infty} \{ u \} u^{t-2} \, du \\\ - &=& \frac{1 - \zeta(1 - t) - \frac{1}{t}}{1 - t} \\\ - &=& \frac{1}{1 - t} \left( 1 - \sum_{n \ge 0} \frac{\gamma_n}{n!} t^n \right). -\end{eqnarray*}$$ -(Generating functions are good for more than combinatorics!) This is equivalent to Julian Rosen's answer, but (I think) packaged slightly more conveniently.<|endoftext|> -TITLE: What does the space induced by this unusual metric(?) on R/Z look like? -QUESTION [8 upvotes]: The motivation for this question comes from music theory. Dmitri -Tymoczko models "good" voice leading as minimizing distance between -pitches in successive chords. While this theory works well for upper -voices, it does not work so well for the bass, which tends to move by -4ths and 5ths quite frequently. (Tymoczko explicitly excludes the -bass from his model.) -Taking log (all logs are base 2 in this question) of frequency and -taking pitches which differ by an octave as equivalent, we get -$\mathbb{R}/\mathbb{Z}$. For the upper voices, we want the standard -metric on this. -For the bass, we want moving by a fifth or a fourth - meaning by $\pm -\log (3/2)$ to be small. So we want $d(x,x\pm\log(3/2))=k_1$, where -$k_1$ is probably somewhere around $0.05$. To make this a metric -space, let's declare that $d(x,y)$ should be the minimum of $|x-y|$, -$k_1+||x-y|-\log(3/2)|$, and $k_1+||x-y|-\log(4/3)|$. -We probably want moving by a major third - meaning by $\pm\log(5/4)$ -to also be small, but not as small. I suspect $2k_1$ would make the -most mathematical sense, but any constant of roughly that magnitude is -fine. Ditto for minor thirds - this would be movement by -$\pm\log(6/5)$, with a slightly larger constant. -If we do this, we might as well make all movements by $\pm\log(p/q)$ -small if $q$ is small. -CLARIFICATION: I also want the standard metric to be one of the options for getting from $x$ to $y$. So the distance between $0$ and $\sqrt{1/500}$ should be $\sqrt{1/500}$, while the distance between $0$ and $7/12$ should be $k_1+|\log(3/2)-7/12|$. (Musically, $|\log(3/2)-7/12|$ is how far off an even-tempered 5th is from Pythagorean tuning.) -Question 1: Can one actually define something along these lines that -satisfies the triangle inequality? (I don't think I actually have; I -probably need to take the minimum (or infimum) of some infinite -sequence, but am not entirely sure that works.) -Question 2: Assuming the answer to (1) is yes, what does this metric -space look like? Can someone help me with a picture that seems less -exotic, perhaps comparing it to the Hawaiian Earring or something of -that sort? In particular, what might the fundamental group look like? -My background: I'm a combinatorialist and algebraic geometer who happens -to be the one least unqualified here to be supervising an -undergraduate independent study on mathematics in music theory. I did -the standard first year graduate courses in point set topology and -algebraic topology, but that was almost a dozen years ago. - -REPLY [3 votes]: I doubt that you can do this in a way which addresses the intended application. For instance, you say you want bass movement by a fifth to be "small" because it is frequent in traditional music. You also want stepwise movement by a full or half step to be "small." But you can combine a fifth and a half step to make a tritone, and the triangle inequality would then force that to be "small." But this is not something you want. It seems to me that the triangle inequality is not an adequate assumption to impose on what you are trying to model.<|endoftext|> -TITLE: When is a complete fan a normal fan? -QUESTION [7 upvotes]: Is there a characterization for when a complete fan in $\mathbf{R}^n$ is the normal fan of a polytope? Thanks! - -REPLY [4 votes]: The characterisation is as follows: There should exist a piecewise linear convex function of the fan, linear at each face of top dimension and having different gradients at all these top-dimensional faces. -Indeed if you have a convex polytope $P$ with vertices $v_i$ in $\mathbb R^n$ this defines you a collection of linear functions $v^*_i$ on $\mathbb R^{n*}$ and the function $\max_i (v^*_i)$ will be a convex function of the dual fan satisfying the above properties.<|endoftext|> -TITLE: Does a closed immersion of an affine scheme in a smooth scheme factor over an open affine subscheme? -QUESTION [5 upvotes]: Assume we have an affine scheme $A$ that comes with a closed immersion into a smooth scheme $i \colon A \hookrightarrow M$, which is not necessarily affine. -Does there exist an affine open subscheme $j \colon V \hookrightarrow M$ such that $A$ already embeds in $V$? -Expressed in Diagrams, does there exist an affine open subscheme $V$ of $M$ such that we have a factorization - - i -A (----> M - \ / - \ / - V ? - -REPLY [5 votes]: There are also examples of smooth proper (nonprojective) varieties $V$ with a finite subset not contained in any affine open subset.<|endoftext|> -TITLE: Homotopy type of the complement to a subvariety of $\mathbb C^n$ -QUESTION [7 upvotes]: Let $V^k\subset \mathbb C^n$ be a sub variety, such that all its irreducible components have dimension $\ge k$. Is it true that $\mathbb C^n\setminus V^k$ has homotopy type of a CW complex of dimension $\le 2n-k-1$? -Comments. 1)This is true for $k=n-1$, since in this case $\mathbb C^n\setminus V^{n-1}$ is affine. Case $k=0$ is trivial. -2) This question would help to answer: -An analogue of Lefschetz hyperplane theorem for complements to subvarieties in $\mathbb C^n$ ? - -REPLY [8 votes]: Take $n=4$ and let $V = \{ z_1=z_2=0 \} \cup \{ z_3=z_4=0 \}$. I claim that $\mathbb{C}^4 \setminus V$ is homotopic to $S^3 \times S^3$, which has nontrivial homology in degree $6$, contrary to your supposed bound, which is in degree $5$. -Note that $\mathbb{C}^4 \setminus V = \left( \mathbb{C}^2 \setminus \{ (0,0) \} \right)^2$. Taking the quotient by $\mathbb{R}_{+}$, we see that $\mathbb{C}^2 \setminus \{ (0,0) \}$ is -homotopic to $S^3$, so $\mathbb{C}^4 \setminus V$ is homotopic to $S^3 \times S^3$. - -I can prove the required cohomology vanishing if you require that $V$ be Cohen-Macaulay. -Write $U$ for $\mathbb{C}^4 \setminus V$. We have the Hodge-de Rham spectral sequence: $H^q(U, \Omega^p) \implies H^{p+q}(U, \mathbb{C})$. Singe $U$ is an open subset of $\mathbb{C}^n$, we have $\Omega^p \cong \mathcal{O}^{\oplus \binom{n}{p}}$ so $H^q(U, \Omega^p) \cong H^q(U, \mathcal{O})^{\bigoplus \binom{n}{p}}$. -We can identify $H^q(U, \mathcal{O})$ with a local cohomology module of $V$, which the Cohen-Macaulay condition should force to be $0$ for $q > n-k-1$. So $H^q(\Omega^p)$'s will be zero for $q>n-k-1$. Then the spectral sequence immediately forces cohomology to vanish for $p+q > n+(n-k-1)$, as you desired. -I have no idea of how to get a statement in homotopy out of the Cohen-Macaulay condition.<|endoftext|> -TITLE: Techniques for computing cup products in singular cohomology -QUESTION [17 upvotes]: Suppose that we are given a CW complex X in terms of the cells and the gluing maps. My understanding is that computing the cup product of the singular cohomology ring from this information is a non-trivial task. I know of two basic strategies that one might take: -1) If the X is homotopy equivalent to a closed oriented manifold, then we can translate from cup product into intersection product and the problem becomes easier to visualize. -2) If X is not too complicated, then we can try to find a simple presentation of X as a finite simplicial complex and compute the cup product explicitly for all the cochains. -My question is: what are other techniques/tricks that can be used to find the cup product? -Surely there must be some general approaches beyond the naive ones I mentioned. Feel free to strengthen the hypotheses or consider specific situations, as I don't expect there to be one trick which works for everything. - -REPLY [29 votes]: This is going to be a perhaps tendentious diatribe. But it is what is. -Naturality, dimensional arguments, and Poincare duality give a -reservoir of elementary examples such as spheres and projective spaces. -In practice, to go from there to more serious examples, one uses spectral -sequences to bootstrap up, and then one uses still more spectral sequences to -bootstrap up to still more serious examples. The dirty secret is that modern -algebraic topologists rarely if ever try to compute cup products by use of -cochains, which means that they rarely if ever use cochains for serious -calculations. The huge range of known calculations show how well this works. -The actual diagonal map $X\longrightarrow X\times X$ can only be helpful -in the very simplest examples, for the obvious reason that explicit -calculations must use cellular cochains (not singular, which are far too -large for explicit computation), and the diagonal map is never cellular: -it takes the n-skeleton to the 2n-skeleton. To compute cup products with -cellular cochains, one must find a cellular map homotopic to the diagonal -map. While such a cellular approximation always exists, it is rarely an -easy task to write one down explicitly. -Peter May<|endoftext|> -TITLE: Codes, lattices, vertex operator algebras -QUESTION [22 upvotes]: At the end of "Notes on Chapter 1" in the Preface to the Third Edition of Sphere packings, lattices and groups, Conway and Sloane write the following: - -Finally, we cannot resist calling attention to the remark of Frenkel, Lepowsky and Meurman, that vertex operator algebras (or conformal field theories) are to lattices as lattices are to codes. - -I would like to understand better what the precise analogy is that is being made here. -Through my attempts to read Frenkel, Lepowsky and Meurman's book, I am aware of the story about how the "exceptional" objects, -Golay code ---> Leech lattice ---> Moonshine module, -form a hierarchy with increasingly large symmetry groups, -Mathieu group M24 ---> Conway group Co1 ---> Monster group, -and how this hierarchy led to the conjecture that uniqueness results for the Golay code and Leech lattice carry over to a uniqueness property of the Moonshine module. Frenkel, Lepowsky and Meurman speak of many analogies between the theories of codes, lattices, and vertex operator algebras. I have some understanding of the connections between codes and lattices, but so far very little understanding of vertex operator algebras and of their connection with lattices (despite having a bit of relevant physics background in conformal field theory). -My questions are - -Are the parallels alluded to above a peculiar feature of these exceptional structures, or something that holds more generally? -Is there a "baby example" one can look at of these correspondences - something based on smaller and more elementary objects? - -REPLY [10 votes]: I think the analogy you describe cannot be made precise with our current technology. For example, the word "functor" doesn't seem to have made an appearance yet in this context. -If you have a code, there are methods to construct lattices using it, but some of the constructions (like the Leech lattice) require special properties of the code. If you have a lattice, there are methods to construct vertex operator algebras using it (e.g., the lattice VOA), but some of the constructions, like orbifolds, depend on properties of the lattice, like the existence of automorphisms of certain orders. In the case of the moonshine module, we need the $-1$ automorphism of Leech, which isn't particularly special for lattices. Conjecturally (see work of Dong, Mason, and Montague 1994-95), we could use any fixed-point free automorphism of Leech to get an isomorphic VOA, and that is somewhat more special. -One class of "baby examples" that arise is when you take the root lattice of a simple (or more generally, reductive) algebraic group. This lattice has an action of the Weyl group. The lattice vertex operator algebra of this lattice has an action of the corresponding Kac-Moody Lie algebra (more generally, I think the centrally extended loop group acts). This is one of the more natural ways to construct $E_8$ from its lattice. -I'm afraid I'm not qualified to describe good baby examples of the transition from codes to lattices.<|endoftext|> -TITLE: A bound on the top homology of a complement to a variety in $\mathbb C^n$ -QUESTION [6 upvotes]: Let $V$ be a subvariety of $\mathbb C^n$ with irreducible components of dimension >$0$. Is $H_{2n-1}(\mathbb C^n\setminus V)=0$? - -REPLY [3 votes]: The sharp bound is this: For any closed algebraic set $V$ of codimension $d$ in ${\Bbb C}^n$, with $U={\Bbb C}^n \setminus V$, one has $\pi_i(U) = 0$ for $0 < i\leq 2d-2$ and $\pi_{2d-1}(U) \neq 0$. Using the Hurewicz isomorphism, you get the same vanishing and non-vanishing for homology. -A simple proof is in the appendix to my notes (with Fulton) on equivariant cohomology, http://www.math.washington.edu/~dandersn/eilenberg/ . A slick reason for vanishing was pointed out by David Speyer: given a (nice) map of an $i$-sphere into $U$, the (real) lines between the points in image of the sphere and points in $V$ sweep out a space of dimension at most $(2n-2d)+i+1$. When $i<2d-1$, you can pick a point in $U$ not lying on any such line, and contract your sphere down to that point. The non-vanishing happens because all algebraic sets have nontrivial fundamental classes in Borel-Moore homology. (Vanishing can also be proved using B-M homology.)<|endoftext|> -TITLE: Probability Problem Involving e -QUESTION [16 upvotes]: I thought of the following probability problem, which seems to have an answer of 1/e, and wonder if someone has an idea as to how to prove this. -Suppose a man has a bottle of vitamin pills and wishes to take a half pill per day. He selects a pill from the bottle at random. If it is a whole pill he cuts it in half, takes a half pill, and puts the other half back in the bottle. If it is a half pill, he takes that. He continues this process until the bottle is empty. What is the expected maximum number of half pills in the bottle? If the bottle starts with n pills, and M is the expected maximum number of half pills, then M/n appears to tend to 1/e as n tends to infinity. - -REPLY [12 votes]: The problem is equivalent to the following: Suppose there are $n$ bins, and repeatedly, we throw balls which fall in one of the bins (uniformly and independently of the history). What is the maximum number of bins with exactly one ball? In the model with the pills, it is explicitly forbidden to draw a pill which has been drawn twice already, but for the current question, this clearly doesn't matter. -We can replace discrete time with continuous time, and throw the balls at the events of a Poisson process. This way, the balls falling in a particular bin arrive according to a Poisson process, and different bins are independent. If the problem was to determine the time $t$ to maximize the expected number of bins with exactly one ball at time $t$, then the answer would clearly be to choose $t$ so that the expected number of balls in a bin is 1, and the probability of having exactly one ball would be $1/e$ (we are maximizing $xe^{-x}$). -By the law of large numbers, the number of bins with exactly one ball will very likely be about $n/e$ at that time. To conclude that $M/n$ converges to $1/e$ in probability, it only remains to show that "exceptional times" with unusually many bins of exactly one ball are not likely to occur. I guess this can be established by quantifying the idea that if at time $t$ there are substantially more than $n/e$ bins with exactly one ball, then most likely there will continue to be so in a time interval after $t$, which is unlikely.<|endoftext|> -TITLE: Is anything known about this braid group quotient? -QUESTION [19 upvotes]: Let $B_n$ be the braid group on $n$ strands. As is well known, if $\sigma_i$ is the operation of crossing the string in position $i$ over the string in position $i+1$, then the elements $\sigma_1,\dots,\sigma_{n-1}$ generate $B_n$ and the relations $\sigma_i\sigma_j=\sigma_j\sigma_i$ ($|i-j|\geq 2$) and $\sigma_i\sigma_{i+1}\sigma_i=\sigma_{i+1}\sigma_i\sigma_{i+1}$ give a presentation of the group. -I am interested in a group that can be obtained from $B_n$ by adding another set of relations. For each $k$, define $T_k$ to be the "twist" of the first $k$ strands. Geometrically, you take hold of the bottom of the first $k$ strands and rotate your hand through 360 degrees in such a way that strands to the left go over strands to the right. In terms of the generators, $T_k=(\sigma_1\dots\sigma_{k-1})^k$, since $\sigma_1\dots\sigma_{k-1}$ takes the $k$th most strand and lays it across the next $k-1$ strands. Let us also define $S_k$ to be the right-over-left twist of the strands from $k+1$ to $n$. That is, $S_k=(\sigma_{k+1}\dots\sigma_{n-1})^{-(n-k)}$. (Also, let us take $T_1$ and $S_{n-1}$ to be the identity.) -I am interested in the group you get if you start with $B_n$ and add in the relation $T_kS_k=1$ for every $k$. Let me make some utterly trivial observations. -If $n=2$ then we get the cyclic group $C_2$. That's because $\sigma_1$ is the only generator and $T_2S_2=\sigma_1^2$. If $n=3$ then we get $S_3$. That's because $T_1S_1=\sigma_2^{-2}$ and $T_2S_2=\sigma_1^{-2}$, so the relations we are adding are $\sigma_1^2$ and $\sigma_2^2$, and it is well known that those, together with the braid relations, give a presentation of the symmetric group. -Beyond that I don't know what to say, though I've convinced myself (without a proof) that when $n=4$ the group is infinite: in general, it seems that the extra relations can be used to do only a limited amount of untwisting. (I do have a proof that there are pure braids that cannot be reduced to the identity once we have four strands. It's a fairly easy exercise and I won't give it here.) -What exactly is my question? Well, I'd be interested to know whether the word problem in this group is soluble in reasonable time. In the service of that, I'd like to know whether this group is one that people have already looked at, or whether it at least belongs to a class of groups that people have already looked at. (E.g., perhaps the solubility of the word problem follows from some general theory.) And is there some nice way of characterizing the subgroup of $B_n$ that we are quotienting by? That is, which braids belong to the normal closure of the set of braids $T_kS_k$? (One way of answering this would be to characterize their normal forms.) -The motivation for the question comes from part of an answer that Thurston gave to a question I asked about unknots. It seems to me that this question ought to be relevant to the untying of unknots, but easier. -One final remark: the word problem in $B_n$ can be solved in polynomial time. (If my understanding is correct, this is a result of Thurston that built on work of Garside.) Since adding more relations makes more braids equal to the identity but also gives more ways of converting a word into another, it is not clear whether the problem I am asking should be easier or harder than the word problem for braid groups. However, my hunch is that it is harder (for large $n$, that is). - -REPLY [4 votes]: THIS ANSWER IS WRONG. See the comment by Gowers. The mapping class group of the n-punctured sphere, which the question was about, is not the same as the spherical braid group. It is obtained from the spherical braid group by quotienting out by a full twist of all the strands. The original answer is below for posterity. -As others have pointed out, this is the braid group on S2. These groups were studied a bit during the 1960's, and then everyone seems to have forgotten all about them. The only book which seems to give braid groups over S2 more than a passing mention is Murasugi and Kurpita's monograph. Only recently does it seem that people have become interested again, because of "braid groups over surfaces" (discussed here for example) and because of the amazing Berrick-Cohen-Wong-Wu paper which connects Brunnian braids over a sphere to homotopy groups of S2! Their result is one of those mysterious alluring connections which makes mathematics worth doing. John Baez wrote a nice blog post about that story. -It is proven by Fadell and van Buskirk that the braid group over S2 is presented by the braid generators $\sigma_1,\ldots,\sigma_n$ modulo the relations: - -$\sigma_i\sigma_j\sigma_i=\sigma_j\sigma_i\sigma_j$ if $\left\vert i-j\right\vert=1$. -$\sigma_i\sigma_j=\sigma_j\sigma_i$ if $\left\vert i-j\right\vert>1$. -$(\sigma_1\sigma_2\cdots\sigma_{n-1})(\sigma_{n-1}\cdots\sigma_2\sigma_1)=1$. - -Its word problem is solved in a very nice paper by Gillette and van Buskirk by modifying the combing algorithm for braids to work for braids over S2. This has exponential run-time, the same as the combing algorithm for the braid group. But first Thurston ("Word Processing in Groups"), and then Dehornoy, modified the combing algorithm to work in polynomial time, I don't know whether their algorithms can be made to work for the braid group over S2; my intuition is that the answer is surely yes, but surely nobody has bothered doing it yet, because braid groups over S2 are well forgotten. -EDIT: This is part of Open Question 9.3.10 in Word Processing in Groups (see also Open Question 9.3.9). I don't know whether it has been solved in the two decades since that book's publication.<|endoftext|> -TITLE: Height of algebraic numbers -QUESTION [9 upvotes]: I would like to find effective upper bound for the height of $a+b$ and $a/b$ and $ab$ knowing the heights of $a$ and $b$. Thanks. - -REPLY [2 votes]: I too was looking for the answer to the same question. It seems necessary to define "height" since there are of course several variants in use. The height $h(a)$ I am interested in is the maximum of the absolute values of the coefficients of the minimal polynomial of the algebraic number $a$. Note that the first answer refers to a different notion of height, since, for example, -$9 = h(9) = h(3 \cdot 3) \not \leq h(3) + h(3) = 3 + 3 = 6.$ -I imagine there must be upper bounds of the form -$h(ab) \leq f(d) h(a)^{g(d)} h(b)^{g(d)}$ -for some simple functions $f$ and $g$, where $d$ is the degree of a field extension of $\mathbb{Q}$ containing both $a$ and $b$, for example. -Are there any such results in the literature, and similarly for $h(a+b)$ and $h(a/b)$?<|endoftext|> -TITLE: Lower bounds on the easier Waring problem -QUESTION [18 upvotes]: The easier Waring problem asks for the least number $v=v(k)$ such that every every integer is a sum of $v$ $k$'th powers with signs, i.e. every $n\in \mathbb{N}$ is of the form $$n=x_1^k\pm x_2^k\pm\dotsb\pm x_v^k.$$ -The problem is ``easier'' because unlike the usual Waring problem (without the signs) the existence of $v(k)$ is easy --- the bound $v(k)\leq 2^{k-1}+\tfrac{1}{2}k!$ follows from the repeated differencing. Of course, the upper bounds on the usual Waring problem apply, and so fact $v(k)=O(k\log k)$. -All the lower bounds on $v(k)$ I have seen come from the congruence considerations. For example, $v(3)\geq 4$ because we need at least four terms modulo $9$. However, if we discard the congruential obstacles is there a non-trivial lower bound? To put bluntly my question is - -Is there $k$ large enough so that the set $\{x_1^k\pm x_2^k\pm x_3^k\pm x_4^k\pm x_5^k\}$has zero density? - -REPLY [19 votes]: As far as I am aware, nothing unconditional is known. The difficulty is that one has no obvious constraint on the size of the variables, so that the $x_i$ could be arbitrarily large in terms of $n$ in a solution. The difficulty of ruling out solutions (when congruence conditions do not rule out solubility) is related to proving insolubility of generalised Fermat equations. -However, there is a conditional approach assuming the truth of the generalised ABC Conjecture (or, as Pomerance describes it, the Alphabet Conjecture). Let's stick with the general situation with $v$ summands. Generalised ABC asserts that in any solution of -$a_1+\ldots +a_s=0$ in which there are no vanishing subsums, one has -$\underset{1\le i\le s}{\max}|a_i|\ll_{s,\epsilon} \left( \prod_{p|a_1\ldots a_s}p\right)^{(s-1)(s-2)/2+\epsilon}.$ There is disagreement about the specific exponent, but this does not matter so much in the conclusion (see a 1986 paper of Brownawell and Masser for a function field version, and I discuss this in a 1994 paper on Quasi-diagonal behaviour). The vanishing subsums in your equation $\pm x_1^k+\ldots +\pm x_v^k=n$ make life easier (treat them separately, and more powerful estimates are possible), so for simplicity suppose that the representations all have no vanishing subsums. Then one obtains $|x_i|^k\ll |nx_1\ldots x_v|^{v(v-1)/2+\epsilon}$. Provided that $k>v^2(v-1)/2$, it follows that $\max |x_i|\ll |n|^{\alpha+\epsilon}$, where $\alpha=v(v-1)/(2k-v^2(v-1))$. OK ... so far so good. What we have shown thus far is that the variables in a representation are bounded by $|n|^{\alpha +\epsilon}$. The total number of variables available to represent the integers $n$ between $N/2$ and $N$ is consequently no larger than a quantity which is $\ll (N^{\alpha+\epsilon})^v$ (there were $v$ variables). Whenever $v(\alpha+\epsilon)<1$, therefore, the set of integers represented must have zero density. If I have not made any computational errors along the way, this leads us to the conclusion that whenever $k>v^2(v-1)$, then the density of integers represented in the easier Waring problem will be zero. (But remember that this is all conditional on the Generalised ABC Conjecture.) For the specific problem with $v=5$, it looks as if $k>100$ conjecturally does the trick (though smaller $k$ should surely also work).<|endoftext|> -TITLE: When are these rings regular? -QUESTION [5 upvotes]: Let $R$ be a noetherian regular domain. Suppose that $a, b \in R$, with $b \neq 0$, and consider the ring $S:=R[\frac{a}{b}]=R[X]/(bX-a)$. Is $S$ regular? If this is not the case are there some conditions on $a$ and $b$ (or on $R$) that imply regularity? For example $a=1$ is enough. - -REPLY [3 votes]: If $R = k[x,y]$ and we throw in $\frac {x^2}y$, we get $k[x,y,z]/(zy-x^2)$, not regular. This shows that it's not enough to assume $a,b$ form a regular sequence. The only sufficient condition I can come up with is that $a$ be outside the square of (every/the) maximal ideal<|endoftext|> -TITLE: Embedding $S_3$ into $Aut(F_2)$ -QUESTION [7 upvotes]: Consider the two (inequivalent) $\mathbb{Z}$-representations $\phi,\psi$ of the symmetric group $S=S_3$ given by -$(1,2)^\phi=\left(\begin{array}{rr}0 &-1\\\ -1 & 0\end{array}\right), \qquad -(1,2,3)^\phi=\left(\begin{array}{rr}0 &1\\\ -1 & -1\end{array}\right);$ -$(1,2)^\psi=\left(\begin{array}{rr}0 &1\\\ 1 & 0\end{array}\right), \qquad -(1,2,3)^\psi=\left(\begin{array}{rr}0 &1\\\ -1 & -1\end{array}\right).$ -Now, let $F=\langle x,y\rangle$ be a free 2-generated group. The representation $\phi$ can be "lifted" to an embedding $\tau:S\to\rm{Aut}(F)$ as follows: -$(1,2)^\tau=[x\mapsto y^{-1};\quad y\mapsto x^{-1}], \qquad -(1,2,3)^\tau=[x\mapsto y;\quad y\mapsto x^{-1}y^{-1}].$ -Question. Can one similarly lift $\psi$? -Remark 1. By "lifting" a representation $\phi:S\to\rm{GL}_2(\mathbb{Z})$ I mean finding an embedding $\tau:S\to\rm{Aut}(F)$ such that $\phi=\tau\alpha$, where $\alpha:\rm{Aut}(F)\to\rm{GL}_2(\mathbb{Z})$ is the natural epimorphism. -Remark 2. A naïve attempt to send -$(1,2)\ \mapsto\ [x\mapsto y;\quad y\mapsto x], \qquad -(1,2,3)\ \mapsto\ [x\mapsto y;\quad y\mapsto x^{-1}y^{-1}]$ -does not give a lifting of $\psi$. - -REPLY [8 votes]: As Tom Goodwillie noted in his comment, $GL(2,\Bbb Z)$ can be identified with $Out(F_2)$, so the question can be rephrased in terms of lifting subgroups of $Out(F_2)$ to $Aut(F_2)$. There is a Realization Theorem for finite subgroups of $Aut(F_n)$ and $Out(F_n)$ which says that such a subgroup can always be realized as a group of symmetries of some finite connected graph with fundamental group $F_n$, where the symmetries fix a basepoint in the graph in the case of $Aut(F_n)$. When $n=2$ there are only two graphs to consider, and the relevant one for $S_3$ is the join of two points with three points. This has two symmetry groups isomorphic to $S_3$, but only one of these two groups fixes a basepoint, so this should answer the question. -The Realization Theorem is discussed in Karen Vogtmann's survey paper "Automorphism groups of free groups and outer space", section II.6. The references given there are to papers by M. Culler, B. Zimmermann, and D. G. Khramtsov from 1981 to 1984.<|endoftext|> -TITLE: Finite graphs that realize all types over $n$-element sets -QUESTION [6 upvotes]: Call a graph $G$ $n$-saturated if for every set $A$ of size $n$ of vertices and all $B\subseteq A$ there is a vertex $v\not\in A$ that forms an edge with all $w\in B$ and -does not form an edge with any $w\in A\setminus B$. -A countably infinite graph is isomorphic to the random graph iff it is $n$-saturated for all $n$. -Here are the questions: - -Is it true that for all $n$ there is a finite graph (of size at least $n$) that is $n$-saturated? -If yes, are there reasonable upper and lower bounds on the size of such a graph? -(An $n$-saturated graph of size $\geq n$ has at least $n+2^n$ vertices, but this is probably far from optimal.) -Is every finite graph an induced subgraph of a finite $n$-saturated graph? -(Edit: As Ori Gurel-Gurevich pointed out, the second question is silly. Clearly, if $G$ is $n$-saturated (and of size at least $n$), then it has every graph with $n$-vertices as an induced subgraph.) - -Example: There is a $2$-saturated graph: Let the vertices of $G$ be the $2$-element subsets of a set with $6$ elements. Two of those vertices form an edge if they (as sets) have a non-empty intersection. It is easily checked that this graph is $2$-saturated. -But it also has 30 vertices. That seems a lot. - -REPLY [7 votes]: Ben Rossman found a nice construction of $n$-saturated graphs; I adjusted it a bit and wrote it up. See http://research.nii.ac.jp/~rossman/k-ec.pdf . Set theorists like Stefan might enjoy the fact that it's reminiscent of Hausdorff's construction of independent families of (infinite) sets.<|endoftext|> -TITLE: Non-vanishing of group cohomology in sufficiently high degree -QUESTION [21 upvotes]: Atiyah in his famous paper , Characters and cohomology of finite groups, after proving completion of representation ring in augmentation ideal is the same as $ K(BG)$, gives bunch of corollaries of this main theorem. One of them that catches my interest is: For any finite non-trivial group $G$ there exists arbitrary large integer $n$ such that $H^n(G,\mathbb{Z})\neq 0 $. I just wonder if anyone can prove this without this powerful theorem. - -REPLY [11 votes]: The first purely algebraic proof of this fact seems to be from Leonard Evens: - -A Generalization of the Transfer Map in the Cohomology of Groups, Trans. Amer. Math. Soc. 108(1963), 54-65 [Theorem 3] - -where he proves the result with help of his norm map. After having established the basic properties of the norm map, the proof is rather elementary: Let $C$ be a cyclic subgroup of prime order of $G$ and let $x$ be a generator of $H^2(C,\mathbb{Z})$. Then the powers of $y = N^G_C(x)$ yield non-trivial cohomology classes of $H^*(G,\mathbb{Z})$ in degrees divisible by $(G:C)$. -From a historical point of view the norm map already occured in disguise in Evens´ paper - -The Cohomology Ring of a Finite Group, Trans. Amer. Math. Soc. 101(1961), 224-239 - -where he proves finte generation of the cohomology ring.<|endoftext|> -TITLE: Realizing the diameter of a finite regular graph -QUESTION [5 upvotes]: Let $X=(V,E)$ be a finite, connected, regular graph with diameter $D$. Is it true that, for every $x\in V$, there exists $y\in V$ such that $d(x,y)=D$? (the answer is clearly yes if $X$ is vertex-transitive). - -REPLY [6 votes]: Counter-example http://www.freeimagehosting.net/uploads/9a4165cab4.png -The diameter is 8, but 1 is centered with at most 5 as distance to every other.<|endoftext|> -TITLE: Smooth proof of Reidemeister theorem -QUESTION [8 upvotes]: In another post Ryan Budney has mentioned a "smooth proof" of the theorem of Reidemeister... -(@Ryan Budney:) Do you know if there is a book (or paper) containing such a proof? I am interested in and can't find one... -Thank you very much, -sincerely, -Johannes Renkl - -REPLY [12 votes]: I don't believe it's written up anywhere. -edit: in the comments Charlie Frohman corrects me: - -MR2128054 (2005m:57041) - Roseman, Dennis(1-IA) - Elementary moves for higher dimensional knots. (English summary) - Fund. Math. 184 (2004), 291–310. - 57R40 (57R45 57R52) DOI: 10.4064/fm184-0-16, eudml - -has a proof similar in flavour to the sketch below. Thanks Charlie! - -The idea is pretty simple, but maybe a bit of a pain to write up completely. Say $f : [0,1] \times S^1 \to \mathbb R^3$ is a smooth isotopy. Let $\pi : \mathbb R^3 \to \mathbb R^2$ be orthogonal projection onto a 2-dimensional subspace. $\pi \circ f : [0,1] \times S^1 \to \mathbb R^2$ is a 1-parameter family of (possibly singular) knot diagrams, but we can assume that at the initial and terminal parts of the family are regular knot diagrams. -The next step is to check the co-dimensions of the various types of singularities such maps can have. -For example, a knot has its derivative. That derivative is "vertical" is a co-dimension 2 condition on a vector. Since a knot is 1-dimensional, it's a co-dimension 1 condition on the knot. Such a vertical derivative for $f$ results in its projection to $\mathbb R^2$ to have a cusp. So a generic knot diagram has no cusps, but a 1-parameter family will have a finite number of cusps, and these appear as Reidemeister moves of "type 1". Technically, this is a formulation of Whitney's theorem that maps between 2-manifolds generically only have fold and cusp-type singularities. -Further, you can ask about double, triple and multiple-points for the projections, non-transverse double points, and so on. Double points are generic but triple-points are co-dimension 2, and $n$-tuple points are co-dimension $2(n-2)$ in general, so you generically can avoid anything worse than triple points. This gives the Reidemeister "type 3" moves. Tangential double points are also co-dimension 2, and generic such ones produce Reidemeister "type 2" moves. -As you can see this isn't quite a full proof using only the technology of Guillemin and Pollack, meaning it's not quite just Sard's theorem that we're using. We're really using the "Multijet transversality" type theorem, like Theorem 4.13 from Golubitsky and Guillemin. But I suspect like Whitney's work "The general type of singularity of a set of $2n-1$ functions of $n$ variables" you should be able to massage the above into a solid argument without dredging up too much formalism. -edit: I do think there is likely a "smart" proof of this that avoids jet transversality. For example, check out Milnor's proof that Morse functions are dense in the space of smooth functions $M \to \mathbb R$ (in Milnor's Morse Theory text). You'd think in principle you'd have to use jet transversality for this but by picking a diverse-enough family of functions to start with, he pulls it all back to the level of Sard's Theorem. I think there should be a similarly slick proof of Reidemeister's theorem, using only Sard's Theorem.<|endoftext|> -TITLE: Smooth structures compatible with a given C^1 structures -QUESTION [10 upvotes]: On a manifold equipped with C^k atlas (with k>0) there is essentially one smooth structure compatible with the atlas. According to Wikipedia, this is a result due to Whitney. This is in stark contrast with a C^0 atlas, where there might exist many smooth structures or none at all. -I was wondering, what is the underlying reason? What makes once-differentiable functions so much better behaved in terms of finding a smooth atlas? -There are many cases where C^1 makes a world of difference - for example, convergence of Fourier series, but maybe there is some geometric explanation? -Also according to Wikipedia, on the long line (not technically a manifold) there are infinitely many smooth structures all compatible with a given C^k structure, so perhaps there is some topology involved... - -REPLY [15 votes]: See my answer to this question. -Consider the space (in some appropriate sense ) of all invertible germs of $C^{\infty}$ maps from $\mathbb R^n$ to itself fixing the origin. This is homotopy equivalent to $GL_n(\mathbb R)$: we can deform the smooth map $f$ to its linear approximation by going through $f_t(x)=\frac{f(tx)}{t}$ as $t$ goes to $0$. The same applies to invertible $C^k$ germs for finite $k>0$, but not to invertible $C^0$ germs. (All of this is equally true for global diffeomorphisms/homeomorphisms $\mathbb R^n\to\mathbb R^n$, too. )<|endoftext|> -TITLE: How should one think about pushforward in cohomology? -QUESTION [47 upvotes]: Suppose f:X→Y. If I decorate that first sentence with appropriate adjectives, then I get a pushforward map in cohomology H*(X)→H*(Y). -For example, suppose that X and Y are oriented manifolds, and f is a submersion. Then such a pushforward map exists. In the de Rham picture, we can see this as integrating a form over fibres. In the sheaf cohomology picture, we can see this via the explication of the exceptional inverse image functor. -The question is how else can we think of this pushforward map. I'd be particularly interested in an answer from the algebraic topology point of view, because I'm hoping that such an answer would eludicate the appropriate level of generality in which a pushforward in cohomology exists (perhaps not only answering the question of for which maps f, but also answering the question of in which cohomology theories can we carry out such a construction). - -REPLY [9 votes]: Parallel to the OP's two examples, if a cohomology class is defined through intersection with a submanifold (or subvariety with fundamental class in locally finite homology) then the pushforward is defined through intersection with the image of this subvariety. Note how immediate it is to see the change in codimension ( while codimension is constant when taking preimage, which corresponds to the natural map). -As others have said, collapse maps can be used to define pushforwards in general. But having models for pushforwards counts towards understanding a particular cohomology theory geometrically. So for example if you can find a good model for the pushforward of tmf then "you'll win a prize"<|endoftext|> -TITLE: Existence of a nice subset of edges in $k-$regular simple graphs? -QUESTION [6 upvotes]: Let $G=(V,E)$ be a finite simple $k-$regular graph ($k\geq 1$). Does $G$ necessarily -contain a subset $E'\subset E$ of edges such that only isolated edges and cycles occur as connected components in $(V,E')$? -(The answer is easily yes for $k=1,2$.) -A counterexample would easily give a counterexample to question "Antipodal" maps on regular graphs? in the case $D=2$ by considering -the complementary graph of $G$ (respectively of two disjoint copies of $G$ if $G$ is -"too small"). - -REPLY [6 votes]: It seems like such a subset should always exist. -Consider the bipartite graph on $2|V(G)|$ vertices corresponding to the adjacency matrix of $G$. Since this graph is regular, by Hall's Theorem it has a perfect matching. In terms of the original $G$, this corresponds to a permutation $\sigma$ on $V(G)$ such that $v$ and $\sigma(v)$ are always adjacent. The cycles of $\sigma$ would then give you the desired decomposition.<|endoftext|> -TITLE: Flatness of normalization -QUESTION [18 upvotes]: Let $X$ be a noetherian integral scheme and let $f \colon X' \to X$ be the normalization morphism. It is known that, if non trivial, $f$ is never flat (see Liu, example 4.3.5). -What happens if we suppose $X$ normal, and we take the normalization in a finite (separable) extension of the function field of $X$? Note that in the easiest case, namely $X=\rm{Spec}(R)$, with $R$ a Dedekind domain, we have that $f$ is flat. - -REPLY [4 votes]: If $f: X' \to X$ is flat, then $X$ tends to inherit nice properties of $X'$ (for example, being regular). So if you arrange for $X'$ to be "nicer" than $X$, as in David and one of Karl's examples, then $f$ can't be flat. -Here is a quick and dirty proof when "nice" = "regular". The claim is that if $R\to S $ is a finite flat local homomorphism of Noetherian local rings and $S$ is regular, then $R$ is regular as well. -Let $m$ be the maximal ideals of $R$. Then as $S$ is regular, $S/mS$ has finite flat dimension (in fact, projective dim) over $S$. But $S$ is flat over $R$, so $S/mS$ has finite flat dimension over $R$. But as $R$-modules, $S/mS$ is direct sum of copies of $k=R/m$, so $k$ has finite flat dimension over $R$, which characterizes regularity.<|endoftext|> -TITLE: Rolling-ball game -QUESTION [23 upvotes]: The analyses -in two recent MO questions -("recent" with respect to the original posting in 2011), -"Rolling a random walk on a sphere" -and -"Maneuvering with limited moves on $S^2$," -suggest a Rolling-Ball Game, as follows. -A unit-radius ball sits on a grid point of -a $\delta \times \delta$ regular grid in the plane, -with $\delta \neq \pi/2$. -Player 1 (Blue) rolls the ball to an adjacent -grid point, and the track of the ball-plane contact point -is drawn on the ball's surface. -Player 2 (Red) rolls to an adjacent grid point. -The two players alternate until each possible -next move would cause the trace-path to touch itself, -at which stage the player who last moved wins. -In the following example, Red wins, as Blue cannot -move without the path self-intersecting. - -           - - -Q1. -What is the shortest possible game, assuming the players cooperate -to end it as quickly as possible? -For $\delta=\pi/4$, the above example suggests 6, but -this min depends on $\delta$. It seems smaller $\delta$ need 8 moves to create -a cul-de-sac? -Q2. -What is the longest possible game, assuming the players cooperate -to extend it as much as possible? -Q3. -Is there any reasonable strategy if the players are truly competing -(as opposed to cooperating)? -Addendum. -We must have $\delta < 2 \pi$ to have -even one legal move, -and the first player wins immediately with one move for $\pi \le \delta < 2\pi$ (left below). - -The right image just shows a non-intersecting path of no particular significance for $\delta=\pi/8$. - -REPLY [7 votes]: For testing potential answers to Q3, let me suggest http://www.math.chalmers.se/~wastlund/Quirks/Game.html. -The appearance as well as the specific set of bugs might depend on your browser, but as I hope will be clear, the length of the longest path as a function of the angle $\delta$ is quite funny (this is part of the reason I spent too much time on this problem, although I'm also interested in similar path-forming games for more "serious" reasons). -After drawing a couple of quick sketches I realized I wasn't even able to figure out the behavior when $\delta$ approaches $\pi$ from below. As it turns out, the number of edges in the longest path is 5 throughout the interval $3\pi/4 \leq \delta < \pi$. For $\pi/2 < \delta < 3\pi/4$ it's 7 (although the game-tree changes also at $2\pi/3$). At $\delta = \pi/2$ it has an isolated local minimum of 5, and for angles just smaller than $\pi/2$, it seems to jump to 23. -Here's why I couldn't let go of this problem: In all the cases where I can visualize the entire game-tree, which is when $\delta\geq \pi/2$, -(1) Alice, who draws the first arc, wins the game-version. -(2) Bob, while losing, can force Alice into a maximal-length path. In other words, Alice cannot force a win in fewer moves than the length of the longest path. This property holds for some similar path-forming games where there are explicit winning strategies. -(3) As a consequence of (1) and (2), the length of the longest path is always odd. -A couple of other observations: If we allow players to cross their own edges but not the opponent's, then an edge is never a liability, and consequently Alice has a non-losing strategy. If on the other hand we allow players to cross the opponent's edges but not their own, then an extra edge cannot be an advantage, and now Bob has a non-losing strategy (two more questions arise here: are these games too necessarily finite, and can the (winning?) strategies be made explicit rather than just exhibited by strategy-stealing?). -Therefore I thought for a while that I was on to something, and that there might be a beautiful reason that Alice must win. -So I let Maple analyze the game for some suitably chosen angles, working symbolically to get reliable results. This is feasible when $\delta$ is such that the ring $\mathbb{Z}[\cos\delta, \sin\delta]$ where the coordinates of the points lie, is either a sub-ring of $\mathbb{Q}$ or of some nice algebraic field (degree 2 or 4). -To summarize, I found counter-examples to each of (1), (2) and (3): For $\delta = \arctan(24/7)$, Bob wins at move 16 but the longest path has length 23. For $\delta = \arctan(12/5)$, the longest path has length 24 (but Alice can force a win in 13). Moreover, for $\delta=\pi/3$ there is a path of length 29 but Alice wins in 17. For $\delta=\arctan(4/3)$, Bob wins in 16 (I don't know the length of the longest path). -Addendum. (by J.O'Rourke). I took the liberty of adding an image of -Johan's $\pi/3$ longest path from his applet, as detailed in his comment below. -Note the near miss where the 28th segment just misses the 1st segment.<|endoftext|> -TITLE: Number of graphs with a given number of nodes, edges and triangles -QUESTION [8 upvotes]: Hi. Does anyone know if it is possible to enumerate the set of labeled/unlabeled graphs (loopless, undirected, only one edge between pairs of nodes) having a given number of nodes, edges and triangles? what about including more information, like number of tetrahedra, etc.? If not, why? -The solution to the unlabeled case with given number of nodes and edges, and to many other enumeration problems can be given in terms of Polya's theorem (by Harary, de Bruijn, Robinson, Read, Polya, etc.). Is it possible to give the resulting generating functions an interpretation in terms of order theory and use this to study their algebraic properties? -Are these generating function methods actually useful for computing these numbers when the graphs are 'large'? Thanks! -Update: Here is the list: -http://www.win.tue.nl/~aeb/graphs/cospectral/triangles.html -I haven't seen it in the OEIS, though. -Does it get any easy if we wanted to enumerate, instead, the set of graphs on given number of vertices and trangles -irrespective- of the number of edges? - -REPLY [6 votes]: Polya theory provides us with an algorithm that allows us to compute the number of isomorphism classes of graphs with $n$ vertices and $m$ edges, for given $n$ and $m$. It does not provide a formula in terms of $n$ and $m$ and it does not even provide a generating function. (Well, we can express it in terms of so-called cycle index, but that is just giving a name to the unknown.) -If now we ask for the number of graphs in terms of $n$, $m$ and the number of triangles then we are lost. I have never seen an enumeration of triangle-free graphs, let alone triangle-free graphs with a given number of edges. And if we cannot count the graphs in a given class, counting isomorphism classes of graphs in the class is usually beyond us as well. -So counting $K_4$'s as well seems even more hopeless. -In the cases where Polya's method works, it can provide asymptotic information for large $n$, but not formulas. So for large $n$ we know that the number of isomorphism classes -of graphs on $n$ vertices is asymptotic to $2^n/n!$, but there is no exact formula. -And why have these enumeration problems not been solved? Perhaps they are too hard, or we're -too stupid. (Me, at least.)<|endoftext|> -TITLE: Is there a 'best' name for a group together with a set it acts on? -QUESTION [5 upvotes]: This is a question of terminology. I want to talk about the category whose... - -...objects are pairs $(G,M)$, where $G$ is a group and $M$ is a $G$-set. -...morphisms $(G,M)\rightarrow (G',M')$ are pairs $(f_G,f_M)$, where $f_G:G\rightarrow G'$ is a group homomorphism, and $f_M:M\rightarrow M'$ is a set map such that -$$ f_M(g\cdot m) = f_G(g)\cdot f_M(m)$$ -for all $g\in G$ and $m\in M$. - -I've been calling these decorated groups (and their morphisms), since they've been arising in connection with decorated local systems. However, I'd prefer a more standard name, hopefully one which evokes the correct idea before explanation. - -REPLY [2 votes]: Angelo's answer above is clearly the correct one. But, somewhat tongue-in-cheek, I would also like to recommend "The category of trivialized groupoids". As I'm sure you're very aware, there's a pretty good analogy - -vector bundles : trivialized vector bundles :: groupoids : group actions.<|endoftext|> -TITLE: Identifying the generating function $ G(a,z) = \sum_{n=0}^{\infty} a^n z^{(n+1)(n+2)/2}. $ -QUESTION [10 upvotes]: I have computed a generating function for a problem involving a particular series, and would like to know if anyone has any references or a categorisation for it? It's -$$ -G(a,z) = \sum_{n=0}^{\infty} a^n z^{(n+1)(n+2)/2}. -$$ -It appears to be related to (mock) theta functions, but seems to be simpler. -In particular, I would like to know whether $G(a,z)$ satisfies any identities? -Many thanks. - -REPLY [5 votes]: Your generating function is related to a simple continued fraction expansion due to Touchard: -$\sum\limits_{k \ge 0} ( - 1)^k q^{k+1\choose2} v^k $ =$ \frac{1}{{1 + v - \frac{{(1 - q)v}}{{1 + v - \frac{{(1 - q^2 )v}}{ \cdots }}}}}.$ -A simple proof can be found in a paper by H. Prodinger - http://de.arxiv.org/abs/1102.5186<|endoftext|> -TITLE: space of homotopy equivalences of $S^1$ -QUESTION [5 upvotes]: Does the space of homotopy equivalences of $S^1$ deformation retract onto the space of homeomorphisms of $S^1$? If so, does anyone have a reference? -I found that Kneser proved that $Homeo(S^1)$ deformation retracts onto $O(2)$ and $Homeo^+(S^1)$ deformation retracts onto $SO(2)$ (orientation preserving homeos deformation retracts onto rotations). I'd like the space $HE^+(S^1)$ of degree 1 homotopy equivlances of $S^1$ to deformation retract onto these. The space I'm calling $HE^+(S^1)$ may go by $HomEq(S^1)$ or $SG_n$ and seems to be of interest to homotopy theorists for higher $n$. - -REPLY [3 votes]: I think Ryan Budney's comment can be made to work. In order to take a straight line homotopy, you need to figure out which rotation you are going to homotope to, and make this choice in a continuous fashion. Here's one possible choice. For a homotopy equivalence $f:S^1\to S^1$, take a lift $F:\mathbb{R}\to\mathbb{R}$. Then $F$ has the property that $F(x+n)=F(x)+n, n\in \mathbb{Z}$ ($\mathbb{Z}$-equivariant). The function $F(x)-x$ is therefore periodic, and attains a minimal value $m(f)$. Take the straightline homotopy from $F(x)$ to $x+m(f)$. This homotopes the continuous $\mathbb{Z}$-equivariant functions to the functions of the form $x+r, r\in\mathbb{R}$, and therefore descends to a homotopy of $HE^+(S^1)$ to $SO(2)$. I think it's clear that this homotopy is continuous with respect to the topology on $HE^+(S^1)$. - -REPLY [3 votes]: Put $$HE^+_1(S^1)=\{f\in HE^+(S^1):f(1)=1\}. $$ -There is an evident homeomorphism $m:S^1\times HE_1^+(S^1)\to HE^+(S^1)$ given by $m(z,f)(x)=z f(x)$, and this restricts to give a homeomorphism $S^1\times Homeo_1^+(S^1)\to Homeo^+(S^1)$. -It will thus suffice to discuss $HE_1^+(S^1)$ and $Homeo_1^+(S^1)$. -Next, define $e:\mathbb{R}\to S^1$ by $e(t)=\exp(2\pi i t)$. Let $X$ denote the space of maps $$ u:[0,1]\to\mathbb{R} $$ -with $u(0)=0$ and $u(1)=1$, and let $Y$ be the subspace of strictly increasing maps. For any $u\in X$ there is a unique map $p(u):S^1\to S^1$ with $p(u)(e(t))=e(u(t))$ for all $t\in [0,1]$. This construction gives a homeomorphism $p:X\to HE^+_1(S^1)$, and restricts to a homeomorphism $p:Y\to Homeo_1^+(S^1)$. This is a fairly straightforward exercise with covering space theory and topologies on mapping spaces. Now $X$ and $Y$ are both convex, so they have obvious contractions to the identity map given by -$$ h(t,u)(x) = tx+(1-t)u(x). $$ -It is not hard to see that $Y$ is not closed in $X$, so it cannot be a retract, so $Homeo_1^+(S^1)$ is not a retract of $HE_1^+(S^1)$.<|endoftext|> -TITLE: Asymptotic Formula for a Mertens Style Sum -QUESTION [7 upvotes]: Hello, -I am wondering if there is a simple asymptotic formula for -$$\sum_{p\leq x}\frac{\left(\log p\right)^{k}}{p},$$ -where $k\geq0$ is some integer. If $k$ is $0,$ by using the Prime Number Theorem we have -$$\sum_{p\leq x}\frac{1}{p}=\log \log x+b+O\left(e^{-c\sqrt{\log x}}\right).$$ -Similarly, the prime number theorem and integration by parts solves the case $k=1$ and gives -$$\sum_{p\leq x}\frac{\left(\log p\right)}{p}=\log x+C+O\left(e^{-c\sqrt{\log x}}\right).$$ -My question is do these integrals have a nice asymptotic formula for every $k$? Specifically, I mean with an error term of the form $O\left(e^{-c\sqrt{\log x}}\right).$ -Thanks! -Remark: This question is related, and in particular if it is solved with a nice enough asymptotic, then so is this. (but not vice versa) - -REPLY [6 votes]: Here is an answer that is similar in spirit to Frank and Peter's answers, but possibly simpler. -Summing by parts, we see that -$$ \sum_{p\le x} \frac{\log^k p}{p} = (\log x)^{k-1} \sum_{p\leq x} \frac{\log p}{p} -(k-1)\int_{2^-}^x (\log u)^{k-2}\sum_{p\le u} \frac{\log p}{p} \frac{du}{u}.$$ -Now use the formula -$$ \sum_{p\leq x} \frac{\log p}{p} = \log x + c_1 + O(\exp(-c_2\sqrt{\log x}))$$ -and it is not hard to derive that -$$ \sum_{p\le x} \frac{\log^k p}{p} = \frac{\log^k x}{k} + c_3 + O(\exp(-c_4\sqrt{\log x})).$$ -This seems easier than dealing with $Li(x)$.<|endoftext|> -TITLE: semisimplicity of p-adic Galois representations -QUESTION [9 upvotes]: Is it true that all continuous finite dimensional $p$-adic representations of $Gal(\bar{K}/K)$ are semisimple, where $K$ is a number field, i.e. if $\rho:Gal(\bar{K}/K) \mapsto GL_n(\bar{\mathbb{Q}}_p)$ is a continuous representation, where $K$ is a number field (even $\mathbb{Q}$ if you want), is it semisimple? -Replacing $\mathbb{Q}_p$ with $\mathbb{C}$, the result is indeed true, using the existence of an invariant Haar measure over the complex numbers. So the natural question (related) is if it exists a $\mathbb{Q}_p$-valued Haar measure in a compact group? Of course some conditions must be removed from the classical formulation (for example it does not make sence to ask that if $f$ is a positive continuous function, then its integral is positive), but the existence of a non-trivial invariant measure such that the volume of the group is 1 should be enough. Any reference is welcome as well - -REPLY [8 votes]: Well, actually, the "motivic Haar measure" LSpice refers to is an analogue of Haar measure -that lives on ${\operatorname{GL}}_n({\mathbb C}((t)))$, not on ${\operatorname{GL}}_n({\mathbb Q}_p)$, and takes values in the Grothendieck ring of varieties, so I think it's not quite relevant here. What is closer to this discussion though is the fact that the usual Haar mesaure on ${\operatorname{GL}}_n({\mathbb Q}_p)$ if it's reasonably normalized (e.g. so that the volume of ${\operatorname{GL}}_n({\mathbb Z}_p)$ -is $1$), actually takes values in $\mathbb Q$ on all reasonable sets that you ever want to consider. More precisely, one can define a sigma-algebra of the so-called "definable sets", and the volumes of definable sets just are in $\mathbb Q$. In this sense they are in -$\mathbb Q_p$ already, so the trick is that for these sets you do not need any completion of $\mathbb Q$ in order to define their volumes, and so you do not need to worry about using the $p$-adic metric... Most sets one works with turn out to be automatically definable, so this fact may be handy in some other situation. -Added some hours later: It was my first post on mathoverflow, and I am still not allowed to add comments to others' posts :) -- so this should be a comment to the comment by LSpice that appears in Emerton's post. -Talking about measure on $\operatorname{GL}_n(\overline{{\mathbb Q}_p})$, there is a paper by E.Hrushovski and D. Kazhdan http://arxiv.org/abs/math/0510133 that talks about integration in algebraically closed valued fields (using logic). As a first approximation, as far as I understand, the values of this measure are something like equivalence classes of definable sets over the residue field (I am certainly being imprecise here). There are several papers by Yimu Yin (the ones to start with are http://arxiv.org/abs/0809.0473, and http://arxiv.org/abs/1006.2467) aimed at clarifying this fundamental work of Hrushovski and Kazhdan in a slightly simplified setting. Unfortunately, I do not know of any non-technical introductory paper about this. There is a short note by Moshe Kamenski http://www.nd.edu/~mkamensk/lectures/motivic.pdf -- maybe this is the best place to start. Also, I hope someone corrects me here if I made any errors in this description.<|endoftext|> -TITLE: Cohomology of the infinite loop space of the affine grassmanian (as in the generalized Mumford conjecture) -QUESTION [6 upvotes]: I've been reading Hatcher's survey "A short exposition of the Madsen-Weiss theorem". In it, he outlines a nice proof of the "generalized Mumford conjecture", which asserts that the stable cohomology of the mapping class group is the same as the cohomology of $\Omega^{\infty} AG_{\infty,2}^{+}$. Here $AG_{n,m}$ is the space of affine $m$-planes in $\mathbb{R}^n$ and the "plus" sign indicates the 1-point compactification. -Now, the Mumford conjecture I know and love asserts that the (rational) stable cohomology ring of the mapping class group is $\mathbb{Q}[e_1,e_2,\ldots]$, where the $e_i$ are the MMM classes. -Can anyone explain to me why the rational cohomology ring of $\Omega^{\infty} AG_{\infty,2}^{+}$ is a polynomial ring with 1 generator in each even dimension? -This appears to be explained in Madsen-Tillmann's paper introducing the generalized Mumford conjecture, but that paper is rather formidable and I have been unable to extract an answer to the above question from it (indeed, it doesn't really appear to be talking about the affine Grassmannian at all!). - -REPLY [7 votes]: Most of this is not special to the case of $AG_{\infty,2}^+$. For any spectrum $X$, we have a Hurewicz map $h:\pi_{\ast}(X)\to H_{\ast}(X)$, which induces a map $h':\mathbb{Q}\otimes\pi_{\ast}(X)\to\mathbb{Q}\otimes H_{\ast}(X)$. Standard calculations show that $h'$ this is an isomorphism when $X=S^n$ for some $n$, and it follows by induction up the skeleta that $h'$ is an isomorphism for all $X$. Next, the homotopy groups $\pi_{\ast}(X)$ are (essentially by definition) the same as the homotopy groups of the space $\Omega^\infty(X)$, so we have an unstable Hurewicz map $h'':\pi_{\ast}(X)=\pi_{\ast}(\Omega^\infty(X))\to H_{\ast}(\Omega^\infty(X))$. By combinining these we get a map $\mathbb{Q}\otimes H_\ast(X)\to \mathbb{Q}\otimes H_\ast(\Omega^\infty(X))$. Next, every infinite loop space is a homotopy-commutative H-space, and this makes $H_\ast(\Omega^\infty(X))$ into a graded-commutative (and graded-cocommutative) Hopf algebra. Now let $A_*$ be the free graded-commutative ring generated by $\mathbb{Q}\otimes H_\ast(X)$, with the Hopf algebra structure for which the generators are primitive. More explicitly, $A_\ast$ is a tensor product of polynomial algebras (one for each even-dimensional generator in $\mathbb{Q}\otimes H_\ast(X)$) and exterior algebras (one for each odd-dimensional generator). It is now quite formal to construct a canonical map $A_\ast\to\mathbb{Q}\otimes H_\ast(\Omega^\infty(X))$ of bicommutative Hopf algebras. One can then show that this map is always an isomorphism. Indeed, it is possible to reduce to the case $X=S^n$ again, and the groups $\mathbb{Q}\otimes H_\ast(\Omega^k S^{n+k})$ can be calculated by repeated use of Serre spectral sequences, and one can then let $k$ tend to infinity. As $\mathbb{Q}\otimes H_\ast(\Omega^\infty(X))$ is free on primitive generators, it is a standard fact that the dual Hopf algebra $\mathbb{Q}\otimes H^\ast(\Omega^\infty(X))$ is also free on primitive generators in the same degrees. -For the Madsen-Weiss case, you now just need to know the groups -$\mathbb{Q}\otimes H_\ast(AG_{\infty,2}^+)$. This is not hard, because $AG^+_{\infty,2}$ is the Thom space of a virtual vector bundle over $BSO(2)=\mathbb{C}P^\infty$, so we can use the Thom isomorphism theorem.<|endoftext|> -TITLE: Is there an intrinsic definition of the topological index map in $K$-theory? -QUESTION [9 upvotes]: In the language of $K$-theory, the Atiyah-Singer index theorem says that for a compact manifold $X$ the topological index map $\text{t-index}: K(TX) \to K(T\mathbb R^n) \simeq \mathbb Z$ induced by embedding $X$ in $\mathbb R^n$ is equal to the analytical index map $K(TX) \to \mathbb Z$ obtained by looking at the index of the elliptic operator whose symbol corresponds to the given element in $K(TX)$. -My question is if there is a definition of the topological index map that does not require an embedding into a euclidean space. Clearly by the index theorem we can take the analytic index as a definition, but is there a more topological/geometric intrinsic definition for t-index that is (relatively) easily seen to be equivalent to the above definition? - -REPLY [9 votes]: I think the best answer you will find is the axiomatic characterization of the topological index in Index of Elliptic Operators I. -One defines an index function to be a map $ind_X: K(TX) \to \mathbb{Z}$ with the properties that $ind_{point}$ is the identity, and for any embedding $i: X \to Y$ the wrong way map $i_!: K(TX) \to K(TY)$ commutes with the index map. It is not hard to show that the index map is actually uniquely characterized by the axioms and that the topological index is an example of an index map. Thus the real content of the index theorem lies in proving that there is an index map which sends the symbol class of an operator to its index. -From this point of view you should think of embedding $X$ in Euclidean space as a computational crutch rather than an essential part of the theorem. If K-theory were invented before De Rham cohomology then we would all be completely satisfied with the topological characterization of the analytic index map explained above because it tells us everything we need to know to compute the analytic index in terms of K-theoretic invariants. But since we tend to prefer cohomology it makes sense to embed our manifold in a space where it is easy to relate K-theory and cohomology. -This all is what I tell myself to calm down before I go to bed at night, but in fact I do not find it totally satisfying. I have no qualms embedding a manifold in Euclidean space when I want to prove the Gauss-Bonnet theorem or the Hirzebruch signature theorem, but I find it rather disturbing that one can prove the Riemann-Roch theorem by starting with an algebraic curve, forgetting all of its beautiful structure, and stuffing it carelessly into Euclidean space. I've always wondered if there is another computational crutch analogous to the topological index which would remember a little more structure.<|endoftext|> -TITLE: How does Tate verify his own conjecture for the Fermat hypersurface? -QUESTION [28 upvotes]: This question is about Tate's 1963 paper "Algebraic Cycles and Poles of Zeta Functions". Here he announces a conjecture (now known as "the Tate conjecture") which states that certain classes in the cohomology of a projective variety are always explained by the existence of algebraic cycles. In the case of a variety $X/\mathbf{F}_q$ of dimension $n$, the conjecture predicts that the subspace of classes in $H^{2d}(X\otimes\overline{\mathbf{F}}_q,\mathbf{Q}_\ell)(d)$ which are Frobenius-invariant is spanned by the image of the space of algebraic cycles in $X$ of codimension $d$. -As an example, Tate gives the projective hypersurface $X$ defined by -$$ X_0^{q+1}+\dots X_r^{q+1} =0,$$ where $r=2i+1$ is odd. Here $X$ admits a large group of automorphisms $U$, namely those projective transformations in the $X_i$ which are unitary with respect to the semilinear form $\sum_i X_iY_i^q$. Using the Lefschetz theorem, it isn't at all hard to compute $H^{2i}(X)$ as a $U$-module: it decomposes as a trivial $U$-module and an irreducible $U$-module of dimension $q(q^r+1)/(q+1)$. And then when you attempt to compute the $q^2$-power Frobenius eigenvalues on the middle cohomology, you find (once again by Lefschetz) that each one is a Gauss sum which in this case is nothing but $\pm q$. (If there is enough demand, I can supply all these calculations here.) Thus, miraculously, all classes in $H^{2i}(X\otimes\overline{\mathbf{F}}_q,\mathbf{Q}_\ell)(i)$ are fixed by some power of Frobenius. -My question is: How did Tate confirm the existence of the necessary cycles? Surely the hyperplane section of codimension $i$ lands in the part of $H^{2i}$ which has trivial $U$-action (for $U$ translates hyperplanes to other hyperplanes, and these all cohomologically equivalent). In order to verify Tate's conjecture, all you need to do is produce a cycle in $X$ whose projection into the big $U$-irreducible part of $H^{2i}$ is nonzero. How did Tate produce this cycle? Did he lift to characteristic zero and appeal to the Hodge conjecture, or what? - -REPLY [24 votes]: I don't know how Tate did it but here is one way. Let $\zeta$ be such that -$\zeta^{q+1}=-1$ and put $a_j=(0\colon\cdots\colon1\colon\zeta:\cdots\colon0)$, -$j=0,\ldots,i$ with the $1$ in coordinate $2j$. Then the linear span $L$ of -these points is contained in the Fermat hypersurfaces and gives a subvariety of -middle dimension. I claim that its class is not a multiple of the $i+1$'st power -of the hyperplane section. Indeed, transform $L$ by the automorphism permuting -the two first coordinates to get $L'$. Then (assuming that $q$ is odd say) $L$ -and $L'$ are disjoint so that $[L]\cdot [L']=0$ but if the class of $L$ would be -a multiple of the linear subspace section this cannot be. Hence, $[L]$ projects -non-trivially to the non-trivial irreducible representation. -Note that Shioda et al have studied cycles on general Fermat hypersurfaces and -verified the Tate conjecture in many (all?) cases. -Addendum: It is tricky to get dimensions and stuff correct so let me expand upon this in the simple case when $i=1$ and $q=2$ (this is a classical example appearing for instance on pp. 176-177 of Mumford: Algebraic Geometry I Complex projective varieties -- everything works in characteristic different from $3$). Then we get the line $(a\colon a\zeta\colon b\colon b\zeta)$ and by letting the monomal automorphisms of the surface act we get all $27$ lines on the Fermat cubic. It is of course true that the line is the intersection of the cubic and a linear subspace but the intersection is not proper so the class of $L$ is not a power of the hyper plane section. -Note incidentally that in comments (p. 180) Mumford points the special nature of characteristic $2$ for this example essentially saying that there an index $2$ subgroup of the automorphism group of the Néron-Severi lattice preserving the canonical class can be realised by automorphisms of the surface.<|endoftext|> -TITLE: Equivalent definitions of topological K-theory over locally compact spaces -QUESTION [14 upvotes]: Hello everyone -My question is the following: -Given a locally compact space $X$, assume it is also connected for simplicity, its K-theory ring is defined as $\tilde{K}(X^+)$, where $X^+$ is the one point compactification and $\tilde{K}$ denotes reduced K-theory (with respect to the point at infinity). Alternatively, we can take differences $[E]-[F]$, where $E$ and $F$ are isomorphism classes of vector bundles over $X$, each one trivial outside a compact set and both with the same rank. Notice that, since both bundles are trivial outside a compact set, they can be extended to $X^+$. -So far so good. There is another definition consisting of equivalence classes of triples $(E,F,\alpha)$, where $E$ and $F$ are now bundles over $X$ and $\alpha:E\to F$ is an isomorphism outside a compact set in $X$. Two triples $(E,F,\alpha)$ and $(E',F',\beta)$ are equivalent if there are ''degenerate'' triples $(G,G,id)$, $(H,H,id)$ and isomorphisms $u:E\oplus G\to E'\oplus H$ and $ v:F\oplus G\to F'\oplus H$ commuting with the maps $\alpha\oplus id$ and $\beta\oplus id$. Notice that in this definition, the bundles are not trivial outside a compact set so there is, a priori, no way to extend them to $X^+$. -Well, both definitions are equivalent. This is easy to prove for compact finite dimensional manifolds (for example Atiyah's book contains a nice argument). For non compact finite dimensional manifolds, it turns out that every vector bundle is a direct summand of a trivial vector bundle (it is even more surprising: for a given bundle, there is a finite covering of the manifold such that the bundle restricted to each set of the covering is trivial). A proof of this is very difficult to find and relies on Dimension theory (Wells, Differential Analysis on Complex Manifolds prop 4.1). Using this property, the proof in this case is immediate. -Looking for the general proof, I realized that every single place where this equivalence is mentioned simply quotes Segal's paper on equivariant K-theory. I have to admit that I don't understand the argument explained there. It is clear that every locally compact space is open and dense in its one point compactification. Take $U$ open in a compact space $Y$, using the first definition of $K(U)$, we should be able to prove that $K(U)\cong K(Y,B)$, where $B=Y-U$, but that would require ''chopping off'' the bundles outside a compact set and gluing something to make the triple be represented by bundles trivial at infinity and, therefore, with a natural extension to $X^+$. This seems to me as a highly invasive process and I don't really see why it is correct. -Could someone please help me with this??? - -REPLY [11 votes]: Let $(E,F,\alpha)$ represent an element of the second group. Wlog $\alpha:E\to F$ is defined globally even though it's only an iso outside $K$. Choose global compactly supported sections $s_1,\dots ,s_n$ of the dual of $E$ such that on $K$ they span the bundle; for large enough $n$ this is possible. Let $E'$ be the trivial rank $n$ bundle and interpret these sections as an injective map $s:E\to E'$. We now have a bundle map $(\alpha,s):E\to F\times E'$ which is injective everywhere. Split off the resulting subbundle of $F\times E'$ and call the other part $F'$. This leads to an isomorphism $E\times F'\cong F\times E'$. Replace $(E,F,\alpha)$ by $(E\times E',F\times E',\alpha\times 1)$. Near infinity, where $\alpha\times 1$ is an iso, the composed iso $E\times E'\to F\times E'\to E\times F'$ looks likes $(1, 0)$ on the $E$ part and therefore looks like $(?,\alpha')$ on the $E'$ part where $\alpha'$ is an iso. Make a little adjustment so that it looks like $(0,\alpha')$ on the $E'$ part. This iso $\alpha':E'\to F'$ near infinity yields $(E',F',\alpha')$ equivalent to $(E,F,\alpha)$, but with the bundles trivial near infinity. (In fact $E'$ is trivial globally.)<|endoftext|> -TITLE: Is there a geometrical interpretation to Fermat's polygonal number theorem? -QUESTION [5 upvotes]: The Polygonal number theorem states that every positive integer is the sum of at most $n$ $n$-gonal numbers. -I do think that expecting a "Proof without words" of this theorem is asking for something unreasonable, but is there any way to "visualize" or to get an intuitive feel for the theorem? -One motivation for this question is the problem asked by Bjorn Poonen here. If there was such an interpretation, then maybe it would help in some way to answer that question? - -REPLY [6 votes]: Let me make some comments on the polygonal number theorem along the theme that things are less thrilling for $k \gt 4$. - -Triangular and square numbers are pleasing to represent as dot patterns, after that $k$-gonal numbers are less attractive. -Around 1796 Gauss proved that every integer is a sum of 3 triangular numbers. This is sometimes called the Eureka Theorem because Gauss was quite pleased with the result. -Evidently, Cauchy showed (around 1813) that from that one can go to the general polygonal theorem that every integer $n$ can be represented as a sum of $k$ $k$-gonal numbers. -Tables due to Peppin and Dickson show how to represent $n \lt 120k-240$ as a sum of at most $k$ $k$-gonal` numbers. -A nice paper by Melyvn Nathanson uses results 2 & 4 to show in a few pages that -a. for $k \ge 5$ and $n \ge 120k-240$, $n$ can be written as a sum of $k-1$ $k$-gonal numbers of which at most four are different than $0$ or $1$. -b. For fixed $k$ every large enough odd $n$ is a sum of four $k$-gonal numbers and every large enough even $n$ is a sum of five $k$-gonal numbers one of which is either $0$ or $1$. - -So it seems possible to me that taking 2 & 4 as given, it might be possible to express at least some of the steps of a transition to the general polygonal theorem in a somewhat geometric way, however if this hypothetical thing is possible it would probably not be a simplification giving insight as much as a somewhat more complicated path which is interesting mainly for the fact that it can be done at all. -I would think of an insight giving proof as one which also showed how to represent $n$. The proofs I know of show that there must be a representation but don't tell you how to find it. Changing $n$ to $n+1$ and/or $k$ to $k+1$ can radically change the representation.<|endoftext|> -TITLE: Irreducible decomposition of tensor product of irreducible $S_n$ representations -QUESTION [7 upvotes]: Are there well known results on the irreducibles in the decomposition of tensor products of irreducible $S_n$ representations? I would also like to know of some references where I can find formulas (if they exist in the literature) for finding multiplicities. - -REPLY [11 votes]: The numbers you want are called Kronecker coefficients. Bürgisser and Ikenmeyer "The complexity of computing Kronecker coefficients" showed that they are hard to compute in general, so in particular there are no "easy" formulas for them. (There are some explicit formulas for simple special cases.)<|endoftext|> -TITLE: Elliptic curves with Mordell-Weil group Z/2Z over Q -QUESTION [9 upvotes]: This question is not very precise; I hope it is suitable for the site. -I have come to a situation where I have to study rational points on an elliptic curve defined over $\mathbb{Q}$. I don't know much about the curve, let alone its equation. I already have one rational point, which sits on a bounded real connected component. What I want to avoid is that this is the only rational point (other than the marked point). -I am not sure what to use about my curve that will help me get there, so I turn the questio the other way round: - -What is known about elliptic curves $E$ over $\mathbb{Q}$ such that $E(\mathbb{Q}) \cong \mathbb{Z}/2 \mathbb{Z}$? - -REPLY [14 votes]: Mazur's theorem ensures that there are exactly 15 possible cases for the torsion part of the Mordell-Weil group of an elliptic curve: the cyclic groups $\mathbb{Z}_n$ (with $1\leq n\leq 10$ or $n=12$) and the groups $\mathbb{Z}_2\times\mathbb{Z}_n$ for $n=2,4,6,8$. -In his paper Universal Bounds on The Torsion of Elliptic Curves, Proc. London. Math. Soc.(1976) 33, 193-237 , Daniel Sion Kubert (who was a student of Mazur) presents in table 3 (page 217) a list of parametrizations for the different possible cases. -In particular, curves with a $\mathbb{Z}_2$ torsion are parametrized by the following family: -$$ -\mathbb{Z}_2\ \text{torsion}:\quad y^2=x(x^2+a x+ b), \quad b^2(a^2-4b)\neq 0. -$$ -The example given in Francesco's answer is a special case with $a=0$. -As another example, the case with torsion $\mathbb{Z}_2\times \mathbb{Z}_2$ is parametrized by the Legendre family: -$$ -\mathbb{Z}_2\times\mathbb{Z}_2 \ \text{torsion}:\quad y^2=x(x+r)(x+s), \quad r\neq 0 \neq s \neq r. -$$ -A slight generalization of the Hesse family parametrizes the curves with torsion $\mathbb{Z}_3$: -$$ -\mathbb{Z}_3 \ \text{torsion}:\quad y^2+a_1 x y +a_3 y =x^3, \quad a_3^3( a_1^3-27 a_3)\neq 0. -$$ -For the other groups you might have to use Tate's normal form -$$ -E(b,c): \quad y^2+(1-c)x y - b y =x^3- b x^2 -$$ -and the condition for a given torsion is expressed as an algebraic condition on $b$ and $c$. -For example for $\mathbb{Z}_4$, we have $c=0$ and $b^4(1+16b)\neq 0$, which gives: -$$ -\mathbb{Z}_4 \ \text{torsion}:\quad E(b,c=0): \quad y^2+x y - b y =x^3- b x^2, \quad b^4(1+16b)\neq 0. -$$ -For a review, you can read chapter 4 of the book of Husemoller . A friendly short review is also available in section 2 of this string theory paper by Aspinwall and Morrison ( they don't present all the 15 cases but for those they analyze, they express everything in Weierstrass form).<|endoftext|> -TITLE: Torsion freeness and birational maps -QUESTION [6 upvotes]: Let $f:X\rightarrow Y$ be a birational morphism between smooth varieties and $F$ a torsion free sheaf. Is $f_{\ast} F$ torsion free? If not, are there conditions on either $F$, $f$, $X$ or $Y$ that could ensure the torsion freeness of $F$? For example if $f$ is surj. and $F$ is the canonical bundle on $X$ a theorem of Kollar ensure the torsion freeness of $f_*(\omega_X)$. - -REPLY [6 votes]: As already mentioned by Donu and Francesco, you don't need such big guns as Kollár's theorem. Also, the Proposition Francesco cites might seem more serious than it is by virtue of being in EGA... -The point is this: For an open set $V\subseteq Y$, the module $f_*\mathscr F(V)$ is the same as the module $\mathscr F(f^{-1}V)$. Being torsion on an integral scheme is equivalent to having a non-empty support that is strictly smaller than the ambient scheme. If $f$ is surjective this already gives you what you want, and if it is only dominant you need to think a little more to see that the statement is true.<|endoftext|> -TITLE: Centralisers in the symmetric group -QUESTION [6 upvotes]: Let $G$ be a $k$-transitive subgroup of the symmetric group $Sym(n)$, $k\geq 2$, $n$ large. (Make $k$ larger if you think it necessary to make the question below non-trivial/interesting.) -Write $C(g)$ for the centraliser of an element and $|C(g)|$ for its number of elements. -What can you say about $\min_{g\in G} |C(g)|$, relative to the size of $G$? Must it be small? -(Note that $k\geq 2$, $n\geq 3$ immediately imply that $G$ is non-abelian (thus making the question possibly non-stupid).) -(I should clarify that I expect (hope?) the answer to be quite a bit smaller than $\ll_{\epsilon} |G|^{\epsilon}$. (Assume $k\geq 3$ or $k\geq 4$ if needed.) In this, a $k$-transitive permutation group feels like a different sort of animal from a linear algebraic group $G$, where $|C(g)|$ is most often no smaller than $|G|^{\dim(T)/\dim(G)}$ ($T$ = a maximal torus of $G$).) - -REPLY [2 votes]: (Written before clarification at end of question was added). Here is a result which seems to be of a somewhat negative nature in the context of your problem, and your suggested line of attack, I think. I will denote the number of conjugacy classes of $G$ by $k(G)$. The group $ G = {\rm SL}(2,2^{n})$ (n>1) is a triply transitive permutation group on $1+2^{n}$ points and has order $(2^{n}+1)2^{n}(2^{n}-1)$. It has $1+2^{n}$ conjugacy classes, so that $k(G) > |G|^{\frac{1}{3}}$. Furthermore, the only orders of centralizers of non-identity elements of $G$ are $2^n$,$2^{n}-1$ and $2^{n}+1$. Hence the minimum centralizer order for an element of $G$ is only marginally smaller than $|G|^{\frac{1}{3}}.$ Admittedly this is a rather special group and a rather rare situation, but it is an infinite family of "bad" examples. -Added later: come to think of it, there are even bad solvable examples. If we take any prime power -$p^{n}$ there is a doubly transitive solvable permutation group $G$ of order $p^{n}(p^{n}-1)$ -and degree $p^{n}$(this is a Frobenius group which is the semidirect product of a vector -space of size $p^n$ acted on by a Singer cycle of order $p^{n}-1$. A point stablizer is -cyclic of order $p^n -1)$.` In this group, the only centralizer orders for non-identity -elements are $p^n$ and $p^{n}-1$. Hence the minimum centralizer order for $G$ is not -much less than $\sqrt{|G|}.$ So it seems that you can't expect to get much less than -$\sqrt{|G|}$ for the smallest centralizer order for $G$. I suspect that even this would -be very difficult to prove without the classification of finite simple groups. -Later remark: It is perhaps worth remarking explicitly here (though I am sure everybody knows it), that $S_n$ contains an element whose centralizer has order $n-1$ (and this is minimal), -and that for $n > 4$, $A_n$ contains an element whose centralizer has order $n-2$, -which is also minimal. Hence the "generic" highly transitive permutation group of degree $n$ behaves as hoped for large $n$ (and for sufficiently large $k$ there are no others, using the Schreir conjecture for $k >7$ or the disallowed Classification of finite simple groups for $k > 5$).<|endoftext|> -TITLE: Values of the Riemann zeta function and the Ramanujan summation - How strong is the connection? -QUESTION [9 upvotes]: (This Question was taken from MSE. As Eric Naslund pointed out there, this question is relevant. The summation method mentioned in this question is actually a good answer to it.) -The Ramanujan Summation of some infinite sums is consistent with a couple of sets of values of the Riemann zeta function. We have, for instance, $$\zeta(-2n)=\sum_{n=1}^{\infty} n^{2k} = 0 (\mathfrak{R}) $$ (for non-negative integer $k$) and $$\zeta(-(2n+1))=-\frac{B_{2k}}{2k} (\mathfrak{R})$$ (again, $k \in \mathbb{N} $). Here, $B_k$ is the $k$'th Bernoulli number. However, it does not hold when, for example, $$\sum_{n=1}^{\infty} \frac{1}{n}=\gamma (\mathfrak{R})$$ (here $\gamma$ denotes the Euler-Mascheroni Constant) as it is not equal to $$\zeta(1)=\infty$$. -Question: Are the first two examples I stated the only instances in which the ramanujan summation of some infinite series coincides with the values of the Riemann zeta function? - -REPLY [7 votes]: The answer can be obtained with the following interpretation of the Ramanujan summation: - -More recently, the use of $C(1)$ has been proposed as Ramanujan's summation, since then it can be assured that one series admits one and only one Ramanujan's summation, defined as the value in 1 of the only solution of the difference equation $R(x) − R(x + 1) = f(x)$ that verifies the condition $\int_1^2 R(x)dx=0$. - -The function $R(x)=\zeta(z,x)+C\;$ satisfy $R(x) − R(x + 1) = x^{-z}$, $x>0$, $z\in \mathbb C$, $z\ne-1$, where -$$ -\zeta(z,x)=\sum_{k=0}^\infty\frac1{(k+x)^z} -$$ -is the Hurwitz zeta function (or its analytic continuation for $\Re z\le 1$.) -The value of $C\;$ can be found with the help of the shift formula for the derivative $\frac{\partial}{\partial x} \zeta(z,x)=-z \zeta(z+1,x)\;$: -$$ -\int_1^2 R(x)dx= -\int_1^2(\zeta(z,x)+C)dx= -\int_1^2 -\left( -\frac1{z-1}\frac{\partial}{\partial x} \zeta(z-1,x)+C\right)dx= -$$ -$$ -C-\frac1{z-1}(\zeta(z-1,2)-\zeta(z-1,1))=C+\frac1{z-1}=0.$$ -Hence $C=-\frac1{z-1}$. Also $\zeta(z,1)=\zeta(z)$ and we have -$$ -\sum_{n=1}^\infty n^{-z}=\zeta(z)-\frac1{z-1}(\mathfrak{R}),\quad z\ne1. -$$ -So Ramanujan summation transforms the Riemann zeta function into the regularized zeta function. It explains why the value $\gamma$ should be expected for the summation of the harmonic series.<|endoftext|> -TITLE: Which book would you like to see "texified"? -QUESTION [19 upvotes]: Let's see if we could use MO to put some pressure on certain publishers... -Although it is wonderful that it has been put -online, I think it would make an even greater read if "Hodge Cycles, Motives and Shimura Varieties" by Deligne, Milne, Ogus and Shih would be (re)written in the latex typesetting (well, if I could understand its content..). -But enough about my opinion, what do you think? Which book(s) would you like to see "texified"? As customary in a CW question, one book per answer please. - -REPLY [4 votes]: Adams - Lectures on Lie Groups - -REPLY [2 votes]: Stong - Notes on cobordism theory<|endoftext|> -TITLE: Occurence of trivial representation in a tensor square. -QUESTION [7 upvotes]: Suppose $G$ is a group and $V$ an irreducible representation of $G$. One has that $V\otimes V\cong \Lambda^2(V)\oplus Sym^2(V)$. It is well-known that if the trivial representation appears as a subrepresentation of $\Lambda^2(V)$ then $V$ is of quaternionic type; while if the trivial representation appears as a subrepresentation of $Sym^2(V)$ then $V$ is a of real type. From this approach, it is clear that the trivial representation cannot appear in both $\Lambda^2(V)$ and $Sym^2(V)$. -What I am curious about is as follows: - - -Question: Is there is some (relatively easy) way to see why the trivial representation cannot appear in both $\Lambda^2(V)$ and $Sym^2(V)$ without introducing the machinery of real/quaternionic types? - - -As a bit of motivation, if one looks at other subrepresentations, then for example if $G = G_2$ and $V_n$ is an $n$-dimensional irreducible representation of $G_2$, then $V_{64}$ appears as a subrepresentation of both $\Lambda^2(V_{27})$ and $Sym^2(V_{27})$. In particular it is possible for the intertwining number of $\Lambda^2(V)$ and $Sym^2(V)$ to be nonzero, but by the real vs. quaternionic characterization, the trivial representation is somehow special in that it cannot contribute to the intertwining number. - -REPLY [14 votes]: This is essentially what Darij wrote, but without mentioning the bilinear forms. (I had written it out before reading far enough into Darij's post to see that he was really doing the same thing, after the part about bilinear forms.) Think of $V\otimes V$ as $\text{Hom}(V^*,V)$. An occurrence of the trivial representation in $V\otimes V$ thus amounts to a $G$-equivariant linear map from $V^*$ to $V$. Since $V$ and therefore also $V^*$ are irreducible, Schur's lemma says that the space of such maps has dimension either 1 (iff $V$ and $V^*$ are isomorphic) or 0. So there's at most one occurrence of the trivial representation in $V\otimes V$.<|endoftext|> -TITLE: Best constant in comparison between Rademacher and gaussian averages? -QUESTION [7 upvotes]: Let $(g_k)$ be a sequence of independent standard gaussians variables on a fixed probability space $\Omega$. Let $(\epsilon_k)$ be a sequence of independent rademacher variables. -What is the best constant in the following inequality: -$$ -\vert\vert\sum_{k}\epsilon_k \otimes x_k \vert\vert_{L^2(\Omega,E)} -\leq -K \vert\vert\sum_{k}g_k\otimes x_k \vert\vert_{L^2(\Omega,E)}\ \ ? -$$ -for any Banach space $E$, any $x_1,\ldots, x_n \in E$. -I know that the best constant is $\leq \sqrt{\frac{\pi}{2}}$ (see Diestel,Jarchow,Tonge "Absolutely summing operators" page 239). - -REPLY [10 votes]: $\sqrt{\pi\over 2}$, the reciprocal of the $L_1$ norm of a standard gaussian, is the best constant. Let $x_k$ be the kth unit basis vector in $\ell_1$ and let the sum go from $1$ to $N$. The square of the left hand side is $N^2$ and the square of the right hand side is $N+N\sqrt{2\over \pi}(N-1)\sqrt{2\over \pi}$ (multiplied by $K^2$).<|endoftext|> -TITLE: Why the BGG category O? -QUESTION [27 upvotes]: Given a finite-dimensional semisimple complex Lie algebra $\mathfrak{g}$, the Bernstein-Gelfand-Gelfand category $\mathcal O$ is the full subcategory of $\mathfrak g$-modules satisfying some finiteness conditions. It contains all finite-dimensional modules as well as all highest-weight modules, it's Noetherian and Artinian, and it's Abelian. It's clear to me why you would want to work in some full subcategory of $\mathfrak g$-modules which has the above properties, but why $\mathcal O$? Is it minimal in some sense with respect to these properties and/or some other important properties? - -REPLY [17 votes]: It might be worth pointing out a different motivation for Category O, namely the theory of Harish Chandra (g,K) modules. These are algebraic models for continuous representations of real reductive groups (the real forms of g). Harish Chandra's amazing theory reduces many questions in the representation theory of real groups to this algebraic theory, which can then be studied geometrically, for example by Beilinson-Bernstein localization. -In any case Category O is essentially the category of Harish Chandra modules associated to the complex reductive group G, when considered as a real Lie group. There are slight subtleties in what kind of semisimplicity/local finiteness etc we require for the center of the enveloping algebra or the maximal torus, but in broad strokes the two coincide, and category O is thus a nice combinatorially accessible model of a very basic object in representation theory of Lie $groups$.<|endoftext|> -TITLE: Relation between Isogeny, Conics and Fermat's method of infinite descent -QUESTION [5 upvotes]: Fermat's proof of FLT(4) is an example of infinite descent as is Euler's (or whoever you attribute it to's) proof of FLT(3). There are similar proofs to Fermat's for Diophantine equations like $x^4 + y^4 = 2z^2$. -I have unsuccessfully tried to view these proofs in terms of group homomorphisms on conics and elliptic curves but it is not at all clear whether this is possible. -Can we reinterpret these infinite descent proofs geometrically, in terms of curves? - -REPLY [8 votes]: (That picture in your avatar is Weil, right? You should start by reading Weil's Number Theory: an approach through history). -FLT(3) is the assertion that the curve $x^3+y^3=1$ has three rational points (including the point at infinity). The standard process of putting an elliptic curve in Weierstrass form shows that this curve is $y^2=x^3-432$ if I remember correctly. Now use descent on this elliptic curve (maybe an isogeny of degree 3) to show that the Mordell-Weil group is finite of order three (see, e.g. Silverman for the general theory). This may be Euler's proof, maybe it's discussed in Weil's book. -FLT(4) is weaker than the statement that the equation $x^4+y^4=z^2$ has only the obvious solutions. Again, this becomes the problem of finding the rational points on $y^2=x^4+1$ which is again an elliptic curve and Fermat's proof is a 2-descent showing that the Mordell-Weil group is finite of order four. I'm pretty sure this is in Weil's book. -I have no idea what conics have to do with any of this.<|endoftext|> -TITLE: What do you do if you find a typo in an equation of a paper? -QUESTION [15 upvotes]: I was reading an interesting paper, and early in the introduction there was an equation with a typo in it. I am absolutely sure of this, and all it was missing was a factor of $n!$. Overall, it was an inconsequential part of the paper, and doesn't affect anything, as it was in the section which tells the reader about the background of the area. But still, the equation as written is not correct, and could be confusing to someone reading who is trying to learn about the area. -In fact, the line right before this particular equation said "it is easily observed that..." which is a little interesting. - -My question is: Should I do anything when I see this? Should I just ignore it, or is it polite to email the author? Or is it impolite to do so? If you published a paper, with an inconsequential, but possibly confusing typo in the introduction, would you want someone to tell you about it or just leave it alone? - -Thanks for the advice! -Also please note, I am not referring to a spelling typo, that is definitely not important! Its just possible that this typo could confuse someone reading the paper. (I was confused at first!) - -REPLY [5 votes]: I usually email the author in case the paper is online in a place controlled by the author (i. e., his website or arXiv). The only downside of this approach (apart from authors not replying) is that some would just take down their papers (rather than correcting them) when they hear about a mistake in a part they consider substantial. This happened to me once (although fortunately the paper is still avaliable at other places in the net). -In case the paper is online but not in such a place, this is a more difficult question, but if the error is substantially confusing, I'd still mail the author. Unfortunately, this sometimes means notifying a clueless author that somebody else has put his paper online (most usually, a university workmate, for his students; I am not talking about pirate sites), and rather than correcting it tries to get it offline. I would be careful here. -In case the paper is offline or behind a paywall, I ignore it unless the mistake really destroys some results from the paper, in which case I rejoice about another little blow to copyrighted literature and the so-called peer review process. -But then again, I am mostly reading openly avaliable texts, so the first case is by far the most frequent. -I, personally, would prefer anyone telling me of any mistake, but I do not have (and do not plan to have) papers outside of open access, so there is no contradiction here.<|endoftext|> -TITLE: Is there a least-fixed-point formulation of inaccessible cardinals? -QUESTION [11 upvotes]: The infinity axiom can be formulated by defining a function $S$ as -$$S(N) = \{0\} \cup \{n+1\\ |\\ n \in N\}$$ -(FWIW, I'm assuming the von Neumann ordinals.) The axiom is then -$$\exists I . I = S(I)$$ -which gives us our first infinite set. Then $\omega$ is the intersection of all the subsets of $I$ that are also fixed points, or the least fixed point of $S$. -I've been curious about this for a while now: can the first (strongly, uncountable) inaccessible cardinal be defined similarly? - -REPLY [5 votes]: One can attain something like a positive solution with the observation that the inaccessible cardinals are precisely the regular fixed points of the beth function $\alpha\mapsto\beth_\alpha$, which is monotone and continuous. In particular, the first inaccessible cardinal is the smallest regular beth fixed point.<|endoftext|> -TITLE: 2-colimits in the category of cocomplete categories -QUESTION [6 upvotes]: Let us denote by $\text{Cat}_c$ the $2$-category, whose objects are cocomplete categories, morphisms are cocontinuous functors and morphisms are natural transformations. Is it then true that $\text{Cat}_c$ is $2$-cocomplete, i.e. has every $2$-functor $C : I \to \text{Cat}_c$, where $I$ is a small $1$-category, a $2$-colimit? -I have the following idea: Let $U :\text{Cat}_c \to \text{Cat}$ be the forgetful $2$-functor. Now $\text{Cat}$ is $2$-cocomplete; a comprehensive reference for this is "The stack of microlocal perverse sheaves" by Ingo Waschkies. Besides, $U$ has a $2$-left adjoint $\widehat{(- )} : \text{Cat} \to \text{Cat}_c$ (see here). So we may consider the cocomplete category $D:=\widehat{\text{colim} UC}$, but of course the functors $C_i \to D$ are not cocontinuous. Thus perhaps we have to define some reflective (and hence cocomplete) subcategory $D'$ of $D$, such that each $C_i \to D$ factors through a cocontinuous functor $C_i \to D'$. But I don't know how to define $D'$. -I'm also interested in colimits in similar $2$-categories, for example $\text{Cat}_{c\otimes}$, which consists of the cocomplete tensor categories. And actually I want something more, namely that these $2$-categories are $2$-locally presentable. Any hints and references are appreciated! - -REPLY [12 votes]: In order not to have to worry about size issues, I'm going to answer the following question instead: - -For a (small) cardinal number $\kappa$, is the category of small categories with $\kappa$-small 2-colimits 2-cocomplete? - -If you take $\kappa$ to be inaccessible, then this will correspond to your question, under a particular choice of foundations. I presume moreover that you mean "2-colimits" in the weak "up-to-equivalence" sense which the nLab uses (which 2-category theorists traditionally call "bicolimits"). -The fact that the 2-category Cat of small categories is 2-cocomplete, in this sense, has been well-known to category theorists for decades. It is obvious that Cat is cocomplete as a 1-category (since it is locally finitely presentable), and since it is closed symmetric (cartesian) monoidal, it follows by general enriched category theory that it is cocomplete as a category enriched over itself. In the nLab terminology, it has all strict 2-colimits. We then observe that strict pseudo 2-limits, which are 2-limits that represent cones commuting up to isomorphism but satisfy their universal property up to isomorphism (rather than equivalence), are particular strict 2-limits. Since any strict pseudo 2-limit is also a (weak) 2-limit, Cat is 2-cocomplete. -Now as Zoran pointed out in the comments, there is a 2-monad on Cat whose algebras are categories with $\kappa$-small colimits; let us call this 2-monad $T$. The strict $T$-morphisms are functors which preserve colimits on-the-nose, while the pseudo $T$-morphisms are those which are $\kappa$-cocontinuous in the usual sense (preserve colimits up to isomorphism). Therefore, the question is whether the 2-category $T$-Alg of $T$-algebras and pseudo $T$-morphisms is 2-cocomplete. -The answer is yes: it was proven by Blackwell, Kelly, and Power in the paper "Two-dimensional monad theory" that for any 2-monad with a rank (preserving $\alpha$-filtered colimits for some $\alpha$) on a strictly 2-cocomplete (strict) 2-category, the 2-category $T$-Alg is (weakly) 2-cocomplete. The 2-monad $T$ has a rank (namely, $\kappa$, more or less), so their theorem applies. I believe this all works just as well in the enriched setting.<|endoftext|> -TITLE: Publishing journals articles without transferring copyright. -QUESTION [45 upvotes]: I'm a grad student getting close to submitting my first journal article (which will be single-authored). My understanding is that it's standard practice for authors to transfer the copyright of their paper to the journal in which it is published. I want my article to be published in a journal, but I don't want to transfer the copyright -- I want the article to be in the public domain. How will editors to behave towards such a request? Also, when should I bring up the topic: when I submit the article, or after it's accepted and I'm asked to sign a copyright transfer? -Some grants apparently have a stipulation that articles written as part of the grant research must be released into the public domain (e.g. grants funded by the US government). In this case, authors presumably sign a consent to publish, instead of a copyright transfer. Hence, there's at least some precedent for what I want to do, though I want my article in the public domain purely because of my personal views on the ethics of copyright. -I couldn't find too much information about this topic by googling. Oleg Pikhurko has a page discussing his attempt to have his articles revert to the public domain after a period of years, as opposed to instantly. It didn't work out particularly well in his case. -I'm not sure how much I'm willing to have my ethical ideals damage my career (e.g. by having publications delayed and/or being banned from submitting to journals). - -REPLY [16 votes]: I would not advise you to give up your ethical ideals. Frankly, I find it offensive that any publishers expect me to assign to them the copyright in an article that I have written. -I think there are also some practical reasons to retain copyright, although they are not as pressing now that we can put our papers on the arXiv (which you should, in my opinion, always do). Kevin's addendum form mentions some of the important rights that some copyright transfer forms might take away from you: the right to create derivative works, use portions of the article in other work, and distribute the final version on the web. -The best solution is, obviously, to publish in journals that always allow the author to retain copyright. Such journals often also have other desirable features, e.g. they make their published articles freely available online, rather than charging huge sums of money to university libraries for subscriptions. Furthermore, publishing good work in these "friendly" journals helps to build up their acceptance in the community, so that future authors will feel less pressure from career considerations to publish in "unfriendly" journals. -Of course, right now it is still the case that some of the "best" journals (in terms of building up your publication record) are "unfriendly". How far you are willing to compromise between ethical ideals and career considerations is a personal decision. My personal choice so far has been to publish in any journal, but insist on retaining copyright. I've had more luck with this than Pikhurko did so far; out of two publications in "unfriendly" journals, in both cases the publisher was willing to allow me to retain copyright (although they have a bit of trouble actually putting the correct copyright notice on the article).<|endoftext|> -TITLE: A Bijection Between the Reals and Infinite Binary Strings -QUESTION [6 upvotes]: Whenever possible, I like to present Cantor's diagonal proof of the uncountability of the reals to my undergraduates. For simplicity, I usually restrict to showing that the subset -$$ -A = \{x \in [0,1) \mid \text{ the decimal representation of $x$ uses only 0's and 1's} \} -$$ -is already uncountable. I was thinking recently that it would be nice to add a quick proof that $A$ is actually of precisely the same cardinality as $\mathbb{R}$. That is, I would like to: -Demonstrate a bijection between $A$ and $\mathbb{R}$. -My first instinct was to use find an injection from $A$ into $\mathbb{R}$ and vice versa, then appeal to Cantor-Bernstein to say that a bijection exists (even if we don't know how to construct it). The identity map suffices from $A$ into $\mathbb{R}$. For the other direction, I thought of something like "for $x \in \mathbb{R}$, map $x$ to its binary representation, disregarding the decimal point". I'm afraid this function fails to be injective, however. For example, 1 (base 10) can be represented as $.\overline{1}$ (base 2), and so 2 (base 10) can be represented as $1.\overline{1}$ (base 2). Thus, 1 and 2 (base 10) will have the same image under my map. -Any methods (not necessarily the one I've attempted to start here) are most welcome. I will accept as "correct" the method which demonstrates the bijection with the greatest level of clarity. - -REPLY [3 votes]: A more concrete way to fix the OP's idea (which is similar to Stefan's but avoids Cantor-Bernstein) is to simply delete $\mathbb{Q}$ from $A$ to produce a new set $B$. Split $B$ into a countable family of sets $B_k$ where $B_k$ consists of all the elements of $B$ with $k$ leading zeros. There is now an obvious bijection between any $B_k$ and any set of the form $[n,n+1)-\mathbb{Q}$ by simply viewing elements of $B_k$ as sequences in binary instead of decimal and ignore the leading zeros and first 1. There is no need to worry about the OP's original concern since $B$ only consists of irrationals. Since there are countably many $B_k$'s and countably many $[n,n+1)-\mathbb{Q}$, pick your favorite way to match them up. The remainder, $A\cap\mathbb{Q}$, is obviously countably infinite, so biject it with $\mathbb{Q}$.<|endoftext|> -TITLE: (almost) statistical independence of nodes degrees in a graph -QUESTION [11 upvotes]: Wireless networks are typically modeled as random geometric graphs. The number of nodes $N$ in the network is drawn from a Poisson distribution with intensity $\lambda$ -$$P(N = n) = \frac{\lambda^n e^{-n}}{n!}.$$ -Once this number has been chosen, the $N$ nodes are placed uniformly at random over a circular (or squared) area of radius $R$. Two nodes are then connected by an edge if their euclidean distance is less than some predefined range, say $r_0$, where $r_0 << R$. -As a result of Penrose's Theorem to ensure that the graph is $k$-connected, it is sufficient to show that the minimum degree is at least $k$, i.e., $d_{\rm min} \geq k$. This holds true asymptotically for a geometric random graph. -I came across two papers dealing with connectivity in (wireless) networks (this is one, and here is another one, but there are others by different Authors in the literature). When posing the condition that the minimum degree is at least $k$, I often find this approximation: -$$ P(G{\rm ~is~k~conn}) \cong P (d_{\rm min} \geq k) \cong P(d \geq k)^n$$ -where $P(d \geq k)$ is the probability that the degree of a node (any node) is at least $k$. The first approximation is essentially true when the number of nodes is at least of a few hundreds. The second approximation is not true since the nodes degrees are correlated. What people say when using it, is that they assume almost statistical independence of the nodes degrees. -In practice there is an excellent match between simulation results and the approximation above. Since I need to use the very same approximation, I was wondering if someone had a better argument to justify its use, rather than assuming its validity from the very beginning. - -REPLY [4 votes]: This is a partial answer concerning the validity of the second approximation. I'll consider the case $k=1$ for simplicity: all the difficulties and tricks will be clear from it already. -First of all, when people claim that some probability estimate agrees well with simulations, they always mean the difference metric. There is no way to distinguish between the probability $10^{-20}$ and $10^{-100}$ in practice though, for most theoretical purposes, the difference here is much larger than between $0.9$ and $0.1$. Thus, we are interested just in showing that the estimate in question is good when it is neither too small, nor too close to $1$. Assuming that we are on the torus (to avoid rather boring discussion of the boundary effects), and that the number of points $N$ is fixed (for the difference metric conditioning on $N$ is not a big problem) we can say that the estimate in question is just $(1-(1-\pi r^2)^{N-1})^N$, which is neither $0$, nor $1$ if $\pi r^2=\frac{\log N+O(1) }{N}$. We shall carry out the computations under this assumption only because the monotonicity of both the true probability and the estimate in $r$ is obvious. Denote $Q=(1-\pi r^2)^{N-1}\asymp \frac 1N$ -Enumerate the points $x_i$ to throw on the torus and let $f_i$ be the random variable that is $1$ if $x_i$ is isolated and $0$ if not. We are interested in $E\prod_i(1-f_i)$. The most natural idea is to use the truncated PIE. The first term after $1$ is $-\sum_i Ef_i=-NQ$, which is just what we need. -To use the truncated PIE, we need to estimate $Ef_1\dots f_m$. The estimate from below is straightforward: $x_{k+1}$ has the area at least $1-\min(k,m)\pi r^2$ to choose from, so we get almost $Q^m$ on the multiplicative scale if $m$ is at most a small power of $N$. To make a (rather crude) estimate from above, consider the connected components of the set $x_1,\dots,x_m$ with respect to the connection at distance $2r$ or less. If each component is single point, we get the lower bound above, i.e., essentially $Q^m$. Suppose that there are $p -TITLE: Cite articles or book where I first found the result? -QUESTION [12 upvotes]: I'm writing up a paper and I'm not sure how to cite a few things. It concerns a conjecture made by Quillen. Some work has been done showing it's true in some cases, false in others. These results I found in a book that had a chapter about the conjecture; of course, the book gave all the references. In the background of my paper, I want to briefly sum up these results. My question is, should I cite the articles, the book, or both? I wasn't sure what the etiquette/rules are in this situation. - -REPLY [11 votes]: The one sided answers neglect the complexity of the issue. First of all most of the papers in the science are nowdays difficult to judge, survey and check for an average reader, not to mention the users like science agencies and general public. The production is huge, there is lots of repetition and formalism differences make it difficult to bring knowledge and fruitfulness to the public and we need to fight this not by messy inclusion of everything but by making honest and informative choices to the reader. So, the main criterion is to write primarily for the reader and the author should have the stand weather the article she cites are likely to be useful to reader or not at all; if one does not take reader as the main criterion but some agenda of pushing the agenda for agencies, why would then the reader respect author's agenda of misleading the reader which reference is good and readable just to fulfill the political agenda ? In mathematics, the pointers from the text to the bibliography, together with the basic standard knowledge in the field should make obvious path how to achieve the results in the paper. This is already a difficult goal and making references only by history and not lead by self-analysis of the paper and its predecessors will make it too difficult to reconstruct the preliminaries used. Of course, one needs to account for balance in making clear what the original resources are either by making clear pointers to other surveys and also citing directly relevant and used work with some emphasis on the "primary" sources, where the latter term is not fully well defined, as each idea has some predecessors. -I disagree with much of the answer and comments of Deane Young, who is on one hand saying cite generously much, cite when in doubt etc. as if we live in an ideal world where the size of bibliography is allowed to be as large as you want, (Example: One of my papers was delayed for 2 years because it was 19 pages and the limit in the journal was 17. Roughly, two pages where bibliography.) and on the other hand saying that we live in a bad world where people depend on our citations, so cite even if you did not read/check and where we need to adapt to "funding agencies". Well, if we all adapt to present treand in funding agencies we agree with a wrong attitude that the citation is proportional to research credit what is not true: often the most cited are often surveys and books, popular digests, and also some original papers which are often only famous but nobody reads, or can read (so many times they are more readable than the public thinks, by inertion). It is better to push the agencies to take this into account and fund according to the full description of the discoveries of the author and description of their impact by listing consequences and new directions opened, and not of numerical quasi-impact. Conforming to numerical quasi-impact as a main principle may exactly push down honest people who cite only papers they trust, they check, they believe the authors and so on, we may also cite references which are written in bad way and not recommendable to the reader. To extend this remark to extreme which appears in practice, I know many people who cite famous people just to show that their own work is relevant for the trends pushed by those. This is the worst kind of citation. -There is also a pressure from referees and editors to include the references of their choice. In such cases one has to follow common sense, and decide if the request is reasonable, even with expense of possibly changing the journal. We should not regret the referees' and editor's time if they abuse their position to make people cite papers of their friends and their own; however we should appreciate it in good cases when their knowledge is channeled to point to the sources we did not use or appreciate and to learn what is the better or more original reference in many cases.<|endoftext|> -TITLE: Which sets of lattice points have rational generating functions? -QUESTION [14 upvotes]: Let $P$ be a subset of $\mathbb N^d$ (or of some normal pointed affine semigroup), and suppose that $f:=\sum_{p\in P}\ t^p\in\mathbb Z[[t_1,\ldots,t_d]]$ is a rational function. What can be said about the structure of $P$? In particular, must $P$ be a finite (disjoint) union of finitely generated modules over affine sub-semigroups? -Equivalently, must $P$ be a finite (disjoint) union of intersections of a rational polyhedron in $\mathbb N^d$ with a sublattice of $\mathbb Z^d$? -(My proximal motivation for asking is the appearance of both conditions in Guo & Miller, Lattice point methods for combinatorial games. One would like to speak on the level of rational generating functions, but the tools they develop only let them get at sets with a given decomposition into modules for sub-semigroups.) - -REPLY [12 votes]: I feel like there has to be an easier proof of this, but I just posted a note on my webpage proving the following Theorem. The key is a paper of Sam Payne's. -Let $f(t_1, \ldots, t_n)/g(t_1, \ldots, t_n) = \sum a(d_1, \ldots, d_n) t_1^{d_1} \cdots t_n^{d_n}$ be a rational function with coefficients in $\mathbb{Q}$. Let $\mathbb{C}_p$ be the completion of the algebraic closure of $\mathbb{Q}_p$, so $\mathbb{C}_{\infty}$ means the standard complex numbers. -We define a function $\phi: \mathbb{Z}_{\geq 0}^n \to \mathbb{Q}$ to be a quasi-polynomial if $\mathbb{Z}_{\geq 0}^n$ can be partitioned into finitely many sets $S_k$, each one the translate -of a finitely generated semi-group, such that the restriction of $\phi$ to each $S_k$ is a polynomial. -Theorem: The following are equivalent: -(1) The polynomial $g$ factors as $\prod_i \Phi_{d_i}\left( t_1^{e^i_1} \cdots t_n^{e^i_n} \right)$ where $\Phi_d$ is the $d$-th cylotomic polynomial and $(e^i_1, e^i_2, \ldots, e^i_n) \in \mathbb{Z}_{\geq 0}^n$, with at least one component of $e^i$ nonzero for each $i$. -(2) The function $(d_1, \ldots, d_n) \mapsto a(d_1, \ldots, d_n)$ is a quasi-polynomial. -(3) There are constants $C$ and $D$ such that -$$|a(d_1, \ldots, d_n)|_{\infty} \leq C \left( \sum d_i \right)^D$$ -and, for every finite prime $p$, there is a constant $C_p$ such that -$$|a(d_1, \ldots, d_n)|_{p} \leq C_p.$$ -(4) For every absolute value $| \ |_p$ on $\mathbb{Q}$ (including the archimedean norm), there are no zeroes of $g(t_1, \ldots, t_n)$ in the open polydisc $\{ (u_1, \ldots, u_n) \in \mathbb{C}_p : |u_1|, |u_2|, \ldots, |u_n| < 1 \}$. -In your setting, suppose that $\sum_{d \in P} t_1^{d_1} \cdots t_n^{d_n}$ is rational. Let $\chi_P$ be the characteristic function of $P$. It clearly obeys condition (3). So the theorem states that $\chi_P$ is a quasi-polynomial. Each of the polynomials making it up must have degree $0$, as it only assumes two values. So the support of $\chi_P$ (that is to say, the set $P$) must be a union of translates of finitely generated semi-groups. -Can someone tell me whether this is new? I think it might be worth publishing, if so.<|endoftext|> -TITLE: Useful tricks in experimental mathematics -QUESTION [30 upvotes]: There are a few computational tricks which are useful in experimental mathematics. -These tricks are mostly very elementary and often only given as exercices in books. -A typical example is the following: -Suppose that a sequence $s_0,s_1,s_2,\dots$ converges exponentially fast. Then the sequence $t_i=s_i-\frac{(s_{i+1}-s_i)^2}{s_{i+2}-2s_{i+1}+s_{i}}$ -converges (generally) faster and has the same limit. Having only access to a few initial -terms of a sequence which seems to converge quickly, this trick improves thus guesses concerning the limit. -This suggests two questions: - -Is there a nice book/article containing a list of useful tricks "ready for use"? -What tricks are useful for you? - -For clarity let me state that I do not count Euclid's algorithm, LLL or such things as -tricks. they are already implemented and ready for use in computer-algebra systems. (A nice -book concerning tricks might have however also ulterior chapters mentioning such useful algorithms and describing them very briefly.) - -REPLY [2 votes]: For the first question ("is there a nice book/article..."), I think the answer ie Yes: Sanjoy Mahajan's Street-Fighting Mathematics, which also exists in a free CC version, summarizes a good number of useful tricks and meta-tricks, some well-known, some less so.<|endoftext|> -TITLE: Example of an amenable finitely generated and presented group with a non-finitely generated subgroup -QUESTION [8 upvotes]: I'm looking for an example of a finitely presented and finitely generated amenable group, that has a subgroup which is not finitely generated. -The question is easy for finitely generated amenable group and an example is the lamp-lighter group $C_2\wr \mathbb{Z}$. -An Abelian and finitely generated group has no such subgroups. There exists a bigger class of groups with this property? - -REPLY [2 votes]: By the way, you may enjoy the fact, due to G. Baumslag, that a standard wreath product $W\wr G$ with $W\neq 1$ and $G$ infinite, is never finitely presented; see Gilbert Baumslag. Wreath products of finitely presented groups. Math. Z. 75 , 22-28, 1961. -For finite presentability of permutational wreath products, see a paper by Cornulier: http://www.normalesup.org/~cornulier/wrea_fp.pdf<|endoftext|> -TITLE: Some questions on Nicolai Reshetikhin's lectures on quantization of gauge theories. -QUESTION [5 upvotes]: This in reference to this fascinating lecture by Nicolai Reshetikhin- -http://staff.science.uva.nl/~nresheti/Holb-Quant-Gauge.pdf - -Given what is said on page 13 in section 4.1 its not clear to me why the ``partition function" should be a vector in the same space of states which is being assigned to the boundary of the space-time manifold. - -I am confused to see the claim that $Z(M) \in H(\partial M)$.I would have thought that the partition function is a linear function on the space of states which takes in a boundary configuration and gives back a number. - -I would like to know what is meant by saying (on the same page) that "..(this vector space of states) may depend on the extra structure at the boundary (it can be a vector bundle over the moduli space of such structures).." -How does the above relate to the claim on page 47 that for Chern-Simons theory the "..space of states assigned to the boundary is the space of holomorphic sections of the geometric quantization line bundle over the moduli space of flat connections in a trivial principal G-bundle over the boundary (provided we made a choice of complex structure)..." - -I would like to know what the above means and I would be happy to get back some further references about this...especially what is a "geometric quantization bundle"? - -Is the above somehow purely an effect of quantization? Thinking classically intuitively i would have thought that the space of states assigned to the boundary is the space of gauge equivalence classes of flat connections on the manifold which can be extended to a flat connection on the whole space-time. Is the above wrong? - -I can't see where the structures of a vector bundle and may be even its sections over the above moduli space seem to be getting involved. - -Is there a general argument to see that the space of states attached to the boundary is always a symplectic manifold and that the subspace of that will ever be picked up by solving the Euler-lagrangian equations in the interior will be some Lagrangian submanifold of it? - -In these lectures some comment seems to be made about how the gauge invariance may complicate the above scenario.I would like to know more about that. - -At the start of the lecture the author seems to demand that the space-times belong to a category of manifolds where the morphism is that of cobordism. I would like to know what was the intuition behind making this choice. Why not some other simpler morphism? - -Does working in this category of cobordisms somehow help in justifying the functoriality demand on the partition function with respect to gluing of manifolds? - -REPLY [9 votes]: This question is really many questions, and so it will be hard for anyone to answer all of them succinctly. I will try to answer some of them. - -Given what is said on page 13 in section 4.1 its not clear to me why the ``partition function" should be a vector in the same space of states which is being assigned to the boundary of the space-time manifold. -I am confused to see the claim that $Z(M)\in H(\partial M)$. I would have thought that the partition function is a linear function on the space of states which takes in a boundary configuration and gives back a number. - -Suppose that I have a functor $(H,Z)$ from the cobordism category to Vect. Then to a manifold $M$ with $\partial M$ decomposed into "incoming boundary" $\partial_{\rm in}M$ and "outgoing boundary" $\partial_{\rm out}M$, I get a morphism of vector spaces $H(\partial_{\rm in}M) \to H(\partial_{\rm out}M)$. In particular, if I take the same manifold but declare the entire boundary to be "incoming", I get what you propose; if I take the entire boundary to be "outgoing", I get what Reshetikhin says. -Moreover, consider the two "elbow macaroni" on a fixed codimension-$1$ manifold $Y$ (i.e. the manifolds $M = I \times Y$, where $I$ is the half-circle, topologically $I = [0,1]$, with either both endpoints marked "incoming" or both endpoints marked "outgoing"). The elbows get mapped under $Z$ to maps $H(Y) \otimes H(Y) \to H(\emptyset)$ and $H(\emptyset) \to H(Y) \otimes H(Y)$, and together identify $H(Y)$ with its dual space. Introduce just a little analysis (so that not all spaces must be finite-dimensional), and you decide that $H(Y)$ must be a Hilbert space. -To avoid the setting in the previous paragraph, one method is to have objects and morphisms in your cobordism category be oriented manifolds. Denote orientation-reversed $Y$ by $\overline Y$. Then any component of $\partial M$ has an induced ("outward", say) orientation, and you declare that a manifold with marked incoming-and-outgoing boundary is a morphism $\overline{\partial_{\rm in}M} \to \partial_{\rm out}M$. Then you do not quantize manifolds to Hilbert spaces but just to vector spaces, but you do have (via the elbows) that $H(\overline{Y})$ is the dual vector space to $H(Y)$. -In fact, most quantum field theories in nature are not define on the cobordism category of oriented manifolds, but only on the category of framed manifolds. - -I would like to know what is meant by saying (on the same page) that "..(this vector space of states) may depend on the extra structure at the boundary (it can be a vector bundle over the moduli space of such structures).." - -Most quantum field theories are not "topological", in the sense that they depend on geometric structure on spacetime. For example, you and I are particles in the Standard Model QFT, but the way that we interact depends sensitively on the distance between us. There are at least two equivalent ways to understand this dependence. One is to use a cobordism category whose objects and morphisms come equipped with extra structure. Another is to use the topological cobordism category, but to take as target not the category of vector spaces, but rather a category of vector bundles. Then the trick is to say: VectorBundles is fibered over Manifolds, and I have a morphism from the cobordism category to Manifolds that sends every object to the moduli space of geometric structures, and I can ask that my QFT intertwine these two maps to Manifolds. - -How does the above relate to the claim on page 47 that for Chern-Simons theory the "..space of states assigned to the boundary is the space of holomorphic sections of the geometric quantization line bundle over the moduli space of flat connections in a trivial principal G-bundle over the boundary (provided we made a choice of complex structure)..." -I would like to know what the above means and I would be happy to get back some further references about this...especially what is a "geometric quantization bundle"? -Is the above somehow purely an effect of quantization? Thinking classically intuitively i would have thought that the space of states assigned to the boundary is the space of gauge equivalence classes of flat connections on the manifold which can be extended to a flat - connection on the whole space-time. Is the above wrong? -I can't see where the structures of a vector bundle and may be even its sections over the above moduli space seem to be getting involved. -Is there a general argument to see that the space of states attached to the boundary is always a symplectic manifold and that the subspace of that will ever be picked up by solving the Euler-lagrangian equations in the interior will be some Lagrangian submanifold of it? - -Yes, you are wrong here. Note that your proposed "assignment" to a boundary manifold depends on what space it is the boundary of. -Glossing some details, classical Chern-Simons theory assigns to a 2-manifold $Y$ the (symplectic --- inherited from CS action) manifold $H(Y)$ of flat $G$-bundles (mod gauge transformations) over $Y$. For a fixed manifold $M$ with $\partial Y = M$, the space of flat bundles that extends over $M$ is a coisotropic submanifold of the space of flat bundles over $Y$. When $M$ is compact, this coisotropic is actually a Lagrangian $Z(M)$. -An analogy may be valuable. Classical Mechanics with configuration space $N$ assigns to a point the symplectic manifold ${\rm T}^* N$, and to an interval a Lagrangian inside $\overline{{\rm T}^* N} \times {{\rm T}^* N}$, where by $\overline \Box$ for $\Box$ a symplectic manifold I mean the same manifold with negative symplectic structure. For well-behaved systems, this Lagrangian correspondence is actually the graph of a symplectomorphism (the "time evolution function"). -Upon quantization, the symplectic manifold $H(Y)$ deforms to a vector space ($C^\infty(H(Y))$ deforms to $\operatorname{End}(H(Y))$), and one way to find/define this -vector space is via geometric quantization. - -At the start of the lecture the author seems to demand that the space-times belong to a category of manifolds where the morphism is that of cobordism. I would like to know what was the intuition behind making this choice. Why not some other simpler morphism? -Does working in this category of cobordisms somehow help in justifying the functoriality demand on the partition function with respect to gluing of manifolds? - -This idea goes back at least to Segal and Atiyah. Ultimately, the notion of "cobordism category" is probably wrong, because it doesn't handle well the more complicated style of cutting and gluing that can occur in higher-dimensional manifolds. (So you should read, e.g., Lurie's classification of TQFTs, and also "Blob Complex" by Morrison and Walker.) In any case, the cobordism category is a good approximation. -The point is simply that you do expect that whatever the "partition function" is, it should behave well under cutting and gluing of manifolds. Let's say you and I are physicists in adjoining laboratories. You do some experiments, so that you completely understand the physics inside your lab: this means in particular that you understand the expectation value of any boundary configuration for your laboratory. (Maybe you mark one end of your lab "incoming" and the other end "outgoing", and fully understand the relationship between incoming and outgoing states. Actually, maybe your lab is in a universe with a distinguished "time" axis, and then "incoming" and "outgoing" are pretty well defined.) This data of expectation values is "the partition function". -Now let's say that my laboratory is next to yours, and I have similar knowledge of expectation values for boundary configurations. Now we get a joint grant and want to merge labs: do we already know the physics for the two labs thought of as a single lab? Certainly yes if the labs do not touch: we just take the tensor product of laboratories. But if they have a common boundary, then when we merge the labs, that common boundary is now part of the interior. How should we work out the expectation values for field configurations on the boundary of the merged lab? We should integrate from our expectation values out all the possible fields on the common boundary. -This makes pretty good sense physically. The problem is to make is make sense mathematically. The cobordism category is well-suited to this (especially if instead you use a cobordism $\infty$-category that allows corners). -Two good warm-ups are: - -quantum mechanics with configuration space $N$ ($H({\rm pt}) = L^2(N)$ and $Z([0,t]) = \exp( i\hat E / \hbar)$, where $\hat E$ is the differential operator acting on $L^2$ normally called the "hamiltonian"; when the action of the system is (momentum)$^2$, $\hat E$ is the Laplacian. Note that in this case, "spacetime" is the interval, and it is a cobordism from a point to a point. Note that $\exp( i\hat E / \hbar)$ "is" the partition function for quantum mechanics. -classical string theory with configuration space $N$ (with Riemannian metric). Spacetimes are surfaces with boundary. $H(S^1) = {\rm T}^*(\operatorname{Maps}(S^1 \to N))$. (Classical theory, so valued in symplectic manifolds and Lagrangian correspondences, not Hilbert spaces and operators.) A cobordism $\Sigma$ ($\partial \Sigma = \coprod S^1$) is mapped to the Lagrangian correspondence that finds the locally-energy(=surface tension)-minimizing immersion $\Sigma \to N$ with given boundary $\coprod S^1 \to N$. Note that this Lagrangian is the graph of exp of a Hamilton-Jacobi "function", which assigns to any point in the Lagrangian the total energy of the corresponding surface. Note also that energy of a surface is additive, so exp of energy is multiplicative, along cuttings of a surface.<|endoftext|> -TITLE: Does the quadratic form $x^2 - 7y^2$ represent infinitely many primes, with the restriction that $0 < y < x/10$? -QUESTION [24 upvotes]: Surely yes, and in more generality, but can it be proved? -It seems that most, if not all, statements about quadratic forms representing primes fall back on algebraic number theory (i.e. splitting of primes in $\mathbb{Q}(\sqrt{7})$) for their proofs, and so are incompatible with the condition that $0 < y < x/10$. -Some related references which didn't lead to a proof: First of all there is this previous MO post, which suggests a negative answer. -There is also this paper of Iwaniec, which uses sieve methods but which also uses the multiplicative structure of solutions to the quadratic form. -There is also the interesting Theorem 5.36 of Iwaniec and Kowalski, which states that the arguments of prime elements of $\mathbb{Z}[i]$ are equidistributed in $(0, 2\pi)$. This is proved using the Hecke $L$-function $\sum_{\alpha \in \mathbb{Z}[i]} \big( \frac{\alpha}{|\alpha|} \big)^{ik} |\alpha|^{-s}$, for all $k$ divisible by 4. This generalizes further, but presumably not to real quadratic fields, where the infinite unit group would foul the construction up. -Finally, using a straight-up sieve (with only the additive structure of solutions to $x^2 - 7 y^2$) seems hopeless, as sieves tend to be bad at finding primes. There is the recent work of Friedlander-Iwaniec on $x^2 + y^4$ and Heath-Brown on $x^3 + 2y^3$, but these use algebraic number theory in $\mathbb{Q}(i)$ and $\mathbb{Q}(\sqrt{-3})$, and seem unlikely to generalize here. -I wonder if there is a promising approach out there which I have overlooked? Thank you! - -REPLY [16 votes]: The units of $k=\mathbf{Q}(\sqrt{7})$ have the form $\pm (8+3 \sqrt{7})^n$ with $n \in \mathbf{Z}$. If $\pi = x+y\sqrt{7}$ is a prime element of $k$, then $\lambda(\pi):= \log |x+y\sqrt{7}|$ is well-defined in $\mathbf{R}/\alpha \mathbf{Z}$ where $\alpha = \log(8+3\sqrt{7})$. Note that $\lambda$ factors as $\lambda = f \circ \sigma$ where $\sigma : k^{\times} \to \mathbf{R}^{\times}$ is a given embedding of $k$ and $f : \mathbf{R}^{\times} \to \mathbf{R}/\alpha \mathbf{Z}$ is a continuous group homomorphism. We can apply Hecke's theory of equidistribution (see Lang, Algebraic number theory, Chap. XV, especially Example 3 at the end of the chapter) to show that the sequence $\lambda(\pi)$ is equidistributed in $\mathbf{R}/\alpha \mathbf{Z}$ where $\pi$ runs through the primes of $k$ (with respect to the usual ordering on the norm of $\pi$). -You want $0 < y < x/10$ which translates into the inequality -\begin{equation*} -\sqrt{p} \leq x+y\sqrt{7} \leq C \sqrt{p} -\end{equation*} -where $C=\frac{10+\sqrt{7}}{\sqrt{93}}>1$ and $p=N_{k/\mathbf{Q}}(x+y\sqrt{7})$. This in turn is equivalent to $\lambda(\pi) \in [\frac12 \log p , \frac12 \log p + \log C]$ inside $\mathbf{R}/\alpha \mathbf{Z}$. -Using the equidistribution result above, the set $X=\{\pi : \lambda(\pi) \in [0,\frac12 \log C]\}$ has a positive natural density (here we consider only primes of $k$ which don't belong to $\mathbf{Q}$, but this is ok because the norm of a rational prime $p$ is equal to $p^2$, so these rational primes are negligible). Moreover, the set $Y=\{\frac12 \log p : \pi \in X\}$ is dense in $\mathbf{R}/\alpha \mathbf{Z}$ because of the prime number theorem. So we can find infinitely many primes $\pi \in X$ with $\frac12 \log p \in [-\frac12 \log C,0]$ inside $\mathbf{R}/\alpha \mathbf{Z}$, which implies what you want using the above discussion.<|endoftext|> -TITLE: Can an action of a compact Lie group be nontrivial if it is trivial on the boundary? -QUESTION [9 upvotes]: Let $G$ be a compact Lie group acting on a connected topological manifold $M$ with boundary. Suppose the action on one boundary component is trivial. Does it follow that the action on the whole of $M$ is trivial as well? -If $M$ and the action map $G\times M\to M$ are smooth, it is not too difficult to show that the answer is positive. Indeed, let $X$ be the set of all fixed points $x\in M$ of the action such that the action of $G$ on $T_xM$ is trivial. This set is closed, so it suffices to show it is open and non-empty (since $M$ is connected). To do so take a Riemannian metric on $M$ and average it to get a $G$-invariant metric. Using this one can show that $X$ contains the boundary component on which $G$ acts identically, so $X$ is non-empty. Moreover, if $x\in X$, then any $g\in G$ acts identically on a neighborhood of $x$ since $g(exp(v))=exp(dg_x v)$ for all $v$ in a sufficiently small neighborhood of $0\in T_xM$. -However, this argument uses smoothness and it is not clear if it can be adapted to the topological case. - -REPLY [11 votes]: Yes, it follows that the action of $G$ on all of $M$ is trivial. In brief this follows from what is known as "local Smith theory." Replace M by the union of $M$ and an open boundary collar on which $G$ acts as the product of the action on the boundary with a trivial action in the collar parameter. Then one would have an action on a connected manifold without boundary that is the identity on an open set. If $G$ is a $p$-group for some prime $p$ then local Smith theory says that each component of the fixed point set has the local mod $p$ (Cech) cohomology $H^{*}(F,F-\{x\};\mathbb{Z}_{p})$ groups of a manifold. It follows that the component of the fixed set containing the fixed boundary collar is all of $M$, plus the collar. For a general compact Lie group $G$ the kernel $K$ of the action is a closed subgroup that contains all elements of prime power order. This is enough to imply that $K=G$, i.e., that $G$ acts trivially.<|endoftext|> -TITLE: An easy proof that S(n) does not embed into A(n+1)? -QUESTION [15 upvotes]: Rotman's book An Introduction to the Theory of Groups (Fourth Edition) asks, on page 22, Exercise 2.8, to show that S(n) cannot be embedded in A(n+1), where S(n) = the symmetric group on n elements, and A(n) = the alternating group on n elements. I have a proof but it uses Bertrand's Postulate, which seems a bit much for page 22 of an introductory text. Does anyone have a more appropriate (i.e., easier) proof? - -REPLY [33 votes]: I think the following is sufficiently elementary: a transposition in $S_n$ is an element of order 2 commuting with at least $2(n-2)!$ elements of the group. But $A_{n+1}$ does not have such an element if $n$ is large enough. Indeed, if $\sigma\in A_{n+1}$ is of order 2, then it is a product of $k$ independent transpositions where $k$ is even and $2\le k\le(n+1)/2$. The number of elements of $A_{n+1}$ commuting with such $\sigma$ equals $2^{k-1}k!(n+1-2k)!$, and this is smaller than $2(n-2)!$ provided that $n\ge 6$. - -REPLY [7 votes]: One could ask Rotman. It may be that in a reorganization of the material in the book that problem ended up earlier than the material needed for the (intended) answer. On the other hand it is not a bad experience for students to see problems where the complete solution seems slightly out of reach. Here, one can prove several small cases and see various potential directions for a general proof. Which will work? which are in the spirit of the subject? Of course it is best to set up the expectation that there might be problems like this.<|endoftext|> -TITLE: If the total Chern class of a vector bundle factors, does it have a sub-bundle? -QUESTION [25 upvotes]: Motivation: $T_{\mathbb P^2}$ isn't an extension of line bundles - -Here's a trick to show that the tangent bundle $T$ of $\mathbb P^2$ is not an extension of line bundles. If it were, we would have a short exact sequence -$$\def\O{\mathcal O} -0\to \O(a)\to T\to \O(b)\to 0 -$$ -for some integers $a$ and $b$. Then we can compute the total Chern class -$$\begin{align*} -c(T)& =c(\O(a))\cdot c(\O(b)) \\ - &= (1+aH)(1+bH) \\ - &= 1+(a+b)H+abH^2, -\end{align*}$$ - -where $H=c_1(\O(1))$ is the class of a hyperplane. - -On the other hand, we have the Euler sequence -$$ -0\to \O\to \O(1)^3\to T\to 0 -$$ -which tells us that -$$\begin{align*} -c(T)&=c(T)\cdot c(\O)=c(\O(1)^3)\\ - &=c(\O(1))^3= 1+3H+3H^2. -\end{align*}$$ - -Now observe that there do not exist integers $a$ and $b$ so that $a+b=ab=3$, so $T$ cannot be an extension of line bundles. - -The Question -More generally, whenever we have an extension of vector bundles $0\to L\to E\to M\to 0$, we have $c(E)=c(L)\cdot c(M)$. So to show that $E$ has no sub-bundles (or equivalently, no quotient bundles), it suffices to show that $c(E)$ doesn't factor. The question is whether the converse is true: - -Suppose $E$ is a rank $r$ vector bundle on a (smooth quasi-projective) scheme (or manifold) $X$ so that $c(E)=c(L)c(M)$ for vector bundles $L$ and $M$ of rank $i$ and $r-i$, respctively. Must $E$ have a sub-bundle or rank $i$ or $r-i$? - -Remark 1: The phrasing is a bit strange compared to the natural-sounding "If the total Chern class of a vector bundle factors, does it have a sub-bundle?" The point is that knowing the rank of $E$ is very important. We showed that $T_{\mathbb P^2}$ has no sub-bundles, but $O(1)^3$ has the same total Chern class and clearly has lots of sub-bundles. -Remark 2: Does either $L$ or $M$ have to be a sub-bundle of $E$? NO! For example, on $\mathbb P^1$, we have that -$$ -c(\O(1)\oplus \O(-1)) = (1+H)(1-H)=1 = c(\O)c(\O) -$$ -but $\O(1)\oplus \O(-1)$ doesn't have a sub-bundle isomorphic to $\O$ (because it has no non-vanishing sections). -Remark 3: What is the answer in the case $X=\mathbb P^n$? - -REPLY [20 votes]: If you are also asking about the case of topological complex vector bundles over manifolds, consider the case $X=S^5$. There are no nontrivial rank $1$ bundles, but there is a nontrivial rank $2$ bundle, and of course its Chern class $1$ factors as $1\times 1$.<|endoftext|> -TITLE: Downsides of using the arXiv? -QUESTION [30 upvotes]: Not to sound like a refusenik or a contrarian, but I have always been a bit of an agnostic when it comes to the arXiv. Somewhere deep in my old Pine folders there is a 15 y.o. discussion with Greg Kuperberg on the subject, and it seems the history proved him right - the arXiv is now incredibly valuable and popular (perhaps, even a little too popular, see below). However, despite all the benefits, I think posting on the arXiv is a serious decision, often enough a good idea, but not always, and definitely not without downside as Matthew Daws writes (to a wide 27+ support): - But, definitely use the arxiv! I don't see any downside to putting a preprint on the arxiv; and it might lead to more people reading your work and hence more recognition. -While benefits of the arXiv are well known and understood, as evidenced by the growth in the volume of submissions, I believe negative aspects are somewhat less known. Thus, my question: -Do you know any downsides of using the arXiv? -If yes, tell us what are they. Is there a story behind? If no, say nothing, of course. I start with a few (mostly minor) quips below to get this started (see also here for a rare critical blog post in another field). - -1) If you are a graduate student or a junior faculty, you might not be as fast as others in developing your own ideas. As soon as you post early results which use your new ideas, they become a fair game. Now someone else can recognize their value and quickly solve your main problem before you are half way through. On the other hand, going the more traditional print publication route would give a couple of years cushion, sufficient in most cases. -2) On a related subject of destructive competition, arXiv can be really unhelpful. One graduate student I know liked a conjecture posed in an arXiv preprint. He solved it in about two months. When he was finishing writing his solution, somebody posted an identical solution. He was quite distressed. Another graduate student I know, discovered two weeks before the Ph.D. defense that the main result in her thesis was just posted on the arXiv in a more general form by a senior faculty elsewhere. Upon insistence of her advisor, she cancelled the defense and left academia without a Ph.D. degree. -3) An obvious point: some/many arXiv preprints are incorrect, leading to questions like this one. This creates a bizarre "neither solved nor open" status: while a solution of an important problem is being checked, no one wants to work on the problem. On rare occasions, two opposite solutions are posted leading to partial paralysis in the field. BTW, plagiarism is yet another variation on this issue (here the authorship is incorrect). -4) The other side of the same coin: if a person (like one friend of mine) posts an incorrect solution of a famous problem, this creates too much attention, potentially destroying a career (esp if in the early stages). -5) The "unaffiliated people problem" which makes it hard from people from third world countries, as well as anonymous authors to contribute (not everyone is as brave as Mnёv, see my answer here). While arXiv's restrictions do help get rid of some cranks, there are other ways to do that, and one can argue that one gem from an unaffiliated author is worth 100 crank papers. -6) Some people apparently read arXiv every day. Really? One friend of mine (in physics) admitted to me he spends 1.5-2 hours every morning doing that. Really??? -7) ArXiv's success also has downsides, as it increases pressure on young mathematicians to post, so as to keep up with others, even if their natural instinct in some cases maybe is to be protective of their ideas and further develop them before making them public. The social pressure can be quite strong, as hiring committees increasingly view arXiv preprints as "near publications". More anecdotally, one senior mathematician I know likes to ask people "What's your most recent arXiv paper?" in place of a more traditional "What are you working on?" as if arXiv posting is a "must do" for everyone. -8) An indirect and less obvious downside: arXiv's success clearly slowed down rather than sped up the natural tendency to bypass traditional print publishers for electronic media, as there is less of a pressure to have all journals nearly free and widely accessible. This is contrary to the early predictions which expected for the transition to happen before 2010. - -REPLY [51 votes]: There is no downside to arXiv. If you don't want to post a paper there, then don't. If you don't want to read papers on arXiv or to study the daily collection of abstracts of papers just posted, then don't. Nor is there any issue of plagiarism of ideas from arXiv that is different from plagiarism from a journal article. To publish an article means to make it public. A manuscript uploaded to arXiv is a publicly accessible and, therefore, a published manuscript, with the author's name attached to all the ideas, theorems, proofs, and conjectures that the paper contains. -There is also an important democratizing aspect of arXiv. In many fields of mathematics, preprints used to be circulated privately among a small group of insiders and their students. It was difficult for outsiders to learn about new results and problems, and to "break into" the club. It was essentially impossible for mathematicians in many poor countries to have access to research papers; the mathematicians were isolated and their universities could not afford journal subscriptions. Now all they need is internet access to find out about current research. The establishment of arXiv as a central repository for preprints in all fields of mathematics is an extraordinarily significant accomplishment, and will have a profound impact on the future of mathematics.<|endoftext|> -TITLE: Uniqueness of loop spaces -QUESTION [7 upvotes]: Suppose X is a loop space; by this we mean there is some space $Y$ with $\Omega Y \simeq X$. -Under what assumptions is (the homotopy type of) $Y$ unique? -As has been pointed out below, the homotopy type of $Y$ being determined uniquely is far from true in general. But for connected $Y$, are there conditions we can impose that make it so? - -REPLY [12 votes]: As others have pointed out, the generic case (whatever that should mean in this case) is that the loop structure on a loop space is not unique. However, things get quite interesting whenever we have a space that actually does have a unique loop structure. I highly recommend looking at: -Dwyer, Miller, Wilkerson: The homotopic uniqueness of $BS^3$, LNM 1298 -and -Dwyer, Miller, Wilkerson: Homotopical uniqueness of classifying spaces. Topology 31 (1992), no. 1, 29–45.<|endoftext|> -TITLE: Fourier analysis, orthogonality, and Plancherel for finite abelian groups -QUESTION [8 upvotes]: I am reading an outstanding paper by Bateman and Katz, improving the best known bounds on the cap set problem (Roth's theorem over $\mathbb{F}_3^N$). -The paper contains some technical lemmas for which I believe there must be an excellent geometric intuition -- which I am afraid I am missing. -Excerpting from, and simplifying, Section 8 of their paper, let -$A \subset Y := \mathbb{F}_3^n$ -be some subset, and also write $A(x)$ for the characteristic function of $A$. Let $H$ be a subspace of $Y$ and let $H^{\perp}$ be its annihilator. For $h \in H$, write $A_{H, h} := A \cap (H^{\perp} + h)$. Then we have a version of Plancherel -$$\sum_{h \in H} |\widehat{A}(h)|^2 = |H| 3^{-2N} \sum_{h \in H} |A_{H, h}|^2,$$ -and further, if $K$ is a subspace of $Y$ containing $H$, -$$\sum_{0 \neq k \in K} |\widehat{A}(k)|^2 = \sum_{0 \neq h \in H} |\widehat{A}(h)|^2 + \frac{1}{|H|} \sum_{h \in H} \sum_{0 \neq k \in K/H} |\widehat{A}_{H, h}(k)|^2.$$ -There are other interesting related formulas as well. The authors remark that the latter equality "can simply be thought of as Plancherel for a 'local Fourier -transform' of $A$. Here, we localize to the translates of $H^{\perp}$." -I can verify the identities readily enough, but I feel like there should be some excellent geometric intuition to be had here, with which all of these equalities are obvious. Is there anything that can be said which will render these equalities transparent? Perhaps some elaboration of the authors' remark? -Thank you! - -REPLY [5 votes]: I don't know whether it would be perceived as "geometric", but an intuition that "works for me" on this and related matters is that "Fourier analysis" on finite abelian groups is "abelian" Fourier analysis (e.g., on products of circles or lines, in the classical analytic scenarios) without the need to "do analysis". -Thus, the once-scandalous fact that the Fourier transform of Dirac delta is the identically-one function is true, without scandal. -More generally/similarly, the Fourier transform of the "delta" of a subgroup H is the collection of characters of the ambient G which restrict to the trivial character on H. -Translations twist by character-values... -Fourier transform of characteristic function of a set $A$ is the sum of the Fourier transforms of the deltas of points in the set. If the set contains or is contained in translates of subgroups, various reasonable identities hold. -Possibly this is all-to-obvious... and doesn't count as "geometric"... but this style of explanation works(-for-me) very well.<|endoftext|> -TITLE: incompleteness in real analysis -QUESTION [7 upvotes]: Godel's theorem tells us that any sufficiently powerful consistent formal theory of the integers is incomplete; but what about formal theories of the real numbers? More precisely, what about theories of the real numbers that are categorical (i.e. have only one model)? One such theory is given by the ordered field axioms plus the least upper bound axiom (every non-empty set of reals that is bounded above has a least upper bound). Note that the usual Archimedean property is not the kind of axiom we can include in a theory of the reals if we want it to be a complete theory, because then we'll need axioms that explain what an integer is, and these will cause us to fall prey to Godel's theorem. -Tarski's decision procedure for real-closed fields is somewhat relevant, but note that it does not answer my question, since the field of real numbers is not the only real-closed field. -Incompleteness is a slippery subject, and I'm glossing over important technicalities (I suspect that a logician would say I should be more specific about what I mean by "every non-empty set of reals"), so experts should feel free to edit my post if it's clear to them that my question is based on some misapprehension (as long as it's also clear to said experts how I would ask the question once my misapprehension were cleared up!). -UPDATE: Maybe my question should have been something more like: Is there a meta-theorem that guarantees that all the questions that are likely to arise in a real analysis course are decidable? Or: Is there a decision-procedure for an interesting fragment of real analysis that includes all the standard theorems from a first course in real analysis? Perhaps the right context for this question would be some first-order theory that has the set of subsets of the reals and the set of functions from the reals to itself as primitives, with enough (but not too many!) axioms. - -REPLY [7 votes]: Let me address the updated version of your question. -There is a philosophical current running through parts of descriptive set theory, and this includes anything that might be described as classical real analysis, to the effect that the realm of Borel mathematics is comparatively immune to the chaos of independence. On this view, one regards the Borel functions, relations and objects as being the most explicitly given, and the land of the Borel is the land of explicit mathematics. -For example, an important emerging field is the theory of equivalence relations under Borel reducibility, arising out of the observation that many of the most natural equivalence relations arising in other parts of mathematics, such as isomorphism relations on classes of algebraic structures, turn out to be Borel equivalence relations on a standard Borel space. Set-theorists seek to understand the comparative difficulty of the corresponding classification problems for these relations by considering the relations under Borel reducibility. This concept provides us with a precise way to measure the comparative difficulty of two classification problems, which then assemble themselves into a complex hiearchy, increasingly revealed to us. To give one example, it falls out of this theory that there can be no Borel classification of the finitely generated groups up to isomorphism by means of countable objects (this relation is not "smooth"). -This theory has been largely immune from the independence phenomenon, for several reasons. Perhaps the best explanation of this is the fact that Borel assertions have complexity $\Delta^1_1$, which lies below the Shoenfield absoluteness theorem. -Theorem(Shoenfield Absoluteness) Any statement of complexity $\Sigma^1_2$ is absolute between any two models of set theory with the same ordinals. -In particular, this implies that the method of forcing is completely unable to affect existence assertions about Borel objects, since such assertions would have complexity $\Sigma^1_1$, as well as more complex assertions. Because forcing is one of the principal tools by which set-theorists have come to exhibit independence, this means that Borel mathematics is completely immune from the forcing technology. -Furthermore, when there are sufficient large cardinals, then one can attain an even greater degree of absoluteness in various senses. For example, in the presence of large cardinals there are various strong senses in which the theory of $L(\mathbb{R})$ is invariant by forcing. Thus, even the realm of projective mathematics ($\Sigma^1_n$ for any $n$) is unaffected by forcing, when there are sufficient large cardinals. -At the same time, we know that it isn't strictly true even that Borel mathematics is immune from independence, since the $\Delta^1_1$ level of complexity includes all of arithmetic, which therefore admits the Gödel incompleteness phenomenon. But because the method of forcing is struck down, however, none of the more spectacular independence results in the realm of analysis, such as the independence results concerning CH and cardinal invariants, arise at the Borel level of complexity. Thus, I believe that the realm of Borel mathematics may be the best, although imperfect, answer to your updated question. -At the same time, it must be said that although the method of forcing is ruled out as a means of proving independence for Borel existence assertions, we have no meta-theorem that says that there will not be some future method that is able to establish independence for such assertions. Surely a major lesson of logic over the past century is the pervasiveness of the independence phenomenon, and I believe that it is only a matter of time for such methods to arise.<|endoftext|> -TITLE: Nilpotent Lie algebras of vector fields -QUESTION [7 upvotes]: Let $L$ be a finite-dimensional nilpotent subalgebra of the Lie algebra $W_n$ of all vector fields in $n$ variables (I am interested both in polynomial and formal vector fields). Does there exist a bound in terms of $n$ on the index of nilpotency of $L$? -For $n=1$ the answer is trivially "yes": every finite dimensional subalgebra is of dimension $\le 3$, concentrated in degrees $-1$, $0$, $1$. For $n>1$, I don't know. -Motivation: the question looks for me interesting in its own right, but also arises in control theory. There, one has a criterion for nonlinear systems to be so-called finitely discretizable (roughly, to admit a polynomial solution of some sort) in terms of nilpotency of the Lie algebra generated by the corresponding vector fields. So, when one applies this criterion on practice, one wants to be sure that it is enough to check the vanishing of the iterated Lie brackets up to a given degree. For the control theory application, the base field is $\mathbb R$, but the question does not seem to be dependent on the base field (as long as it is of characteristic zero at least). - -REPLY [6 votes]: I am afraid not: the subalgebra of $W_2$ spanned by $\partial_x, \partial_y, x\partial_y, \ldots, x^m\partial_y$ has the nilpotency index $m+1$.<|endoftext|> -TITLE: Do abelian spinorial prime three manifolds exist? -QUESTION [6 upvotes]: Does there exist a prime 3-manifold such that its mapping class group has an abelian representation in which the 2$\pi$ rotation is represented by -1? -In detail: -Let $M$ be a closed orientable prime 3-manifold. -Let $D_F(M,p)$ be the group of diffeomorphisms of $M$ that fix a point $p$ of $M$ and a frame there. Define the mapping class group (MCG) of $M$ to be the zeroth homotopy group of $D_F(M,p)$. Then the $2\pi$ rotation is an element of MCG that is the equivalence class of the following diffeo, $R_{2\pi}$: -Consider a coordinate ball of radius 2, $B2$, centred on $p$. $R_{2\pi}$ fixes the ball of radius 1, $B_1$, centred on $p$ and everything outside the sphere of radius 2. In between $B_1$ and $B_2$ the $R_{2\pi}$ maps $(x,y,z)\rightarrow(x\cos\theta+y\sin\theta,y\cos\theta−x\sin\theta,z)$ where $\theta$ is a function of $r=\sqrt(x^2+y^2+z^2)$ which increases smoothly and monotonically from 0 to 2$\pi$ as $r$ increases from 1 to 2. The square of the 2$\pi$ rotation is the identity in MCG. -A manifold is spinorial if $\[R_{2\pi}\]$ is non-trivial in MCG. -Background motivation: -This question is interesting because of the -possibility that fermions can be built on non-trivial -spatial topology. $M$ is the manifold of a 3-D spatial hypersurface in -spacetime. The fixed point is the point at infinity (where the metric -is asymptotically flat) and fixing a frame there has the same effect -on $\pi_0(D_F)$ as requiring some falloff conditions on the -diffeomorphisms at infinity or requiring them to be the identity -outside some ball. The configuration space of General Relativity -in this asymptotically flat setting is (space of asymptotically flat metrics on $M$)/$D_F$ -and its first homotopy group is isomorphic to (what I called above) -MCG, see http://arxiv.org/abs/math-ph/0606066 (I know it is not the usual definition of MCG). -The quantum state, on canonical quantisation of General Relativity, carries a unitary -irreducible representation (UIR) of the MCG and different choices of UIR give different -physics. Prime 3-manifolds are potentially candidates for -elementary particles built from pure geometry: topological geons -(this is speculative!). A prime 3-manifold can -be the basis for a spinorial particle (i.e. spin 1/2, spin 3/2 -....) if $R_{2\pi}$ is nontrivial. Because particles must be -able to be pair produced and annihilated, topology change must be -allowed in the theory which means that the theory should be -quantised in a sum-over-histories framework rather than a canonical -quantisation framework. Within the sum-over-histories -framework it is challenging to realise nonabelian reps of MCG. -Abelian reps on the other hand are more easily accommodated by attaching phases to -topologically distinct sectors of the path integral. -Moreover certain rules that -would result in a spin-statistics correlation for -topological geons would also force the reps to be abelian, http://arxiv.org/abs/gr-qc/9609064 (hence the need for abelian reps). However, if there were no spinorial primes with abelian reps -this would rule out spinorial geons and therefore fermions. - -REPLY [8 votes]: Yes, there exists such a manifold $M$. This follows if there exists aspherical $M$ with $Diff(M)\simeq 0$ (contractible) and $H^2(M;\mathbb{Z}/2\mathbb{Z})=0$. I claim there exists such manifolds. Let's see why such $M$ suffice. -There are two fibrations: -$$ D_F(M,p) \to Diff(M,p) \to GL(3,\mathbb{R})$$ -and -$$ Diff(M,p) \to Diff(M) \to M.$$ -The first comes from considering the derivative at $p$ of a diffeomorphism fixing $p$. The second comes from considering the image of $p$ under a diffeomorphism of $M$. -From the long exact sequence of homotopy for a fibration (e.g. 4.41 Hatcher), we have the exact sequence -$$0=\pi_2(M) \to \pi_1(Diff(M,p)) \to \pi_1(Diff(M))=0 \to \pi_1(M)\to $$ -$$\pi_0(Diff(M,p))\to \pi_0(Diff(M))=0.$$ -We see that $\pi_0(Diff(M,p))=\pi_1(M)$, and $\pi_1(Diff(M,p))=0$. -Similarly, -$$0=\pi_1(Diff(M,p))\to \mathbb{Z}/2\mathbb{Z}=\pi_1(GL(3,\mathbb{R})) \to \pi_0(D_F(M,p))\to \pi_0(Diff(M,p))\to 0 .$$ -From the second sequence, we see that $\pi_0(D_F(M,p))$ is therefore a $\mathbb{Z}/2\mathbb{Z}$ extension of $\pi_1(M)$. These are classified by $H^2(\pi_1(M);\mathbb{Z}/2\mathbb{Z})=H^2(M;\mathbb{Z}/2\mathbb{Z})=0$ (since any such extension is central). Thus, it is a trivial extension, so $\pi_0(D_F(M,p))=\mathbb{Z}/2\mathbb{Z}\times \pi_1(M)$. The $\mathbb{Z}/2\mathbb{Z}$ factor is generated by $[R_{2\pi}]$ in your notation, so clearly the desired abelian representation exists. -Now we need to see that a closed orientable aspherical manifold $M$ with $Diff(M)\simeq 0$ and $H^2(M;\mathbb{Z}/2)=0$ exists. In fact, we may assume $M$ is a hyperbolic homology sphere. For example, take a hyperbolic knot complement with trivial isometry group. Perform Dehn filling of slope $1/k$ for $k$ large to get a closed hyperbolic manifold. This manifold will be an aspherical homology sphere and will have trivial isometry group for $k>>0$ (this follows from Thurston's Dehn surgery theorem and the Margulis lemma, since the isometries must preserve the short core geodesic of the Dehn surgery). By a result of Gabai, $Diff(M)$ will be contractible. -One can work in somewhat greater generality with hyperbolic 3-manifolds $M$ which have the property that $H^2(Aut(\pi_1(M));\mathbb{Z}/2\mathbb{Z})=0$, since the first exact sequence implies that $\pi_0(Diff(M,p))=Aut(\pi_1(M))$. I'm not sure what happens if $H^2(Aut(\pi_1(M));\mathbb{Z}/2\mathbb{Z})\neq 0$; I assumed it $=0$ as a convenient way to see that the group $\pi_0(D_F(M,p))$ splits.<|endoftext|> -TITLE: Is the supremum of continuous functions integrable? -QUESTION [11 upvotes]: Let $f_\alpha$ be a family of continuous positive functions $\mathbb R\to \mathbb R$ -where the index $\alpha$ runs in a compact metric space -and the map $\alpha\to f_\alpha$ is continuous -with respect to compact-open topology on the target. Suppose there is a uniform -upper bound on the integrals of $f_\alpha$'s over $\mathbb R$. -Question. Is $\underset{\alpha}{\sup} f_\alpha$ necessarily an integrable function? -Apology. This sure sounds like a homework level question, but after looking at for a while I am not even sure what the answer is. - -REPLY [4 votes]: Just one more example: let $\phi \in C_c(\mathbb{R})$ with, say, $\sup_{\mathbb{R}} \phi = 1$. For $\alpha \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, set -$$f_\alpha(x) = \begin{cases} \phi(x - \tan \alpha), & \alpha \in (-\frac{\pi}{2}, \frac{\pi}{2}) \\\\ 0, &\alpha = \pm \frac{\pi}{2}. \end{cases}$$ -Then $\int f_\alpha = \int \phi < \infty$ for each $\alpha \ne \pm \frac{\pi}{2}$, and $\int f_\alpha = 0$ otherwise, but $\sup_\alpha f_\alpha = 1$.<|endoftext|> -TITLE: Why are equivariant homotopy groups not RO(G)-graded? -QUESTION [15 upvotes]: I know very little about the fancy equivariant stable homotopy category, so I apologize if this question is silly for one reason or another, but: -I think that stable homotopy, in the non-equivariant case, corresponds to the homology theory associated to the sphere spectrum. And indeed, in the equivariant context we have $RO(G)$-graded homology and cohomology theories for every spectrum. On the other hand, equivariant stable homotopy groups are defined by looking at maps, not from representation spheres, but from spheres smashed with homogeneous spaces. The resulting object has a sort of quirky bigrading, it seems to me, over the integers and subgroups of $G$. -Is there some reason that this choice is more natural than defining homotopy groups as $\pi_{\nu}(E) = [S^{\nu}, E]_G$ for $\nu \in RO(G)$? Is this object somehow trivial or something? -If the answer to this question is that I'm missing something easy or that the objects I gave above are trivial for some reason- then let me know but don't tell me the answer! It'll probably be good for me to figure that out on my own... - -REPLY [17 votes]: Actually, this is a subtle and interesting question, and it is to some extent model dependent. -When one defines the integer graded homotopy groups, one does so using colimits over representation -spheres, so that ``RO(G)-graded information'' is encoded in the definition of homotopy groups. This -is why one can get away with defining weak equivalence in terms of the Z-graded homotopy groups -of all fixed point spectra. -As Justin says, $RO(G)$-graded homotopy groups are of definite interest. Calculating the integer -part of $RO(G)$ graded cohomology theories obscures structure that is only visible in the -$RO(G)$ graded world. But then the coefficients of the theory are the $RO(G)$-graded homotopy -groups of the representing spectrum. -The cells that build up $G$-spectra in the model theoretic sense are model dependent. Using -symmetric or orthogonal $G$-spectra, one must use cells of the form $G\times_H D^V$, and one -can restrict to G-representations. In Lewis-May or EKMM $G$-spectra, where all objects are -fibrant, one can use just cells $G/H_+\wedge D^n$. Here again, the $RO(G)$-graded information -is built into these $G$-spectra, which is why integer grading suffices. -In Tom Goodwillie's space level answer the notion of ``get'' needs interpretation. One can get every weak homotopy type of $G$-space using $G$-CW complexes defined using only cells of the form $G/H \times D^n$, no representations in sight. Intuitively, as Justin says, this works because of the classical theorem that representation spheres are themselves finite $G$-CW complexes in this sense, but that is irrelevant to the proof. If one wants only finitely many cells to get a smooth $G$-manifold, then one must use representations of all subgroups, as Tom says (and even then I don't know a good reference). EDIT added: This last sentence is nonsense. I'll quote from a Math Review: -Illman, S\"oren -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. The main results of the paper are the following. -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. The other results of the paper are the technical preparation for the above results. The paper is written very carefully and has a detailed history of results leading to the above ones.'' -No representation spheres are needed. Where one needs representation spheres is -to construct dual triangulations of smooth compact manifolds and for work in -$RO(G)$-graded homology. See Chapter X, $G-CW(V)$ complexes and $RO(G)$-graded -cohomology, by Stefan Waner, in Equivariant homotopy and cohomology theory -(CBMS Conf. Series 91, AMS). -In any construction of the stable category of $G$-spectra, one wants to invert representation -spheres; for one of many reasons, as Justin says, Poincare' duality requires that. However, -when one inverts representation spheres, one actually inverts all homotopy spheres as well, -whether or not one wants to. See (Fausk, Lewis, May. The Picard group of the equivariant stable -homotopy category. Advances in Math 163(2001), 17--33). Philosophically, one might want to define homotopy groups using the entire Picard group of invertible objects, but that is calculationally unwise, to put it mildly. One can and should feel free to use either $Z$-graded or $RO(G)$-graded -homotopy groups, adapting the choice to the application at hand.<|endoftext|> -TITLE: When is the sum of two quadratic residues modulo a prime again a quadratic residue? -QUESTION [12 upvotes]: Let $p$ be an odd prime. I am interested in how many quadratic residues $a$ sre there such that $a+1$ is also a quadratic residue modulo $p$. I am sure that this number is -$$ -\frac{p-6+\text{mod}(p,4)}{4}, -$$ -but I have neither proof nor reference. It is a particular case of the question in the title: if $a$ and $b$ are quadratic residues modulo $p$, when is $a+b$ also a quadratic residue modulo $p$? -I came into this question when counting the number of diophantine $2$-tuples modulo $p$, that is, the number of pairs $\{ a,b\}\subset \mathbb{Z}^*_p$ such that $ab+1$ is a quadratic residue modulo $p$. - -REPLY [3 votes]: To complement the answers so far let me show using Gauss sums that the number of solutions of $ ax^2+by^2=c $ in $\mathbb{F}_p$ equals $p-\left(\frac{-ab}{p}\right)$ for any $a,b,c\in\mathbb{F}_p^\times$. Indeed, this number equals -$$ \frac{1}{p}\sum_n \sum_{x,y}e\left(n\frac{ax^2+by^2-c}{p}\right) -= \frac{1}{p}\sum_n e\left(\frac{-nc}{p}\right) -\sum_xe\left(\frac{nax^2}{p}\right)\sum_ye\left(\frac{nby^2}{p}\right),$$ -where all sums are over $\mathbb{F}_p$ and $e(t)$ abbreviates $e^{2\pi i t}$. -For $n\neq 0$ we have -$$ \sum_xe\left(\frac{nax^2}{p}\right)\sum_ye\left(\frac{nby^2}{p}\right) -= \left(\frac{na}{p}\right)\left(\frac{nb}{p}\right)\left(\sum_re\left(\frac{r^2}{p}\right)\right)^2 = \left(\frac{-ab}{p}\right)p,$$ -so that the count in question equals -$$ p+\left(\frac{-ab}{p}\right)\sum_{n\neq 0}e\left(\frac{-nc}{p}\right)=p-\left(\frac{-ab}{p}\right). $$<|endoftext|> -TITLE: Does the following Diophantine equation have nontrivial rational solutions? -QUESTION [5 upvotes]: Are there any solutions to the equation $s^{2}(1+t^{2})^{2}+t^{2}(1+s^{2})^{2}=u^2$ where $s,t,u\in \mathbb{Q}$ and $0 < s,t<1$? If so, is there a simple way to parametrize them all? -If I am understanding the geometry behind this problem, even if we pick a specific value for $t$ we are left with an elliptic curve, and it is possible there could be infinitely many solutions. -This question arises from some related (and somewhat esoteric) questions I have about rational points on unit circles. But this question looked "pretty" enough, I thought I'd ask the experts here. - -REPLY [7 votes]: (Edited with more details.) -I know this is a really old question, but this has a nice connection to a problem considered by Euler, what is now known as Euler bricks, and I couldn't resist. The OP's equation is equivalent to finding three rationals $a,b,c$ such that, -$$\begin{aligned} -a^2+b^2\; &= u_1^2\\ -a^2-c^2\; &= u_2^2\\ -b^2-c^2\; &= u_3^2 -\end{aligned}\tag1$$ -A solution (essentially by Euler) is, -$$a = \frac{s^2+1}{2s},\quad b = \frac{t^2+1}{2t},\quad c = 1\tag2$$ -where $s,t$ must satisfy, -$$s^2(t^2+1)^2+t^2(s^2+1)^2 = w^2\tag3$$ -and is the equation considered by the OP. For any solution to $(1)$ with $c\neq0$, then $s,t$ can be recovered as, -$$s = \frac{a\pm\sqrt{a^2-c^2}}{c} = \frac{a\pm u_2}{c}$$ -$$t = \frac{b\pm\sqrt{b^2-c^2}}{c} = \frac{b\pm u_3}{c}$$ -A small solution to $(3)$ is, -$$s = \frac{4p}{p^2-1},\quad t = \frac{3p^2+1}{p(p^2+3)}\tag4$$ -This is the essentially the same one given by Euler for Euler bricks, -$$\begin{aligned} -\alpha^2+\beta^2\; &= v_1^2\\ -\alpha^2+\gamma^2\; &= v_2^2\\ -\beta^2+\gamma^2\; &= v_3^2 -\end{aligned}$$ -where we have used $p \to p\sqrt{-1}$, and have correspondingly tweaked $(1)-(4)$. From this initial rational point $(4)$, one can then generate an infinite more.<|endoftext|> -TITLE: A series of articles should be published all in the same journal, or in different journals? -QUESTION [37 upvotes]: There is a series of papers, which build on each other, to be published. The series starts with purely mathematical articles, and ends with articles oriented to some applications. What would be the best approach: - -Submit all to the same journal. If so, what kind of journal would be best, a purely mathematical one, one of applications, or something between? -Submit them to different journals, possible ranging as well from pure mathematics to applied mathematics. -Try to make two series, one purely mathematical, and the other oriented towards applications, and send them to two journals. -Other solution. - -What advantages and disadvantages has each solution? - -REPLY [3 votes]: An old question, but an interesting example is Robertson & Seymour’s monumental series of papers Graph Minors I, Graph Minors II, Graph Minors III and so on up to Graph Minors XXIII. -All published in Journal of Combinatorial Theory, Series B EXCEPT for Graph Minors II, which was in Journal of Algorithms. -I guess we can count that as a vote for the “all in one journal” option.<|endoftext|> -TITLE: Fundamental groups of Calabi-Yau varieties -QUESTION [15 upvotes]: By a Calabi-Yau variety I mean a smooth projective variety over the complex numbers with numerically trivial canonical divisor. - -For each postive integer $n$ does there exist a finite group $G$ (possibly depending on $n$) which is not the fundamental group of a Calabi-Yau variety of dimension $n$? - -For $n=1,2$ this follows easily from the classification of Calabi-Yau varieties of these dimensions, the only non-trivial finite fundamental group being $\mathbb{Z}/2\mathbb{Z}$ (for Enriques surfaces). For $n=3$ some finite non-abelian groups are known to occur as fundamental groups but I do not know of any non-existence results. -If there are only finitely many families of Calabi-Yau varieties of a given dimension then the question would clearly have a positive answer. However, this is far from being known so I am interested in other possible approaches. - -REPLY [10 votes]: If n is even, then I believe you can use the Atiyah-Bott fixed point formula to rule out many cases. For instance, let G be a simple, non-cyclic group. Consider the action of G on the Hodge group $H^{0,n}(X)$. Since G has only the trivial character, this action must be trivial. Then for every element g in G, the holomorphic Lefschetz number is 2 (if n is odd, the number is 0, which doesn't help). Therefore g has a fixed point.<|endoftext|> -TITLE: Condition on a bipartite graph to have an $m$-factor -QUESTION [7 upvotes]: This might be the most stupid question I am ever posting here: I am asking for a proof or a counterexample to a problem I proposed on MathLinks long ago. -Let $G$ be a bipartite graph, i. e., a graph such that the set of its vertices is the union of two sets $A$ and $B$, and each edge of the graph $G$ connects a vertex from $A$ with a vertex from $B$. For every subset $U$ of $A$ and every integer $k$, let $N_{k}\left(U\right)$ denote the set of all vertices from $B$ which have at least $k$ neighbours in the set $U$. Assume that $\left|A\right|=\left|B\right|$. -Some matchings of $G$ are called disjoint if there is no edge common to two or more of these matchings. -Let $m$ be a positive integer. Prove or disprove that the graph $G$ has $m$ disjoint perfect matchings if and only if the inequality $\left|N_{1}\left(U\right)\right|+\left|N_{2}\left(U\right)\right|+...+\left|N_{m}\left(U\right)\right|\geq m\left|U\right|$ holds for every subset $U$ of $A$. -Note that probably the assumption $\left|A\right|=\left|B\right|$ can be dropped if we replace "perfect matchings" by "$A$-complete matchings" (i. e., every vertex of $A$ is matched in every of these matchings). Also note that the "only if" direction is easy. Finally, of course, for $m=1$, this is Hall's marriage theorem. -The fact that I have posted the above problem on MathLinks in 2007 speaks for it being easy, but the fact that I have always been too lazy to write up my solution speaks for it being wrong. I have tried the obvious induction approach now, but I fail to obtain a reasonable inequality for $m-1$ instead of $m$ after deleting the first perfect matching. Any ideas? - -REPLY [7 votes]: Take the general Ore-Ryser theorem: -Let $G$ be bipartite graph with vertex set $V=X\coprod Y$ ($X$, $Y$ parts) and $f:V\rightarrow \{0,1,2,\dots\}$ be a function such that $f\left(x\right)\leq\left(\text{degree of }x\right)$ for every $x\in X$. Then there exists a subgraph of $G$ (obtained from $G$ by removing some edges, not vertices), for which $f$ is degree function, if and only if for any $Y_1\subset Y$ one has -$$ -\sum_{y\in Y_1} f(y)\leq \sum_{x\in X} \min(f(x),d_{Y_1}(x)), -$$ -and this becomes an equality for $Y=Y_1$ (not necessarily only for $Y=Y_1$). Here, $d_{Y_1}(x)$ denotes the number of neighbors of $x$ in $Y_1$. -If $f(v)\equiv m$, it is easy to verify that we get your problem (or, strictly speaking, we get that one has a $m$-regular subgraph, but it is well known that it factors onto $m$ 1-regular subgraphs). -And the Ore-Ryser theorem may be proved by easy induction, the same as is used for the Hall marriage problem. Either there exists some nonempty $Y_1\ne Y$, for which the equality occurs, then we apply the induction propose for $X$ and $Y_1$ (with $f$ changed on $X$ to $\min(f,d_{Y_1})$) and then to $X$ and $Y\setminus Y_1$. Or we have strict inequalities for all nonempty $Y_1\ne Y$, then take any edge $xy$, remove it and replace $f(x)$ to $f(x)-1$, $f(y)$ to $f(y)-1$.<|endoftext|> -TITLE: Efficient computation of the least fraction with square denominator greater than the square root of 2. -QUESTION [8 upvotes]: The least rational number greater than $\sqrt{2}$ that can be written as a ratio of integers $x/y$ with $y\le10^{100}$ can be found in a moment using a little Python program. Can anyone write a program that finds, in hours rather than centuries, the least rational greater than $\sqrt{2}$ of the form $x/y^2$ with $y^2\le 10^{100}$? -More generally, my question is whether the following computation is known to be feasible or not feasible: -Given $N$, find the least rational greater than $\sqrt{2}$ of the form $x/y^2$, - with $x$ and $y$ integers and $y^2\le N$. For definiteness, let's say that the output should be the required rational written in lowest form. -By a feasible computation I mean one that can be done in $O((\log N)^k)$ bit operations for some constant $k$. -Of course the square root of 2 is not essential here. Any irrational would do, as long as comparisons with rationals are feasible. I don't know of any such irrational for which I can answer the question I've posed. - -REPLY [2 votes]: You should try the algorithms in Elkies' paper (from 2000) "Rational points near curves ..." http://arxiv.org/abs/math/0005139 . His idea is to cover the curve with a bunch of small rectangles, and use lattice basis reduction within each such region. He proves a result which either says that there are small number of solutions or all the solutions lie on a line.<|endoftext|> -TITLE: The Ramanujan Problems. -QUESTION [18 upvotes]: I originally thought of asking this question at the Mathematics Stackexchange, but then I decided that I'd have a better chance of a good discussion here. -In the Wikipedia page on Ramanujan, there is a link to a collection of problems posed by him. The page has a collection of about sixty problems which have appeared in the Journal of the Indian Mathematical Society. -While I was browsing through them at random, I came across this, which I recognized as the Brocard-Ramanujan problem. So, - -Are all these problems solved? Or are there more unsolved problems among them? -Why were the problems posed exactly? Was there some sort of contest in the journal back then? - - -Here is one problem from the list. (I just cannot resist posting it here!) - -Let $AB$ be a diameter and $BC$ be a chord of a circle $ABC$. Bisect the minor arc $BC$ at $M$; and draw a chord $BN$ equal to half the length of the chord $BC$. Join $AM$. Describe two circles with $A$ and $B$ as centers and $AM$ and $BN$ as radii cutting each other at $S$ and $S'$ and cutting the given circle again at the points $M'$ and $N'$ respectively. Join $AN$ and $BM$ intersecting at $R$ and also join $AN'$ and $BM'$ intersecting at $R'$. Through $B$, draw a tangent to the given circle meeting $AM$ and $AM'$ produced at $Q$ and $Q'$ respectively. Produce $AN$ and $M'B$ to meet at $P$ and also produce $AN'$ and $MB$ to meet at $P'$. Show that the eight points $PQRSS'R'Q'P'$ are cyclic and that the circle passing through these points is orthogonal to the given circle $ABC$. - -How does one even begin to prove this? I've tried to construct the entire thing using compass and straightedge, but the resulting mess only added to my confusion. Here is the link to the original problem. -Thanks a lot in advance! -EDIT: This problem is mentioned in the link provided in Andrey Rekalo's answer below in pp. 34-35. The solution is given in the second link in pp. 244-246. - -REPLY [3 votes]: You may want to see this paper as well by Berndt: - -http://www.math.uiuc.edu/~berndt/articles/galway.pdf<|endoftext|> -TITLE: Subspaces isomorphic to $C[0, \omega_1]$ -QUESTION [6 upvotes]: Let $\omega_1$ be smallest uncountable ordinal. I am trying to understand the possible "large" subspaces of $C[0,\omega_1]$, namely those which are isomorphic to the whole space. Therefore I have the following question: -Does every subspace of $C[0,\omega_1]$ isomorphic to $C[0,\omega_1]$ contain a complemented copy isomorphic to itself? The only (complemented) examples that I can "construct by hand", excluding the finite-codimensional ones, are of the form -$\mbox{cl lin}(\mathbf{1}_{[0,\gamma_{\sigma}]}\colon \sigma\leq \omega_1)$ -where $(\gamma_\sigma)_{\sigma<\omega_1}$ is increasing long sequence of limit ordinals and $\sigma_{\omega_1} = \omega_1$ (note that the family $(\{\mathbf{1}_{[0,\alpha]}\colon \alpha \leq \omega_1\})$ forms the long Schauder basis for $C[0,\omega_1]$). -Thank you, -T. - -REPLY [3 votes]: Yes, it can deduced from Lemma 1.2 combined with Proposition 2 of - -D. Alspach and Y. Benyamini, Primariness of spaces of continuous functions on ordinals. Israel J. Math. 27 (1977), 64–92. - -Another proof can be found here (Cor. 1.12).<|endoftext|> -TITLE: Integral representation of higher order derivatives -QUESTION [19 upvotes]: I'm quite curious about the following phenomena, that still puzzle me although I have a proof, and I'd be really glad if someone may shred some light, showing an interpretation or a generalization. I will sketch the computations at request, which consist on a manipulation of the integral formula of the remainder of the Taylor expansion. -1. Let $v\in C^\infty(\mathbb{R})$, vanishing at $0$ with some order $p\in\mathbb{N} _ +$ . In other words, the formal Taylor series of $v$ at $0$ belongs to the ideal $x^p\mathbb{R}[[x]]$ . Then, the function $w(x):=v(x)/x^p$ is also (extensible to) a $C^\infty(\mathbb{R})$ function, and we can express the $k-$ order derivative of $w$ at $x$ (say $x\ge0$) in terms of the derivatives of order $k+p$ of $v$ on $[0,x]$ as follows: -$$\frac{w^{(k)}(x)}{k!}=\frac{\int_0^x (x-s)^{p-1}s^k\, \frac{v^{(k+p)}(s)}{(k+p)!}ds}{\int_0^x (x-s)^{p-1}s^k\, ds} \, . $$ -In other terms, the $k-$th Taylor coefficient of $w$ at $x$ is an integral mean of the $(k+p)-$th Taylor coefficients of $v$, weighted with a Beta distribution on $[0,x]$. This is quite clear if $x=0$ and $v$ is analytic there, and not immediately obvious in general, -but has it a special meaning, or is it an instance of a more general principle? -2. Let $u\in C^\infty(\mathbb{R})$, and assume that the formal Taylor series of $u$ at $0$ belongs to $\mathbb{R}[[x^2]]$ . Then, the function $w(x):=u( \sqrt { x } ) $ is also (extendible to) a $C^\infty(\mathbb{R})$ function, and we can express the $k-$ order derivative of $w$ at $x^2$ in terms of the $2k-$ order derivatives of $v$ on $[0,x]$ as follows: -$$w^{(k)}(x^2)=\frac{\, (2x)^{-2k+1}}{(k-1)!\, }\, \int_0^x (x^2-t^2)^{k-1}u^{(2k)}(t) dt\, $$ -(this may also be written as an equality relating Taylor coefficients by means of an integral mean). -3. There is also a more general statement for a function $w(x):=u(x^{1/p})$ for $p\in\mathbb{N} _ +$, assuming that the formal Taylor series of $u$ is in $\mathbb{R}[[x^p]]$; the $k-$th Taylor coefficient of $w$ at $x^p$ is then an integral mean of the $kp-$th Taylor coefficients of $v$, supported on $[0,x]$, with certain densities depending on $p$ and $n$ recursively defined. -Is there a more general statement connecting analogously operations in $\mathbb{R}[[x]]$ and $C^\infty$ functions via integral means of their Taylor formal series? - -REPLY [10 votes]: This is follows directly from the Taylor development to order $p-1$ with integral remainder: -$$ -f(x) = 0+ \dots + 0+ x^p \int_0^1\frac{(1-t)^{p-1}}{(p-1)!}f^{(p)}(tx)dt. -$$ -Your formulas are differentiated versions of this. -By Whitney (Duke Math J. 10, 1943), if $f$ is invariant under the $\mathbb Z/(2)$-action $x\mapsto -x$ on $\mathbb R$, then $f(x)=g(x^2)$ for some smooth $g$. Gleaser (Ann. Math. 77, 1963) extended this as follows: If $f\in C^\infty(\mathbb R^n)$ is invariant under all permutations of the coordinates, then $f(x)=g(\sigma_1(x),\dots,\sigma_n(x))$ for a smooth function $g$, in the elementary symmetric functions $\sigma_i$. -G. Schwartz (Topology 14, 1975) extended this for any representation of a compact Lie group $G$ on $\mathbb R^n$, for any generating system $\rho_1,\dots,\rho_k$ of the algebra of $G$-invariant polynomials: $\rho^\star: C^\infty(\mathbb R^k)\to C^\infty(\mathbb R^n)^G$ -is surjective. -Mather (Topology 16, 1977) reproved this and showed that there is a linear continuous section of $\rho^\star$. Luna (Ann Inst Fourier 26, 1976) extended this to reductive Lie groups. - -The relation to your question is follows: You require only that the Taylor series of $f$ at zero is invariant under $x\mapsto -x$. So let $\tilde f(x) = \frac12(f(x)+f(-x))$. -Then $\tilde f - f$ is infinitely flat at 0, so $(\tilde f- f)(\sqrt x)$ for $x\ge 0$ makes sense and is smooth and can be extended to a smooth function $g$ on $\mathbb R$ evenly, oddly, or by 0. -$\tilde f(x) = \tilde g(x^2)$ for some smooth $\tilde g$. Then $\tilde g -g$ is what you looked for. -This recipe works for all representations of compact groups by integrating over the group. -Consider $F(z) = f(Re(z))$ and let $\mathbb Z/(p)$ act on $\mathbb C$ by $z\mapsto e^{2\pi i k/p}.z$ where $k=0,1,\dots p-1$. Then we are in the situation of 2. The invariant real polynomials are generated by -$Re(z^p)$ and $Im(z^p)$. Integrate over $\mathbb Z/(p)$ and use flat functions as in 2.<|endoftext|> -TITLE: Global sections of flat scheme also flat? -QUESTION [11 upvotes]: In the most naive form my question would be as follows: If $f:X\to \mathrm{Spec}\;A$ is a flat morphism of schemes is it true that $H^0(X,\mathcal{O}_X)$ is a flat $A$-module? -In general the answer is no: It is a theorem of Chase that a ring $A$ is coherent (meaning that every finitely generated ideal of $A$ is finitely presented) if and only if the product $\prod_I A$ is flat over $A$ for every set $I$. Hence if $A$ is not coherent one can find a set $I$ such that the global sections of $X=\coprod_I \mathrm{Spec}\;A$ are not flat over $A$ while of course $X$ is flat over $\mathrm{Spec}\;A$. An example of a non-coherent ring would be $\mathbb{Q}[X,Y_1,Y_2,\ldots]/(XY_1,XY_2,\ldots)$. -Does anyone know other examples for which the answer is no and which fulfill at least some finiteness conditions? For example is there a quasi-compact example? -Ultimately I'd love to know if it is true if $A$ is noetherian and $f$ is proper. - -REPLY [15 votes]: Here is a example with proper morphism over a noetherian base. We consider a DVR $R$, and we note $S=\mathrm{Spec}(R)$. Let $X/S$ be a proper flat curve $X/R$ which is not cohomologically flat, that is $\mathrm{R}^{1}f_{\ast}\mathcal{O}_X$ has actually torsion, and that $\mathrm{H}^{0}(X,\mathcal{O} _{X})=R$ (such a curve exists, for example, one can consider the proper minimal regular model of some projective smooth curve on $\mathrm{Frac}(R)$ without rational point, see for example Raynaud's paper on picard functor where you can find example for genus one curve). Then I claim that for some integer $n\gg 1$, the reduction $X_n:=X\otimes_R R/\pi^n$ provides an example that we need. Indeed, since $X/S$ is flat, we have the following short exact sequence -$$ -0\rightarrow \mathcal{O}_{X}\rightarrow \mathcal{O}_X\rightarrow \mathcal{O} _{X_n}\rightarrow 0 -$$ -here the first map is multiplication by $\pi^n$. Now the long exact sequence tells us -$$ -0\rightarrow \mathrm{H}^{0}(X,\mathcal{O}_{X})\rightarrow \mathrm{H}^0(X,\mathcal{O}_X)\rightarrow \mathrm{H}^{0}(X,\mathcal{O} _{X_n})\rightarrow \mathrm{H}^{1}(X,\mathcal{O}_{X})[\pi^n]\rightarrow 0 -$$ -Here the first map is the multiplication by $\pi^n$, while the last member is the $\pi^n$-torsion of $\mathrm{H}^{1}(X,\mathcal{O} _{X})$. Now when $n$ becomes sufficiently large, the last member becomes stable. On the other hand, $\mathrm{H}^{0}(X,\mathcal{O} _{X})=R$, hence we get the following exact sequence of $R$-modules -$$ -0\rightarrow R/\pi^n R \rightarrow \mathrm{H}^{0}(X,\mathcal{O} _{X_n})\rightarrow \mathrm{H}^{1}(X,\mathcal{O}_{X})[\pi^n]\rightarrow 0. -$$ -So in this way, we see that $\mathrm{H}^{0}(X,\mathcal{O} _{X_n})$ cannot be flat (hence free) over $R/\pi^n$ as such a free module must be of length a multiple of n, which is impossible since the last member of the previous short exact sequence is stable and non zero for $n\gg 0$.<|endoftext|> -TITLE: Are these two definitions of "uniformly distributed" equivalent? -QUESTION [9 upvotes]: For an article I am writing, I would like to know that two somewhat different -looking conditions are in fact equivalent. Here is the setting. $X$ is a compact -(and first countable) metric space and $\mu$ is a Radon probability measure on $X$. -That is: $\mu$ is a measure on the $\sigma$-algebra of Borel sets of $X$ (the -$\sigma$-algebra generated by the open sets), has total measure one, and is inner -regular, that is the measure of any Borel set $B$ is the sup of the measures of the -compact subsets of $B$. -Now let $\{x_n\}$ be a sequence in $X$. Let's say that this sequence is -"$\mu$-uniformly-distributed-A" if for any open subset $O$ of $X$ -$$ \mu(O) =\lim_{N \to \infty} { \#(O,N) \over N}$$ -where $\#(O,N)$ is the number of $x_k$ in $O$ for $k = 1, \ldots, N$. -(Or, in other words, the measure of $O$ is the "average number of the $x_n$ that are in $O$"). -On the other hand, let's say that the sequence is "$\mu$-uniformly-distributed-B" if for any -continuous real valued function $f : X \to R$, -$$ \int f(x) \, d\mu = \lim_{N \to \infty} {1\over N}\sum_{k = 1}^N f(x_k)$$ -(in other words the integral of $f$ is the "average value of $f$ on the $x_n$"). -(Note that if we assume this equality not for all continuous functions but rather for all -the characteristic functions of open sets, then it reduces to the definition of -"$\mu$-uniformly-distributed-A".) So, as you have no doubt guessed, what I want to know -is if "$\mu$-uniformly-distributed-A" and "$\mu$-uniformly-distributed-B" are in fact always -equivalent. -It is well-known that for the special case where $X = [0,1]$ and $\mu$ is Lebesgue measure -the two are equivalent---see for example Theorem B of section 3.5 of Volume 2 of Knuth's -"Art of computer programming" --- but I don't see how to generalize the argument there. -So does anyone know if this equivalence always does hold, and if so can they direct me -to a proof in the literature. - -REPLY [7 votes]: A sequence is $\mu-$ uniformly distributed -B iff the limit relation A holds for each Borel set $M$ whose boundary has $\mu-$ measure $0.$ -Edit -In the meantime I found the following books which have proofs of this theorem: -"Uniform distribution of sequences" by L. Kuipers and H. Niederreiter, Wiley 1974 -and P. Billingsley, "Convergence of probability measures", Wiley 1999.<|endoftext|> -TITLE: Are any natural examples of Gödel speed-up known? -QUESTION [27 upvotes]: In 1936 Gödel announced a theorem to the effect that proofs of certain theorems $T_1,T_2,\ldots$ become dramatically shorter when one passes from a formal system, such as Peano arithmetic PA, to a stronger one, such as a system in which Con(PA) is provable. More precisely, given any computable function $f$, we can find a sequence $T_1,T_2,\ldots$ of theorems such that $T_k$ has a proof of length of order $k$ in the stronger system, whereas any proof of $T_k$ in PA has length at least $f(k)$. -Various versions of this theorem have been proved, depending on the strengthening of PA chosen, and on the definition of length. See, in particular, this paper. However, I have not found a version with a -natural sequence of theorems $T_k$. For example, it seems plausible that one could use Goodstein's theorem, by taking -$T_k$ = The Goodstein process, starting with input $k$, eventually halts. -Are any such "natural'' examples of Gödel speed-up known? -Update and clarification. Gödel's speed-up theorem gives, for any computable function $f$, a sequence -of theorems $T_1,T_2,\ldots$ of PA such that each $T_k$ has a proof of length $O(k)$ in some strengthening of PA, -while the shortest proof of $T_k$ in PA has length $\ge f(k)$. In this theorem, the sequence $T_1, T_2,\ldots$ -depends on $f$. -If we want a "natural" sequence $T_1,T_2,\ldots$ (in particular, if $T_k=\varphi(k)$ for some fixed formula $\varphi$) -then we can no longer demand that $f$ be an arbitrary computable function, or even of arbitrary computable rate of -growth. This is because (assuming the sequence $T_1,T_2,\ldots$ is c.e.) the function -$g(k)$ = length of the shortest proof of $T_k$ in PA -is computable, so we cannot ask $f$ to grow faster then $g$. -So, since I want the sequence $T_1,T_2,\ldots$ to be fixed, I have to be satisfied if $T_k$ has shortest proof in PA -with length of $O(f(k))$ some reasonably fast-growing $f$. It seems that Harvey Friedman has examples that fit the -bill, as Richard Borcherds has pointed out. However, before I accept Richard's answer, I would like to know a precise -reference. I have pored over Harvey Friedman's Research on the Foundations of Mathematics (North-Holland 1985), -and some other works, without finding a clear statement of speed-up in the above sense. - -REPLY [3 votes]: Take Goodstein's theorem for a particular n. The general case is not provable in PA. So, in higher order logic, the proof is of same length for every n. In PA the proof length grows when n becomes larger.<|endoftext|> -TITLE: Help wanted with Chebotarev condition in characteristic 2 -QUESTION [13 upvotes]: Having promised a longtime collaborator that I would clear my plate to finish up some joint work of ours, I am swallowing my pride and tossing up the following technical point of function field arithmetic to the experts: -Let $k_0 = \mathbb{F}_2(t)$, and let $k(P)$ be a finite separable extension of $k_0$. Let $S$ be a finite set of places $v$ of $k_0$ (thus corresponding to monic irreducible polynomials in $\mathbb{F}_2[t]$ together with possibly the place at infinity). Consider the set of monic irreducible polynomials $w \in \mathbb{F}_2[t]$ with the following properties: -(i) Every place $v$ in $S$ splits completely in the quadratic Artin-Schreier extension of $k_0$ defined -by the polynomial $X^2 + X = \frac{1}{w}$, and -(ii) $w$ splits completely in $k(P)$. -Problem: Show that there are infinitely many such $w$. -The natural strategy is to show that the set of such $w$ satisfies some Chebotarev condition -- presumably it even contains the set of all primes splitting in a certain finite Galois extension of $k_0$ -- and then apply the Chebotarev Density Theorem in this context (e.g. Theorem 9.13A in Rosen's Number Theory in Function Fields). -I have already proved similar statements when $k_0$ is a number field or is $\mathbb{F}_p(t)$ with $p$ an odd prime: for the latter I used the Quadratic Reciprocity Law in such fields. In the present case I was trying to use Hasse's characteristic 2 Quadratic Reciprocity Law -- for which my only exposure is this nice note of K. Conrad -- but after some hours of fiddling around, I am having trouble making this work: in the above formulation at least, what is given is not exactly a reciprocity law -- i.e., it doesn't directly compare $[w,\ell_v)$ to $[\ell_v,w)$ but only establishes a certain periodicity relation which is, in the classical cases, equivalent to QR. -Many thanks if you can help me out with this! I really need it in order to complete revisions on a long overdue paper, so a solution will be worth an acknowledgment in the paper at the least. - -REPLY [16 votes]: By my comments to the question, we can assume the infinite place is not in $S$ (because its splitting in the indicated quadratic extensions is automatic). Let the elements of $S$ be $\pi_1,\dots,\pi_r$, where the $\pi_i$'s are distinct (monic) irreducibles in ${\mathbf F}_2[t]$. For each $\pi_i$, the Artin-Schreier map $\wp(x) = x^2 + x$ on the field ${\mathbf F}_2[t]/(\pi_i)$ has image equal to half of that field. Fix a nonzero element of the image, say $g_i \bmod \pi_i$ (why? you'll see shortly). For a monic irreducible $\pi$ in ${\mathbf F}_2[t]$, if $\pi \equiv 1/g_i \bmod \pi_i$ then the congruence $x^2 + x \equiv 1/\pi \bmod \pi_i$ is solvable in ${\mathbf F}_2[t]/(\pi_i)$ by the choice of $g_i$, and then by Hensel's lemma the polynomial $X^2 + X - 1/\pi$ splits in the completion ${\mathbf F}_2(t)$ at $\pi_i$, so $\pi_i$ splits in the quadratic extension of ${\mathbf F}_2(t)$ obtained by adjoining a root of $X^2 + X - 1/\pi$. -So what you would like is to find infinitely many monic $\pi$ such that $\pi \equiv 1/g_i \bmod \pi_i$ for $i = 1,\dots,r$ and $\pi$ splits completely in a predetermined finite separable extension $L/{\mathbf F}_2(t)$. We can assume $L/{\mathbf F}_2(t)$ is Galois by replacing $L$ with its Galois closure over ${\mathbf F}_2(t)$, and then the hypothesis that $\pi$ splits completely in $L$ is the same as saying $\pi$ is unramified in $L$ with trivial Frobenius conjugacy class in the Galois group of $L/{\mathbf F}_2(t)$. -At the same time, the congruence condition $\pi \equiv 1/g_i \bmod \pi_i$ for monic irred. $\pi$ is also a Frobenius constraint using Carlitz extensions. The group $({\mathbf F}_2[t]/(\pi_i))^\times$ is naturally the Galois group of the splitting field over ${\mathbf F}_2(t)$ of the Carlitz polynomial associated to $\pi_i$. Write this splitting field as $K_{\pi_i}$. The natural isomorphism of the (abelian) Galois group of $K_{\pi_i}/{\mathbf F}_2(t)$ with $({\mathbf F}_2[t]/(\pi_i))^\times$ identifies the Frobenius element attached to $\pi$ with $\pi \bmod \pi_i$ for any monic irred. $\pi$ in ${\mathbf F}_2[t]$ not equal to $\pi_i$. (This is analogous to the way $({\mathbf Z}/(m))^\times$ is identified with the Galois group of a cyclotomic field with the Frobenius at a prime $p$ being $p \bmod m$.) Therefore asking that the monic irred. $\pi$ satisfy $\pi \equiv 1/g_i \bmod \pi_i$ for all $i$ is a bunch of simultaneous Frobenius conditions on $\pi$ in the fields $K_{\pi_1},\dots,K_{\pi_r}$. (Warning: these Frobenius conditions for $\pi$ are usually not conditions for $\pi$ to split completely in $K_{\pi_i}$ because we're working with reciprocals of nonzero elements mod $\pi_i$ that are in the image of the Artin-Schreier map, and the Artin-Schreier map has no good multiplicative properties. Maybe you could get a split completely interpretation using the reciprocal of a nonzero element in the image of the Artin-Schreier map mod $\pi_i$ which is itself in the image of the Artin-Schreier map mod $\pi_i$. A count shows it is very likely that there is such an overlap, but I haven't checked it for certain.) -The Carlitz extensions $K_{\pi_1},\dots,K_{\pi_r}$ are linearly disjoint over $F_2(t)$, so there is no problem finding infinitely many $\pi$ with given Frobenius elements in each of the Galois groups of $K_{\pi_i}/{\mathbf F}_2(t)$. (This is also a consequence of the Chinese remainder theorem and Dirichlet's theorem modulo $\pi_1\cdots\pi_r$.) If the field $L$ is linearly disjoint from all the fields $K_{\pi_i}$ over ${\mathbf F}_2(t)$ then we can express all of Pete's original conditions as a single Frobenius condition in the Galois group of their composite field over ${\mathbf F}_2(t)$ and that has infinitely many solutions by Chebotarev. On the other hand, if $L$ is not linearly disjoint from those Carlitz extensions over ${\mathbf F}_2(t)$ then there could be some subtle compatibility issues to be sure Chebotarev can be applied. I'll stop at this point since I've shown the basic plan for how to interpret everything as a Frobenius condition in a Galois group over ${\mathbf F}_2(t)$.<|endoftext|> -TITLE: Computation of Gromov-Witten invariants for symplectic manifolds -QUESTION [12 upvotes]: According to references, Gromov-Witten invariants were first defined for symplectic manifolds and later for projective varieties algebraically, and they coincide on the overlap. Because I thought Gromov-Witten invariants are important invariants for symplectic manifolds, I looked up papers about the computation. But recent results I found were about algebraic ones. Also simple examples calculated in textbooks don't seem to use symplectic structures. -I wonder why people don't do computations in symplectic category. Is it because of the difficulty, or are there some other reasons? Maybe I am just ignorant about it. I am more familiar with symplectic geometry, so I want to know methods to compute the invariant symplectically. In particular, I want to know if explicit computation was done for some non-K\"ahler manifolds. And is it recommended to study albebraic techniques of computation even if what I am interested in is symplectic manifold? - -REPLY [4 votes]: I noticed this question a few weeks ago but waited to post an answer before I could say: here's a new computation (http://arxiv.org/abs/1106.3959) for a class of non-Kaehler manifolds. -Psychologically one reason that the algebraic category is easier is that gluing works more easily there. For instance, Robbin-Ruan-Salamon have a paper (http://www.math.ethz.ch/~salamon/PREPRINTS/smUW.pdf) that puts a smooth orbifold structure on the (compactified) moduli space of regular stable maps into an integrable almost complex manifold (without using any of the standard gluing techniques). One feels happier about using e.g. equivariant localisation there than in the non-Kaehler case where the smooth structure on any given moduli space (and its obstruction bundle) is maybe not so canonical. -Of course, as the other posters point out, the main reason there are few computations is that it's not so easy to find semipositive (e.g. monotone) non-Kaehler manifolds (for which one can use the "standard" definitions of GW invariants due to Ruan-Tian/McDuff-Salamon).<|endoftext|> -TITLE: $< \aleph_1-$support Product of Cohen forcings -QUESTION [5 upvotes]: Suppose $\kappa$ is an inaccessible cardinal, and let $P$ be the $< \aleph_1-$support product of $Add(\alpha^{++}, 1)$ for singular cardinals $\alpha < \kappa.$ -1- Does this forcing preserve cardinals? -2-(A weaker question) Does $\kappa$ remain inaccessible in the generic extension? - -REPLY [11 votes]: Your forcing notion will collapse all uncountable cardinals -below $\kappa$ to $\aleph_1$. To see this, fix any -uncountable $\gamma\lt\kappa$. Consider the first -$\aleph_1$ many cardinals after $\gamma$ at which forcing -occurs. For the $\xi^{th}$ such cardinal $\alpha$, we are adding a -subset to $\alpha^{++}$. Consider the first ordinal in the set -added at this stage. This can be any ordinal up to -$\alpha^{++}$, including any ordinal below $\gamma$. Because you are using countable support, any condition specifies nontrivial sets only on a countable number of these cardinals, and so it is dense that any particular $\beta\lt\gamma$ appears at least -once in such a way. Thus, in the generic extension, there -will be a surjective map from $\aleph_1$ onto $\gamma$, and -so $\gamma$ is collapsed. -The cardinal $\kappa$ itself is not collapsed, by a -$\Delta$-system argument (and neither is any cardinal above -$\kappa$ collapsed), and so this forcing makes $\kappa$ the -$\aleph_2$ of the extension. In particular, it does not -preserve the inaccessibility of $\kappa$.<|endoftext|> -TITLE: Intrinsically measurable subsets of amenable semigroups. -QUESTION [6 upvotes]: This question is related to the one in https://mathoverflow.net/questions/65322/the-structure-of-certain-maximal-sets-of-means-into-amenable-semigroups. I open a different topic because they can be treated separately. -Let $S$ be a countable amenable semigroup and let $IM(S)$ be the set of the invariant means on $S$. - -Definition. For any subset $W\subseteq S$, we define - $$ -W^-=\inf_{m\in IM(S)}m(\chi_W)\;\;\;\;\;\;\;\;\;\;\;and\;\;\;\;\;\;\;\;\;\;\;\;W^+=\sup_{m\in IM(S)}m(\chi_W) -$$ - -There are many and well-known examples that show that these numbers could be different. I am interested in the case when they are equal. As far as I know it seems to me that this situation has no name in literature, so let me fix the following terminology - -Definition. $W$ is said to have the property IM (Intrinsically Measurable) if $W^-=W^+$. In this case $\mu(W)$ denotes the intrinsic measure of $W$. - -Basic examples of sets with IM are easy. I need to give such examples in order to formulate my first question. First of all, recall the following (almost) classical - -Definition - Let $k$ be a positive integer possibly infinite. A subset $W\subseteq S$ is called $k$-tile if there are $s_1, s_2,... s_k$ elements in $S$ such that - -$s_iW\cap s_jW=\emptyset$, for all $i\neq j$ -$S\setminus\bigcup s_iW$ has the property IM and $\mu(S\setminus\bigcup s_iW)=0$ - - -Tiles obviously have the property IM. Moreover the following operations preserve the property IM: finite union of disjoint sets with IM; if $V\subseteq T$ have the property IM, then also $T\setminus V$ have IM. Let me call elementary those sets with the IM that can be obtained by tiles using the previous operations. -Now I list some basic questions about this property IM. The second question is not just a specification of the first one, but it is interesting for the application that is the motivation for studying this property. - -Question 1. Is there an (explicit) example of a non-elementary IM set $W$ with $\mu(W)>0$? -Question 2. Let $S$ be the multiplicative semigroup of positive integers and let $W$ be the subset of those positive integers with first digit $1$, $2$ or $3$. Does $W$ have the property IM? - -Note that the property IM is not closed under countable union, but even the answer to the following question is not completely evident to me - -Question 3. Is the class of IM sets closed under finite intersection? - [This question has been answered in the negative by Ben Willson (see below)] - -Thanks in advance for any comment, -Valerio - -REPLY [5 votes]: You should look at -Joseph Rosenblatt and Zhuchang Yang, Functions with a unique mean value, Illinois Journal of Mathematics, Volume 34, Number 4, Winter 1990 -where various properties of sets and functions with unique invariant mean a studied. It is a standard fact that Bohr sets in $\mathbb Z$ (or any amenable group) have a unique mean. For example: -$$\mu\left(\left\lbrace n \in \mathbb Z \ \left| \ \min_{m \in \mathbb Z}|\sqrt{2} \cdot n - m| \leq \varepsilon \right.\right\rbrace\right) = 2\varepsilon.$$ -This set is non-elementary in your sense. -EDIT: The rough idea is that if there exist $q$ translates of a set such that the union covers each group element at most $p$ times, then the measure is less that $p/q$. Similarily, if the union of the translates covers each group element at least $p'$ times, then the measure is bigger than $p'/q$. Now, translating the set corresponds (up to some $\delta)$ to translating the interval of length $2 \varepsilon$. It is now easy to see that $p/q$ and $p'/q$ can be made to converge to $2 \varepsilon$.<|endoftext|> -TITLE: fundamental domain of universal covering -QUESTION [6 upvotes]: Let $M$ be a connected compact manifold without boundary, $\pi:\widetilde{M}\to M$ be the universal covering map. A fundamental domain of $(\pi,\widetilde{M}, M)$ is a compact subset $D\subset \widetilde{M}$ such that -1. the union of $\gamma D$ over all $\gamma\in \pi_1(M)$ covers $\widetilde{M}$, -2. the collection $\gamma D^o$ are mutually disjoint, -3. $\pi(D)=M$ and the restriction $\pi|_{D^o}:D^o\to M$ is diffeomorphic onto its image. -My question is: -Does there always exist some simply connected fundamental domain? -Is every fundamental domain simply connected? -Motivation. -I saw the following statement in several papers about dynamical systems: let $B^d(0,1)$ be the unit ball in $\mathbb{R}^d$ and $M$ be a $d$-dimensional compact connected manifold without boundary. Then $M\simeq B^d(0,1)/\sim$ where $\sim $ represents some gluing along $S^{d-1}=\partial B^d(0,1)$. -I think the statement might be related to above question. -Thanks! - -REPLY [5 votes]: The first part of 3 follows from 1. The second part of 3 follows from 2. -There is always a contractible (in particular simply connected) fundamental domain: -Triangulate $M$. Take the union of all the codimension zero open simplices, together with just enough codimension one open simplices to make it connected. This will be $\pi(D^{o})$. It is an open subset of $M$, dense and contractible (homotopy type of a maximal tree in the dual cell structure). Choose a lifting to $\tilde M$ and let $D$ be its closure. -For an example of a non simply connected fundamental domain in dimension $3$, take $M=S^2\times S^1$, $\tilde M=S^2\times \mathbb R$. One fundamental domain is $S^2\times I$, but you can make another by first cutting $S^2\times I$ into two non simply connected pieces meeting along a closed surface and then letting $D$ be the union of the left piece and a translate of the right piece. -Edit: This wrong, as pointed out in the comments. My error ws in imagining that the inclusion of the lifted open thing into its closure was necessarily a homotopy equivalence. I wonder if this approach can be fixed.<|endoftext|> -TITLE: Model category structures on the category of $L_\infty$-algebras -QUESTION [7 upvotes]: Let $k$ be a characteristic zero field. Then it is known that the forgetful functor $dgla(k)\to chain(k)$ from differential graded Lie algebras (over $k$) to cochain complexes induces a model category structure on $dgla(k)$ with "the same" fibrations and weak equivalences as on $chain(k)$, i.e., fibrations are surjective dgla morphisms and weak-equivalences are quasi-isomorphisms. -Also on the category of $L_\infty$-algebras over $k$ there is a forgetful functor $L_\infty(k)\to chain(k)$, which picks the linear part of an $L_\infty$-morphism. - Then, on $L_\infty(k)$ we have two natural functors: the forgetful functor $L_\infty(k)\to chain(k)$ and the embedding $L_\infty(k)\hookrightarrow dgcu(k)$, where $dgcu(k)$ is the category of differential graded counitary cocommutative coalgebras over $k$. -This suggests we could have two natural model category structures on $L_\infty(k)$, and my question is: how are they related? do they coincide? in particular, is a morphism of $L_\infty$-algebras whose linear part is surjective a fibration in the $dgcu(k)$ model structure? is a quasi-isomorphism of $L_\infty$-algebras (i.e., a morphism of $L_\infty$-algebras whose linear part is a quasi-isomorphism) a weak-equivalence in the the $dgcu(k)$ model structure? - -REPLY [7 votes]: If you start with the category of $L_\infty$-algebras with "strict morphisms" ($L_\infty$-morphisms such that the higher components vanish), then you can put a model category structure by the classical operadic means: this category is the category of algebras over the operad $L_\infty:=\Omega( \text{Koszul dual of}\ Lie)$. -Now if you consider the category of $L_\infty$-algebras with $L_\infty$-morphisms, this is not encoded by an operad, but rather by the Koszul dual cooperad of Lie, which is equal to the linear dual of $Com$ up to suspension. One can prove that it is the category of fibrant-cofibrant objects of a certain model category on dg cocommutative coalgebras. -In this case, an $L_\infty$-morphism is an $L_\infty$-quasi-isomorphism if and only if its image under the "bar construction" between $L_\infty$-algebras and dg cocommutative coalgebras is a weak equivalence. -[You can find all the constructions and functors in Chapter 11 of http://math.unice.fr/~brunov/Operads.html. For the model category structure on dg coalgebras over the Koszul dual cooperad of an operad, please wait a little bit; I am typing this these days. :) ]<|endoftext|> -TITLE: Which cluster algebras have been categorified? -QUESTION [17 upvotes]: In "Tilting Theory and Cluster Combinatorics" Buan, Marsh, Reineke, Reiten, and Todorov constructed cluster categories for mutation finite cluster algebras (without coefficients), and Amiot gives a construction of a cluster category given a quiver with potential whose Jacobian algebra is finite dimensional (in particular, this gives a cluster category for cluster algebras coming from unpunctured surfaces with nonempty boundary). -My question is simply in what other instances have cluster categories been constructed? In particular, what about cluster algebras with coefficients? What about other surface cluster algebras? What about tame cluster algebras? - -REPLY [10 votes]: Jan's answer includes many excellent references. I will try to give a few quick comments. -First of all, although the original Buan-Marsh-Reineke-Reiten-Todorov paper contained some results which were restricted to finite type cluster algebras, in subsequent work of them and others, notably Caldero-Keller, it has been shown that the BMRRT cluster category does indeed categorify any acyclic cluster algebra without coefficients (in particular, including those of tame type). -Although categorifications are often done in coefficient-free settings, one can also "cheat" by categorifying the cluster algebra in which the coefficients are treated as variables. For example, this allows one to treat the case of principal coefficients (in some sense, the most important case) for an acyclic cluster algebra using the BMRRT technology. -I should also point out that the categorifications in question here are what are now sometimes called additive categorifications; Hernandez-Leclerc and Nakajima have a multiplicative categorification which is more in the spirit of what people mean by categorification in other settings: they construct a category with direct sum and tensor product, and show the ring structure induced on the Grothendieck group of their categories is naturally a cluster algebra.<|endoftext|> -TITLE: Which maximal closed subgroups of Lie groups are maximal subgroups? -QUESTION [6 upvotes]: Which maximal closed subgroups of Lie groups are maximal subgroups? - -REPLY [2 votes]: There is a paper by M. Golubitsky, "Primitive actions and maximal subgroups of Lie groups", J. Differ. Geom. 7 (1972), 175-191: from the Introduction: "...there exist maximal Lie subgroups whose Lie algebras are not maximal subalgebras".<|endoftext|> -TITLE: Half Cantor-Bernstein Without Choice -QUESTION [22 upvotes]: I had a discussion with one of my teachers the other day, which boiled to the following question: - -Assume ZF. Let $A,B$ be sets such that there exist $f\colon A\to B$ which is injective and $g\colon A\to B$ which is surjective. -Is there $h\colon A\to B$ which is bijective? - -Of course it is enough to show that there is an injection from $B$ into $A$, and by the Cantor-Bernstein theorem (which does not require choice) we finish the proof. -My intuition says that this is true, his intuition says it is false. -Insights, references and possible solutions will be appreciated! - -As Ricky shows below, it depends on $A$ and $B$. -So to a more specific choice of sets (for which I think it is true) we have $A=\mathbb R^\omega$ (i.e. infinite sequences of real numbers) and $B=[\mathbb R]^\omega$ (i.e. countable subsets of real numbers) - -REPLY [3 votes]: To your original question, it seems worth mentioning the point that your hypothesis implies the following: - -If $|A| = |A \times 2|$ and there is a surjection from $A$ onto $B$, then $B$ injects into $A$. - -This really just generalizes the example given by Ricky, but to see this: -Suppose we have a bijection $b: A \rightarrow A \times \{0, 1\}$ and a surjection $g: A \rightarrow B$. $A$ injects into $A \cup B$ by the inclusion map, and we can get a surjection from $A$ onto $A \cup B$ by mapping an $a \in A$ to itself if the second coordinate of $b(a)$ is $0$ or to $g(a)$ if the second coordinate of $b(a)$ is $1$. In other words, we split $A$ into two copies of itself, mapping the first copy surjectively (bijectively) onto itself via the identity and the second copy surjectively onto $B$ via $g$. By your stated principle, this would then mean that we have a bijection $h$ from $A$ onto $A \cup B$ whereby $h^{-1} \upharpoonright B$ would be an injection from $B$ into $A$. -From this observation, I think it becomes intuitively clear how your principle is a form of choice in disguise. In particular, your principle would imply that $\alpha^+$ injects into $\mathcal{P}(\alpha)$ for every infinite ordinal $\alpha$ since we have a surjection from $\mathcal{P}(\alpha)$ onto $\alpha^+$ (even without choice by virtue of the fact that every ordinal of size $|\alpha|$ is encoded by a subset of $\alpha$). But as I'm sure you're already aware, models of ZF have been constructed where this will not be the case.<|endoftext|> -TITLE: Random polycube shapes -QUESTION [16 upvotes]: I am wondering if it is hopeless to obtain any firm results -on the following model of a "random polycube shape." -First, a polycube in $\mathbb{R}^3$ -is a connected face-to-face gluing of unit cubes. -(This is a term prominent in Computer Graphics.) -By a random polycube shape I mean the following. -Start with an $n \times n \times n$ polycube, -forming a cube of side length $n$. -For example, for $n=3$, we start with $3^3=27$ unit cubes. - -   - -Now iterate the following process: -(1) Identify a random exposed cube face. -(2) Adjoin a new cube there. -(3) Remove a randomly selected cube on the boundary of the -shape (i.e., a cube with at least one exposed face), -but only if the resulting polycube remains connected (in the face-to-face dual). -So the shape grows by one cube and shrinks by one cube, -therefore always maintaining $n^3$ cubes, and always -maintaining connectivity. -I am interested in even gross parameters: -What is the mean diameter $d$ (longest cube-to-cube path in the dual) of the shape? -How does the genus $g$ grow as a function of $n$? -Presumably both $d \rightarrow \infty$ and $g \rightarrow \infty$ as $n \rightarrow \infty$, -but it might be difficult to determine the rates of growth. -Pointers to relevant related literature would be appreciated. Thanks! -The (distracting!) animation below (a snapshot every 100 iterations over 10000 iterations) -may not animate in your browser (Also, it, at points, wanders "off-screen"—double apology!): - -REPLY [12 votes]: The objects you are interested in are called "lattice animals". -If you do not mind making slight modifications to your Markov chain, it is possible to make it reversible, so that the stationary distribution will be proportional to the number of boxes that can be removed while keeping the set connected. It is plausible that this number is concentrated, so that you will get approximately a uniformly random lattice animal. -In high dimension (above 8, i believe) these have been studied using lace expansion, and much is known. The graph diameter is of order $n^{1/2}$, and the graph is lose to a random tree. The $Z^n$ diameter is of order $n^{1/4}$. In low dimension these exponents change, and the problem seems much harder. (Compare to self avoiding walk in 3 dimensions.) -As for the genus, it should be linear in $n$, since small loops can appear almost anywhere using just a few boxes, so the entropy cost of adding loops is bounded. This is probably deducible from the lace expansion results in high dimension. This might be doable in lower dimensions as well, but seems hard.<|endoftext|> -TITLE: How do the number of plane curves over a finite field of a fixed genus increase with the degree? -QUESTION [6 upvotes]: Let $k$ be a finite field. We define $N(d,g)$ to be the number of plane curves $f(x,y)$ defined over $k$ of degree $d$ with (geometric) genus $g$. If $D(d) := (d-1)(d-2)/2$ (the maximum possible genus), I would expect that as $d$ goes to infinity that the proportion of curves of degree $d$ with genus $D(d)$ would go to 1. If, on the other hand, we're interested in curves of a fixed small genus (say $g=0$ or 1), I would expect that $N(d,g)$ would still approach infinity, albeit at a much slower rate. The question that I have is how do $N(d,0)$ or $N(d,1)$ approach infinity? Is it a polynomial in $\log d$, faster, slower? -I realize that there has been a lot done with enumerative geometry (Gromon-Witten, Caporaso-Harris), but that seems a bit different, since it always works over an algebraically closed field, and classifies curves by having them pass through some set of generic points, and possibly prescribing, tangency, etc. - -REPLY [7 votes]: The set of singular curves of degree $d$ contains the curves with $f(0,0)=f_x(0,0)=f_y(0,0)=0$. The chance that each of these values ($f(0,0),\ldots$) is zero is $1/q$ over the field of $q$ elements, so the proportion of such curves among all curves of degree $d$ is $1/q^3$, hence the proportion of smooth curves is at most $1-1/q^3$ regardless of $d$ and doesn't go to $1$ as $d$ goes to infinity. (I believe this observation is due to Poonen). -The set of curves of degree $d$ and genus zero is the set of parametrized curves $(P_0(t):P_1(t):P_2(t))$, $P_i$ polynomials of degree $d$, modulo the action of $PGL_2(\mathbb{F}_q)$ on the variable $t$. So there are about $q^{3d-4}$ such curves. That's your $N(d,0)$. You should be able to do $N(d,1)$ using similar ideas. - -REPLY [6 votes]: Fix $g$, the genus, and $q$, the order of $k$. -$N(d,g)$ should be $\approx C q^{3d}$, where $C$ is some constant dependent on $q$ and $g$. (Note that my $C$ has absorbed the $q^{-4}$ in Felipe's answer.) There are some nonrigorous details here. -There are finitely many isomorphism classes of pair $(X, L)$ where $X$ is a genus $g$ curve over $k$ and $L$ is a degree $d$ line bundle. I claim that, for each such pair, there are roughly $C(X,L,q) q^{3d}$ degree $d$ curves in $\mathbb{P}^2$ such that the curve is isomorphic to $X$ and the pullback of $\mathcal{O}(1)$ is $L$, and $C(X,L,q)$ is some constant dependent only on $q$, $X$ and $L$. -Recall that such curves, more or less, come from linear maps $H^0(X, L) \to k^3$, modulo rescaling on the image. The number of such maps is $q^{3 \dim H^0(X, L)}$. By Riemman-Roch, for $d$ larger than $2g$, we have $\dim H^0(X, L) = d-g+1$. So this is where I get $q^{3d}$ from, absorbing everything else into the constant. -To be more precise, we need to (1) discard the maps that have base points (2) discard the maps that correspond to branched covers rather than immersions (3) if $(X, L)$ has nontrivial automorphism group, count orbits under that group. -I claim that (1) reduces our count by a factor of about $Z_X(3)$, where $Z_X$ is the zeta function of $X$. (2) should be negligible for large $d$ -- I get that there are about $q^{2d}$ maps coming from $d$-fold covers of a line, and fewer for every other case. And (3) I would guess just divides by $|\mathrm{Aut}(X,L)|$ -- there probably aren't a lot of cases with nontrivial stabilizers. I haven't checked the claims in this paragraph carefully, but that is what I'd expect the counts to be.<|endoftext|> -TITLE: Effective bounds on Euler's totient -QUESTION [5 upvotes]: Quick question: It's known that -$$\limsup\frac{n}{\varphi(n)\log\log n}=e^\gamma$$ -but are there known C and N such that -$$\varphi(n)>\frac{Cn}{e^\gamma\log\log n}$$ -for all $n>N$? -Failing that, what are good effective bounds on $\varphi$? The square root bound isn't good enough for me. - -REPLY [7 votes]: Yes. Look at -http://en.wikipedia.org/wiki/Euler's_totient_function#Inequalities: -$$\varphi(n)>\frac{n}{e^\gamma\log\log n + \frac{3}{\log\log n}}$$ -for $n>2$.<|endoftext|> -TITLE: There must be a good introductory numerical analysis course out there! -QUESTION [57 upvotes]: Background As a numerical analyst, I've frequently taught the 'Introductory Numerical Analysis' class. Such courses are found in many major universities; the audience typically consists of reluctant engineering majors and some majors of mathematics. -The structure of the course is very similar in many of the institutions whose syllabi I've looked at: one begins with finite-precision arithmetic, then fixed-point methods for root-finding (usually 1-D problems),interpolation by polynomials, quadrature, numerical differentiation, some standard ODE methods, and perhaps some finite difference methods for PDE. Any rationale for this particular sequence of topics is obscured in the course. -The truly deep and interesting aspects - approximation theory, error analysis, computational complexity - are either not discussed, or not dwelt on. Instead, the typical introductory course is a collection of algorithms for problems which seem contrived. -This is a pity. The stronger mathematics student comes away believing numerical analysis is boring and shallow, and the engineer comes away thinking mathematics has nothing to offer a real problem. - The question: Are there examples (links to course outlines or course webpages preferred) of introductory numerical analysis courses which avoid the above-described tedium, and which have a history of attracting strong mathematics students? - The constraints: The courses should be aimed at students with a background in multivariate calculus, linear algebra, undergraduate dynamical systems and PDE. One example per answer, please. - The motivation: The eventual goal is to compile such a list, and based on these courses suggest a better curriculum at my institution. - -REPLY [3 votes]: I am teaching an experimental offering at UVic that goes part-way to addressing your concerns. -The goal of the course is to get 2nd year students comfortable with writing mathematical software in a high-level computer language. This semester we are using Python but the specific language is the choice of the instructor. -The main part of the course is about building students' confidence up, writing small scripts to test mathematical ideas. -But along the way we teach them about various elements from numerical analysis and their limitations. We largely do not teach any theory in this course. The course is about learning by example. So students see first-hand the issues that come from round-off error. They see first-hand arbitrary precision floats and integers, and how they can help (and hinder) an investigation. -We also touch on a variety of topics not specific to numerical analysis.<|endoftext|> -TITLE: Succesful applications of algebra in combinatorics -QUESTION [13 upvotes]: Hi. This may be a very general question. -Are there any examples of problems in combinatorics which were open, but which found a solution when stated in algebraic terms? -If yes, could somebody mention some of these? I'm new to this and don't know many examples yet. -I know about the "Magic Squares", which refers to counting the number of $n\times n$ $\mathbb{N}$-matrices having line sum equal to $r$. This was treated by Anand, Dumir and Gupta, by stating it as the number of ways of distributing $n$ different things, each one replicated $r$ times, among $n$ different persons, in equal numbers. It was solved by R. Stanley (see "Commutative algebra arising from the Anand-Dumir-Gupta conjectures" by Winfried Bruns). -Are there some instances where algebra has been used to enumerate, say, certain sets of graphs? - -REPLY [5 votes]: Stanley's proof of the Upper Bound Conjecture relied on a connection with free resolutions of graded algebras. This has led to the very active area of Stanley--Reisner theory, where combinatorial properties of simplicial complexes are related to algebraic properties of certain graded algebras. -For references, there's a wikipedia page on Stanley--Reisner theory if you're interested: -http://en.wikipedia.org/wiki/Stanley%E2%80%93Reisner_ring -Also, Bruns and Herzog's book "Cohen--Macaulay Rings" has nice a chapter on Stanley--Reisner rings. I'm sure there are other good references as well.<|endoftext|> -TITLE: Why do all incidence theorems follow from Pappus' theorem? -QUESTION [19 upvotes]: In Hilbert and Cohn-Vossen's ``Geometry and the Imagination," -they state in the last paragraph of Chapter 20 that "Any -theorems concerned solely with incidence relations in the -[Euclidean projective] plane can be derived from [Pappus' -Theorem]." -Does anyone know a reference for more details about this claim? -Does anyone have an idea of how to prove it? -I have seen two books by John Stillwell: ``Four Pillars of -Geometry" and "Yearning for the Impossible." These books -discuss the relation between Pappus' theorem in a projective -plane over some algebraic structure and commutativity of the -multiplication in that algebraic structure. In particular -Stillwell describes how all the laws of algebra follow from -Pappus' theorem. -Presumably one then argues that any incidence theorem can be -proved using algebra. I would be happy, for instance, with a -precise statement like this, including in particular a -definition of ``incidence theorem." - -REPLY [10 votes]: All incidence theorems, which hold over any field, follow from Pappus, because Pappus allows to sum up and multiple points on a line, i.e. implies that your plane is over field (Desargues implies that the plane is over skew field). But some specific properties of the field may be encoded as incidence theorems (I think, any algebraic property like $x^2+1=0$ has no solutions'' may be encoded, just use the geometric construction for addition and multiplication), and hold not for all fields, see spectacular examples in David's answer.<|endoftext|> -TITLE: ad-nilpotent degree of a nilpotent Lie Algebra -QUESTION [8 upvotes]: Let $\mathfrak{g}$ be a Lie Algebra (finite dimensional, over $\mathbb{C}$). Engel's theorem tells us that if there exists a $m\in \mathbb{N}$ such that $\mathrm{ad}(x)^m = 0$, $\forall x\in \mathfrak{g}$, then $\mathfrak{g}$ is nilpotent. And if $\mathfrak{g}$ is $(k-1)$-step nilpotent (i.e. the $k$-th term of the lower central series of $\mathfrak{g}$ is the first one that is 0, or equivalently $\mathrm{ad}(x_1)\mathrm{ad}(x_2) \ldots \mathrm{ad}(x_k) = 0$ $\forall x_1, \ldots, x_k \in \mathfrak{g}$), it is clear that $$\min \big\{m\in \mathbb{N} : \mathrm{ad}(x)^m = 0 \forall x\in \mathfrak{g} \big\} \leq k.$$ -Can we find an example where the previous inequality is not an equality? -If this is a very basic fact in the theory, I apologize. -Edit: I've been asked to share de Graaf's example, here it is. - It is a 17-dimensional (nilpotent) Lie algebra over $\mathbb{Q}$ that is 3-Engel and of nilpotency class 4. -(The table has to be read as follows: $\langle 2,4,17|-3\rangle$ means that $[x_2,x_4] = -3 x_{17}$. If there are more tuples starting with $\langle 2,4$ then -one has to take the sum, so if there also was $\langle 2,4,13|-2\rangle$ then -$[x_2,x_4] = -2 x_{13} - 3 x_{17}$.) -$$[ \langle2, 4, 17| -3\rangle,\; \langle2, 6, 3| -3\rangle,\; \langle2, 7, 1| 1\rangle,\; \langle2, 9, 1| 2\rangle,\; \langle2, 12, -9| 1\rangle,\; \langle2, 13, 10| 1\rangle,\; \langle2, 14, 11| 1\rangle,\; \langle2, 15, 13| 1\rangle,\; -\langle2, 16, 14| 1\rangle,\; \langle5, 16, 17| -3\rangle,\; \langle7, 15, 17| 1\rangle,\; \langle7, 16, 3| -2\rangle,\; \langle8, -15, 3| 3\rangle,\; \langle9, 15, 17| -1\rangle,\; \langle9, 16, 3| -1\rangle,\; \langle10, 16, 1| -3\rangle,\; \langle11, 15, 1| -3\rangle,\; \langle12, 13, 17| 4\rangle,\; \langle12, 14, 3| 4\rangle,\; \langle12, 15, 4| -1\rangle,\; -\langle12, 16, 6| -1\rangle,\; \langle13, 14, 1| -4\rangle,\; \langle13, 15, 5| -1\rangle,\; -\langle13, 16, 7| -1\rangle,\; \langle14, 15, 7| -1\rangle,\; \langle14, 15, 9| 1\rangle,\; \langle14, 16, 8| -1\rangle,\; -\langle15, 16, 12| -1\rangle ]$$ - -REPLY [7 votes]: I computed the example of Traustason (see his remark). We start with the free-nilpotent Lie algebra -of class 4 with 3 generators $x_1$, $x_2$, $x_3$. It has dimension 32. Then we divide out the ideal generated by -all brackets containing $x_1$ three times, or containing $x_2$, $x_3$ at least two times. The quotient is -an Engel-3-Lie algebra of nilpotency class 4 and of dimension 11. -It is easy to write down explicit Lie brackets. This should be an example of least possible dimension. -However, the next case, to find an Engel-4-Lie algebra of nilpotency class 7, of least possible -dimension is more complicated, and I am frightened to do the calculation.<|endoftext|> -TITLE: Exponential sums for beginner. -QUESTION [7 upvotes]: What are the good books, online lecture notes or starting material on exponentials sums with applications in number theory for a beginner, apart from N. M. Korobov's book? The book or notes should cover methods of Weyl, van der Corput and Vinogradov, with some details. - -REPLY [3 votes]: K. Chandrasekharan, Exponential sums in the development of number theory, pp. 7-26 in -Proceedings of the International Conference on Number Theory (Moscow, 1971), Trudy Mat. Inst. Steklov 132 (1973).<|endoftext|> -TITLE: Elegant proof that mapping class groups are generated by Dehn twists? -QUESTION [17 upvotes]: One of the foundational results about mapping class groups of surfaces is that they are generated by Dehn twists. A mapping class is a connected component in a space of diffeomorphisms, so another way of stating this is "any diffeomorphism of a surface is generated by diffeomorphisms each of which is supported in an annulus". This was proven by Dehn, and independently in a stronger form by Lickorish. -I'm teaching a summer reading course, and I am toying with the idea of presenting a proof to this statement. But the proof I know is a bit involved- you use the Birman exact sequence to relate the mapping class group of a surface Σ to the mapping class group of $\Sigma-D^2$ (not so trivial), then you use the fact that the complex of curves on a surface is connected (also non-trivial), and finally that for two non-disjoint connected curves α and β there exists a product of Dehn twists T such that $T(\alpha)=\beta$. This proof looks too involved to present properly in a single lecture. -80 or so years after Dehn's proof, and 47 years after Lickorish's: - -Do you know an elegant proof that the mapping class group of a surface is generated by Dehn twists? - -REPLY [12 votes]: I'm pretty sure there doesn't exist a "slicker" proof of this fact in the literature. The proof you describe exists in many forms starting with Dehn and Lickorish -- as I said in a comment, the particular arrangement of it you gave (making use of the complex of curves) is basically due to Ivanov. The only fundamentally different approach I know of is in the paper -MR1425631 (98c:20061) -McCool, James(3-TRNT) -Generating the mapping class group (an algebraic approach). (English summary) -Publ. Mat. 40 (1996), no. 2, 457–468. -This proof is purely algebraic (and rather more complicated than the topological proof). It is based on McCool's 1975 proof that the mapping class group is finitely presentable. By the way, McCool is often written out of the history of the subject, but his work predates Hatcher-Thurston and is the first paper in the literature that explicitly proves that the mapping class group is finitely presentable (though the algebraic geometers proved equivalent facts in the early '60's). -I tend to view the fact that the mapping class is generated by Dehn twists as equivalent to the fact that the complex of curves is connected. For this, there exist several alternate proofs. Many of these proofs also give higher connectivity; I'll try to indicate that as I go. - -There is the combinatorial approach taken in many sources (eg Farb-Margalit's "Primer on Mapping Class Groups"). I don't know who to attribute this to. You could probably prove that the curve complex is simply-connected by this method, but I doubt you could prove higher connectivity. -You could use Teichmuller theory -- this is how Harer originally proved that the curve complex is highly connected. -There is a Morse-theoretic proof due to Ivanov (it is described in his survey "Mapping Class Groups"). It is an adaption of the Cerf theory approach to proving the mapping class group is finitely presentable which is due to Hatcher and Thurston. In his paper "Complexes of curves and Teichmüller modular groups", Ivanov uses this approach to prove that the complex of curves is highly connected. -You could easily deduce that the curve complex is connected from Hatcher's slick proof that the arc complex is contractible, which is located in his paper "On triangulations of surfaces". I don't think they have this written up, but Hatcher and Vogtmann have a proof that the curve complex is highly connected with this as the starting point.<|endoftext|> -TITLE: Do Tamagawa numbers of Galois representations stabilise in the cyclotomic tower? -QUESTION [8 upvotes]: Suppose $T$ is a free finite rank $\mathbb{Z}_p$-module with a continuous action of $\operatorname{Gal}(\overline{K} / K)$, where $K$ is a number field. There is a definition of local Tamagawa numbers $\operatorname{Tam}(T / K_v)$ for each prime $v$ of $K$, going back to Fontaine and Perrin-Riou (or to Bloch and Kato for $K = \mathbb{Q}$). For $v \nmid p$ this is the order of the torsion subgroup of $H^1(I_v, T)^{D_v}$, where $D_v$ and $I_v$ are the decomposition group and intertia group at $v$; for $v \mid p$ it is something more complicated using the Bloch-Kato exponential map. (I'm led to believe that if $T$ is the Tate module of an abelian variety, this recovers the usual description in terms of Neron models.) -If $K_{v, n} = K_v(\mu_{p^n})$, is it true that the factors $\operatorname{Tam}(T / K_{v, n})$ are eventually constant for large enough $n$? -(EDIT: In the light of Rob's comment, maybe I should add the assumption that my Galois representation is crystalline at $p$. I'm chiefly interested in the case of the p-adic representation of a modular forms of level prime to $p$ and non-ordinary at $p$.) - -REPLY [2 votes]: As stated, the answer to your question is certainly no. -For instance, an elliptic curve $E/\mathbb Q$ with split multiplicative ordinary reduction at $p$ will have unbounded Tamagawa number at $p$ in the cyclotomic extension of $\mathbb Q$. To see this, you can check that the Tamagawa number is the order of $H^{2}({\mathbb Q_{\infty,p}},(T_{p}E)_{p}^{+})$. This is what B.Mazur calls the phenomenon of anomalous primes in his Inventiones 18 paper, which is the algebraic counterpart of the existence of exceptional zeroes for $p$-adic $L$-functions. -However, perhaps you meant for the Tamagawa factors outside $p$ to be eventually constant? In that case, I think the answer is yes. -First, $\operatorname {Gal}(K(\mu_{p^{\infty}})/K)$ has finite prime-to-$p$ part so there exists a large enough $n$ so that $\operatorname {Gal}(K(\mu_{p^{\infty}})/K(\mu_{p^n}))$ is unramified outside $p$ (EDIT: Of course this first step is unnecessary). Let $v\nmid p$ be a place of $K(\mu_{p^{n}})$ and let $w|v$ be a place of $K(\mu_{p^{n'}})$ with $n'≥n$. Then $H^{1}(I_{v},T)\simeq H^{1}(I_{w},T)$ so the Tamagawa number is constant.<|endoftext|> -TITLE: Appearances of 'exotic' compact Lie Groups -QUESTION [5 upvotes]: The structure theorem for compact Lie Groups states that all compact Lie groups are finite central quotients of a product of copies of $U(1)$ and simple compact Lie groups. And yet, as easy as arbitrary compact Lie groups are to describe, most Lie Groups one encounters are the various quotients of simple compact Lie groups and maybe some products of these groups (I will herein refer to such groups as standard Lie groups). The only compact examples which one encounters regularly that are not standard Lie groups are the unitary groups $U(n)$ (which are quotients of $U(1)\times SU(n)$) and $SO(4)$ (which is the diagonal $\mathbb{Z}/2\mathbb{Z}$ quotient of $Spin(4) \cong Spin(3)\times Spin(3)$). -I am currently trying to further expand my knowledge and understanding of compact Lie groups, so I am wondering: - - -Question: Has anyone encountered examples of non-standard Lie groups (other than the $U(n)$'s and $SO(4)$) in their research, as the autormorphism group of some object they were studying, or in some other way? If so, would you give a bit of description of the setting you were working in as well as a description of the non-standard group which appeared? - - -Although given a non-standard group, one can easily construct algebraic objects for which it is the automorphism group, I am more interested in instances of the reverse of this process wherein a non-standard group appears in the course of thinking about some other problem. -Edit: Since there still seems to be some misunderstanding of the intent of the question, to clarify the situation I am interested in, I am looking for groups of the form $G_1\times\ldots\times G_k/H$ where each $G_i$ is a compact simple Lie group, $k\geq 2$ and $H\subsetneq Z(G_1\times\ldots\times G_k)$ is not of the form $h_1\times \ldots \times h_k$ with $h_i\subseteq Z(G_i)$. So examples with multiple factors such as the the Structure Group of the Standard Model described by Theo are the sort of thing I'm looking for. - -REPLY [3 votes]: You need the groups $Spin^c(n)=: (Spin(n)\times U(1))/C_2$ in order to define first $Spin^c$-structures on Riemannian manifolds, then Dirac operators. See the book: B. Lawson, L. Michelson: Spin geometry, Princeton University Press (1989).<|endoftext|> -TITLE: Transversal Intersection of Varieties -QUESTION [14 upvotes]: I am having some trouble finding and/or understanding a general definition of subvarieties intersecting transversally. Assume that $Z_1,\ldots,Z_k$ are closed, irreducible subvarieties of a nonsingular algebraic variety $Y$. -Intuitively, I would say that these subvarieties intersect transversally if all varieties have pairwise intersection multiplicity one, i.e. $i(W,Z_i\cdot Z_j; Y)=1$ along any component $W$ of $Z_i\cap Z_j$ for $i\ne j$. -In our scenario, we should have $i(W,Z_i\cdot Z_j;Y)=\mathop{\mathrm{length}}_{\mathscr{O}_{Y,W}}\left(\mathscr{O}_{Z_i\cap Z_j, W}\right)$, if I am not mistaken. -Another intuitive definition would be that the tangent sheaves $\mathscr{T}_{Z_i}$ form a direct sum inside $\mathscr{T}_{Y}$. This seems to agree with the definition in this paper (in 5.1.2), but the author gives an equivalent definition which looks interesting: -For any $y\in Y$, there exists - a system of parameters $x_1,\ldots,x_n$ on $Y$ at $y$ that are regular on an affine neighborhood $U$ of $y$ such that $y$ is defined by the maximal ideal $(x_1,\ldots,x_n)$ as well as - integers $0=r_0 \le r_1 \le \cdots \le r_k \le n$ such that the subvariety $Z_i$ is defined by the ideal $I_i=\left(x_{r_{i-1}+1},\ldots,x_{r_i}\right)$ for all $1\le i\le k$. -Just for the record, what precisely does "defined by" mean here? I assume it means that $U\cap Z_i = Z(I_i)$ ... or do we actually get that $I_i=I(Z_i\cap U)$? -I would like to know if and how these three definitions are equivalent - there is no (general) treatment of this in Hartshorne or even in Fulton's book on Intersection Theory, which befuddled me greatly. - -REPLY [4 votes]: This naive answer is strictly intended to provoke a knowledgeable answer from someone. Transversal intersection at p should mean that all subvarieties are smooth at p, all contain p, and the codimension of the intersection of the tangent spaces equals the sum of the codimensions of the individual tangent spaces. ????<|endoftext|> -TITLE: Hausdorff dimension for invariant measure? -QUESTION [11 upvotes]: A fractal set has a Hausdorff dimension. -In some cases, we may generate a fractal by iterating $f,$ -and let the fractal be the set of starting points $x$ such -that $|f^{\circ n}(x)|$ is bounded as $n$ grows. (The julia set and the sierpinski triangle are such sets, if one allows $f$ to be a Hutchinson operator). -We may also have an invariant measure, $\mu,$ that is, $\mu(A) = \mu(f^{-1}(A)).$ -The support of $\mu$ is the fractal set. -My question is: is there a way to modify the "dimension" notion to take this invariant -measure into account somehow? -Some parts of the fractal might be more dense, and thusly should "contribute more" to the dimension. -An idea would be to use box-counting, but instead of just counting if it is occupied or not, -one uses the invariant measure on the box instead. Has this been studied? - -REPLY [16 votes]: There are a wide variety of notions of dimension of a measure. Your basic intuition is completely correct: for a dynamical system, the dimension of a natural invariant measure provides more relevant information than the dimension of the invariant set, since the system may spend more time in certain parts of the space. -For sufficiently homogeneous measures, all reasonable notions of dimension will agree. By ``sufficiently homogeneous'' I mean something very precise: that -$$ -C^{-1} \, r^s \le \mu(B(x,r)) \le C\, r^s -$$ -for some constant $C\ge 1$, some $s\ge 0$ and all points $x$ in the support of $\mu$. Of course the dimension in this case is $s$. Such measures are often called Ahlfors-regular, and an example is the natural measure on the middle-thirds Cantor set. -For more general measures, the local dimension is one of the most important concepts and has already been mentioned: -$$ -\dim(\mu,x)=\lim_{r\to 0}\frac{\log \mu(B(x,r))}{\log r}. -$$ -But this is really a function of the point $x$ (and not even, as the limit in the definition may not exist, although one can always speak of upper and lower local dimensions). -There are several ways to globalize the information given by the local dimensions. Perhaps the easiest is to take the essential supremum/infimum of the upper/lower local dimensions. This results in four global concepts of dimensions, known as upper/lower packing/Hausdorff dimensions of the measure. They turn out (somewhat surprisingly) to be closely connected to the dimensions of the sets the measure ``sees''. For example, the upper Hausdorff dimension of a probability measure $\mu$ (that is, the essential supremum of the lower local dimensions), is the same as the infimum of the Hausdorff dimension of $A$ over all Borel sets $A$ of full measure. -A finer study is provided by the multifractal spectrum of a measure $\mu$: for each $\alpha$, we form the level set $E_\alpha$ of all points $x$ where $\dim(\mu,x)=\alpha$. Then we try to understand how the size of $E_\alpha$ depends on $\alpha$, for example by studying the function $\alpha\to \dim_H(E_\alpha)$. -There are (many!) other useful concepts of dimension which are not directly related to local dimension. In computing lower bounds for the Hausdorff dimension, the potential method is widely applicable: if a measure $\mu$ satisfies that the energy integral -$$ -I_s(\mu) = \int \frac{d\mu(x)\, d\mu(y)}{|x-y|^s} -$$ -is finite, then the support of $\mu$ has Hausdorff dimension at least $s$. So it makes sense to think of $\sup\{s: I_s(\mu)<\infty\}$ as a notion of dimension of $\mu$. This is often called the (lower) correlation dimension, and is one instance of a more general family of dimensions indexed by a real number $q$ (correlation dimension corresponds to $q=2$, and has several alternative definitions, perhaps pointing to its importance). -Yet another notion of dimension has a dynamical underpinning. Given a probability measure $\mu$ say on the unit cube $[0,1]^d$, we may consider the entropy $H_k(\mu)$ of $\mu$ with respect to the partition into dyadic cubes of side length $2^{-k}$. We then define the entropy (also called information) dimension of $\mu$ as -$$ -\lim_{k\to\infty} \frac{H_k(\mu)}{k\log 2}. -$$ -This is just a sample of the diverse zoo of dimensions of a measure. Which ones to use depends on the context and what you are able to compute/prove. -Coming back to invariant measures, it is very often the case that the local dimension exists and takes a constant value at almost every point. Such measures are called exact dimensional, and have the property that lower and upper Hausdorff dimension, as well as entropy dimension, are all equal to this almost sure value. (But correlation dimension may be strictly smaller, and the multifractal spectrum may still be very rich; in other words, even though attained on a set of measure zero, other local dimensions may still be relevant). -Proving that measures invariant under certain class of dynamics are exact dimensional may be very challenging. Eckmann and Ruelle conjectured in 1984 that hyperbolic measures ergodic a $C^{1+\delta}$ diffeomorphism are exact dimensional. This was proved by Barreira, Pesin and Schmeling in 1999; the paper appeared in Annals. -For invariant measures, there is often a strong connection between their dimension and other dynamical characteristics (at least generically). The conformal expanding case is the easiest: in this case one has the well-known formula ``dimension=entropy/Lyapunov exponent". The nonconformal situation is much harder, but still a lot of deep research has been done, for example Ledrappier-Young theory.<|endoftext|> -TITLE: why are all characters of the maximal torus in a Lie group weights? -QUESTION [7 upvotes]: Let $G$ be a compact connected Lie group, $T$ maximal torus, identified with $\mathbb{R}^n/\mathbb{Z}^n$, $X^*(T)$ -the set of characters of $T$, naturally identified with $\mathbb{Z}^n$. Let next $\Phi$ denote the set of roots, corresponding to $T$, -$$ -\Lambda=\{v\in \mathbb{R}^n: \forall \alpha\in \Phi,\ 2(v,\alpha)/(\alpha,\alpha)\in \mathbb{Z}\}$$ be the lattice of weights. Then $X^*(T)\subset \Lambda$. -What is the easiest and most elementary way to prove it? -It suffices to prove that for any $\lambda\in X^*(T)$ is such a character that for some finite-dimensional complex representation of $G$ there is a non-zero $T$-invariant subspace, on which $T$ acts by multiplying by $\lambda(\cdot)$. -Is there any way to do it without appealing to infinite-dimensional representations (inducing $\lambda$ from $T$ to $G$ and some work after that)? - -REPLY [5 votes]: "Easiest" depends on how you set things up: everything really hinges on how you want to identify $X^\ast(T)$ with $\mathbb Z^n$. It's probably cleanest if you don't work explicitly with $\mathbb Z^n$, but instead state everything in terms of Lie algebras and their duals. I personally like the setup given in Knapp, Lie Groups Beyond an Introduction, IV.6--7, which is fairly standard. In the end it all boils down to associating a copy of $SU(2)$ (or $\mathfrak{su}(2)$ or $\mathfrak{sl}_2(\mathbb C)$ ...) to each root $\alpha$, and then the integrality statement you're after ultimately follows from the fact that $$\exp 2 \pi i x = 1 \, \implies \, x \in \mathbb Z.$$ - -Edit: Here's an outline of the details. Let $\mathfrak t_0$ and $\mathfrak t$ denote, respectively, the real and complex Lie algebras of $T$ and set -$$ L = \{ \xi \in \mathfrak t_0 \mid \exp \xi = 1 \} $$ -for the kernel of the exponential map $\exp \colon \mathfrak t_0 \to T$. -Also define -$$ L^\perp = \{ \lambda \in \mathfrak t^\ast \mid \lambda(L) \subset 2\pi i \mathbb Z \}. $$ -(Admittedly, the $^\perp$ is a slight abuse of notation.) Then there is an isomorphism -$$ L^\perp \stackrel{\sim}{\longrightarrow} X^\ast(T) $$ -given by sending $\lambda$ to the character $e^\lambda$ defined by -$$ e^\lambda(\exp \xi) = e^{\lambda(\xi)}, \qquad \xi \in \mathfrak t_0. $$ -This is basically Proposition 4.58 in Knapp. So our objective now is to show that $L^\perp$ sits in the weight lattice $\Lambda$ --- in other words, we want to show that -$$ 2\frac{(\lambda, \alpha)}{(\alpha,\alpha)} \in \mathbb Z $$ -for all $\lambda \in L^\perp$ and $\alpha \in \Phi$ (Prop 4.59). To this end, let $\psi_\alpha \colon SU(2) \to G$ denote the root morphism corresponding to the root $\alpha \in \Phi$. This morphism has the property that $d \psi_\alpha(h) = 2 \alpha^\vee/(\alpha,\alpha)$, where -$$ h = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \in \mathfrak{su}(2). $$ -Consequently, -$$ 1 = \psi_\alpha(1) = \psi_\alpha(\exp (2\pi i h)) = \exp (2 \pi i \, d \psi_\alpha(h)), $$ -that is, -$$ 2 \pi i \, d \psi_\alpha(h) = 2 \pi i \cdot 2 \frac{\alpha^\vee}{(\alpha,\alpha)} \in L $$ -whence -$$ \lambda(2 \pi i \cdot 2 \frac{\alpha^\vee}{(\alpha,\alpha)}) \in 2 \pi i \mathbb Z \iff 2\frac{(\lambda, \alpha)}{(\alpha,\alpha)} \in \mathbb Z, $$ -as desired. - -REPLY [3 votes]: To answer the last question here, it's certainly not necessary to introduce infinite-dimensional representations into the picture when studying compact Lie groups. On the other hand, I don't quite see the point of trying to relate in isolation the character group of a maximal torus to an abstractly defined "weight" lattice. This is best done within the standard context of finite dimensional representations of a compact Lie group, which of course takes a while to develop from scratch. Look for instance at a standard 1985 textbook by Brocker and tom Dieck Representations of Compact Lie Groups (Springer), where all of the classical theory is laid out systematically. -It's possible of course (as shown by Bourbaki) to treat roots and weights more abstractly without specific reference to Lie groups or Lie algebras, but this is rather artificial here. At any rate, it's risky to start out by identifying the character group of a maximal torus "somehow" with the standard lattice in $\mathbb{R}^n$, since the same could be done with the root lattice or (abstract) weight lattice. It's the placement of one lattice within another that counts. -It's important to have examples at hand, as a reminder that the character group of a maximal torus (say in a semisimple compact group) can vary anywhere between the root lattice (adjoint group) and full weight lattice (simply connected group). -P.S. Added observations: (1) You should assume your group is semisimple (having no central torus of positive dimension) if you want to compare $X(T)$ with the dual lattice of the root lattice, the latter being of finite index in the former. It's also helpful to recall from the general theory that the quotient of these lattices is isomorphic to the center of the simply connected covering group of your given group and to the fundamental group of the adjoint group. (2) As the answers here suggest, your question doesn't have any "easy" answer until you place it in one of the well-established textbook settings for studying structure and representations of compact Lie groups. There are various approaches using analysis, algebraic topology, algebraic geometry, Lie algebras, comparison with complex Lie groups, etc. Your question is mainly pedagogical, so you should first place it in one or more of these traditions. - -REPLY [3 votes]: I'm not sure what I'm allowed to use here. Here is a proof, and I'll see whether or not you are happy with it. -Let $L$ be the lattice of those characters of $T$ which occur in finite dimensional representations of $G$. It is clearly a lattice because, if $\lambda$ occurs in $V$ and $\mu$ occurs in $U$, then $\lambda+\mu$ occurs in $V \otimes U$. As you say, it is enough to show that $L = X^*(T)$. Also, note that the root lattice is in $L$. -Let $\lambda$ be any element of $X^*(T)$. Let $s : T \to \mathbb{C}$ be the function $t \mapsto \sum_{w \in W} \lambda(w(t))$. (Implicitly using that the Weyl group is finite; am I allowed to assume this?) Then $s$ is a $W$-invariant function on $T$, and it is not zero since it is $|W|$ at the identity. Since every conjugacy class of $G$ intersects $T$, and does so in a $W$-orbit, we can extend $s$ to a conjugacy invariant function on $G$. -By the Peter-Weyl theorem, there must be some finite dimensional representation $V$ such that $\langle s, \chi_V \rangle \neq 0$. By the Weyl integral formula, $\langle s, \chi_V \rangle$ is an integral over $T$ of a product of three factors: $s$, which is a certain finite sum of characters of $T$; $\chi_V$, which is a certain finite sum of characters in $L$, and the Weyl integrand, which is a certain sum of characters in the root lattice. -So the product of the last two terms is a sum of characters from $L$. -In order for this integrand to be nonzero, one of the characters in $s$ must be negative a character in the root lattice, so $-w(\lambda) \in L$ for some $w\in W$. Using that $L$ is $W$-stable, this shows that $\lambda \in L$.<|endoftext|> -TITLE: Actions orbit equivalent to profinite ones -QUESTION [7 upvotes]: Let $G$ be a countable discrete residually finite group. -Is there a way to characterise the actions of $G$ that are orbit-equivalent to profinite ones? -Ozawa and Popa introduced the concept of weakly compact actions. Weakly compact actions are stable under orbit equivalence and profinite actions are weakly compact. -Is it possible to find a weakly compact action that is not orbit equivalent to any profinite action? - -REPLY [9 votes]: If a group has a free profinite action then in particular it must be residually finite. On the other hand, any measure preserving action of an amenable group is weakly compact. Thus, any free measure preserving action of infinite amenable group $G$ which is not residually finite cannot be orbit equivalent to a profinite action of $G$. -Of course by Ornstein and Weiss' theorem all such orbit equivalence relations are hyperfinte and so this doesn't work if you allow yourself to change the group $G$. -Update: -For residually finite groups an example will have to be more sophisticated since this will not work with amenable groups. However, there are enough orbit equivalence rigidity theorems around that this is still possible. The relevant theorem here is Ioana's rigidity theorem for profinite actions of property (T) groups (http://arxiv.org/abs/0805.2998): -Theorem[Adrian Ioana]: Let $\Gamma$, and $\Lambda$ be countable discrete groups and suppose that $\Gamma$ has property (T). Let $\Gamma \curvearrowright X$ be a free, ergodic, profinite, probability measure preserving action and suppose that $\Lambda \curvearrowright Y$ is a free probability measure preserving action which is orbit equivalent to the action $\Gamma \curvearrowright X$. Then there exists finite index subgroups $\Gamma_1 < \Gamma$ and $\Lambda_1 < \Lambda$ such that restricting to an $\Gamma_1$ (resp. $\Lambda_1$) ergodic component $X_1$ (resp. $Y_1$) we have that the actions $\Gamma_1 \curvearrowright X_1$ and $\Lambda_1 \curvearrowright Y_1$ are conjugate. -In particular, this shows that for orbit equivalent actions of property (T) groups if one action is profinite then so is the other. This is because a measure preserving action $\Gamma \curvearrowright X$ is profinite if and only if the weak closure $\overline \Gamma$ of $\Gamma$ in Aut$(X)$ is a profinite group, and $\overline \Gamma$ is invariant under conjugation. Thus one just needs to construct a weakly compact action of a residually finite property (T) group which is not profinite. -One such example is given by Dennis Sullivan (http://www.ams.org/mathscinet-getitem?mr=590825) who showed that for $n > 3$ there is a countable dense subgroup $\Gamma_n$ of $O(n + 1)$ which has property (T). $\Gamma_n$ will be residually finite by a classical result of Malcev since it is finitely generated and linear. Also, the Haar measure preserving action $\Gamma_n \curvearrowright O(n + 1)$ given by left multiplication will be free (since $\Gamma_n$ is a subgroup), ergodic (since $\Gamma_n$ is dense), and compact (since $O(n + 1)$ is compact), and hence also weakly compact. However, $\Gamma_n \curvearrowright O(n + 1)$ is not profinite since $O(n + 1) = \overline{\Gamma_n}$ is not profinite. -Note that this gives an example if you consider orbit equivalence with other groups $\Lambda$ as well.<|endoftext|> -TITLE: D-modules on affine space that are regular at infinity -QUESTION [6 upvotes]: If I have a D-module $M$ on $\mathbb{A}^n$ (which is essentially the same thing as a module over the Weyl algebra), then I can push this D-module forward to $\mathbb{P}^n$ to get a D-module $j_*M$ on the projective closure. - -Is there an intrinsic description of those $M$ such that $j_*M$ are regular on the divisor at $\infty$? - -REPLY [3 votes]: In dimension one, $j_*M$ is regular at infinity iff the Fourier transform $F(M)$ has no singularity outside 0 and $\infty$. The reason is obviously that Fourier exchanges the D-modules $\delta_c = D/D(t-c)$ et $Oe^{ct} = D/D(\partial_t-c)$. There's probably a generalization for higher dimensions but I've never seen it.<|endoftext|> -TITLE: Voronoi cell of lattices with the same profile -QUESTION [22 upvotes]: Definition 1. Given a body $V$ in $\mathbb R^n$, -the function $p_V\colon \mathbb R_+\to \mathbb R_+$ -$$p_V(r)=\mathop{\rm vol} [V\cap B_r(0)]$$ -will be called profile of $V$. -Definition 2. Define Voronoi cell of lattice $L$ in $\mathbb R^n$ as -$$V_L=\{\,x\in \mathbb R^n;\,|x|\le |x+\ell| \ \text{for any}\ \ell\in L\,\}.$$ - -Question. Can it happen that Voronoi cells of a pair of lattices have the same profile, but not isometric? - -Comment - -The question is inspired by this one. - -REPLY [4 votes]: My apologies for a trivial comment (not an answer), well-understood by the OP. -But perhaps -this illustration will inspire advancement on this long-unsolved question, -and invite clarification if I am misinterpreting. -I abandon the requirement that $V$ be the Voronoi cell of a lattice, and just -look at convex bodies: -    -Here the two shapes have the same "profile" in Anton's sense: -they intersect origin-centered balls in the same volumes. -But they are not congruent, not equivalent under an isometry. -So the challenge is to arrange something similar for lattice Voronoi cells.<|endoftext|> -TITLE: How quickly did Gödel's Incompleteness Theorem become known and heeded throughout mathematics -QUESTION [11 upvotes]: Does anyone know how news of Gödel's incompleteness theorem spread? Did it do so little by little, or was it shouted in dramatic headlines throughout mathematical literature? If anyone can point me to the history of how people reacted to Gödel's result, I would be grateful. -I am asking this question because I have recently been reading a work that comes from a field very far from logic, namely Lie theory, written in 1940 (ten years after Über formal unentscheidbare...") wherein the author seems to be highly mindful of the different mathematical philosophies - he is at least paying lip service to intuitionism and how his work might make sense in that philosophy (see below). Now I should have thought that Gödel's incompleteness theorem would take much of the heat out of debates as to who had the "best" philosophy. Or did it put wind in the sails of the intuitionists, after Gödel had seemingly demolished the formalists, although I believe Gödel would not have seen his incompleteness theorem validating the intuitionists either, being as he was a strong Platonist. Anyhow, here is the quote: it is "Hauptsatz 1" in the paper and I was fascinated to read these words in the far-removed-from-logic field of Lie theory: -From H. Freudenthal "Die Topologie der Lieschen Gruppen Als Algebraisches Ph\"anomen" Annals of Mathematics vol 42. # 5 (1941) wherein he makes the following statement: -"Main Theorem 1: An isomorphism between two Lie groups, of which one is simple and of the second kind, is needfully continuous. Otherwise put: in the theory of Lie groups, the topology of simple groups of the second kind is a wholly algebraic phenomenon" -Lest you should think that the rewording "otherwise put ..." cannot be construed as a precise statement of a theorem (it does on the surface seem rather vague), Freudenthal goes on to explain: -"In the latter formulation the main theorem also makes sense for someone who outright refuses [the existence of] discontinuous mappings, such as [someone with] intuitionist leanings" - -REPLY [8 votes]: Another possibly relevant article: -Dawson, John W. Jr. (1984). "The Reception of Godel's Incompleteness Theorems". Proceedings of the Biennial Meeting of the Philosophy of Science Association, Vol. 1984,. Volume Two: Symposia and Invited Papers. The University of Chicago Press on behalf of the Philosophy of Science Association. pp. 253-271. JSTOR 192508.<|endoftext|> -TITLE: Infinite products exist *only* for affine schemes? -QUESTION [15 upvotes]: Infinite products don't exist in the category of schemes (see Jonathan Wise's answer here). However, all limits of affine schemes exist in the category of schemes (and are affine). I would like to know if affine schemes are the only ones which have this property, thus even sharpening the result. For example, we may ask the following: -Let $X$ be a scheme such that all powers $X^I$ exist in the category of schemes. Can we deduce that $X$ is affine? -I hope that the answer will be "yes". You may work in the category of $k$-schemes for a field $k$ and assume that $X$ is integral. You may also assume that all limits constructed from $X$ exist, for example all equalizers of morphisms $X^I \to X^J$, etc. Feel free to add other assumptions as well. -Jonathan has already made the following comment (which I cannot fill with details): - -I believe the proof can be modified to show that if a product of a collection of quasi-compact schemes is a scheme then the product of some collection of all but finitely many of them is affine. Assuming they are flat over Z or something, I suspect this will be impossible unless all of those schemes are affine. Certainly an infinite product of projective lines is not affine. - -Some time ago, I've already proven the following result (see here): Assume that $(X_i)$ is a family of $S$-schemes such that their fiber product $P$ exists in the category of schemes. If $Q$ denotes the fiber product of the $X_i$ in the category of locally ringed spaces (which exists and can be described explicitly), then the canonical morphism $P \to Q$ is bijective and the stalk maps are isomorphisms. But I don't know how to get the topology of $P$. This would be very helpful to show that $P$ does not exist. -But as a first step one would have to show that $Q$ is not a scheme, which is already hard in general. Here is an idea: Let $X \neq \emptyset$ be a $k$-scheme and assume that the LRS fiber product $Q=X^I$ ($I$ infinite set) is a scheme. Then the explicit construction implies that there are open subsets $U_i \subseteq X$, such that $U_i = X$ for almost all $i$, and that $\prod_i U_i$ is an affine scheme. Let $U$ be the (finite) product of the $U_i \neq X$. Then $U$ is a scheme, and $U \times X^I$ is an affine scheme. Under suitable finiteness conditions (?), a combination of Serre's criterion and the Künneth formula (see here) would imply that $X^I$ is an affine scheme. But then $X^I \cong X \times X^I$ shows with the same argument that $X$ is affine. But this all works only if $X$ and $X^I$ are quasicompact and quasiseparated. - -REPLY [10 votes]: I will try to argue that if an infinite product of quasi-compact schemes is representable by a scheme, then all but finitely many of them must be affine. -Edit: I rewrote the argument a little to address the comments. I hope the argument might finally be satisfying... -Lemma. If $X_i$, $i \in I$ are quasi-compact schemes over a field $k$, then $\prod X_i$ has a closed point (i.e., there is a map from the spectrum of some field $K$ to $X$ that is representable by closed immersions) whose image in each of the $X_i$ is also closed. -Proof. For each $i$, let $x_i$ be a closed point of $X_i$ (which exists because the $X_i$ are quasi-compact). Let $K$ be a compositum of the residue fields of the $x_i$ (i.e., the residue field at a closed point of $Z = \prod_i x_i$, which exists because $Z$ is affine). Let $X$ be the product of the $X_i$. Then we get a map $\mathrm{Spec} \: K \rightarrow X$, which we have to show is a closed embedding. -It's enough to argue that the map $Z \rightarrow X$ is a closed embedding. But if $Y$ is a scheme over $X$ then $Y_Z$ can be obtained as the intersection of closed subschemes $Y_{x_i} = Y \times_{X_i} x_i$, which is a closed subscheme. $\Box$ -Suppose that the product of schemes $X_i$, indexed by $i \in I$, is representable by a non-empty scheme $X$. Assume that each $X_i$ is quasi-compact, so that for each $i$, we can find a Zariski cover $Y_i \rightarrow X_i$ of $X_i$ by an affine scheme $Y_i$. Then we get a map $Y = \prod Y_i \rightarrow X$ and the source is an affine scheme. Let $x$ be a closed point of $X$. Let $U$ be an affine neighborhood of $x$ in $X$. Let $y$ be a $K$-point of $Y$ above $x$, and let $V$ be a distinguished affine open neighborhood of $y$ in $Y$, contained in the pre-image of $U$, and defined by the invertibility of some $f \in \Gamma(Y, {\cal O}_Y)$. Since $f$ can be expressed using functions defined on finitely many of the $Y_i$, the set $V$ must be the pre-image of a distinguished open affine $V' \subset \prod_{i \in I'} Y_i$ where $I'$ is a finite subset of $I$. Let $I''$ be the complement of $I'$ in $I$ and let $X' = \prod_{i \in I'} X_i$ and define $X''$, $Y'$, $Y''$ analogously. Then $V' \times Y''$ is an affine open neighborhood of $y$ in $Y$, contained in $V$. If $U'$ denotes the image of $V'$ in $X'$ then $U' \times X''$ is an open neighborhood of $x$ in $X$, contained in $U$. Let $x'$ be the projection of $x$ to $X'$, which we can safely assume is a closed point of $X'$, hence a closed point of $U'$, and let $K$ be the residue field of $x'$. Therefore $\{ x' \} \times X''$ is a closed subscheme of $U$ and is therefore affine. But $\{ x' \} \times X'' = K \otimes_k X''$ so by fpqc descent for affine schemes over $k$ (SGA1.VIII.2) it follows that $X''$ is affine. -Replacing $I$ with $I''$ we can now assume that $X$ is affine. Before proving that each $X_i$ is affine, I claim that each $X_i$ is separated. We must show that for any scheme $Y$ and any two maps $Y \rightrightarrows X_i$, the locus in $Y$ where the maps agree is closed. Let $x$ be a closed $K$-point of $\prod_{j \not= i} X_j$ for some extension $K$ of $k$. Then we can extend the maps $Y \rightrightarrows X_i$ to maps $Y \otimes_k K \rightrightarrows X$. Since $X$ is separated, the locus in $Y \otimes_k K$ where these maps agree is closed. But this is the pullback via the map $Y \otimes_k K \rightarrow Y$ of the locus where the maps $Y \rightrightarrows X_i$ agree, so this latter subset of $Y$ must be closed by fpqc descent (see SGA1.VIII.4). -If $X$ is affine and non-empty then it is separated, so each $X_i$ is separated. Therefore $Z := \prod_{j \not= i} X_j$ is separated for each $i$. (Note that we don't yet know that $Z$ is representable by a scheme, but it still makes sense to claim that $Z$ is separated as a functor: for any scheme $T$ and any two elements $a$ and $b$ of $Z(T)$, the locus in $T$ where $a$ and $b$ agree is closed.) Let $z$ be a closed $K$-point of $Z$. Then $X_i \otimes_k K = X \times_Z \{z\}$ is closed in $X$, hence is affine because we assumed $X$ to be affine. This implies $X_i$ is affine by fpqc descent (see SGA1.VIII.2).<|endoftext|> -TITLE: Why the triangle inequality? -QUESTION [26 upvotes]: [Maybe this is asking to be closed; but I thought I'd risk it.] -A metric satisfies the axioms: - -$d(x,y)=0$ if and only if $x=y$. -$d(x,y) = d(y,x)$. -$d(x,y) \leq d(x,z) + d(z,y)$. - -Similarly (and motivationally) a uniformity on $X$ on a filter $F$ on $X\times X$ with: - -$\Delta = \{ (x,x) : x\in X \} \subseteq D$ for all $D\in F$. -$D=\{ (y,x) : (x,y)\in D\}$ for all $D\in F$. -for all $D\in F$ there is $E\in F$ with $E^2\subseteq F$. - -It seems to me that if you drop the triangle-inequality (or the third axiom for a uniformity) then you can still do most basic topology, can still show, for example, that the space of bounded (uniformly) continuous functions $X\rightarrow\mathbb R$ or $\mathbb C$ is a Banach space (of interest to an Analyst like me) and so forth. My question is, why don't we study metrics/uniformities which doesn't satisfy the triangle inequality? (I see from Spaces with a quasi triangle inequality that such a thing is a semi-metric space). -Some reasons I thought of: -i) "Most" metrics arise from other structures-- e.g. norms on a vector space-- and the triangle-inequality comes from an axiom which seems more natural (e.g. the triangle-inequality for a norm really seems important-- it gives some coupling between the additive structure of the vector space and the distance). -ii) If your space is quite nice (e.g. compact) in the topology induced by your semi-metric, then there is an actual metric giving the topology. So why not just assume you have a metric? This is especially true for uniform spaces, as you only need to be completely regular to be uniformisable. -iii) It seems to me that you really do need the triangle-inequality to talk about Cauchy-sequences in a sensible way, and so to talk about completeness. -I suppose I'm slightly motivated by the recent(ish) question on whether definitions in maths are "correct". What makes the triangle-inequality so useful that pretty much everyone assumes it, even though you can do a lot of point-set topology without it? - -REPLY [6 votes]: Here's a real life example of something that fails the triangle inequality, but that is still interesting: the space $L^p(X)$ for $0\lt p\lt 1$. For example, the product $fg$ of two functions $f,g\in L^1(X)$ is in $L^{1/2}(X)$. -This is an example of a quasi-Banach space and -thus satisfies Dick Palais's condition "two things that are close to the same thing are close to each other".<|endoftext|> -TITLE: What is transport of structure in cohomology setting? -QUESTION [10 upvotes]: It seems to me this come up very often when we talk about group action on (étale) cohomology groups. -For example, let $X$ be a scheme over $\mathrm{spec}\mathbb{Z}$, $\mathcal{F}$ an $\ell$-adic sheaf on $X$, and $V_\ell= H^i(X\otimes \overline{\mathbb{Q}}, \mathcal{F})$. The Galois group of $\overline{\mathbb{Q}}$ of $\mathbb{Q}$ acts, by transport of structure, on $V_\ell$. -For an fancier example, in Deligne's Seminar Bourbaki paper "Formes modulaires et représentations $\ell$-adiques", immediately after definition 3.9 in which a $\mathbb{Q}$-vector space of cohomology of modular curves are defined, it says (in English) "This vector space depends only on the universal elliptic curve (up to isogeny) $f_\infty: E\to M_\infty$ so that, by transport of structure, it is equipped with a left action of the adelic group $\mathrm{GL}_2(\mathbb{A}^f)$". (Here $\mathbb{A}^f$ is the ring of "finite" adeles). -As my knowledge of étale cohomology comes mainly from reading J.S. Milne's online notes with a few glances of Freitag-Kiel, I am pretty lost with the mathematics that happens behind the few words "by transport of structure". -Thank you. - -REPLY [8 votes]: I am not sure this is adding anything substantial to the comments, but maybe it helps -to spell things out. At least this used to confuse me in the past. I found a remark in a recent preprint of -Deligne-Flicker helpful: Transport of structures (Bourbaki Ens Ch. IV) is the principle that any isomorphism $Y_1 \to Y_2$ extends to objects constructed from $Y_1$ and $Y_2$. When $Y_1 = Y_2$: symmetries extend. The construction has to be canonical: not involving choices. -Let me now explain why $Aut(\bar k/k)$ acts on $H^i(X \otimes \bar k, \mathbb Q_l)$ (where $X$ is a $k$-scheme and $k \to \bar k$ -is a fixed algebraic closure) by transport of structure. Let's first take a step back and use two different -algebraic closures $\sigma_i : k \to K_i$. (I learnt this from Tate, in fact I think his global class field theory article -in Cassels-Fröhlich.) Then by canonicity of etale cohomology, any isomorphism $K_1 \to K_2$ between $\sigma_1$ and -$\sigma_2$ gives us a canonical isomorphism $H^i(X \otimes K_1, \mathbb Q_l)$ and $H^i(X \otimes K_2, \mathbb Q_l)$ of -$\mathbb Q_l$-vector spaces. (Never mind that it's easy to say what this map actually is, the point is that we know there -has to be one!) In particular, if we take $K_1 = K_2 = \bar k$, we get a canonical $\mathbb Q_l$-linear action of $Aut(\bar -k/k)$ on $H^i(X \otimes \bar k, \mathbb Q_l)$. I think the argument is the same when we replace -$\mathbb Q_l$ by (the pullback to $X \otimes \bar k$ of) a constructible or lisse $l$-adic sheaf over $X$. -Now for your second example, we can play the same game: take two universal elliptic curves $f_{\infty, i} : E_i \to -M_\infty$. Any isogeny between them gives us a canonical isomorphism of Deligne's $\mathbb Q$-vector space $W$ (or rather -$W_1$, $W_2$). Hence if we take $E_1 = E_2 = E$, ... -Finally, let me consider one more example from Deligne's writing that I found very cryptic in -the past. Namely, paragraph 0.2.5 in his article "Valeurs de fonctions et périodes d'intégrales" -(http://publications.ias.edu/deligne/paper/379). Basically what he is claiming is the following. -Suppose $X$ is a smooth projective variety over a field $k$. For an embedding $\sigma : k \to \mathbb C$ -let $H_\sigma := H^*((X \otimes_k \mathbb C)(\mathbb C), \mathbb Q)$, which is a $\mathbb Q$-vector space together -with a bigrading on its complexification (by Hodge theory). He says that "by transport of structure", we obtain -an isomorphism $F_\infty : H_\sigma \to H_{c \sigma}$ such that moreover $F_\infty \otimes c$ -carries $H_\sigma^{pq}$ to $H_{c\sigma}^{pq}$ (where $c$ denotes complex conjugation). -The point is to follow his definitions very carefully: suppose $\sigma : k \to K$ is an embedding -into an algebraic closure $K$ of $\mathbb R$ (i.e., abstractly $\mathbb C$, but it's important not -to work with just this copy). Then $H_\sigma := H^*((X \otimes_{k, \sigma} K) (K), \mathbb Q)$ and the -bigrading is on $H_\sigma \otimes_{\mathbb Q} K = H^*((X \otimes_{k, \sigma} K) (K), K) -= \oplus H_\sigma^{pq}$. (Sanity check: for Hodge theory we indeed want to have the same field of -definition and coefficients.) Now again take $\sigma_i : k \to K_i$. Any isomorphism $\tau:K_1 \to K_2$ between -$\sigma_1$ and $\sigma_2$ gives us a canonical -isomorphism $F : H_{\sigma_1} \to H_{\sigma_2}$ of $\mathbb Q$-vector spaces. This in turn gives -us a canonical isomorphism $F \otimes \tau : H_{\sigma_1} \otimes K_1 \to H_{\sigma_2} \otimes K_2$, and by canonicity of the Hodge structure it -sends $H_{\sigma_1}^{pq}$ to $H_{\sigma_2}^{pq}$. Now specialise to $K_1 = K_2 = \mathbb C$,... -(I find it interesting how Deligne explains in his interview https://www.simonsfoundation.org/science_lives_video/pierre-deligne/ that he is very bad at calculations and therefore needs to work very canonically, -otherwise he gets lost. He also says that he learnt about "transport of structure" from -reading Bourbaki's "Set theory" as teenager.)<|endoftext|> -TITLE: What is the name of this graph? -QUESTION [5 upvotes]: I think this graph has a name: the vertices are bit strings of length $n$, and $(x_1, \ldots , x_n)$ is connected to $(x_2, \ldots, x_n, 0)$, $(x_2, \ldots, x_n, 1)$, $(0,x_1, \ldots , x_{n-1})$ and $(1, x_1, \ldots , x_{n-1})$. I'm wondering (a) what the name is and (b) where I can read more about this graph. Thanks! - -REPLY [9 votes]: They are called De Bruijn graphs (De Bruijn graphs are generally considered directed, and can be defined over any set of symbols, not just $\{0,1\}$).<|endoftext|> -TITLE: Construct a CW complex with prescribed homotopy groups and actions of $\pi_1$. -QUESTION [8 upvotes]: How to construct a CW complex $X$ with prescribed homotopy groups $\pi_i(X)$ and prescribed actions of $\pi_1(X)$ on the $\pi_i(X)$'s? - -REPLY [14 votes]: One method is as follows. - -Construct $Y_i = K(\pi_i, i)$ for $i \geq 2$ as a based space with an action of the group $\pi_1$. You can do this manually (by attaching free orbits of cells along the group action) or canonically (using a functorial construction of Eilenberg-Mac Lane spaces, such as one obtained from the Dold-Kan correspondence and geometric realization). -Take $Y = \prod_{i \geq 2} Y_i$ as a based space with the correct "top" homotopy groups and an action of $\pi_1$ realizing the desired action on the higher homotopy groups. -Take the Borel construction $X = E\pi_1 \times_{\pi_1} Y$. This space, by construction, has fundamental group $\pi_1$, has $E \pi_1 \times Y \simeq Y$ as universal cover, and has the group of deck transformations given by the action of $\pi_1$ on $E\pi_1 \times Y$. - -Of course, you don't have to take $Y$ to be a product of Eilenberg-Mac Lane spaces. Passing between universal covers and Borel constructions reduced your original question to finding a simply-connected space with the given higher homotopy groups and an action of $\pi_1$ on the whole space. -You might also specify more structure; for example, the higher homotopy groups in the outlined construction are not related by Whitehead products or anything similar. This makes the problem harder. If you specify all the structure that the homotopy groups naturally have, you have a so-called $\Pi$-algebra; the paper "The realization space of a $\Pi$-algebra" by Blanc-Dwyer-Goerss studies this problem and how one might go about classifying realizations.<|endoftext|> -TITLE: Random versions of deterministic problems -QUESTION [23 upvotes]: What are the examples of situations where "randomizing" a problem (or some part of it) and analyzing it using probabilistic techniques yields some insight into its deterministic version? -An example of what I have in mind: it is a well-known conjecture that the Hausdorff dimension of the graph of Weierstrass function (everywhere continuous, nowhere differentiable) is given by a certain simple formula, depending on the amplitudes and phases of the cosines in the series. This is still open; however, in the paper "The Hausdorff Dimension of Graphs of Weierstrass Functions" Hunt proved that if you add a uniformly distributed independent random "noise" to each phase, the conjectured formula holds with probability 1. So while the "randomized" approach does not solve the original problem, it somehow lends credibility to the original conjecture and thus gives us some insight about the problem. - -REPLY [5 votes]: One problem that I am aware of is the problem of finding the Poisson boundary of a manifold, that is, the bounded harmonic functions. There is a probabilistic way of approaching this problem using Brownian motion (c.f. here) which has evolved out of the probabilistic proof of Liouville's theorem. -If we let $B=(B_t:t \geq 0)$ be a Brownian motion and $Inv(B)$ be the sigma-algebra of events of the form $\{B_t \in A\}$ iff $\{B_{t+s}\in A\}$ for all $s \geq 0$, then there is a one-to-one correspondence between $Inv(B)$ measurable random variables and bounded harmonic functions. -Liouville's theorem holds on $\mathbb{R}^d$ because if we can start two Brownian motions at $x,y \in \mathbb{R}^d$ and couple them in a finite time, so the invariant sigma algebra is trivial. On manifolds where Liouville's theorem does not hold, then by looking at the invariant algebra, one can work out all of the bounded harmonic functions.<|endoftext|> -TITLE: What results would follow from or imply "randomness" of the primes? -QUESTION [13 upvotes]: This question on random versions of deterministic problems reminded me that many conditional results in number theory hold if the primes are in some sense random, and it is common knowledge that the Riemann hypothesis is tied up in this line of thought. -Now, off the top of my head it's not clear what "randomly distributed" really means in the context of the primes. I know that the RH would imply better versions of the prime number theorem, but that's not quite the same. So I am interested in a clarification of this as well as the following -Main question: what results would imply that, and what results would we gain from, the assumption that the primes are "randomly distributed"? - -REPLY [2 votes]: Billingsley wrote a nice article on the relationship between numbers, random walks and Brownian motion.<|endoftext|> -TITLE: Euler characteristics and characteristic classes for real manifolds? -QUESTION [8 upvotes]: Consider an oriented manifold $X$. To calculate its Euler characteristic, one might integrate the Euler class. Now if $X$ were a complex manifold, and given as a section of some complex bundle $E$ over $Y$, we would have -$\chi(X) = \int\limits_X \mbox{ } c_* (TY) / c_* (E)$ -where $c_*(\cdot)$ is the total Chern class. - - -Can anything of the sort be said if $X$ is a real manifold? - - -Presumably, if one wants only the Euler characteristic modulo 2, one can use the Stiefel-Whitney classes instead of the Chern classes. On the other hand, it seems to me that the topology of $TY$ and $E$ as bundles over $Y$ cannot suffice to carry the information of the Euler characteristic of the zero locus of a section of $E$. So I guess what I'm really asking is: - - -What should I know about a section $\sigma:Y\to E$ in order to know the Euler number of its intersection with the zero section, assuming this is transverse? - -REPLY [3 votes]: As far as I understand the question, it is asking: given a(n oriented) vector bundle $E$ on a(n oriented) manifold, what information on the Euler characteristic of the zero locus of a transversal section of $E$ can we deduce from the characteristic classes of $E$? -I'm afraid the answer to that is "in general, not much, apart from the parity". For instance, every orientable surface $S$ is the zero locus of a function (i.e., a section of the trivial 1-bundle) on the 3-sphere: embed the surface in the sphere and take the "oriented distance" function. In a similar way one can realize $S$ as the zero locus of two functions on $S^4$: realize $S^3$ as the equator in $S^4$, take one of the functions to be the oriented distance function extended to $S^4$ and the other a height function whose zero locus is $S^3$. -The reason one can compute the Chern classes of the zero locus in the holomorphic case is that the total Chern class is invertible, so the total Chern class of tangent bundle of the zero locus of a transversal section comes from the ambient manifold. The Euler class however is not invertible.<|endoftext|> -TITLE: Rationalised K-theory of number fields -QUESTION [12 upvotes]: Let $A$ be the ring of integers in a number field, and consider the rationalised algebraic $K$-theory groups $\mathbb{Q}\otimes K_*(A)$. A theorem of Borel calculates the ranks of these groups; the answer can be described as follows. The tensor product $\mathbb{R}\otimes A$ is isomorphic as a topological $\mathbb{R}$-algebra to $\mathbb{R}^s\times\mathbb{C}^t$ for some $s$ and $t$. It follows that $K_{\ast}(\mathbb{R}\otimes A)$ is isomorphic to $kO_{\ast}^s\oplus kU_{\ast}^t$ (if we use the version of $K$-theory that remembers the topology on $\mathbb{R}\otimes A$). Here $\mathbb{Q}\otimes kO_{\ast}=\mathbb{Q}[x]$ with $|x|=4$, and $\mathbb{Q}\otimes kU_{\ast}=\mathbb{Q}[y]$ with $|y|=2$. Borel's result says that $\mathbb{Q}\otimes K_{\ast}(A)$ is isomorphic, as a graded vector space, to $\Sigma\mathbb{Q}\otimes K_*(\mathbb{R}\otimes A)$ (where $\Sigma$ is the syspension, shifting degrees by one) except that we have to move one of the generators from degree one back down to degree zero. Of course this implies the corresponding relationship between $\mathbb{R}\otimes K_{\ast}(A)$ and $\mathbb{R}\otimes K_{\ast}(\mathbb{R}\otimes A)$. -My question: is there a natural isomorphism between these groups, either in the real case or the rational case? My feeling is that it should be possible to extract this from the work of Borel, if I understood it better. What is the conceptual explanation for the degree shift? - -REPLY [16 votes]: The answer is yes, but only after tensoring with $\mathbb R$. -Thinking of the Beilinson regulator map with values in Deligne cohomology is simpler than thinking about the Borel regulator map; it's been proved that they agree with each other. -The topological Chern character map $ch_n : kU_{i} \to H^{2n-i}(pt,\mathbb Q)$ is an isomorphism $kU_{2n} \otimes \mathbb Q \ \xrightarrow \cong \ \mathbb Q$ when $i=2n$. -The corresponding algebraic Chern character map with values on Deligne cohomology is $ch_n : K_{i}\mathbb C \to H^{2n-i}_{\cal D}(pt,\mathbb Q(n))$. Here $\mathbb Q(n)$ (or $\mathbb Z(n)$) denotes a certain cochain complex of sheaves for the analytic topology on a complex manifold $X$. It starts in degree 0 with $\mathbb Q$ (resp., $\mathbb Z$), in cohomological degree 1 it has $\mathbb C$, and the differential map $d^0 : \mathbb Q \to \mathbb C$ is multiplication by $(2 \pi i)^ n$. The term in degree $i+1$ is the sheaf of holomorphic differentials $\Omega^i$ if $i < n$ and is $0$ if $i \ge n$. The exponential map $\mathbb C \to \mathbb C ^ \times $ given by $z \mapsto e^z$ gives a quasi-isomorphism $ \mathbb Z (1) \to \mathbb C ^ \times [-1]$; the degree shift there answers your second question, partially; another way of saying that is that there is a degree shift in the boundary map $c_1 : H^1(X,\mathbb C^\times) \to H^2(X,\mathbb Z)$. I say "partially", because one must know also that the regulator map involves no further degree shift; in degree 1 it's because the map $\mathbb C ^ \times \to \mathbb R$ given by $z \mapsto {\rm log} |z|$ involves no degree shift. -Now consider the projection $\mathbb C \to \mathbb R$ that sends $ (2 \pi i)^n $ to $0$ and $i^{n-1}$ to $1$; perhaps there is a better normalization for this map, such as choosing to send $(2 \pi i)^{n-1}$ to $1$. It induces a map of cochain complexes $\mathbb Q(n) \to \mathbb R[-1]$; the map it induces on Deligne cohomology, composed with the Chern character map above, is the Beilinson regulator map $$ch_n : K_{i}\mathbb C \to H^{2n-i}(pt,\mathbb R[-1]) = H^{2n-i-1}(pt,\mathbb R),$$whose only nonzero possibility is the map $ch_{n} : K_{2n-1}\mathbb C \to H^{0}(pt,\mathbb R) = \mathbb R$. For the ring of integers $A$, we get a map $K_{2n-1} A \to K_{2n-1}( A \otimes \mathbb C ) \to H^{0}(Spec(A \otimes \mathbb C),\mathbb R) = \mathbb R^{s+2t}$. Borel's theorem is recast as saying that for $n > 1$ the map induces an isomorphism of $(K_{2n-1} A) \otimes \mathbb R$ with the appropriate eigenspace for the action of $G = Gal(\mathbb C/\mathbb R)$ on $$H^0(Spec(A \otimes \mathbb C), \mathbb R (n-1)) = \mathbb R^{s+2t},$$where now I use $\mathbb R (n-1)$ to remind us how $G$ acts on this real vector space of dimension 1. -(Added later: actually, it may be more natural to replace the $\Sigma$ in the question by $\Omega$ and to use the anti-invariants (or invariants, depending on the parity of n) under $G$ acting on $K^{top}(A \otimes \mathbb C) \otimes \mathbb C$ instead of the invariants. Thus the degree shift can be viewed as $-1 = 1 - 2$)<|endoftext|> -TITLE: The algebraic Version of Riemann Hilbert Correspondence -QUESTION [14 upvotes]: It is well known that if I have a differentialable manifold (holomorphic maniford) $M$, then I have a functor from the categroy of vector bundles on $M$ with flat connections to the categroy of local systems on $M$, given by $$(V,\nabla)\mapsto V^{\nabla}$$ and this functor is an equivalence of categories. -Now if I change the setting a little bit. I assume $X$ is a smooth scheme over a field $k$ of characteristic 0. Do I still have a functor like the above? Or in other words, is it true that -for any vector bundle $(V,\nabla)$ on $X$ the sheaf of horizontal sections $V^{\nabla}$ (i.e. the kernel of $\nabla$ as $k$-vector spaces) is still a locally constant sheaf of $k$-vector spaces? -I think it is true, and it should be written somewhere, could you tell me the reference for that? - -REPLY [20 votes]: As the previous answer points out you have to consider local systems for a finer topology than the Zariski topology. It is natural to consider the étale topology. The category of étale local systems of finite dimensional $k$-vector spaces form a tannakian category whose group is the étale fundamental group (more precisely it is a pro-(constant finite) algebraic group). You can find this and much more in Saavedra's book on tannakian categories LNM Volume 265. -see http://www.springerlink.com/content/978-3-540-05844-1/ -The category of bundles with connexion (with a regularity condition at infinity if needed) also form a tannakian category (if $X$ is smooth over an algebraically closed field of characteristic zero say), but the corresponding pro-algebraic group is much larger (Deligne call this the algebraic fundamental group). He computes this group in the example of the affine line in his famous article "Le Groupe Fondamental de la Droite Projective Moins Trois Points" -see http://www.math.ias.edu/files/deligne/GaloisGroups.pdf -So to an étale local system of finite dimensional $k$-vector space you can associate a bundle with connexion, and this will give you a fully faithful Riemann-Hilbert functor, but in this way you will obtain only those bundles with connexion whose monodromy group is finite and étale (in other words those trivialized by an étale cover).<|endoftext|> -TITLE: Symmetric spaces, Horocycle spaces and intertwining operators -QUESTION [7 upvotes]: Let $G=KAN$ be an Iwasawa decomposition of a connected semisimple Lie group with finite center. Let us assume for simplicity that the associated symmetric space $G/K$ has rank 1. -Harish-Chandras plancherel theorem gives an explicit direct integral decomposition of the left-regular representation of $G$ on $L^2(G/K)$. -$L^2(G/K) \cong \int_{i\mathbb{R}} \pi_\mu p(\mu)d\mu$. -Here, $\pi_\mu$ is the spherical principle series representation with parameter $\mu$ and $p$ is an explicitly known density function. Now one has the normalized Knapp-Stein operators $A_\mu$ from $\pi_\mu$ to $\pi_{-\mu}$. These can be pieced together to give a unitary intertwining operator on the direct integral, and hence an operator $T$ on $L^2(G/K)$. Given the prominent role of the Knapp-Stein operators in representation theory I thought that the operator $T$ must have been studied extensively but I wasn't able to find anything on this subject. Specifically I'd like to know if the operator $T$ can be given in purely geometric terms without resorting to the direct integral decomposition. -In a similiar vein, if $M$ is the centralizer of $A$ in $K$ then you can construct an analoguos operator on the space of $L^2$ functions on the horocycle space $G/MN$. Again, is there any way to define this operator in geometric terms without using the direct integral decomposition? -I'd be very happy and grateful if a person more knowledgable then I could shed some light on this and could perhaps give a few references to artciles where these operators have been studied. - -REPLY [7 votes]: An explicit form is in my paper "A duality ... Advan. Math. 1970 and in my book Geometric Analysis on Symmetric Spaces AMS 2008, pp.554-556. Helgason.<|endoftext|> -TITLE: Change in the average geodesic distance of a graph when flipping a single edge -QUESTION [8 upvotes]: Is there a way to determine how the average geodesic distance between nodes of a graph will change just by flipping (1) a single edge without having to traverse the whole graph like in the Djikstra algorithm? -I'm currently doing this by expensively copying the graph, changing the edge, and then calculating the average geodesic distance of the new graph (using Dijkstra's algorithm) and subtracting it from the average geodesic distance of the original graph.. -Is there a more clever way to to this? -notes: -(1) By flipping a edge I mean the following operation: add the edge if it's absent and remove it if it's present. -(2) Good approximations are welcomed. It's part of a Monte Carlo simulation, so I must repeat this calculation many, many times. - -REPLY [2 votes]: With the right data structures (see http://www.ams.org/mathscinet-getitem?mr=2145260), one can maintain a matrix of pairwise distances between vertices in a dynamic graph. Updating the entire matrix after modifying an edge takes $O(n^2\log^3n)$ (amortized). This is at least better than doing a completely new all-pairs shortest path algorithm each time you modify an edge. -I don't, however, see an easy way to take advantage of the fact that you are not interested in the entire matrix of distances, but rather just the average distance.<|endoftext|> -TITLE: topologies on U(H) -QUESTION [6 upvotes]: There are many topologies on the algebra $B(H)$ of bounded operators on Hilbert space: -the weak, strong, ultraweak (also called σ-weak), ultrastrong (also called σ-strong), and some more... -Luckily, the weak and strong topologies agree when restricted to $U(H)\subset B(H)$. -Similarly, the ultraweak and ultrastrong topologies agree on $U(H)$. - -Is it true that the weak and ultraweak topologies agree when restricted to - $U(H)$? - - -Definitions: -A generalized sequence $a_i$ is weakly, strongly, ultraweakly, ultrastrongly convergent if: -• $\langle a_i\xi,\eta\rangle\to\langle a\xi,\eta\rangle\qquad \forall \xi,\eta\in H$ -• $a_i\xi\to a\xi\qquad \forall \xi\in H$ -• $\langle (a_i\otimes 1)\xi,\eta\rangle\to\langle (a\otimes 1)\xi,\eta\rangle\qquad \forall \xi,\eta\in H\otimes \ell^2(\mathbb N)$ -• $(a_i\otimes 1)\xi\to (a\otimes 1)\xi\qquad \forall \xi\in H\otimes \ell^2(\mathbb N)$, - respectively. Here, $H\otimes \ell^2(\mathbb N)$ denotes the Hilbert space tensor product of $H$ and $\ell^2(\mathbb N)$. - -REPLY [9 votes]: The weak and ultraweak topologies coincide on bounded subsets of $B(H)$: see section 3.5 in Pedersen's book "C*-algebras and their automorphism groups".<|endoftext|> -TITLE: The closure of the set of rational points in the Adeles -QUESTION [9 upvotes]: Let $X$ be a smooth geometrically integral projective variety over $\mathbb{Q}$. Then we may consider the closure $\overline{X(\mathbb{Q})}$ of $X(\mathbb{Q})$ inside the adelic points $X(\mathbb{A})=\prod_v X(\mathbb{Q}_v)$ of $X$. However, we may also take the closure $\overline{X(\mathbb{Q})}^v$ of $X(\mathbb{Q})$ inside $X(\mathbb{Q}_v)$ for any place $v$ of $\mathbb{Q}$. -Obviously we have $$\overline{X(\mathbb{Q})} \subset \prod_v \overline{X(\mathbb{Q})}^v \subset X(\mathbb{A}).$$ -My question whether this first inequality is actually an equality? -My motivation is that I am trying to understand better $\overline{X(\mathbb{Q})}$ and what it looks like. I will simply note that the answer to my question is yes in the easy cases where $X$ satisfies weak approximation and when $X(\mathbb{Q})$ is empty. -Edit: To make sure there are not simple counter-examples like the one David pointed out below, I am assuming that $X(\mathbb{Q})$ is Zariski dense. I should also note that I am particularly interested in the case where $X$ is a fano variety. - -REPLY [11 votes]: Here's an example where $X(\mathbf{Q})$ is Zariski-dense but the first inequality is not an equality. -Let $X$ be an elliptic curve over the rationals, such that the group $X(\mathbf{Q})$ is isomorphic to $\mathbf{Z}$. Let me first remind you what $X(\mathbf{Q}_p)$ looks like, for $p$ a prime where the curve has good reduction. There's a natural reduction map $X(\mathbf{Q}_p)\to\overline{X}(\mathbf{F}_p)$, with $\overline{X}$ the reduction of $X$, and this reduction map is surjective onto a finite target. Hence $X(\mathbf{Q}_p)$ naturally breaks up as a finite disjoint union of cosets of the kernel of this map, and the kernel is (and hence all the cosets are) clopen. -Now let $P$ be a generator of $X(\mathbf{Q})$, put the curve into minimal form, choose a random large prime $\ell$ and consider $Q:=\ell.P$. It will be easy to find an example where $Q=(x,y)$ and there are two primes $p_1$ and $p_2$ in the denominator of $x$ but not in the denominator of the coordinates of $P$, and such that $X$ has good reduction at these primes. In short it should be easy to find two primes $p_1$ and $p_2$ such that $P$ has order exactly $\ell$ modulo both $p_1$ and $p_2$. -But now the first inclusion must be strict, because even the closure of $X(\mathbf{Q})$ in $X(\mathbf{Q}_{p_1})\times X(\mathbf{Q}_{p_2})$ is easily seen to be strictly smaller than the product of the closures, for the "same reason" that if $P_1$ has order $\ell$ in the finite abelian group $G$ and $P_2$ has order $\ell$ in the finite abelian group $H$, then the group generated by $(P_1,P_2)$ in $G\times H$ is strictly smaller than the product of the subgroups generated by $P_1$ and $P_2$.<|endoftext|> -TITLE: Is there a midsphere theorem for 4-polytopes? -QUESTION [21 upvotes]: The (remarkable) midsphere theorem says that each combinatorial -type of convex polyhedron may be realized by one all of whose edges are -tangent to a sphere -(and the realization is unique if the center of gravity is specified). -           - -Q1. Is there an analogous theorem for 4-polytopes, - that each combinatorial type may be realized by a polytope - with ridges (or edges?) tangent to a 3-sphere? - -Because the proofs of the midsphere theorem rely on the -Koebe–Andreev–Thurston circle-packing theorem, -a related query is: - -Q2. Is there a generalization of the circle-packing - theorem to sphere-packing? - -Both questions may be generalized to arbitrary dimension. -I suspect the answer to both questions may be No, -in which case a pointer would suffice. Thanks! - -REPLY [5 votes]: In a recent paper of Padrol and me, we studied several generalizations of this problem. http://arxiv.org/pdf/1508.03537v1.pdf -Regarding Q1, Yoav already mentioned Schulte's work, and Gil mentioned that stacked polytopes won't work. In fact, - -For any 0 ≤ k ≤ d − 3, there are stacked d-polytopes that are not k-scribable. - -But the other side of the story is much more joyful: - -Every stacked polytope is (d − 1)-scribable (i.e. circumscribable) and (d − 2)-scribable (i.e. ridge-scribable). Dually, it means that every truncated polytope is edge-scribable and inscribable. In particular, the $1$-skeleton of every truncated polytope is the tangency graph of a sphere packing. - -We also studied the weak scribabilities, also mentioned in Schulte's work. -The most interesting generalization (which I would like to make some advertisement here) is the $(i,j)$-scribability problem: - -Can we realize every $d$-polytope with all $i$-faces "outside" the sphere and all $j$-faces "intersecting" the sphere, $0\le i \le j \le d-1$? - -The case of $i=j$ is just the classical scribability problem. For $i -TITLE: Lagrangian Submanifolds in Deformation Quantization -QUESTION [11 upvotes]: Suppose I have a symplectic manifold $M$, and have a deformation quantization of it, i.e. an associative product $\ast:C(M)[[\hbar]]\otimes C(M)[[\hbar]]\to C(M)[[\hbar]]$ so that $f\ast g=fg+\{f,g\}\hbar+O(\hbar^2)$, where $\{\cdot,\cdot\}:C(M)\otimes C(M)\to C(M)$ is the Poisson bracket coming from the symplectic structure. -According to Lu (http://www.ams.org/mathscinet-getitem?mr=1244874 page 395), a Lagrangian submanifold $L\subseteq M$ should correspond to a left-module over $C(M)[[\hbar]]$, that is, we have a deformed module structure $\ast:C(M)[[\hbar]]\otimes C(L)[[\hbar]]\to C(L)[[\hbar]]$, which reduces to the standard module structure when $\hbar=0$, and is compatible with the start product on $C(M)[[\hbar]]$. (Actually, Lu thought of the left-ideal $I_L\subseteq C(M)[[\hbar]]$ so that $C(L)[[\hbar]]=C(M)[[\hbar]]/I_L$, but I perfer to think of the deformation of the module structure.) -Then we can write, for $f\in C(M)$ and $g\in C(L)$, that $f\ast g=fg+\{f,g\}\hbar+O(\hbar^2)$, for some brackets $\{\cdot,\cdot\}:C(M)\otimes C(L)\to C(L)$. -What are these brackets? How are the related to the symplectic structure? -It seems that in general the existence of a deformed module structure on $C(L)[[\hbar]]$ is nontrivial (perhaps not even known for arbitrary $L$?), but I am hoping that perhaps the deformation to first order is something that one can understand more easily. - -REPLY [8 votes]: Well, in the symplectic case, the situation is somehow much simpler as in the general Poisson case where you only can speak about coisotropic (there is no good meaning of minimal coisotropic as the rank may vary). In the symplectic case you have a theorem of Weinstein which states that a there is a tubular neighbourhood of $L$ which is symplectomorphic to a neighbourhood of the zero section of $T^*L$. Thus the question of a module structure is reduced to the case of a cotangent bundle since star products are local. For cotangent bundles there is a good understanding whether you can have a module structure on the functions on the configuration space $L$: the characteristic class of $\star$ has to be trivial. In fact, together with Martin Bordemann and Nikolai Neumaier we constructed such module structures in a series of papers in the end of the nineties. Also Markus Pflaum has some papers on this. Thus the global statement is that on $L$ you have a module structure for $\star$ iff the characteristic class of $\star$ is trivial in an tubular neighbourhood of $L$. -The module structures have (for particular star products) a very nice interpretation as global symbol calculus for differential operators on $L$. Moreover, if the char. class is not trivial but at least integral (up to some $2\pi$'s) then there is a module structure on the sections of some line bundle over $L$, coming quite close to the functions on $L$. Physically, this is important for the quantization of Dirac's magnetic monopole. -As DamienC already said, the situation in the general Poisson case or even in the general coisotropic case on symplectic manifolds is much more involved. Here my answer to -In the dictionary between Poisson and Quantum, what corresponds to Coisotropic? -might also be of interest for you. -Oh, I forgot: the first order term can be obtained as in the flat case, at least morally. On the configuration space $L$ you chose your favorite connection then the first order term of the module structure is something like "half the Poisson bracket" which means that \begin{equation} -f \bullet \psi = -\iota^* f \psi -+ i \hbar \iota^* -\frac{\partial f}{\partial p_i} \frac{\partial \psi}{\partial q^i} - + \cdots -\end{equation} -(modulo some constants I forgot) where $q^i$ are coordinates on $L$ and $p_i$ are the corresponding fiber coordinates on the cotangent bundle. This has an intrinsic meaning. Here $\iota: L \longrightarrow T^*L$ is the zero section...<|endoftext|> -TITLE: Green function of simple random walk -QUESTION [6 upvotes]: Let $G$ be the Green function of the simple random walk on $\mathbb{Z}^d,\:d\geq 3$; i.e. -$$G(x) = E \sum_{i=0}^{+\infty} 1_{X_i=x},$$ -where $X$ is the simple random walk starting from $0$. The behaviour of $G(x)$ is well know when $x$ is large. But do we know how the function behaves when $x$ is small? -I tried to look in book of Spitzer (Principles of RW) and Lawler, Limic (Random Walk: Modern Introduction). I only found "inverse Fourier transform" formula which I do not have idea how to use (besides some numeric calculations). -May be some of you know other references? -My question stems from esimation of -$$S_d:=\sum_{x\in \mathbb{Z}^d\setminus \lbrace 0 \rbrace} a_x \left(\sqrt{1+\frac{G(x)^2}{G(0)}}-1\right),$$ -where $a_x$ are some coefficients. In the simplest case $a_x =1$ (or $a_x$ = 1 if $\sum_{i=1}^d x_i$ is odd and -1 otherwise; "chessboard of -1, 1"). -In the case of $a_x=1$ one can easily prove that $S=\infty$ if $d=3,4$ and $S<\infty$ if $d\geq5$ (just using the fact that $G(x) \sim |x|^{2-d}$ when $x$ is large.) -But is it possible to say anything more about $S_d$ when $d>5$? E.g. I expect that $S_d \rightarrow 0$ as $d\rightarrow +\infty$. -I made a mistake in the first version the sum was over $\mathbb{Z}^d$ and it should be over $\mathbb{Z}^d\setminus \lbrace0\rbrace$. -Update 24.05.2011 Thanks to observations made in Didier Piau proof I was able to prove that indeed $S_d \rightarrow 0$ (in the case when $a_x=1$). I put the proof here. - -REPLY [2 votes]: In fact $S_d$ does not converge to zero when $d\to\infty$, at least if $a_x=1$ for every $x$. Here is a proof. -For every $x$, $G_d(x)=P_0^{(d)}(\mathtt{hit}\ x)G_d(0)\le G_d(0)$ hence $G_d(x)^2/G_d(0)\le G_d(0)$. For every $z$ in $[0,G_d(0)]$, $\sqrt{1+z}-1\ge c_dz$ with -$$ -c_d=\frac1{G_d(0)}[\sqrt{1+G_d(0)}-1]=\frac1{\sqrt{1+G_d(0)}+1}. -$$ -Hence, for every $(a_x)$, -$$ -S_d\ge c_dT_d,\quad\mbox{with}\ -T_d=\sum_xa_xG_d(x)^2/G_d(0). -$$ -In the special case where $a_x=1$ for every $x$, by reversibility, -$$ -\sum_xG_d(x)^2=\sum_{i,j}\sum_xp_i^{(d)}(0,x)p_j^{(d)}(x,0)=\sum_{i,j}p_{i+j}^{(d)}(0,0)=\sum_{i}(i+1)p_i^{(d)}(0,0), -$$ -hence -$$ -\sum_xG_d(x)^2>\sum_{i}p_i^{(d)}(0,0)=G_d(0). -$$ -In particular, $T_d>1$. Now $G_d(0)$ is a nonincreasing function of $d$ hence $c_d\ge c_5$ for every $d\ge5$ and $S_d>c_5$ for every $d\ge5$. -Edit Much simpler: if $a_x\ge0$ for every $x$, then $S_d>a_0(\sqrt{1+G_d(0)}-1)$ and $G_d(0)>1$ hence $S_d>a_0(\sqrt2-1)$.<|endoftext|> -TITLE: Gluing of manifolds and the Hausdorff condition. -QUESTION [8 upvotes]: Hi! -I apologize in advance if this question is better fit for https://math.stackexchange.com/. -Out of curiosity I'm interested in a particular case of the problem of what properties of a manifold is given by combinatorial information associated to the gluing of the manifold from pieces of $\mathbb{R}^n$. -Let $X$ be a manifold and $K$ be a covering of $X$ by subsets of $X$ homeomorphic to $\mathbb{R}^n$. We may produce a simplicial object $N_\cdot K$ called the Čech nerve of $K$. If the Hausdorff condition on $X$ is a constraint on the allowable gluings of $X$ from pieces of $\mathbb{R}^n$, we might expect to find this as a constraint on the set of nerves $N_\cdot K$ arising from manifolds (contra non-Hausdorff "premanifolds"). -So my question becomes: Does the Hausdorff condition affect the set of possible nerves arising from such coverings? If so by what criterion? Is this criterion both necessary and sufficient? I would be interested in a characterization of such "separatedness" of simplicial objects. -Sincerely, -Eivind - -REPLY [3 votes]: I am leaving this as an answer since I am new here and therefore can't comment on answers. It's not an answer, since I am not saying anything about the nerve of the covering. Apologies to the regulars here! -Here's what I wanted to say: David Carchedi's answer to the question is rather nice. However, it can be stated in much simpler terms, as follows. -Notice that the condition on the map $\coprod U_{\alpha\beta}\to\coprod U_\alpha\times\coprod U_\beta$ to be proper is simply that it be closed, since it is injective and its domain is Hausdorff. Moreover, it's open, so it's a homeomorphism onto its image, so it is closed if and only if its range is closed. The range of the map is just the graph of the equivalence relation $R$ on $X':=\coprod U_\alpha$ such that $X'/R$ is the glued space $X$. -Thus the statement is that $X$ is Hausdorff if and only $R$ has a closed graph in $X'\times X'$. This is true for any equivalence relation $R$ such that the canonical projection $X'\to X'/R=X$ is open. You may find this in Chapter 1, § 8.3 of Bourbaki, General Topology, vol. 1. The type of equivalence relation $R$ coming from such gluings as considered always has this property. This is also in Bourbaki, somewhere in § 4. Anyway, these things are not difficult to check by hand.<|endoftext|> -TITLE: Is there a combinatorial analogue of Ricci flow? -QUESTION [22 upvotes]: The question of generalising circle packing to three dimensions was asked in 65677. There is a clear consensus that there is no obvious three dimensional version of circle packing. -However I have seen a comment that circle packing on surfaces and Ricci flow on surfaces are related. The circle packing here is an extension of circle packing to include intersection angles between the circles with a particular choice for these angles. My initial question is to ask for an explanation of this. -My real question should now be apparent. There is an extension of Ricci flow to three dimensions: so is there some version of circle packing in three dimensions which can be interpreted as a combinatorial version of Ricci flow? - -REPLY [13 votes]: Not yet mentioned is the interesting definition of Ricci curvature by -Yann Ollivier, a definition especially suited to discrete spaces, such as graphs. -His definition "can be used to define a notion of 'curvature at a given scale' for metric -spaces." For example, he shows how the discrete cube $\{ 0,1 \}^n$ -behaves like $\mathbb{S}^n$ in having constant positive curvature, -and possessing an analog of the Lévy "concentration of measure" -(the mass of $\mathbb{S}^n$ is concentrated about its equator). -His definition is used in the recent (April, 2011) paper by -Jürgen Jost and Shiping Liu: -"Ollivier's Ricci curvature, local clustering and curvature dimension inequalities on graphs." -Here are two primary sources: - -Y. Ollivier, Ricci Curvature of Markov Chains on Metric Spaces, J. Funct. Anal. 256 (2009), - No. 3, 810-864. -Y. Ollivier, A survey of Ricci curvature for metric spaces and Markov chains, in Probabilistic - approach to geometry, 343-381, Adv. Stud. Pure Math., 57, Math. Soc. Japan, Tokyo, 2010. - - -Update (7Feb13). Noticed this recent posting to the arXiv: - -Warner A. Miller, Jonathan R. McDonald, Paul M. Alsing, David Gu, Shing-Tung Yau, - "Simplicial Ricci Flow," - arXiv:1302.0804 [math.DG].<|endoftext|> -TITLE: How to calculate the exact differential structure of Brieskorn variety? -QUESTION [16 upvotes]: As Kervaire and Milnor mentioned, an $n$-dim exotic sphere $\Sigma$ which bounds a parallelizable manifold $M$ is totally classified by the signature $\sigma(M)$ modulo the order of $bP_{n+1}$. -Let $n=2m$ be an even integer. Brieskorn had discovered that the singularity of complex hypersurfaces $V_k$, the zero set of $x_0^2+\cdots+x_{n-2}^2+x_{n-1}^3+x_n^{6k-1}=0$ has close relationship with exotic spheres. More precisely, if $\varepsilon>0$ is sufficiently small, let $S_\varepsilon$ be the $(2n+1)$-sphere with center $0$ and radius $\varepsilon$, then $\Sigma_k=S_\varepsilon\cap V_k$ is an exotic sphere, and actually every exotic sphere of dimension $(4m-1)$ which bounds a parallelizable manifold can be obtained in this way. -I want to know which element $\Sigma_k$ represents in $bP_{2n}$. In other words, I want to calculate the signature of the Milnor fibre. Since Brieskorn's original paper was written in German, I couldn't read it. Instead, I've read the papar 'Singularity and Exotic Sphere' written by Hirzebruch. In this paper, Hirzebruch gave the answer: actually $\Sigma_k$ represents $k$th multiple of the generator of $bP_{2n}$. However, he refered the proof to Brieskorn's paper. -Does anyone know a proof? Please tell me, thanks. - -REPLY [3 votes]: An english discussion of Brieskorn exotic spheres can be found in these slides of Ranicki, there is also a collection of links concerning exotic spheres here. -I can give you the gist of Brieskorn's argument, I hope this may help: -First, setting $a=(a_1,\dots,a_n)$, an explicit parallelizable manifold is given by $M_a=\Xi_a\cap D^{2n}$, $\Xi_a=\{(z_1,\dots,z_n)\in\mathbb{C}^n\mid \sum_{i=1}^n z_i^{a_i}=1\}$ and as usual $D^{2n}=\{(z_1,\dots,z_n)\in\mathbb{C}^n\mid \sum_{i=1}^n|z_i|^2\leq 1\}$. The intersection form for $\Xi_a$ is computed in F. Pham: Formules de Picard-Lefschetz généralisées et ramification des integrales. BSMF 93 (1965), 333-367. -Then the steps of the proof are as follows: -1) there are automorphisms $\omega_k$ given by multiplying the $k$-th coordinate with $e^{2\pi i/a_k}$. these generate a finite automorphism group $\Omega_a\cong\prod_{i=1}^n\mathbb{Z}/(a_i)$, let $J_a=\mathbb{Z}[\Omega_a]$ be the group ring and $I_a$ the ideal generated by the elements $1+\omega_k+\cdots+ \omega_k^{a_k-1}$. Pham identified $H_{n-1}(\Xi_a,\mathbb{Z})\cong J_a/I_a$. -2) $H_{n-1}(\Xi_a,\mathbb{C})$ has a basis $v_j=\prod_{k=1}^n\sum_{r=0}^{a_k-1}e^{2\pi i j_k r/a_k}\omega_k^r$. In the basis $v_j+v_{a-j}$ and $i(v_j-v_{a-j})$ of $J_a/I_a\otimes\mathbb{R}$, the intersection form is diagonalized. The matrix of the intersection form can be found on p. 359 of Pham's paper, and implies $\langle v_j+v_{a-j},v_j+v_{a-j}\rangle=\langle i(v_j-v_{a-j}),i(v_j-v_{a-j})\rangle=2\langle v_j,v_{a-j}\rangle$ (and $0$ otherwise). -3) The final computation shows that $\langle v_j,v_{a-j}\rangle>0$ if and only if $0<\sum \frac{j_k}{a_k}<1\mod 2$, and $\langle v_j,v_{a-j}\rangle<0$ if and only if $-1<\sum\frac{j_k}{a_k}<0\mod 2$. -This implies the signature calculation, Satz 3 in Brieskorn's paper resp. the Theorem on p. 20 of Hirzebruch's paper "Singularities and exotic spheres". Actually, Brieskorn credits Hirzebruch for the result and proof.<|endoftext|> -TITLE: To what extent can the following zero-one laws be relaxed? -QUESTION [5 upvotes]: I am interested in what circumstances various zero-one laws in probability theory can be relaxed. In particular, independence is a very important factor in such laws. -1) Borel-Cantelli Lemma: Let $A_1, A_2, \cdots$ be a sequence of events. Then $\mathbb{P}(\limsup_{n \rightarrow \infty} A_n) = 0$ if $\displaystyle \sum_{n = 1}^\infty \mathbb{P}(A_n) < \infty$, and $\mathbb{P}(\limsup A_n) = 1$ if $A_1, A_2, \cdots$ are pairwise independent and $\displaystyle \sum_{n = 1}^\infty \mathbb{P}(A_n) = \infty$. -2) Kolmogorov's zero-one law: If $X_1, X_2, \cdots$ are a sequence of random variables, define $H_n = \sigma(X_{n+1}, X_{n+2}, \cdots)$ to be the smallest sigma algebra for which $X_{n+1}, X_{n+2}, \cdots$ are measureable. Then it is clear that $H_n \supset H_{n+1} \supset \cdots$ Let $H_\infty = \bigcap_{n} H_n$. Now suppose that $X_1, X_2, \cdots$ are independent. Then all events $A \in H_\infty$ satisfy $\mathbb{P}(A) = 0$ or $\mathbb{P}(A) = 1$. -I am particularly interested in cases where independence, which is a rather strong assumption and difficult to verify, can be replaced by estimates on various moments of the random variables, their correlation, etc. For example, the original statement of the Borel-Cantelli Lemma assumed that the sequence of events are independent, but this has since been weakned to pairwise independence. Any help would be greatly appreciated. - -REPLY [2 votes]: The zero-one law is true for extremal Gibbs states.<|endoftext|> -TITLE: "Approximating" $BGL(1)$ by projective spaces -QUESTION [9 upvotes]: Given a representation $V$ of a group $G$, we can think of $V$ as a vector bundle over the classifying stack $BG$, and we can define its index $\chi(BG; V)$ to be the dimension of the $G$-invariant part $V^G$ of the representation. -Now let $G = GL(1)$ (or $\mathbb{G}_m$ or $\mathbb{C}^\ast$ if you like). Then let $\phi_n : \mathbb{P}^n \to BG$ be the map corresponding to the bundle $\mathcal{O}(1)$ (or maybe we should take map corresponding to the bundle $\mathcal{O}(-1)$, I always mix it up). -Let $V$ be a representation of $G$. Then we can compute the index (i.e. sheaf cohomology Euler characteristic) of $\phi_n^\ast V$ over $\mathbb{P}^n$. We can show, just by doing the calculation, that for sufficiently large $n$ (how large depends on the representation $V$), this index is a polynomial in $n$. Then, plugging in $n=-1$ into this polynomial, one can show that the result is equal to $\chi(BG; V)$, defined as above. -So, in this fashion, we can recover the index over the stack $BGL(1)$ if we know the index over each $\mathbb{P}^n$ (for sufficiently large $n$). -This result seems to make sense, because at least in topology we have $B\mathbb{C}^\ast \simeq \mathbb{CP}^\infty$, and at least intuitively we can think of $\mathbb{CP}^n$ as "approximating" $\mathbb{CP}^\infty$. -But so far I don't really have a satisfactory explanation for this, other than "it follows by doing the computation" and "it seems to make intuitive sense" as I've explained above. I wonder if there is a better way to see this; does it follow by some deeper facts?; does it follow by some more general theorems? I'm sorry that my question is not very precise. -I would also be interested in any other theorems or results which relate finite dimensional projective spaces with $BGL(1)$. (There are already a few such results in the link in S. Carnahan's comment below.) - -REPLY [2 votes]: The paper arXiv:0808.2785 of Anderson, Griffeth, and Miller uses precisely this approximation technique to prove a certain positivity result in the torus equivariant k-theory of homogeneous spaces (see their prop. 3.1). AGR reference the paper -Daniel Edidin and William Graham, Riemann–Roch for equivariant Chow groups, Duke -Math. J. 102 (2000), no. 3, 567–594 -for a discussion of the use of this technique in defining equivariant Chow groups.<|endoftext|> -TITLE: What are "perfectoid spaces"? -QUESTION [137 upvotes]: This talk is about a theory of "perfectoid spaces", which "compares objects in characteristic p with objects in characteristic 0". What are those spaces, where can one read about them? -Edit: A bit more info can be found in Peter Scholze's seminar description and in Bhargav Bhatt's. -Edit: Peter Scholze posted yesterday this beautiful overview on the arxiv. -Edit: Peter Scholze posted today this new survey on the arxiv. - -REPLY [66 votes]: Here is a completely different kind of answer to this question. -A perfectoid space is a term of type PerfectoidSpace in the Lean theorem prover. -Here's a quote from the source code: -structure perfectoid_ring (R : Type) [Huber_ring R] extends Tate_ring R : Prop := -(complete : is_complete_hausdorff R) -(uniform : is_uniform R) -(ramified : ∃ ϖ : pseudo_uniformizer R, ϖ^p ∣ p in Rᵒ) -(Frobenius : surjective (Frob Rᵒ∕p)) - -/- -CLVRS ("complete locally valued ringed space") is a category -whose objects are topological spaces with a sheaf of complete topological rings -and an equivalence class of valuation on each stalk, whose support is the unique -maximal ideal of the stalk; in Wedhorn's notes this category is called . -A perfectoid space is an object of CLVRS which is locally isomorphic to Spa(A) with -A a perfectoid ring. Note however that CLVRS is a full subcategory of the category -`PreValuedRingedSpace` of topological spaces equipped with a presheaf of topological -rings and a valuation on each stalk, so the isomorphism can be checked in -PreValuedRingedSpace instead, which is what we do. --/ - -/-- Condition for an object of CLVRS to be perfectoid: every point should have an open -neighbourhood isomorphic to Spa(A) for some perfectoid ring A.-/ -def is_perfectoid (X : CLVRS) : Prop := -∀ x : X, ∃ (U : opens X) (A : Huber_pair) [perfectoid_ring A], - (x ∈ U) ∧ (Spa A ≊ U) - -/-- The category of perfectoid spaces.-/ -def PerfectoidSpace := {X : CLVRS // is_perfectoid X} - -The perfectoid space project home page is here, and the source code is on github. Information for mathematicians on how to read the code above is here, and an informal write-up is here. Johan Commelin, Patrick Massot and I are in the process of writing up a more formal document explaining what we had to do to construct this type. -This definition of a perfectoid space is completely mathematically unambiguous. All terms are defined precisely, and if you don't know what something means then you can right click on it and look at its definition directly, if you compile the software using the Lean theorem prover. Lean is a computer program which checks that definitions and theorems are formally complete and correct. The definitions and theorems have to be written in Lean's language of course. The definition above says that a perfectoid space is a topological space equipped with the appropriate extra structure (sheaf of complete rings, valuation on stalks etc) and which is covered by spectra of perfectoid rings, and a perfectoid ring is a complete uniform Tate ring with the usual perfectoid property. The main work is developing enough of the theory of completions of topological rings and valuations to put the appropriate structure on the adic spectrum of a Huber pair; in total this was over 10,000 lines of code. -Computer scientists have been developing and using formal proof verification tools for decades. I want to make the slightly contentious statement that 99.9% of the time they are working on statements about the kind of mathematical objects which we teach to undergraduates (groups, graphs, 2-spheres etc). There have been some huge successes in this area (computer-checked proofs of Kepler conjecture, odd order theorem, four colour theorem etc). My impression is that on the whole mathematicians are either unaware of, or not remotely interested in, this work, which for the most part concentrates on proving tricky theorems about "low-level" objects such as finite groups. I have now seen what these systems are capable of and in particular I believe that if mathematicians, rather than computer scientists, start working with formal theorem provers then we can build a database of definitions and statements of theorems which would be far far more interesting to the mathematical community than what the computer scientists have managed to do so far. Tom Hales is leading the Formal Abstracts project which aims to do precisely this. Furthermore, I now believe that theorem provers will inevitably play some role in the future of mathematical research. I am not entirely clear about what this role is yet, but one thing I am 100 percent convinced of is that AI will not be proving hard theorems that humans can't do, any time soon. However I am equally convinced that computers will soon be helping us to do research -- perhaps by giving us powerful targeted search through databases of theorems which humans have claimed to prove, and perhaps in the future pointing out gaps in the mathematical literature, as human mathematicians begin to formalise computer proofs more and more.<|endoftext|> -TITLE: High multiplicity eigenvalue implies symmetry? -QUESTION [17 upvotes]: It is well known that on any compact Riemannian symmetric space $X$, the eigenvalues of the Laplacian have very high multiplicity (comparable with the Weyl bound), and the resulting actions $\operatorname{Isom}(X)\to \operatorname{SO}(W_\lambda)$ for eigenspaces $W_\lambda$ give many representations of the Lie group $\operatorname{Isom}(X)$. -Suppose one has an unknown compact Riemannian manifold $X$ ($n=\dim X$), but where the eigenspaces of the Laplacian have large dimension (I don't have a precise definition of "large" here; the weakest definition would probably be something like $\dim W_\lambda>1$ for infinitely many $\lambda$. I'd be happy even with a much stronger assumption, say $\dim W_\lambda$ is at least $\epsilon$ times the Weyl bound $\operatorname{const}\cdot\lambda^{(n-1)/2}$ infinitely often). Can one conclude that $X$ is a symmetric space, or close to one in some sense? -EDIT: I would be interested in any result which takes as a hypothesis some assumption of large multiplicity in the Laplace spectrum, and whose conclusion is some sort of symmetry of the underlying manifold. - -REPLY [9 votes]: One situation where people have thought hard about this issue is the multiplicity of the spectrum of the Laplacian on the modular surface. This is a notoriously difficult problem. -Conjecturally, the (discrete) spectrum is simple, but as far as I know, the best known bounds on the multiplicity of a Maas cusp form of eigenvalue $\lambda$ are somewhere in the neighborhood of $O(\lambda^{1/2}/\log \lambda).$ -(Weyl's law in dimension $2$ says that the number of eigenvalues of size at most $N$ is $O(N)$). -See e.g. section 4 of the following survey of Peter Sarnak: http://www.math.princeton.edu/sarnak/baltimore.pdf -So I would say that some of the stronger versions of your conjecture are almost certainly out of reach.<|endoftext|> -TITLE: Which $\mathbf{Q}_p$-varieties come from $\mathbf{Q}$-varieties? -QUESTION [7 upvotes]: This is a very naive question. Fix a prime $p$ and consider the forgetful map from varieties over $\mathbf{Q}$ to varieties over $\mathbf{Q}_p$. Is there a conjectural "purely $p$-adic" characterization of the image of this map, perhaps in terms of some property of the etale $\pi_1$? -Feel free to restrict to whatever subcategories you like in order for this question to make (more?) sense. - -REPLY [8 votes]: I don't know what you think can happen. Take for example curves of genus at least two or some other class of varieties with a moduli space. Restrict yourself to the Zariski open where the varieties have no non-trivial automorphisms. Then the field of moduli is the field of definition, so a variety is defined over $\mathbb{Q}$ if and only if the corresponding point on moduli is. So you might as well be asking, given a variety defined over $\mathbb{Q}$, how do I distinguish its $\mathbb{Q}$-points among its $\mathbb{Q}_p$-points? Why would you do anything else other than looking at the coordinates? Now, if instead you look at adelic points, then there are other things you can do which may be in the direction you want, like descent obstructions and so on. Being a moduli space may help prove something but, ultimately, it's about points on a variety.<|endoftext|> -TITLE: Tensor product of regular ring (with some conditions) -QUESTION [5 upvotes]: Basically, my question is whether this answer is correct. Here is the point. Let $R$ be a ring, and let $A$ and $B$ be $R$-algebras. Suppose that $A$ is regular and $B \otimes_R B$ is regular too. Does it follow that $A \otimes_R B$ is regular? -What if we suppose that $R$ is regular and $B$ a smooth $R$-algebra? - -REPLY [2 votes]: I think the answer to your first question is "no" and to the second question is "yes". -Let $R = k[x]$ be a polynomial ring in one variable, $A$ the ring $k[y]$ with the map from $R$ to $A$ given by $x \mapsto y^2$. Let $B = k[x]/(x)$ as an $A$-algebra. Then $R,A,B$ are all regular, $B \otimes_R B = k$ is also regular, but $A \otimes_R B = k[y]/(y^2)$ is not regular. -For the second question, the regularity of $R$ is not necessary. If $B$ is smooth over $R$ then $A \otimes_R B$ is smooth over $A$ which is regular. Since regularity is preserved by smooth base change, $A \otimes_R B$ is also regular.<|endoftext|> -TITLE: A mass spring model for hair simulation -QUESTION [14 upvotes]: A strand of hair is represented by a set of particles connected by springs. -The velocity for a particular particle is calculated implicitly using the following formula: -$\boldsymbol{v}^{n+1/2}=\boldsymbol{v}^{n}+\frac{\Delta t}{2}\boldsymbol{a}(t^{n+1/2},\boldsymbol{x}^{n},\boldsymbol{v}^{n+1/2})$ -The force or acceleration ($\boldsymbol{a}$ in the above equation) produced by the spring between two adjacent particles is given by the following: -$\boldsymbol{F}^{n+1}=\frac{k}{l_{0}}\left((\boldsymbol{x}_{2}^{n}-\boldsymbol{x}_{1}^{n})^{\mathrm{T}}\hat{\boldsymbol{d}}^{n}-l_{0}\right)\hat{\boldsymbol{d}}^{n}+\Delta t\frac{k}{l_{0}}(\boldsymbol{v}_{2}^{n+1}-\boldsymbol{v}_{1}^{n+1})^{\mathrm{T}}\hat{\boldsymbol{d}}^{n}\hat{\boldsymbol{d}}^{n}$ -where $\hat{d}^{n}=(x_{2}^{n}-x_{1}^{n})/\left\Vert x_{2}^{n}-x_{1}^{n}\right\Vert$ -My question is, how do you numerically calculate $\boldsymbol{v}_{1}^{n+1}$ and $\boldsymbol{v}_{2}^{n+1}$ in order to calculate the force? -I've tried using Newton's Method but calculating derivative the spring force is just so complicated. -I'm attempting to implement the techniques found in this paper: link - -REPLY [2 votes]: Notice that the only unknown quantities in your force equation are the updated velocities $v_i^{n+1}$, and that force depends linearly on these. Everything else is either known from the last time step, e.g. $x_i^n$, or is a simulation parameter/material constant. -Therefore, after substituting for $\mathbf{a}$ in your first equation, you will end up with a set of linear equations in the unknowns $v_i^{n+1/2}$; in particular the system will be of the form -$$(\mathbf{I} - c\Delta t^2\ \mathbf{d}\mathbf{d}^T)\mathbf{v}^{n+1/2} = \mathbf{b}$$ -for some scalar $c$ and right-hand side $\mathbf{b}$. You can readily solve this system without Newton's method. In particular, for sufficiently small time steps $\Delta t$ the matrix is positive-definite, in which case I recommend using Conjugate Gradients to solve the system iteratively without ever even needing to form the dense matrix $\mathbf{d}\mathbf{d}^T.$<|endoftext|> -TITLE: An extension of Gaussian Isoperimetry -QUESTION [9 upvotes]: The Gaussian isoperimetric inequality (Tsirelson,Sudakov, Borell) states that among all sets of given Gaussian measure in the n-dimensional Euclidean space, half-spaces have the minimal Gaussian boundary measure. -Suppose we put an additional restriction on the set, that it should be symmetric about the origin. Then can we conclude that quarter-spaces (intuitively the first and third quadrant in 2-dimensions, say) have the minimal Gaussian boundary measure? - -REPLY [3 votes]: My guess is that the optimizer is actually a "strip"; i.e., a set of the form {$x : -t \leq x_1 \leq t$}. But I'm somewhat sure that the solution to this problem is not known. You might take a look at the discussion surrounding after Corollary 3.6 in this paper by Klartag and Regev: -http://eccc.hpi-web.de/report/2010/140/ -Barthe may also have some relevant papers.<|endoftext|> -TITLE: When matrices commute -QUESTION [16 upvotes]: Original title: when animals attack. -2016: The short version is that matrices that commute with a fixed matrix $A$ must all be polynomials in $A,$ if and only if the characteristic polynomial of $A$ and the minimal polynomial coincide. Put another way, if and only if each eigenvalue occurs in just one Jordan block. Worth recording the fact that the minimal polynomial of a square matrix does not change when the field of definition is extended. -Inspired by $$ $$ https://mathoverflow.net/questions/65738/when-matrix-multiplication-commutes $$ $$ and $$ $$ http://www.imdb.com/title/tt0293702/ $$ $$ -is it true that, when $$ A \in SL_n(\mathbf Z),$$ -then all integral matrices that commute with $A$ are an integral (or at least rational) polynomial in $A$? I dimly recall proving this for a specific 3 by 3 $A$ that was all 0's and 1's, so calculations were easy. The use of the unit determinant is that $A^{-1}$ is an integral polynomial in $A$ by Cayley-Hamilton. The degree of the polynomial need be no larger than $n-1,$ also by Cayley-Hamilton. -EDIT: as both David Speyer and Tommaso Centeleghe point out in comments below, the statement is true if all eigenvalues are distinct, probably false otherwise. People are smart. And quick. The point being to diagonalize $A$ over $\mathbb C$ and continue. -EDIT TOOO: it seems reasonable to conjecture that the full set of $A$ for which the statement is true is $ A \in SL_n(\mathbf Z)$ such that, should there be any eigenvalue(s) of multiplicity larger than one, all occurrences of that eigenvalue must fit into a single Jordan block. Richard would know. -With or without commutativity, I once made a "multiplicative" function out of -$$ f(x_0, x_1, \ldots, x_{n-1}) = \det (x_0 I + x_1 A + x_2 A^2 + \cdots + x_{n-1} A^{n-1}), $$ -amounting to a kind of fake norm form. The guy I asked about it laughed at me but said that's what I had. -I asked Manjul Bhargava about this: take the matrix $A,$ 3 by 3, to have rows -<0,1,0; 0,0,1; 1,1,1> which I think may actually have had determinant -1, never mind. Then the prime values I got from my fake norm form were all nonresidues mod 11 and all $x^2 + 11 y^2.$ I'm not sure about 2 itself. No proof but presumably a known sort of problem. - -REPLY [6 votes]: (1) Let $A$ be a commutative ring, and let $M$ and $N$ be $A$-modules. If the natural morphism from $A$ to $\text{End}_A(M\oplus N)$ is surjective, then the annihilators of $M$ and $N$ are comaximal. -Indeed, this comaximality is the condition for the projectors attached the given direct sum decomposition to be in the image. -Assume now that $A$ is a principal ideal domain. Let $(G,+)$ be the Grothendieck group of the category $C$ of finitely generated torsion $A$-modules, and let $(H,\cdot)$ be the group of (nonzero) fractional ideals of $A$. -There is a (clearly unique) morphism from $G$ to $H$ which maps $A/\mathfrak a$ to $\mathfrak a$. -It is easy to see that the fractional ideal attached to $M\in C$ is in fact integral. Call it the characteristic ideal of $M$. Moreover we have in view of (1): -(2) The natural morphism from $A$ to $\text{End}_A(M)$ is surjective, if and only if the characteristic ideal of $M$ coincides with the annihilator of $M$. -Assume now that $A=K[X]$, where $K$ is a field and $X$ an indeterminate, and that $V$ is a finite dimensional $K$-vector space equipped with an endomorphism $a$. Then the characteristic ideal of $V$ is generated by the characteristic polynomial of $a$, and (1) and (2) imply: - -The characteristic polynomial of $a$ coincides with its minimal polynomial if and only if any endomorphism of $V$ commuting with $a$ is in $K[a]$.<|endoftext|> -TITLE: Upper bounds on number of vertices of graphs whose complements has no induced cycles of certain lengths -QUESTION [5 upvotes]: Let $G$ be a finite, simple, undirected, connected graph. Suppose that $G$ has maximal degree $d$ and the complement $G^c$ has no induced cycles of lengths $i$, for $4 \leq i \leq l$. My question is: - -What are the best known upper bounds on the number of vertices $n(G)$ of $G$, if we fixed $d$ and $l$? - -For example, if $l=4$ then the condition on $G$ says that $G$ has induced matching number $1$. It is easy to see that $G$ must have diameter at most $3$, so $n(G)$ is of order $d^3$ at best. It seems to me that such bound can be improved, yet I do not see how. - -REPLY [4 votes]: Excluding induced matching of size $2$ appears to be the most restrictive condition and the bound is independent on $l$: - -(a) Let $G$ be a connected graph with maximum degree $d$ and no induced matching of size $2$. Then $|V(G)| \leq \lfloor\frac{d+2}{2} \rfloor \lceil \frac{d+2}{2}\rceil$. -(b) For every $d \geq 1$ there exists a connected graph $G$ with maximum degree $d$, every induced cycle in the complement of $G$ of length $3$, and $|V(G)| = \lfloor\frac{d+2}{2} \rfloor \lceil \frac{d+2}{2}\rceil.$ - -(Updated 5/27/11 to extend the proof to $l=4$.) -Proof: (a) Let $M$ be a maximum (non-induced) matching in $G$ chosen so that the sum of degrees of vertices of $M$ in $G$ is minimal. Let $M$ consist of $m$ edges, let $X:=V(M)$ and let $Y:=V(G)-X$. Note that: -(i) $Y$ is an independent set. (Vertices of $Y$ are pairwise non-adjacent as $M$ is maximal.) -(ii) If $v \in Y$ is adjacent to $u \in X$ and $uw \in M$ is the matching edge incident to $u$ then $\deg(v) \geq \deg(w)$, as otherwise replacing $uw$ by $uv$ in $M$ decreases the sum of degrees of vertices in $M$. -(iii) For every $uw \in M$ at least $m-1$ edges incident with $u$ or $w$ have the other end in $X$, not counting $uw$. (There must be an edge between $uw$ and any other edge in $M$, as otherwise these two edges will form an induced matching.) -Let $n:=|V(G)|$. We use (i),(ii) and (iii) in a counting argument, which is described in terms of discharging as follows. Let every vertex in $Y$ start with a charge $1$. Then the total charge is $|Y|=n-2m$. In a discharging step let every vertex $v \in Y$ distribute its charge uniformly among its neighbors, which are all in $X$ by (i). (Every $v \in Y$ sends a charge $1/\deg(v)$ to each of its neighbors.) -The ends of some edge $uw \in M$ must as a result receive total charge at least $(n-2m)/m$. Let $d_1 :=\deg(u)$, $d_2 :=\deg(w)$ and suppose that $d_1 \geq d_2$, without loss of generality. By (ii) the charge that $u$ and $w$ receive from any single vertex in $Y$ is no greater than $1/d_2$. Finally, by (iii) there are at most $d_1+d_2 - 2 - (m-1)$ neighbors of $u$ and $w$ in $Y$. We get -$$ \frac{n-2m}{m} \leq \frac{1}{d_2}\left( d_1+d_2 -m -1\right).$$ -It is easy to see that the right side is maximized when $d_1=d$ and $d_2=1$. With these choices of degrees we have -$$ n \leq 2m + (d-m)m=(d+2-m)m \leq \lfloor\frac{d+2}{2} \rfloor \lceil \frac{d+2}{2}\rceil.$$ -(b) One can extract an example achieving the bound from the above argument. Let $G$ consist of a set of $\lfloor \frac{d+2}{2} \rfloor$ pairwise adjacent vertices, each of which is joined to $\lceil \frac{d}{2} \rceil$ additional degree one vertices. Then the complement of $G$ consists of a clique and $\lfloor \frac{d+2}{2} \rfloor$ pairwise non-adjacent vertices only having neighbors in this clique. From this description it is not hard to see that all induced cycles in the complement of $G$ are of length $3$.<|endoftext|> -TITLE: Optimal reference for tensor product of symmetric bilinear forms? -QUESTION [7 upvotes]: This is just a reference request on a relatively elementary level (for which I apologize in advance), but every time I bump into this question I suspect I'm missing the "correct" conceptual setting. In the simplest case, one is given two vector spaces $V, W$ over a field of characteristic 0, each endowed with a symmetric bilinear form. Then the tensor product $V \otimes W$ inherits an obvious symmetric bilinear form. A natural result is that nondegeneracy of the given forms implies nondegeneracy of the new form, though the proof seems to require somewhat messy manipulation of bases and indices. Even if the vector spaces are infinite dimensional, the same principle seems valid. Then there is the possibility of working over a field of prime characteristic, as well as passing to free modules over commutative rings, etc. - -Where in the textbook literature can one find the most definitive treatment of nondegeneracy of symmetric bilinear forms on tensor products? (Preferably with few indices to keep track of.) - -REPLY [2 votes]: A rather satisfactory treatment (at least in my opinion) can be found in Greub's book "Multilinear Algebra". -In the copy I have (the 1967 edition) the nondegeneracy of bilinear forms on tensor products is considered in Chapter I, Section 7, Subsection 1.22 (p. 31). -The arguments involved are index-free and work in any characteristic.<|endoftext|> -TITLE: Are these two definitions of nef-ness equivalent for Moishezon manifolds? -QUESTION [7 upvotes]: Recently, I have been learning about nef line bundles. I know that when $X$ is projective or Moishezon, a line bundle $L$ over $X$ is said to be nef iff $$L.C=\int_{C}c_{1}(L)\ge 0$$ for every curve $C$ in $X$. -Demailly gave a definition of nefness that works on an arbitrary compact complex manifold, i.e., a line bundle $L$ over $X$ is said to be nef if for every $\varepsilon >0$ there exists a smooth hermitian metric $h_{\varepsilon}$ on $L$ such that its curvature $\Theta_{h_{\varepsilon}}(L)\ge -\varepsilon\omega$. For projective manifolds, Demailly's definition coincides with the above one given by integration (this is an easy consequence of Seshadri's ampleness criterion). -Question: Is this equivalence also true for Moishezon manifolds? -I don't know of any counterexamples. If it is not true, could someone give me a counterexample? - -REPLY [4 votes]: Yes, the equivalence is true (the second notion used to be called "metric nef" by some). This was an open problem for quite some time until it was solved in -M. Paun "Sur l'effectivité numérique des images inverses de fibrés en droites" Math. Ann. 310 (1998), no. 3, 411–421, see the Corollaire on page 412.<|endoftext|> -TITLE: product of rings -QUESTION [7 upvotes]: I feel a need to apologies for this question, since it seems to be to basic to be asked. -in this question I am primarily concerned with commutative rings and therefore all rings here are assumed to be commutative (and unital of course), though the non-commutative analogue is also interesting. -a ring $R$ is called connected if there are no two non-zero rings $R_1$ and $R_2$ such that $R\cong R_1\times R_2$ (the terminology comes from the fact that $R$ is connected precisely when $Spec R$ is connected as a topological space). the question is about the possibility to represent a general ring $R$ as a (not necessarily finite) product of (non-zero) connected rings. two questions are in place -1) given a ring $R$, does such representation always exist? -2) if it exists, is it unique (in an appropriate sense)? -I think I was able to prove it for noetherian rings (for which such a product will always be finite) by showing the existence of "minimal" non-zero idempotents and that there are finitely many of them, in a rather technical but very routine and straightforward way . this already makes me feel a bit uncomfortable, since in atiyah-macdonald in the proof of the structure theorem for artin rings (p.90), the uniqueness part is proved by a very specialized argument and using "heavy guns" like primary decomposition. since artin rings are notherian and local rings are connected (and this was established in the book before p.90) it would seem more natural to deduce it from the more general and interesting by itself statement with an easy and elementary proof. -so, to conclude, I would like know if I am right that the answer to both questions is positive for noetherian rings and whether it can fail for general rings. - -REPLY [3 votes]: Consider -$$ R = \{a\in\prod_{k=1}^\infty\mathbb{Z} : - \exists n \;\forall j,k>n \quad a_j=a_k\} = - c(\mathbb{Z}) + \bigoplus_{k=1}^\infty \mathbb{Z} -$$ -(where $c:\mathbb{Z}\to R$ is the inclusion of constant sequences). -This is countable, and so cannot split as an infinite product of nontrivial rings. Moreover, -for each $a\in R$ there exist integers $n,m>0$ such that $\prod_{k=-n}^n(ma-k)=0$. This is another reason why $R$ cannot split as an infinite product of copies of $\mathbb{Z}$, -(This is more or less the same as Qiaochu Yuan's comment to Mark Meilstrup's answer.)<|endoftext|> -TITLE: What's difference between 'functional' and 'function'? -QUESTION [6 upvotes]: Hi, I want to know difference that between 'functional' and 'function'. -Of course, in Wikipedia, http://en.wikipedia.org/wiki/Functional_(mathematics), there is many texts. -But what's the simple answer for this question? ;-) - -REPLY [3 votes]: No difference really, just convention. -A functional normally acts on other functions, whereas a function normally acts on some underlying vector space or field. But this is not always true.<|endoftext|> -TITLE: Reference for functors in Kadeishvili's C_\infty paper -QUESTION [12 upvotes]: In his paper Cohomology $C_\infty$-algebra and rational homotopy type, Tornike Kadeishvili describes how the rational cohomology of a simply-connected space carries the structure of a $C_\infty$-algebra, and how this structure determines the rational homotopy type of the space. (The result has been mentioned before on MathOverflow, eg in these answers.) -I'm trying to follow the proofs, which are somewhat light on details. In particular, they rely on an adjoint pair of functors -$$\Gamma\colon CDGAlg \rightleftarrows DGLieCoalg \colon \mathcal{A}$$ introduced in section 4.3, between the categories of commutative differential graded algebras, and differential graded Lie coalgebras. The functor $\Gamma$ is given as the composition $$\Gamma\colon CDGAlg\stackrel{B}{\to}DGBialg\stackrel{Q}{\to}DGLieCoalg$$ where $B$ is a bar construction and $Q$ is the functor of indecomposables. The adjoint functor $\mathcal{A}$ is dual to the Chevalley-Eilenberg functor. There is a standard weak equivalence $\mathcal{A}\Gamma(A)\to A$. -I am struggling to find any reference to these functors in the papers cited in the bibliography. For the proof of Theorem 9.1 we seem to need that the functor $\mathcal{A}\Gamma$ applied to the weak equivalences of $C_\infty$-algebras $$\lbrace f_i\rbrace\colon (H(A),\lbrace m_i\rbrace )\to (A,\lbrace d, \mu, 0,\ldots\rbrace)$$ yields a weak equivalence $\mathcal{A}\Gamma(H(A))\to\mathcal{A}\Gamma(A)$ in $CDGAlg$, but this is not stated anywhere and is not obvious to me. - -Is the above true, and can anyone explain why? -Where can I read more about the functors $\Gamma$ and $\mathcal{A}$ and their properties, in particular in the setting of $C_\infty$-algebras? - -REPLY [5 votes]: Ben Walter and I make the functors $\Gamma$ and $A$ more explicit, by using an explicit model for the cofree Lie Coalgebra functor, in this paper. We do not discuss the application to $\infty$-algebras as Bruno does in much greater generality. (We were interested in using explicit models to be able to compute, in particular in the long exact sequence of a fibration as we do in a sequel to this paper on Hopf invariants.)<|endoftext|> -TITLE: What are the models of Peano Arithmetic plus the negation of the corresponding Gödel sentence like? -QUESTION [6 upvotes]: Since the Godel sentence for PA (G henceforth) is independent of PA, if PA is consistent, so is PA plus not-G. Thus, since PA is a first-order theory, if PA is consistent, PA plus not-G has some models. My question concerns what are these models like. I've read a couple of times that these are 'non-standard models' but I've got the following query. Presumably PA is part of the theory of arithmetic -- call it 'Teo(N)' -- (the whole set of sentences of the language of PA that are true in the standard model) and so is G. Using compactness and downward LS, we show that Teo(N) has models that are non-standard (models containing elements that cannot be reached from 0 by a finte number of applications of the successor function; in fact, these models contain denumerably many galaxies of such elements). Call these models 'NSM1'. If G is part of Theo(N), G is true in NSM1. Thus, if PA plus not-G has models, these must be distinct (non-isomorphic) to NSM1. So, it seems, there must be something specific about these models, not just that they're non-standard. What is it (or is my reasoning flawed somewhere)? -Thanks! - -REPLY [14 votes]: Here is the most "tangible" distinctive feature of models of $PA$ that satisfy the negation of Gödel's true but unprovable sentence. -Let $\phi$ be a true $\Pi^0_1$ arithmetical sentence that is not provable from $PA$, e.g., $\phi$ can be chosen as the sentence expressing "I am unprovable from $PA$" [as in the first incompleteness theorem], or the sentence "$PA$ is consistent" [as in the second incompleteness theorem]. -Thanks to a remarkable theorem of Matiyasevich-Robinson-Davis-Putnam, known as the MRDP theorem [see here], there is a diophantine equation $D_{\phi}$ that has no solutions in $\Bbb{N}$, but has the property that for any model $M$ of $PA$, $D_{\phi}$ has a solution in $M$ iff $M$ satisfies the negation of $\phi$. -[note: $D_{\phi}$ is of the form $E=0$, where $E$ is a polynomial in several variables that is allowed to have negative coefficients, but $D_{\phi}$ can be re-expressed as an equation of the form $P=Q$, where $P$ and $Q$ are polynomials [of several variables] with coefficients in $\Bbb{N}$; hence it makes perfectly good sense to talk about $D_{\phi}$ having a solution in $M$]. -Let me close by recommending two good sources for the study of nonstandard models of $PA$. -Richard Kaye, Models of Peano arithmetic. Oxford Logic Guides, 15. Oxford Science Publications. The Clarendon Press, Oxford University Press, New York, 1991. -Roman Kossak and James H. Schmerl, The structure of models of Peano arithmetic, Oxford Logic Guides, 50. Oxford Science Publications. The Clarendon Press, Oxford University Press, Oxford, 2006.<|endoftext|> -TITLE: Tamagawa numbers of crystalline Galois representations -QUESTION [10 upvotes]: This is a followup to this question. -Let $p \ge 3$ be prime, and let $V$ be a crystalline 2-dimensional representation of $G_{\mathbb{Q}_p}$ and $T$ a lattice in $V$. I'm going to assume just about every niceness condition on $V$ that I can think of: - -$V$ is irreducible; - -$\operatorname{Fil}^0 \mathbb{D}_{\mathrm{cris}}(V)$ is 1-dimensional (so one Hodge-Tate weight of $V$ is $\le 0$ and the other is $> 0$) - -none of the eigenvalues of Frobenius on $\mathbb{D}_\mathrm{cris}(V)$ are integral powers of $p$; - -the Hodge filtration of $V$ has length $< (p-1)$, so $T$ corresponds to a strongly divisible $\mathbb{Z}_p$-lattice $\mathbb{D}(T)$ in $\mathbb{D}_{\mathrm{cris}}(V)$ via Fontaine-Laffaille theory. - - -Let $T$ be a lattice in $V$, and let $\omega$ be a $\mathbb{Z}_p$-basis of the "tangent space" $t_T = \mathbb{D}(T) / \operatorname{Fil}^0 \mathbb{D}(T)$. The Tamagawa number of $T$ over $K_n = \mathbb{Q}_p(\mu_{p^n})$ is given by -$$ \operatorname{Tam}^0_{K_n, \omega}(T) = \frac{[H^1_f(K_n, T) : \exp(\mathcal{O}_{K_n} \omega)]}{[\mathbb{D}(T) : (1- \varphi) \mathbb{D}(T)]}$$ -where $[ A : B ]$ is a generalised index (so $[ \mathbb{Z}_p : \tfrac{1}{p} \mathbb{Z}_p] = \tfrac{1}{p}$ etc). - -Question: Is it true that under the above hypotheses this Tamagawa number is always 1? - -I know this is true for all $n$ if $V$ corresponds to an elliptic curve (because the Tamagawa number has an alternative definition in terms of the index of the nonsingular points in the special fibre of the Neron model) and, if I've understood correctly, for $n = 0$ it is true for any $V$ satisfying the hypotheses above (by a theorem of Bloch and Kato). - -REPLY [8 votes]: I think that for $K_n=\mathbf{Q}_p$ what you're looking for is in my (unpublished) paper -http://perso.ens-lyon.fr/laurent.berger/autrestextes/tamag0919.pdf -see proposition II.2 for instance. -This paper is unpublished because it was rewritten and massively expanded with/by Denis Benois. There is some stuff in the paper with Benois about going up the cyclotomic tower which may help. I'm sorry I don't have time to look in Benois-Berger to see if there's an answer to your full question.<|endoftext|> -TITLE: Homomorphism from $\hat{\mathbb{Z}}$ to $\mathbb{Z}$ -QUESTION [18 upvotes]: I expect this question has a very simple answer. -We all know from primary school that there are no non-trivial continuous homomorphisms from $\hat{\mathbb{Z}}$ to $\mathbb{Z}$. What if we forget continuity: can anybody give an explicit example of a homomorphism? -Note that $\hat{\mathbb{Z}}$ is torsion-free, and not divisible (since it's isomorphic to $\prod_p \mathbb{Z}_p$ and $\mathbb{Z}_p$ is not divisible by $p$). There is the canonical injection $\mathbb{Z} \to \hat{\mathbb{Z}}$; is there some abstract reason why it ought to have a left inverse, and if so can we write it down? - -REPLY [46 votes]: Let $\phi:\hat{\mathbb{Z}}\to\mathbb{Z}$ be a nontrivial homomorphism. As every nontrivial subgroup of $\mathbb{Z}$ is isomorphic to $\mathbb{Z}$, we may suppose that $\phi$ is surjective, with kernel $K$ say. Now $\phi$ induces a surjective homomorphism $\phi_n:\hat{\mathbb{Z}}/n\hat{\mathbb{Z}}\to\mathbb{Z}/n\mathbb{Z}$, but it is standard that $\hat{\mathbb{Z}}/n\hat{\mathbb{Z}}$ has order $n$, so $\phi_n$ must be an isomorphism. This implies that $K\leq n\hat{\mathbb{Z}}$ for all $n$, but $\bigcap_n n\hat{\mathbb{Z}}=0$, so $\phi$ is injective, which is clearly impossible. - -REPLY [4 votes]: Let $\phi$ be such a homomorphism, on additive groups $\hat Z\rightarrow Z$. Write $(\vec x,\vec y)\in\hat Z$ for the element that is $x$ on primes that are 1 mod 3, and $y$ on primes that are 2 mod 3. -Then $\phi(\vec x,\vec 0)=0$ for all $x\in Z$, for $(\vec x,\vec 0)$ is $l$-divisible for any prime $l$ that is 2 mod 3. The symmetrical argument claims $\phi(\vec 0,\vec y)=0$ too. -Without loss of generality, we can assume that a preimage of $1$ is given by $(\vec 1,\vec 1)$. -Next, applying the group law and setting $\alpha=\phi(\vec 1,\vec{-1})$, we derive the system $$1+\alpha=\phi(\vec 1,\vec 1)+\phi(\vec 1,\vec{-1})=\phi(\vec 2,\vec 0)=0$$ -$$1-\alpha=\phi(\vec 1,\vec 1)-\phi(\vec 1,\vec{-1})=\phi(\vec 0,\vec 2)=0$$ -This is impossible, so $\phi$ does not exist.<|endoftext|> -TITLE: A question on the integral of Hilbert valued functions -QUESTION [20 upvotes]: This questions stems from an attempt to recast in a form suitable for teaching some standard computations which are usually proved by handwaving, without much care about the details. My hope is that some expert in this area is hanging around and can give me an immediate answer off the top of their head... -A function $f:I\to H$ from an interval of the reals into a Hilbert space is said to be scalarly integrable if $(f(t),x)$ is in $L^1(I)$ for all $x\in H$. This is sufficient to guarantee that for every measurable subset $E$ of $I$ one can find an $x_E\in H$ with the property $(x_E,x)=\int_E(f(t),x)dt$ for all $x\in H$. The vector $x_E$ is called the Pettis integral of $f$ over $E$. Now if $T:H\to K$ is a bounded map from $H$ to a second Hilbert space $K$, also $Tf$ is Pettis integrable and $T\int_E f=\int_E Tf$. This is quite easy to prove. I need the same property for any closed unbounded operator $T$. This kind of property is well known for stronger notions of integral (for the Bochner integral this is due to Hille), but I guess it is false in such generality for the Pettis integral. -Now my question: assume I know that $f:I\to H$ is scalarly integrable, thus has a Pettis integral, moreover it takes values into the domain $D(T)$, and the integrals $\int_E f$ are all contained in $D(T)$. Can I conclude that $Tf$ has a Pettis integral and $T\int_I f=\int_I Tf$? -EDIT: after an excellent series of posts, it seems that the question is almost settled, in the following sense. Assume in addition that $D(T)$ is dense and that $H$ is separable (or the interval $I$ is replaced by a discrete measure space). Then the result is true. This covers all the relevant applications, where the Hilbert spaces are some $L^2(R^n)$ and $T$ is a closed, densely defined operator like a differential operator possibly with variable coefficients. -I think Bill, Gerald and fedja should wrap up their arguments in a short paper, this can be useful to other people, and the mathematics is not trivial. On the other hand, I do not know which posts should be checked as answer; the final word is Gerald's but Bill's key argument on finite sums is quite cool. Tell me what to do :) - -REPLY [4 votes]: This is some kind of wrap-up of what has already been said. I quite carefully checked all the details involved here, and concluded that the following holds. -Theorem A. Let $\mu$ be a $\sigma$−finite measure on $\Omega$, and let $T:H\supseteq D\to H_1$ be a closed linear operator of real or complex Hilbert spaces with $H$ separable. Let $f:\Omega\to D$ be a function such that $(f;\mu,H)$ is Pettis with $^{\rm Pettis}H$−$\int_Af{\rm d}\mu\in D$ for all $A\in {\rm dom\ }\mu$. Then also $(T\circ f;\mu,H_1)$ is Pettis with $^{\rm Pettis}H_1$−$\int_\Omega T\circ f{\rm d}\mu=T\big({}^{\rm Pettis}H$−$\int_\Omega f{\rm d}\mu\big)$. -The "measure−vector" map $(f;\mu,H)=(f,(\mu,H))$ being Pettis means that for every $A\in{\rm dom\ }\mu$ there is $x\in H$ such that for all continuous linear $u:H\to\boldsymbol K$ the scalar map $(u\circ f;\mu,\boldsymbol K)$ is integrable with $u(x)=\int_Au\circ f{\rm d}\mu$. Here $\boldsymbol K$ is either the standard real or complex topological field and $H$ might be any Hausdorff locally convex space over $\boldsymbol K$. The $x$ when existing is unique, and I then write $x={}^{\rm Pettis}H$−$\int_Af{\rm d}\mu$. -Note that scalar integrability of $(f;\mu,H)$ is a sufficient condition for its being Pettis when $H$ is a reflexive Banach space, in particular, when it is Hilbert. For aesthetical reasons, I did not include this in the formulation of Theorem A above. -To handle the $\sigma$-finite case, one proceeds similarly as in Gerald Edgar's answer but in place of the sets $J_k$ takes the sets $\Omega_{i_k}\cap J_{j_k}$ where $k\mapsto(i_k,j_k)$ is some bijection $\mathbb N\to\mathbb N\times\mathbb N$ , and $\langle\Omega_i:i\in\mathbb N\rangle$ is a measurable partition of $\Omega$ into sets of finite measure. -If I were to write a decent proof of Theorem A, I would divide the proof into a sequence of lemmas e.g. as follows. -First reformulate the problem as follows. Let $H_2$ be the Hilbert space with $D$ as its underlying set and structured so that $[{\rm id},T]:x\mapsto(x,Tx)$ becomes linear and isometric $H_2\to H\times H_1$. So the inner product for $H_2$ is $\varphi_2=\varphi+\varphi_1\circ[T,T]$ when $\varphi,\varphi_1$ are the ones for $H,H_1$, respectively. Let $S$ be the subset of the dual of $H_2$ formed be the linear forms of the form $x\mapsto\varphi(x,y)$ for some $y\in H$. Putting $ E = (H_2)_\sigma(S)$, i.e. the vector space $H_2$ equipped with the weak topology giving the dual $S$, we then assume that $(f;\mu,E)$ is Pettis, and ask whether also $(f;\mu,H_2)$ is such. If it is, it trivially follows that so is $(T\circ f;\mu,H_1)$ since $T:H_2\to H_1$ is a continous linear map. -Lemma 1. In the above setting $S$ is dense in $(H_2)'_\beta$ . -Lemma 2. In the above setting $(f;\mu,H_2)$ is scalarly measurable. -Lemma 3. In the above setting $H_2$ is separable. -Lemma 4. Let $E$ be a Hausdorff locally convex space and let $K$ be compact in $E_\sigma$. Let the non-empty family $\boldsymbol x:I\to E$ be such that $\sum_B\boldsymbol x\in K$ for all non-empty finite $B\subseteq I$. Then $\boldsymbol x$ is summable in $E_\sigma$ with $E_\sigma$−$\sum\boldsymbol x\in K$. -Lemma 5. Let $F$ be a reflexive Banach space with $S$ a dense vector subspace in $F'_\beta$. Let $\boldsymbol x=\langle x_i:i\in\mathbb N_0\rangle$ be a sequence in $F$ such that $\sum_{i\in\mathbb N_0}|u(x_i)|$ is finite, for all $u\in S$. Assume that for every $B\subseteq\mathbb N_0$ there is $s_B \in F$ with $u(s_B)=\sum_{i\in B}u(x_i)$ for all $u\in S$. Then $\mathrm{rng \,} \boldsymbol x$ is a bounded set in $F$ and $\boldsymbol x$ is summable in $F_\sigma$. -Lemma 5 is essentially what is contained in Bill Johnson's answer, and Lemma 4 is the WUC−matter needed there.<|endoftext|> -TITLE: Are there 'analytic' $p$-adic modular forms. -QUESTION [15 upvotes]: The most elementary way to define $p$-adic modular forms is via limits of classical modular forms. -More precisely $f \in \mathbb{Z}_p[[q]]$ is called a $p$-adic modular form -if there are modular forms $f_n$ with integral coefficients such that -$f \equiv f_n \mod p^n$ (as $q$-expansions). Note it does not really make sense to attribute 'a weight' to $f$ since -the $f_n$ are allowed to have different (increasing weights). This is the older definition by Serre. -I know (but I do not understand) a newer definition by Katz, which has a more geometric flavor. -See here math.arizona.edu/~swc/notes/files/01BuzzardL2.pdf. -So we have an approach using the $q$-expansion and we have an approach to $p$-adic modular forms -using geometric ideas. -My question now is, whether there is also a developed theory on analytic p-adic modular forms? -Some ideas what this might mean. For example we could consider the Eisenstein series -$$ E_4(\tau)=\sum_{n,m \in \mathbb{Z}} \frac{1}{(n\tau+m)^4} $$ -as function of $\tau$ not being an element of the upper half-plane but of some subset of $\mathbb{C}_p$. -Does this sum even converge somewhere in $\mathbb{C}_p$. And is it (up to a constant) a classical or Katz $p$-adic modular -form? Does it even equal (mod p) the Eisenstein series $E_4$? -A similar question could be posed for the infinite product usually defining the $\Delta$-function -$$ q\prod_{n=1}^\infty (1-q^n)^{24}. $$ -Does this converge somewhere when $q$ is in some subset of $\mathbb{C}_p$. Is it a $p$-adic modular form? -If there is no such theory? Why not? Is it not interesting? - -REPLY [2 votes]: Since I'm not an expert, this will be just a minor note in addition to the excellent answers already given. -The example modular forms $E_4$ and $\Delta$ you've written down are defined over the integers (after suitable normalization). That is, they eat elliptic curves with nowhere-vanishing differentials over any base, and produce numbers (i.e., functions on the base) in a suitably canonical way. In particular, they are already "classical" $p$-adic modular forms by straightforward base change, and they are analytic forms, because they can be canonically extended to rigid-analytic input data. -The problem with the definition of $E_4$ that you wrote is that it uses the additive uniformization of elliptic curves ($\mathbb{C}/\Lambda \overset{\simeq}{\longrightarrow} E$) in an essential way, and elliptic curves only have an additive uniformization over the complex numbers. However, the $q$-expansion $E_4(q) = 1 + 240 \sum_{n \geq 1} \sigma_3(n) q^n$ makes sense $p$-adic analytically, since the multiplicative uniformization $\mathbb{G}_m/q^{\mathbb{Z}} \overset{\simeq}{\longrightarrow} E$ works in both analytic worlds. -Random points: - -Katz's definition of modular form (see his paper in Springer Lecture Notes 350, scanned on his web page) can be straightforwardly modified to take analytic inputs (real, complex, or $p$-adic). -When people say "$p$-adic modular form", they might mean only non-classical forms, i.e., those that aren't defined in the "too supersingular" locus. The overconvergent forms, namely those that extend from the ordinary locus to the "not too supersingular" locus, are necessarily analytic in nature. -These objects are interesting in part because they can be used to construct interesting Galois representations through some magic I don't understand. I should say something clever using phrases like "$p$-adic Langlands" and "eigenvariety" here.<|endoftext|> -TITLE: Series whose convergence is not known -QUESTION [57 upvotes]: For most of the mathematical concepts I learn, it has more or less always been possible to find (at least google and find) unsolved problems pertaining to that specific concept. Keeping a bag of unsolved problems on most topics I know has been to my benefit in that it reaffirms me that mathematics is a thriving subject. -Coming to point, I am unable to find an elementary series of the kind we know on real analysis courses whose convergence is an unsolved problem. Please, share if you have any. -Thanks. - -REPLY [25 votes]: Convergence of $\sum_{k=1}^\infty (-1)^k \frac{k}{p_k}$ is unknown, where $p_k$ is k-th prime (Guy 1994, p. 203; Erdős 1998; Finch 2003, according to Eric Weisstein's MathWorld).<|endoftext|> -TITLE: Crystalline cohomology via the syntomic site -QUESTION [19 upvotes]: Hello, -Let $k$ be a field of characteristic $p > 0$, and let $Y$ be a $k$-scheme. Consider the -sites $Y_{syn}$ and $(Y/W_n)_{cris}$ (where $W_n$ are the Witt vectors of $k$ of length $n$), of $Y$ with its syntomic topology and its crystalline topology. Then the assignment $\mathcal O_{cris}:Z\mapsto H^0_{cris}(Y/W_n)$ is a sheaf on $Y_{syn}$. It is a fact that -$H^*_{syn}(Y,\mathcal O_{cris})$ is canonically isomorphic to $H^{*}_{cris}(Y/W_n)$, but I don't see how to prove it. -So my question is: How does one prove this fact? -Thanks! - -REPLY [24 votes]: A sketch of the proof is as follows: -Consider the site $Y_{syn-cris}$ where the objects are the same as in $Y_{cris}$ but the -covering families are surjective syntomic families. Then there are maps of topoi: -$\alpha : Sh(Y_{syn-cris})\to Sh(Y_{syn})$ and $\beta : Sh(Y_{syn-cris})\to Sh((Y/W_n)_ {cris})$, -defined by $\beta_*(F)(U,T) = F(U, T)$ and $\alpha_{*}(F)(U) = H^0_{syn-cris}(U/W_n,F)$. -Lemma. $R^i\beta_*\mathcal O_{Y/W_n} = R^i\alpha_*\mathcal O_{Y/W_n} = 0$ for $i > 0$. -This implies the result in your question by a standard application of the Leray spectral -sequence. -As for the lemma: To prove that $R^i\beta_*\mathcal O_{Y/W_n} = 0$ for $i > 0$, it boils -down to checking that if $U$ is an affine in $(Y/W_n)_ {cris}$ then $H^i_{syn}(U,\mathcal O_U) = 0$ -for $i > 0$, and this follows the theory of the Cech complex (it is acyclic because of faithful flatness of the syntomic cover). -To prove that $R^i\alpha_*\mathcal O_{Y/W_n} = 0$, we have to check that if $U$ is an open of -$Y_{syn}$, and $s\in H^i_{cris}(U/W_n)$ (strictly speaking we have to do the computation with syntomic-cristalline cohomology, but by the previous part, the cohomology groups coincide with the crystalline cohomology groups) then there exists a syntomic cover $U_i\to U$ such -that $s\mid U_i = 0\in H^i_{cris}(U_i/W_n)$. Now recall that we can compute this cohomology groups as the hypercohomology groups of the de Rham complex of the divided power envelope of some embedding into a smooth scheme. That means, after shrinking, we can represent $s$ as an $i$-form. -We need to find a syntomic cover such that when we restrict $s$ to this cover, it vanishes. To do this, note that $A[T]\to A[T^{p^{-n}}]$ is a syntomic cover that has the property that the image of -$dT$ is zero.<|endoftext|> -TITLE: Picard number and torsion of Neron-Severi group of abelian varieties over a number field -QUESTION [8 upvotes]: Let $A$ be an abelian variety over a number field $k$ and let $NS_A$ denote its Neron-Severi scheme. Then the group of $k$-rational points of $NS_A$ is a finitely generated abelian group, i.e. $NS_A(k)=H^0(G_k,NS_A(\bar k)) = \mathbb{Z}^\rho \oplus Torsion$. I'd like to know: -What does a torsion element look like? Can it be ample? Is there an easy example, maybe of an elliptic curve, where the torsion of $NS_A(k)$ is non-trivial? -Is the rank $\rho$ of $NS_A(k)$, i.e. the Picard number, bounded by 1 and the square of the dimension of $A$? -Is the group of $\bar k$-rational points of $NS_A$ also a finitely generated abelian group? - -REPLY [6 votes]: Regarding the bounds for the Picard number $\rho$, the lower bound $\rho \geq 1$ comes from the fact that there exist divisors which are not algebraically equivalent to 0 (alternatively, one can argue that $\operatorname{id}_A$ is a symmetric endomorphism of $A$). The upper bound $\rho \leq (\dim A)^2$ (which holds only in characteristic 0) is Exercise 2.6(5) in Birkenhake-Lange's book Complex abelian varieties. -It is possible to show further that if $A$ is an abelian variety over $\mathbf{C}$ of dimension $\geq 2$, then $\rho = (\dim A)^2$ if and only if $A$ is isogenous to a power of some CM elliptic curve (see Exercise 5.6(10) in the same book).<|endoftext|> -TITLE: Bases for infinitely generated free groups -QUESTION [5 upvotes]: Let $F(S)$ be the free group on a (possibly infinite) set $S$. Let $T$ be a subset of $F(S)$ with the following two properties. - -$T$ generates $F(S)$. -$T$ injectively projects to a basis for the free abelian group $H_1(F(S);\mathbb{Z})$. - -Question : Must $T$ be a free basis for $F(S)$? -If $S$ is finite, then this follows from the standard fact that any generating set for $F(S)$ of cardinality $|S|$ is a free basis for $F(S)$ (for example, this is Proposition 2.7 in Lyndon and Schupp's book "Combinatorial Group Theory"). However, I don't see how to adapt this to the case where $S$ is infinite. -I really only care about the case where $S$ is countable, but I can't imagine that this is true for countable $S$ but false for uncountable $S$. - -REPLY [8 votes]: If I understand the question correctly, you are concerned that there might be hidden relations between the elements of $T$. Moreover, since the images of $T$ are linearly independent in the abelianization, these would have to be commutator relations. -This kind of thing can't happen: if there were a nontrivial relation involving elements of $T$, then said relation would involve only finitely many elements of $T$, so it would be a relation on the free subgroup generated by that finite set of elements. Essentially, condition (2) guarantees that the elements of any finite subset of $T$ are algebraically independent, and therefore all of the elements of $T$ are algebraically independent.<|endoftext|> -TITLE: Converse to Banach’s fixed point theorem for ordered fields? -QUESTION [12 upvotes]: Suppose $R$ is an ordered field. Call a continuous map $f: R \rightarrow R$ a contraction if there exists $r < 1$ (in $R$) such that $|f(x)-f(y)| \leq r |x-y|$ for all $x,y \in R$ (where $|x| := \max(x,-x)$). Suppose that every contraction from $R$ to $R$ has a unique fixed point. Must $R$ be the field of real numbers? -For a related question, see Converse to Banach's fixed point theorem? . -Jacek Jachymski's article "A discrete fixed point theorem of Eilenberg as a particular case of the contraction principle" ( http://emis.impa.br/EMIS/journals/HOA/FPTA/2004/131.pdf ) and the references it contains may be relevant. However, the non-Archimedean metric spaces that the article considers are bounded, which non-Archimedean ordered fields certainly are not. Also, my question is not about metric spaces, since my notion of distance lives in $R$ itself, not the real numbers. - -REPLY [15 votes]: Yes, it is true that $R$ must be the field of real numbers. -As $R$ is an ordered field, it is naturally an extension $\mathbb{Q}\hookrightarrow R$. We can prove the following two properties, which characterize the reals among the ordered fields. - -1) $\mathbb{Q}$ has no upper bound in $R$ (i.e., $R$ is Archimedean). - -Proof: Call element $x$ of $R$ infinite if $\vert x\vert$ is an upper bound for $\mathbb{Q}$, and finite otherwise. Then we can define $f\colon R\to R$ by -$$ -f(x)=\begin{cases} -\frac{x}{2}+\frac12\max(x,0)+(2+\max(x,0))^{-1},&\textrm{if }x\textrm{ is finite},\\\\ -x/2,&\textrm{if }x\textrm{ is infinite}. -\end{cases} -$$ -So, - -If $x,y$ are finite then they have an upper bound $a\ge0$ in $\mathbb{Q}$, and it can be seen that $\vert f(x)-f(y)\vert\le(1-(2+a)^{-2})\vert x-y\vert$. -If $x,y$ are both infinite then $\vert f(x)-f(y)\vert=\frac12\vert x-y\vert$. -If $x$ is infinite and $y$ is finite then $\vert f(x)-f(y)\vert\le \frac12\vert x\vert+\vert f(y)\vert\le\frac34\vert x-y\vert$. - -In any case, if $\mathbb{Q}$ had an upper bound $\kappa\in R$ then we have $\vert f(x)-f(y)\vert\le (1-\kappa^{-1})\vert x-y\vert$ so that, by hypothesis, $f$ has a fixed point. But it can be seen that $f(x) > x$ for finite $x$ and $f(x)=\frac x2\not=x$ for infinite $x$. So, it doesn't have a fixed point, giving a contradiction. - -2) Every Cauchy sequence $x_n$ in $R$ converges. - -Proof: Passing to a subsequence1, it can be assumed that $x_n$ is monotonic, and replacing $x_n$ by $-x_n$ if necessary, we can suppose that it is increasing. If it is eventually constant then the result is immediate. Otherwise, by further passing to a subsequence2, we can suppose that $x_{n+2}-x_{n+1}\le\frac12(x_{n+1}-x_n)$ and that $x_{n+1}-x_n < 2^{-n-1}$. Then, $y_n=x_n+2^{-n}$ is a strictly decreasing sequence with $0\le y_n-x_n\le 2^{-n}$. Again, passing to a subsequence, it can be assumed that $y_{n+1}-y_{n+2}\le\frac12(y_n-y_{n+1})$. -We can define $f\colon R\to R$ linearly mapping $(-\infty,x_1]$ onto $(-\infty,x_2]$, $(x_n,x_{n+1}]$ onto $(x_{n+1},x_{n+2}]$, $[y_1,\infty)$ onto $[y_2,\infty)$, and $[y_{n+1},y_n)$ onto $[y_{n+2},y_{n+1})$ ($n\ge1$). This can be done such that $\vert f(x)-f(y)\vert\le\frac12\vert x-y\vert$ on each interval, in which case it does not have any fixed points in these intervals. Furthermore, if $x_n$ had no limit point, then the intervals cover3 $R$ and this defines $f$ everywhere. But, then, $\vert f(x)-f(y)\vert\le\frac12\vert x-y\vert$ for all $x,y\in R$ implying that $f$ has a fixed point, giving a contradiction. - -I'll add a few more details that I passed over rather quickly above. A sequence $x_n$ is Cauchy if, for each $r > 0$ in $R$ then $\vert x_n - x_m\vert < r$ for large enough $m,n$. Any subsequence of a Cauchy sequence is itself Cauchy and tends to a limit $x$ if and only if the orginal sequence tends to $x$. -1 Any sequence in a linearly ordered set has a monotonic subsequence. -2 If $x_n$ is an increasing Cauchy sequence, which is not eventually constant, then it is possible to choose a subsequence $x_{n_k}$ as follows. Once $x_{n_k}$ has been chosen, then there is an $m > n_k$ such that $x_m \not= x_{n_k}$. As it is Cauchy, $n_{k+1}\ge m$ can be chosen such that $\vert x_r-x_s\vert < \min(2^{-k-2},(x_m-x_{n_k})/2)$ for all $r,s\ge n_{k+1}$. This ensures that $x_{n_{k+2}}-x_{n_{k+1}}$ is less than both $2^{-k-2}$ and $(x_{n_{k+1}}-x_{n_k})/2$ for all $k$. -3 If $z\in R$ was not in any of the intervals $(-\infty,x_1]$, $(x_n,x_{n+1}]$, $[y_1,\infty)$, $[y_{n+1},y_n)$ then $x_n < z < y_n$ for all $n$. So, $\vert z-x_n\vert\le y_n-x_n\le 2^{-n}$. Given any $r > 0$ in $R$, the fact that we have already shown $R$ to be Archimedean in (1) implies that $2^n > r^{-1}$ for large $n$. So, $\vert z - x_n\vert < r$ for large $n$, and $x_n\to z$.<|endoftext|> -TITLE: Differential forms on an almost complex manifold -QUESTION [12 upvotes]: Hello! -Let $M$ be an almost complex manifold. Let $TM$ denote its tangent bundle. Then we have the decomposition $TM\otimes\mathbb{C}=T^{1,0}M\oplus T^{0,1}M$ corresponding to the eigenvalues of the almost complex structure. This decomposition yields the decomposition: -$$ -\Lambda^r(T^\star M\otimes\mathbb{C})=\Lambda^r(T^{1,0}M^\star\oplus T^{0,1}M^\star)=\bigoplus_{p+q=r}\Lambda^p(T^{1,0}M^\star)\otimes\Lambda^q(\overline{T^{0,1}M}^\star) -$$ -Now take a section $\omega$ of the complex vector bundle -$$ -\Lambda^{p,q}:=\Lambda^p(T^{1,0}M^\star)\otimes\Lambda^q(\overline{T^{0,1}M}^\star) -$$ -$\omega$ is called a complex differential form of type $(p,q)$. Consider a complex $(p,q)$-form $\omega$ and take its differential. Its differential $\mathrm{d}\omega$ is a section of: -$$ -\Lambda^{p+q+1}(T^\star M\otimes\mathbb{C})=\bigoplus_{m+n=p+q+1}\Lambda^{m,n} -$$ -Therefore $\mathrm{d}\omega$ can be decomposed in a sum of complex differential forms of type $(m,n)$ with $m+n=p+q+1$. However I have read that there are only four terms. My second question is: -How do we prove that in fact $\mathrm{d}\omega$ is a section of: $$\Lambda^{p+2,q-1}\oplus\Lambda^{p+1,q}\oplus\Lambda^{p,q+1}\oplus\Lambda^{p-1,q+2}$$ only? -I am aware that in the case where the almost complex structure is integrable we get only two terms such that finally we have $\mathrm{d}=\partial+\bar{\partial}$. But in fact it seems that in the almost complex case already we do not have so many terms (namely we have only 4 as above). I think this has something to do with the graduation of the algebra of differential forms and the nilpotence of the differential itself but I am not able to prove it. -At last, since I am interesting in the same kind of question concerning Lie and Courant algebroids, I was wondering if this fact could be recast in the language of homotopical algebras (by which I vaguely mean that usual identities on brackets hold up to something else)? This is because the algebra of differential forms is a supercommutative algebra and that we can reformulate $\mathrm{d}^2=0$ by $[\mathrm{d},\mathrm{d}]=0$. Could somebody point me toward an article? -Thank you very much! - -REPLY [9 votes]: Call $C^{\infty}_{p,q}(M)$ the space of smooth complex sections of the bundle $\Lambda^{p,q}T^*_M$ and let $2n$ be the real dimension of $M$. -The fact that -$$ -dC^{\infty}_{p,q}(M)\subset C^{\infty}_{p+2,q-1}(M)+C^{\infty}_{p+1,q}(M)+C^{\infty}_{p,q+1}(M)+C^{\infty}_{p-1,q+2}(M) -$$ -follows immediately from the two following facts: - -The (bigraded) algebra $C^{\infty}_{\bullet,\bullet}(M)=\bigoplus_{p,q=0}^n C^{\infty}_{p,q}(M)$ is locally generated by $C^{\infty}_{0,0}(M)$, $C^{\infty}_{1,0}(M)$ and $C^{\infty}_{0,1}(M)$; -There are (obvious) inclusions -$$ -dC^{\infty}_{0,0}(M)\subset C^{\infty}_{1,0}(M)+C^{\infty}_{0,1}(M), -$$ -$$ -dC^{\infty}_{1,0}(M)\subset C^{\infty}_{2,0}(M)+C^{\infty}_{1,1}(M)+C^{\infty}_{0,2}(M), -$$ -$$ -dC^{\infty}_{0,1}(M)\subset C^{\infty}_{2,0}(M)+C^{\infty}_{1,1}(M)+C^{\infty}_{0,2}(M). -$$ - -Moreover, for an almost complex manifold $M$ with complex structure $J$, the following facts are equivalent: - -$J$ has no torsion (and thus, by Newlander-Nirenberg theorem $J$ is a true complex structure and $M$ a complex analytic manifold); -$dC^{\infty}_{1,0}(M)\subset C^{\infty}_{2,0}(M)+C^{\infty}_{1,1}(M)$ -and $dC^{\infty}_{0,1}(M)\subset C^{\infty}_{1,1}(M)+C^{\infty}_{0,2}(M)$; -$dC^{\infty}_{p,q}(M)\subset C^{\infty}_{p+1,q}(M)+C^{\infty}_{p,q+1}(M)$ for all $p,q=0,1,\dots,n$.<|endoftext|> -TITLE: Proving that a combinatorial sequence has no compact formula -QUESTION [8 upvotes]: Suppose we have a sequence $a_n$ given by some combinatorial formula, e.g. involving a sum of n terms (like ${n \choose k}^{10}3^{-k}$ etc.). Sometimes it is plausible that there is no compact formula for the $a_n$, where one has to adopt a reasonable definition of "compact" (i.e. using a constant, independent of $n$, number of primitive operations). Are there any methods of proving that a certain sequence $a_n$ has no compact formula, in much the same way differential Galois theory allows one to prove that certain integrals are nonelementary? Of course this would relative the choice of "computational primitives" (factorial, ${n \choose k}$ etc.) - -REPLY [14 votes]: Yes, there are. See the book A=B by Petkovsek, Wilf, and Zeilberger. I will quote from it: - -[Petk91] is the Ph.D. thesis of Marko - Petkovsek, in 1991. In it he - discovered the algorithm for deciding - if a given recurrence with polynomial - coefficients has a "simple" solution, - which, together with the algorithms - above, enables the automated discovery - of the simple evaluation of a given - definite sum, if one exists, or a proof of nonexistence, if none exists - (see Chapter 8). A definite - hypergeometric sum is one of the form - $f(n) = \sum_{k=-\infty}^\infty -> F(k,n)$, where $F$ is hypergeometric.<|endoftext|> -TITLE: 2-TQFT are to Frobenius Algebras as ??? are to Hopf Algebras -QUESTION [16 upvotes]: The question arose this morning during a seminar about HAs. -In a few words: can the equivalence $2-TQFT_k \leftrightarrow Frob_k$ be "modified" in a sensible way to give a similar one between the category $HA$ of Hopf algebras and a suitable "topological" category (I mean: a -even functor- category made 'with' topological objects, hopefully in a sufficiently small neighborhood of $2-TQFT$)? In particular i would like to find a visual analogue for the antipode map $s:H\to H$. -Bad thing is that it takes a while to discover there seem to be no way to define it as an arrow in $Cob(2)$: just try to draw in $Cob(2)$ the diagram - -...any sensible choice for $s$ leaves in the manifold one hole more than the minimum. Spending a couple of words about the "sensible choice", it seems to me the only way not to increase the genus of the surface is to take as cobordism a-cap-and-a-cup, namely the [Cob(2)-analogue of the] composition $\eta\circ \epsilon\colon H\to k\to H$ in the former diagram... But I'm not able to characterize it as a Frobenius-Algebra map in any sensible way. -So, help me... -(maybe the person I discussed with this morning is here? His website is this.) - -REPLY [4 votes]: This is my "geometric freshman explanation": -the problem is to put something instead of the "?" doing the job of $s$ in - -...but the composition $\eta\circ\epsilon$ is disconnected, and there is no way to obtain a disconnected manifold starting gluing something to that. :( such a pity.<|endoftext|> -TITLE: Which groups admit a unique Lie group structure? -QUESTION [5 upvotes]: This question is a follow-up on the answer given here Can a Lie group as an abstract group be given more than one topology making it a Lie group? -It is motivated by the following observations: - -If $m,n$ are positive integers, then $\mathbb{R}^m$ is isomorphic to $\mathbb{R}^n$ as abstract groups since they are both $\mathbb{Q}$-vector spaces of the same dimension. So a Lie group structure on $\mathbb{R}^n$ is not unique. -On the other hand, a Lie group structure on a compact $n$-torus $T^n=\mathbb{R}^n/\mathbb{Z}^n$ is unique: if $m\neq n$, then $T^n$ can not be isomorphic to $T^m$ as abstract groups, e.g. since $T^n$ has $2^n-1$ elements of order 2, and $T^m$ has $2^m-1$; nor can $T^n$ be isomorphic to $\mathbb{R}^m$ since $\mathbb{R}^m$ has no elements of finite order at all (apart from 0). -Here is an ad hoc proof that a Lie group structure on $SU(2)$ is unique. Let $G$ be a Lie group isomorphic to $SU(2)$ as an abstract group. Then $G$ semi-simple since semi-simplicity can be described in group theoretic terms (there are no nontrivial solvable normal subgroups). Moreover, all maximal abelian subgroups of $G$ are $T^1$'s by 2. So the complexification of $G$ has Lie algebra $\mathfrak{sl}_2(\mathbb{C})$. There are two Lie semi-simple Lie groups that fit all of the above: $SU(2)$ and $SO(3)$; the first has a center and the second does not, so $G$ must be $SU(2)$. [upd: as pointed out by Claudio, this only shows that $G_e$, the connected component of the unit $G$, is $SU(2)$ or $SO(3)$; but if $G$ has more than one connected component, then it has a normal subgroup, $G_e$, different from $\mathbb{Z}/2$, which $SU(2)$ can't have.] (I believe something similar should work for any semi-simple compact group.) - -So I would like to ask: is there a reasonable way to characterize Lie groups that admit a unique Lie group structure? If not, then what happens if we restrict the attention to (real or complex) semi-simple Lie groups? (I would be particularly interested in a proof that did not rely to much on the classification.) - -REPLY [2 votes]: At least for semisimple groups, there is a close parallel to your question in the way Borel and Tits analyze abstract homomorphisms between algebraic groups: -Homomorphismes “abstraits” de groupes alge ́briques simples. Ann. of Math. (2) 97 (1973), 499–571. For the most part this work doesn't rely on the classification. Certainly there is a common thread here, in the comparison of abstract group notions with algebraic group notions. Look at their Section 9 in particular, where locally compact fields including $\mathbb{R}$ are discussed. -The algebraic group case also suggests strongly that you need to study separately semisimple (or reductive) Lie groups and solvable Lie groups. I'd expect the latter case to be more open-ended.<|endoftext|> -TITLE: A stupid question about Automorphic forms -QUESTION [5 upvotes]: Okay, so an automorphic form $f$ on a reductive group $G/ \mathbb{Q}$ and arithmetic subgroup $\Gamma$ is a smooth function satisfying the following conditions: -(a) invariance with respect to left $\Gamma -$ translations. -(b) Right $K -$ finiteness. -(c) Annihilated with respect to a finite co-dimension ideal of the center $Z(\mathcal{U}(\mathfrak{g}_{\mathbb{C}}))$ ( of the complexified universal enveloping algebra). -(d) Growth conditions. -Now as I understand, it is clear why condition (a) is relevant. Condition (b) comes from the idea that representation of $G$ associated to $f$ is admissible. -My understanding is that condition (d) ensures that we can "extend" the automorphic form to the cusps of the symmetric space. -I would like to know: what exactly is the intuition behind condition (c)? -(I hope it is not just to provide a framework which includes classical modular forms over upper half-plane and the Maass forms.) -Also any comments about "my understanding" of conditions (a), (b) and (d) are appreciated! - -REPLY [16 votes]: These conditions have technical/subtle interactions. It probably suffices to think about automorphic forms on a Lie group, and not think of the interaction with finite places. -For example, on each K-isotype, the Casimir element is an elliptic operator, but it is not elliptic without specifying the behavior under K. From elliptic regularity, such eigenfunctions are real-analytic. -With K-finite and z-finite, moderate growth implies that all derivatives are of moderate growth, etc. (Borel's little book talks about this for SL(2,R), and his Park City notes recapitulate this. The general argument is similar to SL(2,R), anyway.) -K-finite, z-finite, moderate-growth cuspforms are (up to adjustment of central character) provably of rapid decay, so certainly L^2. This and the previous assertion are part of the "theory of the constant term". -Dropping the moderate growth condition is useful occasionally, as in the "weak Maass form" business. -In a more elementary vein, a choice to require annihilation by a finite-codimension ideal in z is akin to looking at functions on the real line annihilated by a product (D^2-lambda_1)...(D^2-lambda_n), namely, polynomial multiples of exponentials. Certainly not every function on the line is annihilated by such, but the spectral theory (Fourier transform) decomposes a general function into a superposition of certain of these special functions. (Here the analogue of K is {1}.)<|endoftext|> -TITLE: GL(V)-representation theory for a Lie bracket kernel -QUESTION [12 upvotes]: Let $V$ be a vector space over a field of characteristic $0$, and let $L_k(V)$ be the degree $k$ part of the free Lie algebra over $V$. There is an exact sequence -$$0\to D_n(V)\to L_1(V)\otimes L_{n+1}(V)\to L_{n+2}V\to 0$$ -where the map on the right is the Lie bracket and $D_n(V)$ is defined as the kernel of this map. Now $D_n(V)$ is a $GL(V)$-module, and I'm curious about how it decomposes as a direct sum of irreducibles. By playing around with dimension formulas, I was able to show that $D_1(V)=S^{(1,1,1)}V$, $D_2(V)=S^{(2,2)}V$ and $D_3(V)=S^{(3,1,1)}V$. Here $S^\lambda(V)$ is the irreducible representation of $GL(V)$ corresponding to partition $\lambda$. So in fact, if I haven't made a mistake, $D_k(V)$ is actually irreducible for $k=1,2,3$, with nice symmetric looking partitions. -Is $D_k(V)$ always irreducible, and is there an easy way to tell what the partition is? If not, does anyone know what is known about this? -If it helps, here is an alternate way of characterizing $D_n$. Consider the Lie operad, and look at the part with a total of $n+2$ input/output slots, denoted $Lie((n+2))$. Then $$D_n\cong [V^{\otimes n+2}\otimes Lie((n+2))]_{Sym(n+2)},$$ which denotes the coinvariants under the action of $Sym(n)$ acting simultaneously on $Lie((n+2))$ and $V^{\otimes n}$, in the latter case by permuting the factors. This is isomorphic to a space of planar unitrivalent trees with leaves labeled by vectors from $V$, modulo antisymmetry and Jacobi (IHX) relations. -I don't know very much representation theory, so it's possible my question has an easy answer. -Added 9/20/2011 -Morita, Sakasai and Suzuki just posted a preprint proving that $D_{4k+2}$ and $D_{4k+3}$ always have symmetric decompositions, in the sense that there is a symmetry in the corresponding Young diagrams exchanging rows and columns. This is a neat result. - -REPLY [7 votes]: Hello. I would like to explain an elementary way to decompose $D_n(V)$ as a $GL(V)$-module. (But for large $n$, it is difficult to calculate by hand.) -We will use the following results. From now, we assume $\dim{V} \gg n$. - -The multiplicities of an irreducible $GL(V)$-module $S^\lambda{V}$ in $L_n(V)$ is equal to the one of an the irreducible $\mathrm{Cyc}_n$-module $\exp(2\pi{i}/n)$ in the $S_n$-module $\mathrm{Res}_{\mathrm{Cyc}_n}^{S_n}D^\lambda$. Here $D^\lambda$ is the irreducible $S_n$-module corresponding to $\lambda$. -There is a combinatorial way to decompose $\mathrm{Res}_{\mathrm{Cyc}_n}^{S_n}D^\lambda$. We use the notion of the "major index" for a standard tableau of shape $\lambda$. This appears in Stanley's slide which is referd by James Griffin. -For more details about 1 and 2, see Garsia's paper "Combinatorics of the free Lie algebra and the symmetric group"(Theorem 8.4) and Reutenauer's book "Free Lie algebras"(Theorem 8.8 and 8.9) etc. -Pieri's rule. We can obtain the irreducible decomposition of the $GL(V)$-module $L_1(V) \otimes S^\lambda{V}$, namely this module is isomorphic to the direct sum of irreducible $GL(V)$-modules $S^\mu{V}$, where $\mu$ runs over the set of partitions by adding a box to $\lambda$. Then, if we know the irreducible decomposition of $L_n(V)$ by 1 (and 2), we can also obtain the irreducible decomposition of $L_1(V) \otimes L_n(V)$. - -Example. -For simplicity, we denote $\lambda$ by $S^\lambda{V}$. -$L_5(V) \cong (4,1) \oplus (3,2) \oplus (3,1^2) \oplus (2^2,1) \oplus (2,1^3)$. -$L_1(V) \otimes L_5(V)$ is isomorphic to -$(5,1) \oplus 2(4,2) \oplus 2(4,1^2) \oplus (3^2) \oplus 3(3,2,1) \oplus 2(3,1^3) \oplus (2^3) \oplus 2(2^2,1^2) \oplus (2,1^4)$. -$L_6(V) \cong (5,1) \oplus (4,2) \oplus 2(4,1^2) \oplus (3^2) \oplus 3(3,2,1) \oplus (3,1^3) \oplus 2(2^2,1^2) \oplus (2,1^4)$. -Then we have $D_4(V) \cong (4,2) \oplus (3,1^3) \oplus (2^3)$. -Similarly, I think, we can obtain $D_5(V) \cong (5,1^2) \oplus (4,2,1) \oplus (3^2,1) \oplus (3,2,1^2) \oplus (2^2,1^3)$.<|endoftext|> -TITLE: Kummer and Fermat's Equation -QUESTION [5 upvotes]: In Report on the Theory of Numbers, H.J.S. Smith writes: -"The impossibility of solving [Fermat's] equation has been demonstrated by M. Kummer, first, for all values of $\lambda$ not included among the exceptional primes; and secondly, for all exceptional primes which satisfy the three following conditions: - -That the first factor of H, though divisible by $\lambda$, is not divisible by $\lambda^2$. -That a complex modulus can be assigned, for which a certain definite complex unit is not congruous to a perfect $\lambda$-th power. -That $B_{\kappa \lambda}$ is not divisible by $\lambda^3$, $B_{\kappa}$ representing that Bernoullian number $[\kappa \leq \mu-1]$ which is divisible by $\lambda$. - -Three numbers below 100, viz. 37, 59, 67, are, as we have seen, exceptional primes. But it has been ascertained by M. Kummer that the three conditions just given are satisfied in the case of each of these three numbers; so that the impossibility of Fermat's equation has been demonstrated for all values of the exponent up to 100. Indeed, it would probably be difficult to find an exceptional prime not satisfying the three conditions, and consequently excluded from M. Kummer's demonstration." -Can anyone cite an exceptional prime that does not satisfy the three conditions? - -REPLY [6 votes]: Washington's book has a table of irregular primes together with the power of p dividing the first factor of the class number. There are plenty for which the first factor is divisible by $p^2$; the first is p=157. So for these Kummer's first condition fails.<|endoftext|> -TITLE: Characterize where the Dirichlet Problem for the Laplacian is always solvable -QUESTION [10 upvotes]: Conway's 1978 textbook Functions of One Complex Variable I gives an unsatisfying characterization of the regions for which the Dirichlet Problem can always be solved, and then comments no cleaner characterization is known. Has any progress been made since then? And what simpler characterization is known today, if one is known? -Here is the problem definition: -An open connected set $G\subseteq \mathbb{C}$ is called a Dirichlet Region if for each continuous function $f:\partial_\infty G\rightarrow \mathbb{R}$ there is a continuous function $u:G^- \rightarrow \mathbb{R}$ such that $u$ is harmonic in $G$ and $u(z)=f(z)$ for all $z$ in $\partial_\infty G$. -(The notation $\partial_\infty G$ refers to the boundary of $G$ in $\mathbb{C}\cup\{\infty\}$, and $G^-$ denotes the closure of $G$ in $\mathbb{C}\cup\{\infty\}$.) -The characterization given in the book is: -Given $a \in \partial_\infty G$, a barrier for $G$ at $a$ is a family $\{\psi_r: r>0\}$ of functions such that: -1. $\psi_r$ is well-defined and superharmonic on $B(a;r) \cap G$ with $0\leq \psi_r(z) \leq 1$ -2. $\lim_{z\rightarrow a}\psi_r(z) = 0$, and -3. $\lim_{z\rightarrow w} \psi_r(z) = 1$ for $w$ in $G \cap \{w:|w-a|=r\}$. -An open connected set $G$ is a Dirichlet Region iff there is a barrier for $G$ at each point of $\partial_\infty G$. - -REPLY [3 votes]: http://eom.springer.de/r/r080680.htm contains some characterizations of domains where the Dirichlet problem is solvable. I believe that a key term in searching for references is "Wiener's criterion."<|endoftext|> -TITLE: The tangent bundle to an infinite-dimensional manifold -QUESTION [19 upvotes]: Suppose that $A,B$ are smooth ($\mathrm C^\infty$) manifolds, and denote by $\hom(A,B)$ the set of $\mathrm C^\infty$-maps $A \to B$. It is a perfectly well-defined set, but often one wants more. For example, I regularly would like to talk about (small, or maybe infinitesimal) variations of a smooth map $A \to B$, and I regularly want to talk about smooth functions on the space of maps. Which is to say, there are many good reasons to want to have some sort of "infinite-dimensional manifold" which I will call $\newcommand\Maps{\underline{\operatorname{Maps}}} \Maps(A,B)$, whose set of points is $\hom(A,B)$. -A standard way of defining $\Maps(A,B)$ is to choose some regularity ($\mathrm C^\infty$ ends up not being a good choice, as the construction will not yield something with an inverse function theorem) and to try to first topologize and then assign a "smooth structure" to the set of functions $A\to B$ with the prescribed regularity. This works well for certain analytic applications, and allows for explicit constructions. But it has the disadvantage of making all facts demand on some real analysis, and the only thing I'm worse at than real analysis is complex analysis. -A different approach is to say that what "possibly-infinite-dimensional manifold" means is presheaf on the category of manifolds, or maybe restrict just to some well-behaved class of presheaves (e.g. sheaves for your favorite Grothendieck topology). Then you could define $\Maps(A,B)$ to be the sheaf -$$ \Maps(A,B) = \hom( - \times A, B) $$ -This definition answers some questions immediately. For example, there is a sheaf -$$ \mathbb R = \hom (-, \mathbb R) $$ -and a "smooth $\mathbb R$-valued function on $\Maps(A,B)$" is precisely a natural transformation of sheaves $\Maps(A,B) \to \mathbb R$. Note that the set of smooth $\mathbb R$-valued functions is actually an $\mathbb R$-algebra, because there is a product of sheaves -$$ \Maps(A,B)^{\times 2} = \hom(- \times A,B)^{\times 2} $$ -(right hand side is product of sets), and also the addition and multiplication maps $\mathbb R^{\times 2} \to \mathbb R$ are smooth. So we get an algebra $\mathrm C^\infty(\Maps(A,B))$ (which is even itself an infinite-dimensional smooth manifold in the same sense). -OK, so I've proposed some definition of "the smooth manifold $\Maps(A,B)$". I have "families of smooth maps", because I can map into $\Maps(A,B)$ from, say, curves, or spaces. What I really want, in addition to infinitesimal variations, are flows. So I'd like a "tangent bundle $\mathrm T (\Maps(A,B))$". -Here are at least three definitions I could think of: - -There is a sheaf on manifolds called "$\operatorname{spec} \mathbb R[\epsilon]/(\epsilon^2)$". Actually, it's easiest to define as a cosheaf; whether by definition or theorem, it satisfies that, for any finite-dimensional manifold $M$, -$$ \Maps( \operatorname{spec} \mathbb R[\epsilon]/(\epsilon^2) \to M ) = \text{the total space of } \mathrm T M $$ -The bundle structure on $\mathrm T M$ corresponds to the unique inclusion $\lbrace \text{pt.}\rbrace \to \operatorname{spec} \mathbb R[\epsilon]/(\epsilon^2)$; the vector bundle structure from the endomorphisms $\Maps(\operatorname{spec} \mathbb R[\epsilon]/(\epsilon^2) \to \operatorname{spec} \mathbb R[\epsilon]/(\epsilon^2)) = \mathbb R$. So one could define: -$$\mathrm T (\Maps(A,B)) = \Maps( \operatorname{spec} \mathbb R[\epsilon]/(\epsilon^2), \Maps(A,B)).$$ -One could reasonably say that a "vector bundle on $\Maps(A,B)$" should be determined by (indeed, the same as) its pulled-back vector bundle along any map $S \to \Maps(A,B)$. Given $f: S \to \Maps(A,B)$, which is to say $f\in \hom(S\times A, B)$, one can define an infinite-dimensional vector bundle $V \to S$ by saying that for any open set $U \subseteq S$, -$$ \Gamma_U(V) = \Gamma_{U \times A}(f^\ast \mathrm T B) $$ -This is the way I usually think about the tangent bundle to $\Maps(A,B)$. More precisely, I usually define $\mathrm T(\Maps(A,B))$ by saying that over each point $f\in \hom(A,B)$, the fiber is $\mathrm T_f\Maps(A,B) = \Gamma_A(f^\ast \mathrm T B)$, and hoping that my readers don't ask me to explain in what sense these fibers glue together into a smooth bundle. -I have above defined an algebra $\mathrm C^\infty( \Maps(A,B))$. A tangent vector could be defined as a "point derivation": a map $v : \mathrm C^\infty(\Maps(A,B)) \to \mathbb R$ that satisfies that $v(\alpha\beta) = v(\alpha) \, \beta(f) + \alpha(f)\, v(\beta)$ for some $f\in \operatorname{spec} \mathrm C^\infty( \Maps(A,B))$ (or maybe just for those $f \in \hom(A,B)$). More generally, you could give this a smooth structure by explaining the notion of "derivation" internal to the world of sheaves. I think equivalently, -for each $f : S \to \Maps(A,B)$, there is a restriction map $f^\ast: \mathrm C^\infty(\Maps(A,B)) \to \mathrm C^\infty(S)$, and you can consider the space of derivations over this map. It would be the space of sections of $f^\ast \mathrm T (\Maps(A,B))$, and this data should glue together into some form of "vector bundle". - -My question is: - -Are definitions 1,2,3 naturally equivalent? If not, which is best, and why? Or should I abandon all three in favor of some other definition? - -My reason for asking, in case it influences your answer, is that I would like to understand in what sense the mapping space between two Q-manifolds is an infinite-dimensional Q-manifold, because one form of perturbative quantum field theory can be understood as the study of deformations of such infinite-dimensional Q-manifolds (in some space of "non-commutative Q-manifolds"). (A good place to read definitions of words like "Q-manifold" is Rajan Mehta's thesis. A good place to read about this version of quantum field theory is a short paper by Dmitry Roytenberg.) I think I know how to answer this "Q-manifold" question in any of approaches 1,2,3, although I haven't thought about the details, and I do not know how to answer it in more analytical approaches. But I'm also interested in related "P-manifold" questions, which are harder, so I'd like to know exactly how cavalier I can be with my infinite-dimensional manifolds. - -REPLY [4 votes]: You said that you'd often like to talk about (small, or maybe infinitesimal) variations of a smooth map. There is a well known (in some circles at least) topology on the space of smooth maps. -It's called the Whitney $C^{\infty}$-topology. -First you define the Whitney $C^k$-topology using the natural projection $\pi : C^{\infty}(A,B) \twoheadrightarrow J^k(A,B),$ where the base space is the k-jet space; which is identified with $\mathbb{R}^p$, for some $p$ depending on $\dim(A)$ and $\dim(B),$ and of course on $k.$ As a basis for the $C^k$-topology on $C^{\infty}(A,B)$ we take the preimages of the open sets in $J^k(A,B)$ under the metric topology induced from $\mathbb{R}^p.$ -Informally, the metric on the jet space measures the difference between derivatives. For example, let $f : \mathbb{R} \to \mathbb{R}$ and, for some $x \in \mathbb{R},$ consider $j_x^kf \in J^k(\mathbb{R},\mathbb{R}) \cong \mathbb{R}^{k+2}.$ We have -$j_x^kf \mapsto (x,f(x),f'(x),\ldots,f^{(k)}(x)) \in \mathbb{R}^{k+2}.$ -If $W^k$ denotes the set of open sets of $C^{\infty}(A,B)$ under the $C^k$-topology, then the Whitney $C^{\infty}$-topology on $C^{\infty}(A,B)$ is defined to be the topology whose basis is $W$, where -$W := \bigcup_{k=0}^{\infty} W^k.$ -The Whitney $C^{\infty}$-topology makes $C^{\infty}(A,B)$ into a Baire space. -See pages 42 - 50 of Golubitsky & Guillemin, "Stable Mappings and their Singularities", (1974).<|endoftext|> -TITLE: Moduli space of curves over $\mathbb Z$ -QUESTION [16 upvotes]: I was wondering if there was a really good place to learn about this topic that actually contains the details of the construction from beginning to end. I'm familiar with the book by Harris and Morrison, and many of the papers about the construction. The recent book by Arbarello, Cornalba, and Griffiths covers the complex case in a very nice and detailed way, but I was wondering if there was a good place to learn this 1) that only uses algebraic geometry and not complex analytic geometry, and 2) that covers the construction over $\mathbb Z$ so one can use it in the arithmetic case. Most books simply leave so many details out that a beginner in the subject finds it hard to fill them all in and truly understand the construction. Any good references would be much appreciated. Thanks - -REPLY [5 votes]: These notes by Dan Edidin: http://www.math.missouri.edu/~edidin/Papers/mfile.pdf give a very readable account of the construction of the moduli stack of curves over $\mathrm{Spec}(\mathbb Z)$ for $g \geq 3$ via the Hilbert scheme. I am not sure if this is what you are looking for, though, depending on what you mean with "from beginning to end": certainly there are several steps left as pointers to the literature.<|endoftext|> -TITLE: homotopy transfer for sheaves of algebras -QUESTION [5 upvotes]: homotopy transfer for algebras -Let $A$ be a differential graded (dg) $k$-algebra, and $H(A)$ its cohomology. $H(A)$ is naturally equipped with the structure of a graded algebra. In general we don't have that $H(A)$ and $A$ are weakly equivalent (i.e. quasi-isomorphic). -Nevertheless, it is well-known that there exists an $A_\infty$-algebra structure $(m_k)_{k\geq1}$ on $H(A)$ with the following properties: - -$H(A)$ equipped with this $A_\infty$-structure is weakly equivalent to $A$. -the first structure map $m_1$ (i.e. the differential) vanishes. -the second structure map $m_2$ coincide with the natural product on $H(A)$. - -This structure is essentially unique: it is unique up to a unique $A_\infty$-isomorphism. -Moreover, there are explicit formulas for for the $A_\infty$-structure and the weak equivalence, in terms of planar trees. The main point is that the formula involves the choice of quasi-isomorphisms $i:H(A)\to A$ and $p:A\to H(A)$, together with an homotopy $h$ between $i\circ p$ and $id_A$. -homotopy transfer for sheaves of dg algebras? -I would be interested to know how to adapt this for sheaves. Namely, if now $A$ is a sheaf of dg algebras and $H(A)$ its cohomology sheaf. First of all one has to assume that $A$ is formal (i.e. quasi-isomorphic to $H(A)$) as a sheaf of $k$-modules. But even in this situation the existence of $i$, $p$ and $h$ is not guarantied. -the question - -How does homotopy transfer works for sheaves of algebras? - -possible (incomplete) answer -One can do things locally and then try to glue, the gluing condition will probably be satisfied only in a weak sens... and then my question might boil down to - -what is the right definition of a (homotopy) sheaf of $A_\infty$-algeras? - -A way to answer this is to use model categories. I was planning to proceed in the following way (very shortly): $A_\infty$-algebras are fibrant objects in the model category of dg coalgebras, then we have a Reedy model structure on presheaves of dg coalgebras, and sheaves of $A_\infty$-algebras can be defined as fibrant objects in this model category. -At this moment I am stuck. I am sure I am not far from the answer, but somehow I can't see the point. - -REPLY [9 votes]: If $R$ is a commutative $k$-algebra, a quasi-coherent sheaf of dg-$k$-algebras on $\operatorname{Spec}R$ would be just a dg-$R$-algebra $A$. It's known that there need not be any $R$-linear A-infinity structure on $H(A)$ quasi-isomorphic to $A$ in the $A$-infinity sense. Therefore the classical transfer theorem (which only works over fields) does not sheafify. Nevertheless, Sagave defined the notion of derived A-infinity algebra in such a way that there is a derived A-infinity structure on $H(A)$ quasi-isomorphic to $A$ in an $R$-linear and derived A-infinity sense, see : -MR2608191 (2011c:16030) -Sagave, Steffen(N-OSLO) -DG-algebras and derived -algebras. (English summary) -J. Reine Angew. Math. 639 (2010), 73–105. -This solves the case of quasi-coherent sheaves on affine schemes, so it could be a good starting point to answer your question...<|endoftext|> -TITLE: Generating the symplectic group -QUESTION [17 upvotes]: The too naive and vague version of my question is the following: given a collection of integer symplectic matrices all of the same size (say 2n by 2n), how can I tell if they generate the full symplectic group Sp(2n,Z)? -Here is the fleshed-out version. Here Sp(2n,Z) means the group of matrices preserving the form -$ J = \left( \begin{array}{cc} 0&I \\ -I&0& \end{array} \right)$ -and I stick to the case of Sp(4,Z). It is a theorem of Stanek that this group is generated by the following three matrices: -$ T = \left( \begin{array}{cccc} 1&0&1&0 \\ 0&1&0&0 \\ 0&0&1&0 \\ 0&0&0&1 \end{array} \right), \quad R = \left( \begin{array}{cccc} 1&0&0&0 \\ 1&1&0&0 \\ 0&0&1&-1 \\ 0&0&0&1 \end{array} \right), - \quad D=\left( \begin{array}{cccc} 0&1&0&0 \\ 0&0&-1&0 \\ 0&0&0&1 \\ 1&0&0&0 \end{array} \right). $ -Now it so happens (in some geometric situation I care about) that I have two other elements A and B of Sp(4,Z), given as follows: -$ A = \left( \begin{array}{cccc} 0&0&-1&0 \\ 0&1&0&0 \\ 1&0&0&0 \\ 0&0&0&1 \end{array} \right), \quad B = \left( \begin{array}{cccc} 0&0&0&-1 \\ 0&0&-1&5 \\ 5&1&2&5 \\ 1&0&0&2 \end{array} \right). $ -Then my more concrete question is the following: - -Is {T,A,B} also a generating set for Sp(4,Z)? - -To make that question seem a little more reasonable, here is some -Evidence: The images of T, A, and B in Sp(4,F_p) do in fact generate Sp(4,F_p) for all primes p up to 47. (My computer stopped cooperating at that point.) -It seems plausible to me that if the subgroup $\left< T, A, B \right>$ was actually proper, then this would be detected by its image in one of the groups Sp(4,F_p), and moreover that the first prime p at which this happened wouldn't be too big (for example, perhaps not bigger than the largest absolute value of coefficients of A or B). But maybe that is completely off the mark. (If so, please tell me so!) -Of course I don't expect a definitive answer to my specific question here, but anything that can be said about approaches to this kind of question would be much appreciated. -Finally let me mention that, for the geometric question I'm interested in, it would be sufficient to know that $\left< T, A, B \right>$ is merely a finite-index subgroup of Sp(4,Z). (Hence the arithmetic groups tag.) But the stronger statement seems to be true, as explained above, so I phrased the question that way. -Edit (May 26): As pointed out by Derek Holt in his answer, it is not true that the reduction of my group mod 2 gives all of Sp(4,F_2). I must have made a mistake in that computation. In any case, he also shows that my group has finite index (indeed, index 6) in Sp(4,Z), settling the question satisfactorily. But any more remarks about generating sets for the symplectic group are still most welcome! - -REPLY [18 votes]: In fact, if you reduce mod 2, then you find that the index of the subgroup generated by $T,A,B$ in ${\rm Sp}(4,2)$ is 6, not 1. -I have now carried out Igor's suggested approach and a coset enumeration (which I did in MAGMA) shows that the subgroup has index 6 in ${\rm Sp}(4,\mathbb{Z})$. -Bender's presentation of ${\rm Sp}(4,\mathbb{Z})$ is on the generators -$K = \left(\begin{array}{rrrr}1&0&0&0\\\\ 1&-1&0&0\\\\ 0&0&1&1\\\\ 0&0&0&-1 \end{array}\right),\ \ \ -L=\left(\begin{array}{rrrr}0&0&-1&0\\\\ 0&0&0&-1\\\\ 1&0&1&0\\\\ 0&1&0&0 \end{array}\right).$ -and the relations are as follows. (I won't typeset this in case anyone wants to cut and paste!) -K^2=1, -L^12=1, -K*L^7*K*L^5*K*L = L*K*L^5*K*L^7*K, -L^2*K*L^4*K*L^5*K*L^7*K = K*L^5*K*L^7*K*L^2*K*L^4, -L^3*K*L^3*K*L^5*K*L^7*K = K*L^5*K*L^7*K*L^3*K*L^3, -(L^2*K*L^5*K*L^7*K)^2 = (K*L^5*K*L^7*K*L^2)^2, -L*(L^6*K*L^5*K*L^7*K)^2 = (L^6*K*L^5*K*L^7*K)^2*L, -(K*L^5)^5 = (L^6*K*L^5*K*L^7*K)^2. -I used a combination of brute force search and useful intermediate matrices (including those used by Tom De Medts in his comment to the original post), to get words for $T,A,B$ in terms of $K,L$. They are easily checked once you have them. -T = (L^-2*K*L^-5*K*L^-5*K*L^2*K)^-2*L, -A = L^-3*(L^-2*K*L^-5*K*L^-5*K*L^2*K)^-2, -B = (L*K*L^-2*K*L^5*K*L^5*K*L^2*L*K*L^-2*K*L^5*K*L^5*K*L^2*L^-1)^5*L^2*K*L^4*(L^-2*K*L^-5*K*L^-5*K*L^2*K)^2*T^-5*L^2*K*L^4*(L^-2*K*L^-5*K*L^-5*K*L^2*K)^-2*L^-4*K*L^-2. -Then a coset enumeration shows almost instantly that the index of the subgroup generated by $T,A,B$ in the group defined by Bender's presentation is 6. -The fact that the index is 6 in the original problem does not actually depend on the fact that this is a complete presentation of ${\rm Sp}(4,\mathbb{Z})$ - only that the relations of the presentation are satisfied in ${\rm Sp}(4,\mathbb{Z})$.<|endoftext|> -TITLE: Billiard dynamics for multiple balls -QUESTION [24 upvotes]: I am interested to learn to what extent results on billiards -in polygons have been extended to multiple balls. -Assume the balls have equal radii and the same mass, -the same initial speed, and all -collisions are perfectly elastic, as depicted in this* image: - -           - - -To be specific, let me start with just the square table. -For single particle billiards, it is well known -that (1) a trajectory of rational slope that avoids the corners -is periodic, and (2) a trajectory of irrational slope that avoids -the corners will be "uniformly distributed" in the sense -that it spends equal times in equal areas. - -Are there analogous results for billiard systems of $n>1$ balls - within a square? - -Perhaps it is necessary to make some assumption concerning -the size of the ball radii and the box dimensions? - -           - - -The literature I've seen on billiard dynamics does not -explore this territory. Likely there are results in the literature, -in which case pointers would be appreciated. Thanks! - -*This impressive Wikipedia image by A.Greg was once the "Picture of the Day." - -Addendum. Following Steve Huntsman's keyword hint, I retrieved Hard Ball Systems and the Lorentz Gas, -which indeed contains many fascinating results. I will mention two: - - In a chapter by Murphy & Cohen, an easy but pleasing result: For $n$ hard spheres moving in unbounded Euclidean -space, with any combination of masses, there exist initial conditions such that -$\binom{n}{2}$ collisions occur. -In a chapter by Burago, Ferleger, & Krononenko, a difficult-to-establish bound: The maximal number of collisions that -may occur for $n$ balls moving in a simply connected Riemannian space of nonpositive -sectional curvature never exceeds -$$(400 n^2 \max{/}\min)^{2 n^4} \;,$$ -where $\max{/}\min$ is the largest ratio of masses. Note there is no dependence on the radii. -Matters are (or were in 2000) unclear without the nonpositive curvature assumption. - -REPLY [4 votes]: The literature on hard sphere interactions is indeed vast, though I think it is appropriate to single out Simanyi's recent completion of the proof of the Boltzmann-Sinai ergodic hypothsis: Nonlinearity 26 1703 (2013). -In my view this question specifically addresses multi-particle billiards in a fixed container, about which much less is known. Chernov's paper "The work of Dmitry Dolgopyat on physical models with moving particles" mentions only three results: Simanyi showed in 1999 that two identical particles in a recangular box is ergodic in any dimension. Chernov and Dolgopyat investigate a two-particle model of Brownian motion (ie with one heavy and one light particle) in a 193-page volume of the Memoirs of the AMS in 2009. The question of ergodicity of more than two particles in a rectangular box is open. -The third result is Bunimovich et al, Commun Math Phys 146 357-396 (1992) which considers many particles in a dispersing container (ie each part of the boundary is concave) which can touch each other but are each confined to their own region. Models of this type have been taken up much later in the physics literature to study heat conduction, see Gaspard and Gilbert, Phys. Rev. Lett. 101 020601 (2008). -A few other, mostly numerical, papers on few-particle billiards are: Chaos 16 013129 (2006), Chaos 18 013127 (2008), Chaos 23 013123 (2013). These are interesting models and results, but I think it's fair to say that the study of two (let alone more) particle billiards is still in its infancy.<|endoftext|> -TITLE: k rational points and base change -QUESTION [6 upvotes]: This could be a tricky question but could help me to better understand these very interesting things. -Let $X$ be an algebraic variety over a field $k$ (in the sense of a k-scheme like in Qing Liu), $K$ an extension of $k$ and define the set $X(K)$ to be the set of morphism of $k$-schemes from $Spec K$ to $X$. -Now is well-known and clear that $X(k)$ are precisely the points in $X$ for which $k(x) = k$. Then one can see that there is a bijection from $X(K)$ to $X_K (K)$ (the base change). -The problem is that someone told me (and I've also found it in Qing Liu's book) that in general $X(K)$ is not the set point such that $k(x) \subset K$. -If I consider $X_K$ as a $K$-scheme then I can use the identification before to get that these points are precisely the one with $k(x) = K$ and corrispond bijectively with $X(K)$. -So there should be an example of a $k$-scheme X where I pick a point $x \in X(K)$ such that $k(x)$ is not contained in $K$ and if I consider the corrispondent point $y$ in $X_K(K)$ via the bijection I'll get a point with $k(y) = K$. -Could someone give some hints to find explicitly this example? -Thank you so much - -REPLY [7 votes]: Let $X$ be a $k$-scheme and $K/k$ be a field extension. Then $X(K)$ can be identified with the pairs $(x,h)$, where $x \in X$ and $h : k(x) \to K$ is a $k$-homomorphism. Remark that $h$ is not uniquely determined and in general it makes no sense at all to ask whether $k(x) \subseteq K$ or not, because these fields don't lie in some common bigger field.<|endoftext|> -TITLE: Which $H$--groups satisfy the rigidity property of abelian varieties? -QUESTION [5 upvotes]: Let us call a group object $G$ in a category $\mathcal C$ rigid, if it has the following property: For every group object $X$ in $\mathcal C$, every morphism $G\to X$ in $\mathcal C$ respecting the unit sections is already a morphism of group objects. -As a formal consequence, rigid group objects are commutative (because the inversion is a group object morphism). -This definition is motivated by the following fact: If $\mathcal C$ is the category of algebraic varieties over $\mathbb C$ (say), then the rigid group objects in $\mathcal C$ are precisely the complex abelian varieties. Indeed, that abelian varieties enjoy the rigidity property is a "standard fact", and the converse follows quickly from Chevalley's structure theorem (every algebraic group is an extension of an abelian variety by an affine group). -If $\mathcal C$ is the category of sets or of hausdorff topological spaces, there are no interesting rigid group objects. I am puzzled with the case where $\mathcal C$ is the category whose objects are the topological spaces which are homotopy equivalent to CW-complexes and morphisms are continuous maps up to homotopy. A group object in this category is commonly called $H$--group. - -Is it true that the rigid $H$--groups are exactly the products of circles? - -REPLY [4 votes]: Even the circle is not rigid in this sense! With such a weak notion of "group up to homotopy" you can make a multiplication on $G=K(\mathbb Z,1)\times K(\mathbb Z,2)$ such that there is no nontrivial homomorphism from $K(\mathbb Z,1)$. -Edit: Let me explain and generalize, showing that a point is the only example. -For $n>0$ and $F$ a field, take the product $G=K(F,n)\times K(F,2n)$ and define a multiplication $\mu:G\times G\to G$ as follows. Let $a\in H^n(G;F)$ and $b\in H^{2n}(G;F)$ be the canonical elements so that maps $f:X\to G$ correspond bijectively with pairs $(f^*a\in H^n(X;F),f^*b\in H^{2n}(X;F))$. Define $\mu$ by -$\mu^*a=a\times 1+1\times a$ -$\mu^*b=b\times 1+1\times b +a\times a$. -This is unital. It is associative up to homotopy since both $\mu\circ (\mu\times 1)$ and $\mu\circ (1\times \mu)$ pull back $a$ to -$a\times 1\times 1+1\times a\times 1+1\times 1\times a$ -and pull back $b$ to -$b\times 1\times 1+1\times b\times 1+1\times 1\times b+a\times a\times 1+a\times 1\times a+1\times a\times a$. -An inverse map exists because the shearing map $(x,y)\mapsto (x,\mu(x,y))$ is a homotopy equivalence. -Now if $(X,m:X\times X\to X)$ is any group up to homotopy and $\alpha\in H^n(X;F)$ any element, then the map $f:X\to G$ given by $f^*a=\alpha$ and $f^*b=0$ cannot be a homomorphism unless $\alpha=0$, since $\mu\circ (f\times f)$ pulls back $b$ to $f^*b\times 1+1\times f^*b+f^*a\times f^*a=\alpha\times \alpha$ while $f\circ m$ pulls back $b$ to $0$. -So $X$ cannot have any cohomology at all in positive dimensions, for any coefficient field. Since $\pi_1(X)$ must be abelian, this makes $X$ homotopically discrete, which as Neil pointed out makes it trivial.<|endoftext|> -TITLE: Deceptively short proof of Regev's $A \otimes B$ theorem -QUESTION [11 upvotes]: The following theorem, due to Regev, is one of the cornerstones of the theory of PI algebras (i.e., associative algebras satisfying a nontrivial polynomial identity): -Let $A$, $B$ be two PI algebras over a field $K$. Then their tensor product $A \otimes_K B$ is PI. -Consider the following "proof" of this theorem. Since $A$ and $B$ are PI, their Jacobson radicals $J(A)$ and $J(B)$ are nilpotent, and $A/J(A)$ and $B/J(B)$ are semisimple PI algebras which are known to be embedded into matrix algebras over a commutative ring, say $M_n(C)$ and $M_m(D)$. Now, $J(A) \otimes B + A \otimes J(B)$ is a nilpotent ideal of $A \otimes B$, quotient by which is isomorphic to $A/J(A) \otimes B/J(B)$ and hence is embedded into $M_n(C) \otimes M_m(D)$, which, in its turn, is embedded into $M_{nm} (C \otimes D)$. Therefore, $A \otimes B$ contains a nilpotent ideal quotient by which is embedded into a matrix algebra over a commutative ring, and hence is PI. -Regev's theorem is a relatively difficult result, first conjectured by Jacobson, and having resisted attempts by a few mathematicians. Thus it hardly admits such a short simple proof. Where is the catch? -The only weak spot a can think of, is that nilpotence of the Jacobson radical of a PI algebra is a relatively new (at least proved long after Regev's theorem in 1970) complicated result whose proof probably involves appeal to Regev's theorem. Is it true? Or am I missing something else? -Edit May 26, 28 2011: As was pointed out by Bugs Bunny, we should require that $A$ and $B$ are finitely generated, as this is the hypothesis of the Razmyslov-Kemer-Brown theorem about nilpotence of the Jacobson radical of a PI algebra (and the theorem does not hold for infinitely-generated algebras). But, the general statement of Regev's theorem obviously reduces to this case. - -REPLY [2 votes]: I spoke about this with Louis Rowen (who, among other, wrote a few books on the subject) and here is what I got from this conversation: - -There is no circular dependency in this argument. -The Razmyslov-Kemer-Brown theorem is more difficult and complicated result than Regev's theorem, so there is no much point to infer the latter from the former.<|endoftext|> -TITLE: Asymptotic Methods in Combinatorics -QUESTION [11 upvotes]: What are good texts to acquaint oneself with standard asymptotic techniques, particularly as they relate to probabilistic combinatorics? - -REPLY [5 votes]: If you want to know about quantities which (1) have nice generating functions and (2) depend on more than one parameter, the most thorough guide will be found in the papers of Robin Pemantle. to the best of my knowledge, he hasn't written a comprehensive guide to his work; his SIAM report might be the best starting point.<|endoftext|> -TITLE: Is the etale fundamental group of Spec(Z)\{p_1,...,p_n} finitely presented? -QUESTION [14 upvotes]: (of course not, it's usually uncountable; I really mean is it the profinite completion of a finitely presented group). -By definition, $\pi_1^{\operatorname{et}}(\operatorname{Spec}(\mathbb Z)\setminus\{p_1,\ldots,p_n\})=\operatorname{Gal}(K/\mathbb Q)$ where $K$ is the maximal extension of $\mathbb Q$ unramified away from $\{p_1,\ldots,p_n\}$. In the standard analogy due to Mazur, this should be thought of as the fundamental group of a $3$-manifold (perhaps $\mathbb S^3$) minus some link $L=K_{p_1}\cup\cdots\cup K_{p_n}$. Now, we know from topology that for any link $L$ in any $3$-manifold $M$, $\pi_1(M\setminus L)$ is finitely presented, however the proof involves lots of "cutting" up of the manifold into simplices, and this type of argument seems unlikely to apply in the case of $\operatorname{Spec}(\mathbb Z)$. -My question is about $\pi_1^{\operatorname{et}}(\operatorname{Spec}(\mathbb Z)\setminus\{p_1,\ldots,p_n\})$: is it the profinite completion of a finitely presented group? What if we replace $\mathbb Q$ with a number field? -I guess I could really just forget about the primes, and ask the same question for $\pi_1^{\operatorname{et}}(\operatorname{Spec}\mathcal O_F)$ for any number field $F$ (it's just that this is trivial when $F=\mathbb Q$, though otherwise probably not). -Also, what types of consequences does/would this have, if true? - -REPLY [9 votes]: The number theorists call this $G_S$, the Galois group of the maximal extension of a number field $k$ unramified outside a finite set of primes $S$. It is known that this group can be topologically generated by a finite number of conjugacy classes (see Neukirch, Schmidt, Wingberg, Cohomology of Number Fields, 10.9.11). -In other words, there exists a finite subset $T$ of $G_S$ such that the only normal closed subgroup of $G_S$ containing $T$ is $G_S$ itself. -The absolute Galois group of Q is not finitely generated, but it is not known whether $G_S$ is finitely generated. According to the discussion ibid. p.532, it is not even known what to expect. The maximal pro-$p$-quotient of $G_S$ is finitely generated for every prime $p$. [This corrects my earlier answer.]<|endoftext|> -TITLE: F→E→B bundle with B,E,F hyperbolic: possible? -QUESTION [12 upvotes]: It would be interesting to me obtain an answer to the following easy to state question: - -Does there exist a (smooth) fibre bundle $\pi\colon E\rightarrow B$ with typical fibre $F$ such that $E$, $B$ and $F$ are hyperbolic (connected, closed) manifolds? - -I am not familiar with hyperbolic manifolds. From a colleague I have learned that "Thurston gave a necessary and sufficient criterion for a surface bundle over the circle to be hyperbolic," which is a similar question (see http://en.wikipedia.org/wiki/Hyperbolic_3-manifold for the quoted part) and could be a sign that this is not an easy question. Nevertheless I would like to ask it. - -REPLY [15 votes]: As Ryan points out, the interesting case is when the fiber is 2-dimensional. As Igor points out, this is a difficult open problem when the fiber has dimension 2. -When the fiber is a surface $F$, the fundamental group of the base $B$ admits a representation into the mapping class group $\mathrm{Mod}(F)$ of $F$. A combination of theorems of Farb-Mosher and Hamenstaedt shows that the fundamental group of the bundle is $\delta$-hyperbolic if and only if the map $\pi_1(B) \to \mathrm{Mod}(F)$ has finite kernel and the image is ``convex cocompact." (A subgroup of $\mathrm{Mod}(F)$ is convex cocompact if it acts on the Teichmueller space of $F$ with quasiconvex orbits.) -The only known examples of convex cocompact subgroups of mapping class groups are all virtually free, and so we are pretty far from knowing if there is a bona fide hyperbolic example like you are asking about. As Igor says, we don't even know if there is a $\delta$-hyperbolic surface-by-surface group. -However, it is conjectured that there is no hyperbolic (meaning truly hyperbolic) surface bundle $E$ over a surface. The reasoning is as follows: by an argument of Thurston, $E$ is symplectic. Using this fact, you can show that $E$ has a finite cover with nonvanishing Seiberg-Witten invariants. On the other hand, it is conjectured that the Seiberg-Witten invariants of a hyperbolic 4-manifold vanish. (See the article "Surface subgroups of mapping class groups" by Alan Reid in "Problems on Mapping Class Groups and Related Topics" edited by Benson Farb.) -For an introduction to convex cocompactness, and more references, see the survey "Subgroups of mapping class groups from the geometrical viewpoint" by me and Leininger. -Added: From the coarse perspective, the case where $F$ and $B$ are surfaces is key. If there were a $\delta$-hyperbolic $E$ with $F$ a surface and $B$ a hyperbolic manifold, then there is a $\delta$-hyperbolic $E'$ with fiber $F$ and base $B'$ a surface. To see this, note that recent work of Kahn-Markovic shows that $\pi_1(B)$ contains a quasiconvex surface group (I believe their theorem holds in all dimensions), and pulling back the bundle to this surface gives an example.<|endoftext|> -TITLE: Can non-homeomorphic spaces have homeomorphic squares? -QUESTION [30 upvotes]: I an wondering if there are non-homeomorphic spaces $X$ and $Y$ such that $X^2$ is homeomorphic to $Y^2$. - -REPLY [45 votes]: Here is an extract from MR0562824 (81d:54005), Trnková, V. Homeomorphisms of products of spaces. (Russian) Uspekhi Mat. Nauk 34 (1979), no. 6(210), 124–138: -S. Ulam raised the following question in 1933: Is there a space $X$ which has nonhomeomorphic square roots, i.e., $X\cong A\times A\cong B\times B$ for some nonhomeomorphic $A,B$? This problem was solved by R. H. Fox in 1947: he constructed two nonhomeomorphic four-dimensional manifolds $A$ and $B$ such that $A\times A\cong B\times B$. -upd: The reference is Fox, R. H. On a problem of S. Ulam concerning Cartesian products. Fund. Math. 34, (1947). 278–287. -The answer to Ulam's question for 3-manifolds is positive as well, see Glimm, James -Two Cartesian products which are Euclidean spaces. Bull. Soc. Math. France 88 1960 131–135. -The answer for 2-polyhedra is negative, see W. Rosicki, "On a problem of S. Ulam concerning Cartesian squares of 2-dimensional polyhedra.", Fund. Math. 127 (1987), no. 2, 101–125. This paper also gives the following elementary example: -Take $A$ to be the disjoint union of the Hilbert cube and $\mathbb{N}$ and $B$ to be the disjoint union of two copies of the HIlbert cube and $\mathbb{N}$. Then both $A^2$ and $B^2$ are homeomorphic to the disjoint union of a countable family of Hilbert cubes and $\mathbb{N}$. -Finally, in this example one can replace the Hilbert cube by any space homeomorphic to its square and not homeomorphic to two copies of itself, e.g., by $\left\{1/n\mid n\in\mathbb{Z}_{>0} \right\}\cup\{0\}$. - -REPLY [24 votes]: Yes. Let $M$ be the Whitehead Manifold, which has the property that $M \not\cong \mathbb{R}^3$, but $M\times\mathbb{R}^3 \cong \mathbb{R}^6$. (In fact $M\times\mathbb{R} \cong \mathbb{R}^4$.) Let -$$ -X \;=\; \mathbb{R}^3 \:\uplus\: M \:\uplus\: M \:\uplus\: M \:\uplus\: M \:\uplus\: \cdots -$$ -and -$$ -Y \;=\; \mathbb{R}^3 \:\uplus\: \mathbb{R}^3 \:\uplus\: M \:\uplus\: M \:\uplus\: M \:\uplus\: \cdots\text{,} -$$ -where $\uplus$ denotes the disjoint union of topological spaces. Then $X$ and $Y$ are not homeomorphic, but -$$ -X^2 \;\cong\; Y^2 \;\cong\; (\mathbb{R}^6 \:\uplus\: \mathbb{R}^6 \:\uplus\: \cdots) \:\uplus\: (M^2 \:\uplus\: M^2 \:\uplus\: \cdots). -$$<|endoftext|> -TITLE: Does recent work of Woodin clash with an older result in Descriptive Set Theory? -QUESTION [5 upvotes]: Background/Motivation -First time posting here, so I give the motivation for the question. -Early on in Descriptive Set Theory Sierpinski proved every - ${\Sigma}^1_2$ set (PCA set in the older nomenclature) is the union of ${\aleph}_1$ Borel sets. Trivial if we assume the Continuum Hypothesis (use singletons!), in a not-CH context it is essentially a result about "How bad can they be?" -An easy corollary is that such sets can only have cardinality that is countable, - ${\aleph}_1$, or that of the continuum. -Around 1970, Solovay sharpened the corollary result to show that if a measurable cardinal exists, such sets enjoy the continuum hypothesis (indeed they have the standard "regularity" properties). Note that Sierpinski's original result stands unimpeached by this. -Then about 1975, D. A. Martin showed every - ${\Sigma}^1_3$ set is the union of - ${\aleph}_2$ Borel sets, again assuming a measurable cardinal. --- -Yet lately I have been reading that Hugh Woodin has changed has opinion about the truth of the CH (now believing it is true), AND is working toward an "Ultimate L" model which admits large cardinals. Would not such a position undercut Martin's result, or am I missing something? - -REPLY [4 votes]: What is the problem? Large cardinals are consistent with CH. -This does not require looking at Ultimate L. But large cardinals are also -consistent with failures of CH. And if you are in a model where the continuum is bigger than $\aleph_2$, only then Martin's result gives you the information that every $\Sigma_3^1$ set is the union of fewer than $2^{\aleph_0}$ Borel sets. -As long as $2^{\aleph_0}\leq\aleph_2$, every set is the union of (not more than) $\aleph_2$ Borel sets. - -Edit: As Andres Caicedo points out in his comment, Martin's result actually say more than just "Assuming a measurable cardinal, every $\Sigma_3^1$-set is the union of $\aleph_2$ Borel sets" and provides nontrivial information even when CH holds.<|endoftext|> -TITLE: Which nonlinear PDEs are of interest to algebraic geometers and why? -QUESTION [50 upvotes]: Motivation - -I have recently started thinking about the interrelations among algebraic geometry and nonlinear PDEs. It is well known that the methods and ideas of algebraic geometry have lead to a number of important achievements in the study of PDEs, suffice it to mention the construction of finite-gap solutions to integrable PDEs (see e.g. this book) and the geometric approach to PDEs developed by A.M. Vinogradov et al. which revolves around the concept of diffiety (the word itself was merged from "differential" and "variety") and which the authors themselves consider, at least to some extent, as a "translation" of ideas from algebraic geometry and commutative algebra to the realm of PDEs, see e.g. these two books. -On the other hand, it appears (as far as my googling skills allow me to tell) that the other way around, i.e., in the applications of nonlinear PDEs in algebraic geometry, the interaction is at least somewhat less intense. I was able to come up with basically just two things: the Novikov conjecture (proved by Shiota) -on the relation of the KP equation to the Schottky problem and the applications of the Monge-Ampere equations in Kahler geometry. - -Question - -Which are the other applications of the nonlinear PDEs in (broadly understood) algebraic geometry? In other words, which nonlinear PDEs are of interest to algebraic geometers and why? -EDIT: It should be obvious, but to play it safe I would like to spell this out loud and clear: please feel free to share not only the already known cases where PDEs have helped the algebraic geometers but also the more open-problem-type cases where, say, there is a PDE that could be of use in algebraic geometry but some crucial bits of information about this PDE (for instance, about the existence of solution(s) with the desired properties) are still missing. - -REPLY [4 votes]: I give two reason to show that nonlinear PDE is of interest to algebraic geometers. -First: Monge-Ampere equation is nonlinear PDE and is of interest to algebraic geometers, Here is one of examples -In fact if $X$ be a projective variety and canonical Ring $$\bigoplus_{m\geq 0}H^0(X,K_X^{m})$$ -is finitely generated and Kodaira dimension is positive then -$$\pi:X\to X_{can}=proj(\bigoplus_{m\geq 0}H^0(X,K_X^{m}))$$ -gives a canonical metric which is twisted by Weil-Petersson metric on $X$ via -$$Ric(g_{can})=-g_{can}+g_{WP}$$ -Which $g_{WP}$ corresponds to moduli space of Calabi-Yau fibers. -In fact we can write twisted Kahler Einstein metric in the form of Monge-Ampere equation by using Yau-Vafa semi Ricci flat metric. - -Second: is the concept of surgery. We believe that surgery in PDE gives - surgery in algebraic geometry and vise versa. Surgery in algebraic - geometry here means flips and flops. -Surgery in PDE<===>Surgery in Geometry<===> Surgery in Algebraic - geometry - -In fact in the proof of Poincare conjecture Perelman used Surgery in PDE<===>Surgery in Geometry , he took PDE surgery by using Ricci flow -and for proof of finding canonical metrics for varieties which do not have definite first chern class (long standing conjecture) we need to use -Surgery in PDE<===> Surgery in Algebraic Geometry -albiet Surgery in Algebraic Geometry may not exists in general.<|endoftext|> -TITLE: Riemann mapping theorem for homeomorphisms -QUESTION [24 upvotes]: How do you prove to any two simply-connected domains in the plane are homeomorphic without using the Riemann mapping theorem? An elementary proof would be nice. - -REPLY [9 votes]: Let me try to solve it completely elementary. My aim is to represent simply-connected domain $\Omega$ as a union $\cup_{i=1}^{\infty} P_i$, where $P_1,P_2,\dots$ are (closed) polygons and $P_i\subset {\rm Int}(P_{i+1})$ ($\rm Int$ is for interior). If such representation is obtained, then we may construct homeomorphisms $f_k:P_k\rightarrow \{z:|z|\leq 1-2^{-k}\}$, with $f_k(\partial P_k)=\{z:|z|=1-2^{-k}\}$, and $f_{k+1}$ extends $f_k$. For $x\in \Omega$ define $f(x)=f_k(x)$ for all $k$ such that $x\in P_k$, we get a homeomorphism from $\Omega$ to the open unit disc $\{z<1\}$. -How to construct our system of polygons? Enumerate all diadic squares $[a\cdot 2^n,(a+1)\cdot 2^n]\times [b\cdot 2^n,(b+1)\cdot 2^n]$, $a,b,n$ are integer, containing in $\Omega$: $S_1,S_2,\dots$. Define $P_1=S_1$. Next, if polygons $P_1,\dots,P_{n-1}$ are constructed, consider the square $S_n$ and denote $F=P_{n-1}\cup S_n$. If necessary, add to $F$ finite number of dyadic squares so that it becomes connected. It may contain holes, but since $\Omega$ is simply connected everything in holes belongs to $\Omega$, add all this to $F$. Finally, add several small dyadic cells to $F$ so that $P_{n-1}\subset {\rm Int}(F)$ and put $P_n:=F$. Our construction garantees that $\cup P_i=\cup S_i=\Omega$ as desired.<|endoftext|> -TITLE: Dissecting a square -QUESTION [6 upvotes]: Edited - some comments may now be out-of-date. -I thought I had a complete set of solutions to this: -Cut a square into identical pieces so -that they all touch the center point. - -It became clear after some discussions that I was very, very wrong. -There are infinite families of solutions, and a sporadic. So I have -two questions: - -What do you think is a complete set of solutions? -What techniques and approaches can I use to prove that the ones -I have are all there are? - -Hope that's clearer. Thanks. - -REPLY [2 votes]: The number of solutions is the maximal possible, namely $2^{2^{\aleph_0}}$. But we need to be specific about some definitions. To begin, what does it mean to cut into identical pieces? There seems to be agreement on this point that it means to partition the square into finitely many pieces each of which can be rotated into the other. The other definition I will assume is that "touching the center point" means having that point in the closure of each set. We also need to assume that we actually have a partition of the square minus the centre point, because otherwise there is no solution except for the square itself because the centre point can belong to only one member of the partition. -Given all of this, let $[(p^0_\xi,p^1_\xi)]_{\xi\in 2^{\aleph_0}}$ enumerate all pairs of points in the square symmetric about the central point. For any function $f:2^{\aleph_0}\to 2$ let $X(f)$ - be the set of all $p_\xi^{f(\xi)}$ and $Y(f)$ the complement. This is a partition of the square into two pieces as desired, and there are $2^{2^{\aleph_0}}$ such partitions.<|endoftext|> -TITLE: On injectivity of Galois representation -QUESTION [8 upvotes]: As we know, the big galois group $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ acts on the $l$-power torsion points of an elliptic curve over Q for some prime $l$, and defines a representation in a natural way. My question is, if we consider the product of such representations for all elliptic curves defined over Q, can we get an faithful one? - -REPLY [8 votes]: No, this isn't possible : it would imply that $\operatorname{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ injects into the group $\mathrm{GL}_2(\mathbf{Z}_{\ell})^{\mathbf{N}}$. But $\operatorname{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ contains pro-$p$-groups for $p \neq \ell$, while $\mathrm{GL}_2(\mathbf{Z}_{\ell})^{\mathbf{N}}$ does not.<|endoftext|> -TITLE: The half-life of a theorem, or Arnold's principle at work -QUESTION [104 upvotes]: Suppose you prove a theorem, and then sleep well at night knowing that future generations will remember your name in conjunction with the great advance in human wisdom. In fact, sadly, it seems that someone will publish the same (or almost the same) thing $n \ll \infty$ years later. I was wondering about what examples of this people might have. Here are two: -Bill Thurston had remarked in the late seventies that Andre'ev's theorem implies the Circle Packing Theorem. The same result was proved half a century earlier by Koebe (so the theorem is now known as the Koebe-Andre'ev-Thurston Circle Packing Theorem). However, in the book -Croft, Hallard T.(4-CAMBP); Falconer, Kenneth J.(4-BRST); Guy, Richard K.(3-CALG) -Unsolved problems in geometry. -Problem Books in Mathematics. Unsolved Problems in Intuitive Mathematics, II. Springer-Verlag, New York, 1991. xvi+198 pp. ISBN: 0-387-97506-3 -the question of existence of mid-scribed polyhedron (which is obviously equivalent to the existence of circle packing) with the prescribed combinatorics is listed as an open problem. -Another example: In the early 2000s, I noticed that every element in ${\frak A}_n$ is actually a commutator, and Henry Cejtin and I proved this in - -arXiv:math/0303036 [pdf, ps, other] -A property of alternating groups -Henry Cejtin, Igor Rivin -Subjects: Group Theory (math.GR) - -However, this result was already published by O. Ore a few years earlier: -Ore, Oystein -Some remarks on commutators. -Proc. Amer. Math. Soc. 2, (1951). 307–314. -But that's not all: in D. Husemoller's thesis, published as: -Husemoller, Dale H. -Ramified coverings of Riemann surfaces. -Duke Math. J. 29 1962 167–174. -Only a few years after Ore's paper, this result is reproved (by Andy Gleason) -- this is actually the key result of the paper. -Another example (which actually inspired me to ask the question): -If you look at the comments to -(un)decidability in matrix groups -, you will find a result proved by S. Humphries in the 1980s reproved by other people in the 2000s (and I believe there are other proofs in between). -It would be interesting to have a list of such occurrences (hopefully made less frequent by the existence of MO). - -REPLY [4 votes]: Here is a very nice example. The abstract of the article "On planarity of compact, locally connected, metric spaces", by R. Bruce Richter, Brendan Rooney and Carsten Thomassen, starts as follows. - -Independently, Claytor [Ann. Math. 35 (1934), 809–835] and Thomassen - [Combinatorica 24 (2004), 699–718] proved that a 2-connected, compact, - locally connected metric space is homeomorphic to a subset of the - sphere if and only if it does not contain $K_5$ or $K_{3,3}$. - -I thought that was quite a witty and honest way of describing this particular case of the phenomenon in question.<|endoftext|> -TITLE: Open problems with monetary rewards -QUESTION [104 upvotes]: Since the old days, many mathematicians have been attaching monetary rewards to problems they admit are difficult. Their reasons could be to draw other mathematicians' attention, to express their belief in the magnitude of the difficulty of the problem, to challenge others, "to elevate in the consciousness of the general public the fact that in mathematics, the frontier is still open and abounds in important unsolved problems.1", etc. -Current major instances are - -The Millennium Prize Problems -Beal's conjecture - -Other problems with money rewards - -Kimberling's list of problems - -Question: What others are there? To put some order into the answers, let's put a threshold prize money of 100 USD. I expect there are more mathematicians who have tucked problems in their web-pages with some prizes. -What this question does not intend to achieve: - -once offered but then collected or withdrawn offers -new pledges of sums of money just here - -P.S. Some may be interested in the psychological aspects of money rewards. However, to keep the question focused, I hope this topic won't be ignited here. One more, I understand that mathematicians do not work merely for money. - -REPLY [4 votes]: Domokos and Várkonyi offer $\\\$\frac{1,000,000}{F + E + V − 2}$ for the polyhedral Gömböc with the minimal number of flat faces, where $F,E,V$ are the number of faces, edges and vertices of the polyhedron.<|endoftext|> -TITLE: Is there a fusion category with an object which does not commute with its dual? -QUESTION [8 upvotes]: Does there exist a fusion category with an object $X$ such that $XX^*\ncong X^*X$ (where the isomorphism need not be natural in any way)? -Feel free to add adjectives such as pivotal, spherical, unitary, etc. - -REPLY [10 votes]: The principal even part of extended Haagerup gives a counterexample. Look at the table in the appendix to our paper http://arxiv.org/pdf/0909.4099 (joint with Stephen Bigelow, Scott Morrison, and Emily Peters) to see that the objects labelled A and B are dual to each other but AB=1+P while BA=1+Q (or maybe the other way around, I'm having trouble remembering our conventions for whether the principal graph is left multiplication or right multiplification).<|endoftext|> -TITLE: Sheaves with isomorphic cohomology, but not quasi-isomorphic -QUESTION [6 upvotes]: Suppose I have two (constructible) sheaves of vector spaces $F$ and $G$ over the same base space that have isomorphic cohomology (degree by degree), but no sheaf map inducing this isomorphism (i.e. they are not quasi-isomorphic). -Now imagine I apply any of Grothendieck's 6 operations (or other functors) to $F$ and $G$, will the resulting sheaves, say $\Psi(F)$ and $\Psi(G)$, have isomorphic cohomology as well? -Thanks! -EDIT: I suppose the answer in general is no. Consider an injective (acyclic) sheaf and any functor that doesn't preserve injectives. A concrete example would be nice though. -EDIT 2: Thanks to Algori's comments I need to substantially limit the sheaves of interest. Assume additionally that the sheaves have identical support and furthermore that when we take the function that assigns to a point the euler characteristic of the stalk over that point (thereby producing a constructible function) to these two sheaves, these functions are identical. If we are using real sub-analytic partitions then a theorem of Kashiwara's says that the Grothendieck group of the bounded derived category of real constructible sheaves is isomorphic to the group of constructible functions. If I understand things correctly this would mean that these two sheaves are equivalent in the Grothendieck group (which seems weird since this function ignores gluing data). I am dealing with particular examples of two constructible (cellular) sheaves over a simplicial complex, which are not isomorphic yet produce identical constructible functions. - -REPLY [4 votes]: Then how about this example: -Let $X$ be the wedge of two circles. To give a local system on $X$ with stalk a complex vector space $V$ is to give two matrices $A_1,A_2\in GL(V)$. If the subgroup $\langle A_1,A_2\rangle$ of $GL(V)$ generated by $A_1,A_2$ has no invariants in $V$, then $H^0$ of the resulting local system will be $0$ and $H^1$ will be of rank $\dim V$ for Euler characteristic reasons. -To give a map in $Sh(X)$ between the local systems that correspond to $(V,A_1,A_2)$ and $(W,B_1,B_2)$ is to give a linear map $f:V\to W$ such that $f A_1=B_1f,f A_2=B_2f$. Now suppose that $V=W$, that both the groups $\langle A_1,A_2\rangle$ and $\langle B_1,B_2\rangle$ have no invariants in $V$, that $A_1$ is diagonalizable and that the eigenvalues of $A_1$ and $B_1$ are pairwise distinct. Then any map between the local systems is zero.<|endoftext|> -TITLE: $H_2$ of a simply connected Lie group vanishes -QUESTION [8 upvotes]: How do I show that the $H_2$ of a simply connected Lie group vanishes? (I don't want to use that $\pi_2(Lie group) = 0$, since this is what I want to prove. And I don't want to use the classification of compact simply-connected Lie groups.) - -REPLY [9 votes]: Start by deformation-retracting $G$ to its maximal compact, so that $G/T$ will be a flag manifold with a Bruhat decomposition, obtainable by Morse theory as in Agol's answer. -If what you want anyway is the $\pi_2$, consider the long exact sequence on -homotopy of $T \to G \to G/T$, giving -$$ \ldots \to \pi_2(T) \to \pi_2(G) \to \pi_2(G/T) \to \pi_1(T) -\to \pi_1(G) \to \ldots -$$ -The first obviously vanishes, and the last does too by assumption. -We know $\pi_2(G/T)$ is free of dimension $\leq rank(G)$ by the -Bruhat decomposition, so the surjection $\pi_2(G/T) \to \pi_1(T)$ -is between free abelian groups of the same finite rank. Hence it's -also injective. So its kernel $\pi_2(G)$ is zero.<|endoftext|> -TITLE: Discrete groups G whose full C*-algebra C*(G) is not quasidiagonal? -QUESTION [15 upvotes]: Is there a known example of a countable discrete group G whose full group C*-algebra C*(G) is not quasidiagonal? -Let us recall that a separable C*-algebra A is quasidiagonal if it admits a faithful -*-representation $\pi:A \to L(H)$ on a separable Hilbert space with the property - that there is an increasing sequence $(p_n)$ of finite rank projections in $L(H)$ which converges strongly to $1_H$ -and such that for each $a\in A$, $\lim_{n\to \infty}\|[\pi(a)p_n-p_n\pi(a)]\|=0$. -$G=SL_3(\mathbb{Z})$ is believed to be a candidate. Bekka has shown -that $C^*(SL_3(\mathbb{Z}))$ is not residually finite dimensional. - -REPLY [5 votes]: I don't know much about the subject, but isn't this an example? There is a discrete property (T) group $G$ without nontrivial finite dimensional representations [2, Remark in the last page]. Then, its max C*-algebra contains a non-QD algebra as its direct summand, namely $C^*(G)(1-z)$ for the Kazhdan projection $z$ (corresponding to the trivial rep) [1, Corollary 17.2.6]. -[1] N. P. Brown and N. Ozawa. -algebras and finite-dimensional approximations, volume 88 of Graduate Studies in Mathematics. American Mathematical Society, Providence, RI, 2008. -[2] P. de la Harpe, A. G. Robertson, and A. Valette. On the spectrum of the sum of generators for a finitely generated group. Israel J. Math., 81(1-2):65–96, 1993.<|endoftext|> -TITLE: Number of triples of roots (of a simply-laced root system) which sum to zero -QUESTION [21 upvotes]: In a paper 1105.5073, the authors took a simply-laced root system $\Delta$ of type $G=A,D,E$, and then counted the number of unordered triples $(\alpha,\beta,\gamma)$ of roots which sum to zero: $\alpha+\beta+\gamma=0$. -They found that there are $rh(h-2)/3$ such triples, where $r$ is the rank of $G$ and $h$ is the Coxeter number of $G$. -Of course we can show this by studying the simply-laced root systems one-by-one (which is what the authors did.) -My question is if there is a nicer way to show this without resorting to a case-by-case analysis, using the general property of the Coxeter element, etc. -update -Let me add a bit of background info why string theorists care about this. -In string theory, there are things called D-branes. If $N$ D-branes are put on top of each other, we have $U(N)$ gauge fields on top of it, and there are of order $\sim N^2$ degrees of freedom on it. There are slightly more involved constructions which give us gauge fields with arbitrary simple group $G$. Then the number of degrees of freedom is $\dim G$. -In M-theory, there are things called M5-branes. If $N$ M5-branes are put on top of each other, we do not know what is on top of it. But there is an indirect way to calculate how many degrees of freedom there should be; and the conclusion is that there should be $\sim N^3$ degrees of freedom (hep-th/9808060). It is one of the big unsolved problem in string/M theory to understand what mathematical structure gives rise to this $N^3$ degrees of freedom. -We can call $N$ M5-branes as the $SU(N)$ version of the construction. There are slightly more involved constructions as in the case of D-branes, but it's believed that there are only simply-laced ones. These are the so-called "6d $\mathcal{N}=(2,0)$" theory which plays an important role in physical approach to geometric Langlands correspondence, etc. See Witten's review. -There are various stringy arguments why there are only simply-laced variants in 6d, but I like this one by Henningson the best. Anyway, the number of degrees of freedom for D and E cases was calculated in this one and this one; it turned out to be given by $h_G \dim G /3$. (As there is only simply-laced ones, there's no distinction between the dual Coxeter number and the Coxeter number.) -This product of the dimension and the (dual) Coxeter number was obtained in a very indirect way, and we'd like to know more. Bolognesi and Lee came up with an interesting numerological idea in the paper cited at the beginning of this question. -There idea is to think of $h_G\dim G$ as -$h_G\dim G/3 = hr + h(h-2)r/3 = $ (number of roots) + (number of unordered triples of roots which sum to zero). -They propose that the first term corresponds to "strings" labeled by roots, and the second term as the "junctions of three strings". To consistently connect three strings at a point, the roots labeling them need to sum to zero. -(They didn't write this way in their paper; they use more "physical" language. I "translated" it when I made the original question here.) This makes a little bit more precise the vague idea that three-string-junctions are important in M5-branes. -I found this fact on the simply-laced root systems quite intriguing, and therefore I wondered a way to prove it nicely. - -REPLY [7 votes]: There is another approach by the strange formula of Freudenthal and de Vries, which states that $h^\vee d = 12 \rho^2$ where the Weyl vector $\rho$ is one half of the sum of positive roots. The long roots have the length square two. Thus -$$\frac13 h^\vee d = \sum_{\alpha>0} \alpha^2+ \sum_{\alpha,\beta>0,\alpha\neq\beta}\alpha\cdot\beta $$ For simple-laced group, the first term of RHS is the number of roots $hr=d-r$, and the second term of RHS should be $rh(h-2)/3$. For the non-simple-laced case, there may be still some interesting interpretation of the above decomposition.<|endoftext|> -TITLE: Is there a version of Seiberg-Witten-Floer or Heegard-Floer homology for 3-manifolds with boundary? -QUESTION [5 upvotes]: Recently, the Seiberg-Witten-Floer homology created by Kronheimer and Mrowka has important applications in Taubes' proofs of Weinstein conjecture and Arnold Chord Conjecture. Also, Cagatay Kutluhan, Yi-Jen Lee, and Clifford Taubes have a series of papers on the arxiv proving the equivalence of Heegard-Floer and Seiberg-Witten-Floer homologies. However, these are only defined for closed 3-manifolds. -My question is if there exists any version of HF or SWFH defined for 3-manifolds with boundary? - -REPLY [2 votes]: A different kind of answer is provided by Juhász's sutured Floer homology, usually denoted with $SFH$. In brief, $SFH$ is an invariant of a pair $(M,\Gamma)$, where $M$ is a manifold with nonempty boundary $\Gamma$ is a collection of curves (called sutures) $\partial M$ satisfying certain conditions; the complex and its differential are very similar to the ones used to compute $\widehat{HF}$. -A good reference for the construction of $SFH$ is Juhász's paper Holomorphic discs and sutured manifolds; there are some applications to contact topology, for example Honda, Kazez and Matic's papers The contact invariant in sutured Floer homology and Contact structures, sutured Floer homology and TQFT. -Let me just make a couple of remarks: - -Bordered Floer homology depends on some choice on the boundary, too (namely, a parametrisation $F \simeq \partial M$). -Bordered Floer homology determines sutured Floer homology for any choice of the sutures.<|endoftext|> -TITLE: Geometric meaning of derivatives of the curvature -QUESTION [8 upvotes]: Let $\Gamma$ be a sufficiently smooth, closed, convex curve in the plane parametrized by arclength $s$. Further assume that $\kappa>0$ and that the second derivative of $\kappa$ with respect to $s$ (denoted $\kappa''$) is such that -\begin{equation} -(\kappa''\cdot \kappa^{-3})(p)< C -\end{equation} -for every point $p\in\Gamma$ at which $\kappa$ attains a global minimum (for some absolute constant $C<\infty$). -What can be said about the curve $\Gamma$? In particular, is the quantity $\kappa''\cdot \kappa^{-3}$ of geometric significance and has it been considered before? Can it for instance be expressed in terms of the (centro-)affine curvature of $\Gamma$? References would be most appreciated! -Motivation comes from trying to reduce several local problems involving the restriction of the Fourier transform to $\Gamma$ to the corresponding (more tractable) problem for an appropriate osculating conic. -Thank you. - -REPLY [12 votes]: This is not really an answer, but it's too long for a comment, so I'm posting it this way. One way in which the expression you are considering has appeared in recent years is via the heat equation shrinking plane curves. (See, for example, Gage and Hamilton, The heat equation shrinking convex plane curves, J. Differential Geometry 23 (1986), 69–96. They show that, in this case, one has -$$ -\kappa_t = \kappa'' + \kappa^3 = \kappa^3\bigl(1 + \kappa''/\kappa^3\bigr). -$$ -Whether this might be helpful to you, I don't know. It does appear to say that your inequality says something about how fast the curve will move under the heat shrinking flow. -As for a relation with centro-affine curvature, yes, this can be worked out. For abstract reasons, there is a polynomial relation among the quantities $\kappa$, $\kappa'$, $\kappa''$, $\lambda$, $\dot \lambda$, and $\ddot\lambda$, where $\lambda$ is the centro-affine curvature of the curve and the overset dot represents the derivative with respect to the centro-affine arclength $d\sigma$. I don't know a reference for this; it's classical (but messy). It starts with the easily-derived (and well-known) relation $\kappa (ds)^3 = \lambda (d\sigma)^3$ between the differentials and goes on from there. Actually, I think that the relation is better expressed between the quantities $\mu = \kappa^{1/3}$ and $\nu = \lambda^{1/3}$, but even this formula, when you work it out, isn't so nice. -Afterthought: It occurred to me that you put the 'centro' in parentheses because you might be interested in the affine curvature and the affine arclength, too. Of course, the affine arclength is $\kappa^{1/3}ds$, and the affine curvature $\tau$ is given by the formula -$$ -\tau = \frac{\kappa\kappa'' + 3\kappa^4 - \tfrac{5}{3} (\kappa')^2}{3\kappa^{8/3}} . -$$ -Thus, it's not quite what you want. As for a reference, you could consult any good differential geometry book that discusses affine geometry.<|endoftext|> -TITLE: Pattern avoiding permutations and zig-zags -QUESTION [13 upvotes]: In a recent talk (in fact today, 26 May 2011) at the W80 conference celebrating the 80th birthday of Herbert Wilf http://www.cargo.wlu.ca/W80/, Doron Zeilberger gave a talk on pattern avoiding permutations. Given a permutation $\sigma \in \mathfrak{S}_n$, the symmetric group on $n%$ letters, we say $\sigma$ avoids a pattern $\tau$ if no substring of $\sigma$is configured in the order of $\tau$. For example, if $\tau = 123$, then $35421$ avoids $123$ since no substring of $a_1 a_2 a_3$ of $35421$ satisfies $a_1 < a_2 < a_3$. On the contrary, $13254$ does not avoid $123$ since for $a_1 = 1, a_2 = 2, a_3 = 5$ we have $a_1a_2a_3$hits the pattern $123$. Zeilberger mentioned that this is an enormously difficult problem even for simple patterns like short cycles. Explicit answers are known for the pattern $1432$ and $1342$, but Zeilberger claimed that enumerating the permutations avoiding $1324$ is an incredibly difficult problem that will still be completely unknown in 200 years (he also claimed that in 200 years the Riemann Hypothesis and the P vs. NP problem will both be exercises in undergraduate textbooks, as to illustrate the relative difficulty of the problem). -I remarked to him that $1324$ is a zig-zag pattern, namely a pattern $a_1a_2\cdots a_n$ such that $a_1 < a_2, a_2 > a_3, a_3 < a_4, \cdots$ I asked if the fact that $1324$ is a zig-zag pattern is what makes it difficulty. He couldn't answer; but thought the remark was worthy of pursuit. -And so I turn the question back to MO: Does anyone have any results on avoiding any zig-zag patterns? If that is not available, does anyone have a heuristic as to why enumerating permutations avoiding zig-zag patterns would be difficult, or is it just completely unknown? - -REPLY [13 votes]: I tend to think that it is more unusual when we can count permutations that avoid a given pattern than when we can't. -Consider the case of patterns of length 4. Thanks to the work of Stankova [4] and Backelin, West, and Xin [1], we know that there are essentially three different cases: avoiding 1234, avoiding 1342, and avoiding 1324. The permutations that avoid 1234 correspond to standard Young tableau with at most 3 rows, which allowed Gessel [3] to count them. Note that even in this "nice" case, the generating function is not algebraic, which does not bode well for other patterns. In the 1342 case, there was another fortunate accident in that the "indecomposable" 1342-avoiding permutations are in bijection with "β(0,1) trees", which Bóna [2] proved and used to enumerate this class. As for 1324, it appears that there simply isn't a corresponding happy coincidence. -The case for patterns of length 5 or more is even worse. To the best of my knowledge, the only patterns for which we have formulas are the monotone patterns and the patterns equivalent to monotone patterns. In short, it seems that counting pattern avoiding permutations is always hard, except in a very small number of cases. -[1] Backelin, Jörgen; West, Julian; Xin, Guoce (2007), "Wilf-equivalence for singleton classes", Adv. In Appl. Math. 38 (2): 133–149, doi:10.1016/j.aam.2004.11.006, MR2290807. -[2] Bóna, Miklós (1997), "Exact enumeration of 1342-avoiding permutations: a close link with labeled trees and planar maps", J. Combin. Theory Ser. A 80 (2): 257–272, doi:10.1006/jcta.1997.2800, MR1485138. -[3] Gessel, Ira M. (1990), "Symmetric functions and P-recursiveness", J. Combin. Theory Ser. A 53 (2): 257–285, doi:10.1016/0097-3165(90)90060-A, MR1041448. -[4] Stankova, Zvezdelina (1994), "Forbidden subsequences", Discrete Math. 132 (1–3): 291–316, doi:10.1016/0012-365X(94)90242-9, MR1297387.<|endoftext|> -TITLE: Nonstandard Reals in the Complex Plane -QUESTION [16 upvotes]: [Edited so as to reflect answers and comments given so far] -Let $F$ be a real closed field. Then by Artin-Schreier theory, $F[i]$ is algebraically closed. If $F$ is further assumed to have cardinality at most continnum, then by a classical theorem of Steinitz [stating the isomorphism of any two uncountable algebraically closed fields of the same cardinality and characteristic], we can conclude that $F[i]$ is isomorphic to a subfield of $\Bbb{C}$ of complex numbers. -In particular, if $F$ is a non-archimedean real closed field of cardinality continuum, then $F[i]$ is isomorphic to $\Bbb{C}$ [The proof uses the axiom of choice in a serious way, by the way]. -We can therefore conclude: -Theorem. Every nonarchimedean real closed field of power at most continuum is isomorphic to a subfield of $\Bbb{C}$. -As a special case, we may conclude that there is a subfield $F$ of $\Bbb{C}$ such that $F$ is a non-archimdean real closed field that, furthermore, has a subfield isomorphic to the field $\Bbb{R}$ of real numbers. -The above considerations allow me to state my questions. -Questions. -(a) [UNANSWERED] Is there an uncountable Borel non-Archimedean real closed field $F$ of $\Bbb{C}$? -NOTE: In his comment below Dave Marker asks whether this question has a negative answer if we further assume that $F[i]=\Bbb{C}$, then Gerald Edgar pointed out in his comment that this is indeed the case; based on a result that appears in a joint paper of his with Chris Miller. -(b) [ANSWERED] Is it possible for an uncountable such $F$ to be at least Lebesgue measurable ?* -NOTE. (b) has been answered. First Martin Goldstern pointed out that (b) follows from $MA + \lnot CH$; and that (b) is also true in any universe of set theory obtained by adding a Cohen real. Then Gerald Edgar pointed out that (b) is provable outright in $ZFC$ [see the answers below]. -(c) [UNANSWERED] If the answer to (a) is positive, does the answer change if we insist for $F$ to have a subfield isomorphic to $\Bbb{R}$. - -REPLY [4 votes]: Is there a Cantor set $K$ in $\mathbb R$ of algebraically independent elements? -Thinning it out, if necessary, we can assume $K^n$ has Hausdorff dimension $0$ for all $n$, and there are still other things algebraically independent of it. Then $F_1 = \mathbb Q(K)$ is at least an analytic subfield, still of dimension zero, and its algebraic completion $F_2$ is again analytic, now isomorphic to $\mathbb C$, but not all of $\mathbb C$ and not equal to $\mathbb R$. So $F_2$ is still of Hausdorff dimension $0$. Of course it has a subfield $F_3$ isomorphic to $\mathbb R$. (Probably many different ones?) So if we now add something algebraically independent of that, specifying that it is infinitely large, say, compared to $F_3$, then can't we now do a real-completion to get something still analytic and nonarchimedean real closed?<|endoftext|> -TITLE: Appendices to papers authored by others -QUESTION [18 upvotes]: Every now and then, one sees in mathematical papers appendices, authored by person(s) different from the authors of the main body of the paper. - -What is the rationale behind such appendices? In most of the cases I have seen, it can be a bona fide short separate paper (granted, tight closely to the main paper in question, but many papers are tight closely to each other). -What is the complete list of authors of a paper with an appendix? I belong to the school of thought claiming that references should be given according to the first letters of authors names. Each time, when citing a paper with appendix, I am agonizing over whether the authors of the appendix should be included or not in this abbreviation. More seriously, if a person is a (co)author of such an appendix (but not of the main body of the paper), how it is reflected in his list of publications? -My impression is that the number of such appendices has been increased drastically in the recent years. Is it true? If yes, what may be reason(s) for that? - -Edit May 27, 2011: DamienC made an interesting (and looking plausible to me) suggestion that the purported recent increase of separately-authored appendices is a way to cope with no longer fully adequate convention of authors ordering. If so, it would be interesting to compare with how the things evolved, say, in life sciences, from the situation where coauthors were not acceptable - a situation, as far as I understand, common to all scientific papers at the beginning of XX century - till today ugly state of affairs with convoluted and self-contradictory rules who goes first, who goes last, how the authors should be clustered, etc. This may suggest that in mathematics things evolve in a different direction, which, though not perfect, is sort of reassuring. -2nd edit May 27, 28 2011. My question is not about appendices in general, including differences in various subjects and subcultures (for example, in statistics it is common to defer all proofs to appendices) -- this is an entirely different topic. My question is about the sutuation when the set of the authors of an appenix is different from the set of the authors of the main paper -- and, bibliographic (and bibliometric, and social, if you wish) difficulties (or at least what I see as bibliographic, etc. difficulties) arising from such situations. Sorry if that was not clear enough from the first place. - -REPLY [13 votes]: I surmise that the speed of communication now exceed the speed of publication (or even the speed of the typesetting/proofreading cycle). So, you prove something, send the rough draft to people who, in your opinion, may be interested, start preparing it for publication, and in the middle somebody answers one of the questions in your paper. The best thing is if his answer makes your paper obsolete. Then you throw it to the garbage and waste no more time on the boring writing routine. But more often than not the answer is given to some not so important question that doesn't really change much but still adds an interesting twist. To publish it separately would be ridiculous. To ignore it would be strange. So what to do? Well, it becomes an "appendix". Now, the gentleman's way is to offer the full co-authorship. Also the gentleman's way is to reject it. Now we are stuck with either an extended thanks, or a separate "appendix authorship", and here goes. I believe we'll get even weirder things like "This paper supercedes the work of X who improved the result of Y, which was a generalization of theorem by Z proved in responce to my question (all 4 private communication)" if the speed goes up a bit more. I personally am in favor of pushing the gas pedal to the floor and shaking the current ideas of priority and authorship off entirely but I know that my mathematical opinions are very non-orthodox...<|endoftext|> -TITLE: Derived categories of (coherent) sheaves of modules: exceptional images, gluing, and proper descent? -QUESTION [12 upvotes]: I am interested in the properties of (the derived categories) of various categories of (coherent) sheaves of modules (over varieties). I would like to understand in what extent these properties are similar to those of (etale) constructible sheaves and what are the differences. I have looked at Derived categories of coherent sheaves: suggested references? -but I still do not know the answers to the following questions: -1. Do the categories in question become 'very bad' if the base variety is not proper or is not smooth? -2. Are there (derived) exceptional images (i.e. $Rf^!$ and $Rf_!$) defined in this setting? -3. Could one 'glue' somehow (some version) of the derived category of sheaves of modules over $X$ from those over $Z$ and over $X\setminus Z$ ($Z$ is a closed subvariety of $X$)? -4. Does there exist a version of proper descent for this setting? -Any comments or references would be very welcome! - -REPLY [5 votes]: This is just a slight addition to previous answers of Chris and Anatoly. -There are some very beautiful, more-recent-than-the-classical-references papers of Neeman and Lipman-Neeman on (roughly) the subject of question 2. Warning though: in the first (Neeman only) paper, $f^!$ means the right adjoint of $\mathbb{R}f_*$, which won't agree with the $f^!$ defined e.g. in Deligne's appendix to Hartshorne's "Residues and Duality" without a properness hypothesis. [By the way, this appendix is a classical reference.] -Also, as in Anatoly's answer, you shouldn't ask any questions in this area at the abelian level. :-) And indeed, without some additional hypotheses you will have to be more careful about which derived category you want to work in. EDIT: If you are interested mainly in a right adjoint to $\mathbb{R}f_*$ then quasicompact quasiseparated will do; this is the setting of Neeman/Lipman-Neeman. Thanks to Anatoly's help, though, I have finally noticed that the original post doesn't seem to indicate caring about this, so perhaps the observation is a useless one to make!<|endoftext|> -TITLE: Which Turing machines accept the language of trivial words in a finitely presented group? -QUESTION [11 upvotes]: Let $G$ be a finitely presented group with generators $g_1, g_1^{-1},\ldots, g_n, g_n^{-1}$. Let $L(G)$ be the language of all those words in $g_1, \ldots, g_n$ which represent the trivial element of $G$. It's well known that there exists a Turing machine $T$ which accepts $L(G)$ (it doesn't necessary always stop). -Conversely, given an alphabet $A$ consisting of symbols $g_1, g_1^{-1}, \ldots, g_n, g_n^{-1}$, and a language $L$ on $A$ accepted by a Turing machine $T$ it's easy to give neccesary and sufficient conditions on $L$ so that for some group $G$ we have $L=L(G)$. Namely $L$ should be closed under (1) concatanation (2) reductions and additions of the terms $g_ig_i^{-1}$ and $g_i^{-1}g_i$, (3) "conjugation" i.e. given $w\in L$ the words $gwg^{-1}$ and $g^{-1}wg$ are also in $L$. - -Question 1. Is there a set of conditions on a Turing machine $T$ which assures that the language $L(T)$ accepted by $T$ fulfills the conditions (1)-(3) above? - -For the purpose of this question "a set of conditions" means an algorithm which always stops, which takes as the input a Turing machine $T$, and if $L(T)$ fulfills (1)-(3) then the algorithm returns YES (if it returns NO then it can be either way). -Of course I'm interested in algorithms which output YES on a possibly big set of Turing machines. - -Question 2. Is there an algorithm as above which returns YES exactly on the set of those machines $T$ such that $L(T)$ fulfills conditions (1)-(3). - -REPLY [2 votes]: Further to Derek Holt's answer, the negative answer would appear to still apply to the conditions stated in his answer. In fact, you can't even decide if a PDA (let alone a TM) accepts the word problem of a group, as stated in the introduction to: -Stephen R. Laken and Richard M. Thomas, Space Complexity and Word Problems of Groups, Groups-Complexity-Cryptology Volume 1 (2009), No. 2, 261-273 -The proof is simple and relies on the well known fact that one can't decide whether the language a PDA accepts over an alphabet $\Sigma$ is equal to $\Sigma^{*}$<|endoftext|> -TITLE: Why isn't the perfect closure separable? -QUESTION [12 upvotes]: This question has escalated from math.stackexchange. I'm doing this because it has been a while since the question has been open, receiving no satisfactory answers, even when subject to a bounty. I hope it is adequate for MO, I apologize if it is not. -Let $F\subset K$ be an algebraic extension of fields. By taking the separable closure $K_s$, we obtain a tower $F\subset K_s \subset K$ such that $F\subset K_s$ is separable and $K_s\subset K$ is purely inseparable. -Wikipedia, following Isaacs, Algebra, A Graduate Course p.301, says: - -On the other hand, an arbitrary algebraic extension $F\subset K$ may not possess an intermediate extension $E$ that is purely inseparable over $F$ and over which $K$ is separable. - -The question is: why? And more explicitly, had I not seen this this soon, I would surely have conjectured that the perfect closure, $K_p$, which satisfies $F\subset K_p$ purely inseparable, also satisfied $K_p\subset K$ separable... But why doesn't it? -To clarify, I'm looking both for a counterexample and for intuition regarding the impediment of the situation to be symmetrical. -ADDED: After posting the link to this answer on Math.SE, Georges Elencwajg kindly answered there also, providing further intuition on this subject. - -REPLY [16 votes]: A counterexample -Take as ground field $F=\mathbb F_2(u,v)$ and consider the polynomial $f(X)=X^6+uvX^2+u\in F[X]$. This polynomial is irreducible by Eisenstein. Let $F\subset K$ be the extension obtained by adjoining a root $a$ of $f(X)$ to F, so that $K=F[a]$, $[K:F]=6$ and $f(a)=0$ .The element $a^2\in K$ has minimal polynomial over $F$ the separable (because of degree $3$) polynomial $g(X)=X^3+uvX+u$ . -From this follows that the separable closure of $F$ inside $K$ is $F^{sep}=F[a^2]$ and that the extension $F^{sep}=F[a^2] \subset K=F[a]$ is purely inseparable, as expected. -And now comes the surprise: there are no elements in $K$ purely inseparable over $F$ !(except the elements of $F$, of course) and so $K$ is not separable over the purely inseparable closure $F^{perf}=F$ of $F$. The proof that an element of $r\in K$ purely inseparable over $F$ belongs to $F$ is easy, once you realize that such an element must satisfy $r^2\in F$ . [Expand $r$ in the $F$-basis $a^i, 0\leq i\leq5$ and calculate $r^2$ (easy in characteristic $2$ !), remembering that $f(a)=0$.You will see that $r$ was already in $F$.] -(Edit: Kevin's comment and link have reminded that I had already given analogous examples in characteristic $p>2$, and completely forgotten about them! The two answers now cover all positive characteristics.) -A positive result -If $F\subset K$ is normal , then indeed $K$ is separable over $F^{perf}$, the extensions -$F^{sep}$ and $F^{perf}$ are linearly disjoint over $F$ and the canonical map - $F^{sep} \otimes_F F^{perf} \to K$ is an isomorphism. (As an aside observe that this is one of the rare cases where the tensor product of two fields is a field)<|endoftext|> -TITLE: Density of a set of natural numbers whose differences are not bounded. -QUESTION [11 upvotes]: I guess this is quite standard and probably easy for experts or young lovers of number theory. -For $A\subseteq\mathbb N$, denote by $d^+(A)$ its upper density, which is -$$ -d^+(A)=\lim\sup_{n\rightarrow\infty}\frac{|A\cap[1,n]|}{n} -$$ -Now let $A=(a_n)$ be an increasing sequence of natural numbers and let $d_n$ be the $n$-th difference $d_n=a_{n}-a_{n-1}$. - -Question: Suppose that $d_n$ is not bounded, is it true that $d^+(A)=0$? - -More generally: if $d_n$ is not bounded, is it true that any (additive) invariant mean on $\mathbb N$ takes value $0$ over $A$? -Thanks in advance, -Valerio - -REPLY [11 votes]: Remove from $\mathbb N$ the subset $\cup_{n=1}^\infty \lbrace n!,n!+1,\dots,n!+n\rbrace$. -The result has density $1$ and arbitrarily large gaps.<|endoftext|> -TITLE: Fourier transforms of characteristic functions -QUESTION [8 upvotes]: I am wondering how badly summable the Fourier transform of the characteristic function of a measurable subset of $S^1$ can be. - -Question: Let $\alpha \colon \mathbb N \to [1,\infty)$ be a monotone increasing function with $\lim_{n \to \infty} \alpha(n) = \infty$. Is there a measurable subset $E \subset S^1$, such that - $$\sum_{n \in \mathbb Z} | \widehat \chi_E(n)|^2 \cdot \alpha(|n|) = \infty \ ?$$ - Here, $\widehat \chi(n)$ are the usual moments $$\widehat \chi_E(n):= \int_E z^n \ dz.$$ - -The only example I know is the Fourier transform of the characteristic function of an interval, which grows like $1/n$. On the other hand, one can easily see that the growth cannot be better than $1/n$ (something like $1/n^{1 + \varepsilon}$), since $\ell^1 \mathbb Z \subset C(S^1)$. -More concretely: - -Question: Can anyone compute the growth of the Fourier transform of the characteristic function of something like a Cantor set of non-zero measure? - -Again, more abstractly: - -Question: What can be said about the growth of the Fourier transform of the characteristic function of a generic subset of $S^1$? - -REPLY [2 votes]: It is a difficut problem to compute the Fourier transform of the characteristic function of a union of open intervals (in the general case) and it is known that such Fourier transform can converge to 0 with a very slow growth rate. I am currently working on that problem. Probably the growth rate may depend on some arithmetical properties of the boundary of the set as it is the case in studying the Fourier dimension of sets.<|endoftext|> -TITLE: The cone of positive semidefinite matrices is self-dual? (reference needed) -QUESTION [17 upvotes]: I'm seeking a reference for the following fact. - -The cone of positive semidefinite matrices is self-dual (a.k.a. self-polar). - -This result is relatively easy to prove, has been known for a long time, and is fundamental to things like semidefinite programming. Ideally, I would like a reference that reflects all three of those properties. Unfortunately, the properties themselves make it hard to find a good reference to cite. (Many sources I've looked at consider this result elementary and well-known enough to simply state without proof or reference. That was sort of my plan as well, but a referee is now asking for a reference, and seeing as how our paper is outside of optimization theory, I think that's probably reasonable.) -By the way, this result is occasionally referred to as Fejer's Trace Theorem, although I have never encountered an actual reference to any publication of Fejer. So if anyone knows the source of this attribution, that would be interesting. -Any help would be greatly appreciated! - -REPLY [6 votes]: I know you are looking for a reference and you probably know how to prove it (and that the post is old). However, I want to include a short form of the proof for those coming to the post for this reason. -Note that -$$\def\<{\langle}\def\>{\rangle}\DeclareMathOperator{\tr}{tr} -\=\tr(XY)=\sum_{i,j=1}^n\lambda_i^X \lambda_j^Y\^2,$$ -where $\lambda_i^X$ is the the $i$-th eigenvalue of $X$ corresponding to the (normalized) eigenvector $v_i^X$ (and equivalently for $Y$). Consider the eigenvectors to be choosen as an orthonormal system. From this, it is easy to see $\\ge 0$ for positive semi-definite matrices $X$ and $Y$. However, if e.g. $X$ is not positive semi-definite but has a negative eigenvalue, say $\lambda_1^X<0$, then we can choose -$$Y := v_1^X (v_1^X)^\top \in \mathbf S^n_+.$$ -This gives $\<0$. And this is exactly what proves that $\mathbf S^n_+$ is self-dual: -$$X\in\mathbf S^n_+ \qquad\Longleftrightarrow \qquad \\ge 0\text{ for all $Y\in\mathbf S^n_+$}.$$<|endoftext|> -TITLE: Drinfeld's equivalence of quantized function algebras and quantized universal enveloping algebras -QUESTION [8 upvotes]: In his 1986 ICM address, Drinfeld discusses a way of producing a quantized function algebra (or more precisely a quantized formal series Hopf algebra) from a quantized universal enveloping algebra -- the construction appears at the end of section 7. -Are there any other discussions about this construction in the literature? - -REPLY [8 votes]: Fabio Gavarini has many papers on this result sketch by Drinfeld, and some extensions of it. -In The quantum duality principle he proves Drinfeld's statement in details. Later on Gavarini proved a global version, and with Nicolas Ciccoli he got a version for subgroups and homogeneous spaces.<|endoftext|> -TITLE: A Pascal's-triangle -like random process -QUESTION [8 upvotes]: I was exploring Pascal's triangle on a cylinder when I encountered this puzzle-like problem. -It is surely elementary, but perhaps weekend-entertaining. -Start with a permutation of $(1,2,3, \ldots, n)$, call it $(x_1,x_2,x_3,\ldots,x_n)$. -Perform two operations in sequence: Sum, and Shuffle. -Sum computes a new sequence $x'_i = (x_i + x_{i+1})$, with cylindrical index wraparound. -(This is the Pascal's-triangle aspect.) -Shuffle permutes the resulting numbers (below I used a random permutation). -Then the Sum/Shuffle-transformations are repeated. -Here is an example, for $n=4$: - -       - -I hope the process is clear without further formal definitions. -The numbers in the sequence grow by powers of 2, but approach the same normalized ratio, -and so in that sense "approach" equality. -One could ask many questions (I'll resist, although I welcome speculation!), but this is perhaps the primary question: - -For which $n$ could the described process lead to a sequence of identical numbers? - -For $n=2$, $(1,2)$ is the only option, which ends in $(3,3)$ after just one step. -For $n=3$, all starting permutations lead to a permutation of $(3,4,5)$, which leads to -a permutation of $(7,8,9)$, which leads to a permutation of $(15,16,17)$, and -then to $(31,32,33)$, and so on. This is all I know: $n=2$ must end in a constant -sequence, $n=3$ cannot, and $n=4$ might. -Addendum. After Douglas Zare's (nice!) analysis answered the question -for all but $n = 0 \pmod 4$, -I concentrated on $n=8$ and found these two examples: -$$ -\left( -\begin{array}{cccccccc} - 7 & 2 & 8 & 1 & 5 & 4 & 6 & 3 - \\ - 10 & 9 & 10 & 9 & 6 & 9 & 10 - & 9 \\ - 9 & 6 & 9 & 10 & 9 & 10 & 9 & - 10 \\ - 15 & 15 & 19 & 19 & 19 & 19 & - 19 & 19 \\ - 19 & 19 & 15 & 19 & 19 & 19 & - 19 & 15 \\ - 34 & 38 & 34 & 34 & 38 & 38 & - 38 & 34 \\ - 34 & 38 & 34 & 38 & 34 & 38 & - 34 & 38 \\ - 72 & 72 & 72 & 72 & 72 & 72 & - 72 & 72 -\end{array} -\right) -$$ -$$ -\left( -\begin{array}{cccccccc} - 4 & 3 & 6 & 1 & 8 & 7 & 2 & 5 - \\ - 9 & 7 & 9 & 7 & 9 & 15 & 9 & - 7 \\ - 9 & 7 & 9 & 15 & 9 & 7 & 9 & - 7 \\ - 16 & 16 & 24 & 24 & 16 & 16 & - 16 & 16 \\ - 24 & 16 & 16 & 16 & 24 & 16 & - 16 & 16 \\ - 40 & 40 & 32 & 32 & 40 & 40 & - 32 & 32 \\ - 40 & 32 & 40 & 32 & 40 & 32 & - 40 & 32 \\ - 72 & 72 & 72 & 72 & 72 & 72 & - 72 & 72 -\end{array} -\right) -$$ -The rows alternate between permutation and sum, i.e., -each odd row is a permutation of the even row above. - -REPLY [3 votes]: It is possible if and only if 4 divides n. Douglas Zare showed that if 4 does not divide n, then it is impossible (btw, his argument might be simpler to state mod 2 - if 4 does not divide n, then it is already impossible to get all numbers divisible by 4). Now, if n=4k, do the following: -1, 12, 2, 10, 4, 8, 6, 7, 5, 9, 3, 11. -This will give 2k-1 n's, 2k-1 n+2's and two n+1's. Next, do -12, 14, 12, 14, 12, 14, 12, 14, 12, 14, 13, 13. -This gives 4k-2 2n+2's, one 2n+3 and one 2n+1. -Now, for simplicity, subtract 2n+2 from each number. We have all zeros but one +1 and one -1. Make the permutation so that we double the number of +1's and -1's in each step, until they become >n/4, then "mix them a bit"* so that in the next step we can exactly n/2 of both. Then an alternating sequence will give all equal numbers. -$^*Edit$: As Douglas pointed out this mixing is not that clear how to do. So instead, I claim that we for any i from 1 to n/2 it is possible to get exactly i +1's and i -1's. The proof is by induction - we either double or double -1, by putting a +1 and a -1 next to each other. Eventually we can get n/2 of each and we are done. -However, this my method only works for the numbers 1, 2, .. , n. What if we start from another sequence? Mod 2 the problem is always solvable, maybe that helps, but over Z already 1, 0, .. , 0 is not clear how to solve. -Edit: As Doublas pointed out this is possible if and only if n is a power of 2. So to summarize, I think if 4 does not divide n>1, then you can not get all equal numbers starting from 1, 2, .. , n, if 4 divides n you can, while if is n>2 is a power of 2, then it seems you can start from any sequence. (Though we have not seen a full proof of this last part yet.)<|endoftext|> -TITLE: 1-related groups -QUESTION [11 upvotes]: Every 1-related group $G$ with at least 2 generators is an HNN extension of another 1-related group $G_1$ with free associated subgroups. Indeed, if the total exponent of one letter in the relator is 0, then one can take that letter as the free letter. If there are two letters $a$ and $b$ in the relator $r$ and $a$ occurs with the total exponent $m$, $b$ occurs with total exponent $n$, and, say $m\ge n>0$ (the other cases are similar), then make a substitution $a=a$, $b=b'a^{-1}, c=c,...$ (i.e. change the generating set accordingly). In the resulting 1-related presentation $\langle a',b,c,...| r(a,b'a^{-1},c...)\rangle$ the total exponent of $a$ is $m - n < m$, and the total exponents of other letters stay the same. By iterating this procedure, one can reduce the general case to the case of total exponent 0. Note that the relator of $G_1$ can be longer than the relator of $G$. I think it is almost obvious then that starting with a cyclic group, and using HNN extensions with free associated subgroups, one can get any 1-related group (and all intermediate groups will be 1-related also). -Question. Given a 1-related group $G$, what is the shortest representation of $G$ as a sequence of HNN extensions as above? Can one estimate that length from above in terms of the length of the relator? - Update I always thought that the length of the sequence should be linear in $|r|$. Ian Agol's answer below suggests a quadratic upper bound. Is there a linear upper bound? - -REPLY [7 votes]: Check out the proof of Theorem 4.1 of Joe Masters' paper. Given a 1-relator group presentation, realize the free group as the fundamental group of a compact surface (with boundary), and the relator as an immersed loop in this surface. Choose such a surface so that the self-intersection number of the loop representing the relator is minimal. Then Masters' proof shows that this self-intersection number decreases after each (non-trivial) HNN extension in the Magnus-Moldavansky hierarchy (for the base case of an embedded loop on a surface, notice that the 1-relator group is either free or a surface group if the loop is parallel to the only boundary component). I think one can at least bound this self-intersection number from above quadratically in the length of the relator, so this gives a crude estimate.<|endoftext|> -TITLE: Is the Leopoldt conjecture almost always true? -QUESTION [25 upvotes]: The famous Leopoldt conjecture asserts that for any number field $F$ and any prime $p$, the $p$-adic regulator of $F$ is nonzero. This is known to be equivalent to the vanishing of $H^2(G_{F'/F},\mathbf{Q}_p/\mathbf{Z}_p)$, where $F'$ is the maximal pro-$p$ extension of $F$ unramified outside $p$. When $F/\mathbf{Q}$ is abelian, the conjecture was proven by Brumer. -My question: is there any reasonable sense in which the Leopoldt conjecture is "usually" true - e.g., is it known for any fixed $F$ at almost all primes $p$, or (say) for almost all quartic extensions of $\mathbf{Q}$ with $p$ fixed? A glance through the mathscinet reviews of all the papers with "Leopoldt conjecture" in their title didn't reveal anything, but perhaps this is well-known to experts. - -REPLY [2 votes]: Fix $p$. Leopoldt's conjecture is that the $p$-adic regulator does not vanish. Since that is an open condition in the $p$-adic numbers, the set of fields on which the conjecture is true ought to be open. That sounds like a pretty good avatar of your question, although it doesn’t give density like the Zariski topology. Also, this is only a conjecture. -What does that even mean? Define the $p$-adic topology on the set of number fields by way of test curves, which are parameterized by finite extensions of $\mathbb Q(t)$. Each unramified rational value of $t$ gives an extension of $\mathbb Q$. So each such test curve defines a set of number fields and identifies it with a cofinite subset of $\mathbb Q$. We can transfer the $p$-adic topology on $\mathbb Q$ to the set of number fields. A subset of fields is $p$-adically open if its preimage in each test curve is open. A function of number fields is $p$-adically continuous if its pullback to each test curve is. -Is the $p$-adic regulator a continuous function of number fields? I doubt it. Consider the test curve $x^2=t$. For negative $t$ there are no units, so the regulator is 1. For positive $t$, it isn’t. So probably the test curves are too big and we should make the finite partition according to Archimedean behavior. But even after restricting to real quadratics, it is still unclear because the fundamental unit is hard to control because of the class group. -But there are some test curves for which we can control specific units to make a continuous proxy for the regulator, and thus prove that the set of points satisfying Leopoldt’s conjecture is open (though maybe empty). Specifically, the Ankeny-Brauer-Chowla family is the extension of $\mathbb Q(a_1,\ldots,a_n)$ defined by $\prod (x-a_i)=1$. Each specialization is a degree $n$ totally real field with $n-1$ independent units easily described in terms of the generator. The behavior of the extension of $\mathbb Q_p$ is locally constant, so we can locally think of the $n-1$ units as varying continuously through a single $p$-adic algebra. Thus the logarithms of the units vary continuously and so do their minors. This isn’t quite the regulator, since these units don’t necessarily generate all units, but a maximal minor avoiding zero is equivalent to Leopoldt’s conjecture, so the conjecture is open on this test variety. -I got this all from here, which also discusses a generalization (indeed, a deformation) of ABC fields to fields with imaginary places.<|endoftext|> -TITLE: Abel summation of the alternating series of primes? -QUESTION [10 upvotes]: Consider the ordinary generating function of the sequence of primes ($2+3x+5x^2+7x^3 + ...$); by the ratio test and the prime number theorem, its radius of convergence is $1$. Thus, we might well ask about the limit of its value as $x$ approaches $-1$ from above (i.e., about the Abel summation of the alternating series of primes $2 - 3 + 5 - 7 + ...$). So I do! Is this limit well-defined? Is it finite? If so, what is known about its value? (I pessimistically suspect I will be told it is actually disappointingly infinite, or even worse, that the limit diverges by oscillation, but I haven't enough experience in this sort of thing to confidently rule out more interesting possibilities...) -[I realize the alternating series of primes has partial sums with arbitrarily large magnitudes, by the existence of arbitrarily large prime gaps, so there's no hope for it to be convergent in the standard sense...] - -REPLY [5 votes]: I've a summation-method based on the triangle of Eulerian-numbers, (just for reference: -$ \qquad \qquad \tiny \begin{array} {rrrrr} - 1 & . & . & . & . & . \\\ - 1 & 0 & . & . & . & . \\\ - 1 & 1 & 0 & . & . & . \\\ - 1 & 4 & 1 & 0 & . & . \\\ - 1 & 11 & 11 & 1 & 0 & . \\\ - 1 & 26 & 66 & 26 & 1 & 0 \\\ - ... & ... & ... - \end{array} $ -Call this (infinite) matrix E. The row-sums are the factorials and scaling the rows by the reciprocal factorials make the rowsums the unit; actually we can construct a regular matrix-summation method on this. We use the column-sums of E, more precisely the dot-products of the infinite vector containing our primes with alternating signs and the reciprocal factorials (call it X) -$ \qquad \qquad \small X=\text{ [ 2 , -3 , 5/2! , -7/3! , 11/4! , -13/5! , 17/6! , -19/7! , ... ]} $ -with E in a resulting vector T, which contains then the sequence of the Eulerian-transforms -$ \qquad \qquad \small X * E = T $ -as instance of an "Eulerian transformation" (as different to the other name of "Euler-transformation" which uses the pascal matrix instead). Finally the sum of the entries in T is the acceptable value for the alternating sum of the primes under "Eulerian summation", if this sum converges. We want finally write, using the lower triangular matrix of ones D for the partial sums in the vector S : -$ \qquad \qquad \small D * T^{\tiny \text{ T}} = S $ -and if the sequence of entries in S converges, then $s =\lim_{n\to \infty} s_n $ will be our limit for the alternating sum of primes. -Because of the reciprocal factorial which is involved in the dotproducts, because of the little growth rate of the sequence of prime numbers and because of the simple composition of the eulerian numbers each of that columnwise dotproduct occurring in the multiplication $X *E$ is convergent, so each entry in T is well defined. Here are the first couple entries of T: -$ \qquad \qquad \small T = \text{[ 0.7036728 , 1.059633 , 0.9470500 , 0.006954269 , -0.9466667 , -0.6131519 , 0.1756172, 0.5145809,... ]} $ -and the partial sums from T show a nicer behave than that of the sequence of alternating signed primes. -However, this is still not conclusive - maybe with some more terms (I computed this up to n=127) one can give a more conclusive result. Here is a plot of the entries in T - -(source) -It impresses me, that this Eulerian transforms are all of small absolute value, and this let's me hope, that we might be on the right track here. -On the other hand, in the following plot of the partial sums, the (directly summed) partial sums still show oscillating behave (its summands still having alternating signs), so I've also included three variants using additional Eulersums of small orders (0.2,1 and 2) to hopefully flatten that oscillating curves down to convergence. - -(source) -However these Eulersums still does not suffice to arrive at a convergent sequence of partial sums - at least not with 128 terms only, perhaps we should look up to 256 terms or more. -[update]: After optimizing the procedures I've got the first 256 partial sums of the T-vector: - -(source) -(Well, from the pictures I begin to suspect that we need still another method, possibly fourier-series or even a completely different approach.) - -[update]: to answer on Greg's comment I compared the profile of partial sums of that summation using the Eulerian numbers and that of the common, ordinary Euler-summation. Surprise: the profile is nearly identical, only that the method using Eulerian summation arrives at the same profile using only the forth part of coefficients: - -(source)<|endoftext|> -TITLE: Categorical characterization of quasi-compact schemes -QUESTION [12 upvotes]: I would like to know if it is possible to characterize the property "quasi-compact" in the category of schemes by means of a pure categorical language. This would imply in particular that every equivalence of categories $\text{Sch} \to \text{Sch}$ preserves quasi-compact schemes. Together with Jonathan's answer here, this would answer affirmatively my question about the rigidity of the category of schemes, at least over a field $k$. -For example the property of being empty is categorical, because it just says that the scheme is initial. The terminal scheme is $\text{Spec}(\mathbb{Z})$, so this is also categorical. Further examples of categorical properties or schemes: Spectra of fields, the underlying set of a scheme (in particular surjective morphisms), connected schemes, $\text{Spec}(\mathbb{Z}_p)$, and much more, see here. The usual definition of quasi-compact involves open immersions, which are, a priori, not categorical. - -REPLY [3 votes]: (This is a paraphrased version of something I learned from BCnrd. Consequently I'm posting it as a CW answer.) -There is currently no known useful criterion for quasi-compactness from the functor of points. Nonetheless "locally of finite presentation" does have one (cf. EGA IV-8.14). -This means in particular that to check for properness via the valuative criterion (which is purely functorial, of course), the extra steps of ensuring finite type need additional work---locally of finite type can be checked thanks to the criterion just listed, but one also wants quasi-compactness. -So in practice what one often does is to take something known to be quasi-compact (e.g. a concrete scheme) and surject this onto the moduli space in question. Surjectiveness can be checked functorially (using the spectra of large algebraically closed fields). This is the main way of establishing that an Artin stack or algebraic space is quasi-compact.<|endoftext|> -TITLE: What are the applications of immanants? -QUESTION [36 upvotes]: Definitions of determinant: -$\det(A) = \sum_{\sigma \in S_n}\operatorname{sgn} \sigma\prod_{i}a_{i, \sigma(i)}$ -and permanent: -$\mathrm{per}(A) = \sum_{\sigma \in S_n}\prod_{i}a_{i, \sigma(i)}$ -admit a generalization in the form of immanant: -$\mathrm{Imm}_{\lambda}(A) = \sum_{\sigma \in S_n}\chi_{\lambda}(\sigma)\prod_{i}a_{i, \sigma(i)}$ -where $\lambda$ labels irreducible representations of $S_n$ and $\chi_{\lambda}$ is the character. Determinant and permanent are easily seen to be special cases of $\mathrm{Imm}_{\lambda}$. -While determinants are ubiquitous in mathematics and permanents also have many application, esp. in combinatorial problems, other kinds of immanants seem to be rarely used. Are there any problems where use of $\mathrm{Imm}_{\lambda}$ other than $\det$ and $\operatorname{per}$ is natural? - -REPLY [12 votes]: A problem with applying immanants is that after expressing something in terms of immanants, you might not be closer to calculating them and you can't easily manipulate expressions of immanants. For example, suppose you want to count Hamiltonian cycles in a graph (positivity solves the Hamiltonian cycle problem). You notice that this is expressible as a linear combination of immanants of the adjacency matrix: -$$\sum_{\sigma \in S_n} f(\sigma)\prod_{i}a_{i, \sigma(i)} $$ -where $f(\sigma) = 1$ when $\sigma$ is an $n$-cycle and $0$ otherwise. Since $f$ is constant on conjugacy classes, we can decompose it as a linear combination of characters. However, the Hamiltonian cycle problem is NP-complete, so it would be surprising if there were a quick way to compute this linear combination of immanants. Either there are many terms to compute, or these terms are hard to compute, or both, even if you are just trying to determine whether the result is positive.<|endoftext|> -TITLE: Accumulation of algebraic subvarieties: Near one subvariety there are many others (?) -QUESTION [27 upvotes]: Let's work over the field $\mathbb{C}$ of complex numbers, and let $X\subset \mathbb{P}^n$ be a projective variety. Let $\tilde{X}\subset \mathbb{P}^n$ be any small open neighborhood of $X$, in the complex topology. - -Question: Do there exist many algebraic varieties $\subset \tilde{X}$ of dimension equal to the dimension of $X$? - -Of course, $X$ itself lies in there, and one may apply some automorphism of $\mathbb{P}^n$ close to the identity to obtain some further examples. I would be interested in any construction of more subvarieties in $\tilde{X}$, or examples showing that in general these may not exist. Note that the degree of the subvarieties is allowed to be arbitrarily large, and that I do not want to consider only deformations of $X$ (which might be rigid). - -REPLY [20 votes]: The answer is yes (as in the case of Sasha's answer we use ramified covers) -Proof. Let $X$ be any variety in $\mathbb CP^n$. Take a section $s_m$ of $O(m)$ on $X$ such that $s_m$ is not equal to $m$-th tensor power $s^{\otimes m}$ of any section $s$ of $O(1)$ restricted on $X$. Now, let $s_m^{\frac{1}{m}}$ be the multi-section of $O(1)$ on $X$. This multi-section defines a subvarity $X_m$ in the total space of $O(1)$ on $X$, that is the cover of $X$ of degree $m$. -Finally notice that there is a family of maps from the total space of $O(1)$ on $X$ to $\mathbb CP^n$ that sends the zero section of $O(1)$ on $X$ to $X$. The image of such a map in $\mathbb CP^n$ is just the union of all lines in $\mathbb CP^n$ that join a fixed point $p$ with all points of $X$, the point $p$ itself does not belong to the image. -Then the image of $X_m$ in $\mathbb CP^n$ is the desired variety. END. -We used here the fact that $O(1)$ on $\mathbb CP^n$ can be embedded in $T\mathbb CP^n$ as a subsheaf (in various ways).<|endoftext|> -TITLE: Sylow subgroups invariant under an automorphism -QUESTION [7 upvotes]: Let $G$ be a finite group and $\sigma$ an automorphism of $G$. Suppose $p$ is a prime and $\sigma$ has prime order $q \neq p$. It's easy to see that $\sigma$ fixes a Sylow $p$-subgroup of $G$ if $\sigma$ is fixed-point free or if $q$ does not divide the order of $G$. -Suppose that $q$ does divide the order of $G$. What reasonable assumptions on $G$ or on $H$, the fixed point subgroup of $\sigma$, would we have to make to guarantee that $\sigma$ fixes a Sylow $p$-subgroup of $G$? Any ideas? Thanks. - -REPLY [6 votes]: This is a very general question, perhaps too general for a definitive answer, and as Jack Schmidt pointed out (in a now deleted comment), it is already a delicate question when $\sigma$ is an inner automorphism. There are bad examples (a little different from Jack's) for all symmetric groups of prime degree greater than $3$. If $p$ is a prime, and $G$ is the symmetric -group $S_{p}$, then a Sylow $p$-subgroup $P$ of $G$ is self centralizing of order $p$, -so $N_{G}(P)$ has order dividing $p(p-1)$. In fact the order is $p(p-1)$. Hence the only elements -of order prime to $p$ which normalize a Sylow $p$-subgroup are powers of a $p-1$-cycle. -Such elements (apart from the identity) have a unique fixed point, and all other cycles of equal -length dividing $p-1$. There are many elements of prime order $q \neq p$ in $S_p$ which are not -of this form, for example any element $\sigma$ of prime order $q$ dividing $p-2$. -Another type of example is provided by ${\rm GL}(n,p)$. If we take a prime $q$ such that -$q$ divides $p^{n}-1$ but does not divide $p^{m}-1$ for any $m -TITLE: Ramanujan's Incorrect formula -QUESTION [14 upvotes]: I actually looked at one of my Questions (posted at MATH.SE) again and found a formula which actually Ramanujan had discovered. - -Ramanujan: If $\alpha$ and $\beta$ are positive numbers such that $\alpha \cdot \beta = \pi^{2}$ then, $$\alpha \cdot \sum\limits_{n=1}^{\infty} \frac{n}{e^{2n\alpha} -1} + \beta \cdot\sum\limits_{n=1}^{\infty} \frac{n}{e^{2n\beta}-1} = \frac{\alpha+\beta}{24} -\frac{1}{4}$$ - -I actually heard that this result is not true. I would like to know where the mistake is and whether something can be rectified in this proof so that, my above problem can be summed by using this result. - -I would also like to know the Intuitive idea behind discovering such mysterious formulas. - -REPLY [44 votes]: I believe this formula is true, provided the $\alpha$ in the second sum is changed to a $\beta$, as suggested by Todd Trimble's comment. Let -$$ P(x) = \prod_{n=1}^\infty \frac{1}{1-x^n} $$ -be the generating function for the number of partitions of a non-negative integer $n$. Dedekind proved that $P$ satisfies the transformation formula -$$ \log P(e^{-2\pi t}) - \log P(e^{-2\pi /t}) = \frac{\pi}{12} \Bigl( \frac{1}{t} - t \Bigr) -+ \frac{1}{2} \log t $$ -for $t > 0$. -Differentiating this formula with respect to $t$ gives -$$ -\sum_{n=1}^\infty \frac{2\pi n}{e^{2\pi n t}-1} - \frac{1}{t^2} \sum_{n=1}^\infty \frac{2\pi n}{e^{2\pi n/t} -1} = \frac{\pi}{12} \Bigl( -\frac{1}{t^2} - 1\Bigr) + \frac{1}{2t} $$ -Now multiply through by $-t/2$ and substitute $\alpha = \pi t$, $\beta = \pi /t$ to get -$$ \sum_{n=1}^\infty \frac{\alpha n}{e^{2n\alpha}-1} + \sum_{n=1}^\infty \frac{\beta n}{e^{2n\beta}-1} = \frac{1}{24}(\beta+\alpha) - \frac{1}{4}$$ -which is Ramanujan's formula. -The transformation formula for $P$ is related to the theory of modular forms, of which -the Eisenstein series mentioned in Derek Jennings' answer to your question on math.stackexchange are important examples. Briefly, if we define -$$ \eta(\tau) = \frac{e^{2\pi i \tau/24}}{P(e^{2\pi i \tau})} = e^{2\pi i \tau/24} \prod_{n=1}^\infty (1-e^{2\pi i n \tau}), $$ -then $\eta(\tau)^{24}$ is a modular form of weight $12$. As such, $\eta$ satisfies the identity -$$ \eta(-1/\tau) = \sqrt{-i \tau}\; \eta(\tau). $$ -The transformation formula for $P$ follows by setting $\tau = it$ and taking logs.<|endoftext|> -TITLE: How nice are representation varieties of Fuchsian groups? -QUESTION [11 upvotes]: Background -Let $S_{g,n}$ be an oriented surface of genus $g$, with $n$ punctures. We explicitly prohibit the non-hyperbolic cases: - -$g=0$, $n=0,1,2$. -$g=1$, $n=0$. - -Let $\Gamma$ be the fundamental group of $S_{g,n}$; a group arising this way is called a Fuchsian group (as opposed to some authors, we don't require that $\Gamma$ comes with an embedding into $PSL_2(\mathbb{R})$). -Let $G$ be a complete reductive algebraic group over a field $k$. The representation algebra $Rep(\Gamma,G)$ is defined such that $k$-algebra maps -$$ Rep(\Gamma,G)\rightarrow R$$ -are naturally equivalent to group maps -$$ \Gamma\rightarrow G(R)$$ -The representation variety is then $X_{\Gamma,G}:=Spec(Rep(\Gamma,G))$, despite the fact that this scheme can be non-reduced and hence not really a 'variety'. -Question -For arbitrary groups $\Gamma$, the scheme $X_{\Gamma,G}$ can be quite bad. It is non-reduced $G=PSL_2(\mathbb{C})$ and for $\Gamma$ some Artinian groups (Kapovich-Millson, 1999) or for $\Gamma$ the fundamental group of some 3-manifolds (Kapovich, 2001). -For $\Gamma=\mathbb{Z}^2=\pi_{1,0}$ (one of the excluded cases), the representation variety $X_{\mathbb{Z}^2,G}$ is the commuting scheme of $G$. The reducedness of the commuting scheme is still an open question. -Despite this, it seems like it might be possible there are general theorems about nice properties of $X_{\Gamma,G}$, when $\Gamma$ is Fuchsian. For example, if $g=0$, then $\pi_{0,n}=F_{n-1}$, the free group on $n-1$ generators. Then $X_{F_{n-1},G}=G^{n-1}$, which is a smooth variety. -For $\Gamma$ Fuchsian, is it known whether $X_{\Gamma,G}$ is - -reduced? -normal? -smooth? - -I have a vested interest in their normality, but that seems like the question least likely to be addressed directly. - -REPLY [11 votes]: Answers to all questions are negative, however, $Hom(\Gamma, G)$ is smooth away from representations $\rho$ such that the centralizer of $\rho(\Gamma)$ in $G$ is finite. -(This is due to Andre Weil, "Remarks on cohomology of groups", Annals of Math., 1964, but since then it was reproven by many others in a variety of ways. For instance, Goldman has a very clear proof using Poincare duality and vanishing of $H^2$. However, if you allow Fuchsian groups with torsion instead of just surface groups, Weil's result is the best. Incidentally, there is a way to have an interesting theory of representations of fundamental groups of surfaces with boundary: You just have to consider relative representation varieties where you fix the conjugacy classes of images of boundary loops. Weil deals with these as well.) -i. There are (real) non-reduced examples: $G=PU(2,1)$ and $\rho$ is a discrete and faithful representation that lands in $PU(1,1)$, see e.g. paper of Goldman and Millson http://www2.math.umd.edu/~wmg//LocalRigidity.pdf -The key reason is that the dimension of Zariski tangent space to $Hom(\Gamma, G)$ at all these representations is larger than the actual dimension of the representation variety. On the other hand, over algebraically closed fields it will be reduced. -ii. Even if you look at the reduced scheme, both smoothness and normality will fail at the trivial representation to a nonabelian group, like $SL(2)$. The analytical germ of $Hom(\Gamma, G)$ at the trivial representation is given by: Take the vector space $Z^1(\Gamma, {\mathfrak g})$ (with trivial action of $\Gamma$ on the Lie algebra) and impose the quadratic equations given by vanishing of the cup-products $[\omega \cup \omega], \omega\in Z^1$. See -Goldman's paper "Representations of fundamental groups of surfaces", reference [5] in the paper by Goldman and Millson linked above. There is actually a much stronger result by Goldman-Millson and Simpson which describes local singularities of representation varieties of Kahler groups in terms of cup product on $H^1$. -However, it is an open problem if $Hom(\Gamma, G)$ is smooth at representations with finite centralizers, where $\Gamma$ is a Kahler group.<|endoftext|> -TITLE: Calculating the geodesic equation for a particular set of phase-space coordinates -QUESTION [8 upvotes]: Let $g$ be a Riemannian metric on the $d$-dimensional flat space $\mathbb R^d$, and consider the usual Lagrangian $$L(x, \dot x) = \tfrac 1 2 g_{ij}(x) \dot x^i \dot x^j.$$ Let $\hat g := \sqrt g$ denote the square root of the metric $g$, implicitly defined by the formula $\hat g_{ai} \hat g_{bj} \delta^{ab} = g_{ij}$, where $\delta^{ab}$ is the identity 2-tensor. I want to introduce phase-space-type coordinates $$u_i = \hat g_{ij} \dot x^j.$$ In the coordinates $(x,u)$, the metric on $u$ is just the Euclidean metric: $\langle u, u \rangle := \delta^{ij} u_i u_j$. -Let $x = x(t)$ denote a geodesic for the metric $g$, and define $u(t) := \hat g_{ij} \dot x^j$. These coordinates are convenient, because along the geodesic, $u(t)$ remains on the sphere of radius $|u(0)|$: $$\tfrac{d}{dt} \langle u, u \rangle = \tfrac{d}{dt} \delta^{ij} u_i u_j = \tfrac{d}{dt} g_{ij} \dot x^i \dot x^j = 0,$$ since geodesics are parametrized by unit speed. Conveniently, this means that $\langle u, \dot u \rangle \equiv 0$. -(You may wonder why don't I just use Hamiltonian phase-space coordinates $(x,p)$. In my research, I consider $g$ as a parameter, ranging over all possible Riemannian metrics on the plane $\mathbb R^2$. Hamiltonian coordinates have the nice property that for a fixed metric, the energy shells $\{ g^{ij}(x) p_i p_j = \mathrm{constant} \}$ are invariant under the geodesic flow. Unfortunately, these energy shells are not independent of the metric parameter $g$. In the coordinates $(x,u)$, on the other hand, the shells $\{ \langle u, u \rangle = \mathrm{constant} \}$ are just spheres in Euclidean space, and do not depend on $g$. In particular, it is important to me that these spherical shells are invariant under rotations in the phase space $\mathbb R^d \times \mathbb R^d$). -I want to calculate the geodesic equation in the coordinates $(x,u)$, particularly for the case that $d=2$. It is easy to see that $\dot x^j = \hat g^{ji} u_i$, where the superscripts denote the inverse of $\hat g$. When I calculate $\dot u$, though, I get a mess: $$\dot u_a = \big( \hat g_{ab,c} \hat g^{cj} \hat g^{bi} - \hat g_{ab} \Gamma_{uv}^b \hat g^{ui} \hat g^{vj} \big) u_i u_j,$$ where $\Gamma_{uv}^b$ are the Christoffel symbols for the metric $g$. I tried simplifying this expression, to no effect. There is plenty of symmetry around (e.g., $\langle u, \dot u \rangle = 0$), and I'm sure that the formula for $\dot u$ takes a much, much simpler form. -Question: Is there a simple expression in these coordinates for the evolution of $u(t)$? -Let me explain why the above expression is inadequate. For the metric $g$, let $U_g$ denote the vector field given by $U_g(x,u) = (u, \dot u)$ (where $\dot u$ is the expression above), so that solutions to the differential equation $(\dot x, \dot u) = U_g(x,u)$ are geodesics for the metric $g$. I need to calculate the (Euclidean) divergence $\operatorname{div} U$. I am pretty sure that in the end, $\operatorname{div} U$ can be expressed in some simple geometric quantities involving the metric (like the Riemannian divergence $\operatorname{div}_g$ of some vector field, scalar curvature $K_g$, etc.). For the messy $\dot u$ above, though, it is impossible for me to see what the true character of $\operatorname{div} U$ is. - -REPLY [4 votes]: The answer that you want, namely div $U_g$, is not going to expressible in terms of a geometrically invariant quantity (such as, say, the scalar curvature of $g$) because it depends on the underlying coordinates $x$ in which you have presented the metric. For example, if $\det g(x)$ is constant (which is a coordinate dependent thing), then the divergence of $U_g$ (computed in the $xu$-coordinates, which, I assume, is what you mean by the 'Euclidean divergence') will vanish identically (and conversely, as a matter of fact). -I'll try to explain this in the symplectic formulation, since that's the version I find the clearest, but I think that you can make the translation on your own. You start with a Riemannian metric $g = dx\cdot G(x) dx$, where $G$ is a function on $\mathbb{R}^n$ with values in positive definite symmetric matrices. Let's write $G = F^TF$, where $F$ is invertible (but not necessarily symmetric; you can take $F$ to be the positive definite square root of $G$ if you like, but that's not necessary for my argument). The Lagrangian is then $L = \tfrac12 u\cdot u$, where $u = F(x)\ dx$, regarded as an $\mathbb{R}^n$-valued function on the tangent bundle of $\mathbb{R}^n$ (i.e., an orthonormal coframing of the underlying manifold). Applying the Legendre tranform, the symplectic form on the cotangent bundle pulls back to the tangent bundle to be -$$ -\Omega = dp\wedge dx = d((F(x)\ u)^T) \wedge dx - = (du)^T\ F(x)^T \wedge dx + u^T\ d(F(x)^T)\wedge dx -$$ -The vector field $U_g$ is the $\Omega$-Hamiltonian vector field associated to $L$, i.e., it satisfies -$$ -\iota(U_g)\ \Omega = -dL = - u\cdot du -$$ -(where $\iota(X)$ denotes interior product, what I normally call 'lefthook'). By the standard identity, the divergence of $U_g$ with respect to the Liouville volume form, i.e., $\mu = \tfrac1{n!}\Omega^n$, vanishes identically. Now, you can compare the Liouville volume form with the 'Euclidean' volume form in $xu$-coordinates by noting that, by exterior algebra, we have -$$ -\tfrac1{n!}\Omega^n = \det(F(x))\ \tfrac1{n!}\bigl((du)^T\wedge dx\bigr)^n -$$ -It follows that the 'Euclidean' divergence of $U_g$ in the $xu$-coordinates is given by the formula -$$ --U_g\bigl(\log|\det(F(x))|\bigr) = -\tfrac12\ U_g\bigl(\log|\det(G(x))|\bigr). -$$ -Since, as you have already computed, $U_g(x) = F^{-1} u$, it follows that the divergence you want is, as a function, linear in the $u$-coordinates. You might write it as -$$ --\tfrac12 \nabla \bigl(\log|\det(G(x))|\bigr)\cdot (F^{-1} u). -$$<|endoftext|> -TITLE: Is the Characteristic of a Field Detectable from the Topology of a Topological Vector Space? -QUESTION [12 upvotes]: Motivation -A topological vector space is a vector space over a (topological) field, K, that carries a topology such that addition and scalar multiplication are continuous maps, e.g., all normed vector spaces. Since $\mathbb{C} \approx \mathbb{R}^2$ as topological spaces (with their standard topologies), it's clear that the actual field cannot be detected by the topology alone. Also, cardinality considerations show that any topology on a finite field cannot be homeomorphic to a vector space over a field of characteristic $0$. But that still says nothing about whether there are some topologies, such that, for instance $\mathbb{Z}_p(t) \approx \mathbb{Q}$. I can't find any topological invariants that will distinguish just the fields (as vectors spaces of dimension 1). I suspect that if this problem is solvable, it will be just be topological group considerations, but I can't figure out how to do it. - -So here are my two well-posed questions: - -a) Suppose K and F are topological fields. If K and F are homeomorphic, then is it necessarily true that char K = char F? -b) Suppose K and F are topological fields. Further suppose that $V$ is a topological vector space over K and $W$ is a topological vector space over F. If $V$ and $W$ are homeomorphic, then is it necessarily true that char K = char F? - -Obviously, an answer to b) implies an answer for a). I'm not sure how much of my thinking I should put here. But I do know that if char K = p, then we get a homeomorphism of $V$ onto itself by translation for each $v \in V$, which each have finite order p. These are mapped to self-homeomorphisms of $W$ which also have order p, but a priori, they don't "see" any of the algebraic structure of $W$, so I'm not sure this is actually an obstruction. Any thoughts (or references) to proofs or counterexamples would be greatly appreciated. - -REPLY [10 votes]: Torsten Ekedahl answered your first question in a nontrivial set of cases. As a concrete example, the characteristic two field $\mathbb{F}_2((t))$ is homeomorphic to the characteristic zero field $\mathbb{Q}_2$ by the map $$\sum_{k \gg -\infty} a_k t^k \mapsto \sum_{k \gg -\infty} a_k 2^k.$$ -This also provides a counterexample to your second question in dimension greater than zero. -The answer to the question in the title is a bit more subtle. That is, there are topological vector spaces that only allow one characteristic. This is obvious for finite dimensional vector spaces over finite fields, but it's also true that there are no fields of positive characteristic that are homeomorphic to $\mathbb{R}$, because the additive structure would allow one to produce a finite group of automorphisms of the order topology of size greater than 2.<|endoftext|> -TITLE: Kernel of a bundle map -QUESTION [7 upvotes]: Hello! -Let $E$ and $F$ be two vector bundles and let $f:E\rightarrow F$ be a bundle map. Then the kernel of $f$ is not always a subbundle of $E$. Does somebody have a simple example? Does there exist any simple conditions for which the kernel is a subbundle? -Thank you! - -REPLY [5 votes]: Here is a standard example (which you can find also in the big book of Demailly) in the complex category with the language of sheaves which perhaps may be of some interest for you. -Take $X=\mathbb C^3$, $\mathcal F=\mathcal O^{\oplus 3}$, $\mathcal G=\mathcal O$ and -$$ -\varphi\colon\mathcal O^{\oplus 3}\to\mathcal O,\quad (u_1,u_2,u_3)\mapsto\sum_{j=1}^3z_j\,u_j(z_1,z_2,z_3). -$$ -Since $\varphi$ yelds a surjective bundle morphism on $\mathbb C^3\setminus\{0\}$, it is easy to see taht $\ker\varphi$ is locally free of rank $2$ over $\mathbb C^3\setminus\{0\}$. However, by looking at the Taylor expansion of the $u_j$'s at $0$, you can check that $\ker\varphi$ is the $\mathcal O$-submodule of $\mathcal O^{\oplus 3}$ generated by the three sections $(-z_2,z_1,0)$, $(-z_3,0,z_1)$ and $(0,z_3,-z_2)$, and that any two of these three sections cannot generate the $0$-stalk $(\ker\varphi)_0$. -Thus, $\ker\varphi$ is not locally free. -On the other hand, if you have a surjective bundle morphism, the kernel is always locally free.<|endoftext|> -TITLE: Commutators in the reduced C*-algebra of the free group -QUESTION [19 upvotes]: Is it known whether any element of trace 0 in the reduced $C^*$-algebra of a non-abelian free group, is a limit of sums of (additive) commutators? - -REPLY [10 votes]: More is true: an element of a C*-algebra is a norm limit of sums of commutators if and only if it is 0 on any bounded trace. For selfadjoints, this was proven by Cuntz and Pedersen (in their only paper together I think). One reduces from arbitrary elements to selfadjoints by writing $c=a+bi$, with $a$ and $b$ selfadjoint. So if $C^*_r(G)$ has a unique tracial state then any element vanishing on it is a limit of sums of commutators. -A couple more facts proven by Cuntz and Pedersen: (1) the commutators can be arranged to form a convergent series, (2) for selfadjoints, the commutators can be chosen of the form $x^* x-x x^*$.<|endoftext|> -TITLE: Example of a compact set that isn't the spectrum of an operator -QUESTION [46 upvotes]: This question is somewhat ill-posed (due to the word easy) and is triggered by idle curiosity: - -Is there an easy example of a (separable, infinite-dimensional) Banach space $X$ and a nonempty compact set $K \subset \mathbb{C}$ such that $K$ is not the spectrum of a bounded linear operator $T: X \to X$? - -As I'm not at all knowledgeable beyond the the very first basics in the geometry of Banach spaces I apologize if the following notes completely miss the point. I add them to give some background and in order to clarify what I consider "easy": -Notes: - -If $X$ is a Hilbert space (or more generally, if $X$ admits an unconditional basis $\{e_{n}\}$), it is easy to construct a diagonal operator with spectrum $K$ by choosing a countable dense subset $\{\lambda_{n}\} \subset K$ and letting $T$ be the diagonal operator sending $\sum x_n e_n$ to $\sum (\lambda_n x_n)e_n$. -Standard examples for spaces without an unconditional basis are $L^1[0,1]$ and $C[0,1]$. I think I convinced myself that in both cases every non-empty compact set of $\mathbb{C}$ arises as the spectrum of an operator, so these obvious candidates don't seem to answer my question. (If this should be wrong, please tell me!) -Variants of the Gowers-Maurey space and the Argyros-Haydon space afford examples such that the spectrum $K$ must be countable with at most one accumulation point. See Gowers's blog for background on that. For the Argyros-Haydon space this is easy to see by the very motivation for its construction: It has the remarkable property that a bounded linear operator is of the form $\lambda \cdot \operatorname{id} + C$, where $C$ is compact (thus solving the long-standing scalar-plus-compact problem). -I asked a version of this question a few weeks ago on math.SE but with no answers so far. In view of the illuminating comments by Robert Israel and Jonas Meyer I got there I updated it a bit. -The present question is related to Pietro Majer's question Banach spaces with few linear operators ? here on MO. I looked at Maurey's chapter Banach spaces with few operators in the Handbook of the Geometry of Banach spaces Vol. 2, Elsevier 2003, (Johnson, Lindenstrauss, eds) but the examples discussed there are way beyond what I would count as easy. -It may well be (as I'm rather ignorant on this topic) that the level of difficulty of an example must be comparable to the one of the construction of the Gowers-Maurey space or even the Argyros-Haydon space, so if there's a compelling reason pointing in this direction, please let me know. - -REPLY [4 votes]: I don't know if this question is still interesting for someone, but in 2008 Argyron and Haydon constructed in "A hereditarily indecomposable L∞-space that solves the scalar-plus-compact problem. Acta Math. 206 (2011), no. 1, 1–54" a infinite-dimensional Banach space in which every operator is the sum of a compact operator and a multiple of the identity. Since the spectrum of compact operators consist of a sequence of point that accumulates only at zero and a escalar perturbation only translate this spectrum, I think this kind of space is pretty suitable for your cuestion. Just take a connected compact set $K$ that is not a singleton and you cannot have operators $T$ with $\sigma(T) = K$. Moreover, it is enough your compact set to have two or more accumulation points to have this property. -It is interesting to mention that one of the other "applications" these space has is to solve a conjectute about the invariant subspace problem. By Lomonosov theorem every compact operator has a non-trivial hyperinvariant subspace, so every escalar perturbation of a compact operator also has a hyperinvariant subspace too. This is the first non-trivial example of an infinite-dimensional space such that every operator has a non-trivial hyperinvariant susbpace.<|endoftext|> -TITLE: Accumulation of algebraic subvarieties: Near one subvariety there are many others (?), 2 -QUESTION [12 upvotes]: This is a sequel to the question Accumulation of algebraic subvarieties: Near one subvariety there are many others (?) . -Let $Y$ be some projective variety, over $\mathbb{C}$. Let $X\subset Y$ be some closed subvariety, and let $\tilde{X}$ be some small open neighborhood of $X$, in the complex topology. - -Naive Question: Are there many varieties $\subset \tilde{X}$ of dimension equal to the dimension of $X$? - -In the previous question, this was answered positively if $Y=\mathbb{P}^n$, using crucially the fact that the tangent sheaf of $\mathbb{P}^n$ is ample, which allows to construct deformations of infinitesimal thickenings of $X$. -On other hand, the answer in the general case is obviously no, because one might take as $Y$ the blow-up of $\mathbb{P}^2$ at a point, and for $X$ the exceptional divisor. In that case, any $X^\prime\subset \tilde{X}$ has to project to the point in $\mathbb{P}^2$, for otherwise it meets any line in $\mathbb{P}^2$, which however may be outside of the image of $\tilde{X}$. But then $X^\prime$ is set-theoretically equal to $X$. -But in this case, the obstruction already exists if we take $\tilde{X}$ to be a small neighborhood in the Zariski topology. - -Question: Assume that within any Zariski neighborhood of $X$, there are many projective subvarieties of the same dimension. Then is the same true for $\tilde{X}$? - -Finally, note that the condition that within Zariski neighborhoods, there are many subvarieties of the same dimension is e.g. guaranteed by the following numerical criterion. - -Lemma: Assume that $X$ meets every subvariety of complementary dimension. Then within any Zariski neighborhood of $X$, there are many projective subvarieties of the same dimension. - -For the proof, just note that the complement of any Zariski neighborhood of $X$ has codimension smaller than $\mathrm{dim} X$, and hence by taking an intersection of generic hyperplane sections, you get subvarieties which do not meet the boundary. Also note that the condition of the Lemma is trivially verified if $Y=\mathbb{P}^n$. -However, note that one can not in general expect to have complete intersections inside of $\tilde{X}$, already if $Y=\mathbb{P}^n$ and $X$ is smooth. For the proof, note that if we take $\tilde{X}$ small enough, it will have the same (singular, say) cohomology as $X$; in particular, for any $X^\prime\subset \tilde{X}$, we get a map $H^i(X,\mathbb{Q})\rightarrow H^i(X^\prime,\mathbb{Q})$. It is compatible with cup-product, and injective in top-degree (compare with cohomology of $\mathbb{P}^n$), thus injective. But $H^i(X^\prime,\mathbb{Q})$ is zero below the middle dimension if $X^\prime$ is a complete intersection. -One might also ask the following weaker question (say, $Y$ is smooth in order to have a nice cup-product): - -Question: Assume that the intersection pairing of $X$ with any subvariety of complementary dimension is positive. Are there many subvarieties in $\tilde{X}$ of the same dimension? - -Counterexamples to these questions are most welcome, on the other hand I would be particularly interested in any positive result or method to construct such subvarieties, so feel free to make additional assumptions. -EDIT: Both questions above are shown to have negative answers by the nice example of Dmitri. Therefore, let me ask the following, pretty vague, question: - -Question: Is it possible to formulate any positive result in this direction, possibly assuming that 1) the tangent or normal sheaf is in some sense positive or ample, 2) some positivity condition as already required above + 3) whatever you need. - -REPLY [14 votes]: One little counterexample (to both versions of the question). On a quintic $3$-fold in $\mathbb CP^4$ there are $2875$ lines. You can take any of these lines. An analytic neighbourhood of such a line is the one of the small resolution of the cone $x_1^2+x_2^2+x_3^2+x_4^2=0$. So there can be no curves in the neighbourhood of such a line, since the cone is affine. -This example is not surprising, since the normal bundle to such a line is $O(-1)\oplus O(-1)$, one needs at least a bit of positivity...<|endoftext|> -TITLE: An inequality on concave functions -QUESTION [5 upvotes]: Could somebody help me to answer the following question? -Let $f:R_+ \rightarrow R_+$ -be a nonindentically zero, nondecreasing, continuous, concave function with $f(0)=0$. Do we have that for any $s,t \in [0,1]$, -$$f(x)f(stx)\leq f(sx)f(tx), \quad \forall x \geq 0.$$ -or equivalently, do we have that for any $t\in (0,1)$, $\frac{f(x)}{f(tx)}$ is nonincreasing on $x >0$. -Thanks! - -REPLY [5 votes]: The answer is no. -Mikael de la Salle has given a counterexample. In a more general setting, the necessary and sufficient condition for the inequality to hold is $g = \log \circ f \circ \exp$ is concave. The inequality can be rewritten into $g(y) + g(a + b + u) \le g(a + y) + g(b + y)$, where $y = \log x$, $a = \log s \le 0$ and $b = \log t \le 0$. This is precisely the necessary and sufficient condition for concavity of $\log \circ f \circ \exp$, which is certainly not the same as concavity of $f$.<|endoftext|> -TITLE: How do I get the correct long exact sequence for relative group cohomology in terms of derived functors? -QUESTION [15 upvotes]: Background: -I want to consider relative group cohomology: the construction is as follows. I have a subgroup $H\subseteq G$ (and note that I don't want to assume that $H$ is normal in $G$), and a $\mathbb Z[G]$-module $M$. Then we have the standard chain complxes $C^\ast(G;M)$ and $C^\ast(H,M)$, and there is a natural morphism $C^\ast(G,M)\to C^\ast(H,M)$, which induces the "restriction homomorphism" on group cohomology $\operatorname{res}:H^\ast(G,M)\to H^\ast(H,M)$. Let us define the "relative group cohomology" as the cohomology of the chain complex which fits into the exact sequence: -$$ -0\to C^\ast(G,H;M)\to C^\ast(G;M)\to C^\ast(H;M)\to 0 -$$ -(i.e. $C^\ast(G,H;M)$ is defined to be the kernel of the second map). I haven't ever heard of these "relative group cohomology" groups, but it seems like a very natural idea to me, and what I'm trying to do is define algebraically the cellular cohomology groups $H^\ast(K(G,1),K(H,1);M)$ (in case we have a $K(H,1)$ which is naturally a subcomplex of a $K(G,1)$). If anyone has a good reference for these I'd like to know! Note that by definition, the relative group cohomology groups $H^\ast(G,H;M)$ fit into a natural long exact sequence: -$$ -\cdots\to H^\ast(G,H;M)\to H^\ast(G;M)\to H^\ast(H;M)\to\cdots -$$ -and this is what one would expect the cellular cohomology groups $H^\ast(K(G,1),K(H,1);M)$ to satisfy. If I've messed this construction up, please tell me. -Question: -How do I understand the relative group cohomology in terms of derived functors? We know that $H^\ast(G;M)=\operatorname{Ext}^\ast_{\mathbb Z[G]}(\mathbb Z,M)$ and $H^\ast(H;M)=\operatorname{Ext}^\ast_{\mathbb Z[H]}(\mathbb Z,M)$. But since these are $\operatorname{Ext}$'s in different categories, it doesn't seem clear how to fit a third into the exact sequence. What I'd like is some $\operatorname{Ext}$ definition of the relative group cohomology groups I've defined above. -More Info: -I've tried the following, but it seems to give the "wrong" answer. We can get everything into the same category by observing that $M^H=\operatorname{Hom}_{\mathbb Z[G]}(\mathbb Z[G/H],M)$, and thus the cohomology is given by $H^\ast(H;M)=\operatorname{Ext}^\ast(\mathbb Z[G/H],M)$ (from now on, all $\operatorname{Ext}$'s are in the category of $\mathbb Z[G]$-modules). Furthermore (and correct me if I am wrong), the restriction homomorphism $H^\ast(G,M)\to H^\ast(H,M)$ is induced by the "sum coefficients" morphism $\mathbb Z[G/H]\to\mathbb Z$ (giving the map $\operatorname{Ext}^\ast(\mathbb Z,M)\to\operatorname{Ext}^\ast(\mathbb Z[G/H],M)$). So, now it looks like we get what we want, but now comes a surprise. The "first argument"s of the $\operatorname{Ext}$'s fit into a short exact sequence: -$$ -0\to\ker\to\mathbb Z[G/H]\to\mathbb Z\to 0 -$$ -and thus we have a long exact sequence of $\operatorname{Ext}$: -$$ -\cdots\to\operatorname{Ext}^\ast(\mathbb Z,M)\to\operatorname{Ext}^\ast(\mathbb Z[G/H],M)\to\operatorname{Ext}^\ast(\ker,M)\to\cdots -$$ -But now it looks like $\operatorname{Ext}^\ast(\ker,M)$ is not giving the relative group cohomology groups we want: the long exact sequence isn't the same as the one above, it's gotten flipped around. I guess this doesn't entirely disqualify the construction, since perhaps we have $H^\ast(G,H;M)=\operatorname{Ext}^{\ast-1}(\ker,M)$, but in this case I'd still like an explanation for why this dimension shifting happens. - -REPLY [5 votes]: Here are some remarks: the situation we would like to model algebraically is this: suppose $H$ is a subgroup of a group $G$ and $M$ is a $G$-module. Let $L_M$ be the corresponding local system on $BG$. Then we have a map of classifying spaces $f:BH\to BG$ obtained as follows: take $EG$, a contractible space on which $G$ acts freely and quotient it by $H$; the result will be $BH$ and it maps to $BG=EG/G$. The local system $L_M$ pulls back to $BH$, which gives a map $f^*:H^*(BG,L_M)\to H^*(BH,f^{-1}L_M)$. This is, of course, the map $H^*(G,M)\to H^*(H,M)$. -A side remark: if $H$ is normal, then $f$ is the projection of a principal $G/H$-bundle, which gives the classifying map $g:BG\to B(G/H)$. Now, if one replaces $g$ with a Serre fibration, the fiber will be $BH$ [this is not completely obvious; perhaps I'll add a reference later] and the local system restricted to the fiber will be isomorphic to $f^{-1}L_M$. Taking the Leray spectral sequence of this fibration one gets the Serre-Hochschild spectral sequence. -Now, coming to the derived functors picture: the cohomology $H^i(G,M)$ is the $i$-th derived functor of the invariants functor from the category of $G$-modules to the category of abelian groups. In other words, we have to derive the functor $M\mapsto Hom(\mathbb{Z},M)$. There are two ways to do that. First, we can replace $\mathbb{Z}$ with a projective resulotion; this is what people usually do since this resolution can be written down explicitly and the resulting $Hom$ complex is the standard cochain complex $C^*(G,M)$. But we can take an injective resolution of $M$; it would work just as well. So $H^i(G,M)$ is nothing but $Hom_{D G-mod}(\mathbb{Z},M[i])$ (unless I've messed it up and there should be minuses somewhere). -Every $G$-module is an $H$-module, which gives a functor $f^{-1}:D G-mod\to D H-mod$. This functor corresponds to pulling sheaves from $BG$ back to $BH$ and it has a right adjoint $f_*$, which corresponds to pushing sheaves forward from $BH$ to $BG$. The construction of $f_*$ is not too difficult, but requires some work, see Bernstein-Lunts, Equivariant sheaves and functors, part I. We have a natural adjunction morphism $M\to f_* f^{-1}M$; if we apply $Hom_{DG-mod}(\mathbb{Z},-)$ to the shifts of this morphism, we get the maps $H^*(G,M)\to H^*(H,M)$. Now, the relative cochain complex of the posting computes $Hom_{DG-mod}(\mathbb{Z},-)$ of an object $A$ of $DG-mod$ such that $A\to M\to f_* f^{-1}M\to A[1]$ is a distinguished triangle.<|endoftext|> -TITLE: What about the empty torsor? -QUESTION [5 upvotes]: Let $G$ be a group. A $G$-torsor is a set $X$ together with an action of $G$ such that for all $x,y \in X$ there is exactly one $g \in G$ such that $gx=y$. This looks like a group which has forgotten its identity (no pun intended). -Usually it is assumed that $X$ is nonempty (or more generally an inhabited object according to the nlab article), but it seems to me that it makes perfect sense also to allow $X = \emptyset$ because of the equivalent definition using Heaps. This equivalence also suggests that $X = \emptyset$ should imply that $G$ is trivial. But this is not guaranteed by the definition given above. -Question: What is a natural definition of torsors which also includes the empty set with the action of the trivial group (the empty torsor)? -In the case of sets as above, we may just add that the action is faithful, i.e. the homomorphism $G \to \text{Sym}(X)$ is injective. But how can we give a definition for, say, group schemes acting on schemes, without making a nasty case distinction? -Also I would like to know if you agree with me that it is natural to include the empty torsor. It will be an initial object in the category of torsors and as I said, especially in the definition of a heap the assumption of being nonempty seems to be artificial according to universal algebra. -[added] Thank you for all the good answers. I agree that it's not natural to consider an empty torsor. A $G$-torsor should be something which is locally isomorphic to $G$. - -REPLY [8 votes]: A torsor is a map to the classifying stack BG. Of course in algebraic geometry this -is (in one form or another) the definition of BG (and this answer is equivalent to the local nontriviality in the above answers), but in any case there's an underlying -intuition based on topology of what this means, and there isn't room in this picture (as far as I can see) for the empty torsor. If you want there to exist a universal G-torsor (ie the point mapping to BG) it seems to me that you have to continue to exclude empty torsors. -(Edit: or you could define BG as a simplicial scheme and get a marginally less tautologous identification of G-torsors with maps.)<|endoftext|> -TITLE: Probability of having a "perfect" game of Set -QUESTION [12 upvotes]: The card game Set has very simple rules (see here for rules), and it has prompted mathematicians to ask several questions. I will describe one of these questions. When the game ends, there are usually some left over cards, none of which form a Set; but occassionally it happens that the remaining cards all match up, and there are no left over cards. Call such a situation a "perfect" game. My question is: - -What is the probability of having a perfect game? What are the best known upper and lower bounds for this probability? - -Here we assume that if there are multiple sets available, then one is selected at random (with uniform probability among all distinct available sets). According to this handout the exact value for the probability of a perfect game is "very open". Anecdotally, I've probably played Set hundreds (maybe thousands?) of times, and only a handful have been perfect games (one of which was earlier today!). -A related question one could ask is whether or not the actual gameplay changes the probability. What I mean is: - -Say all 81 cards are laid down face up, and Sets are removed randomly until no Sets remain. Is the probability of having no left over cards the same as it is when the game is played normally? - -I could remark that the actual game of Set is a special case of a larger class of games, where the cards have n attributes each with r possible values; the usual game is the case n=4 and r=3. One could ask the same questions for this larger class of games. Also, these questions are more naturally viewed as questions about configurations of lines in affine space, but I've chosen to stick with the card game terminology. - -REPLY [9 votes]: I have simulated playing 10 million games of Set, and for those simulations 1.2% of the games were perfect games (when selecting randomly among the available Sets the whole game). The full results from the simulations are here.<|endoftext|> -TITLE: Singular Homology/Cohomology as a derived functor? -QUESTION [22 upvotes]: Hello, -Learning some Alg.geometry and Sheaf theory, I got used to the notion that cohomology arises naturally as a derived functor of some sort. -This has led me thinking, singular cohomology, from algebraic topology, was never defined (In all books i've checked) as a derived functor, but just by giving cycles and boundaries. -I could not figure out by myself any reasonable functor whose derived functors yield singular cohomology, So I pose this question out here. -I hope this might shed some more insight on what singular cohomology actually measures. -Thanks - -REPLY [10 votes]: Here's an elaboration of David Roberts' answer, a perspective that can be found in Quillen's 'Homotopical Algebra'. Quillen thought of homology as "derived abelianisation", and the abelianisation of a space (by which I mean simplicial set) can only be the free simplicial abelian group on that space, seeing as it's left adjoint to the forgetful functor $sAb \to sSet$. -Composing the free simplicial abelian group functor with the Dold-Kan correspondence - an equivalence of categories between simplicial abelian groups and chain complexes - gives a functor $sSet \to Ch$, which happens to coincide (more or less) with the singular complex functor. If we endow $Ch$ with the injective model structure this is a left Quillen functor. By Quillen's philosophy, homology of spaces should be the total left derived functor of this functor. But the word 'derived' is sort of vacuous here, because all simplicial sets are cofibrant, and this negates the need to explicitly use any model category language. This is why we think of the singular complex as the "homology object" of a space.<|endoftext|> -TITLE: Commutativity and Kostant sections -QUESTION [8 upvotes]: Let $(e, h, f)$ be an $\operatorname{SL}_2$-triple in $\mathfrak g$. My understanding is that $e + C_{\mathfrak g}(f)$ is called a Kostant section only in case $e$ is regular; but I don't impose this restriction. (EDIT: It seems that, in general, it's called a Slodowy slice.) Given such a datum, there is a unique decomposition (*) $\mathfrak g = [e, \mathfrak g] + C_{\mathfrak g}(e, f) + C_{\mathfrak g}(f)(< 0)$, where “$< 0$” in the last term refers to the grading coming from the action of $h$. -Now suppose that we are given only a nilpotent element $e \in \mathfrak g$, and an arbitrary element $X \in \mathfrak g$. Can we complete $e$ to an $\operatorname{sl}_2$-triple $(e, h, f)$ so that, in the decomposition $X_1 + X_2 + X_3$ of $X$ according to (*), we have that $X_2$, or at least its semisimple part, commutes with $X_3$? The obvious answer is “No, because, in certain characteristic, you can't complete $e$ to an $\operatorname{sl}_2$-triple at all.”  To avoid such trivialities, assume the characteristic is as large as necessary. -As a very mild hint that the answer is ‘yes’, the weak form of the question works whenever $e$ is distinguished (since then the semisimple part of $X_2$ is central in $\mathfrak g$); and, by explicit calculation, it works when $\mathfrak g = \mathfrak{gl}_3$ and $e$ is the $(1, 2)$-nilpotent, as long as $p > 3$. Because of the first hint, I'd like to make an appeal to Bala–Carter(–Pommerening–Premet …) theory, but I can't figure out how. - -REPLY [9 votes]: In some cases the answer to the weaker version of the question (involving the semisimple -part of $X_2$) is YES. This will happen if $C_g(e)$ is self-dual which is the case, for instance, when $g=gl_N$, $N=nm$, and $e$ has $n$ Jordan blocks of size $m$. -To see this, one can use the fact that all maximal toral subalgebras of $C_g(e)$ are conjugate. -However, in general the answer to the question is NO as the following example shows. Let $g=sp_4$ and $e$ a long root vector in $g$. Then the action of $ad\ h$ on $g$ gives us a $\mathbb{Z}$-grading $g=g(-2)\oplus g(-1)\oplus g(0)\oplus g(1)\oplus g(2)$ with $g(-2)=\mathbb{C}f$, -$g(2)=\mathbb{C}e$, $g(0)\cong gl(2)$, $g(-1)$ the standard $2$-dimensional module for $g(0)$, -and $g(1)\cong g(-1)^*$ as $g(0)$-modules. We also have that $C_g(f)= g(0)'\oplus g(-1)\oplus g(-2)$ and $C_g(e,f)=g(0)'$. -Let {$x,H,y$} be an $sl_2$-triple in $g(0)'$ which spans $g(0)'\cong sl_2$ and let -{$v_1,v_{-1}$} and {$w_1,w_{-1}$} be eigenbases for $ad\ H$ of $g(1)$ and $g(-1)$, respectively, so that $[H,v_1]=v_1$, $[H,w_{-1}]=-w_{-1}$, etc. -We choose $X=X_1+X_2+X_3$ such that $X_1=0$, $X_2=H+y$ and $X_3=w_{1}$. Then $X_2$ is semisimple and $[X_2,X_3]=w_1+w_{-1}$. -We may adjust $X$ effectively only by applying automorphisms of the form $Ad\ g$ with -$g$ in the unipotent radical of the centraliser of $e$ in $Sp_4$. Then -$Ad\ g=\exp(a\ ad\ e)\exp(b\ ad\ v_1)\exp(c\ ad\ v_{-1})$ for some $a,b,c\in\mathbb{C}$. Such an adjustment is not going to change -$X_3=w_1$, but it will replace $X_2=H+y$ by $H+y+b[v_1,w_1]+c[v_{-1},w_1]=H+y+\lambda x+\mu H$ for some $\lambda,\mu\in\mathbb{C}$. (It will change $X_1$ as well, but this is not important.) Since $[x,w_1]=0$ and $[y,w_1]=w_{-1}$, this is not going to help us to improve $X$ the way we wanted.<|endoftext|> -TITLE: Does every knot contain all four vertices of an isosceles trapezoid? -QUESTION [9 upvotes]: I ask this question with some trepidation, because it may be trivial and/or of entirely recreational interest. -Erika Pannwitz proved in 1933 that every non-trivial knot contains a quadrisecant (four points which lie along a common line). Generically, a knot has finitely many quadrisecants. But what is so special about the condition of collinearity? In a dream I had a few nights ago (I apologise that this question arose in such a context), somebody asked me whether every knot contains all four vertices of an isosceles trapezoid (not in those words). I could answer the question neither in my sleep nor after I woke up. The claim sounded plausible to me (the codimension seems about right), and if it's true, then one might dream that a signed count of such trapezoids might give rise to a knot invariant along the lines of Budney-Conant-Scannell-Sinha's New perspectives on self-linking. - -Does every knot contain all four vertices of an isosceles trapezoid? More generally, is there a nice description of the subclass $\mathcal{C}$ of quadrilaterals such that every knot contains all four vertices of at least one, and generically finitely many, quadrilaterals in $\mathcal{C}$? -EDIT: Does every knot contain all four vertices of a rectangle? - -I have thought about this problem a bit (trying to put it in the framework people use for dealing with colinearity problems), and it feels like it should be easy (and well-known to experts), but I'm a bit stuck. - -REPLY [9 votes]: In fact, there is a large number of inscribed quadrilateral results (rhombi, rectangles, etc.), mostly inspired by the square peg problem of Toeplitz and Schnirelmann. Here are some references (in alphabetical order): by Griffiths, Makeev, Stromquist, and Wu. There are other related papers, but these are all on quadrilaterals inscribed into space curves. A small warning: not all of these are correct and precise everywhere (please forgive me for being cryptic - this is not the place to elaborate). Finally, if you would like to see more context, see Nielsen's site, Jordan Ellenberg's blog post, my short preprint, and my book, Chapter 5 (sorry for the self-promotion).<|endoftext|> -TITLE: Detect compactness on unit sphere -QUESTION [5 upvotes]: In Pedersen's Analysis Now, for example, you learn that a bounded operator on Hilbert space $T: \mathcal{H} \to \mathcal{H}$ is compact if and only if the image $T(B)$ of the unit ball is compact. It is pretty easy to see therefore that any bounded operator which carries the unit sphere to a compact set is compact. Does the converse hold, that is, if $T: \mathcal{H} \to \mathcal{H}$ is compact, is $T(S)$ compact, where $S$ is the unit sphere in $\mathcal{H}$? -With apologies if this is extremely obvious. - -REPLY [8 votes]: It may fail to be closed. If T is injective and compact on an infinite dimensional Hilbert space, then $0$ is in the closure of $T(S)$, and not in it.<|endoftext|> -TITLE: On the physics background of p-Laplacian equation -QUESTION [6 upvotes]: Could you tell me the physics background of p-Laplacian equation? Thank you! -Actually, I know nothing about this. But I am curious about the original of these PDEs or where they come from. Could you introduce some related books or articles to me? - -REPLY [12 votes]: It is used to model Newtonian and non Newtonian Fluids. You generally have different values for $p$ depending on the type of fluid: - -$p=2$ (Newtonian) for fluids like water and air -$p>2$ for viscous ("sticky") fluids -$p<2$ for non-viscous ("runny") Fluids - -Edit: I just realized that this alone probably doesn't satisfy your question. Let me clearify the actual physics behind the above thinking: -The viscous property of those fluids arises from shear stress ("drag") inside of the fluid. To describe this effect in fluids, we use the speed the particles travel in relation to each other (the factor being $\vert \nabla u \vert ^{p-2}$ ). -Newtonian fluids are purely linear, so the correct approximation is $p=2$. For non-newtonian the viscosity is non-linear, so $p\not= 2$.<|endoftext|> -TITLE: Is $k[x_1, \ldots, x_n]$ always an integral extension of $k[f_1, \ldots, f_n]$ for a regular sequence $(f_1, \ldots, f_n)$? -QUESTION [9 upvotes]: The elements of a regular sequence in $k[x_1, \ldots, x_n]$ are algebraically independent over $k$ (see for example Matsumura ex. 16.6), and so for a length n regular sequence $(f_i)$ of homogeneous elements, $k(x_1, \ldots, x_n)$ will be algebraic over $k(f_1, \ldots, f_n)$. -My question is whether it's also true that $k[x_1, \ldots, x_n]$ is integral over $k[f_1, \ldots, f_n]$. Aside from experimental evidence (dozens of randomly generated regular sequences in Macaulay with n=3 and low-ish degrees), I don't really have any reason to think that it should be true, and I haven't been able to prove or disprove it (or find a mention of it anywhere). -Thanks. -Edit: added homogeneous condition - -REPLY [7 votes]: Stumbled by coincidence over this question. Therefore the late answer. -In the graded case it is true that $k[x_1,...,x_n]$ is integral over $k[f_1,...,f_n]$ ($k$ a field). But even more is true: -Let $k$ be a field and let $R = \oplus_{i\ge 0}R_i $ be a finitely generated graded commutative $k$-algebra with $R_0= k$. Assume moreover that $R$ is Cohen-Macaulay and that -$y_1,...,y_n$ is a regular sequence in $R$. Set $I := \oplus_{i> 0}R_i$. Then - -$\sqrt{I} = (y_1,...,y_n)$ -There is $k > 0$ with $I^k \subseteq (y_1,...,y_n)$. -$R$ is a free $k[y_1,...,y_n]$-module. - -Proof: 2) It's well-known that if $y \in R$ is a homogeneous regular element of positive degree, then Krull-dim $R/(y) =$ Krull-dim$(R) - 1$. Applying this repeatedly to $R/(y_1,...,y_i)$ one obtains that $\bar{R} := R/(y_1,...,y_n)$ has dimension $0$. Hence $\bar{R}$ is Artinian and thus there is $k > 0$ such that $\bar{R}_i = 0$ for all $i \ge k$. So, if $x \in I$ then $\bar{x}^k = 0$ in $\bar{R}$, i.e. $x^k \in (y_1,...,y_n)$. -1) Follows from 2) -3) Induction on $n$. Case $n=0$: $R$ is Artinian and hence a finite dim. $k$ vector space. -Assume the case $n-1$ is true. Since $\tilde{R} := R/(y_1)$ is again CM and $\tilde{y}_2,...,\tilde{y}_n$ a regular sequence, there are $\tilde{a}_i \in \tilde{R}$ with $\tilde{R} = \oplus_i k[\tilde{y}_2,...,\tilde{y}_n]\tilde{a}_i$. -Let's show $R = \oplus_i k[y_1,...,y_n]a_i$: Let $f_i \in k[y_1,...,y_n]$ with $\sum_i f_i a_i = 0$. -We proceed by induction on $m := \max_i\; \deg_{y_1}(f_i)$: Reduction modulo $y_1$ shows $f_i \in (y_1)$ for all $i$. If $m=0$, then $f_i = 0$ for all $i$. If $m> 0$ let $f'_i = f_i/x_1$. Thus $y_1\sum_i f'_i a_i = 0$ and since $y_1$ is regular, we conclude $\sum_i f'_i a_i = 0$ with $m' < m$. Then by induction hypothesis $f'_i = 0$ and hence $f_i = 0$. -q.e.d.<|endoftext|> -TITLE: A 2F1 Hypergeometric identity from a Feynman integral -QUESTION [11 upvotes]: Using two different approaches to evaluating the dimensionally regularized ($d=4-2\epsilon$ dimensional Euclidean space), equal mass ($x=m^2$), 2-loop vacuum Feynman diagram -$$ -\begin{align} -I(x) &= \int\frac{\mathrm{d}^dp}{\pi^{d/2}}\frac{\mathrm{d}^dk}{\pi^{d/2}} - \frac1{(k^2+x)(p^2+x)((k+p)^2+x)} \\\ - &= \int _0^{\infty }\int _0^{\infty }\int _0^{\infty } - \frac{e^{-x(s_1+s_2+s_3)}}{\left(s_1s_2+s_2s_3+s_3s_1\right)^{d/2}} - \mathrm{d}s_1\mathrm{d}s_2\mathrm{d}s_3 \\\ - &= x^{d-3}\,\Gamma\left(2-\frac{d}{2}\right)\Gamma\left(1-\frac{d}{2}\right) - \,f(d)\,, -\end{align} -$$ -the following hypergeometric identity arises -$$ -\begin{align} -f(d) &=f_1(d) = -2\, {}_2F_1\left(1,\frac{d-1}{2};\frac{3}{2};-\frac{1}{3}\right) --4^{2-d} 3^{(d - 3)/2} B\left(\frac{3 - d}2, \frac{3 - d}2\right) \\ -&= f_2(d) = \frac{4}{3} \left( -{}_2F_1\left(1,\frac{d-1}{2};\frac{3}{2};-\frac{1}{3}\right)+\frac{1 }{d-3} {}_2F_1\left(1,\frac{d-1}{2};\frac{5-d}{2};-\frac{1}{3}\right) \right)\,, -\end{align} -$$ -where the second term in $f_1(d)$ can be reduced with the identity -$B(x,x)=2^{1-2x}B(x,1/2)$. -The identity $f_1(d)=f_2(d)$ checks out numerically and (provided no mistakes have been made in the calculations) it should be identically true. So far I have been unable to find a direct proof of the identity. -Can anyone here prove this identity or suggest a good approach? -A computer proof (using human checkable code/steps) is acceptable. - -For convenience I've provided the Mathematica InputForm of the two functions: -f1[d_] := 2 Hypergeometric2F1[1, (d - 1)/2, 3/2, -1/3] - - 2^(4 - 2 d) 3^((d - 3)/2) Beta[(3 - d)/2, (3 - d)/2] - -f2[d_] := 4/3 (Hypergeometric2F1[1, (d - 1)/2, 3/2, -1/3] + - 1/(d - 3) Hypergeometric2F1[1, (d - 1)/2, (5 - d)/2, -1/3]) - - -Aside: -$f_1(d)$ comes from direct integration using Feynman parameters (my own calculation, I don't know of a reference that includes it). -$f_2(d)$ comes from direct integration using the Mellin-Barnes representation (the result presented above is a version of eqn(33) of hep-ph/9304303, see also references within) . - -Edit: -I just noticed this MO answer that mentions the HolonomicFunctions package for Mathematica. It shows that both functions obey the recursion -$(4+4 d)f_i(d+4)+(4-7 d)f_i(d+2)+(-6+3 d)f_i(d)=0$, -but of course, the integral diverges for integer $d\geq4$ and I need to prove the relation for all $d<4$ (dimensional reduction) or for all complex $d$ (dimensional regularization). - -Aside #2: -Maybe I've been viewing this problem backwards, and I should not be using hypergeometric identities to check the Feynman integrals, but rather using the Feynman integrals as inspiration for new hypergeometric identities. See the new paper: -Finding new relationships between hypergeometric functions by evaluating Feynman integrals - -REPLY [6 votes]: Since you are using Mathematica, you definitely want to take a look at the extremely useful package HolonomicFunctions by Christoph Koutschan. -In your particular example, -Annihilator[f1[d], {S[d]}] - -shows that this function satisfies the recurrence -\begin{equation} - (4+4d)f_1(d+4)+(4-7d)f_1(d+2)-(6-3d)f_1(d)=0. -\end{equation} -Once known, Mathematica itself can check symbolically that both of your functions satisfy this recurrence: -(4+4d)f1[d+4] + (4-7d)f1[d+2] - (6-3d)f1[d] // FullSimplify -(4+4d)f2[d+4] + (4-7d)f2[d+2] - (6-3d)f2[d] // FullSimplify - -After checking initial conditions (which Mathematica can do) it follows that $f_1(d)=f_2(d)$ for all even integers $d$ -But as I'm typing I see that the OP just figured all of this out by himself... ;) So let me just mention that one strategy now could be to look at $f_1-f_2$, show that it satisfies the necessary exponential growth conditions (should be alright after combining the poles; apart from these each function seems to be good by itself), and apply Carlson's Theorem. I hope that helps...<|endoftext|> -TITLE: Is there a canonical map from the cohomology of orbifold chiral de Rham on an orbifold to the cohomology of chiral de Rham on a crepant resolution? -QUESTION [11 upvotes]: The two-variable elliptic genus is a topological invariant of almost-complex manifolds that takes values in power series. These power series turn out to describe weak Jacobi forms when the manifold is Calabi-Yau. It is defined by constructing a formal power series in K-theory out of exterior and symmetric powers of tangent and cotangent bundles, and taking its Euler characteristic. It seems to capture some stringy geometry of a manifold $X$, in the sense that it is (under the assumption that such objects can be precisely defined) the partition function of the half-twisted $\mathcal{N}=(2,2)$ superconformal $\sigma$-model with target $X$. The modularity properties seem to reflect some correspondence between traces of certain operators and small deformations of a nodal genus 1 curve. -Work by Borisov and Libgober, McKay correspondence for elliptic genera in the turn of the century yielded two generalizations of elliptic genus: - -a notion of orbifold elliptic genus, where one adds contributions from fixed loci of conjugacy classes. This seems to be related to the fact that the space of small loops on a stack is more or less its inertia stack. -a notion of singular elliptic genus, defined on certain singular varieties by taking a resolution. It is independent of the resolution. - -The authors showed that when the coarse moduli space of an orbifold has a crepant resolution, the elliptic genus of that resolution is equal (up to a factor involving a theta function and its derivative) to the orbifold elliptic genus. For example, one could calculate the elliptic genus of a K3 surface by computing the orbifold elliptic genus of the $[\pm 1]$-quotient of an abelian surface, since any K3 is diffeomorphic to the minimal resolution of a Kummer surface. -There is a way to interpret the elliptic genus mathematically in a way that is closer to the physicists' method, by the chiral de Rham complex (paper of the same name by Malikov, Schechtman, and Vaintrob—see also Yuji's question). It is defined as a sheaf of vertex superalgebras on the manifold $X$, and its global cohomology yields the elliptic genus as graded characters of certain operators. Physically, according to Kapustin in Chiral de Rham complex and the half-twisted sigma-model, the chiral de Rham complex is the perturbative part of the half-twisted SCFT. Naturally, one can construct this complex on a crepant resolution of an orbifold, and Frenkel and Szczesny, Chiral de Rham complex and orbifolds constructed a version of chiral de Rham on orbifolds, and showed that its cohomology yields the orbifold elliptic genus. -Question: Is there a vertex superalgebra isomorphism between the cohomology of the orbifold chiral de Rham (possibly tensored with some superalgebra whose character involves a theta function and its derivative) and the cohomology of the chiral de Rham of a crepant resolution? -The character equality makes it clear that some map should exist on the level of vector spaces, but it would be nice if there were a more categorified correspondence. Vague physical explanations for such a map and references to partial answers would also be greatly appreciated. - -REPLY [4 votes]: The short answer is no, one does not expect such isomorphism. What one does expect is some kind of "flat" family of super-vertex algebras that interpolates from one to the other. The base of the family should be the Kahler parameters of the model, i.e. complex parameter of the mirror, if such exists. In the case of CY hypersurfaces in toric varieties, this seems plausible, up to some technicalities. -If one passes to the chiral rings, then this pretty much becomes the Ruan's conjecture about small quantum cohomology of different crepant resolutions.<|endoftext|> -TITLE: Why is the mapping class group of hyperbolic manifolds finite? -QUESTION [6 upvotes]: Hi! I'm trying to understand why a hyperbolic n-manifold has finite mapping class group if $n \geq 3 $. In books I'm reading it's said it's a consequence of Mostow's rigidity theorem: -"If M and N are complete hyperbolic manifolds with finite total volume, any isomorphism of fundamental groups is realized by a unique isometry." -A corollary of this is that: -" If M is hyperbolic (complete, with finite total volume) and $n \geq 3 $, then Out($\pi_{1}(M)$) is a finite group, isomorphic to the group of isometries of M ". -But how could this could solve my problem? I mean, I know there is Dehn-Nielsen Theorem which states that Out($\pi_{1}(M)$) is isomorphic to MCG(M), but I know this to be true only in dimension 2...what can I say in dimension (at least) 3? -Thank you. - -REPLY [8 votes]: In dimension three, this was proven by Gabai, Meyerhoff, and N. Thurston. This was proven orginally for Haken 3-manifolds by Hatcher. Gabai extended this to hyperbolic 3-manifolds satisfying a certain technical condition, which was then verified for all closed hyperbolic 3-manifolds in the above paper. Gabai extended this result to prove that $Diff(M) \simeq Isom(M)$. -The analogous result in higher dimensions was proved by Farrell and Jones (see Theorem 5, I think this is only for dimension $>5$, but this isn't explicitly stated). Proofs are given here. I don't think dimensions $4$ or $5$ have been worked out.<|endoftext|> -TITLE: Which functions of one variable are derivatives ? -QUESTION [44 upvotes]: This is motivated by this recent MO question. - -Is there a complete characterization of those functions $f:(a,b)\rightarrow\mathbb R$ that are pointwise derivative of some everywhere differentiable function $g:(a,b)\rightarrow\mathbb R$ ? - -Of course, continuity is a sufficient condition. Integrability is not, because the integral defines an absolutely continuous function, which needs not be differentiable everywhere. A. Denjoy designed a procedure of reconstruction of $g$, where he used transfinite induction. But I don't know whether he assumed that $f$ is a derivative, or if he had the answer to the above question. - -REPLY [15 votes]: Here are a few characterizations of derivatives: - -D. Preiss and M. Tartaglia -On Characterizing Derivatives -Proceedings of the American Mathematical Society, Vol. 123, No. 8 (Aug., 1995), 2417-2420. -Chris Freiling, On the problem of characterizing derivatives, -Real Analysis Exchange 23 (1997/98), no. 2, 805-812. -Brian S. Thomson, On Riemann Sums -Real Analysis Exchange 37 (2011/12), 1-22. [You can download the PDF file here.] - -The problem was first posed by W. H. Young. We include in our article about the Youngs a full quote stating his problem; -Bruckner, Andrew M. and Thomson, Brian S. -Real variable contributions of G. C. Young and W. H. Young. -Expo. Math. 19 (2001), no. 4, 337–358. [You can download the PDF file here.]<|endoftext|> -TITLE: Estimating the size of reduction of rational points on $\mathbb{G}_m^2$ -QUESTION [7 upvotes]: Hi, -Let $\Gamma$ be a free subgroup of rank 2 in $\mathbb{G}_m^2(\mathbb{Q})$. For all but finitely many primes p we can reduce $\Gamma$ modulo p. Let $S$ be the of primes for which $\Gamma$ does not reduce modulo p, and for any $p$ not in $S$, let $\gamma_p$ be the size of $\Gamma \mod p$. My question is what is known about the function - -$f(x)= \sum_{p\not\in S,\ p\leq x}\frac{\log p }{\gamma_p}$ - -In particular what is the asymptotic behavior of $f$? Is the corresponding infinite series convergent whenever $\Gamma$ is not contained in an algebraic subgroup of $\mathbb{G}_m^2$? Do you know of any references that might be relevant to those questions? -Thanks in advance, - -REPLY [3 votes]: I would just like to give a small update for the question. In my thesis https://epub.uni-bayreuth.de/1721/1/thesis.pdf I showed that the group $\Gamma\ \mod{p}$ has two generators for almost all primes p. So I would conjecture that on average $\gamma_p \sim p^2$ which would imply that the sum above indeed converges.<|endoftext|> -TITLE: Generalizations of Dehn-Nielsen-Baer -QUESTION [8 upvotes]: For a manifold $M$, define the mapping class group $Mod(M)$ to be the set of self-diffeomorphisms of $M$, modulo isotopy. In symbols, $Mod(M) = \pi_0 Diff(M)$. Of course, every self-diffeomorphism gives an automorphism of $\pi_1(M)$, well-defined up to conjugacy because of basepoint issues. Thus we have a homomorphism $\sigma: Mod(M) \to Out ( \pi_1 M)$. -When $M$ is a surface, the Dehn-Nielsen-Baer theorem says $\sigma: Mod(M) \to Out ( \pi_1 M)$ is an isomorphism. My question is: what can be said in higher dimensions? -Assuming $M$ is a $K(\pi, 1)$, one can identify $Out ( \pi_1 M)$ with the set of homotopy classes of homotopy equivalences of $M$. Through this lens, injectivity of $\sigma$ is the question of whether two homotopic diffeomorphisms need to be isotopic. Surjectivity of $\sigma$ is the question of whether every homotopy equivalence is homotopic to a self-diffeomorphism. -From a naive point of view, both injectivity and surjectivity seem hard. What is known about them? - -REPLY [2 votes]: For $M$ oriented closed hyperbolic, I guess the surjectivity follow from Mostow's rigidity. In fact, $Out(\pi_1(M))\cong Isom(M)$ so the map $\sigma$ should be a projection $Mod^{\pm}(M)\to Isom(M)$ in disguise.<|endoftext|> -TITLE: Why is Class Field Theory the same as Langlands for GL_1? -QUESTION [41 upvotes]: I've heard many people say that class field theory is the same as the Langlands conjectures for GL_1 (and more specifically, that local Langlands for GL_1 is the same as local class field theory). Could someone please explain why this is true? -My background is as follows: I understand the statements of class field theory (in other words, that abelian extensions correspond to open subgroups of the idele class group, and the quotient is the Galois group of that abelian extension). I know what modular forms are and what a group representation is, but not much more than that. So I'm looking to see why the statement of class field theory that I know is essentially the same as a certain statement about L-functions, representations, or automorphic forms, in such a way that a more advanced mathematician could easily recognize the latter statement as Langlands in dimension 1. - -REPLY [25 votes]: What you are looking for is the correspondence between algebraic Hecke characters over a number field $F$ and compatible families of $l$-adic characters of the absolute Galois group of $F$. This is laid out beautifully in the first section of Laurent Fargues's notes here. -EDIT: In more detail, as Kevin notes in the comments above, an automorphic representation of $GL(1)$ over $F$ is nothing but a Hecke character; that is, a continuous character -$$\chi:F^\times\setminus\mathbb{A}_F^\times\to\mathbb{C}^\times$$ -of the idele class group of $F$. You can associate $L$-functions to these things: they admit analytic continuation and satisfy a functional equation. This is the automorphic side of global Langlands for $GL(1)$. -How to go from here to the Galois side? Well, let's start with the local story. Fix some prime $v$ of $F$; then the automorphic side is concerned with characters -$$\chi_v:F_v^\times\to\mathbb{C}^\times$$ -Local class field theory gives you the reciprocity isomorphism -$$rec_v:W_{F_v}\to F_v^\times,$$ -where $W_{F_v}$ is the Weil group of $F_v$. Then $\chi_v\circ rec_v$ gives you a character of $W_{F_v}$. This is local Langlands for $GL(1)$. The matching up local $L$-functions and $\epsilon$-factors is basically tautological. -We return to our global Hecke character $\chi$. Recall that global class field theory can be interpreted as giving a map (the Artin reciprocity map) -$$Art_F:F^\times\setminus\mathbb{A}_F^\times\to Gal(F^{ab}/F),$$ -where $F^{ab}$ is the maximal abelian extension of $F$. Local-global compatibility here means that, for each prime $v$ of $F$, the restriction $Art_F\vert_{F_v^\times}$ agrees with the inverse of the local reciprocity map $rec_v$. -Since $Art_F$ is not an isomorphism, we do not expect every Hecke character to be associated with a Galois representation. What is true is that $Art_F$ induces an isomorphism from the group of connected components of the idele class group to $Gal(F^{ab}/F)$. In particular, any Hecke character with finite image will factor through the reciprocity map, and so will give rise to a character of $Gal(F^{ab}/F)$. This is global Langlands for Dirichlet characters (or abelian Artin motives). -But we can say more, supposing that we have a certain algebraicity (or arithmeticity) condition on our Hecke character $\chi$ at infinity. The notes of Fargues referenced above have a precise definition of this condition; I believe the original idea is due to Weil. The basic idea is that the obstruction to $\chi$ factoring through the group of connected components of the idele class group (and hence through the abelianized Galois group) lies entirely at infinity. The algebraicity condition lets us "move" this persnickety infinite part over to the $l$-primary ideles (for some prime $l$), at the cost of replacing our field of coefficients $\mathbb{C}$ by some finite extension $E_\lambda$ of $\mathbb{Q}_l$. This produces a character -$$\chi_l:F^\times\setminus\mathbb{A}_F^\times\to E_\lambda^\times$$ -that shares its local factors away from $l$ and $\infty$ with $\chi$, but now factors through $Art_F$. Varying over $l$ gives us a compatible family of $l$-adic characters associated with our automorphic representation $\chi$ of $GL(1)$. The $L$-functions match up since their local factors do.<|endoftext|> -TITLE: infinite configuration of lines -QUESTION [12 upvotes]: I was looking at some random problems and questions I liked when I was in high school and I found this one which I still cannot prove. -Does there exist a configuration of a countable number of straight lines in the plane such that: -1) no two are parallel -2) no three are concurrent -3) any bounded subset of the plane is intersected by a finite number of lines -4) the area of every minimal polygon is equal, where a minimal polygon is a polygon formed by a finite subset of the set of lines such that no lines pass through the inside of the polygon. -The answer is certainly no, but it is not that easy to prove. Any ideas? - -REPLY [13 votes]: I can sketch a proof based on assuming this "finite" result: -A). For any pentagonal star one of the 5 triangles will have area strictly -smaller than that of the central pentagon. -(I think a brute force attack should yield a proof here.) - -The proof of the original problem would then go as follows. -b). A) generalizes to n-agons by considering the pentagon -spanned by any 5 vertices. - -c). b) implies that a tiling with polygons of EQUAL AREA -is not possible unless all polygons are either triangles or -quadrilaterals. -d). Take one 4-tile and continue tiling next to it inside -the cone enclosed by the converging lines of 2 opposite edges; -we have a sequence of quadrilaterals which must end with a -triangle were the the 2 lines meet. This shows that the tiling -must contain a 3-tile somewhere. - -e). By d) take a 3-tile and continue tiling outwards, inside each of the -3 beams generated by the lines of each pair of edges; the original 3-tile -will be the first tile in each beam, but every other tile after it must be -a 4-tile (build them one at a time and keep using c)). -We can ignore what happens in the 3 leftover cones radiating from the 3 vertices. - -f). In one of the 3 beams (which now look like ladders) take any one of the new rungs -from step e) and extend it - that line will then collide with one of the other -2 beams (but cannot overlap with any of its rungs). That will cut -one of the 4-tiles, creating a 5-tile. - -Apologies for bumping up the question repeatedly while trying to edit my answer.<|endoftext|> -TITLE: Why can Diophantine equations represent exponential growth? -QUESTION [11 upvotes]: The wikipedia page on Matiyasevich's theorem challenges: - -Unfortunately there seems to be as yet no short intuitive explanation as to why Diophantine equations can represent exponential growth only a daunting list of number theoretic lemmas (Note: I've only heard explanations from logicians so if any number theorist wants to show me I'm wrong please do). - -Anyone care to have a go? - -REPLY [2 votes]: The intuition I have is that relations definable by polynomials is complicated enough that projections can give rise to very complicated sets. The essential point here is not exponential growth, that comes from the unboundedness of the quantifiers, the essential point is that the graph of a very fast growing function can be represented by a polynomial. -On the other hand, if the question is about how polynomial equations can represent graphs of complicated function, then I guess the best explanation is by the people who came up with the idea of how to use these equations to represent the graph of a fast growing function, which can find in My Collaboration with JULIA ROBINSON - by Yuri MATIYASEVICH (skip to the line numbered (6) in the text) or in his book. - -"I saw at once that Julia Robinson had a fresh and wonderful idea. It was connected with the special form of Pell's equation -(6) -$$x^2-(a^2-1)y^2 = 1. $$ -Solutions $<\chi_0, \psi_0>, <\chi_1, \psi_1>,\cdots, <\chi_n, \psi_n>,\cdots$ of this equation listed in the order of growth satisfy the recurrence relations -(7) -$$\chi_{n+1}=2a\chi_n-\chi_{n-1}$$ -$$\psi_{n+1}=2a\psi_n-\psi_{n-1}$$ -It is easy to see that for any $m$ the sequences $\chi_0,\chi_1,\cdots, \psi_0,\psi_1,\cdots$ are purely periodic modulo $m$ and hence so are their linear combinations. Further, it is easy to check by induction that the period of the sequence -(8) $$\psi_0,\psi_1,\cdots,\psi_n,\cdots (\mod a-1)$$ -is -(9) $$0, 1, 2,\cdots, a - 2,$$ -whereas the period of the sequence -(10) $$\chi_0-(a-2)\psi_0,\chi_1-(a-2)\psi_1,\cdots, -\chi_n-(a-2)\psi_n,\cdots (\mod 4a-5)$$ -begins with -(11) $$2^0, 2^1, 2^2,\cdots$$ -The main new idea of Julia Robinson was to synchronize the two sequences by imposing a condition $G(a)$ which would guarantee that -(12) $$\text{the length of the period of (8) is a multiple of the length of the period of (10).}$$ -If such a condition is Diophantine and is valid for infinitely many values of $a$, then one can easily show that the relation $a = 2^c$ is Diophantine. Julia Robinson, however, was unable to find such a $G$ and, even today, we have no direct method for finding one. -I liked the idea of synchronization very much and tried to implement it in a slightly different situation. When, in 1966, I had started my investigations on Hilbert's tenth problem, I had begun to use Fibonacci numbers and had discovered (for myself) the equation -(13) $$x^2 - xy - y^2=\pm 1$$ -which plays a role similar to that of the above Pell's equation; namely, Fibonacci numbers $\phi_n$ and only they are solutions of (13). The arithmetical properties of the sequences $\psi_n$ and $\phi_n$ are very similar. In particular, the sequence -(14) $$0, 1, 3, 8, 21, \cdots $$ -of Fibonacci numbers with even indices satisfies the recurrence relation -(15)$$ \phi_{n+1}=3\phi_n-\phi_{n-1}$$ -similar to (7). This sequence grows like $[(3+\sqrt 5)/2]^n$ and can be used instead of (11) for constructing a relation of exponential growth. The role of (10) can be played by the sequence -(16) $$\psi_0,\psi_1,\cdots,\psi_n,\cdots (\mod a-3)$$ -because it begins like (14). Moreover, for special values of a the period can be determined explicitly; namely, if -(17) $$a = \phi_{2k}+\phi_{2k+2},$$ -then the period of (16) is exactly -(18) $$0,1,3,\cdots,\phi_{2k},-\phi_{2k},\cdots,-3,1.$$ -The simple structure of the period looked very promising. -I was thinking intensively in this direction, even on the night of New Year's Eve of 1970, and contributed to the stories about absentminded mathematicians by leaving my uncle's home on New Year's Day wearing his coat. On the morning of January 3, I believed I had found a polynomial B as in (5) but by the end of that day I had discovered a flaw in my work. But the next morning I managed to mend the construction. -What was to be done next? As a student I had had a bad experience when once I had claimed to have proved unsolvability of Hilbert's tenth problem, but during my talk found a mistake. I did not want to repeat such an embarrassment, and something in my new proof seemed rather suspicious to me. I thought at first that I had just managed to implement Julia Robinson's idea in a slightly different situation. However, in her construction an essential role was played by a special equation that implied one variable was exponentially greater than another. My supposed proof did not need to use such an equation at all, and that was strange. Later I realized that my construction was a dual of Julia Robinson's. In fact, I had found a Diophantine condition $H(a)$ which implied that -(19) $$\text{the length of the period of (16) is a multiple of the length of the period of (8).}$$ -This $H$, however, could not play the role of Julia Robinson's $G$, which resulted in an essentially different construction."<|endoftext|> -TITLE: Geometric interpretation of the argument of the Fubini-Study bilinear form on projective space? -QUESTION [7 upvotes]: Let $s_0$ and $s_1$ be the holomorphic sections of the tautological bundle $O(1)$ over the complex projective line ${\mathbb{CP}}^1$ which correspond to the functions $1$ and $\frac{x_1}{x_0}$ in the open set $U_0= \{x_0\neq 0\}$. Let $U(z,w)=s_0(z)\overline{s_0(w)} + s_1(z)\overline{s_1(w)}$ be the bilinear form corresponding to a Fubini-Study metric. Normalizing, we obtain a function $\tilde{U}(z,w)=\frac{u(z,w)}{\sqrt{U(z,z)U(w,w)}}$. -The modulus of $\tilde{U}$ has a geometric interpretation. I asked about this in this question. By choosing a suitably normalized metric on ${\mathbb{CP}}^1$, we have the identity $|\tilde{U}(z,w)|^2= cos^2 d(z,w)$. -My question is whether the argument of $\tilde{U}$ has an anologous geometric interpretation in terms of "complex-angular" measure between the points $z$ and $w$? - -REPLY [5 votes]: I'm not sure what you mean by complex-angular measure, but there is indeed a geometric interpretation. -Suppose $z$ and $w$ have homogeneous coordinates $[z_0,z_1]$ and $[w_0,w_1]$, respectively. Then -$$ -U(z,w) = 1 + \frac{z_1}{z_0} \frac{\overline{w}_1}{\overline{w}_0}, -$$ -and hence -$$ -\widetilde{U}(z,w) = \frac{\overline{z}_0}{|z_0|} \frac{w_0}{|w_0|} -\frac{z_0 \overline{w}_0 + z_1 \overline{w}_1}{\sqrt{(|z_0|^2+|z_1|^2)(|w_0|^2+|w_1|^2)}}. -$$ -So the question is really asking what the complex phase of $\overline{z}_0 w_0 (z_0 \overline{w}_0 + z_1 \overline{w}_1)$ means. -To put this in a more general setting, let's look at $\mathbb{C}\mathbb{P}^{n-1}$. We'll represent points by unit vectors in $\mathbb{C}^n$, and we'll use the Hermitian form $\langle \cdot, \cdot \rangle$ on $\mathbb{C}^n$. As you pointed out, the absolute value of the inner product $\langle x,y \rangle$ measures the distance between $x$ and $y$ in $\mathbb{C} \mathbb{P}^{n-1}$. The complex phase of $\langle x,y \rangle$ is not well-defined, because we can rotate $x$ and $y$ by unit complex numbers without changing the points in projective space. However, the phase change along a cycle of points is invariant: in the expression -$$ -\langle x,y \rangle \langle y,z \rangle \langle z,x \rangle, -$$ -phase rotations for $x$, $y$, or $z$ won't change the answer (because they occur once on the linear side of the Hermitian form and once on the conjugate-linear side). -I'm not sure what to call this quantity. I think it's more or less the Pancharatnam phase from physics, but that's a little far afield from what I know. In any case, it gives an invariant for $k$-tuples of points under the action of the unitary group $U(n)$. For two points, you just get $\langle x,y \rangle \langle y,x \rangle = |\langle x,y \rangle|^2$, so it gives the Fubini-Study metric again. For more than three points, you get a perfectly good invariant, but it's not as essential because it can generically be reduced to $2$- and $3$-point invariants. For example, -$$ -\langle w,x \rangle \langle x,y \rangle -\langle y,z \rangle \langle z,w \rangle = -\frac{\langle w,x \rangle \langle x,y \rangle -\langle y,w \rangle - \cdot \langle y,z \rangle \langle z,w \rangle -\langle w,y \rangle -}{\langle y,w \rangle \langle w,y \rangle}, -$$ -as long as $w$ and $y$ aren't orthogonal. However, when $\langle w,y \rangle = \langle x,z \rangle = 0$, the $4$-point invariant is not determined by lower-order invariants (the $3$-point invariants all vanish, and the $2$-point invariants don't determine the complex phases at all). -The unitary group $U(n)$ maps one $k$-tuple of points in $\mathbb{C} \mathbb{P}^{n-1}$ to another if and only if all the invariants of corresponding sub-tuples agree, and as long as there are no orthogonal pairs of points, it suffices to get agreement for pairs and triples. So these invariants completely classify the orbits of $U(n)$ acting on finite subsets of $\mathbb{C}\mathbb{P}^{n-1}$. -The phase of $\widetilde{U}(z,w)$ in the original example is a special case of this invariant. It's the $3$-point invariant of $[z_0,z_1]$, $[w_0,w_1]$, and $[1,0]$ (assuming the Hermitian form is conjugate-linear in the second variable). Specifically, that gives -$$ -(z_0 \overline{w}_0 + z_1 \overline{w}_1) w_0 \overline{z}_0, -$$ -as desired. The appearance of $[1,0]$ here is a little arbitrary, but that's just a consequence of how the original problem was set up. -I've cheated a bit in one respect: the question asked about $\mathbb{C}\mathbb{P}^1$, and on $\mathbb{C} \mathbb{P}^1$ the higher-order invariants are not so exciting, because the orbits of $U(2)$ on arbitrary $k$-tuples of points are simply determined by the pairwise distances between the points. (It's essentially the same as $SO(3)$ acting on $S^2$, in which case this is a familiar fact.) So if we restrict attention to $\mathbb{C}\mathbb{P}^1$, then it's still true that the answer is a $3$-point invariant, but it's not really giving any new information beyond the Fubini-Study metric. However, in higher dimensions it does.<|endoftext|> -TITLE: Positive matrices matrices over commutative rings -QUESTION [6 upvotes]: Assume that $R$ is a commutative ring with a ring compatible ordering and let $A$ and $B$ be symmetric $n\times n$ matrices with entries in $R$ such that $\sum x_iA_{ij}x_j\geq 0$ and $\sum x_iB_{ij}x_j\geq 0$ for all $x=(x_1,\ldots,x_n)\in R^n$. Is it true that $\operatorname{tr}(AB)=\sum A_{ij}B_{ji}\geq 0$? -This is true for matrices, and the proof usually involves the Kronecker product and diagonalization of the matrices, which is problematic over rings (well, not the Kronecker product of course). Does anyone know how to find a theorem like this for rings? -In general, I wonder how much of the standard properties of positive matrices that can be extended to rings? - -REPLY [4 votes]: I tend to believe that the answer to your question, in the generality you want, is negative. Since I am not sure of the proof (and have not written it up in detail), I am making this answer community wiki. Feel free to edit. -We will consider the case $n=2$, and let $R$ be the polynomial ring $\mathbb Z\left[x,y,z,x',y',z'\right]$. Define the nonnegative cone on $R$ to consist of all polynomials which can be written as sums of products of squares and polynomials of the form $a^2x+2aby+b^2z$ with $a,b\in R$ or of the form $a^2x'+2aby'+b^2z'$ with $a,b\in R$. -Let $A$ be the matrix $\left(\begin{array}{cc}x&y\\ y& z\end{array}\right)$, and let $B$ be the matrix $\left(\begin{array}{cc}x'& y'\\ y'& z'\end{array}\right)$. Then, $A$ and $B$ are nonnegative definite (meaning that your condition is satisfied), but I claim that $\mathrm{Tr}\left(AB\right)$ does not lie in the nonnegative cone. Why? -A polynomial in $\mathbb Z\left[x,y,z,x',y',z'\right]$ is said to be positively led if its leading monomial with respect to the lexicographic ordering ($x > y > z > x' > y' > z'$) is a positive integer. The positively led polynomials form a cone, which contains our nonnegative cone because: -(1) the sum and the product of two positively led polynomials are positively led (this is trivial); -(2) squares of polynomials are positively led (this is very easy); -(3) polynomials of the form $a^2x+2aby+b^2z$ with $a,b\in R$ or of the form $a^2x'+2aby'+b^2z'$ with $a,b\in R$ are positively led. (Proving this requires some work. For $a^2x+2aby+b^2z$, we wlog assume that $a$ and $b$ are monomials (because we can always restrict our concentration to the leading monomials of $a$ and $b$; all the other monomials don't contribute anything to the leading monomial of $a^2x+2aby+b^2z$), then show that the monomial $aby$ cannot be $\geq$ to each of the monomials $a^2x$ and $b^2z$ at the same time. Similarly for $a^2x'+2aby'+b^2z'$.) -Now, assume that $\mathrm{Tr}\left(AB\right)$ is nonnegative. Then, $\mathrm{Tr}\left(AB\right) = xx'+2yy'+zz'$ can be written as a sum of products of squares and polynomials of the form $a^2x+2aby+b^2z$ with $a,b\in R$ or of the form $a^2x'+2aby'+b^2z'$ with $a,b\in R$. Now, each of the addends in this sum must have degree $\leq 2$ (where "degree" means "total degree"). This is because the degree of the sum of some positively led polynomials is always equal to the highest of their degrees (and not just smaller or equal to it!), so if we had some terms of degree $\geq 3$, they could not cancel out, contradicting to $\deg\left(xx'+2yy'+zz'\right)=2$. -Now another minor lemma: -(4) If a polynomial of the form $a^2x+2aby+b^2z$ with $a,b\in R$ has degree $\leq 2$, then $a$ and $b$ must be integers (and the polynomial has degree $1$). -This is easy to see by the method we used to prove (3): We assume WLOG that $a$ and $b$ are just monomials (because otherwise, we just replace the polynomials $a$ and $b$ by their leading monomials; this does not change the degree of $a^2x+2aby+b^2z$). Now as in (3) we show that the monomial $aby$ cannot be $\geq$ to each of the monomials $a^2x$ and $b^2z$ at the same time. Hence, the leading monomial in $a^2x+2aby+b^2z$ must be either $a^2x$ or $b^2z$ or $a^2x+b^2z$. In each of these cases, we conclude that at least one of $a$ and $b$ must be an integer (i. e., a monomial of degree $0$), let's say that it's $a$. Now this yields that the leading monomial in $a^2x+2aby+b^2z$ must be $b^2z$, so that $b$ too is an integer. This proves (4). -Similarly: -(5) If a polynomial of the form $a^2x'+2aby'+b^2z'$ with $a,b\in R$ has degree $\leq 2$, then $a$ and $b$ must be integers (and the polynomial has degree $1$). -So let us conclude: -We know that $xx'+2yy'+zz'$ can be written as a sum of products of squares and polynomials of the form $a^2x+2aby+b^2z$ with $a,b\in R$ or of the form $a^2x'+2aby'+b^2z'$ with $a,b\in R$. -But we know that each of the addends has degree $\leq 2$. Thus each of these addends is either a square or the product of two polynomials of the form $a^2x+2aby+b^2z$ with $a,b\in \mathbb Z$ or of the form $a^2x'+2aby'+b^2z'$ with $a,b\in \mathbb Z$. (In fact, any other combination would make the degree too high; in particular, multiplying with a square increases the degree by $\geq 2$ unless the square is just an integer square, and multiplying by a polynomial of the form $a^2x+2aby+b^2z$ with $a,b\in R$ or of the form $a^2x'+2aby'+b^2z'$ with $a,b\in R$ increases the degree by $1$ if $a,b\in\mathbb Z$ or by $\geq 3$ otherwise (due to (4) and (5)).) -We can WLOG assume that all squares occuring in the sum are squares of homogeneous linear polynomials (because we can simply remove the constant terms; here we use that $xx'+2yy'+zz'$ is homogeneous of degree $2$). So we have -(6) $xx'+2yy'+zz' = \left(\text{sum of squares of some homogeneous linear polynomials}\right)$ -$ + \sum_{i\in I} \left(a_i^2x+2a_ib_iy+b_i^2z\right)\left(A_i^2x+2A_iB_iy+B_i^2z\right)$ -$ + \sum_{j\in J} \left(a_j^2x+2a_jb_jy+b_j^2z\right)\left(A_j^2x'+2A_jB_jy'+B_j^2z'\right)$ -$ + \sum_{k\in K} \left(a_k^2x'+2a_kb_ky'+b_k^2z'\right)\left(A_k^2x'+2A_kB_ky'+B_k^2z'\right)$, -where $I$, $J$, $K$ are three disjoint finite sets, and $a_i$, $b_i$, $A_i$, $B_i$, $a_j$, $b_j$, ... are integers. -Now, forget about the lexiographic ordering I introduced (the one that had $x > y > z > x' > y' > z'$), and introduce a new one, with $y > \text{all other variables}$. With the respect to this new ordering, the left hand side of (6) has leading monomial $yy'$. Thus, the right hand side also must have leading monomial $yy'$. Therefore, none of the homogeneous linear polynomials whose squares appear on the right hand side of (6) can contain the variable $y$ (because the square of any such polynomial would contain $y^2$, and thus the leading monomial of the right hand side (6) would be $y^2$ (here we are using again the fact that the degree of the sum of some positively led polynomials is always equal to the highest of their degrees)). But this means that none of the squares on the right hand side (6) can contain the monomial $yy'$ (because such a monomial could only come from a $y$ inside the square, but we have ruled out this possibility). Therefore, the coefficient of $yy'$ on the right hand side of (6) is $\sum_{j\in J}2a_jb_j\cdot 2A_jB_j$. This is divisible by $4$. The coefficient of $yy'$ on the left hand side of (6) is not divisible by $4$. Contradiction. -At least if I didn't mess anything up. Given the length of the proof, this is rather improbable. -Anyway it still keeps the question open whether we are in more luck if we require $R$ to be a $\mathbb Q$-algebra. -EDIT: I think that even if $R$ is supposed to be a $\mathbb Q$-algebra, then your answer is No. Let me try to prove it: -Replace $\mathbb Z$ by $\mathbb Q$, and "integers" by "rationals" throughout the above. We can still get to (6), but we don't get the contradiction through divisibility by $4$ anymore. -Consider (6) again. I have showed that none of the homogeneous linear polynomials whose squares appear on the right hand side of (6) can contain the variable $y$. But the same argument works for any other variable instead of $y$ (just consider the lexicographic order where this variable is higher than all others). This shows that none of the homogeneous linear polynomials whose squares appear on the right hand side of (6) can contain any variables. In other words, these squares are $0$. This simplifies (6) to -(7) $xx'+2yy'+zz' = \sum_{i\in I} \left(a_i^2x+2a_ib_iy+b_i^2z\right)\left(A_i^2x+2A_iB_iy+B_i^2z\right)$ -$ + \sum_{j\in J} \left(a_j^2x+2a_jb_jy+b_j^2z\right)\left(A_j^2x'+2A_jB_jy'+B_j^2z'\right)$ -$ + \sum_{k\in K} \left(a_k^2x'+2a_kb_ky'+b_k^2z'\right)\left(A_k^2x'+2A_kB_ky'+B_k^2z'\right)$. -Unless the sum $\sum_{i\in I} \left(a_i^2x+2a_ib_iy+b_i^2z\right)\left(A_i^2x+2A_iB_iy+B_i^2z\right)$ on the right hand side of (7) is identically zero, it contributes at least one of the monomials $x^2,xy,xz,y^2,yz,yx,z^2,zx,zy$ with nonzero coefficient to the right hand side of (7), and no other term on the right hand side of (7) can kill this monomial. But this is impossible, as none of these monomials appears on the left hand side of (7) ! Thus, the sum $\sum_{i\in I} \left(a_i^2x+2a_ib_iy+b_i^2z\right)\left(A_i^2x+2A_iB_iy+B_i^2z\right)$ must be identically zero. Similarly, the same holds for the sum $\sum_{k\in K} \left(a_k^2x'+2a_kb_ky'+b_k^2z'\right)\left(A_k^2x'+2A_kB_ky'+B_k^2z'\right)$. Now (7) simplifies to -(8) $xx'+2yy'+zz' = \sum_{j\in J} \left(a_j^2x+2a_jb_jy+b_j^2z\right)\left(A_j^2x'+2A_jB_jy'+B_j^2z'\right)$. -Now, the coefficient of $xz'$ on the right hand side of (8) is $\sum_{j\in J}a_j^2B_j^2$. But the coefficient of $xz'$ on the left hand side of (8) is zero. Thus, $\sum_{j\in J}a_j^2B_j^2=0$. This means that every $j\in J$ satisfies either $a_j=0$ or $B_j=0$. Similarly, every $j\in J$ satisfies either $b_j=0$ or $A_j=0$. Therefore, every $j\in J$ must belong to one of the following pigeonholes: -Pigeonhole 1: $j$'s satisfying $a_j=0\text{ and }b_j=0$. -Pigeonhole 2: $j$'s satisfying $B_j=0\text{ and }b_j=0$. -Pigeonhole 3: $j$'s satisfying $a_j=0\text{ and }A_j=0$. -Pigeonhole 4: $j$'s satisfying $B_j=0\text{ and }A_j=0$. -Any $j$ lying in Pigeonhole 1 can be removed from $J$ without invalidating (8) (because if $j$ lies in Pigeonhole 1, then the addend corresponding to $j$ on the right hand side of (8) is zero and contributes nothing to the sum). Similarly, any $j$ lying in Pigeonhole 4 can be removed from $J$ without invalidating (8). Thus, what remains of (8) is -$xx'+2yy'+zz' = \sum_{j\in \text{Pigeonhole 2}} \left(a_j^2x+2a_jb_jy+b_j^2z\right)\left(A_j^2x'+2A_jB_jy'+B_j^2z'\right)$ -$+ \sum_{j\in \text{Pigeonhole 3}} \left(a_j^2x+2a_jb_jy+b_j^2z\right)\left(A_j^2x'+2A_jB_jy'+B_j^2z'\right)$ -$= \sum_{j\in \text{Pigeonhole 2}} \left(a_j^2x\right)\left(A_j^2x'\right) + \sum_{j\in \text{Pigeonhole 3}} \left(b_j^2z\right)\left(B_j^2z'\right)$. -This is absurd.<|endoftext|> -TITLE: local-global principle for units -QUESTION [10 upvotes]: Say that $L/K$ is a quadratic extension of number fields with $K$ totally real and $L$ totally imaginary. -Then the Hasse norm theorem says that an element of $K$ that is everywhere a local norm is the global norm of something in $L$. (In fact, this is more generally true, whenever $L/K$ is cyclic.) -Is it also the case that an element of $\mathcal{O}_K^*$ that is everywhere a local norm is the global norm of something in $\mathcal{O}_L^*$? - -REPLY [17 votes]: No: there are units that are norms of elements but not norms of units. The simplest example -are real quadratic number fields $Q(\sqrt{m}\,)$ with $m$ a sum of two squares such that the -fundamental unit has positive norm, for example $m = 34$. -For finding other examples, one may look at the ambiguous class number formulas for cyclic extensions $L/K$ of prime degree $p$: -$$ Am(L/K) = h_K \frac{\prod e(P)}{p(E_K : E_K \cap NL^\times)}, \qquad - Am_s(L/K) = h_K \frac{\prod e(P)}{p(E_K : N E_L)}. $$ -Here $Am$ denotes the subgroup of ideal classes fixed by the Galois group $G$, -$Am_s$ the subgroup of classes generated by ideals fixed by $G$, and $e(P)$ is the -ramification index of the prime $P$. The index -$$ (Am:Am_s) = (E_K \cap NL^\times : NE_L) $$ -is the obstruction to the local-global principle for units. -Edit. Let me, however, point out that D. Folk (When are global units -norms of units?, Acta Arith. 76 (1996), 145-147) has proved the following -result: if $L/K$ is normal and if $H$ denotes the Hilbert class field of $L$, -then a unit from $K$ that is a local norm in all completions of $H/K$ is the -norm of a unit from $L$. This suggests the following question: given a cyclic -extension $L/K$, is there an unramified abelian extension $E/L$ with the -property that a unit from $K$ is the norm of a unit from $L$ if and only if -it is a local norm in all completions of $E/K$?<|endoftext|> -TITLE: How did Birch and Swinnerton Dyer arrive at their conjecture? -QUESTION [6 upvotes]: I suspect that they knew that the $L-$function is defined only for $Re(s) \gt 3/2$. Did they attempt to evaluate the $L-$function at $s=1$ by plugging $s=1$ in the infinite product $\prod_p (\frac{1}{1-a_pp^{-s}+p^{1-2s}})$? I think that it gives $\prod (\frac{N_p}{p})^{-1}$, which does not make sense under the usual definition of infinite product. -Note: I have posted the same at math.stackexchange but I would like further explanation beyond what was given there. - -REPLY [40 votes]: For what it's worth, here are some historical comments. -Both Birch and S-D spoke in Cambridge a few weeks ago, on the history of their conjecture. To my surprise, both of them emphasized the role not of the $L$-function, but of the Tamagawa number, in their comments on how it all came about. -Tamagawa had introduced this invariant associated to a semisimple algebraic group over a number field, and one can interpret it adelically or as a product of local factors. B and S--D were trying to "compute the Tamagawa number of an elliptic curve" -- or more precisely, to see what the analogue should be in this situation. The local factors came out to be $N_p/p$, at least at the good primes. Historically, the $L$-function came later. I asked S--D why this might have been, and he said something like "Weil was pushing the $L$-function as being of central importance, so, naturally, everyone else was avoiding it like the plague". I am not sure this comment is to be taken so seriously -- this is perhaps more a reflection of S-D's sense of humour (he'd made some rather caustic comments about Mordell's bridge playing skills earlier, again probably just to get laughs, and he succeeded admirably in getting them). But you have to remember the resources available to them at the time: they were initially not thinking about the $L$-function, but they did have this access to this gigantic machine, the size of a lecture theatre, that was capable of computing the product of $N_p/p$ for all primes less than 1000, and this for them was basically a miracle, because ten years prior to that if you wanted to do this calculation then you'd better have a lot of pencils/paper handy. -So they used what they had, they were focussed initially on Tamagawa numbers (this is I guess the reason that the fudge factors at the bad primes became known as Tamagawa factors?), they had access to a computing machine and S--D knew how to use it, and perhaps crucially one should stress that whilst we now know the $L$-function to be of central importance, it was perhaps not so clear at that time. There was no Langlands programme, there was no converse theorem -- this was the late 50s. It was only a couple of years later, when Birch was talking to Shimura, that Shimura told him that in the CM case one could actually evaluate the $L$-function at 1 in concrete cases and get concrete numbers, and then Davenport told Birch a concrete algorithm which would work and could be done by hand. I have seen with my own eyes the piece of paper in which Davenport sketched the idea to Birch, and Birch has written "keep this" on the top and underlined it! Birch then proceeds to compute various explicit examples of special values of $L$-functions on the next few pages, in the CM case, but maybe this was already after the first work had been published. -I mentioned all this to Rene Schoof yesterday and he claimed that there were pictures of the pieces of paper on youtube of all places (I know William Stein was taking pictures frantically -- my eyes were just popping out of my head -- all these really important historical documents, that Birch claimed were just gathering dust in a wardrobe at home!).<|endoftext|> -TITLE: Destroying the P-filter-property -QUESTION [8 upvotes]: It is known that if a forcing notion is proper, then every P-filter will generate a P-filter in the generic extension (see, e.g., Shelah, Proper and Improper Forcing, VI.5) -On the other hand, if we start collapsing cardinals, we can destroy the P-filter property. For example, making a base of a P-point countable will add a sequence of elements from the filter (namely, a complete enumeration of that base) that serves as a counterexample for the P-filter property in the extension. -So my question is: - -Are there examples of "nicer" (e.g., not collapsing cardinals) forcing notions that destroy P-filters in this way, i.e., add a sequence in the filter that the filter cannot decide? -More spectacularly, is there maybe a forcing notion that could preserve a P-point as an ultrafilter while destroying the P-filter property? - -EDIT: As Martin Goldstern pointed out, I should add that I'm interested in filters on $\omega$. - -REPLY [2 votes]: A second partial (and not very deep, sorry) answer: Assume that at least one of the following holds: - -zero sharp does not exist, i.e., Jensen's covering lemma holds. -The continuum is below $\aleph_\omega$ (or at least: your filter is generated by less than $\aleph_\omega$ many sets). - -Then the answer is again "no" (if the forcing is not allowed to collapse cardinals). -Proof: Every new countable subset of the filter base is contained in an old countable subset of the filter base. (This is well known, but for the sake of completeness I give a sketch of the proof. Fix a bijection between the filter base and some ordinal $\alpha$. Every new countable set $A \subseteq \alpha$ is contained in an old set $B$ of size $\aleph_n$, for some $n$. [Under assumption 1, the covering lemma gives $n\le 1$; assumption 2 just says outright that there is some $n$.] Now fix $A$ and choose $n$ as small as possible. Using a bijection from $B$ to $\aleph_n$ in $V$, we may wlog assume that the original set $A$ was a subset of $\omega_n$. If $n>0$, then the countable set $A$ is bounded in $\omega_n$, so we can cover $A$ by an ordinal of cardinality $\aleph_{n-1}$, contradiction.)<|endoftext|> -TITLE: Uniruled + Picard number 1 = Fano? -QUESTION [6 upvotes]: Hello, I'm a newbie to mathoverflow. I reading a paper about Fano varieties (over C) and there is an assumption that uniruled varieties with picard number 1 are Fano...why is this true? Sorry if this is obvious thing. I am just learning the definitions still. - -REPLY [2 votes]: Just to give yet another argument using more or less the definition of uniruledness. -$X$ is uniruled if there exists a dominant morphism from a variety $Y\times \mathbb P^1$ where $\dim Y=\dim X-1$. Therefore the image of $\{y\}\times \mathbb P^1$ for a general $y\in Y$ moves in a family of dimension $\dim Y$ in $X$ and hence has a semi-positive normal bundle and in particular the determinant of its normal bundle has to be a non-negative line bundle. Then the adjunction formula shows that $-K_X$ is positive on this curve. -If the Picard number is $1$, then $\mathrm{Pic}X\otimes \mathbb Q\simeq \mathbb Q$. If $L$ is a fixed ample divisor, then for any divisor $D$ there exists $a,b\in \mathbb N$ such that $aD\sim bL$. Since there exists at least one curve on which $-K_X$ is positive, it follows that $-K_X\sim \alpha L$ for some $\alpha\in \mathbb Q_+$ and hence it is ample itself. It follows that $X$ is Fano.<|endoftext|> -TITLE: Car movement - differential geometry interpretation. -QUESTION [15 upvotes]: I've posted this on Math Stack Exchange and I didn't get any answer in a couple of days, so I'll try and post it here too. -The problem presented below is from my differential geometry course. The initial reference is Nelson, Tensor Analysis 1967. The car is modelled as follows: - -Denote by $C(x,y)$ the center of the back wheel line, $\theta$ the angle of the direction of the car with the horizontal direction, $\phi$ the angle made by the front wheels with the direction of the car and $L$ the length of the car. -The possible movements of the car are denoted as follows: - -steering: $S=\displaystyle\frac{\partial}{\partial \phi}$; -drive: $D=\displaystyle\cos \theta \frac{\partial}{\partial x}+\sin\theta \frac{\partial}{\partial y}+\frac{\tan \phi}{L}\frac{\partial}{\partial \theta}$; -rotation: $R=[S,D]=\displaystyle\frac{1}{L\cos^2 \phi}\frac{\partial }{\partial \theta}$; -translation: $T=[R,D]=\displaystyle\frac{\cos \theta}{L\cos^2 \phi}\frac{\partial}{\partial y}-\frac{\sin\theta}{L\cos^2\phi}\frac{\partial}{\partial x}$ - -Where $[X,Y]=XY-YX$ (I can't remember the English word now). All these transformations seem very logical. My question is: - -How can we justify the mathematical interpretation made above, especially the part with the rotations and translations? - -The interpretations are quite interesting: - -from the expression of $D$, when the car is shorter, you can change the orientation of the car very easily, but when it is longer, like a truck, you it is not that easy ( see the term with $\frac{\partial}{\partial \theta}$) -the rotation is faster for smaller cars, and for greater steering angle -translation is easier for smaller cars. - -REPLY [23 votes]: I'm not sure what you mean by "justify", but one way of looking at this is through the Chow theorem. -You've calculated that $R = [S, D]$. This is a statement about vector fields, which are infinitesimal flows. So if you steer by a small amount, followed by driving over a small amount, followed by turning the wheel back into the original position and driving backward, you end up performing a small rotation. You can interpret $T = [R, D]$ in the same way: if you want to parallel park the car (performing a translation), you turn ($R$), drive forward ($D$), turn back ($-R$) and drive backward ($-D$). -A normal car has four degrees of freedom, the angles $\theta$ and $\phi$ on your figure and the coordinates $x, y$ of the center of mass. However, you as the driver can only be drive and steer the car, you can't just translate the car or rotate it around its axis. So you can only control $S$ and $D$. We say that this system is underactuated, meaning that you have less control inputs than there are degrees of freedom in the system. -However, differential geometry comes to the rescue: if you consider the distribution spanned by $S$ and $D$, you'll find that it is totally nonholonomic, meaning that you can create any possible motion (any linear combination of $R$, $D$, $T$ and $S$) by a linear combination of $S$ and $D$ and their (iterated) commutators. Physically, this means that you can perform any motion with the car by driving and steering, although you will sometimes have to execute a nontrivial motion in $S$ and $D$ (look up "control by sinusoids") -If now you have a metric on the distribution spanned by $S$ and $D$ (telling you, for instance, how much it costs to steer resp. drive the car) you have an instance of what is called a sub-Riemannian geometry. (I hope this will entice Richard Montgomery to comment on this question). -I can polish/extend this answer a great deal, so please feel free to ask follow-up questions.<|endoftext|> -TITLE: Number of spanning forests in a graph -QUESTION [9 upvotes]: Hello, -I have two questions that have been bugging me recently. The first is about the number of spanning forests in a graph and the second is about enumerating these with edge labels. -Q1: I am aware of Kirchhoff's Matrix-Tree theorem regarding the number of spanning trees in a graph. I was wondering if there is a generalization to this theorem that counts the number of spanning k-forests in a graph. What I am mostly interested in is this: is there a method of finding the number of k-forests in a graph by taking a determinant of some matrix? -Q2: Suppose you label each edge as $e_{i,j}$ meaning that you are taking the undirected edge from $v_i$ to $v_j$ in the graph. Then in the Laplacian matrix if you plug in the sum of $e_{i,j}$'s instead of $\deg(v_i)$ and $-e_{i,j}$ instead of -1 when that edge connects vertices $i$ and $j$, you get the combinatorial Laplacian. Taking the determinant of a minor of this matrix gives the Kirchhoff polynomial which is an enumeration of the spanning trees of the graph, where each monomial contains the variables for all the edges in the given tree. My question is whether we can generalize this to spanning forests. - -REPLY [3 votes]: I am surprised no one mentioned the work of Alan Sokal and coworkers on precisely this issue -of weighted enumeration of spanning forests which is related to the $q\rightarrow 0$ limit of the -Potts model as well as the multivariate Tutte polynomial. -A determinant expression corresponds to a Fermionic (Grassmann/Berezin) integral with a quadratic -`action' in the exponential. There is an analogue of the matrix-tree theorem for spanning forests -with a Fermionic integral with quartic action see: -http://arxiv.org/abs/cond-mat/0403271 -which appeared in PRL. -Follow ups such as -http://arxiv.org/abs/0706.1509 -can be found by looking at Sokal's papers on arxiv: -http://arxiv.org/find/grp_math/1/au:+sokal/0/1/0/all/0/1 -Note that there was a whole semester at the Isaac Newton Institute revolving around this -topic: -http://www.newton.ac.uk/programmes/CSM/ -One can even watch the videos of the talks. -The one perhaps most relevant to this question is the talk by Andrea Sportiello in the fourth -workshop.<|endoftext|> -TITLE: Counting subtrees of a random tree ("random Catalan numbers") -QUESTION [13 upvotes]: Given a rooted tree $T$ and an integer $k \geq 1$, let $N_k(T)$ be the number -of subtrees of $T$ containing the root and having exactly $k$ nodes (take $N_k(T)=0$ if $T$ has less than $k$ nodes). -Next, fix an integer $d \geq 2$, and let $T_d$ be the infinite $d$-ary rooted tree (every node has $d$ children). It is well-known (see, e.g. Stanley's "Enumerative combinatorics", theorem 5.3.10) that -$$ N_k(T_d) = \frac{1}{k}{dk \choose k-1} < (ed)^{k-1} ~. $$ -When $d=2$, these are simply the Catalan numbers. -Now suppose that $\mathcal{T}$ is a Galton–Watson tree with offspring distribution $B$ and $\mathbb{E}(B)=\mu \in (1,\infty)$. - -What can be said about the behavior of $N_k(\mathcal{T})$, either in probability or in expectation, when the branching distribution $B$ may be unbounded? - -In particular, it seems likely that under suitable assumptions on $B$, $N_k$ again grows exponentially in $k$. Is it the case, for example, that $N_k/(2e\mu)^{k-1} \to 0$ in expectation (or in probability), perhaps assuming that $B$ has sufficiently large exponential moments? -Perhaps the problem is more combinatorially tractable if one assumes that $B$ has a Poisson distribution? This special case is interesting to me. - -REPLY [8 votes]: Let $p_n$ be the offspring distribution for $B$ and define $q_n = \sum_{m\ge n} p_m (m)_n$, where $(m)_n = m!/(m-n)!$ is the descending factorial. -(new paragraph to appease MO latex scripts). Then the expected number of copies of a rooted tree $\theta$ inside the Galton Watson tree $\mathcal{T}$ is given by $\prod_{v\in\theta} q_{d_v}$ (where $d_v$ does not count the parent of $v$. -This is seen by induction on $\theta$. Given the root degree in $\mathcal{T}$ is $m$, the number of ways to select $d$ children of the root is $(m)_d$. For each of these, the expected number of ways to embed the sub-trees of $\theta$ is given by the formula (induction hypothesis). Since $\mathcal{T}$ is Galton-Watson, these are independent, and the expectation is the product of expectations. -This gives an identity $F(z) = Q(zF(z)$ for the generating function of the expected number of trees with weight $z$ for each edge. It seems that for nice $p$'s the singularity should have the same algebraic type, and so the expected number of trees in $\mathcal{T}$ grows as $C n^{-3/2} z_c^{-n}$. -In the case of the Poisson-Galton-Watson tree, it is easy to see either from the above or by staring into (probability) space that the expected number of copies of any tree $\theta$ is just $\lambda^{|\theta|}$ (still counting edges), so the expectation is just $\lambda^n C_n\sim Cn^{-3/2}(4\lambda)^n$. -Computing higher moments is probably doable in the Poisson case, but seems less fun. I will wait for additional motivation before delving into computations, but if staringinto space yields anything I'll report here.<|endoftext|> -TITLE: Interpreting the Garnir relations in terms of the Yang-Baxter equation -QUESTION [8 upvotes]: One possible construction of the Specht modules goes as follows. -Given a partition $\lambda$ of $n$, we can write down Young's seminormal form for the representation of $S_n$ corresponding to $\lambda$. Writing a basis $e_1, \ldots, e_l$ in bijection with the set of standard Young tableaux $T_1, \ldots, T_l$ of shape $\lambda$, one defines Young's seminormal representation by: -$ e_i \cdot s_j = e_i$ if entries $j$ and $j+1$ are in the same row of $T_i$. -$ e_i \cdot s_j = - e_i$ if entries $j$ and $j+1$ are in the same column of $T_i$. -$s_j$ acts by the matrix $\begin{pmatrix} -1/k & 1 \\ 1-1/k^2 & 1/k \end{pmatrix}$ on the subspace with basis $(e_i, e_{i'})$, where $T_{i'}$ is obtained from $T_i$ by switching entries $j$ and $j+1$, and the entry $j$ appears higher up in $T_i$ than in $T_{i'}$. Here $k$ is the axial distance between boxes with entries $j$ and $j+1$, also equal to the difference in contents (where the content of box $(i,j)$ is $j-i$). -Using this definition, and looking at the Hasse diagram for the dominance ordering on standard Young tableaux of shape $\lambda$, one sees that this is the same as formally writing in the basis $v_i = v R(w)$, where $v$ corresponds to the Young tableau $T$ filled in along rows (so largest in the dominance ordering), $w$ is a reduced word taking $T$ to $T_i$ through the Hasse diagram (i.e. only passing through standard Young tableaux), and if $w = s_{i_1} s_{i_2} \cdots s_{i_k}$, then $R(w) = R_{i_1}(k_1) R_{i_2} (k_2) \cdots R_{i_t}(k_t)$ where` $R_i(k) = \frac{-(k+1)}{k} \cdot \frac{1}{k+1}(1- ks_i)$ and the entries $k_1, \ldots, k_t$ are given by drawing a wiring diagram for $w$ (like a braid), writing the contents of $T$ in order at the top, and putting $b-a$ at the intersection of two strands labelled $a, b$. -The Yang-Baxter equation -$R_i(s) R_{i+1}(s+t) R_i(t) = R_{i+1}(t) R_i(s+t) R_{i+1}(s)$ -guarantees that this is well defined, i.e. that $R(w)$ doesn't depend on which reduced word $w$ we pick to represent our tableau. (This also allows one to prove that Young's seminormal form is actually a representation, i.e. the braid relation is satisfied by the representing matrices) -Now, instead of writing in the basis $v R(w_i)$ and multiplying elements on the right and rearranging, one can instead use the basis $u_i = v w_i$, the natural basis, which defines the Specht module corresponding to $\lambda$. However, if we forget the above construction and try to write down the representation from scratch in this way, we will not be able to write down $u_i s_j$ if $s_j$ doesn't take $T_i$ to another standard tableau. The standard way around this is to use Garnir relations, but is there an interpretation possible in terms of the above $R(w)$? Ideally, one would like to make a claim like $u_i R(w) = 0$ if $w$ doesn't take $T_i$ to another standard tableau, and this would allow us to read off the $u_i w$ inductively, but these $R(w)$ run into problems (i.e. coefficients of $0$ or $-1$) precisely when $w$ doesn't take $T_i$ to another standard tableau. -As an example, consider $\lambda = (4,2)$; this gives the following picture for the Hasse diagram: - -Suppose we have numbered the tableaux $T_1, \ldots, T_9$ in order, row by row from left to right. -We have $T = T_1$, and for instance $w_9 = s_4 s_5 s_3 s_4 s_2$ takes $T$ to $T_9$. Writing out the wiring diagram with the above procedure shows that -$R(w_9) = (s_4 + 1/4)(s_3+1/3)(s_4+1/2)(s_2+1/2)$ -Given the seminormal representation in the basis $e_1, \ldots, e_9$, we can figure out the change of basis matrix and write down the corresponding Specht module. For instance, one finds that $u_9 \cdot s_3 = -u_1-u_2-u_3-u_4-u_5-u_6-u_7-u_8-u_9$. How does one see this only using the $R(w)$ elements, without appealing to the seminormal representation or another construction of the Specht module (say, using polytabloids and straightening)? - -REPLY [4 votes]: You will find this explained in section 5. of: -MR1988991 (2004f:20014) Ram, Arun . -Skew shape representations are irreducible. - Combinatorial and geometric representation theory (Seoul, 2001), - 161--189, Contemp. Math., 325, Amer. Math. Soc., Providence, RI, 2003.<|endoftext|> -TITLE: A toy model for the t-section problem -QUESTION [6 upvotes]: Let $S(x)$ be the area of the yellow curvilinear triangle. I'd like to find a graph for which $S(x)=H(x)$ where $H $ is some prescribed function (small, smooth, vanishing near the endpoints to any order you wish, etc.). Is it always possible or there are some non-obvious hidden restrictions? -alt text http://www.artofproblemsolving.com/Forum/latexrender/pictures/b/4/6/b46dbb397d96f17cce1e421bfaf29b9a80d26c81.png -The question comes from the infamous t-section problem (if you know the areas of all sections of a symmetric convex body by the hyperplanes at some fixed small distance $t$ from the origin (so small that all sections are non-empty), can you recover the body?). The problem is open even on the plane. I do not say that this toy question is directly relevant here but an answer to it will certainly make a few things clearer for me. - -REPLY [2 votes]: There are restrictions. -At most points, $S'(x)$ is the length of the right leg minus the length of the left leg of the curvilinear triangle, perhaps with exceptions on a null set where there are tangencies. If $S(0)=S(1)=0$ then the lengths of these legs are at most $\sqrt{2}(1-x)$ and $\sqrt{2}x$. For almost all $0\le x \le 1$, $S'(x)$ satisfies $-\sqrt{2}\le -\sqrt{2} x \lt S'(x) \lt \sqrt(2) (1-x) \le \sqrt{2}$. This is an extra condition on $H$ which rules out some smooth small functions which have large derivatives near some points, such as $10^6 \exp(-1/(x (1-x))^2)$ for $0\lt x \lt 1$, which has a derivative of $1.132$ at $x=0.436$ although the value of the function is small.<|endoftext|> -TITLE: Ergodic Theorem and Nonstandard Analysis -QUESTION [32 upvotes]: Here is a quote from Lectures on Ergodic Theory by Halmos: - -I cannot resist the temptation of - concluding these comments with an - alternative "proof" of the ergodic - theorem. If $f$ is a complex valued - function on the nonnegative integers, - write $\int f(n)dn=\lim -> \frac{1}{n}\sum_{j=1}^nf(n)$ and whenever - the limit exists call such - functions integrable. If $T$ is a - measure preserving transformation on a - space $X$, then $$ -> \int\int|f(T^nx)|dndx=\int\int|f(T^nx)|dxdn=\int\int|f(x)|dxdn=\int|f(x)|dx<\infty. -> $$ Hence, by "Fubini's theorem" (!), - $f(T^nx)$ is an integrable function of - its two arguments, and therefore, for - almost every fixed $x$ it is an - integrable function of $n$. Can any of - this nonsense be made meaningful? - -Can any of this nonsense be made meaningful using nonstandard analysis? I know that Kamae gave a short proof of the ergodic theorem using nonstandard analysis in A simple proof of the ergodic theorem using nonstandard analysis, Israel Journal of Mathematics, Vol. 42, No. 4, 1982. However, I have to say that I am not satisfied with his proof. It is tricky and not very illuminating, at least for me. Besides, it does not look anything like the so called proof proposed by Halmos. Actually, Kamae's idea can be made standard in a very straightforward manner. See for instance A simple proof of some ergodic theorems by Katznelson and Weiss in the same issue of the Israel Journal of Mathematics. By the way, Kamae's paper is 7 pages and Katznelson-Weiss paper is 6 pages. -To summarize, is there a not necessarily short but conceptually clear and illuminating proof of the ergodic theorem using nonstandard analysis, possibly based on the intuition of Halmos? - -REPLY [13 votes]: I feel the answer is "no", at least while staying true to the spirit of Halmos's text. Halmos's "proof", if valid, would imply something far stronger (and false), namely that $\lim_{N \to \infty} \frac{1}{N} \sum_{n=1}^N f(T_n x)$ converged for almost every x, where $T_1, T_2, \ldots$ are an arbitrary sequence of measure-preserving transformations. This is not true even in the case when X is a two-element set. So any proof of the ergodic theorem must somehow take advantage of the group law $T^n T^m = T^{n+m}$ in some non-trivial way. -That said, though, nonstandard analysis does certainly generate a Banach limit functional $\lambda: \ell^\infty({\bf N}) \to {\bf C}$ which one does induce something resembling an integral, namely the Cesaro-Banach functional -$f \mapsto \lambda( (\frac{1}{N} \sum_{n=1}^N f(n) )_{N \in {\bf N}} ).$ -This does somewhat resemble an integration functional on the natural numbers, in that it is finitely additive and translation invariant. But it is not countably additive, and tools such as Fubini's theorem do not directly apply to it; also, this functional makes sense even when the averages don't converge, so it doesn't seem like an obvious tool in order to demonstrate convergence.<|endoftext|> -TITLE: Relation between monads, operads and algebraic theories -QUESTION [20 upvotes]: I've begun to interest in algebraic theories and their categorical models: in particular monads, generalized multicategories and operads, lawvere theories and their generalization. Is there any reference that treat systematically the relation between such models of theories, where model means a presentation of theory? - -REPLY [10 votes]: All nice recommendations; some more -Lawvere theories and monads -For the connection between monads and Lawvere theories, I've found this a really nice exposition of the $\mathbf{Set}$ case: - -Martin Hyland, John Power - The category theoretic understanding of universal algebra: Lawvere theories and monads - Electronic Notes in Theoretical Computer Science (pdf at M Hyland's website) - -and if what you want is something like the ultimate monads <-> Lawvere theories correspondence: - -Clemens Berger, Paul-André Melliès, Mark Weber - Monads with arities and their associated theories (arXiv) - -Basically all of the (Lawvere and the like) theories <-> (special) monads equivalences can be seen as special cases of their general monad with arities <-> theories with arities equivalence. This paper is a real joy to read, and how all sort of nerve theorems can be viewed as an instance of the equiv between a monad with arities and the corresponding (generalized) algebraic theory is just wonderful! -Also, on the operadic side, their notion of homogeneous theory is related with $T$-operads (with $T$ cartesian and local right adjoint), and yields a nice account of symmetric operads as homogeneous theories with arities the Segal category $\Gamma$.<|endoftext|> -TITLE: Connectifications? -QUESTION [11 upvotes]: Like many of my questions, this question is actually aimed at $p$-adic analysis. -One of the main obstacles in doing analysis $p$-adically ist that the $\mathbb{Q}_p$ is totally disconnected. -From previous answers and reading I learned that one tries to circumvent these problems -and the result are things like "rigid analytic spaces" or "Berkovich spaces". -Then I recently thought. Ok $\mathbb{Q}_p$ if it is not connected why not do what often is done when some -property does not hold: -incomplete -> take the completion -not compact -> compactify -not algebraically closed -> take algebraic closure. -So why not just connectify $\mathbb{Q}_p$? Now my questions is twofold. - -Searching on the for connectify or connectification I only got a few results, which -only seem to be from point set topology without applications in other areas of mathematics. -Why is that so? Is connectification somehow bad behaved, or does it not exist in general? -By connectification I would understand something like for algebraic closure or completion or -the stone-cech compactification which can all be defined by a universal property. -Does this make sense at all to define connectification like that. -Coming back to the case $\mathbb{Q}_p$, I guess that one also wants the connectification -of $\mathbb{Q}_p$ to be a field. Is this maybe not satisfied, or are there other reasons -not to consider the connectification of $\mathbb{Q}_p$. - -REPLY [13 votes]: After seeing wood's last comment (comment #2 under his question), I've decided to add a few words (a bit too many for a comment) which hopefully make clear the force of Qiaochu's answer. -Generally speaking, the categorical meaning of "completion" refers to taking a left adjoint of a full inclusion of categories; in our situation we are considering the full subcategory -$$\text{Conn} \hookrightarrow \text{Top}$$ -from connected spaces to general topological spaces. Examples where such completions exist are: the inclusion of complete metric spaces into the category of metric spaces and continuous maps with Lipschitz constant 1 (Cauchy completion), the inclusion of compact Hausdorff spaces into the category of all spaces (Stone-Cech compactification), and the inclusion of fields into the category of integral domains and injective ring maps (field of fractions construction). -(There's a bit of fine print here: sometimes one also demands that the unit of the adjunction, here the universal map of an object to its completion, be injective. For example, the inclusion of abelian groups into the category of groups does have a left adjoint (the abelianization), but this isn't injective. Similarly, to get the map from a space to its Stone-Cech compactification to be injective, one should really consider the inclusion of compact Hausdorff spaces in the category of completely regular spaces. Sometimes the suffix -ization or -ification is used in cases where the unit is not injective.) -The salient point behind Qiaochu's answer is that a left adjoint, if it exists, must preserve coproducts (or in fact colimits generally). Now, supposing that the left adjoint to the inclusion $\text{Conn} \to \text{Top}$ exists, it would first of all take a one-point space to a one-point space (the proof is easy), and it would take a coproduct of two one-point spaces in $\text{Top}$, viz. a two-point discrete space, to a coproduct of two one-point spaces in $\text{Conn}$. But Qiaochu's example shows this cannot possibly exist. -The only remedy that I can think of in this situation is to change things up a bit, in a way that I don't think will be at all useful to the OP. There are for example situations where an algebraic structure on a space forces it to be connected (and then some), where one can construct the corresponding free algebraic structures to get a left adjoint to the (non-full) forgetful functor mapping to $\text{Top}$. The most obvious example might be to consider spaces equipped with a contraction: consider spaces $X$ equipped with a basepoint $x_0: 1 \to X$ and with an action $\alpha: [0, 1] \times X \to X$ of the multiplicative monoid $[0, 1]$, such that $\alpha(0, x) = x_0$ for all $x \in X$. Here the free algebra on a general space $Y$ is just the cone $CY$ with the obvious algebraic structure. But this isn't likely to be useful to the OP.<|endoftext|> -TITLE: $C_0$-semigroups applications -QUESTION [5 upvotes]: My graduation thesis was about stability theorems for $C_0$-semigroups (see the Wikipedia article for the definitions: http://en.wikipedia.org/wiki/C0-semigroup). I would like to know if there is some applicability of the stability theorems I know in this field. The only applications I found for my thesis were about the Hille-Yosida theorem and some of its applications to existence and uniqueness of solutions of partial differential equations. -I will not put any names to my theorems, since maybe they are not known to the world as my teachers name them. Here are some of them: - -The $C_0$-semigroup $\{T(t)\}_{t \geq 0}$ is exponentially stable if and only if there exists $p \geq 1$ such that $\int_0^\infty \|T(t)\|^pdt <\infty$. -The $C_0$-semigroup $\{T(t)\}_{t \geq 0}$ is exponentially stable if and only if it satisfies the following condition: -For any $f \in \mathcal{C}$ it follows that $x_f \in \mathcal{C}$ where $x_f: \Bbb{R}_+ \to X,\ x_f(t)=\int_0^t T(t-s)f(s)ds$, and $\mathcal{C} = \{ f : \Bbb{R}_+ \to X,\ f \text{ continuous and bounded } \}$. - -The last theorem can be formulated and proved in some cases for $(L^p,L^q)$ spaces with $(p,q) \neq (1,\infty)$. A more general concept, dichotomy can be formulated (the space splits into two spaces, on one of them there is stability, and on the other one there is instability. -All these sound very nice, and have quite beautiful proofs, but are they applicable to some branches of applied math, such as ordinary or partial differential equations, or they are just pure math, and thats it? - -REPLY [6 votes]: The first theorem (proved by Richard Datko in the Hilbert space case and by Pazy in the Banach space case) is very useful to establish controllability for systems governed by hyperbolic differential equations. Examples include the boundary control of plates, rods, and other elastic structures. There is a book by Jack Lagnese and Jacques-Louis Lions on this general topic. -There, I have done it and mentioned the names of two of my former colleagues<|endoftext|> -TITLE: Quotient singularities with no crepant resolution? -QUESTION [10 upvotes]: I read that in dimension $\geq 4$ there are Gorenstein abelian quotient singularities that have no crepant resolutions. What is the simplest example? I immagine that there should be toric examples. Is it the case that the cone does not admit a suitable subdivision in simplicial cones, corresponding to a crepant resolution? In your answers, please consider that I am only an amateur algebraic geometer, but I know some toric geometry. - -REPLY [23 votes]: [EDIT: added proof that $\mathbb Q$-factoriality implies that the exceptional set of any resolution is a divisor.] -Definition A variety is called $\mathbb Q$-factorial if every Weil divisor on it is $\mathbb Q$-Cartier, i.e., some multiple of it is a Cartier divisor. -The general statement you might want is that - -Claim A $\mathbb Q$-factorial, terminal singularity does not admit a non-trivial crepant resolution. - -Proof $\mathbb Q$-factoriality implies that the exceptional set of any resolution is a divisor, and being terminal implies that all the discrepancies are positive. $\square$ - -Remark One might think that a Gorenstein terminal singularity does not admit a non-trivial crepant resolution. Here is an example that this is not true. Consider a cone over a smooth quadric surface in $\mathbb P^3$. This is a hypersurface in $\mathbb A^4$, so it is clearly Gorenstein. Blowing up the vertex and a simple calculation using adjunction shows that this is a terminal singularity. However, blowing up a divisor that is a cone over a line on the quadric surface gives a small resolution which will be crepant by being an isomorphism in codimension $1$. This shows that it is necessary to add the $\mathbb Q$-factoriality condition for the above Claim. - -- - -Example - For the example in Jim Bryan's answer, $\mathbb C^4/\pm$, or more generally, $\mathbb C^{m}/\pm$, the point to notice is that this is just the cone over the Veronese embedding of $\mathbb P^{m-1}$. Isolated quotient singularities are $\mathbb Q$-factorial and an easy computation shows that the discrepancy of the single exceptional divisor of the blow up of the vertex is $\dfrac m2-1$. This implies that it is terminal as soon as $m>2$, but for $m$ odd it will not be Gorenstein, so the first example of the desired kind is for $m=4$. - -Addendum Here is a proof that $\mathbb Q$-factoriality implies that the exceptional set of any resolution is a divisor: - -Claim Let $X$ be a $\mathbb Q$-factorial variety and $f:Y\to X$ a proper birational morphism. Let $E=\mathrm{Exc}(f)$ denote the exceptional set of $f$, i.e., the largest (closed) subset of $Y$ such that $f|_{Y\setminus E}:Y\setminus E\to X\setminus f(E)$ is an isomorphism. Then $E$ is of pure codimension $1$ in $Y$. - -Proof Let $y\in E$ and suppose that $\mathrm{codim}_YE\geq 2$ in a neighborhood of $y$. Let $C\subseteq E$ be an arbitrary proper curve such that $f(C)$ is a point and $y\in C$ and let $H\subseteq Y$ be an effective divisor such that $y\in H$, but $C\not\subseteq H$. This implies that $H\cdot C>0$. Consider the Weil(!) divisor $f_*H$ on $X$ (the push-forward is meant as a cycle). As $X$ is $\mathbb Q$-factorial, some multiple of $f_*H$ will be Cartier, so replacing $H$ with that multiple we may assume that actually $f_*H$ is Cartier. Then it makes sense to pull it back (as a Cartier divisor). So we get a (Cartier) divisor $f^*f_*H$ which agrees with $H$ on $Y\setminus E$. In particular, if $\mathrm{codim}_YE\geq 2$ in a neighborhood $U$ of $y$, then $H|_U=(f^*f_*H)|_U$. Now by construction $y\in C\cap U\neq\emptyset$, so along $C$, $f^*f_*H=H+F$ where $F$ is an effective (exceptional) divisor that does not contain $C$. Finally, this leads to a contradiction, because we get that -$$ 0=f^*f_*H\cdot C \geq H\cdot C>0$$ -since $f(C)$ is a point.<|endoftext|> -TITLE: Basic results in bounded geometry -QUESTION [5 upvotes]: I'm doing analysis (dynamical systems) in the context of Riemannian manifolds of bounded geometry and I find myself reproving quite a few standard results/tools from standard differential geometry, such as locally finite covers and subordinate partitions of unity, a tubular neighborhood theorem, smoothing of submanifolds... -The main difference from the standard results is that I require uniformly bounded estimates, so for example the tubular neighborhood must have a uniformly finite size and a uniformly bounded diffeomorphism. This means that I cannot simply generalize the standard proofs. -I'm not familiar with bounded geometry. The only reference with explicit details I found is Schick: Manifolds with Boundary and of Bounded Geometry where a uniformly locally finite cover with subordinate partition of unity is proven. -Are there other books or articles which include similar results in the context of bounded geometry? - -REPLY [2 votes]: Many results, in particular about Sobolev spaces, for Riemannian manifolds with bounded geometry, are in: - -J. Eichhorn. -Global Analysis on Open Manifolds. -Nova Science Publishers Inc., New York, 2007. -H. Triebel. -Theory of Function Spaces. II, Volume 84 of Monographs -in Mathematics. -Birkhauser Verlag, Basel, 1992.<|endoftext|> -TITLE: Proof of Krylov-Bogoliubov Theorem -QUESTION [22 upvotes]: Where can I find a proof (in English) of the Krylov-Bogoliubov theorem, which states if $X$ is a compact metric space and $T\colon X \to X$ is continuous, then there is a $T$-invariant Borel probability measure? The only reference I've seen is on the Wikipedia page, but that reference is to a journal that I cannot find. -Of course, feel free to answer this question by providing your own proof. - -REPLY [3 votes]: Then the theorem also doesn't hold. Take $X = \mathbb{R}$, and $T\colon x \mapsto x+1$. Then $T$ has no invariant probability measure. (all the measures "escape to infinity").<|endoftext|> -TITLE: Classical geometric interpretation of spinors -QUESTION [46 upvotes]: A lot of notions in differential geometry have direct meaning in Physics. For example: - -A Riemannian metric is a way to encode distances on a manifold and in Physics it is the gravitational field. The curvature of the Levi-Civita connection gives the strength of the gravitation in a certain sense, -A principal $G$-connection is a object that allows us to do parallel transport conveniently with respect to an action of a certain Lie group $G$, and in Physics it is a gauge field, that is a field that is related to a fundamental interaction, for instance a principal $U(1)$-connection can be seen as the electromagnetic field. The curvature of the connection gives the field strength, in a way. - -I would like to have an interpretation of what is a spinor field (when the manifold on which we are working admits a spin structure) in classical differential geometry, that is a section of the spinor bundle. By classical differential geometry I mean typical manifolds, not supermanifolds. This is because, for me, spinors in the theory of supermanifolds, play a different role, since in a way they are "odd spacetime coordinates". I am interested in the geometry of classical fields: a spinor field represents "matter" (fermions) whereas gauge fields (that is, principal connections) represent "forces" (bosons). But this is Physics. I am interested in a mathematical interpretation like: - -Riemannian metric = gravitational field = a way to measure distances, -Principal connection = gauge field = a way to do parallel transport, -Spinor field = matter field = what in Mathematics? - -So my questions are: -In classical differential geometry (that is, ordinary manifolds), how can we interpret geometrically spinor fields? How can we interpret the spin connection and its curvature? -Thanks. -EDIT: In a comment below I was saying that spinor geometry is of fundamental importance to the Atiyah-Singer theorem. So perhaps this gives a lead to other people to help me with the interpretation of spinors in classical differential geometry. - -REPLY [26 votes]: As far as I know, this sort of structure was first invoked by Dirac in order to take a square root of the Laplacian, and this he was doing in order to write down Lorentz invariant Klein-Gordon equations. It is a useful exercise to try to solve the equation $D^2 = \Delta$ on a Euclidean space $V$ for a first order operator $D$; you will find that the coefficients have to satisfy certain relations that cannot be satisfied by ordinary real or complex numbers. The algebraic structure required to obtain these relations is provided by an algebra $A$ with $V$ as a linear subspace such that $v^2 = -||v||^2 1$ in the algebra. In other words, you need to take a "square root" of your quadratic form. -In brief, a spinor bundle on a Riemannian manifold is a setting for taking a square root of the Riemannian metric. To be precise, it is a bundle $S$ on which tangent vectors act as bundle morphisms in such a way that $v^2 s = -||v||^2 s$. In Dirac's equation, the coefficients of $D$ were given by certain matrices (the "Pauli spin matrices"), and thus he was thinking of $D$ as taking values in a vector space which carries a representation of the algebra $A$. Thus the spinor bundle is a global version of that vector space. -That tells you what properties the spinor bundle is supposed to have, but it doesn't tell you what the bundle actually is. If you look it up in a book, you will find that the spinor bundle is an associated bundle to a principal $Spin(n)$ (or $Spin^c(n)$) bundle via the spin representation, but to me that is only a little more helpful than defining a Riemannian metric to be a reduction of structure group from the principal $GL(n)$-bundle of frames to a $O(n)$-bundle. -Here is what I would consider to be a more concrete and well-motivated description. Let us return to the algebra $A$ associated to a Euclidean space $V = \mathbb{R}^n$ as above. The universal example of such an algebra is the Clifford algebra $Cl(V)$, equipped with a natural left action of $V$. Choosing an orthonormal basis for $V$, one can describe $\mathbb{R}_n := Cl(V)$ as the universal algebra over $\mathbb{R}$ generated by symbols $e_1, \ldots, e_n$ subject to the relations $e_j^2 = -1$ and $e_j e_k = -e_k e_j$ for $i \neq j$. It is not hard to see that $Cl(V)$ is isomorphic as a vector space (but not as an algebra) to the exterior algebra of $V$, and thus $Cl(V)$ inherits a natural $\mathbb{Z}/2\mathbb{Z}$ grading, given by products of even / odd numbers of generators. Notice that right multiplication by the $j$th generator is an odd anti-involution, so a choice of orthonormal basis for $V$ gives $Cl(V)$ the structure of a $n$-multigraded super algebra. -We can define a (real) spinor bundle of a $n$-manifold to be a bundle which is locally isomorphic to the trivial bundle whose fibers are given by $\mathbb{R}_n$ equipped with a left action of the tangent bundle and a $n$-multigrading structure coming from a choice of local orthonormal frame. There is an obvious notion of complex spinor bundle as well: just use the complex Clifford algebra $\mathbb{C}_n$. Note that the fiber dimension of this bundle will be twice that of the bundle obtained via the spin representation, but the multigrading operators can be used to "reduce" my version of the spinor bundle down to the usual version. There are lots of reasons why I believe it is more convenient to think of a spinor bundle as a bundle of Clifford algebras with extra supersymmetry data, but I will briefly focus on a topological reason that I think cuts to the heart of the matter. -The existence of a real spinor bundle on a manifold $M$ (a "Spin structure") is a rather severe condition. The complexification of a real spinor bundle is a complex spinor bundle, but not all complex spinor bundles ("Spin$^c$ structures") arise in this way. For example, any complex manifold has a spin$^c$ structure, but even $\mathbb{C}P^2$ fails to have a spin structure. An orientation on $M$ can be recovered from a choice of spin$^c$ structure, and indeed "spin$^c$-able" is only a little bit stronger than orientable - most orientable manifolds that you can name are probably spin$^c$-able. My point in bringing this up is to relate spinor bundles to K-homology, the generalized homology theory dual to topological K-theory. In ordinary homology theory, a choice of orientation on an $n$-manifold $M$ is the same thing as a choice of fundamental class in $H_n(M)$. Similarly, a choice of real / complex spinor bundle on a $n$-manifold $M$ is the same thing as a choice of fundamental class in the $n$th degree real / complex K-homology of $M$ (the multigrading data are crucial here). This observation is the starting point for some of the more conceptual proofs of the Atiyah-Singer index theorem, but this answer has gone on long enough. I hope it helps!<|endoftext|> -TITLE: I don't get a part of Bernstein's / Deligne-Morgan's proof of Poincaré-Birkhoff-Witt -QUESTION [9 upvotes]: Question: I am talking about the proof given on pages 50-52 of Pierre Deligne, Pavel Etingof, Daniel S. Freed, Lisa C. Jeffrey, David Kazhdan, John W. Morgan, David R. Morrison, and Edward Witten (editors), Quantum Fields and Strings: A Course for Mathematicians, Volume 1, AMS 1999 (on google books and in the usual internet sources). -The problem is easily described: In the middle of page 52, the authors say "and (1.3.7.7) gives that [...]". But I don't see how (1.3.7.7) gives the equation that follows. -Sidenotes: The proof was rather readable and well-written up to that point, so I assume the blindness is on my side. If anyone wishes to read the proof (or reprint the book ;) ), here are a few minor mistakes to watch out for: - -On page 51, $\left[xy\right]$ should be $\left[x,y\right]$ in "while the second term $\frac12\left[xy\right]$ is antisymmetric". -On page 51, in the definition of the map $\left\lbrace x_1,...,x_{n+1}\right\rbrace$, all three terms on the right hand side should end with $x_{n+1}$ rather than $x_n$. -On page 52, in the first formula of this page, the commutators $\left[x\left[y,z\right]\right]$ and $\left[z\left[x,y\right]\right]$ should be $\left[x,\left[y,z\right]\right]$ and $\left[z,\left[x,y\right]\right]$ instead. -On page 52, in the middle of this page, "and the $\left\lbrace x_1,...,x_n\right\rbrace$ vanish" should probably be "and the $\left\lbrace x_1,...,x_{n+1}\right\rbrace$ vanish". -On page 52, in the middle of this page, "1.3.7.4" should be "(1.3.7.4)". - -REPLY [5 votes]: Here is an explanation Pavel Etingof has given to me in email. Thanks Pavel! -Every $\sigma\in S_{n}$ satisfies -$\sum\limits_{i}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast\left[ x,y_{\sigma i}\right] \ast\cdots\ast y_{\sigma n}$ -$=n\left[ x,y_{\sigma1}\right] \ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}$ -$+\sum\limits_{i>1}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast\left[ x,y_{\sigma i}\right] \ast\cdots\ast y_{\sigma n}$. -But since -$\sum\limits_{i>1}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast\underbrace{\displaystyle \left[ -x,y_{\sigma i}\right] }_{\displaystyle\substack{\displaystyle =x\ast y_{\sigma i}-y_{\sigma i}\ast -x\\\displaystyle \text{(since the inclusion of }\mathfrak{L}\\\displaystyle \text{into }\operatorname*{Sym} -\nolimits^{\ast}\mathfrak{L}\text{ is a morphism}\\\displaystyle \text{of Lie algebras)} -}}\ast\cdots\ast y_{\sigma n}$ -$=\sum\limits_{i>1}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast\underbrace{\displaystyle \left( -x\ast y_{\sigma i}-y_{\sigma i}\ast x\right) \ast\cdots\ast y_{\sigma n} -}_{\substack{\displaystyle =x\ast y_{\sigma i}\ast\cdots\ast y_{\sigma n}-y_{\sigma i}\ast -x\ast\cdots\ast y_{\sigma n}\\\displaystyle \text{(by the induction hypothesis, since }i>1\text{)}}}$ -$=\sum\limits_{i>1}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast x\ast y_{\sigma -i}\ast\cdots\ast y_{\sigma n}$ -$-\sum\limits_{i>1}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast -x\ast\cdots\ast y_{\sigma n}$ -$= \sum\limits_{i>0}\left( n-\left( i+1\right) +1\right) \underbrace{\displaystyle y_{\sigma1}\ast\cdots\ast x\ast y_{\sigma -\left(i+1\right)}\ast\cdots\ast y_{\sigma n}}_{\displaystyle =y_{\sigma1}\ast\cdots\ast -y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}}$ -$-\sum\limits_{i>1}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast -x\ast\cdots\ast y_{\sigma n}$ (here we substituted $i+1$ for $i$ in the first sum) -$=\sum\limits_{i>0}\left( n-\left( i+1\right) +1\right) y_{\sigma1}\ast\cdots\ast -y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}$ -$-\sum\limits_{i>1}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast -x\ast\cdots\ast y_{\sigma n}$ -$=\left( n-1\right) y_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}-\sum\limits -_{i>1}y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}$, -this becomes -$\sum\limits_{i}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast\left[ x,y_{\sigma -i}\right] \ast\cdots\ast y_{\sigma n}$ -$=n\left[ x,y_{\sigma1}\right] \ast y_{\sigma2}\ast\cdots\ast y_{\sigma -n}+\left( n-1\right) y_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$ -$-\sum\limits_{i>1}y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast -y_{\sigma n}$ -$=n\left[ x,y_{\sigma1}\right] \ast y_{\sigma2}\ast\cdots\ast y_{\sigma -n}+ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$ -$-y_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}-\sum\limits_{i>1}y_{\sigma1}\ast -\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}$ -$=n\left[ x,y_{\sigma1}\right] \ast y_{\sigma2}\ast\cdots\ast y_{\sigma -n}+ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$ -$-\sum\limits_{i>0}y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma -n}$. -Thus, (1.3.7.7) rewrites as -$\dfrac{1}{n!}x\ast\sum\limits_{\sigma}y_{\sigma1}\ast\cdots\ast y_{\sigma n}=\left( -\text{symmetrized product of }x,y_{1},...,y_{n}\right) $ -$+\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}n\left[ x,y_{\sigma1}\right] -\ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+\dfrac{1}{\left( n+1\right) -!}\sum\limits_{\sigma}ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$ -$-\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}\sum\limits_{i>0}y_{\sigma1}\ast\cdots\ast y_{\sigma -i}\ast x\ast\cdots\ast y_{\sigma n}$. -Since -$\left( \text{symmetrized product of }x,y_{1},...,y_{n}\right) $ -$=\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}\left( \sum\limits_{i>0}y_{\sigma1} -\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}+x\ast y_{\sigma -1}\ast\cdots\ast y_{\sigma n}\right) $, -this simplifies to -$\dfrac{1}{n!}x\ast\sum\limits_{\sigma}y_{\sigma1}\ast\cdots\ast y_{\sigma n} -=\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}x\ast y_{\sigma1}\ast\cdots\ast -y_{\sigma n}$ -$+\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}n\left[ x,y_{\sigma1}\right] -\ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+\dfrac{1}{\left( n+1\right) -!}\sum\limits_{\sigma}ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$. -Thus -$\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}n\left[ x,y_{\sigma1}\right] -\ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+\dfrac{1}{\left( n+1\right) -!}\sum\limits_{\sigma}ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$ -$=\dfrac{1}{n!}x\ast\sum\limits_{\sigma}y_{\sigma1}\ast\cdots\ast y_{\sigma n} --\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}x\ast y_{\sigma1}\ast\cdots\ast -y_{\sigma n}$ -$=\underbrace{\displaystyle \left( \dfrac{1}{n!}-\dfrac{1}{\left( n+1\right) !}\right) -}_{\displaystyle =\dfrac{n}{\left( n+1\right) !}}\sum\limits_{\sigma}x\ast y_{\sigma1}\ast -\cdots\ast y_{\sigma n}=\dfrac{n}{\left( n+1\right) !}\sum\limits_{\sigma}x\ast -y_{\sigma1}\ast\cdots\ast y_{\sigma n}$. -Divide this by $\dfrac{n}{\left( n+1\right) !}$ to obtain -$\sum\limits_{\sigma}\left[ x,y_{\sigma1}\right] \ast y_{\sigma2}\ast\cdots\ast -y_{\sigma n}+\sum\limits_{\sigma}y_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$ -$=\sum\limits_{\sigma}x\ast y_{\sigma1}\ast\cdots\ast y_{\sigma n}$. -In other words, -$0=\sum\limits_{\sigma}\left( x\ast y_{\sigma1}\ast\cdots\ast y_{\sigma n} --y_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}-\left[ x,y_{\sigma1}\right] -\ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}\right) $ -$=\sum\limits_{\sigma}\left\lbrace x,y_{\sigma1},...,y_{\sigma n}\right\rbrace =\left( -n-1\right) !\sum\limits_{i}\left\lbrace x,y_{i},y_{1},...,\widehat{y_{i}},...,y_{n} -\right\rbrace $ -(here we used that $\left\lbrace x_{1},...,x_{n+1}\right\rbrace $ is symmetric in the -last $n-1$ variables, so that every $\sigma\in S_{n}$ satisfies $\left\lbrace -x,y_{\sigma1},...,y_{\sigma n}\right\rbrace =\left\lbrace x,y_{i},y_{1} -,...,\widehat{y_{i}},...,y_{n}\right\rbrace $ for $i=\sigma1$). -Thus, $\sum\limits_{i}\left\lbrace x,y_{i},y_{1},...,\widehat{y_{i}},...,y_{n}\right\rbrace -=0$, qed.<|endoftext|> -TITLE: Historical question about modularity of CM curves -QUESTION [11 upvotes]: I'm looking for the answer of who first proved modularity of CM curves? That is if $E$ is an elliptic curve over $\mathbb{Q}$ which has complex multiplication then there's a non-constant morphism from $X_0(N)$ to $E$ for some $N$ (not necessarily the conductor). The possible names that I've thought of are Hecke, Deuring, Weil or Shimura. Does anybody know something more definite? - -REPLY [12 votes]: Dear Victor, -I believe it was Shimura in the paper On elliptic curves with complex multiplication as factors of the Jacobians of modular function fields, Nagoya Math. J. 43 (1971), 199–208. -Regards, -Matthew<|endoftext|> -TITLE: Why is a 2d TQFT formulated as a functor? -QUESTION [28 upvotes]: Usual mathematical formulation of a 2d (closed) TQFT is as a functor from the category of 2-dim cobordisms between 1-dim manifolds to the category of vector spaces (satisfying various properties.) -For example, a pair of pants (a morphism from $S^1$ to $S^1 \times S^1$) is mapped to a linear map $f:V\to V\otimes V$; similarly a pair of pants in the other direction, a morphism from $S^1 \times S^1$ to $S^1$ is mapped to a linear map $g:V\otimes V\to V$. Then there are axioms (of the symmetric monoidal category) which say that $f$ and $g$ are essentially the same, reflecting the fact that both $f$ and $g$ came from the same pair of pants. -For me (as a quantum field theorist) all this seems very roundabout. The extra axioms are there to ensure what is obvious from the point of view of the two-dimensional field theory; the extra axioms were necessary because the boundaries are arbitrarily grouped into "the source" and "the target" of a morphism, by picking the direction of time inside the 2d surface. (It's called "the Hamiltonian formulation" in physics.) -I think you shouldn't introduce the time direction in the first place, or in the physics terminology, you should just use the "Lagrangian formulation". -In some sense, the idea of "morphism" itself implies an implicit choice of the direction of time. However, you shouldn't introduce the direction of time in an Euclidean quantum field theory. So, you shouldn't use the concept of morphism. The idea of "arrow" itself is so passé, it's a pre-relativity concept which put paramount importance to "time" as something distinct from "space". -So, I would just formulate a 2d TQFT as an association of $f_k:Sym^k V \to K $ to a Riemann surface having $k$ $S^1$ boundaries, and an axiom relating $f_{k}$ and $f_{l}$ to $f_{k+l-2}$. -Why is this not preferred in mathematics? Yes in the physics literature too, the transition from the Hamiltonian framework (pre Feynman) to the Lagrangian framework (post Feynman) took quite a long time... -Or is the higher-category theory (of which I don't know anything) exactly the "Lagrangian formulation" of the TQFT? - -REPLY [36 votes]: Mathematicians have sometimes defined TQFTs in the way Yuji suggests. Indeed, Getzler and Kapranov define the notion of "modular operad" for precisely this purpose (it formalizes the relations between $f_k$, $f_l$ and $f_{k+l-2}$, as well as between $f_k$ and $f_{k-2}$). Earlier, Kontsevich and Manin axiomatized Gromov-Witten invariants along these lines (without distinguisning between incoming and outgoing). -Perhaps the main reason that mathematicians use the language of symmetric monoidal categories is that this is very familiar to them. If you want to explain the idea of a TQFT to the average mathematician, it's easier to say "it's a functor" than to say "it's a collection of linear maps $f_k$ satisfying these relations..." -In addition, there are many very basic examples where the distinction between incoming and outgoing is really important. For example, if $A$ is any associative algebra, then the Hochschild cohomology $HH(A)$ of $A$ carries maps $HH(A)^{\otimes n} \to HH(A)$ indexed by Riemann surfaces of genus $0$, with $n$ incoming and one outgoing boundary components. However, $A$ needs to have a great deal of additional structure -- it needs to be a Calabi-Yau algebra -- in order for this to extend to a fully-fledged TQFT. -As for Yuji's last point, I wouldn't think of the higher-categorical formulation of TQFT as a version of the Lagrangian formalism. After all, for $0+1$ dimensional TQFTs, the higher-categorical formulation reduces to the usual Hamiltonian formalism. - -REPLY [22 votes]: The fact that you can "change the arrow of time" in this way is essentially the fact that bordism category "has duals." So what you're suggesting is that you should directly axiomatize "n-categories with duals" all in one go instead of first axiomatizing n-categories and then saying that they have duals. -This has been done in the TQFT literature. In particular, Jones's Planar Algebras are a version of 2-categories with the duals built in, Kevin Walker's version of n-categories have duals built in, and Dror Bar-Natan's "canopolis" has some duals built in (see this blog post of mine). -The main disadvantage of this approach is that it is often convenient to first show that something is an n-category and then afterward check that it has duals. In Walker's formalism it's difficult to show that something non-topological is actually an n-category with duals, and thus sometimes harder to actually construct TQFTs. - -REPLY [15 votes]: There are interesting examples where the duality you're trying to bake in does not hold. In particular, the 3+1 dimensional (pseudo-)TQFTs coming from Seiberg-Witten theory in 4 dimensions do not have such duality maps. This is possible since the TQFT is only defined for a subset of the 4-d cobordism, for instance cobordisms where both the input and the output are connected 3-manifolds. (The actual TQFT allows one of the two sides to be disconnected, depending on which flavor you look at.) -Presumably this is related to some strangeness from the physics point of view.<|endoftext|> -TITLE: Bounds on squarefree numbers -QUESTION [15 upvotes]: Let $q_1,q_2,\ldots$ denote the squarefree integers 1, 2, 3, 5, .... What effective bounds are known for $q_n$? Clearly -$$q_n\sim\zeta(2)n$$ -but I need hard inequalities. Of course from the above there exist $\varepsilon,N$ with -$$(\zeta(2)-\varepsilon)n < q_n < (\zeta(2)+\varepsilon)n$$ -for all $n>N,$ but I do not have proven values for $\varepsilon,N$. Of course -$$|q_n-\zeta(2)n| < f(n)$$ -for sublinear $f$ would be preferable (and should be possible; squarefree numbers are fairly well-behaved). -I'm sure this is in some standard reference but I haven't found it. Ideas? - -For those interested, my actual goal is to find a reasonable bound for the powerful (2-full) numbers for computational purposes. Their asymptotic growth is tightly constrained but for numerical computations I would prefer worst-case bounds that I can trust rather than a heuristic that says the error probably won't be much more than, say, 10 times the O-term. - -REPLY [11 votes]: Result -$$ -\zeta(2)n - 0.058377\sqrt n < q_n < \zeta(2)n + 0.058377\sqrt n -$$ -for all $n\ge268293.$ For smaller values of $n$ you can use -$$ -\zeta(2)n - 36 < q_n < \zeta(2)n + 33 -$$ -which actually holds for $1 \le n \le 696017.$ -Here's a chart of the residuals below 268293. Yellow and gray are the upper and lower bounds for this range (-36 to +33) and hence never cross 0. Red and blue and the upper and lower bounds for larger numbers ($\pm\,0.058377\sqrt n$) which cross 0 several times. There are tick marks on the x-axis every 25,000. - -Proof -Let -$$ -Q(x):=\sum_{n\le x}\mu(n)^2 -$$ -be the number of squarefree numbers up to $x.$ Cohen, Dress, & El Marraki [2] (improving on Cohen & Dress 1) prove that -$$ -\left|Q(x)-\frac{x}{\zeta(2)}\right| < 0.02767\sqrt x -$$ -for $x\ge438653.$ So -$$ -\left|Q(\zeta(2)x)-\frac{\zeta(2)x}{\zeta(2)}\right| < 0.02767\sqrt{\zeta(2)x} < 0.0354882\sqrt x -$$ -for $x\ge266079.$ We want $Q(x) \ge n$ so we need a slightly larger argument. Choosing $\zeta(2)n + k\sqrt n$ (where $k$ is to be chosen later) we get -$$ -Q\left(\zeta(2)n + k\sqrt n\right) > n + \sqrt n\left(\frac{k}{\zeta(2)} - 0.02767\sqrt{\zeta(2) + k/\sqrt n}\right) -$$ -and so the goal is to choose a small $k$ such that -$$ -\frac{k}{\zeta(2)} \ge 0.02767\sqrt{\zeta(2) + k/\sqrt N} -$$ -for all $N\ge n$, which happens just when -$$ -k \ge \frac{c}{2\sqrt n}\left(c\zeta(2)^2+\zeta(2)^2\sqrt{4n/\zeta(2)+c^2}\right) -$$ -(with $c=0.02767$ for brevity). Choosing $n=10^6$ shows that $k=0.058377$ is admissible; substituting this value gives -$$ -Q\left(\zeta(2)n + 0.058377\sqrt n\right) > n + \sqrt n\left(\frac{0.058377}{\zeta(2)} - 0.02767\sqrt{\zeta(2) + 0.058377/\sqrt n}\right) > n -$$ -as desired. Checking up to that bound reveals that this works for all $n\ge268293.$ Similar calculations in the other direction complete the proof; in that direction only $n\ge217502$ is needed. - -1 H. Cohen and F. Dress, Estimations numériques du reste de la fonction sommatoire relative aux entiers sans facteur carré, Colloque de théorie analytique des nombres "Jean Coquet": proceedings journées SMF-CNRS, CIRM, Marseille (Septembre 1985), pp. 73–76. -[2] Henri Cohen, Francois Dress, and Mohamed El Marraki, Explicit estimates for summatory functions linked to the Möbius μ-function, Functiones et Approximatio Commentarii Mathematici 37 (2007), part 1, pp. 51–63.<|endoftext|> -TITLE: Determining a recurrence relation -QUESTION [5 upvotes]: I would like to solve the general problem of determining a linear recurrence relation that fits a given integer sequence of length $n$, or stating that none exists (with fewer than $n/2-k$ coefficients for some reasonable fixed $k$, perhaps 2, to ensure that I'm not overfitting). Actually, I'd like to find the smallest one. -Of course the problem is apparently easy: just search for one of length 1, 2, ..., n/2-k until one is found. At each step all that is needed is to solve the appropriate matrix equation (tried to write it out, no tabular environment here, but it's obvious) and check the result against the unused members of the sequence. But for reasonably large $n$ this is impractical: the linear algebra is too difficult. -Unfortunately, it's not rare for me to work with a sequence that is clearly a recurrence relation, but which appears to have a large order, so I can't simply assume that not finding a relation with order below 100 means that none exists. This raises two questions for me: - -Is there a fast way to calculate these, compared to the naive approach above? -If one recurrence is known, can this be used to speed the search for recurrences of smaller order (or to prove that none exist)? - -One unenlightened approach that comes to mind for #1 is to solve the matrix over the real numbers (using floating-point approximations) rather than solving it exactly over $\mathbb{Q}$. This seems reasonable, but it's not obvious how much precision is needed nor how far the numbers could be if a solution was actually found (in which case, presumably, the system should be re-solved exactly). Although solving systems this way results in serious speedup even using quadruple precision, without appropriate numerical analysis I don't think it's usable. Hopefully there is a better way. -On #2, consider a (periodic) sequence with recurrence relation $a_n=a_{n-6}.$ Basic algebra suffices to show that any recurrence of the form $a_n=2a_{n-1}-2a_{n-2}+a_{n-3}$ is also of that form, so some sequences of order 6 can be simplified (nontrivially, that is not just removing trailing zeros) to order 3. Does it suffice, for example, to check only orders dividing 6? - -REPLY [12 votes]: This problem comes up often in coding theory, and it can be solved efficiently by the Berlekamp-Massey algorithm (Wikipedia has pseudo-code). This is more or less equivalent to using continued fractions, although many expositions don't present it that way: given a sequence $a_0,a_1,\dots,a_N$, look at the rational function $\sum_{i=0}^N a_i x^{-i}$ and compute its continued fraction expansion, with coefficients that are polynomials in $x$. That just amounts to applying Euclid's algorithm to the polynomials $\sum_{i=0}^{N} a_i x^{N-i}$ and $x^N$. -A degree $d$ linear recurrence must be satisfied by more than $2d$ terms of the sequence to mean anything, and any such recurrence will be detected by this method. Specifically, that means the corresponding rational function with degree $d$ denominator must be a convergent to the continued fraction. -In practice, as Charles Matthews suggested, you can generally speed up the arithmetic substantially by working modulo a prime (say, a fairly large random prime). This is particularly an issue when there isn't a low-degree recurrence, since in that case generically you'll get a lot of partial quotients of degree $1$ with rapidly growing denominators. Checking that there's no low-degree recurrence modulo a prime will be much faster, since it will avoid the huge denominators.<|endoftext|> -TITLE: Monotonicity of complete homogeneous symmetric polynomials -QUESTION [5 upvotes]: The complete homogeneous symmetric polynomials are defined as -$$ -h_k (x_1, \dots,x_n) = \sum_{1 \leq i_1 \leq i_2 \leq \cdots \leq i_k \leq n} x_{i_1} x_{i_2} \cdots x_{i_k}. -$$ -For example, -$$ -h_3(x_1,x_2,x_3) = x_1^3 + x_2^3 + x_3^3 + x_1^2 x_2 + x_1^2 x_3 + x_2^2 x_3 + x_2^2 x_1 + x_3^2 x_1 + x_3^2 x_2 + x_1 x_2 x_3. -$$ -I'm interested in conditions as simple as possible (and ideally independent of $n$) that will suffice to prove that -$$ -h_1(x_1,\dots,x_n) > h_2(x_1,\dots,x_n) > h_3(x_1,\dots,x_n) > \cdots -$$ -for some given set of positive real numbers $\{ x_1, \dots, x_n \}$. -For example, do these inequalities hold if $h_1(x_1,\dots,x_n) \le 1$? Do they all hold if the first one $h_1(x_1,\dots,x_n) \ge h_2(x_1,\dots,x_n)$ holds? - -REPLY [14 votes]: I claim that, if $h_k > h_{k+1}$, then $h_j > h_{j+1}$ for all $j \geq k$. This includes both your assertions above. (Note that $h_0$ should be considered to be $1$.) -More precisely, I will prove that, for any positive reals $x_j$, we have -$$\frac{h_0}{h_1} \leq \frac{h_1}{h_2} \leq \frac{h_2}{h_3} \leq \cdots $$ -So, if any term is greater than $1$, so all all the following terms. -Proof: The inequality $h_{k-1}/h_k \leq h_k/h_{k+1}$ is equivalent to $\det \left( \begin{smallmatrix} h_{k} & h_{k-1} \\ h_{k+1} & h_k \end{smallmatrix} \right) \geq 0$. -By the Jacobi-Trudi identity, this determinant is equal to the Schur function $s_{k,k}$. In particular, since all the monomials in $s_{k,k}$ have nonnegative coefficients, the determinant is nonnegative as desired. QED - -A few comments: This paper by Bruce Sagan looks relevant, although I don't have JSTOR access right now to check. -As a general heuristic, when presented with an inhomogenous inequality between symmetric functions, it is a good idea to try to think what homogenous inequality it might be a special case of. So trying to show "$1 \geq h_1$ implies $h_k \geq h_{k+1}$" suggests instead showing "$h_1 h_k \geq h_{k+1}$", since the latter is homogenous.<|endoftext|> -TITLE: word problem in free Burnside groups (and other torsion groups) -QUESTION [6 upvotes]: Question 1. Is it known that for some free Burnside groups the word problem is undecidable? - -Provided that the answer is negative, what about the following easier question. - -Question 2. Is there a known example of a finitely generated (and preferably finitely presented) group $G$ and an integer $k$ such that all elements of $G$ have order at most $k$ and the word problem in $G$ is undecidable? - -REPLY [9 votes]: Concerning question 1: for free Burnside groups of odd exponent $n\geq 665$ the decidability was shown by S.I.Adian. Lysionok proved the same in the case of even exponent -$n=16k\geq 8000$. The corresponding deciding procedure is just Dehn's algorithm. -It will be interesting to know how to solve the word problem for groups $B(m,n)$ when $n=2k$, where $k\geqslant 665$ and odd. These groups are infinite (it follows from the cited result of Adian), but it seems that the decidability of the word problem for these groups is an open question. -As to the question 2: -The pure existence of non-finitely presented group with undecidable w.p. easily follows from the result of S.I.Adian who proved that there are continuum non-isomorphic periodic groups with fixed number of generators $m\geqslant 2$ satisfying the periodic law $X^n=1$ if $n$ is odd and $n \geq 665$. -One should mention, that the number of "additional relations" (besides all periodic) is not necessarily finite. -Later M.Sapir constructed an example of periodic group with finite number of additional relations. The paper is "On the word problem in periodic group varieties", IJAC, 1991, V.1, No.1.<|endoftext|> -TITLE: Is every closed embedded codimension-n submanifold cut out by a section of a rank-n vector bundle? -QUESTION [10 upvotes]: Let $M$ be a smooth manifold (over $\mathbb R$) and $N \hookrightarrow M$ a closed embedding. Locally near any point in $N$, I can find coordinates $x^1,\dots,x^{\dim M}$ on $M$ so that $N$ is the vanishing locus of $x^{1+\dim N},\dots,x^{\dim M}$. My question is essentially whether I can do this globally. More precisely, I would like: - -Does there exist a smooth vector bundle $V \to M$ with $\operatorname{rank} V = \dim M - \dim N$ and a global section $s\in \Gamma(V)$ so that $N = \lbrace s = 0 \rbrace$ and this critical locus is nondegenerate in the following sense: for $n\in N\hookrightarrow M$, the matrix $\mathrm d s: \mathrm T_n M \to \mathrm T_{(n,0)}V = V_n$ is full rank (i.e. it has kernel only $\mathrm T_nN \hookrightarrow \mathrm T_nM$)? - -My intuition is strongly "yes", but I have been unable to come up with a construction of $V,s$. Note that, in $C^\infty$-land, every closed set is the vanishing locus of a single smooth function, but said function tends to vanish to very high (often infinite) order on the closed set. On the other hand, for my question, one cannot hope to win with only trivializable bundles. For example, when $M = S^1$ is a circle and $N$ is an odd number of points, the vector bundle $V$ is necessarily the "other" rank-$1$ vector bundle, i.e. the non-orientable one. -The reason I would like this is because the data $(V,s)$ is essentially a "projective resolution" of the embedding $N \hookrightarrow M$. From this point of view, I can see larger more general constructions that will achieve what I need for my application, but (being larger and more general) those constructions are harder to compute with. -Provided that $(V,s)$ does exist, my follow-up question is how unique it is (at best, unique up to unique isomorphism, but this cannot be right if $M$ has interesting topology far away from the image of $N$, so instead you expect that the choices of $(V,s)$ should be parametrized by some measure of the relative topology)? - -REPLY [13 votes]: The answer in general is no. Take e.g. $M=\mathbb{R}^n$. There are no non-trivial vector bundles on it. So, if $N$ is the zero locus of a transversal section of a bundle $E$ (which is necessarily trivial), then using the exact sequence $$0\to T N\to T M|_N\to \nu_M N\to 0,$$ where $\nu_M N$ is the normal bundle of $N$ in $M$, and the fact that $\nu_M N\cong E|_N$ we conclude that the tangent Stiefel-Whitney classes of $N$ are all zero. -[upd: the same strategy works for $M=S^n$; the tangent Stiefel-Whitney class of $S^n$ is zero; the top Stiefel-Whitney class of $E$ need not be zero, but it gives zero when restricted to $N$.]<|endoftext|> -TITLE: Schur multipliers over non-algebraically closed ground fields? -QUESTION [11 upvotes]: Recently some arithmetic dynamicists came to town, bringing with them some interesting problems in arithmetic geometry. -I started thinking a bit about one of their problems, and it got me wondering about Schur multiplier groups over an arbitrary field. Traditionally, if $G$ is a group -- let us say it is finite -- then the Schur multiplier group $M(G)$ is $H^2(G,\mathbb{C}^{\times})$, i.e., group cohomology, with $\mathbb{C}^{\times}$ viewed as a trivial $G$-module. This group is also Brauer-like in that it measures obstructions to projective representations of $G$ -- i.e., homomorphisms $\rho: G \rightarrow \operatorname{PGL}_N(\mathbb{C})$ -- to be liftable to honest representations of $G$ -- i.e., homomorphisms $\tilde{\rho}: G \rightarrow \operatorname{GL}_N(\mathbb{C})$. It is not hard to see that you don't actually need to work over $\mathbb{C}$: if $\# G = n$, you can work over any field $K$ such that $K^{\times} = K^{\times n}$ and has primitive $n$th roots of unity. -But now suppose I have an arbitrary ground field $K$ and a homomorphism $\rho: G \rightarrow \operatorname{PGL}_N(K)$ which I am wondering lifts to a representation of $G$. What is the theory of this? Two basic questions: -1) Is it still true that the appropriate group to look at is $M_K(G) = H^2(G,K^{\times})$? -Added: Let me sharpen this question. The answer below shows that a projective representation gives rise to a class in $M_K(G)$ no matter what the ground field may be. But in the classical case the converse is also true: every element of $M_{\overline{K}}(G)$ arises in this way from a projective representation, uniquely up to projective equivalence. Does that still hold over an arbitrary ground field? I am a bit skeptical at the moment... -2) If the answer to 1) is yes, then it seems that the theory will have a much different flavor over an arbitrary field. (Here I say arbitrary but I am quite willing to assume for the moment that the characteristic of $K$ does not divide the order of $G$, so that we are in the setting of classical representation theory. This assumption will be in force in what follows.) For instance, if $G$ is cyclic of order $n$, then $M_K(G) \cong K^{\times}/K^{\times n}$. This means that over something like a number field there will be many projective representations of finite cyclic groups which do not lift. I think this is correct. In particular, I believe that for the cyclic group of order $2$, the map $G \rightarrow \operatorname{PGL}_2(K)$ associated with the order $2$ linear fractional transformation $z \mapsto \frac{\alpha}{z}$ is liftable to $\operatorname{GL}_2(K)$ iff $\alpha \in K^{\times 2}$. On the other hand I would like to deduce from the theory of "rational Schur multiplier groups" facts like the following: if $G$ is cyclic of odd order $n$ then every projective representation $\rho: G \rightarrow \operatorname{PGL}_2(K)$ lifts to a representation. (Again, in this case, if I am not mistaken, this can be shown by hand without much trouble, but I would like to see it come out of some general Schur-like theory.) In particular are there examples of computations of $M_K(G)$ in the literature for simple easy finite groups $G$, as there are for the usual $M(G)$? - -REPLY [6 votes]: As far as 1) is concerned : -Exercise 6.6.5 in Weibel's Introduction to Homological Algebra book : -For any field k and any n, let $\gamma$ denote the class in $H^2(PGL(n,k), k^*)$ -corresponding to the extension $$1 \rightarrow k^* \rightarrow GL(n,k) \rightarrow PGL(n,k) \rightarrow 1$$ -If $\rho : G \rightarrow PGL(n,k)$ is a projective representation, show that $\rho$ lifts to a linear representation $G \rightarrow GL(n,k)$ if and only if $\rho^{\ast}(\gamma) = 0$ in $H^2(G,k^{\ast})$. Here, $\rho^{\ast}$ is the obvious map from $H^2(PGL(n,k), k^{\ast})$ to $H^2(G, k^*)$ induced by $G \rightarrow PGL(n,k)$.<|endoftext|> -TITLE: morphisms of affine schemes question -QUESTION [5 upvotes]: So, in chapter 2, section 2 of Hartshorne, (prop 2.3), he describes how if $\varphi : A\rightarrow B$ is a homomorphism of rings, then you get a morphism of (affine schemes): -$\newcommand{\Spec}{\textrm{Spec }}$ -$\newcommand{\oO}{\mathcal{O}}$ -$\newcommand{\mf}[1]{\mathfrak{#1}}$ -$(f,f^\sharp) : (\Spec B, \oO_{\Spec B})\rightarrow (\Spec A, \oO_{\Spec A})$ -where $f:\Spec B\rightarrow\Spec A$ is given by $f(\mf{p}) = \varphi^{-1}(\mf{p})$. Here, for each $\mf{p}\in\Spec B$, $\varphi$ also gives us a map of localizations $\varphi_\mf{p} : A_{\varphi^{-1}(\mf{p})}\rightarrow B_\mf{p}$. -Also, $f^\sharp : \oO_{\Spec A}\rightarrow f_*\oO_{\Spec B}$ is given by sending $\sigma\in\oO_{\Spec A}(V)$ (for each open $V$), to the function $[\mf{p}\mapsto \varphi_\mf{p}(\sigma(f(\mf{p})))] \in \oO_{\Spec B}(f^{-1}(V)) = f_*\oO_{\Spec B}(V)$ -So, my question is... What are the stalks of the direct image sheaf $f_*\oO_{\Spec B}$? -This is clearly a sheaf on $\Spec A$, so there should be a stalk for each $\mf{p}\in\Spec A$. There are two cases. -Firstly, suppose $\mf{p}\in\Spec A$ is in the image of $f$, then by definition, the stalk of $(f_*\oO_{\Spec B})_{\mf{p}}$ is the direct limit: -$\lim_{U\supset f^{-1}(\mf{p})}\oO_{\Spec B}(U)$ -But this is not quite a stalk of $\oO_{\Spec B}$ (since $f$ may not be injective). However, Hartshorne seems to suggest that this is actually just $(\oO_{\Spec B})_{\mf{q}}$ which is just the localization of $B$ at $\mf{q}$, where $\mf{q}$ is any point of $f^{-1}(\mf{p})$. I don't really see why this must be true. (Especially since he seems to suggest that all the localizations at $\mf{q}$ are the same, for any $\mf{q}\in f^{-1}(\mf{p})$. -Secondly, suppose $\mf{p}\in\Spec A$ is not in the image of $f$. Then what? I can imagine that if there is some neighborhood $V$ of $\mf{p}$ such that $f^{-1}(V)$ is empty, then the stalk would be zero. But suppose there is no such $V$? Then What? (Alternatively, must there always exist such a $V$ in this case?) -Thanks guys - -will - -REPLY [3 votes]: You consider $B$ as an $A$-module. Then $f_*O_X$ ($X = Spec B$) is the quasi-coherent $O_Y$-module ($Y = Spec A$) corresponding to the $A$-module $B$. Thus for a prime ideal $y \in Y$, $(f_*O_X)_y$ is the localization of the $A$-module $B$ in $y$.<|endoftext|> -TITLE: when is a section of vector bundle determined by its zero locus? -QUESTION [6 upvotes]: Let $V_X$ be a vector bundle of rank $r>1$ on a smooth (connected) projective variety of dimension $r$. Let s be the global holomorphic section, whose zero locus is a zero dimensional subscheme $Z\subset X$. Is there some result that the section is determined (up to scaling) by its zero locus? At least for some nice vector bundles on nice manifolds? -upd. I meant $X_\Bbb C$ and some conditions on some resolution of $V_X$ by some v.a. line bundles on $X$. jvp cites an excellent result below. - -REPLY [3 votes]: Take a look at On sections with isolated singularities of twisted bundles and applications to foliations by curves by Campillo and Olivares. -There you will find the following result. - -Theorem. Let $X$ be a projective manifold of dimension $n$, $\mathcal L$ be an ample >line-bundle on $X$, and $E$ a rank $n$ vector bundle over $X$. If $k\gg 0$ and $s, s'$ >are sections of $E \otimes \mathcal L^{\otimes k}$ such that - -the zero scheme $Z(s)$ of $s$ has dimension zero, and -the zero scheme $Z(s')$ of $s'$ contains $Z(s)$ - -then there exists $\varphi \in H^0(X,\mathrm{End}(E))$ such that $s' = \varphi(s)$. In particular, if $E$ is simple ( $H^0(X,\mathrm{End}(E))=\mathbb C$ ) then $s = \lambda s'$ for a suitable complex number $\lambda$. - -In the particular case $E=T\mathbb P^n$, it suffices to take $\mathcal L= \mathcal O_{\mathbb P^n}(1)$ and $k\ge 1$.<|endoftext|> -TITLE: is residue field ever flat over its local ring? -QUESTION [14 upvotes]: Let R be a local ring with maximal ideal m and residue field k. Is k ever flat over R? What conditions are needed on R? -Sorry, it's not a very profound question. It came up in a derived functor calculation. - -REPLY [35 votes]: Consider the exact sequence $0 \to \mathfrak m \to R \to k \to 0.$ Tensoring with $k$ gives -$0 \to \mathfrak m/\mathfrak m^2 \to k = k \to 0.$ Thus if $k$ is flat over $R$, then -$\mathfrak m = \mathfrak m^2$. If furthermore $R$ is Noetherian, this implies that -$\mathfrak m = 0$, and hence that $R = k$. -Conclusion: For Noetherian $R$, the desired flatness hold only if $R = k$. -Added: A colleague points out that flat local maps of local rings are always faithfully flat, -hence injective. Thus even in the non-Noetherian case, the only way for $k$ to be -flat over $R$ is if $R = k$. -In fact, one can directly extend the above argument to the not-necessarily-Noetherian case, as follows: -Let $I$ be any finitely generated ideal contained in $\mathfrak m$. Since $k$ is -flat, $k\otimes_R I \hookrightarrow k\otimes_R \mathfrak m,$ the target of which vanishes, as noted above. -Thus $k\otimes _R I$ vanishes, and so by Nakayama's lemma, $I = 0$. Since $\mathfrak m$ -is the union of such $I$, we see that $\mathfrak m = 0$, and thus $R = k$.<|endoftext|> -TITLE: Learning roadmap for harmonic analysis -QUESTION [54 upvotes]: In short, I am interested to know of the various approaches one could take to learn modern harmonic analysis in depth. However, the question deserves additional details. Currently, I am reading Loukas Grafakos' "Classical Fourier Analysis" (I have progressed to chapter 3). My intention is to read this book and then proceed to the second volume (by the same author) "Modern Fourier Analysis". I have also studied general analysis at the level of Walter Rudin's "Real and Complex Analysis" (first 15 chapters). In particular, if additional prerequisites are required for recommended references, it would be helpful if you could state them. -My request is to know how one should proceed after reading these two volumes and whether there are additional sources that one could use that are helpful to get a deeper understanding of the subject. Also, it would be nice to hear suggestions of some important topics in the subject of harmonic analysis that are current interests of research and references one could use to better understand these topics. -However, I understand that as one gets deeper into a subject such as harmonic analysis, one would need to understand several related areas in greater depth such as functional analysis, PDE's and several complex variables. Therefore, suggestions of how one can incorporate these subjects into one's learning of harmonic analysis are welcome. (Of course, since this is mainly a request for a roadmap in harmonic analysis, it might be better to keep any recommendations of references in these subjects at least a little related to harmonic analysis.) -In particular, I am interested in various connections between PDE's and harmonic analysis and functional analysis and harmonic analysis. It would be nice to know about references that discuss these connections. -Thank you very much! -Additional Details: Thank you for suggesting Stein's books on harmonic analysis! However, I am not sure how one should read these books. For example, there seems to be overlap between Grafakos and Stein's books but Stein's "Harmonic Analysis" seems very much like a research monograph and although it is, needless to say, an excellent book, I am not very sure what prerequisites one must have to tackle it. In contrast, the other two books by Stein are more elementary but it would be nice to know of the sort of material that can be found in these two books but that cannot be found in Grafakos. - -REPLY [2 votes]: Here are a few relevant books that appeared since this question was asked. - -Barry Simon's Harmonic analysis volume -E. M. Stein and R. Shakarchi, Functional analysis. -C. Muscalu and W. Schlag, Classical and multilinear harmonic analysis -P. Mattila, Fourier Analysis and Hausdorff Dimension<|endoftext|> -TITLE: Minimal models with local coefficients -QUESTION [7 upvotes]: Let $X$ be a path-connected nilpotent space (meaning $\pi_1(X)$ is nilpotent and acts nilpotently on the higher homotopy groups). Let $\rho\colon\thinspace\pi_1(X)\to \mathrm{Gl}(V)$ be a representation, where $V$ is a rational vector space. Then $\rho$ defines a local coefficient system on $X$, and we have the cohomology with local coefficients $H^\ast(X;\rho)$. -It is well known that Sullivan's theory of minimal models works well for nilpotent spaces, and one can (more or less) easily obtain a minimal, nilpotent cdga $M_X$ with $H(M_X)\cong H^*(X;\mathbb{Q})$, whose isomorphism type determines the rational homotopy type of $X$. (Here I am assuming some finiteness hypotheses, such as are satisfied if $X$ is a manifold or finite CW complex.) - -My question is, is there a (more or less) easy way to obtain a minimal (in some sense) dg-module $M_{X,\rho}$ over $M_X$ which has $H(M_{X,\rho})\cong H^\ast(X;\rho)$ as modules over $H^\ast(X;\mathbb{Q})$? - -Ideally I would like to be able to calculate $H^1(X;\rho)$ when $X$ is a nilmanifold (so $\pi_1(X)$ is nilpotent with trivial higher homotopy groups) and calculate cup products $a_1\cup a_2\in H^2(X;\rho_1\otimes\rho_2)$. So far I have looked at -Gómez-Tato, Antonio Théorie de Sullivan pour la cohomologie à coefficients locaux.[Sullivan's theory for cohomology with local coefficients] -Trans. Amer. Math. Soc. 330 (1992), no. 1, 235–305. -and parts of Sullivan's original paper -Sullivan, Dennis Infinitesimal computations in topology. Inst. Hautes Études Sci. Publ. Math. No. 47 (1977), 269–331. - -REPLY [3 votes]: I think one could argue as follows: one can start by constructing a cochain complex $A^*(X,\rho)$ that computes the twisted cohomology so that it will be a module over the Sullivan cochains $A^*(X)$ of $X$ (with constant coefficients). Then one can plug it in Hinich's machinery: see http://arxiv.org/PS_cache/q-alg/pdf/9702/9702015v1.pdf , section 3. A minimal model would then be a cofibrant replacement of $A^*(X,\rho)$ as an $A^*(X)$-module. -Let me also mention a reference that may be useful: Gomez-Tato, Halperin, Tanr\'e, Rational homotopy theory for non-simply connected spaces. Trans. Amer. Math. Soc. 352 (2000), no. 4, 1493–1525.<|endoftext|> -TITLE: Stratified Pseudomanifold -QUESTION [6 upvotes]: Hi there, -I have a, I guess, simple question. -In the definition of an n-dimensional stratified pseudomanifold one demands the following filtration -$X=X_n \supset X_{n-1}=X_{n-2} \supset X_{n-3}\supset ... \supset X_0 \supset X_{-1}=\emptyset$. where all $X_i$'s are closed subspaces of X. -Why is it required that $X_{n-1}=X_{n-2}$? -Thanks! -Levi - -REPLY [8 votes]: It's not! -Okay, that's a little too glib. You're right that it's often required that $X^{n-1}=X^{n-2}$, but it depends somewhat on your purpose. In most of my recent papers, including those I'm current working on with Jim McClure, and in the papers of Saralegi (which look at intersection homology from an analytic point of view), there's freedom to include codimension one strata, but you need to be more careful about your definition of intersection (co)homology. -In even more detail: there are a variety of connected reasons why one would classically not work with codimension one strata. One reason has to do with the properties of classical intersection homology theory. Many of the results of intersection homology, including the ones that are approached sheaf theoretically, depend, at heart, on having a nice formula for the intersection homology on cones. If there are no codimension one strata around and you're working with perversities as defined by Goresky and MacPherson, then this cone formula works out, the sheaf of intersection chains is quasi-isomorphic to the Deligne sheaf, and away you go. However, with codimension one strata, and Goresky-MacPherson perversities, the cone formula doesn't quite work out. I have a detailed exposition about this in the following paper: -Greg Friedman, An introduction to intersection homology with general perversity functions, in Topology of Stratified Spaces; -Greg Friedman, Eugénie Hunsicker, Anatoly Libgober, Laurentiu Maxim (editors) -Mathematical Sciences Research Institute Publications 58, Cambridge University Press 2011, 177-222 -Here's the link to the copy on my web site: -http://faculty.tcu.edu/gfriedman/papers/MSRI-revised-2.pdf -One argument that I make there looks carefully at what the perversity should be on such a stratum: if the perversity is 0 (as one might expect), then the perversity is greater than the top perversity allowed by Goresky and MacPherson (which would have to be -1 for a codimension one stratum), and that leads to the cone formula issues. On the other hand, if the perversity assigned is $\leq -1$, now you're below the bottom perversity $\bar 0$ allowed by Goresky and MacPherson, and that causes other issues. -Another way to think about codimension one strata is that they could arise if you took a manifold with boundary and let the boundary be a codimension one stratum. Then, as algori notes, you have a problem getting your fundamental class as a geometrically defined cycle (unless you sufficiently modify your definition of intersection homology). And that causes trouble with Poincare duality. -So, after all that, in summary: a pseudomanifold can have codimension one strata, but much of intersection homology breaks down. Unless you modify your intersection homology a bit, and then everything does work! See my papers or feel free to e-mail if you have further questions.<|endoftext|> -TITLE: Ramanujan's eccentric Integral formula -QUESTION [19 upvotes]: The wikipedia page on Srinivasa Ramanujan gives a very strange formula: - -Ramanujan: If $0 < a < b + \frac{1}{2}$ then, $$\int\limits_{0}^{\infty} \frac{ 1 + x^{2}/(b+1)^{2}}{ 1 + x^{2}/a^{2}} \times \frac{ 1 + x^{2}/(b+2)^{2}}{ 1 + x^{2}/(a+1)^{2}} \times \cdots \ \textrm{dx} = \frac{\sqrt{\pi}}{2} \small{\frac{ \Gamma(a+\frac{1}{2}) \cdot \Gamma(b+1)\: \Gamma(b-a+\frac{1}{2})}{\Gamma(a) \cdot \Gamma(b+\frac{1}{2}) \cdot \Gamma(b-a+1)}}$$ - - -Question I would like to pose to this community is: What could be the Intuition behind discovering this formula. -Next, I see that Ramanujan has discovered a lot of formulas for expressing $\pi$ as series. May I know what is the advantage of having a same number expressed as a series in a different way. Is it useful at all? -From what I know Ramanujan basically worked on Infinite series, Continued fractions, $\cdots$ etc. I have never seen applications of continued fractions, in the real world. I would also like to know if continued fractions has any applications. - -Hope I haven't asked too many questions. As I was posting this question the last question on application of continued fractions popped up and I thought it would be a good idea to pose it here, instead of posing it as a new question. - -REPLY [8 votes]: Another proof of the Ramanujan's identity is given in https://arxiv.org/abs/1309.3455 (On gamma quotients and infinite products, by Marc Chamberland, Armin Straub). It is based on the formula -$$\prod_{k=0}^\infty \frac{(k+\alpha_1)\cdots (k+\alpha_n)}{(k+\beta_1)\cdots (k+\beta_n)}=\frac{\Gamma(\beta_1)\cdots\Gamma(\beta_n)}{\Gamma(\alpha_1)\cdots\Gamma(\alpha_n)}, \tag{1} $$ -where $n\ge 1$ is an integer, $\alpha_1,\ldots,\alpha_n$ and $\beta_1,\ldots,\beta_n$ are nonzero complex numbers, none of which are negative integers, and $\alpha_1+\ldots +\alpha_n=\beta_1+\ldots +\beta_n$. -Using the factorization -$$\frac{1+x^2/(b+k)^2}{1+x^2/(a+k)^2}=\frac{(k+a)^2(k+b+ix)(k+b-ix)}{(k+b)^2(k+a+ix)(k+a-ix)},$$ -and (1), Ramanujan's product transforms to -$$ \prod_{k=0}^\infty\frac{1+x^2/(b+k)^2}{1+x^2/(a+k)^2}=\frac{\Gamma(b)^2\Gamma(a+ix)\Gamma(a-ix)}{\Gamma(a)^2\Gamma(b+ix)\Gamma(b-ix)},$$ -and then the Ramanujan's identity is equivalent to the Mellin–Barnes integral (in this respect, see also David Hansen's answer below) -$$\frac{1}{2\pi i}\int_\limits{-i\infty}^{i\infty}\frac{\Gamma(a+s)\Gamma(a-s)}{\Gamma(b+s)\Gamma(b-s)}ds=\frac{1}{2\sqrt{\pi}}\frac{\Gamma(a)}{\Gamma(b)}\frac{\Gamma\left(a+1/2\right)\Gamma\left(b-a-1/2\right)}{\Gamma\left(b-1/2\right)\Gamma(b-a)}.$$<|endoftext|> -TITLE: Generators of the graded ring of modular forms -QUESTION [19 upvotes]: Let $\Gamma$ be a finite-index subgroup of $\operatorname{SL}_2(\mathbb{Z})$. I've seen it stated (in a comment in the code of a computer program) that the graded ring -$$ M(\Gamma, \mathbb{C}) = \bigoplus_{k \ge 0} M_k(\Gamma, \mathbb{C}),$$ -where $M_k(\Gamma, \mathbb{C})$ is the space of modular forms of weight $k$ and level $\Gamma$, -is always generated as a $\mathbb{C}$-algebra by forms of weight $\le 12$. -Why is this true? Moreover, can one improve on the bound of 12? (For the subgroups $\Gamma_0(N)$, weight $\le 6$ always seems to be sufficient.) - -REPLY [3 votes]: In the special case of the principal congruence subgroup $\Gamma=\Gamma(N)$ with $N \geq 3$, Khuri-Makdisi has proved that the graded algebra of modular forms on $\Gamma$ is generated by the modular forms of weight 1. In fact, the algebra generated by the Eisenstein series of weight 1 contains all modular forms in weight $\geq 2$, so misses only the cusp forms of weight 1. This is proved in his article Moduli interpretation of Eisenstein series. -In the case $N=2$, then $M_*(\Gamma(2))$ is freely generated by the weight 2 Eisenstein series $e_1,e_2$ obtained by putting the universal elliptic curve in the form $y^2=(x-e_1)(x-e_2)(x-e_3)$.<|endoftext|> -TITLE: Example of a locally presentable $2$-category -QUESTION [7 upvotes]: Background -Let $R$ be a (as always commutative) ring. Consider the $2$-category $\text{Cat}_{c\otimes}(R)$ of cocomplete $R$-linear tensor categories. There are various reasons why $\text{Cat}_{c\otimes}(R)$ can be seen as a categorified version of the category of $R$-algebras. For example, coproducts are categorified sums, tensor products are categorified products, and the imposed cocontinuity of the tensor product can be seen as a sort of distributive law. There is a $2$-functor $\text{Alg}(R) \to \text{Cat}_{c\otimes}(R), A \mapsto Mod(A)$ ("categorification"), which is $2$-left adjoint to $\text{Cat}_{c\otimes}(R) \to Alg(R), C \mapsto \text{End}(1_C)$ ("decategorification"). See also Lurie's article Tannaka duality for geometric stacks for evidence that that this category is important in algebraic geometry: The category of geometric stacks over $R$ embeds via quasi-coherent sheaves fully faithful into $\text{Cat}_{c\otimes}(R)$ (if we restrict to so called tame tensor functors). See also this entry in Todd Trimble's blog. -Question -Is $\text{Cat}_{c\otimes}(R)$ a locally presentable $2$-category? -Note that I don't want to consider it as a $(2,1)$-category here, but you may assume this if it is really necessary. To avoid set-theoretic difficulties in the following, perhaps we should restrict to categories with $\kappa$-small colimits for a fixed regular cardinal $\kappa$. Then $\text{Cat}_{c\otimes}(R)$ is probably $2$-cocomplete, see Mike Shulman's answer here. It already seems to be hard to describe colimits explicitly (see here). -There is a free cocomplete $R$-linear tensor category on one object, explicitly given by $\text{Mod}(R)^{\mathbb{N}}$ with a convolution tensor product. Imagine this as the categorified ring of power series over $R$. Thus $Hom(\text{Mod}(R)^{\mathbb{N}},-)$ is identified with the forgetful $2$-functor $\text{Cat}_{c\otimes}(R) \to \text{Cat}$, which is, however, not faithful since the data of a cocont. tensor functor does not only consist of a funcor, there is also the natural isomorphism expressing the compatibility with the tensor structure. Thus $\text{Mod}(R)^{\mathbb{N}}$ is not a generator. Remark that in the decategorified setting, the free $R$-algebra on one object, namely the polynomial algebra $R[x]$ is a generator of $\text{Alg}(R)$. -But maybe $\text{Cat}_{c\otimes}(R)$ is too big to be generated by a set of $\lambda$-presentable objects? If this is not the case, what is a reasonable full subcategory which contains the categories of quasi-coherent sheaves and is locally presentable as a $2$-category? -EDIT: Motivated by Jacob Lurie's answer, I would like to change the question to the following: Let $\lambda$ be a regular cardinal and $\text{Cat}_{c\otimes}^\lambda(R)$ denote the $2$-category of locally $\lambda$-presentable $R$-linear tensor categories (of course the definition of this includes that the $\lambda$-presentable objects form a submonoid with respect to $\otimes$). Is it a locally presentable $2$-category? - -REPLY [8 votes]: Your 2-category can be described as commutative algebras over "R-mod" in the setting of cocomplete categories (at least this is one meaning of the word "R-linear", which does not imply that the underlying category is abelian. If you want to stick to abelian categories the answer will be the same but require some more words.) -So a more basic question is: is the collection of cocomplete categories (maybe with some adjectives attached) locally presentable? If you fix a regular cardinal $\kappa$, require all categories to be generated by a set of $\kappa$-compact objects, and all functors to preserve $\kappa$-compact objects, then the answer is yes. For example, if $\kappa$ is uncountable, then this 2-category is equivalent (via the functor which formally adjoins -$\kappa$-filtered colimits) to the 2-category of small categories which admit $\kappa$-small colimits, and functors which preserve $\kappa$-small colimits (this 2-category is "algebraic" in nature, albeit with respect to operations of (fixed) infinite arity). -If you don't restrict your functors to preserve $\kappa$-compact objects, then it's not reasonable for the underlying $(2,1)$-category to be locally presentable because it is not even locally small. Your post contains an example: there's a free $R$-linear tensor category on one generator, and the category of $R$-linear tensor functors from that to some target -category is equivalent to the target category, which is typically not small. -(There is probably some reasonable way to salvage the situation if you take advantage of the full 2-category structure: the morphism categories in your example will not be small but they are nevertheless accessible.)<|endoftext|> -TITLE: Friedman and proof of Hanna Neumann Conjecture -QUESTION [8 upvotes]: Two years ago, Joel Friedman submitted a paper purporting to prove the Hanna Neumann Conjecture, which eventually turned out to contain a fatal bug and was withdrawn. Quite recently, Friedman repeated his attempt at proof with paper "Sheaves on Graphs and a Proof of the Hanna Neumann Conjecture": http://arxiv.org/abs/1105.0129 . Has this attempt been verified by anyone or is still under review? - -REPLY [4 votes]: I don't think that this latest round has been independently verified. -Igor Mineyev has also announced a solution. See http://www.math.uiuc.edu/~mineyev/math/<|endoftext|> -TITLE: O-linear Weil-pairing on abelian varieties with real multiplication -QUESTION [7 upvotes]: Let $A/k$ be an abelian variety with real multiplication by some ring of integers $\mathcal O \subset F$. Let $n$ be an integer prime to the characteristic of $k$. -We have the standart $e_n$ pairing $A[n] \times A^\vee[n] \to \mu_n$. -It satisfies $e_n(a \cdot x, y) = e_n(x, a \cdot y)$ for any $a \in \mathcal O$. -It is desirable to "extend" this pairing to an $\mathcal O$-linear pairing with respect to the $\mathcal O$-action on $A[n] \times A^\vee[n]$. -In Rapoprt's "Compactifications de l'espace de modules de Hilbert-Blumenthal" 1.21 we find such an extension: -$e_{n_{\mathcal O}} \colon A[n] \times A^\vee[n] \to (\mathcal D^{-1}/n\mathcal D^{-1})(1)$ -where $\mathcal D$ is the different of $F$ and with $(1)$ we denote the Tate-twist. -Concerning the definition of the pairing it is only said that $Tr(e_{n_{\mathcal O}}) = e_n$. -Can someone make this definition more explicit? - -REPLY [7 votes]: Yes, it's just a linear algebra stuff. Probably, it's more convenient to look at the Weil pairing between the $\ell$-adic Tate modules of $A$ and its dual, taking into account that these modules are free $O\otimes Z_{\ell}$-modules of the same rank. -From the linear algebra point of view the situation is as follows. -Let $e: M \times N \to P$ be a perfect pairing of free $Z_{\ell}$-modules $M$ and $N$ of the same rank that takes values in a free $Z_{\ell}$-module $P$ of rank $1$. Assume also that $M$ and $N$ are free $O\otimes Z_{\ell}$-modules of the same rank. Then we get a natural $Z_{\ell}$-linear map -$M \times N \to Hom_{Z_{\ell}}(O\otimes Z_{\ell}, P),$ -$(m,n) \mapsto [a \mapsto e(ax,y)]$ for all $a \in O$. -Taking into account that -$$Hom_{Z_{\ell}}(O\otimes Z_{\ell}, Z_{\ell})=D^{-1}\otimes Z_{\ell}$$ (via the trace map), - we get the natural pairing -$$M \times N \to P \otimes_{Z_{\ell}}[D^{-1}\otimes Z_{\ell}].$$ -Here $M$ and $N$ are the Tate modules of $A$ and its dual while $P$ is $Z_{\ell}(1)$ and $e$ is the Weil pairing.<|endoftext|> -TITLE: (Real) algebraic geometry for (real) trigonometric polynomials? -QUESTION [8 upvotes]: Has somebody developed a comprehensive theory of the algebraic structure of trigonometric polynomials in several variables? If yes, where? - -Background: -By a (real) trigonometric polynomial in $d$-variables, I mean a map $\mathbb{T}^d \to \mathbb{R}$ that is given by an expression of the form -$$ - f(x) = \sum_{|k| \leq K} \hat{f}(k) \exp(2\pi\mathrm{i} k\cdot x) -$$ -where $k \in \mathbb{Z}^d$ and $|k| = \sup_{j=1,\dots,d} |k_j|$. Also $\mathbb{T} = \mathbb{R}/\mathbb{Z}$. -These trigonometric polynomials have many of the properties of usual polynomials, but are NOT polynomials. So as far as I know it, one cannot apply the usual algebraic-geometry constructions. -An example of a result, I would be interested in is: Given polynomials $f_1, \dots, f_{\ell}$ how does the dimension of their zero locus -$$ - \{x \in \mathbb{T}^d:\quad f_j(x) = 0,\quad j=1,\dots,\ell\} -$$ -relate to the ideal generated by these polynomials? - -One approach -In the Annals paper by Bourgain and Goldstein, a hint of how to do this is given. Write -$$ - \exp(2\pi\mathrm{i} k \cdot x) = \prod_{j=1}^{d} \exp(2\pi\mathrm{i} x_j)^{k_j}. -$$ -Using that $\exp(2\pi\mathrm{i} x_j) = \cos(2\pi x_j) + \mathrm{i} \sin(2\pi x_j)$, one can write a trigonometric polynomial as a honest polynomial in the $2 d$ variables $C_j = \cos(2\pi x_j)$ and $S_j = \sin(2\pi x_j)$. A computation shows that this is a honest polynomial with real coefficients. -Call this polynomial $\tilde{f}$. -These set from the previous example can then be described as the zero locus of the polynomials $\tilde{f}_j$ and the polynomials -$$ - (C_j)^2 + (S_j)^2 = 1. -$$ -It seems to me that using this approach one can more or less carry over most results, but I am not very good at algebra, so I might miss subtleties. It would be nice if there was some work out of these things by somebody in the field. - -REPLY [2 votes]: The Bézout theorem as you describe should work in $\mathbb{T}^2$ for the reasons you outlined (do the change of variables, add in the Pythagorean conditions and apply the regular Bézout). I am more familiar with the situation over $\mathbb{R}^2$, where, since the functions are periodic, you cannot expect finitely many solutions. -However, the natural setup in that case is given by Khovanskii's theory of fewnomials: if you restrict the arguments of your sines and cosines to some bounded interval, you can represent your trigonometric polynomials as an instance of the more general Pfaffian functions. These functions have a natural complexity associated to them, which is degree-like, though it's actually a vector of positive integers. The larger the restriction intervals, the bigger the complexity. (Any elementary function would work here, you don't need them to be trigonometric polynomials). -Khovanskii's theorem is an explicit upper bound on the number of isolated solutions of such a system (with as many equations as variables). Unfortunately, this is only an upper-bound (the problem is conjectured to be decidable, but this is not known), and the bounds are very big, probably much too big. -Khovanksii's theorem deals with more general functions than only trigonometric polynomials, so it may be possible to improve somewhat.<|endoftext|> -TITLE: What is known about the centraliser of the Hecke algebra in the affine Hecke algebra? -QUESTION [5 upvotes]: This question is a sequel to -66602 -The Hecke algebra is the quotient of the group algebra of the braid group of type $A_n$ by quadratic relations and the affine Hecke algebra is the quotient of the group algebra of the braid group of type $B_n$ by quadratic relations. This is well known and the relations can be found in the references in the answer and comment to the above question. -This gives an inclusion of the Hecke algebra in the affine Hecke algebra and I am interested in the centraliser algebra. The obvious elements in the centraliser are the central elements in the affine Hecke algebra. This centre is (I don't have the reference) the algebra of symmetric polynomials in the $x_1,\ldots ,x_n$. The other obvious elements are the central elements in the Hecke algebra. -These are all the elements I can construct in the centraliser. Is there any construction which gives more elements in the centraliser? Of course I would really like to have a set of generators. -The motivation is to understand the $6j$ symbols of $GL(n)$ which was also the subject of my question -15800. - -REPLY [4 votes]: This answer is being completely rewritten because it wasn't very comprehensible. We will show that there are more elements in the centraliser than the obvious ones mentioned in the question. At best, we will have an inelegant and impractical way to construct these extra elements. -For most of this answer we'll just talk about the q=1 case, ie group algebras. We need some notation. Let W be the finite Weyl group and W' be the affine Weyl group. This is the semidirect product of W and the root lattice Q. k=ground field (char 0). -The centraliser is the set of elements in the group algebra k[W'] invariant under the conjugation action of W. The general construction is: Given a vector space V with an action of W, then for any v ∈ V, the average $\sum_{w\in W} w.v$ is in the invariant space VW. -So now in the affine Hecke algebra, fix an isomorphism between the finite Hecke algebra and the group algebra k[W] (In type E7 and E8 you'll need a square root of q in k). Then you can find elements in the centraliser using the above construction. -The rest of this answer will show that at q=1, there are more elements in the centraliser than described by Bruce. We need more notation. Z(?)=centre(?). For λ in Q, we write eλ for the corresponding element in the group algebra k[W'] and wλ for the action of w on λ. -We aim to show there are elements in the centraliser not in Z(k[W]) ⊗ Z(k[W']). -Note that Z(k[W']) is spanned by $\sum_{w\in W}e^{w \lambda}$. -Pick λ in a free W-orbit and u ∈ W. We apply our averaging construction to v=ueλ. -So $y=\sum_{w\in W}wxw^{-1}e^{w\lambda}$ lies in the centraliser. If y is also in the tensor product, then we must have $y=z\sum_{w\in W}e^{w \lambda}$ for some $z\in Z(k[W])$. -Let u vary over W and fix λ. We get |W| linearly independent choices of y, yet there is a smaller dimensional space of choices of z (Well if |W|=2, this is false, but a different argument is concoctable). Thus there exist elements in the centraliser that are not constructed in the question. -Initially there was a question about the braid group Bn here whose existence makes David Hill's first comment understandable.<|endoftext|> -TITLE: Matrix representation of real *-algebras -QUESTION [6 upvotes]: It is a standard fact that everyt real $n$-dimensional algebra is a subalgebra of $M_n(\mathbb R)$. -The transposition map, operating in $M_n(\mathbb R)$, is an involutive ($(A^t)^t=A$) antiautomorphism (an $\mathbb R$-linear isomorphism satisfying $(AB)^t=B^tA^t$). This makes $M_n(\mathbb R)$ a real *-algebra. The same is true in every transpose-closed subalgebra of $M_n(\mathbb R)$. -Is every real *-algebra of this kind? (a transpose-closed subalgebra of $M_n(\mathbb R)$, with the * represented by transposition) -This is true at least in the famous real *-algebras $\mathbb C$ and $\mathbb H$. - -REPLY [11 votes]: You have to add the condition -$$\sum_{i} a_i^*a_i =0 \quad \Rightarrow \quad a_i =0 \quad \forall i.$$ -If you consider $\ast$-algebras satisfying this condition, then $-1$ is not a sum of hermitean squares and you find a positive linear functional $\varphi$ on the algebra which satisfies $\varphi(1)=1$. You can now perform the GNS-construction (in the real setting if you want) to obtain a $\ast$-representation on a real Hilbert space. Since $A$ is finite-dimensional, you obtain a $\ast$-homomorphism to a real matrix-algebra.<|endoftext|> -TITLE: Action of PGL(2) on Projective Space -QUESTION [35 upvotes]: Let $k$ be a field, let $G = PGL_2(k)$ be the projective general linear group of $k$, and let -$X = k \cup \{ \infty \}$ be one-dimensional projective space over $k$. Then $G$ acts on $X$ (via fractional linear transformations). This action has the following properties: -1) The action of $G$ on $X$ is simply 3-transitive. That is, it acts simply transitively on -the set of 3-tuples of distinct elements of $X$. (Edited as indicated in the comments.) -2) Suppose that $x,y \in X$ are distinct elements and that $g \in G$ satisfies -$gx = y$, $gy = x$. Then $g$ has order $2$. -Is the converse true? (That is, if we are given an action of a group $G$ on a set $X$ -satisfying 1) and 2), does it follow that $G = PGL_2(k)$ for some field $k$, -with its natural action on $k \cup \{ \infty \}$? -(This is true at least when $G$ and $X$ are finite: it can be deduced from the theorem of Frobenius on Frobenius groups.) - -REPLY [28 votes]: A KT-field $(F,+,\times,\sigma)$ -consists of a neardomain $(F,+,\times)$ together with an involutionary -automorphism $\sigma$ satisfying -$$\sigma(1 + \sigma(x)) = 1 - \sigma(1 + x)$$ -for all $x \in F \setminus \{0,1\}$. (My impression is that neardomains are quite weak entities, e.g. $F^{\times}$ is required to be a group but it may not be commutative, $(F,+)$ is not even necessarily a group. Industrious MO reader adds the definition of a neardomain to this answer if they wish.) -Sharply $3$-transitive groups are determined up to isomorphism -as permutation groups on $\mathbf{P}^1(F) = F \cup \{ \infty \}$ consisting of maps of the -form: -(i): $x \mapsto a + m x, \quad \infty \mapsto \infty$ -(ii): $x \mapsto a + \sigma(b + m x), \quad \infty \mapsto a, \quad - m^{-1} b -\mapsto \infty$, -where $a,b \in F$ and $m \in F^{\times}$. -Consider the set of elements $\gamma \in G$ such that -$\gamma(0) = \infty$ and $\gamma(\infty) = 0$. They are given exactly -by mappings of the form: -$$\gamma: x \mapsto \sigma( \lambda x)$$ -for any $\lambda \in F^{\times}$. -If all such $\gamma$ have order two, then -$$\sigma(\lambda \sigma(\lambda x)) = x$$ - for all $x, \lambda \in F^{\times}$. Setting $x = \lambda^{-1}$, it follows -that $\sigma(\lambda) = \lambda^{-1}$ for all $\lambda \in F^{\times}$. -Since $\sigma$ is an automorphism, it follows that $F^{\times}$ is -commutative. From a Theorem of Kirby (see below), it follows that $(F,+,\times)$ -is actually a commutative field, and $G = \mathrm{PGL}_2(F)$. -All the results and definitions of this answer can be gleamed from -the math review: MR0997066 (91b:20004a) of a paper by William Kerby - -A class of canonical sharply 3-transitive groups, Results Math. 16 (1989), no. 1-2, 89–106 (doi:10.1007/BF03322647). - -The paper is only $3$-pages long, so I assume that is is relatively elementary - although I can't access it myself, and it may refer to previous results. (Full disclosure, all I did was type "sharply 3-transitive" into mathscinet, -I don't actually know what a neardomain actually is.) -In case your actual purpose is to generalize this result to $(\infty,\pi)$-whatzit categories with creamy rice pudding centres, you might want to take a glance at the actual paper.<|endoftext|> -TITLE: Torsion for Lie algebras and Lie groups -QUESTION [13 upvotes]: This question is about the relationship (rather, whether there is or ought to be a relationship) between torsion for the cohomology of certain Lie algebras over the integers, and torsion for associated Lie groups. -Let $\mathfrak{g}$ be a complex simple Lie algebra. Then we can choose a basis of Chevalley generators for $\mathfrak{g}$. The structure constants describing the Lie brackets among the Chevalley generators are all integers, so the Chevalley basis spans a Lie algebra over the integers, which I shall denote by $\mathfrak{g}_\mathbb{Z}$. We can compute the (Chevalley-Eilenberg) Lie algebra cohomology of $\mathfrak{g}_\mathbb{Z}$ with coefficients in $\mathbb{Z}$ using the usual Koszul complex, and thus get for each $1 \leq n \leq \dim \mathfrak{g}$ a finitely-generated abelian group $H^n(\mathfrak{g}_\mathbb{Z},\mathbb{Z})$. -On the other hand, (as I understand it) the Lie algebra cohomology of $\mathfrak{g}$ with coefficients in $\mathbb{C}$, denoted $H^*(\mathfrak{g},\mathbb{C})$, is the same as the cohomology of some associated (compact?) connected Lie group $G$. This philosophy is described in the section "Motivation" of the Wikipedia article on Lie algebra cohomology. (I'm not an expert on this matter, so I invite someone more knowledgable to correct me if I've got something incorrect. I think that the associated Lie group $G$ will have $\mathfrak{g}$ as its Lie algebra, or perhaps $\mathfrak{g}$ will be the complexification of the Lie algebra of $G$.) We can also compute $H^n(G,\mathbb{Z})$, the cohomology of $G$ with coefficients in $\mathbb{Z}$. - -Is there, or ought there to be, a connection between the torsion of $H^n(G,\mathbb{Z})$ and the torsion of $H^n(\mathfrak{g}_\mathbb{Z},\mathbb{Z})$? Should the torsion primes of the two abelian groups be the same? Should the torsion primes of one be a subset of the torsion primes for the other? - -To the best of my knowledge, the torsion primes of the compact connected simple Lie groups were worked out in the 1950s and/or 1960s, especially by Armand Borel, but at present no such list seems readily available for the Lie algebra cohomology side of the picture. - -REPLY [3 votes]: There is a simple observation about some odd behaviour of the prime $2$. The Lie group $SL_2(\mathbb{C})$ has torsion-free cohomology, because as a space it is weakly equivalent to $S^3$. -Consider the Lie algebra $\mathfrak{sl}_2(\mathbb{Z})$ of $2\times 2$-matrices with trace zero, and the usual commutator. This is a $\mathbb{Z}$-form of $\mathfrak{sl}_2$. Since the Cartan acts with weight $2$, the reduction of the above $\mathbb{Z}$-form to $\mathbb{F}_2$ is in fact not simple but solvable - the Cartan is an ideal and the quotient is abelian. This is still visible for the $\mathbb{Z}$-form, which has abelianization isomorphic to $(\mathbb{Z}/2)^{\oplus 2}$, generated by the matrices -$$ -\left(\begin{array}{cc} -0&1\\0&0\end{array}\right) \qquad\textrm{ and }\qquad -\left(\begin{array}{cc} -0&0\\1&0\end{array}\right). -$$ -The universal coefficient formula then implies $H^2(\mathfrak{sl}_2(\mathbb{Z}),\mathbb{Z})\cong(\mathbb{Z}/2)^{\oplus 2}$. -In this case, $2$ would be a torsion prime for the Lie algebra, but not for the Lie group. For all other primes, the cohomology of the Lie algebra and of the Lie group coincide. At least the reason for this behaviour (the weight $2$ action of the Cartan on the weight spaces) persists for all the $\mathfrak{sl}_n$, but I have not made the calculations to see if this really implies the existence of non-trivial $2$-torsion in the homology of $\mathfrak{sl}_n$, $n\geq 3$. -Whether or not a relation should exist between torsion in the homology of the Lie algebra and the Lie group seems to be related to the mod $p$ splitting behaviour of the Lie group. In some cases, the mod $p$ homotopy type of the Lie group splits as a product of spheres. In these cases, I would expect stronger relationships (maybe excluding the odd behaviour of the prime $2$). However, I do not really know how to possibly relate these things - the $\mathbb{Z}$-form of the Lie algebra is related to differential forms, the splitting of the Lie groups is serious algebraic topology, and these only seem to interact nicely with rational coefficients via rational homotopy theory.<|endoftext|> -TITLE: How to decide this function takes integer values? -QUESTION [16 upvotes]: I am working a combinatorial question. I need to decide when the following function -$$\frac{m(7m^2 - 22m +7)}{ \sqrt{ (m^2- 2m + 9)(9m^2 - 2m +1) } } $$takes integer values with following -assumption: 1) m takes all positive integer values; 2) $(m^2- 2m + 9)(9m^2 - 2m +1)$ is a perfect square (= $n^2$ for some integer $n$). Limited computation for $m < 20000$ shows $m = 1, 5$. -Many thanks, -Jianmin - -REPLY [41 votes]: The OP's specific Diophantine problem has been answered, but it's still worth pointing out a general technique that applies to problems of this kind: - -Given an algebraic function $f(x)$, find all $x$ such that both $x$ and $f(x)$ are integers - -whenever $f$ is not itself a polynomial but can be expanded in a power series about $x=\infty$ with rational coefficients. Such is the case for the function -$$ -f(m) = \frac{m (7m^2-22m+7)}{\sqrt{(m^2-2m+9)(9m^2-2m+1)}} -= \frac{7}{3} m - \frac{128}{27} + O(1/m) -$$ -of the combinatorics problem, or indeed its denominator $$\sqrt{(m^2-2m+9)(9m^2-2m+1)} = 3m^2−\frac{10}{3}m+\frac{337}{27}+O(1/m)$$ (as I noted in my comment to C.Matthew's answer). Often $f$ is a rational function, which automatically satisfies the condition as long as it is not a polynomial. The technique is simply: - -Use the power-series expansion to write $f(x) = P(x) + O(1/x)$ with $P \in {\bf Q}[x]$, find a common denominator $D$ so that $DP \in {\bf Z}[x]$, and observe that once $x$ is large enough that the $O(1/x)$ error drops below $1/D$ in absolute value the only way that $Df(x)$ can be integral is for $x$ to be a root of $f(x) = P(x)$. This reduces the problem to a finite search. - -The OP didn't say where exactly his question came from, but such problems arise often in the theory of combinatorial designs and strongly regular graphs. For example, in a Moore graph of girth 5 and degree $d$, every vertex has $d$ neighbors and any two distinct vertices are xeither adjacent xor have a common neighbor, in which case that neighbor is unique. Using the fact that every eigenvalue of the adjacency matrix has integral multiplicity, one shows that either $d=2$ or $d=m^2+m+1$ for some integer $m$ such that $(m^2+m+1)(m^2+m-1) / (2m+1) \in {\bf Z}$. So we write -$$ -16 \frac{(m^2+m+1) (m^2+m-1)} {2m+1} = 8m^3+12m^2+2m-1 - \frac{15}{2m+1} -$$ -and (since $15/(2m+1)$ cannot vanish) deduce that $2m+1 \leq 15$, so $m \leq 7$. We then find that of the remaining candidates only $m=1,2,7$ work. Hence $d$ is one of 2, 3, 7, or 57. [See for instance Cameron and Van Lint's Designs, Graphs, Codes and their Links (LMS 1991, 1996) for this and many more examples. As it happens, each of $d=2,3,7$ occurs for a unique Moore graph, namely the 5-cycle, Petersen graph, and Hoffman-Singleton graph respectively, while the existence of a Moore graph of degree 57 is a famous open problem.] -Further shortcuts may be available when $D$ and/or the implicit constant in $O(1/x)$ are large enough that the concluding search, though finite, is still inconvenient or infeasible to do exhaustively. For example, in the Moore graph problem, $15/(2m+1)$ must be integral, so $2m+1$ must be a factor of $15$, and this immediately yields the solutions $m=1,2,7$, corresponding to factors $3,5,15$. The numerator $15$ could have been predicted by from the value $15/16$ taken by $(m^2+m+1) (m^2+m-1)$ at $m=-1/2$, the root of the denominator $2m+1$. More generally if $f(x)$ is a rational function $A(x)/B(x)$ with $A,B \in {\bf Z}[x]$ relatively prime, we can use the Euclidean algorithm for polynomials to find $M,N \in {\bf Z}[x]$ such that -$MA-NB$ is a nonzero integer, say $R$; then if $f(x)$ is an integer then so is -$$ - M(x) f(x) - N(x) = \frac{M(x)A(x) - N(x)B(x)}{B(x)} = \frac{R}{B(x)} , -$$ -and we need only find integer solutions of $B(x) = r$ for each of the factors $r$ of $R$ and test each of these candidates. $R$ is a factor of the resultant of $A$ and $B$. This is actually not far from G.Robinson's analysis: if we square to get the rational function $m^2 (7m^2-22m+7)^2 / ((m^2-2m+9)(9m^2-2m+1))^2$, the resultant is $2^{36} 3^{12}$, explaining the factors of 2 and 3 that arose in that solution. There are still $2 \cdot 37 \cdot 13 = 962$ factors to try (allowing negative as well as positive $r$), but that's much less than the number of candidates that the generic method would require testing.<|endoftext|> -TITLE: Sets as Combinatorial Games -QUESTION [8 upvotes]: Just a few days ago my seemingly eternal and recurrent fascination for Conway's combinatorial game theory (CGT) & surreal numbers had a recrudescence, so I grabbed this excellent survey, and began reading. -Some old thoughts came to the surface from the archives of my memory. Here they are: -the class $SURREAL$ contains the class $ON$, and ordinals are the spine of $V$, the "universe of sets". So, pushing the analogy, can I say that combinatorial games generalize sets, or conversely sets are (special) combinatorial games? -If the answer is yes, can I even go further, and develop some foundational theory which starts from games, not sets, and then define ordinary sets as those special games? -This question can be broken down into 3 sub-questions: - -does there exist a treatment of combinatorial games as a first order axiomatic theory, presented without the recourse to sets? -what kind of games are ordinary ZF sets? (perhaps "solitaire" games, where the opponent doesn't do anything, or perhaps perfectly symmetric games). In other words, assuming 1) above, which interpretations of ZF are available inside CGT? -could one reformulate some familiar constructions of classical set theory in the language of CGT? - -Any material, thoughts, refs, on 1) -3)? -PS In this daydreaming I saw a picture of an extended universe where there is a double-cone of sets, V and -V, as in SURREAL there are positive and negative ordinals.... - -REPLY [4 votes]: I wrote a little something about sets as 1-player games here. A set is the collection of moves you can make in a game. Well-foundedness means you always lose. So the only thing you can do is try to put off the inevitable for as long as you can. If you can model the state of a computer program as a game, you can use this to prove termination. As Todd says, there isn't really an advantage to be had by looking at things this way, except it's fun to think of sets in a different way, and I think it might help motivate some parts of set theory to some students who might otherwise find the concepts daunting.<|endoftext|> -TITLE: conformally flat manifold with positive scalar curvature -QUESTION [16 upvotes]: In the paper Conformal Deformation of a Riemannian metric to a constant scalar curvature of Richard Schoen (J. Differential Geom. 20(2) (1984) 479-495, doi:10.4310/jdg/1214439291), in the first page, it says that - -"Note that the class of conformally flat manifolds of positive scalar curvature is closed under the operation of connected sum, and hence contains connected sums of spherical space forms with copies of $S^1\times S^{n-1}$." - -My question is: Is there any other conformally flat manifold with positive scalar curvature which is not in these forms? Or the manifolds Schoen mentioned exhaust the list of all conformally flat manifolds of positive scalar curvature? - -REPLY [12 votes]: Products $M^m \times S^{n-m}$ will be conformally flat, where $M^m$ is a compact manifold of curvature $-1$ and $S^{n-m}$ has curvature $1$. If $n>2m$ then the scalar curvature of the product will be positive (positive curvature of the sphere dominates the negative curvature of the hyperbolic manifold, so the scalar curvature of the product will be positive). You can also (frequently) deform these metrics a bit so that they no longer (locally) split as products and still have positive scalar curvature and remain conformally flat. Thus, you have counter-examples in all dimensions $n\ge 5$. See e.g. this paper by R.Mazzeo and N.Smale http://intlpress.com/JDG/archive/1991/34-3-581.pdf for further discussion.<|endoftext|> -TITLE: Is the Laplacian on a manifold the limit of graph Laplacians? -QUESTION [39 upvotes]: Here's the sort of thing I have in mind. Let $M$ be a Riemannian manifold, compact if it helps, and let $\Delta_M$ be the Laplace-Beltrami operator. Choose a sequence of triangulations of $M$ so that the triangles in the $n$th triangulation all have diameter smaller than $1/n$, and let $\Delta_n$ denote the corresponding graph Laplacian - that is, the operator which sends a function $f$ on the vertices of a graph to the function whose value at a vertex is the average of $f$ on all neighboring vertices. -Is there a precise sense in which $\Delta_n$ converges to $\Delta_M$? If not, can we at least calculate spectral invariants for $\Delta_M$ using spectral invariants for $\Delta_n$? -The question is a little strange since the graph Laplacian acts on a finite dimensional space (for a finite graph) while the ordinary Laplacian is an unbounded operator on infinite dimensional Hilbert space. It is also not clear that the two operators capture the same kind of geometric information; for example, the dimension of $M$ can be recovered from $\Delta_M$, but it is not obvious to me how to calculate the dimension of $M$ from the $\Delta_n$'s. Nevertheless, my intuition tells me that there's something out there. - -Added: Thanks everyone for all the links / references - I'll need a few more days to chase them down. I'm glad there is so much literature on this question! - -REPLY [2 votes]: The idea is quite old. The finite element method was designed exactly to approximate eigenfunctions with finite dimensional operator defined -via triangulations (see papers of Babushka and Osborn). -There is a recent paper by Burago and Ivanov https://arxiv.org/abs/1301.2222. The references -of this paper include some earlier work, for example, the work of -Fujiwara.<|endoftext|> -TITLE: Does a variety contain a cartesian product of two curves? -QUESTION [6 upvotes]: We are given an affine variety $V\subset \mathbb{A}^n\times\mathbb{A}^n$, and wish to know if it contains a product of the form $C_1\times C_2$, where $C_1$ and $C_2$ are two curves in $\mathbb{A}^n$. First, is there an algorithm to decide this? Second, is it true that if $V$ is of degree $d$ and does contain a product of the form $C_1\times C_2$, then $V$ contains a product of the form $C_1\times C_2$ with $\deg C_1,\deg C_2\leq f(d,n)$, for some function $f$? - -REPLY [5 votes]: Yes, there is such an algorithm. There is an effectively computable constant $N$ such that if $V$ contains a product $S\times T$ where $S,T$ are $N$-point sets, then $V$ contains product of two curves. It is actually true even in the semialgebraic setting. The result is Theorem 1.9 from http://arxiv.org/abs/1207.0705 -[ Apologies for answering my own question with a reference to my own paper. When I asked the question, I did not know the answer. ]<|endoftext|> -TITLE: For what finite groups is the cardinality of a minimal generating set well defined? -QUESTION [5 upvotes]: Recently I learned that the cardinality of a minimal set of generators of a finite $p$-group -$G$ is well defined namely it is equal to the dimension of $H^1(G,\mathbb{F}_p)$. Moreover, if -$S:=\{g_1,\ldots,g_s\}$ is a minimal generating set of $G$ then the cardinality of a minimal set of relations with respect to $S$ is also a well defined integer, namely it is equal to the -dimension of $H^2(G,\mathbb{F}_p)$. -For a general group, the cardinality of a minimal set of generators is not well defined. Take for example the symmetric group of degree $n$. For instance you may get a cardinality equal to $2$ or $n-1$. -Q1: So for what class of finite groups do we expect the cardinality of a minimal generating set to be well defined? -P.S. By "a minimal set of generators" I mean "irredundant", that is the set that generates the group, but no proper subset does. - -REPLY [8 votes]: This is an answer for the abelian case only. -For a finite abelian group (let us exclude the trivial one) there are two standard ways to decompose it as a direct sum of cyclic groups. One into cyclic groups of order $n_1,\dots,n_r$ (not $1$) such that $n_i \mid n_{i+1}$ and the other one into cyclic groups of order $m_1, \dots, m_s$ such that each $m_i$ is a prime power (not $1$). [Under each of the assumptions 'divisibility' and 'prime power' the repective decomposition is unique, in the latter case of course up to the ordering.] -The parameter $r$ is sometimes called the rank and the parameter $s$ the total rank of the group (although terminology here is not completely uniform). -Now, it is known that the rank is the minimal cardinality of a generating set, in the sense that there does not exists a set of a smaller cardinality that generates the group. -And, that the total rank is the maximal cardinality of a minimal generating set, that is there exists a generating set of cardinailty $s$ such that no subset of this set generates the group. -Thus, the cardinality of all minimal/irredundant generating set is uniquely determined if and only if the rank equals the total rank. -This is the case if and only if the group is an (abelian) $p$-groups. -Note that for a non-$p$-group on can first consider the first decomposition into cycylic groups, and then decompose each cyclic component as the sum of cyclic groups of prime power order (more or less Chines Remainder Theorem); so only if all the orders in the first decomposition are also prime powers (and thus necessarily powers of the same prime) does one not get a different value for rank and total rank.<|endoftext|> -TITLE: Carayol via the trace formula -QUESTION [10 upvotes]: Hi, -Is there a proof of the result that Carayol proves in "Sur les representations l-adiques..." -using the Langlands-Kottwitz method of comparing the Lefschetz trace formula and the Selberg trace formula? As far as I can see, the proof in Carayol's paper is done in a more case-by-case fashion... -Thanks - -REPLY [6 votes]: In Langlands's Antwerp article (LNM 349) he proves local-global compatibility (what you are refererring to as "Carayol's result") in the case of principal series or special representations (so not just good reduction), using a comparison of trace formulas. He wasn't able to get the supercuspidal case in this way (and that is what led Deligne to write his letter to P.S.).<|endoftext|> -TITLE: Classification of PDE -QUESTION [30 upvotes]: Recently I have been attending a course on PDE's. I was totally ignorant of the subject and wasn't that motivated to be honest. But I was intrigued and felt I had to take the course seriously both for exams and because I was lead to it through a very convoluted path of research (from arithmetic through algebraic geometry to D-modules = holomorphic linear differential equations). -Anyway, I was told that PDEs were classified into 3 families: hyperbolic, elliptic, parabolic respectively. I saw examples for each: Wave, Laplace and Heat equations. I also saw lots of different methods to solve these examples: using symmetry to reduce to an ODE, Fourier transform, Distributions, Fourier series (in dimension 1 to deal with boundary conditions). Each time it seemed the answer was "reduce the problem to a polynomial equation or an ODE by any way you can". So my geometric brain kicked in and tried to give a unified geometric interpretation to all this. -I think I understand the distinction between those equations. We are considering 2nd order PDEs with real coefficients. And the idea is to reduce the classification to that of their principal symbols seen as quadratic forms on $\mathbb{R}^n$. -So I asked this kind a questions: let's consider a LINEAR PDE $Pf = 0$ where $P$ is a linear differential operator of degree $d$ on $\mathbb{R}^n$. -Question 1: Let's say a "naive elliptic PDE" is any PDE given by a differential operator $P$ whose principal symbol $\sigma(P)$ satisfies $\sigma(P)(x,\xi) \neq 0$ for $\xi\neq 0$. Is this definition any good? -If answer to question 1 is yes, -Question 2: What is the analogue of a parabolic or hyperbolic operator? -The obvious perfectly nice answer would be: "PDE's are classified by the hypersurfaces defined by their principal symbols". Unfortunately the answer I got was something like "Don't over-think it, the classification is more heuristic than anything.". Does that mean "there is a classification along theses lines but it is a bit more subtle" or that "things are much much more complicated in higher dimensions/degrees"? -Anyway... -Question 3: Is it at least true that the classification of PDE with constant coefficient is related to the classification of real algebraic projective hypersurfaces $\{\sigma(P) = 0 \} \subset \mathbb{P}^{n-1}_{\mathbb{R}}$? -Let's assume the previous questions aren't completely wrong for trivial reasons. Let's consider a PDE with non-constant coefficients. We should then classify those according to "families of algebraic projective hypersurfaces" in $\mathbb{P}(T^*\mathbb{R}^n)$. -Question 4: Which kind of families can we expect? Is that related to Gabber's theorem on involutivity of the characteristic variety? -I am now assuming someone answered all these questions without thinking the words "this is so completely wrong". I have a final question (at least before the next one): -Question 5: Why is this all so hard to learn/teach? -PS: Thanks to the people who took time to answer (lots of food for thoughts). I'd be happy to read more especially if you have some references. - -REPLY [5 votes]: If you are interested in a real scalar linear PDE with constant coefficients, then it has all been worked out, primarily by Ehrenpreis, using the Fourier transform. Unfortunately, I don't know of a decent reference. -If you are interested in variable coefficient complex scalar linear PDE's, then there is extensive work by many people, including Hormander, Nirenberg, Treves. -If you're interested in analytic or hyperfunction (dual to analytic functions, analogous to distributions for smooth functions) solutions to constant coefficient systems, then I believe that leads to D-modules and therefore territory unknown to me. I believe there are ties to algebraic geometry. -But as other have mentioned, PDE's, whether a scalar equation or a system of equations, do not succumb to any kind of neat classification or general approach. Only a few special types (the ones you cite) are useful and tractable. Most are not useful and more or less impossible to understand. -Also, many if not most of the most important and interesting are nonlinear. And of first or (more often) second order. There are higher order PDE's that are worth studying but relatively few.<|endoftext|> -TITLE: Serre and Tate's conjectures on étale cohomology -QUESTION [16 upvotes]: In the appendix of Serre and Tate "Good Reduction of Abelian Varieties" [Annals of Mathematics 88 (1968), 492-517], the authors make the following conjectures. -Suppose that $X$ is a smooth proper variety over a local field (which does not necessarily have good reduction). Write $\rho_\ell$ for the Galois representation given by $H^n_{et}(X, \mathbb{Z}_\ell)$, for some prime $\ell$ not equal to the residue characteristic. Suppose the residue field is finite. Then Serre and Tate conjectured: -(1) The restriction of $\operatorname{Tr} \rho_\ell$ to the inertia subgroup is locally constant, takes values in $\mathbb{Z}$, and is independent of $\ell$. -(2) The characteristic polynomial of $\rho_\ell(\pi)$ has rational coefficients independent of $\ell$, and that its roots have absolute value $q^{-k/2}$ for some $0 \leq k \leq 2n$. (For any choice of $\pi$ of Frobenius element in the Galois group; here, $q$ is the cardinality of the residue field.) -Are these conjectures now proven, or are they still open? And if they are proven, what is a reference for this? - -REPLY [2 votes]: The article of Katz "review of $\ell$-adic cohomology" in Motives (PSPM 55, 1990) contains a good survey of the partial results in the direction of those conjectures that were known then. -There hasn't been many progresses since.<|endoftext|> -TITLE: Conjugate points in Lie groups with left-invariant metrics -QUESTION [19 upvotes]: For any Lie group $G$ there exist many left-invariant Riemannian metrics, namely, one just takes any inner product on the tangent space at the identity $T_eG$ and then left translate it to the other tangent spaces. If $G$ is compact, one can do better. Namely, one can start with an inner product at $T_eG\cong\mathfrak g$ which is also invariant under the adjoint representation $\mathrm{Ad}:G\to GL(\mathfrak g)$ (obtained by averaging an arbitrary inner product with respect to a Haar measure) and it turns out that the resulting Riemannnian metric on $G$ is also right-invariant. Since it is invariant -under left and right translations, it is called bi-invariant. -The Riemannian geometry of bi-invariant metrics is very nice. For instance, geodesics -through the identity coincide with one-parameter groups, so the Riemannian exponential map coincides with the Lie group exponential. The Riemann curvature tensor has a simple formula -$R(X,Y)X=-[[X,Y],X]$ for unit vectors $X$, $Y\in\mathfrak g$, from which follows that the sectional curvature is nonnegative. Actually, compact semisimple Lie groups equipped with bi-invariant Riemannian metrics are symmetric spaces of compact type (the geodesic symmetry at the identity is the inversion map $g\mapsto g^{-1}$) and hence its geometric and topological invariants are amenable to explicit computations. -Fix a bi-invariant metric in a compact semisimple Lie group $G$. As for any other symmetric space, the Jacobi equation along a fixed geodesic $\gamma$ (say starting at $e$) has constant coefficients with respect to a parallel orthonormal frame (since the sectional curvature is parallel). -Invoking the real root space decomposition of the Lie algebra $\mathfrak g = \mathfrak t+\sum_{\alpha\in\Delta^+} \mathfrak g_\alpha$ with respect to a maximal torus with Lie algebra $\mathfrak t$, we have $R ( H , X_\alpha)H = \mathrm{ad}_H^2 X_\alpha = -\alpha(H)^2 X_\alpha$ and hence a typical Jacobi field $J$ along $\gamma(t)=\exp tH$ (for a unit vector $H\in\mathfrak t$) vanishing at $t=0$ is -of the form $J(t)=\sin(\alpha(H)t)X_\alpha(t)$, where $X_\alpha(t)$ is the parallel -vector field along $\gamma$ with $X_\alpha(0)=X_\alpha\in\mathfrak g_\alpha$. -It follows that $\gamma(t)$ is a conjugate point to $\gamma(0)=e$ along $\gamma$ if $\alpha(H)t\in \pi\mathbf Z$, and then the contribution to the multiplicity is $\dim\mathfrak g_\alpha=2$. We see that the total multiplicity of the conjugate point $\gamma(t)$ is twice the number of roots $\alpha\in\Delta^+$ such that $\alpha(H)t\in\mathbf Z$, hence it is even. In other words, in bi-invariant metrics conjugate points always have even multiplicity (in particular, due to the Morse index theorem also the index of geodesics is always even). -My question is whether this property characterizes bi-invariant metrics among left-invariant ones. Namely, assume we have a left-invariant Riemannian metric on $G$ such that for every point $g\in G$ conjugate to the identity element along a geodesic the multiplicity is even. Is it true that the metric must be bi-invariant? - -REPLY [2 votes]: A left-invariant Riemannian metric on Lie group is a special case of homogeneous Riemannian manifold, and its differential geometry (geodesics and curvature) can be described in a quite compact form. I am most familiar with the description in 28.2 and 28.3 of here of covariant derivative and curvature. -But on a Lie group itself there is an explicit description of Jacobi fields available for -right invariant metrics (even on infinite dimensional Lie groups) in section 3 of: - -Peter W. Michor: Some Geometric Evolution Equations Arising as Geodesic Equations on Groups of Diffeomorphism, Including the Hamiltonian Approach. IN: Phase space analysis of Partial Differential Equations. Series: Progress in Non Linear Differential Equations and Their Applications, Vol. 69. Birkhauser Verlag 2006. Pages 133-215. (pdf). - -I shall now describe the results (which go back to Milnor and Arnold). For detailed computations, see the paper. -Let $G$ be a Lie group with Lie algebra -$\def\g{\mathfrak g}\g$. -Let $\def\x{\times}\mu:G\x G\to G$ be the multiplication, let $\mu_x$ be left -translation and $\mu^y$ be right translation, -given by $\mu_x(y)=\mu^y(x)=xy=\mu(x,y)$. -Let $\langle \;,\;\rangle:\g\x\g\to\Bbb R$ be a positive -definite inner product. Then -$$\def\i{^{-1}} -G_x(\xi,\eta)=\langle T(\mu^{x\i})\cdot\xi, - T(\mu^{x\i})\cdot\eta)\rangle -$$ -is a right invariant Riemannian metric on $G$, and any -right invariant Riemannian metric is of this form, for -some $\langle \;,\;\rangle$. -Let $g:[a,b]\to G$ be a smooth curve. -In terms of the right logarithmic derivative $u:[a,b]\to \g$ of $g:[a,b]\to G$, given by -$u(t):= T_{g(t)}(\mu^{g(t)\i}) g_t(t)$, -the geodesic equation has the expression -$$ \def\ad{\text{ad}} -\partial_t u = u_t = - \ad(u)^{\top}u\,, -$$ -where $\ad(X)^{\top}:\g\to\g$ is the adjoint of $\ad(X)$ with respect to the inner product $\langle \;,\; \rangle$ on $\g$, i.e., -$\langle \ad(X)^\top Y,Z\rangle = \langle Y, [X,Z]\rangle$. -A curve $y:[a,b]\to \g$ is the right trivialized version of a Jacobi field along the geodesic $g(t)$ described by $u(t)$ as above iff -$$ -y_{tt}= [\ad(y)^\top+\ad(y),\ad(u)^\top]u - - \ad(u)^\top y_t -\ad(y_t)^\top u + \ad(u)y_t\,. -$$ - -Continued: - -For connected $G$, the right invariant metric is biinvariant iff $\ad(X)^\top = -\ad(X)$. -Then the geodesic equation and the Jacobi equation reduces to -$$ -u_t = \ad(u)u = 0,\qquad y_{tt} = \ad(u)y_t -$$ -Now we can look at examples. -If $G=SU(2)$ then $\g=\mathfrak{sl}(3,\mathbb R)$ and we can take an arbitrary inner product on it. -(Maybe, I will continue if I have more time in the next few days).<|endoftext|> -TITLE: number of weighted trivalent trees -QUESTION [6 upvotes]: Given a trivalent tree (graph without loops with valence of 3+ at each vertex) on N marked points, let's assign to each vertex the number (its valence - 3)! (note the ! at the end). Take the product of these numbers over all vertices in the graph. Sum the resulting products over all trivalent graphs on N marked points. The answer seems to be $(N-2)^{N-2}$. I would guess this is well-known. Is there something I can reference? Thanks. - -REPLY [3 votes]: This is just building off the previous discussion -- I will just give some indication of why Dan's $f$ and $g$ are Legendre transforms of each other. This is essentially a fleshed out version of some of what Frédéric was saying that I wanted to work out in detail -- it would look too small and ugly as a comment. -We're going to use some basic facts about the Lambert W-function.Lambert W-function and the Legendre transformation, but nothing past what you can find in those Wikipedia links. -First, showing that $f$ and $g$ are Legendre transforms of each other is essentially showing that their derivatives are compositional inverses of each other. So we will show this for $f^\prime$ and $g^\prime$, and not use anything else about the Legendre transformation. -The Lambert $W$ function $W(x)$ is defined by the functional equation $x=W e^W$. You recognize it's coming into play when you start seeing $n^n$ terms, as they show up in the taylor expansion of it and simple expressions related to it. In particular, wikipedia tells us that: -$$\left(\frac{W}{x}\right)^r = e^{-rW}=\sum_{n\geq 0} r(n+r)^{(n-1)} \frac{(-x)^n}{n!}$$ -The first equation is the functional equation, and in general to get the taylor expression you want to use Lagrange inversion and the functional equation. -In any case, if we set $r=-1$, take $0^0=1$ and hopefully not make any sign errors, we should obtain -$$g^\prime(x)=1-e^{W(-x)}.$$ -Simple calculus gives -$$f^\prime=-(1-x)\ln(1-x).$$ -From these expressions it is immediate that $f^\prime(g^\prime(x))=x$ by using the functional equation for $W$.<|endoftext|> -TITLE: What is the "symplectic duality" between holomorphic symplectic manifolds? Where can I read more about it? -QUESTION [11 upvotes]: I'm recently working on something called 3d mirror symmetry in QFT literature, which involves two hyperkähler manifolds. -There seems to be a corresponding(?) mathematical theory called symplectic duality, pursued by Braden, Licata, Proudfoot and Webster. -Where can I read about it? The only thing I could find so far is the proposal by Proudfoot et al. I'm particularly intrigued by the fifth example in page. 7, which says - -More generally, the moduli space of $G$-instantons on a crepant resolution of $\mathbb{C}^2/\Gamma$ is dual to the moduli space of $G'$-instantons on a crepant resolution of $\mathbb{C}^2/\Gamma'$, where $G$ is matched to $\Gamma'$ and $G'$ is matched to $\Gamma$ via the McKay correspondence. - -Where can I read about this duality, in particular the case when neither $G$ nor $G'$ is of type $A$? - -REPLY [7 votes]: Nowhere. The paper is still in preparation, and looks to be for a few more months at least. Probably the best document at the moment is this (extremely long) set of talk slides of mine. -I should note: even when there is a paper, there won't be a definition that will tell you (by which I mean someone who studies mirror symmetry from the string theory side) anything new. Just an observation of a lot of very striking coincidences. -Also, it's not going to discuss the non-type A Mackay example in detail; it's not one of the ones we understand well, we just included it because the physicists told us to.<|endoftext|> -TITLE: Generalization of Tamarkin's ARO 1993, final round, problem 10/8: still a conjecture? -QUESTION [32 upvotes]: This is from the category "problems I cannot believe that are still open". But then again, I don't know whether it is still open; it seems to have escaped the attention of most number theorists and algebraists except for those in olympiad circles. This is the reason I am posting it here. -Let $p$ be a prime. Define a linear operator $F_p:\mathbb R^{\mathbb Z}\to\mathbb R^{\mathbb Z}$ by -$\left(F_p f\right)\left(n\right) = \dfrac{f\left(n\right)+f\left(n+1\right)+...+f\left(n+p-1\right)}{p}$ for every $n\in\mathbb Z$ and every $f\in\mathbb R^{\mathbb Z}$. -(Of course, elements of $\mathbb R^{\mathbb Z}$ are just two-sided infinite sequences of reals, written as functions from $\mathbb Z$ to $\mathbb R$. The operator $F_p$ replaces a sequence by the sequence of the arithmetic means of its $p$-windows.) -An element $f\in \mathbb Z^{\mathbb Z}$ is said to be average-integral if it satisfies $F_p^kf\in\mathbb Z^{\mathbb Z}$ for every nonnegative integer $k$. -For any $f\in\mathbb R^{\mathbb Z}$, define $f^p\in\mathbb R^{\mathbb Z}$ by -$f^p\left(n\right)=\left(f\left(n\right)\right)^p$ for every $n\in\mathbb Z$. -Conjecture: If $f\in \mathbb Z^{\mathbb Z}$ is average-integral, then so is $f^p$. -Remarks: For $p=2$, this was problem 8 for grade 10 in the Allrussian Mathematical Olympiad 1993, proposed by D. Tamarkin (the one of operad theory fame?). There is a discussion with several proofs of the $p=2$ case on MathLinks, and it shows that the $p=2$ case is actually a tip of an iceberg (namely, for $p=2$, the average-integral elements of $\mathbb Z^{\mathbb Z}$ form a ring, so not only squares but also pointwise products of average-integral elements are average-integral). For $p=3$, the conjecture is still true, but the iceberg apparently is not anmyore; it took me a lengthy computation with combinatorial divisibilities to verify the conjecture. For higher $p$, I don't know of any results at all. Has anything been done since 1993 at all? - -REPLY [16 votes]: Ok, here's how to show that $(1+n^{p+1})^p$ always gives a counterexample for $p > 3$. -Set $d(k) = \lfloor \frac{k}{p-1} \rfloor - v_p(k!)$. Note that $d(k)$ is equal to the sum of the base-$p$ digits of $k$, divided by $p-1$, and rounded down, so for instance if $k$ is a nonzero multiple of $p-1$ then we have $d(k) \ge 1$. We have -$p^{d(k)}(F_pn^k - n^k) = \sum_{j=0}^{k-1}(p^{d(k)-1}{k\choose j}\sum_{i=0}^{p-1}i^{k-j})n^j,$ -and we have -$d(k)-1 + v_p({k\choose j}) + v_p(\sum_{i=0}^{p-1}i^{k-j}) - d(j) $ -$= \lfloor \frac{k}{p-1} \rfloor - \lfloor \frac{j}{p-1} \rfloor - v_p((k-j)!) - 1 + v_p(\sum_{i=0}^{p-1}i^{k-j})$ -$ = (d(k-j) + v_p(\sum_{i=0}^{p-1}i^{k-j})-1) + (\lfloor \frac{k}{p-1} \rfloor-\lfloor \frac{j}{p-1} \rfloor-\lfloor \frac{k-j}{p-1} \rfloor) \ge 0$, -so by induction on $k$ we can show that $p^{d(k)}n^k$ is average-integral for any $k$. -We have $d(k(p+1)) \le 1$ for $k = 1, ..., p-2$, $d(p^2-1) = 2$, and $d(p^2+p) = 0$. Thus, we easily see that $(1+n^{p+1})^p = 1 + \sum_{k=1}^{p-2}{p \choose k}n^{k(p+1)} + pn^{p^2-1} + n^{p^2+p}$ is average-integral if and only if $pn^{p^2-1}$ is. Next, note that $pn^{p^2-1}-pn^{\underline{p^2-1}}$ is a sum of multiples of $p$ times monomials $n^k$ with $k < p^2-1$, and that for $k < p^2-1$ we have $d(k) \le 1$, so $pn^{p^2-1}$ is average-integral if and only if $pn^{\underline{p^2-1}}$ is. -Now note that we can express $\frac{\Delta^{p-1}f}{p}$ as an integral linear combination of $F_pf$ and the shifts of $f$, since we have $(-1)^k{p-1 \choose k} \equiv 1 \pmod{p}$. Thus, if $pn^{\underline{p^2-1}}$ was average-integral then we would have $\frac{\Delta^{p^2-1}pn^{\underline{p^2-1}}}{p^{p+1}} = \frac{p(p^2-1)!}{p^{p+1}}$ an integer, but this is obviously not the case, so we're done. (And thus, $(1+n^{p+1})^p$ fails to be average-integral by the $p+1$st iteration.) -Edit: In fact, we can show that a polynomial in $n$ produces an average-integral sequence if and only if it can be expressed as a $\mathbb{Z}_{(p)}$-linear combination of terms of the form $p^{d(k)}n^k$, or equivalently if it can be expressed as a $\mathbb{Z}_{(p)}$-linear combination of terms of the form $p^{d(k)}n^{\underline{k}}$. -The second claim is much easier to prove: we have $\frac{\Delta^{p-1}p^{d(k)}n^{\underline{k}}}{p} = p^{d(k)-1}k^{\underline{p-1}}n^{\underline{k-(p-1)}}$, which is an integer relatively prime to $p$ times $p^{d(k-(p-1))}n^{\underline{k-(p-1)}}$. Also, if $k < p-1$, then $d(k) = 0$ and we can recover the coefficient of $n^{\underline{k}}$ from the first $p-1$ values of the polynomial without doing any multiplication or division by $p$. -For the first claim, take a polynomial which is not an integer combination of terms of the form $p^{d(k)}n^k$, and look at the largest $k$ such that the coefficient on $n^k$ is not a multiple of $p^{d(k)}$. By subtracting off a polynomial that we already know to be average-integral (ignoring denominators other than $p$), we can assume that this is the leading term of the polynomial. Now convert to the falling power basis, and note that the leading term remains the same, to show that this polynomial is not average-integer.<|endoftext|> -TITLE: Inverting products of matrices -QUESTION [5 upvotes]: I need to compute a large number of inverses of the following form: -$(A \Lambda_k A^\top)^{-1}$ -where $A \in \mathbb{R}^{m \times n}$, $n > m$ and $\Lambda_k = \text{diag}(\lambda_1, ..., \lambda_n)$ with $\lambda_i > 0$. Is there an efficient way to do this? -In the end, I want to sample from Gaussians with means $(A \Lambda_k A^\top)^{-1} \mu_k$ and covariances $(A\Lambda_kA)^{-1}$. -I actually can choose $A$ freely, as long as the rows of $A$ are orthogonal to the rows of a matrix $B \in \mathbb{R}^{(n - m) \times n}$, i.e. $A \cdot B^\top = 0$. Oh, and $\text{rank}(A) + \text{rank}(B) = n$. - -REPLY [3 votes]: Now that you've provided some more information, I think I can make some useful suggestions. -First, a quick review of linear transformations of multivariate normal random vectors. If $z$ is an MVN vector with mean $\mu$ and covariance matrix $C$, and $M$ is a matrix of the appropriate size and $y=Mz$, then $y$ is MVN with $E[y]=M\mu$ and $Cov(y)=MCM^{T}$. -This is a key fact that can be very useful in algorithms for generating MVN random numbers with desired mean and covariance. If you want to generate an MVN vector $x$ with mean $\mu$ and covariance $C$, then you can do this by - -Compute the Cholesky factorization of $C$, $C=R^{T}R$. -Let $z$ be an N(0,I) random vector. -Let $x=R^{T}z+\mu$. -Then $E[x]=R^{T}E[z]+\mu=\mu$ and $Cov(x)=R^{T}IR=C$. - -You could simply apply this algorithm to your problem by computing the Cholesky factorization of the covariance matrix: -$(A\Lambda_{k} A^{T})=R^{T}R$. -Then -$(A\Lambda_{k} A^{T})^{-1}=R^{-1}R^{-T}$. -Then you could generate the desired random vector $x$ with -$x=R^{-1}z+R^{-1}R^{-T}\mu_{k}$. -Computing the matrix $A\Lambda_{k} A^{T}$ takes $O(m^2n)$ time. Computing the Cholesky factorization takes $O(m^3)$ time, with a somewhat larger constant factor. Computing the inverse of $R$ can be done quickly by backsolving, but still takes $O(m^3)$ time. However, if $n$ is much bigger than $m$, you'll end up spending most of your time computing -$A\Lambda_{k} A^{T}$. -Because of the orthogonality constraints on the rows of $A$, you can assume that $A$ and $A\Lambda_{k} A^{T}$ will be fully dense. Thus there doesn't appear to be anything you can gain here by using sparse matrix techniques. -My first version of this answer suggested using the $QR$ factorization of $A$. This would leave you with -$A\Lambda A^{T}=QR\Lambda_{k} R^{T}Q$ -where $Q$ is an $m$ by $m$ orthogonal matrix and $R$ is an $m$ by $n$ upper triangular matrix. Unfortunately, you'd then have to go on and compute a Cholesky factorization of -$R\Lambda_{k} R^{T}$, which is just as much work as computing the Cholesky factorization of $A\Lambda_{k} A^{T}$. So I don't think the QR factorization is worth while after all. However, if it happened that $m=n$, then the QR factorization approach would be very helpful!<|endoftext|> -TITLE: Pull-push in Godin's HCFT for string topology -QUESTION [9 upvotes]: I am reading Veronique Godin's famous article "Higher string topology operations" (http://arxiv.org/abs/0711.4859) that demonstrates that the string topology operations on $(H_\bullet(L X), H_\bullet(P X))$ for $X$ a compact oriented smooth manifold are part of a homological 2d TQFT with target space $X$. -The idea of the article is clear and simple, but the realization is quite technical. I have the following question on a technical detail. -The idea is that for $\Gamma$ some suitable surface with incoming and outgoing boundary inclusions -$$ - \partial_{in} \Gamma \stackrel{in}{\to} \Gamma \stackrel{out}{\leftarrow} \partial_{out} \Gamma -$$ -we form the mapping space correspondence -$$ - X^{\partial_{in} \Gamma} \stackrel{X^{in}}{\leftarrow} X^\Gamma - \stackrel{X^{out}}{\to} X^{\partial_{out} \Gamma} -$$ -Then the string operations and their generalizations are supposed to be given by pull-push of homology classes through this span -$$ - out_* \circ in^! : H_\bullet(X^{\partial_{in} \Gamma}) \to H_{\bullet+\chi\Sigma-dim X}(X^{\partial_{out} \Gamma}) - \,, -$$ -where the wrong-way map $in^!$ is "dual fiber integration" via Thom collaps and Thom isomorphism. Or rather, the idea is to have these operations be parameterized by the moduli space of surfaces. -While nice and simple, this idea is maybe a bit too simple: it is hard to get the wrong-way map on homology for $X^{\Gamma} \to X^{\partial_{in} \Gamma}$ under control. Concretely, we'd need to fatten it to an embedding and then find a tubular neighbourhood, both of which is subtle for these infinite-dimensional mapping spaces. -So instead, in Godin's article on p. 22, around diagram (6), this morphism is replaced itself by a zig-zag. The idea is to first fatten $X^{in}$ by remembering where the extra edges $eE$ and extra vertices $eV$ land, separately, that are not in the incoming boundary, then contract these edges to points, and then finally embed the spaces of these extra vertical and contracted edges into a contractible mapping space $W^{eV \coprod eE}$ induced by an embedding of the manifold $X$ into a vector space $W$. -In total then this produces a long zig-zag replacing the above simple span, that in the labelling used in the article reads -$$ - X^{\partial_{in} \Gamma} \times W^{eE \coprod eV} - \leftarrow - X^{\partial_{in} \Gamma} \times X^{eE \coprod eV} - \to - X^{\partial_{in} \Gamma} \times P X^{eE} \times X^{eV} - \stackrel{X^{in} \times \cdots }{\leftarrow} - X^{X^\Gamma} - \stackrel{X^{out}}{\to} - X^{\partial_{out} \Gamma} - \,. -$$ -We can now pull-push-pull-push homology through this longer zig-zag. The contractible factor $W^{eE \coprod eV}$ on the far left doesn't disturb the form of this (if one does it right) and so this produces some map. -I think I understand Godin's article, all the constructions involved in this and the result. Though it is a bit of a tour-de-force, due to some technicalities, not the least of which do arise precisely because of all the extra structure to be taken care of in this longer zig-zag. -So my question finally is: while I see that we can get pull-push through the longer zig-zag to work, how do we know that this longer zig-zag is a good replacement for the naive short and simple zig-zag? Is this justified "only" (not that I doubt that this is a big achievement) by the fact that it works, produces and HQFT and reproduces the string topology operations? -What if I came up with a different long zig-zag, that also makes all these things work? Will it necessarily give the same result? -What if we tried to lift the pull-push construction to the chain level by a chain-level version of the Thom isomorphism, producing a genuine "TCFT" instead of just an "HCFT". Wouldn't that make us want to fall back to the simple one-step pull-push? - -REPLY [9 votes]: Urs, although I answered some of these questions privately, I will post them here as well. -At the moment there is no technology to do the naive short zig-zag you suggest. Godin's techniques for umkehr maps of mapping spaces of maps into a manifold $M$ can do essentially two things: (1) create isolated points in the domain, (2) glue some points in the domain together. The first is done by adding some contractible padding in the form of the Euclidean space $W$ to the domain of the wrong-way arrow and embedding the maps from the new points into $M$ into this Euclidean space; it is then in the correct form to do the standard Thom collapse construction of the umkehr map. The second uses propagating flows, which have as a necessary condition that the map between mapping spaces is giving by the pullback of a finite-dimensional embedding, such that heuristically is of finite codimension. If one tries to identify for examples intervals, this is no longer the case. In Godin's zig-zag, the first wrong-way arrow uses (1), the second uses (2). -We don't know how to do your short zig-zag yet, because if the complement of the incoming boundary graph contains a loop, then we cannot create it directly by creating a point and expanding it. One would get -\begin{equation}M^{\partial_{in} \Gamma} \times W \leftarrow M^{\partial_{in} \Gamma \sqcup *} \leftarrow M^{(\partial_{in} \Gamma) \sqcup (\Gamma \backslash \partial_{in} \Gamma)}\end{equation} -There is no known method to do the umkehr map for the second wrong-way arrow, because it is not a homotopy equivalence and hence we cannot produce an inverse up to homotopy. What Godin does is break the complement into a disjoint union of points and lies graphs, each of which is contractible and can be created by method (1). (Coincidentally, this particular way of breaking up the graph is natural and a good choice to get a grip on the local system appears due to the Thom isomorphism for unoriented virtual bundles.) So Godin's construction seems to me be the minimal one with the current technology that does give HCFT operations which coincide with all previously known string topology operations. This could be taken as justification. -However, it is legitimate to ask the questions: Should this construction work? Why this zig-zag? I think the answer lies in the direction that umkehr maps should be functorial: if $f^!$, $g^!$ and $(g \circ f)^!$ make sense then they should coincide. In the case of Godin's zig-zag, the umkehr maps that appear are in the following order: -\begin{equation}(f_{glue})^! \circ (f_{expand})^* \circ (f_{create})^!\end{equation} -where $f_{create}: M^{\partial_{in} \Gamma} \times M^{eV \sqcup eE} \to M^{\partial_{in} \Gamma} \times W^{eV \sqcup eE}$ creates the additional points, $f_{expand}: M^{\partial_{in} \Gamma} \times M^{eV \sqcup eE} \to M^{\partial_{in} \Gamma} \times M^{eV} \times PM^{eE}$ expands those corresponding to edges into intervals, $f_{glue}:M^\Gamma \to M^{\partial_{in} \Gamma} \times M^{eV} \times PM^{eE}$ glues the new vertices and edges together and to the incoming boundary. Now, $f_{expand}$ is a homotopy equivalence, so we are really looking at a composition of two umkehr maps $(f_{glue})^!$ and $(f_{create})^!$, which together do the same thing the umkehr map for your short zig-zag should do. If the construction of that umkehr map is functorial (as it should be), then they should coincide. This is why I think the construction based on Godin's longer zig-zag gives the same answer as a conjectural simpler zig-zag construction. -For the same reason, any other zig-zag should be the give the same operations as the conjectural construction for a simpler zig-zag and hence be equal to Godin's construction. Of course, this is not a proof. If you give me a different zig-zag, I can probably find a zig-zag of zig-zag's between that and Godin's, which shows that the corresponding operations are equal.<|endoftext|> -TITLE: Blowing-up and direct image sheaf. -QUESTION [7 upvotes]: Let $\pi:Z=Bl_{Y}(X)\rightarrow X$ be the blowing-up of a smooth projective variety X along a subvariety $Y$, $E$ the closed subscheme defined by $\pi^{−1}I_{Y,X}\cdot O_Z$. -Is it true that (without assume $Y$ smooth) $\pi_{\ast}(O_Z(-E))=I_{Y,X}$? -Thanks for your help. - -REPLY [7 votes]: As Sándor's nice answer shows, $\pi_{\ast}(O_Z(-E))=I_{Y}$ is not neccesarily true, even for $Y$ a normal subvariety of $X$. On the other hand, the following statement -$$\pi_{\ast}(O_Z(-mE))=I_{Y}^m \quad\mbox{ for $m\gg 0$}$$ -always holds, even without any assumptions on the subscheme $Y$. Here is a short explaination why: -It suffices to deal with the case $X=\mbox{Spec} A$ is affine and $I=I_Y=(g_1,\ldots,g_n)\subset A$. The $g_i$'s determine a surjection $A^r \to I$, and hence a surjection of Rees algebras $\mbox{Sym}^*(O_X^r) \to \bigoplus_{m \ge 0} I^m$. Taking Proj this means that there is an embedding $Z=Bl_Y X \subset \mathbb{P}=\mathbb{P}(O_X^n)$ where the exceptional divisor $E$ corresponds to $O(-1)\big|_Z$. Now, if $p:\mathbb{P}(O_X^n)\to X$ is the projection, we have $$p_* O_{\mathbb{P}}(m) \to p_*O_Z(-mE)$$ is surjective for $m$ large by relative Serre vanishing. Moreover, since $p_*O_{\mathbb{P}}(m)=Sym^m(O_X^n)$ we can identify the image of this map with $I^m$ and hence we have $\pi_{\ast}(O_Z(-mE))=I^m$. -In particular, if $\pi_* O_E=O_Y$ holds (as in Sándor's answer) the above map will always be surjective and $\pi_{\ast}(O_Z(-mE))=I^m$ for all $m\ge 0$.<|endoftext|> -TITLE: Shortest Path in Plane -QUESTION [8 upvotes]: I thought about the following problem: -Given a polygonal subdivision of the euclidian plane where each of the polygons has a speed associated with it, and given two points s,t, I'm interested in the fastest path from s to t. -I don't know if the problem is NP-hard or not. -If the regions are convex, I have an easy 2-approximation algorithm. -(Maybe there is an PTAS algorithm for the convex/general case?) -I'm pretty sure that this problem has been studied before. -Does anybody know any publications about it and/or has an idea how to show NP-hardness. -(But after all, maybe the problem is easy.) -Thank you -Andy - -REPLY [12 votes]: The problem you posed is known in the literature as the weighted region problem. -It was the focus of Joe Mitchell's Ph.D. thesis, under the direction of -Papadimitriou, and their results were eventually published in the Journal of the ACM: -"The weighted region problem: finding shortest paths through a weighted planar subdivision," -Joseph S. B. Mitchell, -Christos H. Papadimitriou, Volume 38, Issue 1, Jan. 1991. -As maproom noted, Snell's law applies and helps. -Under the interpretation in this paper, the problem can be solved in polynomial time, -$O(n^8 L)$, where $L$ reflects the input precision (assumed to be integer coordinates) -and the output error tolerance $\epsilon$. -Here is their abstract. - -Abstract. - The problem of determining shortest paths through a weighted planar polygonal subdivision with $n$ vertices is considered. Distances are measured according to a weighted Euclidean metric: The length of a path is defined to be the weighted sum of (Euclidean) lengths of the subpaths within each region. An algorithm that constructs a (restricted) “shortest path map” with respect to a given source point is presented. The output is a partitioning of each edge of the subdivion into intervals of $\epsilon$-optimality, allowing an $\epsilon$-optimal path to be traced from the source to any query point along any edge. The algorithm runs in worst-case time $O(ES)$ and requires $O(E)$ space, where $E$ is the number of “events” in our algorithm and $S$ is the time it takes to run a numerical search procedure. In the worst case, $E$ is bounded above by $O(n^4)$ (and we give an $\Omega(n^4)$ lower bound), but it is likeky that $E$ will be much smaller in practice. We also show that $S$ is bounded by $O(n^4L)$, where $L$ is the precision of the problem instance (including the number of bits in the user-specified tolerance $\epsilon$). Again, the value of $S$ should be smaller in practice. The algorithm applies the “continuous Dijkstra” paradigm and exploits the fact that shortest paths obey Snell's Law of Refraction at region boundaries, a local optimaly property of shortest paths that is well known from the analogous optics model. The algorithm generalizes to the multi-source case to compute Voronoi diagrams. - -Here is their Fig. 4, illustrating that the shortest path between two points $x$ and $x'$ within -one region $f$ is not always the segment $x x'$:<|endoftext|> -TITLE: Finite subgroups of $PGL_2(K)$ in characteristic $p$ -QUESTION [12 upvotes]: Let $K$ be a field of characteristic $p$. What are the finite subgroups of $PGL_2(K)$ whose orders are divisible by $p$? And if $G$ and $H$ are two such subgroups that are isomorphic, can one say when they are conjugate inside $PGL_2(K)$? -While the finite subgroups of $PGL_2(\mathbb{C})$ are well understood from a variety of different viewpoints, the answer to the above question does not appear to be well known. See, e.g., this article of Beauville for a pleasant discussion of the case in which $p$ does not divide the order of the group. - -REPLY [7 votes]: After thinking about this question for a few months, I've managed to give a complete answer. The article is posted on the arXiv (arXiv:1112.1999v1 [math.NT]). The short version is that a finite $p$-irregular subgroup of $\mathrm{PGL}_2(k)$ is isomorphic to $\mathrm{PSL}_2(\mathbb{F}_q)$ or $\mathrm{PGL}_2(\mathbb{F}_q)$ for $q$ a power of the characteristic of $k$, to a $p$-semi-elementary group (a semi-direct product of a $p$-group and a cyclic group), to a dihedral group, or to $\mathfrak{A}_5$. Thanks to Geoff Robinson's suggestion above, I was able to modify Dickson's arguments to give a complete classification up to conjugacy over an an algebraically closed field. The same arguments go through for separably closed fields, except when the characteristic of $k$ is 2, where a little extra work is required. Finally, I use Galois descent (following Beauville) to pass to arbitrary fields.<|endoftext|> -TITLE: Extension theory with bump function -QUESTION [8 upvotes]: Let $B_t(0)$ denote the $n$ dimensional ball of radius $t$ centered at the origin. Does there exist a $\phi\in C(\mathbb{R}^n)$ function with the properties: -$ -\phi (x) = -\begin{cases} -1&x\in B_r(0) -\\\ -0&x\not\in B_{r+3}(0) -\end{cases} -$ -and for any real-valued function $f\in \mathcal{H}^\tau(\mathbb{R^n})$ ($\tau\in\mathbb{R}$, $\tau>d/2$) we have -$ -\left\|\phi(\cdot) f(\cdot)\right\|_{\mathcal{H}^\tau(A_1)}\leq C\left\|f\right\|_{\mathcal{H}^\tau(A_2)} -$ -where $C$ is a constant independent of $f$, $A_1 = B_{r+2}(0)\setminus B_{r+1}(0)$ and $A_2 = B_{r+3}(0)\setminus B_{r}(0)$. -This has a similar feel to extension theory results if we think of $\phi$ as an extension operator which preserves $f$ on $B_r(0)$. - -REPLY [4 votes]: Short answer: yes. -Let $\psi_\epsilon(x):=\frac{1}{\epsilon^n}\exp{\epsilon^2/(\epsilon^2-|x|^2)}$ for $|x|<\epsilon$, and $\psi_{\epsilon}(x)=0$ for $|x|\geq \epsilon$. Set $\epsilon=2$, and define -$\phi$ is the convolution of $C\phi_{\epsilon}$ with the characteristic function of $B_{r+3/2}(0)$, that is, -$\phi(x):= C\psi_\epsilon(x)* \chi_{B_{r+3/2}(0)}$. Here $C$ is a normalizing constant (this may not be needed, but I haven't checked). -This yields a smooth cut-off function which is 1 in the ball $B_{r+1}(0)$, and zero outside $B_{r+2}(x)$. -To see this does the trick, one can use a localization theorem, for example, Theorem 3.20 in 'Strongly Elliptic Systems and Boundary Integral Equations' by W. McLean. This theorem states: -'Suppose that $\phi \in C^r_{comp}(\mathbb{R}^n)$ for some integer $r\geq 1$, and let $|s|\leq r$. If $u\in H^s(\Omega)$ then $\phi u \in H^s(\Omega)$, and $||\phi u||_{H^s(\Omega)} \leq C_r||\phi||_{W^{r,\infty}(\mathbb{R}^n)}Q_u$ where $Q_u=||u||_{H^s(\Omega). -}$ (Apologies, I encountered trouble while trying to typeset the LaTeX here). -The same result holds with $H^s(\Omega)$ replaced with $\tilde{H}^s(\Omega)$.' -The proof proceeds using $\Omega = A_2$, and then -either by -(a) considering the situation for $s=r$, using duality to see it holds for $s=-r$, and the intermediate $s$ by interpolation. This is suggested by Yakov above. -or (b) by examining $\hat{\phi u}$ and using Peetre's inequality. -Since the constructed $\phi \in C^\infty$ and has compact support, it will satisfy the inequality you seek. In my comment I asked whether you wanted a $\phi$ of minimal regularity (relative to $\tau$); my construction works but may be overkill.<|endoftext|> -TITLE: If compact connected Lie groups are homeomorphic as topological space, are they isomorphic as Lie groups? -QUESTION [13 upvotes]: Let $G_{1}$ and $G_{2}$ be compact connected Lie groups. -If $G_{1}$ and $G_{2}$ are homeomorphic as topological spaces, are -they isomorphic as Lie groups? - -REPLY [31 votes]: No. The simplest example I can think of is that $SO(4)$ is homeomorphic to $SO(3)\times Sp(1)$ as topological spaces, but they are not isomorphic Lie groups. In fact, there is a double covering $Sp(1)\times Sp(1)\to SO(4)$, $(q_1,q_2)\cdot x = q_1xq_2^{-1}$, where $Sp(1)$ -is viewed as the unit quaternions and $x\in\mathbf H\cong\mathbf R^4$, which gives -a Lie group isomorphism $Sp(1)\times Sp(1)/\{\pm1\}\cong SO(4)$. -On the other hand, every continuous homomorphism between Lie groups is automatically smooth, -so if the homeomorphism is a homomorphism, then it must be a Lie group isomorphism. -Edit: I was puzzled by the fact that all known examples are given by pairs of locally isomorphic groups. Then I found the following papers: -Toda, H., A note on compact semi-simple Lie groups, Japan J. Math. 2 -(1976), 355-358. -in which he proves that "two simply connected, compact (and hence semi-simple) Lie -groups are isomorphic to each other if and only if they have isomorphic homotopy -groups for each dimension". Later this was generalized by S. Boekholt -(Journal of Lie Theory Volume 8 (1998) 183-185) as follows: -"Let $G_1$ and $G_2$ be two compact, connected Lie groups with -isomorphic homotopy groups in each dimension. Then $G_1$ and $G_2$ are locally -isomorphic." -So, going back to the question, we cannot avoid local isomorphism. - -REPLY [16 votes]: No. -For instance, topologically $U(2) = SU(2) \times U(1)$, since both are homeomorphic to $S^3 \times S^1$, but the group structures are different. Another example is given by $SO(3) \times SU(2)$ which is diffeomorphic to $SO(4)$. -On the other hand, any commutative connected compact real Lie group of dimension $n$ is isomorphic (as a real Lie group) to the real torus $\mathbb{T}^n:=(\mathbb{S}^1)^n$. -Analogously, any connected compact complex Lie group of dimension $n$ is isomorphic (as a complex Lie group) to a complex torus, i.e. a quotient of the form $\mathbb{C}^n/\Gamma$, where $\Gamma \subset \mathbb{C}^n$ is a lattice. Notice that in the compact complex case commutativity comes for free. Two complex tori $\mathbb{C}^n /\Gamma_1$ and $\mathbb{C}^n / \Gamma_2$ are isomorphic as complex Lie groups if and only if there exists $g \in \textrm{GL}_n (\mathbb{C})$ such that $\Gamma_2=g (\Gamma_1)$, but of course they are always both isomorphic to $\mathbb{T}^{2n}$ as real Lie groups.<|endoftext|> -TITLE: Lower bound on # spanning trees in a connected graph -QUESTION [10 upvotes]: Are there are any good lower bounds for the number of spanning trees for a connected graph $G$ in terms of (for example) number of edges $E$ or number of vertices $V$ ? -Are improved bounds available if one knows the number of bridges of the graph? (A bridge, aka a cut-edge, is an edge whose removal disconnects the graph) -EDIT: -Here is a simpler statement of the question. What is the minimum number of spanning trees possible for a graph with $m$ edges and $n$ vertices which is (1) 1-edge-connected or (2) 2-edge-connected? - -REPLY [3 votes]: I don't now if this is trivial or not but given a connected graph $G=(V,E)$, there are are at least $|E|-|V|+2$ spanning trees. -Take any spanning tree T. There are $|E|-|V|+1$ edges $e$ that are not in this spanning tree. For any such $e$, $T+e$ has exactly one cycle. There exists an edge $f \in T$ s.t. by removing $f$ the cycle is broken and the tree is still connected. Each $T+e-f$ spanning tree is distinct as they include a different edge $e$. -I also would like to add something implicit to David's answer (but apparently I do not have reputation for commenting). The graph $G$ constructed that way includes the $K_{n-k}$ clique as a subgraph. By Cayley's Theorem, that clique has ${(n-k)}^{n-k}$ spanning trees and therefore $G$ has at least ${(n-k)}^{n-k}$ spanning trees.<|endoftext|> -TITLE: Metrically singular Alexandrov space. -QUESTION [10 upvotes]: Perelman's stability theorem shows in particular that a finite dimensional compact Alexandrov space $(X,d)$ such that $X$ is not a topological manifold cannot be approximated in the Gromov-Hausdorf topology by Riemannian manifolds of the same dimension whose sectional curvature is bounded from below. -My questions are : -(1) Are there examples of compact Alexandrov spaces (say non-negatively curved) $(X,d)$ such that $X$ is a topological manifold but $(X,d)$ cannot be approximated by Riemannian manifolds of non-negative sectional curvature ? -(2) Does it change something if we allow the manifolds $(M_n,g_n)$ in the sequence to have a lower bound on the sectional curvature which is only going to zero as $n$ goes to infinity ? -Thanks. - -REPLY [9 votes]: Suspicious example: -Take a "funny" manifold with sectional curvature $\ge 1$ say $X$; -funny means Cayley flag or Aloff--Walach/Eschenburg/Bazaikin space, (not $S^n$ or $\mathbb{C}\mathrm{P}^n$ or $\mathbb{H}\mathrm{P}^n$). -Consider spherical suspension $\Sigma$ over $X$. -The space $\Sigma$ has curvature $\ge 1$. -It is expected that $\Sigma$ can not be approximated by any Riemannian manifolds with lower curvature bound, but no one can prove it. -The question has a lot in common with stabilized version of the Soul Theorem. -Partial results: - -Petersen--Wilhelm--Zhu: $\Sigma$ can not be approximated by Riemannian manifolds with curvature $\ge 1/4+\varepsilon$ -Kapovitch: for Cayley flag, if $\Sigma$ can be approximated then it is collapsing and the dimension drops down by 15 at least. - -P.S. -It seems you forget to say "same dimension" in your questions. -In this case, in addition to Perelman's result you have a result of V. Kapovitch, -which says that space of directions, as well as iterated spaces of directions have to be homeomorphic to a sphere. This was conjectured by P. Petersen. -(There are examples of Alexandrov spaces which homeomorphic to $\mathbb R^n$, -but space of directions at some point are not homeomorphic to spheres.)<|endoftext|> -TITLE: roots of polynomials outside the unit disc -QUESTION [10 upvotes]: Is there any "sufficient and necessary" condition for a comlpex polynomial P(z)=a_0+...a_n z^n, in terms of its coefficinets, to have all its roots outside the unit disc(or equivalently inside the unit disc)? -If not, is there any such conditions where the coefficinets are real or rational or integers? - -REPLY [2 votes]: There is a well-known Routh-Hurwitz stability condition (condition in terms of coefficients for the roots of a polynomial -to lie in the left half-plane). -By composing with a fractional linear transformation which maps the left half-plane onto -the exterior of the unit circle, you obtain a condition for the roots to be in the exterior. -(Of course, composition is a rational function, but just take its numerator). -Similarly, you can obtain necessary and sufficient conditions for the roots to be in the unit disc, -or on the unit circle, or in any disc or a circle. -The stability condition is written in terms of certain determinants made of coefficients; -it can be found in books and papers on polynomials and on control theory.<|endoftext|> -TITLE: Is the category of affine fppf groups closed under normal quotients? -QUESTION [9 upvotes]: Let $S$ be a scheme and let $N$, $G$ be affine flat group schemes of finite presentation over $S$. -If we assume that $N$ is a closed normal subgroup of $G$, we may form the fppf quotient sheaf $G/N$, which -is a sheaf of groups. By using descent and Artin's representability result for algebraic spaces, it follows that $G/N$ is an algebraic space which is -flat, separated and of finite presentation over $S$. By a classic result in the case where $S$ is the -spectrum of a field, it also follows formally that $G/N$ has affine fibres. My question is: -Is $G/N$ always affine over $S$? -Phrased in another way: Is the category of affine fppf group schemes over $S$ -a semi-abelian subcategory of the semi-abelian category of fppf group sheaves over $S$? -Another related question is if a flat, affine, finitely presented group scheme always may be embedded in -the automorphism group of a locally free coherent sheaf on the base. There is a remark about this in -SGAIII_1 EXPOSE VI_B 11.11.1, where it is stated without proof that this is supposed to be true for some -base schemes under certain regularity conditions. -EDIT: -A third related question is if someone knows of an example of a separated, flat, finitely presented group algebraic space with affine fibres which is not affine. - -REPLY [9 votes]: In general, the quotient $G/N$ is not representable. Lemma X.14 of Raynaud's book "Faisceaux amples sur les schémas en groupes et les espaces homogènes" gives a counter-example with $S=\mathbb{A}^2_k$ (which is regular 2-dimensional), $G=(\mathbb{G}_{a,S})^2$ and $N\subset G$ étale over $S$. -However if $S$ is locally noetherian of dimension at most 1, then $G/N$ is representable by a scheme : this is theorem 4.C of Anantharaman's thesis "Schémas en groupes, espaces homogènes et espaces algébriques sur une base de dimension 1". If you assume moreover that $S$ has finite normalization (e.g. $S$ excellent), then $G/N$ is affine if $G$ is. Indeed by Serre's cohomological criterion, affineness is not affected by nilpotents hence you may base-change by $S_{red}\to S$ and hence assume that $S$ is reduced (note that the formation of $G/N$ commutes with base change). Then let $S'\to S$ be the normalization, a finite morphism, and let $G',N'$ be the pullbacks to $S'$. The restriction of $(G/N)'=G'/N'$ to the generic points of $S'$ is affine (since we are then over a field) and because $S'$ is Dedekind, it follows that $(G/N)'$ is affine (Anantharaman prop. 2.3.1). Finally $(G/N)'\to G/N$ is finite surjective hence $G/N$ is affine by Chevalley's theorem.<|endoftext|> -TITLE: Kempf Vanishing theorem and Representation of Lie algebras. -QUESTION [7 upvotes]: Let $G$ be a reductive connected algebraic group and let $B$ a Borel subgroup. One of central themes of the representation theory of $G$ is the study of the induction functor $H^0$ from $B$ representations to $G$ representations. Many of the features of $H^0$ in the characteristic zero case also hold in the modular case. On the other hand the Borel-Weil-Bott theorem fails in general in the modular case and hence the simplicity of $H^0(λ)$ also breaks down in general. Still, we consider the $H^0(λ)$’s to be the fundamental objects of study, the reason being tha their characters, like in the characteristic zero case, are given by the Weyl character formula. This fact in turn relies on the Kempf Vanishing theorem, i.e. $$H^i(λ)=0 \text{ for } i>0 \text{ and } λ∈P^+.$$ -Beside this introduction, in the realm of representation theory of Lie algebras, I've heard few times that Kempf's Vanishing theorem is used to prove some universal properties of Weyl modules in the family of finite-dimensional highest-weight modules. -QUESTION: How is used this Kempf theorem to get this conclusions about universality? -Thanks, - -REPLY [12 votes]: As usual the history is a little complicated to track, but it should be understood first that the term Weyl module and notation $V(\lambda)$ were first used to describe the module obtained by a natural reduction modulo $p$ process from the usual finite dimensional simple module (of dominant highest weight $\lambda$). Here one uses integral bases derived by Steinberg from a Chevalley basis of the relevant complex Lie algebra. From the construction one knows that the formal character of $V(\lambda)$ is given by Weyl's formula. -On the other hand, there are intrinsic finite dimensional modules $H^i(G/B, \mathcal{L}(\lambda))$ for the algebraic group $G$ in characteristic $p$ given in terms of a line bundle for the weight $\lambda$ (or its negative depending on conventions about $B$ and positive roots). For dominant weights Kempf's theorem now implies that only $H^0$ survives; its Euler character is then the Weyl character for that weight, so the dimension of the module is the same as that of the corresponding Weyl module. -The key point to establish is that the Weyl module is the Serre dual of such a module of global sections, by naturally identifying it with the top cohomology relative to the weight "linked" to $\lambda$ by the longest element of the Weyl group. This follows by dimension comparison because there is a natural embedding of $V(\lambda)$ (or any other "highest weight module") into this highest cohomology. So the universal property of $V(\lambda)$ follows indirectly from Kempf's vanishing theorem. -This was first written down as Satz 1 in a 1977 paper by J.C. Jantzen in J. Reine angew. Math.. His book Representations of Algebraic Groups (AMS 2003 edition) has a more systematic treatment, with Kempf's theorem proved in Chapter II.4 and with "Weyl modules" redefined as the top cohomology groups. So the universal property is a bit hidden in the background. -One other comment on the question: the word "hence" is inappropriate, since the failure of Borel-Weil-Bott is not directly the reason for Weyl modules to be in most cases non-simple. Here you have to get into the linkage principle and other parts of the development, to see the analogy with (infinite dimensional) Verma modules in characteristic 0 and the Kazhdan-Lusztig conjecture, etc. -P.S. I didn't want to reproduce the short argument in Jantzen's paper because of the special notation needed, but the embedding idea (which goes back at least to Chevalley) involves realizing global sections of a line bundle inside the infinite dimensional algebra of regular functions on the algebraic group. Then the classical ideas about "representative functions" on a group come into play for finite dimensional representations.<|endoftext|> -TITLE: Reebless and taut foliations -QUESTION [18 upvotes]: Suppose we are given a closed oriented 3-manifold. -It is well known that taut foliations are Reebless, and if a Reebless foliation isn't taut then the leaves which don't admit a closed transversal are tori. Furthermore, it is straightforward that in a taut foliation all the closed leaves are homologically non trivial. -Conversely, is there any complete criterion to determine if a Reebless foliation is taut? - -REPLY [24 votes]: For simplicity assume the foliation is transversely oriented (otherwise you can think in terms of the transversely oriented double cover). Therefore each leaf has a canonical orientation. Then a foliation (Reebless or not) is taut if and only if there is no positive linear combination of the oriented torus leaves that is trivial in homology. This is equivalent to the -condition that every leaf admits a closed transversal. This basic theory was - essentially developed by Novikov in his famous article on the topology of foliations -(although Novikov did not develop all the properties of taut foliations).<|endoftext|> -TITLE: Extending state space to make a process Feller -QUESTION [10 upvotes]: Let $X$ be a locally compact Hausdorff space, and let $Y_t$ be a continuous Markov process on $X$ with transition function $P(t, x, \Gamma) := \mathbb{P}_x (Y_t \in \Gamma)$. Let $T_t$ be the corresponding transition semigroup, i.e. $T_t f (x) = \int_X f(y) P(t,x,dy)$. Let $C_0$ be the set of continuous functions on $X$ vanishing at infinity, and $C_b$ the bounded continuous functions on $X$. -If $T_t C_0 \subset C_0$, then the process $Y_t$ is Feller. (Since the process is continuous, it follows that $T_t$ is a strongly continuous semigroup on $C_0$.) - -Suppose we only know that $T_t C_0 \subset C_b$. (For instance, this happens if the process has a continuous transition density.) Can we extend the state space $X$ to a larger space $\tilde{X}$ on which a suitably extended version of the process $Y_t$ is Feller? - -The example I have in mind is something like Brownian motion on $X = \mathbb{R}^3 \backslash \{0\}$. If $f \in C_0(X)$, then $T_t f$ need not vanish near 0, so $T_t f \notin C_0$ and the process is not Feller. However, if we take $\tilde{X} = \mathbb{R}^3$, then Brownian motion on $\mathbb{R}^3$ is Feller. Moreover, $\tilde{X} \backslash X = \{0\}$ is an exceptional set, so the process started at a point of $X$ doesn't really see the point that we added. -To generalize this, one might consider the $C^*$-subalgebra of $C_b$ generated by $\{T_t C_0 : t \ge 0\}$ and take $\tilde{X}$ to be its spectrum. However, there are a lot of details that don't seem entirely clear. -This seems like it should be known, so a reference would be much appreciated. - -REPLY [10 votes]: Yes, it is possible to extend the state space with respect to which $Y$ is a Feller process. Then, $X$ will be a dense open subset of the extension $\hat X$. Furthermore, for any initial distribution of $Y_0\in\hat X$, then $Y$ will have a continuous modification which necessarily satisfies $Y_t\not\in\hat X\setminus X$ for all positive times (almost surely). In your example with $Y$ being a Brownian motion and $\hat X=\mathbb{R}^3\setminus\{0\}$ then this process corresponds to adding back the origin. -This is actually a special case of a more general method called Ray-Knight compactification, which applies to right-continuous Markov processes taking values in a Suslin space. Ray-Knight compactification does not always lead to processes which are Feller though, as they can have branch points in the extended state space. The type of processes obtained by this method are called Ray processes. See the 1975 paper by Getoor & Sharpe detailing Ray-Knight compactification, or, any reasonably comprehensive textbook on Markov processes should describe this method. In your question the conditions that $X$ is locally compact, $T_tC_0\subseteq C_b$ and $Y$ is continuous are enough to ensure that we obtain a Feller process so, in particular, there will be no branch points in the extended domain. One point before continuing; I'm assuming, as is standard in the definition of Feller processes, that the space $X$ has a countable base (locally compact, Hausdorff with a countable base, aka an lccb space). -Constructing the extension is a rather natural application of the Gelfand-Naimark theorem, although showing that the resulting process is Feller will still require some work. Let $\mathcal{A}$ be the smallest closed $C^*$-subalgebra of $C_b(X)$ containing $C_0(X)$ and closed under application of $T_t$. We can define $\hat X$ to be its spectrum (the set of nonzero $C^*$ homomorphisms $\mathcal{A}\to\mathbb{C}$), under the weak topology. Then, we identify $X$ as an (open, dense) subset of the (locally compact) space $\hat X$ in the same way as in the construction of the Stone-Chech compactification. By the Gelfand-Naimark theorem, every $f\in\mathcal{A}$ extends uniquely to an $\hat f\in C_0(\hat X)$, giving an isometry $\mathcal{A}\to C_0(\hat X)$, $f\mapsto\hat f$. Then, $T_t$ extends to a map $\hat T_t\colon C_0(\hat X)\to C_0(\hat X)$, $\hat T_t\hat f=\hat{T_tf}$. This is automatically a Markov transition function, but we can show that it is also Feller in your case. It is a bit of work, and I'll break it up into smaller statements. - -1) $T_tC_b(X)\subseteq C_b(X)$. - -Any nonnegative $f\in C_b(X)$ is a pointwise limit of an increasing sequence $f_n\in C_0(X)$ and, therefore $T_tf=\lim_{n\to\infty}T_tf_n$ is also a limit of an increasing sequence in $C_0(X)$, so is lower semicontinuous. Applying the same statement to $\Vert f\Vert - f$ implies that $f$ is also upper semi-continuous, so is continuous. - -2) If $f\in C_0(X)$ is nonnegative and $t_n\to0$ then $(T_{t_n}f-f)_+\to0$ in the compact-open topology. - -(Note: This doesn't use continuity of $Y$. And, in general, it is not necessary that $T_{t_n}f\to f$ in the compact open topology, even for $Y$ right-continuous.) -It is enough to consider $f$ with compact support and, by right-continuity of the process $Y$, we have $T_{t_n}f(x)\to f(x)$ as $n\to\infty$ for all $x\in X$. Then, setting $g_m=m\int_0^{1/m}T_sf\,ds$, it can be seen that $\Vert T_{t_n}g_m-g_m\Vert\le 2mt_n\Vert f\Vert\to0$ as $n\to\infty$. Now, letting $S_m$ be the space of convex combinations of $\{g_m,g_{m+1},\ldots\}$, $f$ is a pointwise limit of a sequence in $S_m$, so is in its closure under the compact-open topology (this is a consequence of the Hahn-Banach theorem and the Riesz representation theorem). Therefore, there exists $f_m\in S_m$ converging to $f$ in the compact-open topology. For any $\epsilon > 0$ and compact $K\subseteq X$, take $m$ large enough such that $\vert f-f_m\vert\le\epsilon$ on the union of $K$ and the support of $f$. So, -$$ -T_tf-f \le T_tf_m-f_m+T_t(f-f_m)_++(f_m-f)\le T_tf_m-f_m + 2\epsilon. -$$ -on $K$. Letting $t$ decrease to zero gives $\limsup_{t\to 0}(T_tf-f)\le2\epsilon$ uniformly on $K$. - -3) For any $f\in\mathcal{A}$ and $t_n\to0$, $T_{t_n}f\to f$ uniformly as $n\to\infty$. - -This is the point where continuity is required. It is not too hard to show that the space of $f\in C_b(X)$ satisfying the conclusion forms a $C^*$-algebra and is closed under $T_t$, so it is enough to prove it for nonnegative $f\in C_b(X)$ with compact support $K$. Also, the fact that $T_tC_0(X)\subseteq C_b(X)$ can be used to prove the strong-Markov property. If $Y_0\not\in K$ and, letting $\tau$ be the first time at which $Y$ hits $S$, continuity implies that $f(Y_{\tau})=0$ and $Y_\tau\in K$. So, for $x\not\in K$, -$$ -\begin{align} -T_tf(x)=\mathbb{E}_x[f(Y_t)]&=\mathbb{E}_x\left[T_{(t-\tau)_+}f(Y_{\tau\wedge t})\right]\\\\ -&\le\sup_{y\in K,s\le t}\left(T_sf(y)-f(y)\right)_+. -\end{align} -$$ -The previous statement says that the right-hand-side tends to zero as $t\to0$ and, as it is independent of $x$, $T_tf\to0$ uniformly on $X\setminus K$. -Now let $f_n$ be the sequence converging uniformly on compacts to $f$ constructed in the proof of 2. Then, we have $f_n\to f$ uniformly on $X\setminus K$ as $n\to\infty$ so, in fact, $f_n$ tends uniformly to $f$. Taking the limits $t\to0$ and $n\to\infty$ in -$$ -\Vert T_tf-f\Vert\le\Vert T_tf_n-f_n\Vert+2\Vert f_n-f\Vert -$$ -shows that $\Vert T_tf-f\Vert\to0$. -This almost shows that $\hat T$ is Feller, just the following technical lemma is left. - -4) $\hat X$ is an lccb space. - -As $X$ is lccb, $C_0(X)$ has a countable dense subset $S$. The closure of $S$ under products, taking $\mathbb{Q}$-linear combinations and applying $T_t$ for $t\in\mathbb{Q}_+$ is a countable dense subset of $\mathcal{A}$ (use 3). So $C_0(\hat X)\cong\mathcal{A}$ is separable. This implies that $\hat X$ has a countable base. -So, $\hat T_t$ is a Feller transition function on $\hat X$. Therefore, any Markov process $\hat Y$ w.r.t. this transition function has a right-continuous modification. We can also show that, regardless of the initial distribution, $\hat Y$ will be continuous and $\hat Y_t\not\in\hat X\setminus X$ for all times $t > 0$ (almost-surely). For any time $s > 0$ at which $Y_s\in X$ then the statement of the question implies that $\hat Y$ has a continuous modification lying in $X$ at all times $t\ge s$ (which is unique). So, it is enough to show that $\mathbb{P}(\hat Y_s\in X)=1$ for each time $s > 0$. But, this is standard (see Prop. 4.6 (iii) from the paper of Getoor & Sharpe).<|endoftext|> -TITLE: Can one find the size of a Sylow normalizer from the character table? -QUESTION [8 upvotes]: Is the size of the normalizer of a Sylow p-subgroup determined by the ordinary character table of the group? - -And if so, how does one calculate it? -In a solvable group, apparently one can compute the prime divisors of the Sylow normalizers from the character table (Isaacs–Navarro, 2002), but I don't see any discussion of the entire order. I suppose it must be harder to compute the order, and I somewhat hope it is too hard, that is, the character table does not determine the Sylow normalizer's order. If it helps to prove you can find the order, then I am happy to assume one also knows the power maps (and so element orders). - -Isaacs, I. M.; Navarro, Gabriel. - "Character tables and Sylow normalizers." - Arch. Math. (Basel) 78 (2002), no. 6, 430–434. - MR1921731 - DOI:10.1007/s00013-002-8267-4 - -REPLY [3 votes]: I don't see a full answer to this at present, but here are some thoughts on the $p$-solvable case. -I think it is equivalent in that case to the question: given a Sylow $p$-subgroup $P$ of a finite -$p$-solvable group $G$, can we determine the order of $N = O_{p'}(C_{G}(P))$ from the character -table of $G$? If we can do always do this, then we can find $|N_G(P)|$ by an inductive -argument. It is well known that for such $G$, the subgroup $N$ is contained in $O_{p'}(G)=M$, say, -and, in fact, $N = M \cap N_{G}(P).$ Since the character table of $G$ contains that of $G/M$, -we can work by induction if we can determine $|N|$. However, on the negative side, while it -is possible to determine which are the (necessarily $p$-regular) conjugacy classes of $G$ -which meet $N$, it may not be so easy to determine $|N|$ from the character table of $G$. -But we can see the equivalence of the questions in this case, because if we can determine -$|N_G(P)|$ from the character table of $G$ and $|N_{G/M}(MP/M)|$ from the character table -of $G/M$, then we can determine $|N| = |M \cap N_G(P)|$ from the character table of $G$. -(Added later: I should have said that if $M = 1$ we can calculate $|N_G(P)|$ by working -with $G/O_p(G)$).<|endoftext|> -TITLE: to what extent does the category Cov(X) determine a topological space X? -QUESTION [8 upvotes]: Given a topological space $X$, form the category $\mathrm{Cov}(X)$ consisting of open subsets $U \subset X$ as objects and inclusions as morphisms. To what extend can we recover (properties of) $X$ from $\mathrm{Cov}(X)$? -Example: $\mathrm{Cov}(X)$ determines the Cech cohomology groups of $X$, and so for nice $X$ its homology groups, and thus for $\pi_1(X)$ abelian by the Hurewicz theorem the homotopy type of $X$. -Can we also read of $\pi_1(X)$ from $\mathrm{Cov}(X)$? This should be possible if there is a nice covering of $X$ by open contractible subsets with intersections connected (by the Seifert-van-Kampen theorem). -How is $B\mathrm{Cov}(X) = |N\mathrm{Cov}(X)|$ related to $X$? - -REPLY [11 votes]: In order to have enough freedom, I would prefer to consider a basis $\mathcal{U}$ of open subsets of $X$. In this case, considering $\mathcal{U}$ as a partially ordered set (for the inclusion), there is a canonical inclusion functor into topological spaces -$$\mathcal{U}\to \mathit{Top} \ , \ U\mapsto U$$ -whose colimit is precisely (and obviously) the space $X$. But, in fact, we have a better property. the natural map -$$hocolim_{U\in\mathcal{U}} \ U\to colim_{U\in\mathcal{U}} \ U=X$$ -is a weak homotopy equivalence (i.e. induces an isomorphism of homotopy groups); for a sketch of proof, see below. -Of course, we might consider the case where $\mathcal{U}$ consists of all the open subsets of $X$, but, as noticed in Todd's answer, we then get a partially ordered set with initial and terminal object, which is not very interesting, homotopy theoretically. However, if $X$ is locally contractible (e.g. if $X$ is a CW-complex), then it might be interesting to consider for $\mathcal{U}$ the contractible open subsets of $X$. In that case, as the maps $U\to pt$ are weak homotopy equivalences, then the natural map -$$hocolim_{U\in\mathcal{U}} \ U \to hocolim_{U\in\mathcal{U}} \ pt=: B\mathcal{U}$$ -is a weak homotopy equivalence as well. In other words, in this case, $X$ and $B\mathcal{U}$ are canonically isomorphic in $Ho(Top)$, which is a nice way of seeing that partially ordered sets are models for homotopy types. - -Edit: -Here is a sketch of proof of the fact that the map $hocolim_{U\in\mathcal{U}} \ U\to colim_{U\in\mathcal{U}} \ U$ is a weak homotopy equivalence for any $X$. The point is that we may consider the model category $P(X)$ obtained as the left Bousfield localization of the projective model category of simplicial presheaves on $X$ by the class of hypercovers; see -1 D. Dugger, S. Hollander and D. Isaksen, Hypercovers and simplicial presheaves, Math. Proc. Cambridge Philos. Soc. 136 (2004), no. 1, 9-51. -Then, there is an obvious left Quillen functor from $P(X)$ to the model category of topological spaces which sends a representable $U$ to the corresponding subspace of $X$: indeed, we have a left Quillen functor from the projective model structure on $P(X)$, and we conclude using Theorem 1.3 of -2 D. Dugger and D. Isaksen, Topological hypercovers and $\mathbf{A}^1$-realizations, Math. Z. 246 (2004), no. 4, 667-689. -To finish the proof, as left Quillen functors preserve homotopy colimits (up to weak equivalences), it is thus sufficient to prove that $hocolim_{U\in\mathcal{U}} \ U$ (seen as a simplicial preasheaf) is weakly equivalent to the terminal object in $P(X)$ (for the local model structure). This follows from Theorem 6.2 of 1 (which allows to replace $P(X)$ by the model category of simplicial presheaves on $\mathcal{U}$) and from from the fact that the homotopy colimit of all representable presheaves is always weakly equivalent to the terminal presheaf; see for instance Lemma 3.4.27 and Theorem 3.4.34 in my book Les préfaisceaux comme modèles des types d'homotopie, Astérisque 308, 2006; this can also be obtained easily from Proposition 2.9 in -3 D. Dugger, Universal homotopy theories, Adv. Math. 164 (2001), no. 1, 144-176.<|endoftext|> -TITLE: Sets of divergence of Fourier series -QUESTION [13 upvotes]: Carleson theorem (later extended by Hunt) states that given an $L^2$ function $f:{\mathbb R}/{\mathbb Z}\to{\mathbb C}$, the set of points $x$ where the Fourier series $$\lim_{n\to\infty}\sum_{k=-n}^n\hat f(k)e^{2\pi ik x}$$ does not converge to $f(x)$ has measure 0. -Kahane and Katznelson proved that given any measure zero set $E$ there is a continuous function $f:{\mathbb R}/{\mathbb Z}\to{\mathbb C}$ whose Fourier series diverges at all points of $E$. -These two results leave a little gap. -What is known about those sets $E$ for which there is an $L^2$ (or even continuous?) function $f$ whose Fourier series diverges at all points of $E$ and pointwise converges to $f$ at all points not in $E$? -There is some ambiguity with the question as currently stated, as it depends on the representative of the $L^2$-class of $f$ that one chooses. I would hope an answer would help clarify the effect of specific representatives. Let me point out that, once we pick representatives, not every measure zero set can be such an $E$. If $f$ is continuous, this is easy to see; in fact, $E$ must be Borel (of low complexity; and this of course seems related to this question). As pointed out below in a comment by Juris Steprans, just on cardinality grounds we know not every measure zero set can appear, even for $L^2$ functions. Hunt's extension of Carleson's result says that we may assume $f\in L^p$ for any $p\in(1,\infty)$; I do not even know whether the sets $E$ will vary with $p$. - -REPLY [5 votes]: I believe that the problem of characterizing the sets of divergence for classical Fourier series is more or less open for all interesting classes ($C$, $L^\infty$, $L^p$ with $p>1$). -The strongest result that I'm aware of is due to Buzdalin who showed that any null-set $E\in F_\sigma\cap G_\delta$ is a set of divergence for the Fourier series of some continuous complex-valued function ("Trigonometric Fourier series of continuous functions diverging on a given set", Math. USSR Sbornik, 24 (1974)). -The characterization problem is mostly solved however for several other orthogonal systems, including the Haar and Franklin systems. There is also a very recent paper by Karagulyan where it is proved, in particular, that - -A necessary and sufficient condition for a set $E \subset [0, 1]$ to be a set of divergence -for the sequence of $(C, \alpha)$-means ($\alpha>0$) of the Fourier series of some function $f \in L^\infty[0, 1]$ is that $E$ is a $G_{\delta\sigma}$-set of measure $0$. - -(See G.A. Karagulyan, "Characterization of the sets of divergence for sequences of operators with the localization property", Sbornik: Mathematics, 202 (2011), pp. 9–33.) - -To complicate things further, people tend to distinguish between the sets of divergence and unbounded divergence. A set $E \subset [0, 1]$ is said to be a set of divergence (resp. unbounded divergence) for a series of functions -$$\sum_{n=1}^{\infty}f_n(x),\qquad x\in[0,1],$$ -if the series diverges for $x ∈ E$ and converges for $x \in [0, 1] \backslash E$ (resp. diverges unboundedly for $x ∈ E$). -One may think of the two optimistic working conjectures. - - -Every $G_{\delta\sigma}$-set $E $ of measure $0$ is a set of divergence for the Fourier series of some function $f \in C[0, 1]$. - -Every $G_{\delta}$-set $E$ of measure $0$ is a set of unbounded divergence for the Fourier series of some function $f \in C[0, 1]$. - - - -Conjecture 2 was explicitly formulated by P.L. Ul'yanov in the late 1960s. Both conjectures seem to be open.<|endoftext|> -TITLE: Relations between sums of powers -QUESTION [23 upvotes]: This question is so naive that it could have been asked before on this site. If so, I'll delete it. -Among beautiful formula, I like a lot this one: -$$\left(\sum_{n=1}^Nn\right)^2=\sum_{n=1}^Nn^3.$$ -Is there any other algebraic relation between the polynomials $P_k$ defined by -$$P_k(N):=\sum_{n=1}^Nn^k \qquad?$$ -I suspect yes, because $1,P_0,P_1,\ldots$ is a basis of ${\mathbb Q}[X]$ (but not a basis of the $\mathbb Z$-module ${\mathbb Z}[X]$), and if one replace $P_{\ell m}$ by $P_\ell P_m$, we get another basis. But are there nice relations? - -REPLY [2 votes]: http://en.wikipedia.org/wiki/Faulhaber%27s_formula -QUOTE: -Faulhaber observed that if $p$ is odd, then -$$1^p + 2^p + 3^p + \cdots + n^p$$ -is a polynomial function of -$$a=1+2+3+\cdots+n= \frac{n(n+1)}{2}.$$ -END OF QUOTE -QUOTE: -Donald E. Knuth (1993). "Johann Faulhaber and sums of powers". Math. Comp. (American Mathematical Society) 61 (203): 277–294. arXiv:math.CA/9207222. doi:10.2307/2152953. JSTOR 2152953. The arxiv.org paper has a misprint in the formula for the sum of 11th powers, which was corrected in the printed version. -END OF QUOTE -CORRECT VERSION: -http://www-cs-faculty.stanford.edu/~knuth/papers/jfsp.tex.gz<|endoftext|> -TITLE: Sections of 2-vector bundles -QUESTION [9 upvotes]: If we work over a field $k$, and take the recursive definition of $n$-vector spaces (as, e.g. in Topological Quantum Field Theories from Compact Lie Groups, arXiv:0905.0731) then a $2$-vector space is a $k$-algebra $A$, to be thought as a placeholder for its categoty of modules; morphisms between $A$ and $B$ are $B$-$A$-bimodules, ad 2-morphisms are morphisms of bimodules. -Now consider a 2-vector bundle over some space $X$. How does one see that its global sections are a 2-vector space? -The answer somehow depends on the notion of section one adopt, but in any case the relation between the various definitions should be investigated. Basically there are two notions coming to my mind: -i) natural transformations from the trivial 2-bundle to the given bundle. This is a very neat object, but it is not clear (to me) that this is a 2-vector space: which is the underlying algebra? -ii) the limit in 2-Vect of the functor from the nerve of a good open cover of $X$ to 2Vect defining the 2-bundle. This is manifestly a 2-vector space, but it is not clear (to me) that this limit exists. -Clearly the dream statement here would be that i) has a natural structure of 2-vector space, and that this 2-vector space represents the limit ii), but I'm unable to prove this. -(or the dual version of the above, under suitably finiteness assumptions) - -REPLY [2 votes]: In the special case $k=\mathbb C$ (or $\mathbb R$), you can take "2-bundle" to mean -continuous field of C*-algebras (possibily with the extra condition that it be locally trivial, or locally trivial-up-to-Morita-equivalence). If you don't know what acontinuous field of C*-algebras, you can safely think of it as a bundle of C*-algebras. -The Čech data that you mention in your comment indeed provide examples of continuous fields of C*-algebras. So, in that sense, continuous fields of C*-algebras really behave like 2-categorical bundles (they also behave like 1-categorical bundles, but that's irrelevant). -Now, you can simply take the global sections to be the C*-algebra of global sections. -In other words, take the 1-categorical global sections. -Starting with the simplest kind of non-trivial Čech data: all the algebra bundles are trivial (with fiber $\mathbb C$); all the bimodules are trivial (with fiber $\mathbb C$); the coherences are given by a two-cocycle with values in $S^1$, then the corresponding algebra of global sections is a continuous trace C*-algebra: a non-unital C*-algebra that is locally Morita equivalent to an abelian C*-algebra.<|endoftext|> -TITLE: Uniform proof of dimension formula for minimal special nilpotent orbit? -QUESTION [13 upvotes]: Given a simple Lie algebra over an algebraically closed field of good characteristic such -as $\mathbb{C}$, its subvariety $\mathcal{N}$ of nilpotent elements has dimension $2N$ (where $N$ is the number of positive roots) and is the union of finitely many orbits under the adjoint group $G$. These orbits are partially ordered by $\mathcal{O'} \leq \mathcal{O}$ iff $\mathcal{O'} \subset \overline{\mathcal{O}}$. Some pictures are available here. -Using the Killing-Cartan classification of simple types $A$ - $G$, the orbits have been well studied. One general fact is that all orbits have even dimension, but for detailed results case-by-case work is usually essential. There is a unique dense regular orbit and a unique orbit just beneath it in the partial ordering: the subregular orbit, of dimension $2N-2$. Similarly, there is a unique minimal nonzero orbit (determined by any root vector for a long root), whose dimension was shown by Weiqiang Wang here to be $2 h^\vee -2$ with $h^\vee$ the dual Coxeter number of the root system: one greater than the sum of coefficients of the highest short coroot. Wang's proof uses standard facts about root systems and avoids the classification. -Meanwhile, Lusztig's refinement of Springer's work on Weyl group representations in the context of a -desingularization of $\mathcal{N}$ led to a notion of special nilpotent orbit, not readily characterized within the Lie algebra setting. (But it has shown up independently in other active areas of representation theory.) -The regular and zero orbits are always special; all Richardson orbits (including the subregular orbit) are special, but there are sometimes other special ones. There is always a unique minimal (nonzero) special orbit; it is determined by any root vector for a short root except in type $G_2$ where it is instead the subregular orbit. -The Lusztig-Spaltenstein duality map on the set of all orbits (generalizing the transpose map for partitions which parametrize orbits in type $A$) has as image the special ones, on which it induces a duality involution which interchanges the regular and zero orbits, etc. It turns out that only in types $A, D, E$ (where all roots have equal length) is the minimal nilpotent orbit special, thus in duality with the subregular orbit. Wang's result recently led me to observe case-by-case: - -The minimal special nilpotent orbit has dimension $2h-2$. - -Here $h$ is the Coxeter number (order of a Coxeter element in the Weyl group, or one greater than the sum of coefficients of the highest root). The formula seems not to be written down anywhere (?) - -Is there a uniform proof of this dimension formula within the Lie algebra framework, avoiding the classification and using as little information as possible from Springer theory or other areas of representation theory? - -UPDATE: After observing the dimension formula a week or so ago, I checked around with people close to nilpotent orbits but no one seemed to recognize this from the literature or have a classification-free approach to suggest. Lusztig himself replied briefly that he didn't know any uniform proof, but later thought about it more and pointed out a few hours ago the remarks (b) and (c) at the end of his short 1981 Advances in Mathematics paper "Green polynomials and singularities of unipotent classes" (which I wrote a review of at the time but haven't revisited). So it's clear that he had found a unique orbit in each case of dimension $2h-2$ and had begun to fit these into his work on special orbits and what he later called "special pieces" of the unipotent variety. He was thinking about finite fields of definition, where Kostant had told him a polynomial formula for the number of rational points in the minimal orbit for simply-laced types. Lusztig himself extends the combinatorics to other cases by using the entire special piece. So there are other interesting connections here than what I started with in the parallel study of $\mathcal{N}$. But still no uniform explanation. - -REPLY [4 votes]: This isn't an answer to your question, but is a remark which is a bit too long to be a comment. Brylinski and Kostant observed (Nilpotent orbits, normality and Hamiltonian group actions, JAMS 7, 1994) that the minimal special orbit closures in the non-simply-laced cases can be covered by minimal nilpotent orbit closures in simply-laced cases, as follows. -Start with a Lie algebra ${\mathfrak g}$ with simply-laced Dynkin diagram $A_{2n-1}$, $D_n$ or $E_6$, and a graph automorphism (an involution except perhaps in type $D_4$) of ${\mathfrak g}$ with fixed points ${\mathfrak g}_0$. Then there is a ${\mathfrak g}_0$-stable complement to ${\mathfrak g}_0$ in ${\mathfrak g}$, and the ${\mathfrak g}_0$-equivariant projection ${\mathfrak g}\rightarrow{\mathfrak g}_0$ restricts to a finite map $\overline{{\rm O}_{min}({\mathfrak g})}\rightarrow\overline{{\rm O}_{min-sp}({\mathfrak g}_0)}$, which is a quotient by ${\mathfrak S}_2$ or ${\mathfrak S}_3$ and is a universal cover over the minimal special orbit. Note that the Coxeter number of ${\mathfrak g}$ is the same as that of ${\mathfrak g}_0$ (since we exclude the cases $A_{2n}$), so to a certain extent this brings us back to the question you asked (though it certainly isn't the sort of answer you were looking for).<|endoftext|> -TITLE: Question concerning h-cobordisms -QUESTION [7 upvotes]: Suppose we have a cobordism $W$ of manifolds $M_0$ and $M_1$ and suppose the inclusion of $M_0$ into $W$ is a homotopy equivalence. Is the same true for the inclusion of $M_1$ (ie. is $W$ already an h-cobordism)? -Using Poincare Lefschetz duality one can show that this map induces isomorphisms on homology. -Hence it suffices to show that the inclusion $M_1\rightarrow W$ induces an isomorphism on $\pi_1$. - -REPLY [5 votes]: However, the answer is "yes" after stabilizing three times: The product $W \times J^3$ (where $J^3$ is a $3$-cube) is an $h$-cobordism from $M_0 \times J^3$ to the closure of the remaining part of the boundary. There are details in Remark 1.1.3 of my book project "Spaces of PL manifolds and categories of simple maps" with Jahren and Waldhausen.<|endoftext|> -TITLE: How is the differential in complex cobordism defined? -QUESTION [12 upvotes]: This is my first MO question...hopefully it's not a bad one... -Background: As a stable homotopy theorist, I like to think of complex cobordism $MU$ as a ring spectrum. If I needed to get my hands dirty I could look at the representing spaces or go through the Thom construction of $MU$. -I would like to give a talk aimed at a more general mathematical audience discussing how formal group laws and Quillen's Theorem get involved with the story of complex cobordism. To do this I'm going to want to introduce complex cobordism in the more classical way, e.g. following Poincare's original definition or the comment in Ravenel's Complex Cobordism and Stable Homotopy (page 10) that $MU_{*}(X)$ can be defined completely analogously to $H_{*}(X)$. So here's what the start of the talk would look like: -(1) Define $\Omega_n = \mathcal{M}^n/\sim$ where $\mathcal{M}^n$ is a particular collection of $n$-dimensional manifolds$^!$ and $\sim$ is the cobordism relation. -(2) Define $C_n(X) = \langle \sigma:M^n\rightarrow X \;|\; M^n\in \mathcal{M}^n \rangle /\sim$ where $\sim$ is the bordism relation. Then there must be some differential $C_n(X)\rightarrow C_{n-1}(X)$ which gives rise to a homology theory $MU_n(X)$ (or dually to $MU^n(X)$) -(3) Mention Thom's theorem that $\Omega_* \cong \pi_*(MU)$, and proceed from there. -My question is, how is that differential in (2) defined? Is it easy to show that $d^2 = 0$? More generally, do people think this is a reasonable way to introduce $MU$ to an audience containing no stable homotopy theorists? -$^!$: By manifold here I mean even dimensional real manifold with a map $J:TM\rightarrow TM$ such that $p\circ J \simeq p$ and $J^2 = -1$. So $J$ gives an $\mathbb{R}$-linear action of $i$ on $T_xM$ for all $x$. This seems to capture a larger class of manifolds than just complex manifolds, and I can't see any way to get a larger class than this. - -REPLY [18 votes]: For a general audience it is much better to treat $MO$ rather than $MU$, because the complex orientation creates many unpleasant subtleties. (See papers of Buchstaber and Ray for interesting examples where these subtleties make a concrete and computable difference.) Let us say that a geometric chain of dimension n in X is an equivalence class of pairs $(M,f)$, where M is a compact smooth manifold (possibly with boundary) and $f:M\to X$ is a continuous map. Here $(M,f)$ is equivalent to $(M',f')$ if there is a diffeomorphism $u:M\to M'$ with $f'u=f$. We write $GC_{\ast}(X)$ for the graded abelian monoid of geometric chains. This has a differential $\partial[M,f]=[\partial M,f|_{\partial M}]$, and the homology is $MO_{\ast}(X)$. (This needs a few remarks about homology of complexes of monoids, but there is nothing very subtle going on.) I have a general audience talk about $MO$ at http://neil-strickland.staff.shef.ac.uk/talks/durham.pdf<|endoftext|> -TITLE: Hilbert-Mumford criterion and closedness -QUESTION [8 upvotes]: A version of the Hilbert-Mumford criterion states the following: Let $G$ be a linearly reductive group and $V$ a representation of $G$ over a field $k$ (alg. closed, char. zero). Suppose that $y \in \overline{Gx} - Gx$. Then, there is a one-parameter subgroup $\lambda : k^\times \to G$ such that -$$ -\lim_{t\to 0} \lambda(t)x \in \overline{Gy}. -$$ -My question is: Is there an example where every one parameter subgroup misses the orbit of $y$? I.e., is there an example where, for every $\lambda: k^\times \to G$ -$$ -\lim_{t\to 0} \lambda(t)x \in \overline{Gy} \implies -\lim_{t\to 0} \lambda(t)x \in \overline{Gy}-Gy? -$$ -If $G$ is a torus the answer is "no". What if $V$ is replaced by a more general scheme $X$ that is not itself a representation? - -REPLY [9 votes]: I have a counterexample now, thanks to some notes of Zinovy Reichstein I found. I think the counterexample is paraphrased as follows: Let $V$ be an irreducible representation of $G$ and suppose $x$ does not have a highest weight vector $y$ in its orbit, but $y \in \overline{Gx}$. There will be no way to get to $y$ from $x$ using only semi-simple elements. -To be precise (and to take the example from Reichstein) let $G = {\rm SL}_2(\mathbb{C})$ and $V = {\rm{Sym}} ^n \mathbb{C}^2$, $n \geq 2$. If $a$ and $b$ are the standard basis vectors of $\mathbb{C}^2$ then $a^{n-1} b \in V$ has the highest weight vector $a^n$ in its orbit closure, but not its orbit. Now check that there is no one-parameter subgroup $\lambda(t)$ such that $\lambda(t) a^{n-1} b \in G a^n$.<|endoftext|> -TITLE: Lifting local compactness to a covering space -QUESTION [5 upvotes]: (I decided to repost this from MathSE, since the question seems to not be as easy as I had thought) -NB: In this question, local compactness is used in its weak form, i.e. in a locally compact space, every point has a compact neighbourhood. Also, T3 is the weaker condition of the pair T3/regular. -Let $p:\tilde{X}\to X$ be a covering map. Since $p$ is a local homeomorphism, $\tilde{X}$ and $X$ share the "standard" local properties: local (path) connectedness, T1, etc. -The black sheep of the family of local properties is local compactness. Of course, a suitable separation axiom (T2 for instance) makes it into a proper local property with suitable local bases and everything. Therefore, if $X$ is locally compact Hausdorff, so is $\tilde{X}$. Interestingly, if $X$ is locally compact and T3, then $\tilde{X}$ is again locally compact. -My question is, how can this fail. More precisely, given a locally compact space $X$, does there exist a non-locally compact cover $\tilde{X}$ of $X$? -I strongly believe this to be true, because the compact neighbourhoods in $X$ might be too big to be seen by the covering map, but I haven't been able to find an example. I've been trying to construct a cover of the one-point compactification of the rationals, this being the nastiest space with the required properties I could think of. I've come up with this: -pick a proper neighbourhood $U$ of $\infty$ in $\mathbb{Q}^+$. Then take as a cover the space $\mathbb{Q}^+\times\{1/n;n\in\mathbb{N}\}\cup U\times\{0\}$ with the obvious projection. I think I've managed to prove this is a covering map and it looks to be not locally compact at the 0-th level, but that could just be my intuition being dead wrong. -I'd be interested in any thoughts on this. Also, somewhat less importantly, does local compactness descend from the covering space? Under what conditions do $X$ and $\tilde{X}$ share local compactness (can we in some way force the covering map to be proper, open, etc.)? - -REPLY [5 votes]: A suitable counterexample can be constructed as follows. Let $\mathbb I=\{x\in\mathbb R:0 -TITLE: Recovering Hecke L-series from Artin L-functions -QUESTION [9 upvotes]: Let $K$ be a number field, $\chi : C_K \to \mathbb{C}^\ast$ a Hecke character (that is, a character of the idèle class group), and $L(\chi,s)$ the corresponding Hecke $L$-series. I wish to understand how one may construct a Galois extension $G = Gal(L/K)$ and a complex representation $\rho : G \to \mathbb{C}^\ast$ such that $\mathcal{L}(L/K,\rho,s) = L(\chi,s)$, where $\mathcal{L}$ here denotes the Artin L-function. -I know how this works when $\chi$ factors through a congruence subgroup mod $\mathfrak{m}$, or equivalently, if $\chi$ is a Dirichlet character mod $\mathfrak{m}$; namely, take $L = K^\mathfrak{m}$, the ray class field mod $\mathfrak{m}$, and use the Artin symbol to turn the given character into a Galois character. -But I am worried that trying to do the same thing for general $\chi$, replacing the Artin symbol with the map $\phi_K : C_K \to Gal(K^{ab}/K)$, will not work. Indeed, the Wikipedia article on 'Hecke Character' suggests that only the Dirichlet characters are accounted for by Class Field Theory (see the last paragraph in the section 'Definition using ideals'). This worries me. - -REPLY [8 votes]: Dear Barinder, -Re. your comment "there cannot be a common generalization of Artin and Hecke $L$-series", -to the contrary, there is such a common generalization, namely the $L$-series of a representation of the global Weil group. These will (conjecturally) have an analytic continuation and functional equation, and they include all Hecke $L$-series (Hecke characters, by which I mean idele class characters, are just one-dimensional reps. of the global Weil group), and all Artin $L$-series (which are reps. of the global Weil group which factor -through the map to $G_K$). -Regards, -Matthew<|endoftext|> -TITLE: exactly simulating a random walk from infinity -QUESTION [10 upvotes]: In diffusion-limited aggregation on the square lattice, one lets a particle do "random walk from infinity" until it hits the current aggregate, at which point the site occupied by the particle is added to the aggregate; then the particle is started from infinity again, and so on. -This begs the question: What does it mean to let a particle do random walk from infinity? Or, in a more practical vein, how does one simulate such a walk? -The answer to the first question is, Random walk from infinity is just the limit as $v$ goes to infinity of random walk started at $v$. More precisely, for each $v$, we can look at random walk that starts from $v$ and stops when it hits (0,0) $T$ time-steps later (where the hitting-time $T$ is random), and we can re-index it so that it starts at $v$ at time $-T$ and hits (0,0) at time 0. This gives us a probability measure on paths that end at (0,0). Now we take the limit as $v$ gets farther and farther from (0,0), and we do some work and show that the law of the random path approaches a limit, and that this limit doesn't depend on how $v$ goes off to infinity. Actually I'm bluffing; I don't know how to prove this. I assume it's in the literature; can anyone point me to a relevant book or article? -Now we come to the problem of simulation. What we want is a black box that prints out -the path in reverse, chugging away until we turn it off. Note that this is not the same as having a black box that, for any $n$, prints out the last $n$ steps of the path; for, after the box has printed out that path, if we decide we want to know "What happened before that?", we can't find out by increasing $n$ and consulting the box again, because then the black box will give us the last $n+1$ steps of a DIFFERENT random path. On the other hand, if we had a box that, given the last $n$ steps of the path, could sample from its (up to 4 different) continuations one step back into the past with the appropriate conditional probabilities, one could use this box iteratively to build the sort of black box I want. It might be very hard to compute these conditional probabilities in the case of the square grid, but I suppose that would be one way to do it. A more robust (and, to my way of thinking, prettier) solution would be something more combinatorial, using ideas like coupling and domination. - -REPLY [6 votes]: To answer your first question (regarding existence of the limit): I never remember references, but all you have to do here is show that for any two far enough starting points, there is a coupling of the RWs started at them, such that with high probability the two paths hit the aggregate at the same point (for example, because they start walking together before hitting the aggregate). In $\mathbb{Z}^2$ it's pretty straightforward to do: if you let one walker walk till it hits the aggregate, then with high probability its path will separate the aggregate from the starting point of the second walker. Then you let the second walker walk till it hits the first path and follow this path thereafter. -As for the second question: as I said, I never remember references, but I believe that you can find how to calculate the harmonic measure exactly using the 2d potential kernel in Spitzer's "Principles of Random Walks". -To elaborate on George's comment and your reply (again, I'm pretty sure all this appears in Spitzer): Let $A$ be a finite set of vertices and start a SRW at $X_0=x$ and for some $y\in A$ we look at $\mathbb{P}(X_{\tau_A}=y)$, where $\tau_A$ is the hitting time of $A$. Then this probability is equal to $\sum_w \mathbb{P}(w)$ where the sum is over all paths starting at $x$ and ending at $y$ and not going through $A$. Since the SRW on $\mathbb{Z}^2$ is reversible this is equal to the sum over paths strating at $y$ and ending at $x$ and not going through $A$. This is exactly the expected number of visits to $x$ for a SRW started at $y$ and killed upon returning to $A$. This is proportional to the probability of hitting $x$ before returning to $A$ (again, when starting at $y$). If we take $x$ to be far away we see that conditioning on hitting $x$ before returning to $A$ is the same (asymptotically) as conditioning on the walk not returning to $A$ for a long time. -More can be said about the distribution of the conditioned RW, but right now I have to go.<|endoftext|> -TITLE: Homeomorphism type of union of two balls intersecting in a ball. -QUESTION [14 upvotes]: Let $d$ be an integer. Let $A,B \subseteq \mathbb R^d$ be two sets homeomorphic to an open $d$-ball such that their intersection is again homeomorphic to an open $d$-ball. Does it follow that their union is homeomorphic to an open $d$-ball? - -Motivation/Background -The motivation comes from nerve type theorems. For instance it follows in the above-mentioned case that $A \cup B$ has to be nullhomotopic. -In more general setting, let $A_1, \dots, A_n \subseteq \mathbb R^d$ be a collection of sets homeomorphic to an open $d$-ball such that the intersection of every subcollection is either empty or homeomorphic to an open $d$-ball. Moreover, let us assume that it is determined which subcollections are supposed to have a nonempty intersection. This data can be "stored" as a simplicial $K$ with vertices $A_1, \dots, A_n$ and whose faces are exactly that subcollections which have a nonempty intersection. -By standard nerve theorems, $K$ has to be homotopy equivalent to $A_1 \cup \cdots \cup A_k$; however I wonder whether anyone is aware of any "homeomorphism-type" nerve theorem: - -Let $A_1, \dots, A_n \subseteq \mathbb R^d$ and $K$ be described as above. Does it follow that the homeomorphism type of $A_1 \cup \cdots \cup A_n$ is determined by $K$? Does it follow if $K$ is at most $d$-dimensional? - -Even if the answer to the question above is negative, an interesting specific case occurs when $A_1, \dots, A_n$ are assumed to be convex. (Note that the answer to first question is positive in this case, since $A \cup B$ is star-convex.) - -I came to these questions when I was considering a certain algorithmic result on the collections of sets as described above. With a coauthor, we finally circumvent these questions (it showed up to be more convenient). However, I would be still very curious about the answers. - -REPLY [12 votes]: No to the first question. You can make examples where the "fundamental group at infinity" of $A\cup B$ is nontrivial. -Start with a finite complex $X$ that has trivial homology but nontrivial fundamental group. Embed the suspension $\Sigma X$ in $S^d=\mathbb R^d\cup\infty$. The suspension is contractible and is the union of two cones, also contractible. Let $A$ and $B$ be the complements of the two cones. If $d$ is big enough then $A$, $B$, and $A\cap B$ (the complement of $\Sigma X$) can be shown to be diffeomorphic to $\mathbb R^d$ by the $h$-cobordism theorem. But $A\cup B$, the complement of $X$, is such that the complement of a compact set in it is never simply connected. -EDIT: -To be a bit more precise, the open set $A$ (or $B$ or $A\cap B$) will be simply connected if the codimension of its complement in $S^d$ is at least $3$. (This is clear at least if the complement is embedded nicely enough.) And $A$ will have trivial homology, by Alexander duality, so it will be contractible. Now, Siebenmann's thesis gives sufficient conditions on a noncompact smooth manifold (of not too small dimension) for there to be a compact manifold with boundary having that as its interior. If there are arbitrarily small simply connected neighborhoods of infinity then these conditions are satisfied. Using this you see that $A$ is the interior of a contractible manifold with simply connected boundary. The $h$-cobordism theorem implies that that compact thing is a closed disk, so that $A$ is an open disk. -Or you can take a slightly different approach and avoid Siebenmann's thesis. Embed that suspension of $X$ piecewise linearly, make a regular neighborhood that is the union of regular neighborhoods of the two cones intersecting in a regular neighborhood of $X$. Use the complements of these compact neighborhoods. -Or, better yet, let $Q$ be a compact codimension zero acyclic non simply connected manifold (with boundary) in the interior of $D^d$. The complement in $S^{d+1}=(\mathbb R^d\times \mathbb R)\cup\infty$ of $D^d\times [0,1]\cup Q\times [1,2]$ is an open ball, as is the complement of $Q\times [1,2]\cup D^d\times [2,3]$, as is the complement of their union. But the complement of the intersection is not.<|endoftext|> -TITLE: Subdivision of triangles into congruent triangles -QUESTION [12 upvotes]: Recently some old notes of mine have gotten me to thinking about the problem of subdividing a triangle into $N$ smaller triangles, all congruent to one another. A little thought shows the following are possible values of $N$: -$\bullet$ If $N$ is a perfect square then any triangle can be subdivided into $N$ such smaller triangles -$\bullet$ If $N$ is a sum of two squares, say $N = a^2+b^2$ then a right triangle with side lengths in the ratio $a:b:\sqrt{a^2+b^2}$ can be subdivided into $N$ smaller triangles -$\bullet$ If $N=3$ then a $30-60-90$ triangle admits a decomposition into 3 congruent triangles -$\bullet$ Furthermore, by iterating the subdivisions indicated above, one can also obtain such subdivisions for any $N$ of the form $3^k\cdot m^2$ - - -Question 1: Are the values of $N$ listed above the only ones possible if one requires the subtriangles be similar to the original triangle? -Question 2: In each of the examples above, the subtriangles formed are always similar to the original triangle. Does the answer to Question 1 change if one no longer requires that subtriangles be similar to the original triangle. (For example, if one does not require that the subtriangles are congruent to the original triangle, then an equilateral triangle may be subdivided into 3 congruent copies of a 30-30-120 triangle) - -REPLY [8 votes]: Regarding the first question, yes, those are the only options. Proof can be found in: -S. L. Snover, C. Waiveris, and J. K. Williams. Rep-tiling for triangles. Discrete -Math., 91(2):193–200, 1991.<|endoftext|> -TITLE: strong approximation and one-dimensional automorphic representations -QUESTION [7 upvotes]: Hi, -Let $D$ be a quaternion algebra over $\mathbf Q$ such that $D\otimes\mathbf R = M_2(\mathbf R)$. -Let $\pi = \pi_\infty \otimes_p \pi_p$ be an irreducible automorphic representation of $D^\times$. -Supposedly if $\pi_\infty$ is one-dimensional, then every $\pi_p$ is one-dimensional, and it should follow from strong approximation, but how to prove it? -Thanks - -REPLY [11 votes]: Let $G$ be the group of norm one elements in $D^{\times}$. An easy argument -shows that it suffices to prove the claim for $G$ in place of $D^{\times}$. -In other words, I will let $\pi$ be an automorphic rep. of $G$. -Now suppose that $\pi_{\infty}$ -is one-dimensional. This means that in fact $\pi_{\infty}$ is trivial (because $G(\mathbb Q_v) -= SL_2(\mathbb R)$ has no non-trivial characters). -Now we use the fact that the automorphic representation lives inside the space of automorphic forms -on $G(\mathbb Q)\backslash G(\mathbb A).$ Thus $\pi$ is a space -of functions $f(g)$ on $G(\mathbb A)$ such that $f(\gamma g u) = f(g)$ for any -$\gamma \in G(\mathbb Q)$ and $u \in U$, a compact open subgroup of $G(\mathbb A^{\infty})$ -(with $U$ depending on $f$). -Our assumption on $\pi$ shows that furthermore $g_{\infty}f = f$ for all $g_{\infty} \in G(\mathbb R)$. -Thus if $\gamma \in G(\mathbb Q)$, $g \in G(\mathbb A)$, -$g_{\infty} \in G(\mathbb R)$, and $u \in U$, then -$$f(\gamma g ug_{\infty}) = (g_{\infty} f)(\gamma g u) = f(\gamma g u) -= f(g).$$ -Now strong approximation says that the double coset space -$G(\mathbb Q)\backslash G(\mathbb A)/ U G(\mathbb R)$ is a point, -and combined with the above calculation, this shows that $f$ is constant, and -thus generates the trivial representation under the action -of $G(\mathbb A)$. -Since $f$ was an arbitrary element of $\pi,$ we see that $\pi$ is trivial, -i.e. that all $\pi_v$ are one-dimesional. - -Note that it is important here that $\pi$ was a subspace of automorphic forms, -and not just a subquotient. This is automatic when $D$ is a (non-trivial) -quaternion algebra, because then $G$ is anisotropic, hence all automorphic forms -are automatically $L^2$, and so the space of automorphic forms is semi-simple. -On the other hand, if we consider $GL_2$, then while the above proof goes -through for cuspidal representations (which are necessarily subreps., and not -just subquotients, of automorphic forms), one can find (non-cuspidal) automorphic reps. -of $GL_2$ which are trivial at $\infty$ but infinite-dimensional at the other places. -(Of course, these are then subquotients of automorphic forms which can't be split off -as subrepresentations.)<|endoftext|> -TITLE: Spaces that are both homotopically and cohomologically finite -QUESTION [18 upvotes]: Is it true that every connected space with -1) just finitely many nontrivial homotopy groups, all finite, -and -2) just finitely many nontrivial rational cohomology groups, all finite rank, -is weakly homotopy equivalent to a point? -In 1953 Serre proved that any noncontractible simply-connected finite CW-complex has infinitely many nontrivial homotopy groups. That kills off a lot of possible counterexamples. -In 1998, Carles Casacuberta wrote: - - -However, we do not know any example of a finite CW-complex with finitely many nonzero homotopy groups which is not a $K(G, 1)$, and the results of this paper suggest that it is unlikely that there exist any. - - -I'm interested in my question because the spaces it asks about are the connected spaces whose homotopy cardinality and Euler characteristic are both well-defined. These concepts are morally 'the same', but it seems the spaces on which they're both defined are in very short supply, unless we stretch the rules of the game and use tricks for calculating divergent alternating products or sums. -For some further discussion of these issues see the comments starting here: -http://golem.ph.utexas.edu/category/2011/05/mbius_inversion_for_categories.html#c038299 -and also these slides and references: -http://math.ucr.edu/home/baez/counting/ -Edit: Condition 1) was supposed to say our space is "cohomologically finite", while 2) was supposed to say it's "homotopically finite". It's been pointed out that condition 1) is too weak: spaces like $\mathbb{R}P^\infty = K(\mathbb{Z}/2,1)$ exploit this weakness and serve as easy counterexamples to my question. They are cohomologically infinite in some sense, but not in a way detected by rational cohomology. -So let me try again. I can think of two ways: -Fix #1: Is it true that every connected space with -1) just finitely many nontrivial homotopy groups, all finite, -and -2) just finitely many nontrivial integral cohomology groups, all finitely generated, -is weakly homotopy equivalent to a point? -Fix #2: Is it true that every connected finite CW complex with just finitely many nontrivial homotopy groups, all finite, is homotopy equivalent to a point? - -REPLY [7 votes]: Yes to Fix #2. -A space with finitely many homotopy groups, all finite, is in the Bousfield class generated by the Eilenberg-MacLane spaces for those groups in those dimensions. The Sullivan Conjecture---Miller's Theorem---implies that any such space has no nontrivial maps into any finite-dimensional space. Thus if the space is finite (-dimensional) its identity function is homotopic to the constant function. -There's a lot more to say about this point of view. -The fundamental relation is that the space of pointed maps $\mathrm{map}_*(X,Y)$ is weakly contractible. Thus $X$ can't 'see' $Y$ at all from a homotopy-theoretical perspective. -Sullivan Conjecture/Miller's theorem: If $G$ is a (locally) finite group, then $BG$ cannot see any finite-dimensional space. -Then we have -Zabrodsky Lemma: If $F \to E \to B$ is a fibration sequence (with $B$ path-connected) and if $F$ cannot see $Y$, then $B$ can see $Y$ if and only if $E$ can see $Y$. -Now a simple induction shows that for any $n\geq 1$, $K(G,n)$ cannot see any finite-dimensional space; and then that any space $X$ with finitely many nonzero homotopy groups, all finite, also cannot see finite-dimensional spaces.<|endoftext|> -TITLE: Adding large sets by countable conditions preserving the GCH -QUESTION [5 upvotes]: Let $\kappa$ be an inaccessible cardinal. Is there any forcing notion $P$ with the following properties: -1-$P$ preserves GCH and the strong inaccessibility of $\kappa$, -2-$P$ adds a subset of $\kappa$ of size $\kappa$, -3-$P$ is the $< \aleph_1-$support product of some forcing notions $P_{\alpha}, \alpha < \kappa.$ -4- The generic filter for $P$ can be reconstructed from the subset of $\kappa$ added in 2. - -REPLY [2 votes]: Here is my answer to the updated question: -If you allow the forcing $P_\alpha$ to be trivial, then the Cohen real forcing example still works: add a Cohen real, and then perform $\kappa$ many stages of trivial forcing, with countable support. Overall, this is the same as just adding a Cohen real, since the later stages don't do anything, and it will still have the properties you request as in the original question. -But if you insist that every $P_\alpha$ is nontrivial as a forcing notion, then the inaccessibility of $\kappa$ will necessarily be destroyed. To see this, suppose that each $P_\alpha$ has incompatible conditions. Pick a condition $p_\alpha$ in $P_\alpha$ such that there are other conditions in $P_\alpha$ incompatible with $p_\alpha$. Consider any block $B$ of $\omega_1$ many coordinates $\alpha$ at which such nontrivial forcing is to be done. Let $G$ be $V$-generic for the iteration. For each such $\alpha\in B$, let $s_B(\alpha)$ be $0$ or $1$ accordingly as to whether $p_\alpha\in G$ or not. Thus, for each block $B$, the binary pattern of $s_B$ determines a subset of $\omega_1$. Since your iteration uses only countable support, it is dense that the binary pattern of $s_B$ is different from that of $s_{D}$, if $B$ is disjoint from $D$, since we may introduce such a difference by moving beyond the support of any given condition. Since we may divide $\kappa$ into $\kappa$ many disjoint blocks $B$, it follows that in $V[G]$ we have $\kappa$ many subsets of $\omega_1$, and so $\kappa$ is no longer inaccessible, contrary to 1. - -Here is my answer to the original question: -Stefan's comment provides an easy example if you don't intend that the forcing should not add bounded sets to $\kappa$. Just add a Cohen real. This of course preserves GCH, has countable conditions, and adds a subset to $\kappa$ (since every subset of $\omega$ is already a subset of $\kappa$); if you throw in all infinite ordinals as well, then you've got a new subset of $\kappa$ of size $\kappa$. -But even if you don't want to add bounded sets to $\kappa$, there are some easy examples showing that the question is probably not quite what you want to ask. This is because the size of conditions is not a fundamental feature of a forcing notion. For any partial order $\mathbb{P}$, let's make an isomorphic copy $\mathbb{P}'$ by replacing each condition $p$ with its singleton $p'=\{p\}$. We order $p'\leq q'$ just in case $p\leq q$. So now we have a partial order all of whose conditions have size $1$, and hence are countable. So any forcing notion is isomorphic and hence forcing equivalent to a forcing notion with countable conditions. Thus, this property is probably not really what you want to ask about.<|endoftext|> -TITLE: Pi1-sentence independent of ZF, ZF+Con(ZF), ZF+Con(ZF)+Con(ZF+Con(ZF)), etc.? -QUESTION [20 upvotes]: Let -ZF1 = ZF, -ZFk+1 = ZF + the assumption that ZF1,...,ZFk are consistent, -ZFω = ZF + the assumption that ZFk is consistent for every positive integer k, -... and similarly define ZFα for every computable ordinal α. -Then a commenter on my blog asked a question that boils down to the following: can we give an example of a Π1-sentence (i.e., a universally-quantified sentence about integers) that's provably independent of ZFα for every computable ordinal α? (AC and CH don't count, since they're not Π1-sentences.) -An equivalent question is whether, for every positive integer k, there exists a computable ordinal α such that the value of BB(k) (the kth Busy Beaver number) is provable in ZFα. -I apologize if I'm overlooking something obvious. -Update: I'm grateful to François Dorais and the other answerers for pointing out the ambiguity in even defining ZFα, as well as the fact that this issue was investigated in Turing's thesis. Emil Jeřábek writes: "Basically, the executive summary is that once you manage to make the question sufficiently formal to make sense, then every true Π1 formula follows from some iterated consistency statement." -So, I now have a followup question: given a positive integer k, can we say something concrete about which iterated consistency statements suffice to prove the halting or non-halting of every k-state Turing machine? (For example, would it suffice to use ZFα for some encoding of α, where α is the largest computable ordinal that can be defined using a k-state Turing machine?) - -REPLY [25 votes]: In 1939, Alan Turing investigated such questions [Systems of logic based on ordinals, Proc. London Math. Soc. 45, 161-228]. It turns out that one runs into problems rather quickly due to the fact that the $(\omega+1)$-th such theory is not completely well-defined. Indeed, there are many ordinal notations for $\omega+1$ and these can be used to code a lot of information. -Turing's Completeness Theorem. If $\phi$ is a true $\Pi_1$ sentence in the language of arithmetic, then there is an ordinal notation $a$ such that $|a| = \omega+1$ and $T_a$ proves $\phi$. -This result applies to any sound recursively axiomatized extension $T$ of $PA$. In particular, this applies to (the arithmetical part of) $ZF$. -To avoid this, one might carefully choose a path through the ordinal notations, but this leads to a variety of other problems [S. Feferman and C. Spector, Incompleteness along paths in progressions of theories, J. Symbolic Logic 27 (1962), 383–390]. - -REPLY [15 votes]: I think there is some vagueness inherent in the "similarly define ...". How is one to assign the consistency statement $Con(ZF_\lambda)$ for computable $\lambda$? This looks trivial but it is not. -I think also ZF is something of a red herring here. The question arises in PA (since we are looking at $\Pi_1$ sentences quantifying over natural numbers.) -Feferman has shown ("Transfinite Recursive Progression of Theories" JSL 1962) that it is possible to assign for every n in an effective manner a $\Sigma_1$-formula $\varphi_n(v_0)$ where each of the latter is to be thought of as enumerating (integer codes of) axiom sets (which I'll call "theories."). This is done in such a fashion so that if $a,b$ are integers with $b = 2^a$ that $T_b$ is $T_a$ together with the statement -$$\forall \psi \in \Sigma_1\forall x [ Prov_{T_a}\psi(x) \longrightarrow \psi(x)]$$ -(This is thus a "1-Reflection Principle" - for $\psi\in\Sigma_1$ here). He does this -with a view to considering those integers $a$ that are notations for recursive ordinals (in the sense of the notation system devised by Kleene - "Kleene's $O$".) - (There are clauses for $a$ representing a notation for a limit ordinal, when $a = 3^e$). -He proves that there are linear paths through the system of notations of computable ordinals, going through all recursive ordinals $\alpha$,so that -Every true $\Pi_2$ -sentence in arithmetic is proven by one of the theories along the path. -The starting -theory $T_0$ here can be PA (or ZFC if you want). Such a path gives -a definite meaning to $ZF_0, \ldots, ZF_\alpha, \ldots$ etc. for recursive $\alpha$. -Moreover for such a particular progression of theories one would would construe the answer to the question to be "No". -Feferman's starting point was the 1939 paper of Turing ("On Systems of Logic Based on Ordinal"). Turing also considered such paths through Kleene's $O$, but could just prove a theorem for $\Pi_1$ sentences, (using simpler "Consistency" statements"). Feferman shows that if one takes "$n$-Reflection" statements -for every $n$ each time one extends the theory then there are paths along which every true statement of -arithmetic is proven. -The moral of the story is that there are very complex ways of simply defining sequences -of theories, (because there are infinitely many ways, or Turing programs, of representing a recursive ordinal) which can hide/disguise all sorts of information. -A very readable survey is Franzen: "On Transfinite Progressions" BSL 2004. -Update (This is an answer to Scott Aaronson's Update.) -He asks: -given a positive integer k, can we say something concrete about which iterated consistency statements suffice to prove the halting or non-halting of every k-state Turing machine? -Let $M_0, \ldots ,M_{n-1}$ enumerate the $k$-state TM's. Let $P$ be the subset of $n$ -of those indices of TM's in the list that halt. -The statement -$\forall i (i \in P \rightarrow M_i$ halts $ \wedge \, - i \notin P \rightarrow M_i $ does not halt $)$ -is a $\Pi_2$ statement. In Feferman's paper (op.cit.) he shows that every true -$\Pi_2$ statement is proven by a theory $T_a$ in a 1-Reflection sequence, where $a$ is a notation for an ordinal of rank equal to $\omega^2 + \omega + 1 $. -So in terms of the question we do not need to vary the $\alpha$ depending on what -ordinals a $k$-state machine can produce. (Just fix $\alpha$ as given -above.) Of course it gives us zero practical information: there are infinitely many such notations of that rank, and we may not know which one to look at.<|endoftext|> -TITLE: Riemannian manifolds that are scalar flat but not Ricci flat -QUESTION [7 upvotes]: What are the examples of Riemannian manifolds that have zero scalar curvature but non-zero Ricci curvature? Is there any sort of classification of such manifolds? - -REPLY [2 votes]: On a compact connected manifold that does not carry a metric of positive scalar curvature, every scalar flat metric is Ricci-flat. Thus on such manifolds there are -no such metrics. -If a manifold carries a metric of positive scalar curvature then it also carries a metric of zero scalar curvature. I assume that for dimension at least 3 one could extend this statement to saying that it even admits a scalar flat metric with non-zero Ricci curvature, but I have no proof at hand currently. What I know for sure is: there are many manifolds having an obstruction against Ricci-flat metrics and admitting a metric of positive scalar curvature. On Ricci-flat manifolds, the first Betti number is at most the dimension, and if it is the dimension then the manifold is flat. This yields manifolds without Ricci-flat metrics and many of them carry a metric of positive scalar curvature.<|endoftext|> -TITLE: Is Margolis's axiomatisation conjecture still alive? -QUESTION [8 upvotes]: The construction of the category of finite spectra is easy, but there are different constructions of the whole homotopy category of spectra, all of which leading to the same result up to an equivalence. In his book, Spectra and the Steenrod Algebra, North-Holland, Amsterdam, N.Y., 1983, H.R. Margolis conjectured that there is a system of axioms, which characterizes the homotopy category in therm of finite spectra. If I do understand correctly, the results of M. Porta arXiv:0706.4458v2 and A. Heider arXiv:0707.0707v1 solve this problem for algebraic, respectively topological triangulated categories. I'm most familiar with algebraic triangulated categories, -so it is very possible to occur misunderstanding in the topological case. But for the algebraic case it seems to me that the problem is settled: If $\mathcal T$ is an algebraic $\alpha$-well generated triangulated category, then it is equivalent to a fixed quotient of the derived category of the dg-algebra with several objects (these objects are the $\alpha$-compact objects of $\mathcal T$). Or, am I wrong? - -REPLY [16 votes]: Margolis' conjecture is as follows: Let $\mathcal{S}$ be the stable homotopy category and $\mathcal{T}$ any compactly generated triangulated category. If the subcategories of compact objects are equivalent as triangulated categories $\mathcal{S}^c\simeq\mathcal{T}^c$ then $\mathcal{S}\simeq\mathcal{T}$. This conjecture is open. If $\mathcal{T}$ is known to be topological then it is a theorem in Schwede's Annals paper. The category $\mathcal{T}$ cannot be algebraic because $\mathcal{S}$ is not.<|endoftext|> -TITLE: Where do all these projection formulas come from? -QUESTION [62 upvotes]: I have been intrigued for a long time by the formal similarity of results from different areas of mathematics. Here are some examples. -Set theory Given a map $f:X\to Y$ and subsets $X' \subset X, Y'\subset Y$, we have $$f(f^{-1}(Y')\cap X')=Y'\cap f(X')$$ -Ringed spaces Given a morphism of ringed spaces $f:X\to Y$, an $\mathcal O_X$-module $\mathcal F$ and a locally free module of finite type $\mathcal L$, we have -$$f_\ast(f^\ast{\mathcal L}\otimes_{\mathcal O_X} \mathcal F)=\mathcal L \otimes_{\mathcal O_Y} f_{\ast}\mathcal F$$ -Topology Consider a proper continuous map of connected oriented manifolds $f:X\to Y$, then for $x\in H^\ast _c(X,\mathbb Z)$ and $y\in H^\ast _c(Y ,\mathbb Z)$ we have (Dold, p.314) -$$ f_!(f^\ast y . x)=y. f_!(x)$$ -Chow rings Given a proper map $f:X\to Y$ between nonsingular algebraic varieties and cycle classes $a\in CH^\ast(X), \beta \in CH^\ast(Y)$ we have -$$ f_\ast(f^\ast \beta . \alpha)=\beta. f_\ast(\alpha) $$ -K-theory Given a proper morphism of finite Tor dimension $f:X\to Y$ between schemes (and assuming $X$ and $Y$ have suitable ample line bundles), Quillen proved in his fundamental article on higher K-theory (Springer LNM 341, page 126) that for $x\in K_0(X)$ and $y\in K'_0(Y)$ -$$f_\ast(f^\ast y . x)=y. f_\ast (x) $$ -Derived categories Given a ring morphism $f:R\to S$, a bounded above complex $A$ of $R$-modules and a complex $B$ of $S$-modules we obtain in $\mathbb D(R)$ (Weibel, p.404) -$$ f_\ast(\mathbb L f^\ast( A) \otimes_S^{\mathbb L} B)=A \otimes_R^{\mathbb L} (f_\ast B)$$ -The question Of course I'm well aware that there are strong links between say K-theory and Chow rings and that the examples of projection formulas are not independent. What I would like to know is whether there is some general context of which these examples could be said to be illustrations, even if not particular cases in the strict sense. An analogy would be that Grothendieck's Galois theory explains the similarity between the traditional Galois theory of fields and the theory of covering spaces although it is not true that the general theory of topologiclal coverings is a special case of Grothendieck's results. -Edit After seeing several comments and an answer, I'd like to clarify my question. It is not principally to find a general formulation of which all those results would be a special case (although that certainly would be nice). But rather to know if there is a powerful, presumably tough, result or theory which would imply a good deal of the examples mentioned above. Perhaps a bit like K-theory used for Riemann-Roch , Bott periodicity, classification of vector bundles... - -REPLY [5 votes]: Perhaps some of these formulae have allegory theory (or a generalization thereof) lurking in the background? In particular, recall Freyd's modularity law: -$$R(R^\dagger A \cap B) \supseteq A \cap RB$$ -Now suppose that $f$ is a partial map, meaning that $ff^\dagger \subseteq \mathrm{id}_Y.$ Then -$$f(f^\dagger A \cap B) \subseteq ff^\dagger A \cap fB \subseteq A \cap fB$$ -So, for $f$ a partial map between sets, we have: -$$f(f^\dagger A \cap B) = A \cap fB$$ -for any relations $A$ and $B$ with appropriate domains and codomains. Taking the domains of these relations to be the set $1$ yields the desired result.<|endoftext|> -TITLE: Is 2-sylow subgroup of a rational group also a rational group? -QUESTION [10 upvotes]: As we know, a finite group $G$ is a rational group if $\chi (g)\in\mathbb{Q}$, where $\chi$ is every irreducible charahter and $g\in G$. I have an interesting question that is "Is 2-Sylow subgroup of a rational group also a rational group?" -Any hints will be appreciated :) - -REPLY [6 votes]: This had been a long standing conjecture, but it has now been answered negatively. Isaacs and Navarro have found a counterexample.<|endoftext|> -TITLE: Is the Galois x Hecke action on cohomology of Shimura varieties semi-simple? -QUESTION [8 upvotes]: Given a reductive group $G/\mathbf Q$ (+ additional data), and a compact open subgroup $K\subset G(\mathbf A^\infty)$, there is a standard construction that produces a Shimura variety $S$ and if we choose a rational representation $\xi : G\to Aut(V)$, we obtain a certain $l$-adic local system $\mathcal L$ on $S$. -Moreover, all of this can be done over some fixed number field $E$. -Question 1: If $S$ is proper, is it true that the $Gal(\overline E/E)\times H(G,K)$ action on -$H^*(S\otimes\overline E, L)$ is semisimple? (where $H(G,K)$ is the corresponding Hecke algebra). -Question 2: What happens if $S$ is not proper? Is the action semi-simple at least if we restrict to parabolic cohomology (the image of compactly supported cohomology inside usual cohomology)? -References for places where this is discussed would also be great. -Thanks - -REPLY [8 votes]: To get a Shimura variety, the reductive group $G$ should satisfy some axioms. In fact, -you should begin not just with $G$, but with a Shimura datum for $G$. -Leaving that aside, the Hecke action will be semi-simple (if we omit Hecke operators at -primes dividing the level); more generally, one could take the limit over all levels, -and then get a $G(\mathbb A^{\infty})$-action, which will be semi-simple (e.g. by -comparison with automorphic forms; and here I am supposing $S$ compact for the moment). -Each $G(\mathbb A^{\infty})$-rep. will appear with some multiplicity, and this multiplicity -space carries the $Gal(\overline{E}/E)$-action. If this action is irreducible, which is the -case e.g. for modular curves (admittedly non-compact, but ignore that for the moment!) or -Shimura curves, then it is certainly semi-simple. But in more general situations it -need not be irreducible, and then its semi-simplicity is a certain case of the Tate conjecture, and I'm pretty sure that nothing will be known about it. -E.g. already in the case of Hilbert modular varieties (or their compact cousins, coming -from quat. algebras, if you prefer), in some situations one expects a reducible Galois rep'n, and these are not known to be semi-simple, I believe. -The case of open varieties will be similar, in that not much will be known there either, -unless it is forced by irreducibility.<|endoftext|> -TITLE: PSD matrix with non-negative entries -QUESTION [6 upvotes]: We have convex sets $C_1=Conv(yy^{T}|y^{T}y=a,y\in R^{M})$ and $C_2=Conv(yy^{T}|y^{T}y=a,y\in R_{\geq 0}^{M})$. Clearly $C_2\subset C_1$. Does there exist a PSD matrix $A$ having $tr(A)=a,A(i,j)\geq 0$ $\forall i,j$ and $A\in C_1\setminus C_2$? - -REPLY [5 votes]: Maxfield and Minc (1962) in their paper entitled On the matrix equation $X'X=A$, quote an example due to Hall (1958), A survey of combinatorial analysis, which shows that for $M\ge 5$ we can find counterexamples of the desired kind. Here is their example: - -The matrix $$A = \begin{bmatrix} -1 & 0 & 0 &1/2 & 1/2\\\\ -0 & 1 & 3/4 & 0 & 1/2\\\ -0 & 3/4 & 1 & 1/2 & 0\\\\ -1/2 & 0 & 1/2 & 1 & 0\\\\ -1/2 & 1/2 & 0 & 0 & 1 -\end{bmatrix}$$ - is positive semidefinite, yet there is no matrix $X$ with nonnegative entries such that $X^TX=A$. - -The eigenvalues of the above matrix are approximately (2.12,1.42,1.25,.20,0), where the $0$ is exact as this matrix has rank-4.<|endoftext|> -TITLE: Norm of tridiagonal operator -QUESTION [7 upvotes]: Recently, I needed to estimate the operator norm of the tridiagonal operator, but I am sure answers much more refined than my simple observations must be known. -Let $T$ be the linear operator that maps a square matrix to its tridiagonal part. Thus, the action of $T$ on a matrix $X$ can be defined by the Hadamard product $M \circ X$, where $m_{ij}=1$ if $|i-j| \le 1$, and $m_{ij}=0$ otherwise. - -What is the operator 2-norm of $T$? (or what is a good approximation thereof?) - -The observation $\|M \circ X\| \le \|M\| \|X\|$ shows that $\|T\| \le 3$. A more refined estimate follows from Theorem 5.5.3 of Horn and Johnson's Topics in Matrix Analysis, which says that $\|M\circ X\| \le r_1(M)c_1(X)$, where $r_1$ is maximum row-length (Euclidean norm) and $c_1$ is max column length. This result then implies that $\|T\| \le \sqrt{3}$. -I am sure that significantly more refined estimates of $\|T\|$ are available, and will be thankful if you can provide me a reference, or maybe a short proof itself. - -REPLY [9 votes]: A good answer is given by R. Bhatia: Pinching, trimming, truncating, and averaging of matrices. Amer. Math. Monthly 107 (2000), no. 7, 602–608. -If you consider the operator $T_r$ that retains the diagonals defined by $|i-j|\le r$ (yours is $T_1$), its norm is accurately bounded by -$$L_r=\frac{1}{2\pi}\int_{-\pi}^\pi |D_r(\theta)|d\theta,$$ -where $D_r$ is the Dirichlet kernel.<|endoftext|> -TITLE: Countable Dense Sub-Groups of the Reals... -QUESTION [5 upvotes]: Can countable dense additive subgroups of the reals be well-ordered up to isomorphism by inclusion? -If so, is $\mathbb{Q}$ the smallest (up to isomorphism) countable dense subgroup of the reals, and what is the second smallest (up to isomorphism)? - -REPLY [5 votes]: This should be a comment to Joel's answer, but it got too long. Joel has exhibited a good part, but not quite all, of the classification of non-trivial subgroups of the rationals (also known as the classification of rank-one, torsion-free, abelian groups), a classification which, if I remember correctly, goes back to Reinhold Baer in the 1930's. For the record, here's the classification. Let $s$ be a function from a subset $D$ of the set of primes into the non-negative integers. Associated to $s$ is the group $G_s$ of those rational numbers expressible as $a/b$ with integers $a$ and $b$ such that, for each prime $p\in D$, the power of $p$ in the prime decomposition of $b$ is at most $p^{s(p)}$. (Primes not in $D$ can occur arbitrarily often in denominators.) Then the non-zero subgroups of $\mathbb{Q}$ that contain the integers are exactly these $G_s$'s. (Note that, up to isomorphism, containing the integers is unimportant, as it can always be achieved by rescaling.) Two of them, say $G_s$ and $G_t$, are isomorphic iff $s$ and $t$ have the same domain and agree at all but finitely many points in that domain. All these groups are dense in the reals, except for those isomorphic to $\mathbb Z$, i.e., those of the form $G_s$ where the domain of $s$ consists of all the primes and $s(p)=0$ for all but finitely many $p$. -Note that all of this concerns only groups of rank 1, i.e., those in which every two elements are linearly dependent over the rationals. For groups of higher but still finite rank, things get more complicated --- in a precise sense: If I remember correctly, the complexity of the isomorphism problem for torsion-free abelian groups of rank $n$ is known to be strictly increasing (in the sense of Borel reducibility) as $n$ increases.<|endoftext|> -TITLE: When is a general projection of $d^2$ points in $\mathbb{P}^3$ a complete intersection? -QUESTION [9 upvotes]: It is well known that $4$ general points in $\mathbb{P}^2$ are complete intersection of two conics. On the other hand, if $d \geq 3$, $d^2$ general points are not a complete intersection of two curves of degree $d$. More precisely, if $d =3$ there is only one cubic passing through $9$ general points, whereas if $d \geq 4$ there is no curve of degree $d$ passing through $d^2$ general points. -While investigating some questions about factoriality of singular hypersurfaces of $\mathbb{P}^n$, I ran across the following problem, which seems quite natural to state. - -Let $d \geq 3$ be a positive integer and let $Q \subset \mathbb{P}^3$ be a subset made of $d^2$ distinct points, with the following property: for a general projection $\pi \colon \mathbb{P}^3 \to \mathbb{P}^2$, the subset $\pi(Q) \subset \mathbb{P}^2$ is the complete intersection of two plane curves of degree $d$. -Is it true that $Q$ itself is contained in a plane (and is the complete intersection of two curves of degree $d$)? -If not, what is a counterexample? - -Any answer or reference to the existing literature will be appreciated. Thank you. -EDIT. Dimitri's answer below provides a counterexample given by $d^2$ points on a quadric surface. Are there other configurations of points with the same property? It is possible to classify them up to projective transformations (at least for small values of $d$)? - -REPLY [14 votes]: Counterexample. Consider a $Q$ quadric in $\mathbb CP^3$, let $L_1...,L_n$, $M_1,...,M_n$ be lines on $Q$ so that $L_i\cap L_j=\emptyset$, $M_i\cap M_j=\emptyset$, while $L_i$ intersect $M_j$. Take $n^2$ points $L_i\cap M_j$. -Proof. For a generic projection $\pi: \mathbb P^3\to \mathbb P^2$ both collection of lines $L_i$ and $M_j$ project to (reducible) curves of degree $d$ on $\mathbb P^2$. Their intersection are exactly the projections of the collection of points $L_i\cap M_j$. This gives us a pencil of degree $d$ curves, and in a generic situation a generic curve from the pencil will be smooth. -It is interesting to notice that in the case $n=3$ the above construction is rigid, i.e., it produces a unique example up to projective equivalence of $\mathbb P^3$. Are there some further examples of $9$ points in $\mathbb P^3$ having this property? Are they rigid as well?<|endoftext|> -TITLE: Does a finite suspension spectrum make a space finite? -QUESTION [6 upvotes]: Suppose that $X$ is a space whose suspension spectrum $\Sigma_+^\infty(X)$ is dualizable in the stable homotopy category. I believe this is equivalent to saying that $\Sigma_+^\infty(X)$ is (weakly) homotopy equivalent to a finite cell spectrum. What does this imply about $X$? In particular, does it imply that $X$ is weakly equivalent to a finite cell complex, as a space? - -REPLY [11 votes]: No. In the stable homotopy category a retract of a finite cell spectrum is again a finite cell spectrum, but in the weak homotopy category of spaces a retract of a finite cell complex is not necessarily a finite cell complex; there is an obstruction in the kernel of $K_0\mathbb Z[\pi_1(X)]\to K_0\mathbb Z$. -EDIT: -For a simply connected space, finite generation of the direct sum of its integral homology groups implies that it is equivalent to a finite complex. Thus a connected space must become finite after one suspension if its suspension spectrum is finite. The same then follows without assuming connected. -Therefore finiteness of $\Sigma^\infty X$ is equivalent to finiteness of $\Sigma X$, and (as shown by Fernando's answer) this is strictly weaker than finiteness of $X$. - -REPLY [9 votes]: Take an acyclic group, i.e. a group $G$ with trivial homology. Choose $G$ which is not finitely presented, so that its classifying space $X=BG=K(G,1)$ cannot be finite. Since $\Sigma X\simeq \star$ is contractible, $\Sigma^{\infty}_{+}(X)=S$ is the sphere spectrum (dualizable).<|endoftext|> -TITLE: Determinant of a $4n \times 4n$ block matrix where every block is singular -QUESTION [7 upvotes]: I have a 4n$\times$4n matrix, which can be written as -\begin{pmatrix} -0 & A &B &C \cr -D& 0& E & F \cr -G& H & 0 & J \cr -K& L& M& 0 -\end{pmatrix} -each entry being an n$\times$n matrix with vanishing determinant. Is there a rule for checking if the full matrix has zero determinant? How about the special case -\begin{pmatrix} -0 & A &B &C \cr --A^T & 0& E & F \cr --B^T & E^T & 0 & J \cr --C^T & F^T & J^T & 0 -\end{pmatrix} -still with vanishing determinants for each n$\times$n matrix? -(The n is the dimension of an SU group -- I can probably work out the SU(2) or n=3 case by brute force, but I would like to know if there is some method that does not require explicit calculation.) -Many thanks in advance for any help or suggestion. - -REPLY [2 votes]: It would be nice if the rule for determinants for $2\times2$ matrices generalized to the case of $2n\times 2n$ matrices: -$\det \begin{pmatrix} -A & B \cr -C & D -\end{pmatrix} -=\det A \det D - \det B\det C$, -but this is sadly not true. -Nonetheless, the familiar Laplace expansion theorem for minors of order $n-1$ does have a generalization to minors of any order, including, in this case, minors of order $2n$ of a $4n \times 4n$ matrix, see -http://www.proofwiki.org/wiki/Laplace's_Expansion_Theorem -This might help.<|endoftext|> -TITLE: Space of compact operators -QUESTION [6 upvotes]: I am interested in the Banach space $\mathcal{K}=\mathcal{K}(\ell^2)$ of compact operators on $\ell^2$, however my questions can be stated for any $\mathcal{K}(E)$, where $E$ is an arbitrary Banach space. I think that everyone who tries to study "classical" operator spaces like $\mathcal{K}$, Schatten $p$-class operators etc. immediately discovers the similarity with "commutative" counterparts, i.e. $c_0$ and $\ell^p$. This phenomenon is visible when one uses (generalised) singular numbers for certain classes of operators. Again, I have got plenty of questions concerning this stuff, let me list at least two of them: -1) what are the complemented subspaces of $\mathcal{K}$? Is $\mathcal{K}$ complemented in $\mathcal{B}(\ell^2)$? Recently, Haydon and Argyros constructed an HI-space $E$ such that $\mathcal{K}(E)$ has codimension 1 $\mathcal{B}(E)$, thus complemented. -2) is every bounded operator from $p$-Schatten class to $\mathcal{K}$ compact? -What other properties $\mathcal{K}$ shares with $c_0$? - -REPLY [3 votes]: This post is not intended as an answer to a question that was asked & answered 10 years ago, but as an appendix for newcomers to provide background about the DPP in the context, which the question owner and the commentators (and the other experts in the field) already know about. - -Every reflexive complemented subspace of a C*-algebra is isomorphic to a Hilbert space. - -TFAE for a C*-algebra $A$: - - - -$A^{\ast}$ has DPP (Dunford-Pettis property) -$A$ has DPP -$A$ contains no complemented copy of $\ell^2$. -$A^{\ast\ast}$ is a finite type I von Neumann algebra. - -Perhaps the major difference between $c_0$ & $K(\ell^2)$, $\ell^1$ & trace class operators on $\ell^2$, is the lack of DPP for the latter. - -On the contrary, if a C*-algebra $A$ does not have DPP, then there exists a complete isometry $K(\ell^2)\to A$. - -If $E$ is reflexive, then - - - -a. $K(E)$ is Arens regular. (so is $c_0$) -b. $K(E)$ contains no copy of $\ell^1$. (so does $c_0$) -c. $K(E)$ does not have DPP. - -As a consequence of (b) and (c), if $X$ is a subspace of $K(E)$, then both $X$ and $K(E)/X$ cannot have DPP simultaneously. - -If $E$ is reflexive and has approximation property, then $K(E)^{\ast\ast}= B(E)$. Thus, $K(E)$ is an ideal in its bidual $B(E)$. Equivalently, every left & right multiplication operator $L_a,R_a:K(E)\to K(E)$, defined by $L_ax = ax$ & $R_ax = xa$ are weakly compact. - -On the other hand, the left & right multiplication operators on $c_0$ are compact by DPP.<|endoftext|> -TITLE: Image of projective 1-space contained in projective 1-space over a smaller field? -QUESTION [12 upvotes]: This is inspired by -Does "all points rational" imply "constant" for this "cubic" curve over an arbitrary field? . -Say $K/F$ is a finite separable extension of fields. Assume $F$ is infinite (or else there are trivial counterexamples to the question below). Say -$$t:\mathbf{P}^1_K\to\mathbf{P}^1_K$$ -is a non-constant morphism defined over $K$, with the property that -$$t(\mathbf{P}^1(K))\subseteq\mathbf{P}^1(F).$$ -Does this imply that $K=F$? -Comments: (1) the question linked to above is a special case of this, written in a slightly more hands-on way. (2) There are counterexamples in the inseparable case of the form $t(x)=x^p$ with $K/F$ purely inseparable of degree $p$. (3) If $K$ is algebraically closed then $F=K$ trivially because $t$ is surjective on points. (4) If $F$ is a number field then I guess one can use the theory of "thin sets" (the ideas used in the proof of Hilbert irreducibility) to prove that $F=K$. (5) If you generalise to other projective curves then again the result fails, because e.g. the curves could have no $K$-points at all and $t$ could be the identity. (6) On the other hand one could ask about projective $n$-space or even affine 1-space -- one just needs a lot of $K$-points to make the question interesting... - -REPLY [7 votes]: This proof was inspired by an idea of Francois Brunault and appears piecemeal in the comments to the question; it's probably not materially different from David's proof. Choose some $F$-basis $\{a_1,\ldots,a_n\}$ for $K$ with $a_1=1$, and let $p:\mathbf{A}^n_K\to\mathbf{A}^1_K$ be the map $(X_1,X_2,\ldots,X_n)\mapsto \sum_ia_iX_i$. -Lemma Suppose that $X$ and $Y$ are varieties over a field $F$ such that $X(F)$ is dense in $X$, that $K/F$ is a separable extension and that $f:X_K\to Y_K$ is a morphism such that $f(X(F))\subset Y(F)$. Then $f$ is defined over $F$. -Proof: Indeed, we can replace $K$ by its Galois closure and assume that $K/F$ is Galois. For every automorphism $g$ of $K/F$, $gfg^{-1}\vert_{X(F)}=f\vert_{X(F)}$; since $X(F)$ is dense in $X$, this implies that $gfg^{-1}=f$, and we conclude by Galois descent. -Corollary 1 $t$ is defined over $F$. -Corollary 2 $t\circ p$ is defined over $F$. -Let $L/F$ be the Galois closure of $K/F$. -Claim The base changed map $p:\mathbf{A}^n_L\to\mathbf{A}^1_L$ is Galois-equivariant. -End of proof assuming claim: For any automorphism $g$ of $L/F$ and any $1\leq i\leq n$, $$ga_i=g(p(e_i))=p(g(e_i)=p(e_i)=a_i,$$ -where $e_i$ is the $i^{\text{th}}$ standard basis vector for $\mathbf{A}^n_L$. So all the $a_i$ lie in $F$ (and, in particular, $n=1$). -Proof of claim: For any automorphism $g$ of $L/F$, we have -$$t\circ p=g(t\circ p)g^{-1}=gtg^{-1}\circ gpg^{-1}=t\circ gpg^{-1}.$$ -So the pair $(p,gpg^{-1})$ defines a map $f:\mathbf{A}^n_L\to Z_L$, where $Z$ is the self fiber-product of $\mathbf{P}^1_F$ over $\mathbf{P}^1_F$ via $t$. $Z$ is a curve (not necessarily reduced or irreducible), and $f$ maps onto an irreducible component of $Z$. One of the irreducible components of $Z$ is the diagonal $\Delta$. Showing that $p=gpg^{-1}$ amounts to showing that $f$ maps onto $\Delta$. But $f$ maps the $F$-points of the first co-ordinate axis of $\mathbf{A}^n$ into the diagonal: -$$f(x,0,\ldots,0)=(x,gx)=(x,x),$$ -for all $x\in\mathbf{A}^1(F)$. So it must map all of $\mathbf{A}^n_K$ into the diagonal, and we are done. -EDIT: Here is one way to view this proof in a more general framework. Let $f:X\to Y$ be a map of $F$-varieties, with $X$ non-empty. Suppose that: - -$f$ is finite. -$f(X(K))\subset Y(F)$. -$X(F)$ is dense in $X$. - -We also suppose that the Weil restriction $X'=\text{Res}_{K/F}X_K$ is representable. Let $p:X'_K\to X_K$ be the natural adjunction map. The hypotheses imply (see Lemma above) that both $f$ and $f\circ p$ are defined over $F$. The same proof as in the claim above shows that $p$ is also defined over $F$. This shows that $K=F$.<|endoftext|> -TITLE: Does adding degeneracies to a semi-simplicial diagram change the homotopy colimit? -QUESTION [5 upvotes]: Let $\Delta_{+}$ be the sub-category of the simplex category $\Delta$ containing only injective functions, and take $M$ to be a nice model category. I'll write $i \colon \Delta_{+} \hookrightarrow \Delta$ for the inclusion. -Now assume we have a semi-simplicial diagram $X \colon \Delta_{+}^{\text{op}} \to M$. We can then form the left Kan extension $i_! X \colon \Delta^{\text{op}} \to M$, which adds the degeneracies to the semi-simplicial diagram $X$. Now here's my question: -Is $$\text{hocolim}_{\Delta_{+}^{\text{op}}}X = \text{hocolim}_{\Delta^{\text{op}}}i_{!}X,$$ or are they at least equivalent? What about if I take a homotopy left Kan extension instead? -I've got another question, but I'm afraid this might be too specific. A Segal groupoid in a model category $M$ is a simplicial object $X \colon \Delta^{\text{op}} \to M$ which satisfies the Segal condition, i.e. $$X_n \to X_1 \times ^h _{X_0} \dots \times ^h _{X_0} X_1$$ is a weak equivalence, and also $$d_0 \times d_1 \colon X_2 \to X_1 \times ^h _{d_0,X_0, d_0} X_1 $$ is a weak equivalence. The second condition says that every horn of type $b \leftarrow a \to c$ can be filled. Since both conditions only involve face maps, they make perfect sense for a semi-simplicial object too. So I guess it makes sense to ask, given a semisimplicial object $Y \colon \Delta_{+}^{\text{op}} \to M$ satisfying the two conditions above, is the (homotopy) left Kan extension $i_! Y$ a Segal groupoid in $M$? - -REPLY [10 votes]: As I understand it, homotopy left Kan extension along any functor $i:\mathcal C\to\mathcal D$ preserves hocolim. -Left Kan extension $i_!$ is left adjoint to restriction $i^\star$, i.e. composition with $i$. (Of course, if $i$ is not fully faithful, "extension" is a bit of a misnomer: restriction composed with "extension" is not isomorphic to the identity.) Colimit over $\mathcal D$ is left Kan extension along the functor $p:\mathcal D\to \star$. Since $(p\circ i)^\star=i^\star\circ p^\star$, the left adjoints also compose as they should. -Surely the same is true in a derived sense. -Kan extension (as opposed to homotopy Kan extension) will preserve hocolim in cases where it coincides with homotopy Kan extension, i.e. in cases where it preserves weak equivalences. In the case of $i:\Delta_+\to\Delta$, this requires only that coproduct in $\mathcal M$ preserve weak equivalences, which is true for example for spaces and for based simplicial sets but not for based spaces.<|endoftext|> -TITLE: Map from simplex to itself that preserves sub-simplices -QUESTION [10 upvotes]: I believe this may be a standard algebraic topology problem, so I apologize in advance if this belongs in stackexchange (it's not a homework problem, however, and came about in a research context). I've got a continuous map $f$ from the $n$-simplex to itself, such that the image of every strict sub-simplex is itself. So, each vertex gets mapped to itself, as does each edge, and so on and so forth. Does it follow that $f$ must be surjective? -Thank you! - -REPLY [2 votes]: Recently, in search for an early reference for this statement I came across this 2009 preprint by Roman Karasev: -KKM-type theorems for products of simplices and cutting sets and measures by straight lines. -It has the statement as Lemma 1: - -In the following pdf, also by Karasev: Geometry of measures: partitions and convex bodies, 2013, the statement appears as Lemma 9.2. -There is a proof, very simliar to my other answer (and to the proof in the preprint from 2009): - -Karasev also provides a very nice application: Brouwer's fixed-point theorem! - -I wonder whether 2009 is the first time this Lemma was written down somewhere, my feeling is that it should have been used by somebody before that time.<|endoftext|> -TITLE: homological algebra with spectra -QUESTION [9 upvotes]: I'm reading Mike Hopkins' COCTALOS notes and having trouble with some pretty basic statements about (naive) spectra. Basically I'm nervous doing homological algebra with them, although in these problems I'm having I couldn't imagine the proofs being anything other than simple diagram chases. The relevant definitions are in section 4. -First, I'm having trouble with the proof of Prop. 4.12, which claims that every E-Adams resolution arises from an Adams tower. The proof is based on the diagram at the bottom of page 13 (sorry, I'm not sure how to put diagrams in an MO problem) and shows how to build the first level of the tower. To continue the tower, we need that two composite maps (at the top of page 14) are null, and this is what's confusing me. For (i), I have no idea, although $h$ must be E-mono since it's the first map in a factorization of an E-mono. For (ii), the problem is that we know both $\Sigma^{-1} I_0\rightarrow \Sigma^{-1}I_3$ and $\Sigma^{-1}I_1\rightarrow \Sigma^{-1}I_3$ are null, but the cofiber $X_1 = \Sigma^{-1}I_1/\Sigma^{-1}I_0$ may not map E-nully to $\Sigma^{-1}I_3$. In fact, there isn't necessarily a single choice of map anyways; it depends on the choice of nullhomotopy of $\Sigma^{-1}I_0\rightarrow \Sigma^{-1}I_3$ (factoring through $\Sigma^{-1}I_2$). All I know is that I need to find a map -$C (E\wedge X_1) \simeq I\wedge E\wedge I_0 \rightarrow E\wedge \Sigma^{-1}I_3$ extending the map from $E\wedge X_1$. (Here $I$ is the pointed interval.) -Second, I don't understand the statement that "these maps are null after smashing with E and hence must actually be null as [their images] are E-injective". Suppose $A\rightarrow J$ is E-null and $J$ is E-injective. The only thing to do here is start with an E-mono $A\rightarrow B$ and obtain a lift $B\rightarrow J$ in the diagram $B\leftarrow A \rightarrow J$. Then the lift will have to be E-null too. But I don't see how this would imply that $A\rightarrow J$ is actually null. And there's no obvious E-mono I've got to work with anyways, besides maybe ${Id}_A$ (which doesn't help at all). -And lastly, I'm having trouble proving the $\Leftarrow$ direction of Lemma 5.2, that if for $E$ a ring spectrum the map $J=S\wedge J\rightarrow E\wedge J$ is the inclusion of a retract then $J$ is E-injective. I'd prove this by taking an E-mono $X\rightarrow Y$ and trying to find a lift in $Y\leftarrow X \rightarrow J$ for any $X\rightarrow J$. The only thing to do here is smash with $E$ and obtain $E\wedge B\leftarrow E\wedge A \rightarrow E\wedge J \rightarrow J$ (where the first map is mono), but I can't see what this buys me. There's one fact we've got about monos (Lemma 4.3), but not sure whether it helps here: $A \rightarrow B$ is mono iff there is some $C\rightarrow B$ such that $A\vee C\rightarrow B$ is a weak equivalence. -Since these seem like they should be such routine verifications, I'm feeling that there must be some basic technique I'm missing. Does anyone have any suggestions? - -REPLY [5 votes]: Suppose that the inclusion $\eta\wedge 1:I\to E\wedge I$ admits a retraction, say $r$. Consider an $E$-monomorphism $f:A\to B$ and an arbitrary map $g:A\to I$. As $f$ is an $E$-monomorphism we can choose a retraction $s:E\wedge B\to E\wedge A$, and then the composite -$$ h = (B \xrightarrow{\eta\wedge 1} E\wedge B \xrightarrow{s} E\wedge A - \xrightarrow{1\wedge g} E\wedge I \xrightarrow{r} I) -$$ -satisfies $hf=g$. This shows that $I$ is $E$-injective. -Now suppose we have an $E$-null map $f:A\to I$, where $I$ is $E$-injective. This gives a commutative square - -\begin{array}{ccccc} - & A & \xrightarrow{f} & I & \\ - \eta\wedge 1 & \downarrow & & \downarrow &\eta\wedge 1 \\ - & E\wedge A & \xrightarrow{f\wedge 1} & E\wedge I -\end{array} - -The bottom map is zero, so if we go down then across we get zero, so if we go across and then down we also get zero. However, the right hand vertical map is a split monomorphism, so we can conclude that $f$ itself is zero. -This covers your second and third questions. I'm not going to discuss the first one because it is too painful to do the diagrams, but hopefully the other two will give you some ideas.<|endoftext|> -TITLE: Bertini theorems for base-point-free linear systems in positive characteristics -QUESTION [17 upvotes]: Suppose that $X$ is a smooth algebraic variety over an algebraically closed (uncountable if it helps) field of characteristic $p > 0$. Suppose that $L$ is a line bundle, probably ample or at least positive, and that $\delta \subseteq |H^0(X, L)|$ is a linear system. -It is well known that just because $\delta$ is base-point-free, it does not mean that a general member defines a smooth subscheme (it need not even be reduced, for example Frobenius pull-backs of linear systems, Remark 10.9.3 in Hartshorne's algebraic geometry). -However, let's suppose the following: -$f : X \to Y$ is a map, $M$ is a very ample line bundle on $Y$ and $L = f^* M$. Suppose further that $\delta$ is the pull-back of the complete linear system $|H^0(Y, M)|$. Are there any (separability?) conditions on the map $f$ which still guarantee that Bertini holds for $\delta$? I imagine this must be well known, but I don't know the right references. -In particular, I am looking for conditions weaker than etale (/ etale outside a finite set of points)? Say in the birational case, or the finite case? - -REPLY [2 votes]: I thought it might be useful to post a statement of an answer. This is contained within Cumino-Greco-Manaresi and also Spreafico's paper which Anton mentioned earlier, but can't really fit in a comment and I think might be useful to someone else. I probably should have posted this earlier. -Indeed, in the context of my original question, if $f : X \to Y$ induces separable residue field extensions (or separably generated not necessarily algebraic residue field extensions), then everything is fine (say over an algebraically closed field). I should note that even really nice maps may fail to have separably generated residue field extensions. -Indeed, consider the projection map $\mathbb{A}^2 \to \mathbb{A}^1$ (if you want, this can be an open chart from the projection $\mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^1$) over a field $k$ of characteristic $p > 0$. This corresponds to the inclusion $k[x] \subseteq k[x,y]$. This does not have separable residue field extensions. -Consider the ideal $(x-y^p) \subseteq k[x,y]$. This pulls back to the zero ideal in $k[x]$ and induces the residue field extension $k(x) \subseteq k(x^{1/p})$ which is clearly non-separable. -This indicates to me that even birational maps can fail this property horribly (for example, if I blow up a curve on a 3-fold similar things can be arranged). Finite maps are much better behaved of course.<|endoftext|> -TITLE: Quantization of conjugacy classes in a Lie group -QUESTION [10 upvotes]: Let $G$ be a Lie group (and to be safe, let's assume it is semisimple). Consider the action of $G$ on itself by conjugation, and form the GIT (algebro-geometric) quotient $G/\!/G$. Then let $\pi:G\to G/\!/G$ denote the quotient map. For example, if $G=SL(2,\mathbb C)$, then $G=\operatorname{Spec}\mathbb C[a,b,c,d]/(ad-bc-1)$ and $G/\!/G=\operatorname{Spec}\mathbb C[t]$, and the projection $\pi$ is induced from the map of rings which sends $t$ to $a+d$ (all I'm saying is that the trace determines the conjugacy class in $SL(2,\mathbb C)$, at least for semisimple ones). -There is a well-known closed $2$-form on $G$ whose symplectic leaves are exactly the fibers of $\pi$ (i.e. the conjugacy classes). It can be defined, for example, as follows. Let $g\in G$, and denote its conjugacy class by $G_g$. Then $T_gG_g$ is exactly the image of the map $R:\mathfrak g\to\mathfrak g$ given by $\alpha\mapsto\alpha-\operatorname{ad}_g\alpha$. Now the map $\mathfrak g\otimes\mathfrak g\to\mathbb C$ given by $\alpha\otimes\beta\mapsto\operatorname{tr}(g[\alpha,\beta])$ has exactly the same kernel as $R$, and thus it descends to give an alternating bilinear form on $T_gG_g$ (by $\operatorname{tr}$ I mean trace in the adjoint representation). Some additional arguments show that this form is closed. -This construction is sometimes done instead with $\mathfrak g^\ast\to\mathfrak g^\ast/\!/G$, though I'm more interested in the case $G\to G/\!/G$. -There are many papers which study the deformation quantization (or geometric quantization) of (the leaves of) this Poisson structure on $G$, but everything is very VERY abstract. Is there any place where I can find explicit formulae for things like: -1) the Poisson bracket of standard functions on $G$ (e.g. the coordinates $a,b,c,d$ in the case $G=SL(2,\mathbb C)$). -or even better -2) the deformed product (star product) on $\mathcal O(G)$? -I feel like these should be known or easy to calculate, but an answer has eluded my search. -I've tried doing some explicit calculations myself, but everything seems to hinge on getting a succinct formula for the $2$-form on $G$, and I always end up with something horrendous. - -REPLY [5 votes]: Joseph Donin and Andrei Mudrov have worked a lot on this question... the quantization of conjugacy classes seems to be related to dynamical r-matrices and and the so-called reflection equation. See e.g. this paper of Mudrov (Quantum conjugacy classes of simple matrix groups)... there are explicit formulae in the end of the paper, but I am not sure this answers your question (other explicit formulae for $GL_n$ can also be found in this older paper by Donin and Mudrov).<|endoftext|> -TITLE: categorifying induction in homotopy type theory -QUESTION [7 upvotes]: In trying to understand homotopy type theory, I stumbled upon the following silly question, which is likely to be trivial for the experts. -Let $B=\sqcup_{n\in\Bbb N} BS_n$, which I'd like to think of as the categorified version of the natural numbers $\Bbb N$. There is an obvious map $\sigma:B\to B$ that covers the successor map $s:\Bbb N\to\Bbb N$, $s(n)=n+1$. -On the other hand, in Martin-Löf's type theory, there is the inductive type of natural numbers, written as -Inductive nat : Type := O : nat | S : nat -> nat. - -in the syntax of Coq. The induction rule reads -nat_rect: forall P: nat -> Type, -x: PO -> ((forall n: nat, Pn -> P(Sn)) -> (forall n: nat, Pn)). - - -Question: Can nat and S be interpreted as $B$ and $\sigma$? Presumably not, but why so? - -In fact, such interpretation is impossible, according to Voevodsky: (1) because nat has contractible components (see Theorem isasetnat here); (2) because nat_rect would cause the fibration $\sigma$ to have a section. -Both arguments elude me, however. (1) just needs better knowledge of Coq and more patience with this new way of writing down proofs than I've developed so far (so I don't follow the proof of his Theorem isasetifdeceq). In (2), I fail to see how nat_rect manages to mention a section of $\sigma$, or even that $\sigma$ must be a fibration. Indeed, let me parse the second line of the induction rule. -P: nat -> Type - -I'm reading this as $p\in Map(B, U)$, where $U$ is a universe. -forall n: nat, Pn - -This is the space of sections of $p^*\xi$, where $\xi$ is the universal fibration over $U$. -S: nat -> nat - -This says $\sigma\in Map(B, B)$. -Pn -> P(Sn) - -And this is $Map(\xi^{-1}(p(x)),\xi^{-1}(p(\sigma(x)))$. -forall n: nat, Pn -> P(Sn) - -Here the previous space of maps needs to be understood as the homotopy fiber of a fibration -over $B$. This fibration is $G_\sigma^*(p\times p)^*Map(\xi,\xi)$, where $G_\sigma: B\to B\times B$ is the graph of $\sigma$, $p\times p: B\times B\to B\times B$, and $Map(\xi,\xi)$ is the fibration over $U\times U$ whose fiber over $(X,Y)$ is $Map(\xi^{-1}(X),\xi^{-1}(Y))$. (I understand that fibrations like $Map(\xi,\xi)$ and $\xi\times\xi$ are implicitly postulated by saying that $U$ is closed under products, dependent products, etc.; and these postulates correspond to Martin-Löf's universe formation rules.) -So we end up with the space of sections $Sect(G_\sigma^*(p\times p)^*Map(\xi,\xi))$. -(forall n: nat, Pn -> P(Sn)) -> (forall n: nat, Pn) - -This is $Map(Sect(p^*\xi), Sect(G_\sigma^*(p\times p)^*Map(\xi,\xi)))$. Let me call it $M(p)$. -x: PO -> [(forall n: nat, Pn -> P(Sn)) -> (forall n: nat, Pn)] - -and this is just $Map(\xi^{-1}(p(0)),M(p))$. -It seems to be a bit harder to parse the entire nat_rect; but I don't see how on earth this could help one to find a section of $\sigma$. - -REPLY [7 votes]: The first reason you give is sufficient to answer your question: any interpretation of nat (and any other type with decidable equality) must have contractible components. Let me try to unpack the proof: -The proof of isasetifdeceq goes as follows: Fixing $x:X$, we must show that $\text{Paths}(x,x)$ is contractible. We know that $\Sigma_{x':X}\text{Paths}(x,x')$ is contractible, so we just need the natural map $f:\text{Paths}(x,x)\to\Sigma_{x':X}\text{Paths}(x,x')$ to be a weak equivalence. This follows from the hypothesis using the theorem onefiber, which establishes that for a fibration which is empty over all but one path-component of the base, the total space is equivalent to a fiber over the remaining component. -Regarding (2). I think there's some confusion here: Let me try to unpack nat_rect in a more direct way: For any fibration $\Sigma_{\text{nat}} P\to\text{nat}$ over nat, given a point in $P_0$ and a section of $\Pi_{n:\text{nat}}\text{Map}(P_n,P_{n+1})\to\text{nat}$, you get a section of $P$. That is the interpretation of primitive recursion/induction.<|endoftext|> -TITLE: Heller operator without left adjoint? -QUESTION [9 upvotes]: Suppose given a noetherian ring $R$. On the stable category $R\text{-}\underline{\text{mod}} := R\text{-mod}/R\text{-proj}$, we have the Heller operator -$$ -\Omega : R\text{-}\underline{\text{mod}} \to R\text{-}\underline{\text{mod}} \; , -$$ -defined by a choice of short exact sequences $\Omega X \to P \to X$ with $P$ projective, accordingly on morphisms. -Does there exist a ring $R$ for which $\Omega$ does not have a left adjoint? -Of course, $R$ should neither be self-injective ($\Omega$ an equivalence) nor hereditary ($\Omega = 0$). -If $\Omega$ has a left adjoint $S$, then $(X,\Omega f) = (SX,f)$ is injective for $X$ an $R$-module and $f$ a monomorphism. So I've been searching (without success) for a monomorphism $f$ in the stable category that is not mapped to a monomorphism under $\Omega$. First of all, $f$ should be a non-split mono, but this is possible, since we are not in the triangulated case. -It was pointed out to me that in "On a theorem of E. Green on the dual -of the transpose" by M. Auslander and I. Reiten, it is shown that that -Omega lifts to an exact functor isomorphic to tensoring with a bimodule if $R$ is an artin algebra (MR1143845). -The question is related to this question ("Brown, but not Quillen?"). - -REPLY [5 votes]: This is answered by Proposition 1.7 in -[Auslander, Maurice; Reiten, Idun. Homologically finite subcategories. Representations of algebras and related topics (Kyoto, 1990), 1--42, London Math. Soc. Lecture Note Ser., 168, Cambridge Univ. Press, Cambridge, 1992.], - -If $R$ is a two-sided noetherian ring and $d \geq 1$ an integer, then - the functor $$\operatorname{Tr}\Omega^d\operatorname{Tr} \colon - R\text{-}\underline{\text{mod}} \to R\text{-}\underline{\text{mod}}$$ - is a left adjoint of $\Omega^d \colon R\text{-}\underline{\text{mod}} \to R\text{-}\underline{\text{mod}}$. - -In particular $\operatorname{Tr}\Omega\operatorname{Tr}$ is a left adjoint of $\Omega$. -Here $\operatorname{Tr} \colon R\text{-}\underline{\text{mod}} \to R^{\operatorname{op}}\text{-}\underline{\text{mod}}$ denotes the transpose, if $M$ is an $R$-module with projective presentation $$P_1 \to P_0 \to M \to 0,$$ then $\operatorname{Tr} M=\operatorname{coker} (\operatorname{Hom}_R(P_0,R) \to \operatorname{Hom}_R(P_1,R))$.<|endoftext|> -TITLE: Trace of the identity map in a projective module -QUESTION [6 upvotes]: Let $A$ be a commutative algebra (over the complex numbers, with a unit) and let $M$ be a finitely generated projective $A$-module, and let $m_1,\ldots,m_n$ be a set of generators of $M$. The Dual Basis Theorem states that there exists $m_1^\ast,\ldots,m_n^\ast\in M^\ast$ such that $x=\sum m_i^\ast(x)m_i$ for all $x\in M$. This implies that one can write the identity map on $M$ as the tensor $\mathbf{1}=\sum m_i^\ast\otimes m_i$ and one defines $\operatorname{tr}\mathbf{1}=\sum m_i^\ast(m_i)$. What are the properties of $\operatorname{tr}\mathbf{1}$? Is there any relation to, for instance, the rank of $M$? When is it proportional to the unit element in $A$? - -REPLY [2 votes]: Let $A$ be the matrix with entries $m_i^*(m_j)$, and put $B(x)=xA+I-A$ and $f(x)=\det(B(x))$. We find that $A^2=A$, and $B(xy)=B(x)B(y)$, and $f(xy)=f(x)f(y)\in R[x,y]$. We also have $f(1)=1$. It follows that $f(x)=\sum_{i=0}^ne_ix^i$ for some elements $e_i\in R$ which are orthogonal idempotents, ie $e_ie_j=\delta_{ij}e_i$ and $\sum_ie_i=1$. If we put $R_i=R/(1-e_i)\simeq Re_i$ and $M=M/(1-e_i)M\simeq e_iM$ we see that $R=\prod_iR_i$ as rings and $M=\prod_iM_i$ as modules. Moreover, $M_i$ is locally free of rank $i$ over $R_i$. -We also have $\text{tr}(1)=\text{tr}(A)=f'(1)=\sum_iie_i$, so $\text{tr}(1)$ maps to $i$ in $R_i$. It is cleaner to work with the elements $e_i$ because they distinguish the different $R_i$ even in a finite characteristic situation, which is not true for $\text{tr}(1)$.<|endoftext|> -TITLE: a game on numbers -QUESTION [11 upvotes]: Hello, here is a little two-players game. -Players A and B choose three numbers : a, b and c for A, a', b' and c' for B. The values are numbers between 0 and 1, their sum is 1, and they are ordered: $a \geq b \geq c$ and $a' \geq b' \geq c'$. -Then, the players A and B compare their choices : $a$ vs $a'$, $b$ vs $b'$ and $c$ vs $c'$. -If a player has 2 values bigger than the other, he wins (otherwise it's a tie). -I would like to study whether there is a good strategy in this game but I don't know how to start. Do you have an idea on the general way of studying this kind of game? Any reference of book/article is welcomed :) - -REPLY [6 votes]: https://arxiv.org/abs/1708.07916 -This arxiv paper might be helpful. The abstract is reproduced below: - -This paper explores the Nash equilibria of a variant of the Colonel Blotto game, which we call the Asymmetric Colonel Blotto game. In the Colonel Blotto game, two players simultaneously distribute forces across n battlefields. Within each battlefield, the player that allocates the higher level of force wins. The payoff of the game is the proportion of wins on the individual battlefields. In the asymmetric version, the levels of force distributed to the battlefields must be nondecreasing. In this paper, we find a family of Nash equilibria for the case with three battlefields and equal levels of force and prove the uniqueness of the marginal distributions. We also find the unique equilibrium payoff for all possible levels of force in the case with two battlefields, and obtain partial results for the unique equilibrium payoff for asymmetric levels of force in the case with three battlefields. - -Theorem 3.3 on page 5 characterized all Nash Equilibrium Strategies for the game described in the question.<|endoftext|> -TITLE: Automorphisms of a weighted projective space -QUESTION [23 upvotes]: What is the automorphisms group of the weighted projective space $\mathbb{P}(a_{0},...,a_{n})$ ? -Consider the simplest case of a weighted projective plane, take for instance $\mathbb{P}(2,3,4)$; any automorphism has to fix the two singular points. Consider a smooth point $p\in\mathbb{P}(2,3,4)$. What is the subgroup of the automorphisms of $\mathbb{P}(2,3,4)$ fixing $p$ ? - -REPLY [2 votes]: Just to complement Al-Amrani's answer (and to self-promote a little bit): the automorphism 2-group of the weighted weighted projective stack $\mathbb{P}_S(a_1,\ldots,a_n)$ over an arbitrary base scheme $S$ has been computed, and its structure elaborated, in -https://doi.org/10.1016/j.jalgebra.2017.05.002 -If I am not misunderstanding something, Al-Amrani's result can be interpreted as saying that the $\pi_0$ of the 2-group of automorphisms of the stack is isomorphic to the group of automorphisms of its coarse moduli space (namely, the corresponding weighted projective space).<|endoftext|> -TITLE: Distributing points with respect to a concave function -QUESTION [5 upvotes]: Suppose I have a concave function defined on the unit interval such that $f(0) = f(1) = 0$ and $\int_0^1 f(t) dt = \alpha$, where $\alpha$ is "small" (say $0.01$ or thereabouts). Say I distribute $n$ points $x_1,\dots,x_n$ on the unit interval and consider the function $F(x_1,\dots,x_n) = \int_0^1 f(t) \cdot \min_i\{|x_i - t|\} dt$. Is there a lower bound on $F$ as a function of $\alpha$ and $n$? If $n=1$, I can show that a lower bound is $\alpha/6$, so I'm curious if something like $\alpha/(6n)$ holds in general. - -REPLY [2 votes]: It seems that for large $n$ there will be an upper bound $c_n \alpha$ with $c_n$ asymptotic to $\frac{2}{9n}$ (and thus a bit better than the $\frac{1}{6n}$ suggested by the value of $c_1$). The asymptotic equality condition should be: $f$ is a triangular function $\tau_z$ for some $z \in (0,1)$, defined by -$$ -\tau_z(x) = \cases{ - x/z,& {\rm if}\phantom0x \leq z,\cr - (1-x)/(1-t), & {\rm if}\phantom0 x \geq z - } -$$ -(i.e. $\tau_z$ is piecewise linear with a corner only at $x=z$ where $\tau_z$ attains its maximal value of $1$), and $x_1,\ldots,x_n$ are distributed on $(0,1)$ with density proportional to $\sqrt{f\phantom.}$ (which is probably not what most of us would have guessed at first). -The condition that $\alpha = \int_0^1 f(x) \phantom. dx$ be "small" is a red herring because the problem is linear in $f$, so that what works for small $\alpha$ works equally for all $\alpha$. It is also enough to work with $f = \tau_z$ because the convex hull of functions $c \tau_z$ is precisely the concave functions $f$ on $[0,1]$ vanishing at the endpoints; e.g. if such $f$ has a continuous second derivative then -$f(x) = \int_{z=0}^1 (z-z^2) \phantom. (-f''(z)) \phantom. \tau_z(x) \phantom. dz$, and $-f''(z) \geq 0$ for $f$ concave. -Now the idea is that if $x_1,\ldots,x_n$ are regularly distributed with density $\delta(\cdot)$ on $[0,1]$ then $\min_i |x_i - t|$ behaves like $1/(4\delta(t))$, because it oscillates between $0$ and $1/(2\delta(t))$ like a nonnegative triangle wave. The only restriction on $\delta$ is that it be a positive function with $\int_0^1 \delta(t) \phantom. dt = n$. By Cauchy-Schwarz, -$$ -\int_0^1 \frac{\tau_z(t)}{\delta(t)}dt \cdot \int_0^1 \delta(t) \phantom.dt \geq \left( \int_0^1 \sqrt{\tau_z(t)} \phantom. dt \right)^2, -$$ -with equality iff $\delta^2$ is proportional to $\tau_z$. Now we know $\int_0^1 \delta(t) \phantom. dt = n$, and compute $\int_0^1 \sqrt{\tau_z(t)} \phantom. dt = 2/3$ for all $z$. -Hence $\int_0^1 (\tau_z(t) / \delta(t)) \phantom. dt \geq \frac4{9n}$. Since $\int_0^1 \tau_z(t) \phantom. dt = 1/2$, we deduce that -$$ -\int_0^1 \frac{\tau_z(t)}{4\delta(t)}dt \geq \frac2{9n} \int_0^1 \tau_z(t) \phantom. dt, -$$ -from which the claimed asymptotic should follow after some epsilon-chasing.<|endoftext|> -TITLE: Proof of Bott Periodicity in twisted K-theory -QUESTION [15 upvotes]: I have a question about the Proof of Bott Periodicity in twisted K-theory -by Atiyah and Segal in their paper Twisted K-theory. -Following their notation, to prove Bott periodicity in this context it is enough -to provide a $U(H)$-equivariant homotopy equivalence -$$ -Fred^{(0)}(H)\to \Omega^{2}Fred^{0}(H). -$$ -One may assume that all the spaces in sight have the norm topology for simplicity. -This is done in two steps. -Step 1. Take $S_{n}$ an irreducible graded module for -the complexified Clifford algebra $C_{n}$. Then for $n$ even, tensoring with $S_{n}$ gives an isomorphism -$$ -Fred^{0}(H)\to Fred^{n}(S_{n}\otimes H). -$$ -This map is clearly $U(H)$-equivariant. -Step 2.There is a map -$$ -Fred^{n}(S_{n}\otimes H)\to \Omega^{n}Fred^{0}(S_{n}\otimes H). -$$ -which was constructed explicitly by Atiyah and Singer and it is easy to see that it is a $U(H)$-equivariant homotopy equivalence. -However, one would like to get back to $\Omega^{n}Fred^{0}(H)$. The spaces -$$ -\Omega^{n}Fred^{0}(S_{n}\otimes H) \text{ and } \Omega^{n}Fred^{0}(H) -$$ -are homotopy equivalent but all the maps I seem to be able to construct don't preserve $U(H)$-equivariance and this is taken as granted in the proof by Atiyah and Segal. -Can anyone tell me what I am missing? - -REPLY [10 votes]: Twisted K-theory is just a particular case of K-theory of Banach algebras. Therefore, Bott periodicity is a consequence of general results. See for instance Max Karoubi. Twisted K-theory old and new.<|endoftext|> -TITLE: Minimal primes and zero-divisors -QUESTION [15 upvotes]: The non-zero elements of the minimal prime ideals of a noetherian commutative ring are zero-divisors. -The proof of this fact I know of uses primary decomposition. Are there alternative (e.g. more direct) proofs ? - -REPLY [12 votes]: Let $A$ be your ring and $X=\mathrm{Spec} A$. The minimal primes of $A$ correspond to the irreducible components of $X$. An element of $f\in A$ induces a function $\widehat f:X\to \coprod_{\mathfrak p\in \mathrm{Spec} A}\kappa(\mathfrak p)$ and this function vanishes everywhere if and only if $f\in\mathfrak p$ for all $\mathfrak p\in \mathrm{Spec} A$, that is, when it is nilpotent. For an $f\in A$ contained in a minimal prime the induced function $\widehat f$ vanishes on the irreducible component corresponding to the minimal prime containing $f$. -If $X$ has a single irreducible component, then the functions induced by the elements of $A$ in the corresponding single prime ideal are vanishing everywhere hence they are nilpotent (in particular zero-divisors). -If $X$ has more than one irreducible component, then do the following: Choose $f_1,\dots,f_m$ such that each $f_i$ is contained in exactly one minimal prime ideal and for each minimal prime there is (exactly) one $f_i$ contained in it. This choice ensures that the product of any proper subset of these functions will not vanish on at least one irreducible component and hence it is not nilpotent. (The point is, that if their product vanished on an irreducible component, then one of them would have to, but then it would be the one that vanishes on that component). Now take an arbitrary element $g$ in any of the minimal primes and for simplicity assume it is from the one corresponding to $f_1$. (We allow $g$ to be contained in other primes as well, but that has no consequence). Then the following claim implies that $g$ is a zero-divisor. -Claim Let $g,f_2,\dots,f_t$ be such that $g\cdot f_2\cdots f_t$ is nilpotent, but $f_2\cdots f_t$ is not nilpotent, then $g$ is a zero-divisor. -Proof Let $n$ be the smallest non-negative integer for which there exists a $t-1$-uple -$(n_2,\dots,n_t)\in\mathbb N^{t-1}$ such that $g^n\cdot f_2^{n_2}\cdots f_t^{n_t}=0$. Observe that $n$ exists and $n\geq 1$ by the assumptions. -Then $g\cdot (g^{n-1}\cdot f_2^{n_2}\cdots f_t^{n_t})=0$, but $g^{n-1}\cdot f_2^{n_2}\cdots f_t^{n_t}\neq 0$. $\square$ -Comment Apparently this is essentially the same proof as the one Graham included in the comments, but I can't let go any chance of giving a geometric proof of an algebra question. -Also, it clarifies the unclear step pointed out by Georges. In fact, it seems one needs to do a little yoga to get the result. -In any case algebra=geometry. :)<|endoftext|> -TITLE: Source and context of $\frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 dx/(1+x^2)$? -QUESTION [138 upvotes]: Possibly the most striking proof of Archimedes's inequality $\pi < 22/7$ is an integral formula for the difference: -$$ -\frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 \frac{dx}{1+x^2}, -$$ -where the integrand is manifestly positive. This formula is "well-known" but its origin remains somewhat mysterious. I ask: - -Who discovered this integral, and in what context? - -The earliest reference I know of is Problem A-1 on the 29th Putnam Exam (1968). According to J.H.McKay's report in the American Math. Monthly (Vol.76 (1969) #8, 909-915), the Questions Committee consisted of N.D.Kazarinoff, Leo Moser, and Albert Wilansky. Is one of them the discoverer, and if so which one? -The printed solution, both in the Monthly article and in the book by Klosinski, Alexanderson, and Larson, says only "The standard approach, from elementary calculus, applies. By division, rewrite the integrand as a polynomial plus a rational function with numerator of degree less than 2. The solution follows easily." But surely there's more to be said, because this integral is a minor miracle of mathematics: -$\bullet$ Not only is the integrand manifestly positive, but it is always small: $x-x^2 \in [0,1/4]$ for $x \in [0,1]$, and the denominator $1+x^2$ is at least 1, so $(x-x^2)^4/(x^2+1) < 1/4^4 = 1/256$. A better upper bound on the integral is $\int_0^1 (x-x^2)^4 dx$, which comes to $1/630$ either by direct expansion or by recognizing the Beta integral $B(5,5)=4!^2/9!$. Hence $\frac{22}{7} - \pi < 1/630$, which also yields Archimedes's lower bound $\pi > 3\frac{10}{71}$. -$\bullet$ The "standard approach" explains how to evaluate the integral, but not why the answer is so simple. When we expand -$$ -\frac{(x-x^2)^4}{1+x^2} = x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - \frac4{x^2+1}, -$$ -the coefficient of $x/(x^2+1)$ vanishes, so there's no $\log 2$ term in the integral. [This much I can understand: the numerator $(x-x^2)^4$ takes the same value $(1\pm i)^4 = -4$ at both roots of the denominator $x^2+1$.] When we integrate the polynomial part, we might -expect to combine fractions with denominators of 2, 3, 4, 5, 6, and 7, obtaining a complicated rational number. But only 7 appears: there's no $x$ or $x^3$ term; the $x^4$ coefficient 5 kills the denominator of 5; and the terms $-4x^5-4x^2$ might have contributed denominators of 6 and 3 combine to yield the integer $-2$. -Compare this with the next such integrals -$$ -\int_0^1 (x-x^2)^6 \frac{dx}{1+x^2} = \frac{38429}{13860} - 4 \log 2 -$$ -and -$$ -\int_0^1 (x-x^2)^8 \frac{dx}{1+x^2} = 4\pi - \frac{188684}{15015}, -$$ -which yield better but much more complicated approximations to $\log 2$ and $\pi$... -This suggests a refinement of the "in what context" part of the question: - -Does that integral for $(22/7)-\pi$ generalize to give further approximations to $\pi$ (or $\log 2$ or similar constants) that are useful for the study of Diophantine properties of $\pi$ (or $\log 2$ etc.)? - -REPLY [19 votes]: This integral has a series counterpart -$$\sum_{k=0}^\infty \frac{240}{(4k+5)(4k+6)(4k+7)(4k+9)(4k+10)(4k+11)}=\frac{22}{7}-\pi$$ -https://math.stackexchange.com/a/1657416/134791 -(UPDATE Peter Bala New series for old functions https://oeis.org/A002117/a002117.pdf, 2009, formula 5.1) -Equivalently, -$$\sum_{k=1}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\frac{22}{7}-\pi$$ -which may be seen as the first truncation of -$$\sum_{k=0}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\frac{10}{3}-\pi$$ -Therefore, this series shows the following path to $\frac{22}{7}$ -$$\frac{10}{3}-\frac{240}{1·2·3·5·6·7}=\frac{10}{3}-\frac{2·5!}{\frac{7!}{4}}=\frac{10}{3}-\frac{8·5!}{5!6·7}=\frac{1}{3}\left(10-\frac{4}{7}\right)=\frac{1}{3}·\frac{66}{7}=\frac{22}{7}$$ -also illustrating how only $7$ remains.<|endoftext|> -TITLE: When is the pushforward of a vector bundle still a vector bundle? -QUESTION [24 upvotes]: Let $X$ and $Y$ be varieties. Let $E$ be a locally free sheaf over $X$. Let $f: X \to Y$. Is there some nice criteria which ensures that $f_\ast E$ is still locally free? Sorry, if this is a very standard question. - -REPLY [27 votes]: Under reasonable hypotheses on $X$, $Y$ and $f$, the answer is that $f_*E$ is locally free if and only if $\dim H^0(X_y, \, E_y)$ is a constant function, where -$$X_y:=f^{-1}(y), \quad E_y:=E|_{X_y}.$$ -More precisely, there is the following result, whose proof can be found in [Mumford, Abelian Varieties, Chapter II]: - -Theorem (Base Change). Let $f \colon X \to Y$ be a proper morphism of Noetherian schemes, with $Y$ reduced and connected, and $E$ a coherent sheaf on $X$, flat over $Y$. Then for all integers $p \geq 0$ the following conditions are equivalent: -$\boldsymbol{(i)}$ $y \to \dim H^p(X_y, E_y)$ is a constant function; -$\boldsymbol{(ii)}$ $F:=R^pf_*E$ is a locally free sheaf on $Y$ and, for all $y \in Y$, the natural map -$$F \otimes_{\mathcal{O}_Y} k(y) \to H^p(X_y, E_y)$$ -is an isomorphism. - -For instance, if $f$ is a finite map and $E$ is a line bundle we obtain that $f_*E$ is a vector bundle with $\operatorname{rank} f_*E=\deg f$, whereas $R^p f_*E=0$ for $p >0$. - -REPLY [7 votes]: Check out Grauert's Theorem, in Hartshorne III.12.<|endoftext|> -TITLE: Torsion - subgroup and quotient -QUESTION [9 upvotes]: Let $G$ be a finitely generated pro-$p$ group. -Let $T$ be the set of all torsion elements in $G$. - - -Is it possible for $T$ to be a non-closed subgroup? - - -Anyway, - - -Can $G/\overline{\langle T\rangle}$ have torsion? - - -For any finitely generated abelian (more generally, powerful) pro-$p$ group $G$, I know that 1. and 2. have negative answers: $T$ is finite and $G/T$ is torsion-free. -Thanks. - -REPLY [2 votes]: It seems to me that the answer on both questions is Yes. -Let $F$ be a finitely generated free non-abelian pro-$p$ group, $N=\overline{F^{\prime\prime}}$ and $K=\overline {[N,F]}$. Then $N/K$ is torsion free and of infinite rank. -Let $a_1,a_2,\ldots$ be a free $\mathbb Z_p$-generating set of $N/K$. Put $G_1=F/\langle K, a_1^{p},a_2^{p^2},\ldots\rangle$. Let $a=\prod a_i$ and let $\bar a$ be its image in $G_1$. Then $\bar a$ is not a torsion element but lies in the clousure of the subgroup of torsion elements. This answers the first question. -Let $G_2$ be the quotient of $G_1\times \langle z\rangle $ by the subgroup generated by $z^p\bar a$. Then the image of $z$ in $G_2$ is not a torsion element.<|endoftext|> -TITLE: Signed factors of harmonic polynomials -QUESTION [10 upvotes]: Let ${\rm Harm}_n^d$ be the space of real harmonic polynomials in $n$ variables, homogeneous of degree $d$. If $P\in{\rm Harm}_n^d$, then -$$\left(\frac{\partial^2}{\partial x_1^2}+\cdots+\frac{\partial^2}{\partial x_n^2}\right)P=0.$$ -A harmonic polynomial is not necessarily irreducible in ${\mathbb R}[X_1,\ldots,X_n]$. For instance every non-zero $P\in{\rm Harm}_2^4$ splits as the product of two quadratic forms; it turns out that none of them is positive definite. Besides, it is not two difficult to show that if $X_1^2+\cdots+X_n^2$ divides a harmonic polynomial $P$, then $P=0$. These observations lead me to me following question: - -Is it possible that a non-zero harmonic polynomial (say homogeneous) factorizes $P=QR$ in ${\mathbb R}[X_1,\ldots,X_n]$, with the factor $Q$ being non-constant and positive definite (i.e. $Q(x)>0$ for every $x\ne0$) ? - -I incline toward a negative answer, of course. - -REPLY [5 votes]: S. Kharlamov pointed out to me that it is a consequence of the diagonalization of the Laplacian $\Delta_S$ over the unit sphere. Its eigenvalues are the integers $\lambda_d=d(d+n-2)$, and the corresponding eigenspace $E_d$ is given by the trace over $S$ of the harmonic polynomials of degree $d$. Finally, the space of traces of polynomials of degree $\le d$ is the sum of the $E_{d-2k}$ for $k=0,\ldots,[d/2]$. This implies that -${\rm Harm}_n^d$ is orthogonal to ${\rm Hom}_n^{d-2k}$, the space of homogeneous polynomials of degree $d-2k$, whenever $k=1,\cdots,[d/2]$. The orthogonality refers to the $L^2$-scalar product over $S$: -$$(f,g):=\int_Sf(x)g(x)d\sigma(x).$$ -Suppose now that $P$ is harmonic of degree $d$ and it splits as $QR$, with $Q$ positive definite and non constant. Then $Q$ has degree $2k$ for some $k\ge1$. We have seen that $(P,R)=0$, which means -$$\int_SQ(x)R(x)^2d\sigma(x)=0.$$ -Because $Q$ has a constant sign, this implies $QR^2\equiv0$, that is $P=0$. -Of course, instead of integrating over the unit sphere, one may integrate against the standard Gaussian measure. A classical trick in several topics, including random matrices theory.<|endoftext|> -TITLE: Is there any sequence $a_n$ of nonnegative numbers for which $\sum_{n \geq 1}a_n^2 <\infty$ and $\sum_{n \geq 1}\left(\sum_{k \geq 1}\frac{a_{kn}}{k}\right)^2=\infty$? -QUESTION [25 upvotes]: Is there any sequence $a_n$ of nonnegative numbers for which $\displaystyle\sum_{n \geq 1}a_n^2 <\infty$ and -$$\sum_{n \geq 1}\left(\sum_{k \geq 1}\frac{a_{kn}}{k}\right)^2=\infty\quad?$$ -See also https://math.stackexchange.com/questions/42624/double-sum-miklos-schweitzer-2010 - -REPLY [5 votes]: In the recent issue of Matematikai Lapok, the report on the Schweitzer contest attributes the problem to Zoltan Buczolich and Julien Bremont.<|endoftext|> -TITLE: Ricci curvature of the symplectic group -QUESTION [6 upvotes]: Is the Ricci curvature of the compact symplectic group $Sp(n)$ bounded below by $cn$ for some constant $c > 0$ independent of $n$? -For $O(n)$ and $U(n)$ I know many references which state such a bound on Ricci curvature, although none of them actually include complete proofs. (Pointers to such proofs for $O(n)$ and $U(n)$ would also be appreciated.) The usual practice for these groups is merely to refer to Cheeger and Ebin's book, which develops enough general theory of curvature of Lie groups that carrying out the calculations is presumably a straightforward exercise, for those who are on top of such things. (As far as I can see, Cheeger and Ebin don't even state the results in these particular cases.) It's been about ten years since I've been up on such things, which is why I'm hoping someone here knows the answer instead of just trying to work it out myself. - -REPLY [7 votes]: The groups $SO(n)$, $SU(n)$, and $Sp(n)$ all have Ricci tensor equal to a constant times the metric tensor. (Note: contrary to what I wrote in the question and what one may find stated in several places in the literature, this is false for $U(n)$. This is easy to see from Claudio's answer: if a Lie group has nontrivial center then its Ricci tensor cannot be nondegenerate.) -With the normalization induced by the standard embedding in $\mathbb{R}^{\beta n^2}$ (where $\beta = 1,2,4$ in the three cases above, respectively), the constant is -$$ -\frac{\beta(n+2)}{4} - 1. -$$ -Reference: Appendix F of An Introduction to Random Matrices by Anderson, Guionnet, and Zeitouni.<|endoftext|> -TITLE: Nice example of a topologically trivial bundle with nontrivial connection -QUESTION [13 upvotes]: So, I've been trying to understand what exactly an anomaly is, and how they arise in physics. Apparently an anomalous theory is some theory whose action is given by a section of some bundle (rather than a function). Hence, only if the bundle is topologically trivial, thus allowing one to write the action as a function, can we then integrate the action over the moduli space; "giving" the quantum theory. Now, there is a paper by Freed "Determinants, Torsion and Strings" where he calls this a global anomaly (perhaps first coined by Witten, not sure), and goes on to say that there are also local anomalies due to the fact that a bundle can be topologically trivial without having a nontrivial connection. So, I have a question: -(1) What's a nice (nontrivial) example of a trivial bundle with nontrivial connection? - -REPLY [9 votes]: Let $M = {\mathbb R}^1$ be the time-line, with the trivialized bundle $M \times {\mathbb R}^1$ on it. There's a connection on it, called "inflation", such that parallel transport of $N$ at time $t$ to time $t'$ says how much $N$ dollars in year $t$ would be worth in year $t'$. This example comes from the frankly hilarious The Physics of Finance.<|endoftext|> -TITLE: Where does the primary obstruction of a fibration show up in its spectral sequence? -QUESTION [15 upvotes]: Let $f\colon\thinspace E\to B$ be a Serre fibration whose fibre $F$ is $k-1$-connected, $k\geq 1$. Assume $B$ is a connected CW complex. Then the primary obstruction to the existence of a cross section of $f$ is defined; it is a cohomology class $$\mathfrak{o}(f)\in H^{k+1}(B;\tilde{H}_{k}(F)).$$ -Here the coefficients may be twisted by $\pi_1(B)$. The definition involves choosing a section on the $k$-skeleton which you then try to extend, but the class itself is canonical (depends only on the fibration). -Meanwhile, there is the cohomology Leray-Serre spectral sequence of the fibration, with $$E_2^{p,q}=H^p(B;H^q(F))\implies H^*(E),$$ where again the coefficients in the $E_2$ term may be twisted by the action of $\pi_1(B)$. -Here is my question, which I'm a little embarrassed to ask: - -Is there a canonical class in the $E_2$ term which relates somehow to $\mathfrak{o}(f)$? - -Sorry for being (intentionally) vague. -Edit: As Grigory M points out in his answer, if we work over a field and assume the local system on the base formed from the homology of the fibres is trivial, then the first non-trivial differential $$d_{k+1}\in \mathrm{Hom}(H^k(F),H^{k+1}(B))$$ -is the linear dual of an element $$d_{k+1}^\ast\in\mathrm{Hom}(H_{k+1}(B),H_k(F))\cong H^{k+1}(B;H_{k}(F))$$ which should equal the obstruction class. - -Has anyone seen a reference for this? -Can anyone give a more general statement when the local coefficient system is non-trivial? - -Thanks. - -REPLY [9 votes]: In the general case of integral coefficients and possibly non-trivial local coefficient system, let $\pi=\pi_1(B)$. -A cocycle for the obstruction class is an element $o\in Hom_{\mathbb Z\pi}(C_{k+1}\tilde{B},\pi_kF)$. Now $o$ induces a group homomorphism -$$ H^0\left(B;{H^k(F)}\right)\cong Hom_{\mathbb Z}(\pi_kF,\mathbb Z)^\pi \cong Hom_{\mathbb Z\pi}(\pi_kF,\mathbb Z)\to Hom_{\mathbb Z\pi}(C_{k+1}\tilde{B},\mathbb Z).$$ -Here we use the universal coefficient theorem and the fact that $H^0(B,M)=M^\pi$ for connected $B$. -Since $o$ is a cocycle, the group homomorphism factors through cocycles $C^{k+1}(B;\mathbb Z)$, so, since $H^0(F)=\mathbb Z$ there is an induced map $$ H^0\left(B;{H^k(F)}\right) \to H^{k+1}(B;H^0(F))$$ -which is the $d_{k+1}$-differential $$E_{2}^{0,k}= E_{k+1}^{0,k}\to E_{k+1}^{k+1,0}=E_2^{k+1,0}$$ in the spectral sequence. -Or to put it differently, there is a map -$$H^{k+1}(B,\pi_k F) \to Hom(H^0(B;H^k(F)),H^{k+1}(B;H^0(F)))$$ -adjoint to the composition of cup product and Hurewicz and Kronecker maps -$$H^{k+1}(B,\pi_k F) \otimes H^0(B;H^k(F)) \to H^{k+1}(B;\pi_k F \otimes H^k(F))\to H^{k+1}(B;\mathbb Z)$$ which sends the obstruction class to the $d_{k+1}$-differential. -Moreover if we consider the spectral sequence $H^*(B;H^*(F;\pi_k(F)))\to H^*(E;\pi_k(F))$ the connection is even more direct: then $H^0(B;H^k(F;\pi_k(F))$ contains the distinguished element corresponding to the identity of $\pi_k F$ under the universal coefficient theorem, whose image under $d_{k+1}$ is the obstruction class in $H^{k+1}(B;H^0(F;\pi_k(F))$.<|endoftext|> -TITLE: A "meta-mathematical principle" of MacPherson -QUESTION [27 upvotes]: In an appendix to his notes on intersection homology and perverse sheaves, MacPherson writes - -Why do we want to consider only spaces $V$ that admit a decomposition into manifolds? The intuitive answer is found by considering the group of all self homeomorphisms of $V$. Certainly if $V$ is to be of "finite topological type", then this group should have finitely many orbits. It is these orbits that should be the natural strata of $V$. That the orbits of this group should be manifolds results from the meta-mathematical pricinple that a space of "finite topological type" whose group of self-homeomorphisms acts transitively must be a manifold. I don't know a precise mathematical statement that realizes this meta-mathematical principle, but I expect that there is one. - -As these notes are dated from 1990, I was wondering if the past twenty years have seen any work done towards a precise formulation of this meta-mathematical principle. - -REPLY [13 votes]: Let me point out that this appendix in MacPherson's notes is explicitly mentioned in a wonderful and highly accessible book by the model theorist Lou van den Dries, Tame Topology and O-minimal Structures (page 8). Indeed, the entire corpus of o-minimal geometry can be viewed as giving a precise response to the frequently expressed desire, perhaps most eloquently enunciated in Grothendieck's Esquisse d'un Programme, to put the sort of "tame topology" that MacPherson is pointing to on firm theoretical ground. -Where MacPherson says that only finitely many data are required to define a finite topological type (FTT), he says he means subsets of manifolds -- probably we can assume the manifolds are Euclidean spaces $\mathbb{R}^n$ without any real loss of generality -- and a reasonable guess is that he means the data are specified by finitely many conditions, for example a subset carved out by finitely many equalities and inequalities involving some basic staple functions like polynomials should qualify as an FTT. Which functions can be admitted is presumably open to discussion, so long as finite expressions involving them do not lead to things like the Cantor set being "finitely definable", which for the purposes of this discussion will be considered "pathological". -There are a number of formalisms which capture this intuition in one way or another; the best known or most investigated is probably that of o-minimal structures (there are also the $\mathcal{X}$-sets of Shiota, among others). Rather than spell out the precise definition, let me roughly describe an o-minimal structure as consisting of subsets of $\mathbb{R}^n$ (where $n = 0, 1, 2, \ldots$) which - -Are closed under all first-order logical operations: unions, intersections, relative complements, cartesian products, and closed under taking direct images along coordinate projections $\mathbb{R}^{n+1} \to \mathbb{R}^n$ (in order to accommodate existential quantification); -Include the equality and inequality relations $\{(x, y) \in \mathbb{R}^2: x = y\}$ and $\{(x, y) \in \mathbb{R}^2: x < y\}$; -Do not include any subsets of $\mathbb{R}^1$ except for arbitrary finite unions of points and intervals $(a, b)$, allowing $a = -\infty$ and $b = \infty$. In other words, that contain only those subsets of $\mathbb{R}$ which have to be there by the preceding axioms, and no more. This is the famous order-minimality or o-minimality condition; it rules out for example the set of natural numbers as belonging to an o-minimal structure. - -Regarding this last axiom: logicians know how to exploit the arithmetic of natural numbers to define all sorts of chaotic and pathological structures. For example, if graphs of polynomial functions are admitted, and if $\mathbb{N} \subseteq \mathbb{R}$ is also admitted, then chaos ensues: a logician can write down some complicated finite formula in these predicates to define any Borel set you jolly well please, or for that matter any set in the projective hierarchy. Thus, the o-minimality axiom is there to ensure a level of respectable tameness. -The archetypal example of an o-minimal structure is the class of semi-algebraic sets, but many others are known. In any o-minimal structure, the "definable sets" (i.e., the sets belonging to the structure) admit Whitney stratifications into definable manifolds, and quite a rich theory has been developed; I refer to the book by van den Dries for details. -It would be very interesting if model theorists had looked into this business of the Bing-Borsuk conjecture; it could be that adding in o-minimality hypotheses would help address technical difficulties people have experienced in trying to prove this (but I am hardly qualified to say anything about this). A quick Google search didn't turn up much that I could see, but a paper by Frank Quinn in these proceedings seemed to touch ever so briefly on such issues. -Come to think of it, there are some resident experts on o-minimal theory here at MO (Thierry Zell and Dave Marker come to mind), and it would be wonderful if they could weigh in here.<|endoftext|> -TITLE: First Quantization is a mystery... but de-quantizing perhaps not -QUESTION [6 upvotes]: There is an well-known infamous DICTUM: --Second Quantization is a functor, First Quantization is a mystery-. -Indeed, second quantization is the "Fock functor", which builds the Fock space in a canonical way out of the Hilbert space of a single particle. -But, what about first quantization? There is probably no hope to canonically associate an Hilbert space to the manifold of states of a classical particle (mathematically, there seem to be an inherent element of choice as far as turning functions into operators). -However, there is (I suspect) some functorial description for going the other way around, FROM the quantum scenario INTO the classical one (corresponding to the limit $h\rightarrow 0$). If this is true, there maybe a "fiber" of candidate quantum descriptions, all collapsing into the same classical one. -Any place where this has been worked out clearly? - -REPLY [5 votes]: You may be interested in the answers to a question on physics.SE, "Quantum Mechanics on Manifold". -The gist of it is that there is a plethora of different quantization schemes (canonical quantization, geometric quantization, ...). As you note, they all have the same classical theory as a limit. Overview: -Quantization Methods: A Guide for Physicists and Analysts - -That said, speaking as a physicist (caveat emptor), I think the paper -http://link.aps.org/doi/10.1103/PhysRevA.23.1982 -pretty much covers everything that is of practical relevance. Unsurprisingly, the Schrödinger equation on an embedded manifold $M \subseteq \mathbb{R}^3$ depends not only on intrinsic data, like the metric or curvature, but also on extrinsic data about the embedding of the manifold inside 3D space. That's to be expected because an electron can tunnel through the full 3D space and hence take "extrinsic short-cuts". I think that's also the reason why there is so much ambiguity in quantization schemes: it's simply not something that you can do completely intrinsically.<|endoftext|> -TITLE: Equality of Cardinality of Power Set -QUESTION [12 upvotes]: For two sets A and B. Suppose|2^A| = |2^B| (cardinality of power sets of A and B), does |A|=|B| ? -(It is easy to see that|A|=|B| if we assume generalized continuity hypothesis. Do we have the same result without it?) - -REPLY [26 votes]: The answer is that if the axioms of set theory are consistent, then you cannot prove that conclusion. Although it seems very reasonable to expect that a smaller -set must have strictly fewer subsets, which is another way -of stating your property, in fact this property is -independent of ZFC. -(The fact that many people find this surprising is the reason I posted this -answer -to the MO question requesting examples of -reasonable-sounding statements that are independent of -ZFC.) -You are asking whether the continuum function -$\kappa\mapsto 2^\kappa$ is injective, and it turns out that if ZFC is consistent, then this assertion is neither provable nor refutable in ZFC. -On the one hand, the property is relatively consistent with -ZFC, as you observe, since it follows easily from the GCH. -On the other hand, it is known by the method of forcing to be relatively consistent that the property fails. Specifically, in Cohen's original model of -$\text{ZFC}+\neg\text{CH}$, he -forced over a model of GCH to add $\omega_2$ many Cohen -reals, and the result is $2^\omega=2^{\omega_1}=\omega_2$ in his model, -which violates your property. Cohen's model has an uncountable set $B\subset\mathbb{R}$ that has the same number of subsets as the countable set $A=\mathbb{N}$. This is usually one of the -first nontrivial forcing arguments that set-theorists -learn, when first exposed to the technique, and when teaching this, I invariably find the situation somewhat magical. -It is also a consequence of Martin's -Axiom that -$2^\kappa=2^\omega$ for all $\kappa\lt\frak{c}$, and if one -has MA plus $\neg\text{CH}$, which is known to be -relatively consistent by the forcing method, then there are -again counterexamples to the requested property. -Meanwhile, one can show by forcing that the injectivity of -the continuum function is not equivalent to GCH, since by -Easton's -theorem, -one can find a forcing extension (of any model of GCH) in -which $2^{\aleph_n}=\aleph_{n+2}$ for every natural number -$n$, and otherwise the GCH holds. Such a model exhibits the -desired injectivity property, but does not satisfy GCH. One -can use Easton's theorem more generally to make even more -extravagant violations of GCH, while still ensuring an -injective continuum function.<|endoftext|> -TITLE: Can proper-smooth base change be used to show that varieties cannot be lifted to characteristic zero? -QUESTION [8 upvotes]: Recall the following corollary to the proper and smooth base change theorems: - -Let $\pi: X \to S$ be a proper, smooth morphism. Then the direct images $R^i \pi_* \mathcal{F}$ are locally constant constructible for any l.c.c. sheaf $\mathcal{F}$ (with torsion prime to the characteristic of the residue fields) on $X_{et}$. - -It follows in particular that if $S$ is the Spec of a DVR (say with an algebraically closed residue field), then the cohomology of the generic fiber (at least base-changed to a separable closure) is the same as that of the specific fiber. Consider the case where $S$ has unequal characteristic --- then essentially, this means that the special fiber $X_0$ admits a (smooth) lifting to characteristic zero. Then the above observations says something about what the cohomology of $X_0$ has to be (by comparing to the cohomology of the generic fiber, which is also the complex cohomology). -Can this be used to show that schemes in characteristic $p$ aren't liftable to characteristic zero? I don't know if this is easy, because etale cohomology groups seem to satisfy many of the same properties (e.g. Poincare duality, dimensional vanishing) of complex cohomology. Milne's book seems to list this as an application of proper-smooth base change but does not actually give an example. - -REPLY [12 votes]: There is an example due to Hirokado of a Calabi-Yau threefold in characteristic 3 with third Betti number zero which implies that it cannot be lifted to characteristic zero. See: Hirokado, Masayuki - -A non-liftable Calabi-Yau threefold in characteristic 3. -Tohoku Math. J. (2) 51 (1999), no. 4, 479–487.<|endoftext|> -TITLE: Euler characteristic and generating series -QUESTION [11 upvotes]: This question was inspired by the question posed by John Baez here: https://mathoverflow.net/questions/67209?sort=votes#sort-top and Neil Strickland's answer to that question. -Let $X$ be a CW complex. If $X$ is finite, there are no problems defining its Euler characteristic whatsoever. However, when $X$ is infinite, there are at least two different ways which give similar but not identical results: -If the number $c(d)$ of cells of $X$ of dimension $d$ is finite for each $d$, one can form a generating function $f(t)=\sum_d c(d)t^d$ and define $\chi(X)=\lim_{t\to -1^+}f(t)$ (assuming the limit makes sense and exists). -If $X$ has a finite $d$-fold cover $\tilde X$ which is homotopy equivalent to a finite CW complex, then we can set $\chi(X)=\frac{1}{d}\sum_{i=1}^\infty (-1)^i\dim H^i(\tilde X,k)$ where $k$ is a field. This is the Wall characteristic (a.k.a. the rational Euler characteristic) of $X$ and it is a homotopy invariant and does not depend on $k$. -Now, if we try to calculate the characteristic of $\mathbb{R} P^\infty$ with its standard cell decomposition (one cell in each dimension from $0$ to $\infty$), then both definitions agree: the first one gives the generating function $\frac{1}{1-t}$, which gives $\frac{1}{2}$, when evaluated at $t=-1$; the second one also gives $\frac{1}{2}$ using the double cover $pt\cong S^\infty\to \mathbb{R} P^\infty$. -However, if we take a finite group $G$ and try to compute the characteristic of the Milnor construction of $BG$, then the first definition will give $\frac{1}{1+\# G}$, since the number of simplices of dimension $d$ is $(\# G)^d$, while the second one gives $\frac{1}{\# G}$. -I would like to ask: is there a conceptual reason the two definitions agree in the first case and almost (but not quite) agree in the second? This is, of course, a rather vague question, so here is a slightly more specific version: if the number of cells of $X$ of each dimension is finite, then is there a way to deduce the Wall characteristic of $X$ from the generating series for those numbers, assuming the Wall characteristic exists? - -REPLY [7 votes]: I hope a topologist provides an answer. I'll make some general comments about sums of series. -Something to keep in mind when working with divergent series is that they do diverge. In particular, divergent sums are not associative. (This is similar to the fact that conditionally convergent sums are not commutative, as is standardly explained in freshman calculus.) Perhaps the most famous manifestation of the non-associativity is the "proof" that $0=1$ by writing $0 = \sum 0 = \sum (1-1) = \dots$. But a better example, because it's more illustrative, is the proof that $\frac12 = \frac13$: as you point out in the case of $\mathbb R P^\infty$, one should predict that $1 - 1 + 1 - 1 + \dots = \frac12$, because if you translate the sum by one step and then add the translated series to the original series in columns, all terms cancel except the first one. But if you instead consider $1 - 1 + 0 + 1 - 1 + 0 + \dots$, you see that you need three shifts the sum to $1$. Put another way, the Abel sum of $1 - 1 + 0 + 1 - 1 + 0 + \dots$ is $\sum t^{3n} - \sum t^{3n+1} = \frac1{1-t^3} - \frac t{1-t^3} = \frac1{1+t+t^2} \to \frac13$. -I think you're finding some similar behavior in the case of the Milnor $\mathrm BG$. Of course, the "correct" Euler characteristic is $1/\chi(G)$. What I think needs to happen is that the "new" simplices when you construct $\mathrm B G$ should have the even spacing in the sum, the same as they have in the reduced bar construction. But the degenerate simplices in Milnor's cell complex should be associated in such a way as to cancel out like in $(1-1) + (1-1) + \dots$. -Of course, this is a bit ad hoc. A possibility is that the dimensions of the homology groups do have the correct Abel sum? That would be the answer if you believe me that only the nondegenerate simplices should contribute. But I haven't thought whether this is correct on the nose.<|endoftext|> -TITLE: Is any K3 surface of degree 8 in P^5 the complete intersection of quadrics? -QUESTION [5 upvotes]: Here the base field is the field of complex numbers. - -REPLY [11 votes]: Any $K3$ surface $S$ of degree $8$ in $\mathbb{P}^5$ is contained in $3$ linearly independent quadrics. It can be seen that in the general case $S$ is a complete intersection of $3$ quadrics. However, there are some special cases where this is not true, but they can be completely described. -The point is that $S$ is a complete intersection of $3$ quadrics if and only if a general hyperplane section (which is a smooth curve of genus $5$) has no $g_3^1$. -You can find all details in [Beauville, Complex Algebraic Surfaces, Chapter VIII, Example 14 and Exercise 11].<|endoftext|> -TITLE: Concentration of measure for arbitrary convex bodies? -QUESTION [9 upvotes]: There are various "concentration-of-measure" theorems, -the best known that due to Lévy, -which is this (informally): the volume of a sphere $S^d$ in $d$ dimensions is largely -concentrated around an $\epsilon$-tubular neighborhood of an equitorial hyperplane $H$. -Here the tubular neighborhood is the set of points within distance $\epsilon$ of $H$. -I believe there is an analogous theorem for centro-symmetric convex bodies -(although I have no reference for this). -My question is: - -Is there a concentration-of-measure theorem for arbitrary convex bodies $K$, - something along these lines: There exists a hyperplane $H$ such that - most of the volume of $K$ is concentrated - in a tubular neighborhood of $H$? - -Or are there convex bodies so "skewed" that they resist any such section? -Likely this is known, in which case a reference would suffice. Thanks! - -REPLY [11 votes]: There are many results, and an active research industry, along these lines. In general the Euclidean ball is the best-behaved convex body in this respect, and just how similar an arbitrary convex body is depends on how you try to make your question more precise. -Here's one very general result, a special case of what is sometimes called Borell's lemma: - -Let $K \subset \mathbb{R}^n$ be a convex body with volume 1 and let $A \subset \mathbb{R}^n$ be a convex symmetric set such that $vol(K\cap A) \ge 3/4$. Then for $r > 1$, - $$ vol(K \setminus r A) \le \frac{3}{4} 3^{-(r+1)/2}. -$$ - -Thus in particular the volume of $K$ far from a hyperplane decays exponentially. -Much sharper results exist. I mentioned a couple recent references in a comment on Gil's answer. For much more, I suggest you start by perusing the papers of Grigoris Paouris.<|endoftext|> -TITLE: a paradoxical decomposition of a group -QUESTION [6 upvotes]: Just my curiosity... Are there proofs the following fact, which does not involve Hall's matching theorem: -A group $\Gamma$ is amenable if and only if it does not admit a paradoxical decomposition. -Def: A group $\Gamma$ has a paradoxical decomposition if there are pairwise disjoint subsets $F_1,\ldots, F_n$, $E_1,\ldots, E_m$ of $\Gamma$, and elements $g_1,\ldots g_n$, $h_1,\ldots,h_m\in \Gamma$ such that $\Gamma$ can be expressed as -$$\Gamma= \bigsqcup_{i} g_i F_i= \bigsqcup_{j} h_j E_j$$ - -REPLY [3 votes]: According to a note in Grigorchuk's and Sunic's Self-Similarity and Branching in Group Theory, there is a proof not using the matching theorem in the book The Banch-Tarski Paradox by Stan Wagon. -By the way, you have to mention that $\Gamma$ is also the union of $E_1,...,E_m,F_1,...,F_n$.<|endoftext|> -TITLE: Compact Quantum Groups from Hopf Algebras -QUESTION [7 upvotes]: For a compact quantum group $C_q[G]$, it was shown by Woronowicz that $C_q[G]$ contains a dense Hopf algebra generalising the algebra of representations of $G$. I am interested in the other way around, ie given a Hopf algebra $H$ (say a Drinfeld--Jimbo algebra if it makes things easier) can it always be completed to give a compact quantum group? If so, is this completion unique in anyway, or are there many ways to get a cqg from a Hopf algebra? - -REPLY [7 votes]: A little side remark, concerning the last part of your question: Marc Rosso describes in -Marc Ross, Algèbres enveloppantes quantifiées, groupes quantiques compacts de matrices et calcul différentiel non commutatif, Duke Math. J. Volume 61, Number 1 (1990), 11-40. -another way to get compact quantum groups from Hopf algebras. He defines inner products on the representation spaces of the deformed enveloping algebras of simple Lie groups (Drinfeld and Jimbo's $U_q(g)$'s), such that they form a concrete monoidal $W^*$ category. Applying Woronowicz' Tannaka-Krein duality he can recover compact quantum groups from this: $C(G_q)$, a deformation of the algebra of continuous functions on the simple Lie group $G$. The Hopf algebras $U_q(g)$ do not sit inside $C(G_q)$, but their restricted duals can be identified with the dense Hopf *-algebra contained in $C(G_q)$.<|endoftext|> -TITLE: Hecke algebra generated by a single element -QUESTION [8 upvotes]: Let $\mathbb{T}_{\mathbb{Z}}$ be a $\mathbb{Z}$-module -generated by Hecke operators $T_n$ acting on the space of cups forms $S_{k}(\Gamma,\mathbb{C})$ for the congruent subgroup satisfying $\Gamma_1(N)\subset\Gamma$ for some positive $N$. We know that the $\mathbb{Z}$-module $\mathbb{T}_{\mathbb{Z}}$ is finitely generated and the upper bound for the number of generators is given by the Sturm bound. When we compute some examples like for $\Gamma=\Gamma_0(33)$ and $k=2$ we get $\mathbb{T}_{\mathbb{Z}} = \mathbb{Z}[T_3]/(T_3+1)(T_3^2+T_3+3)$. In all other cases I have computed it always happens that the algebra looks similarly,i.e. $\mathbb{T}_{\mathbb{Z}} = \mathbb{Z}[T_k]/f(T_k)$ for some polynomial with integer coefficients. Is it always the case or are there any know examples that $\mathbb{T}_{\mathbb{Z}}$ as a ring is generated minimally by more than one Hecke operator ? - -REPLY [5 votes]: In certain cases a reason that causes the subring generated by a single Hecke operator $T_\ell$ in $\mathbb{T}_\mathbb{Z}$ to have index greater than one is the existence of exceptional primes for the normalised eigenforms $f\in S_k(\Gamma,\mathbb{C})$ (we shall say that $p$ is exceptional for $f$ if (one of) the $p$-adic Galois representation $\rho_{f,\lambda}$ associated to $f$ has small residual image, i.e., the image of $\bar\rho_{f,\lambda}$ does not contain $SL_2(F)$, for any finite extension $F/F_p$). -The special cases that I have in mind are instances of the inner twist phenomena that Kevin Buzzard has already mentioned in one of his comments, and you can see this also in level one, if I am right. Let me elaborate on that. -Let $p$ be a prime $\equiv 3$ mod $4$, and set $k=(p+1)/2$. Let $h$ be the class number of $Q(\sqrt{-p})$. The space $S_k(\Gamma(1))$ gives rise to $n=(h-1)/2$ distinct systems of mod $p$ Hecke eigenvalues so that the associated mod $p$ Galois representations $\rho_1,\ldots,\rho_n$ (which are unramified outside $p$) have dihedral image. This implies that if $\ell$ is a prime that is not a quadratic residue mod $p$, then the trace of $\rho_i$ at a Frobenius element at $\ell$ is zero. -In particular, if $n>1$ (i.e. $h>3$) then there are at least two DISTINCT systems of mod $p$ Hecke eigenvalues $(a_q)$ and $(b_q)$ such that their $\ell$-th members are equal. This implies that the integral Hecke ring $\mathbb{T}$ has more ring homomorphisms valued in $\bar{F_p}$ than the subring generated by $T_\ell$ alone does. In this case one can show that $p$ divides the index of the latter in the former. -With the previous argument you see that for almost all primes $p\equiv 3$ mod $4$ in weight $(p+1)/2$ half of the primes $\ell$ are such that $T_\ell$ does not generate $\mathbb{T}$. But what about the other half of the primes $\ell$? -By relating the traces of $\rho_i$ at Frobenius elements over primes $\ell$ that are split in $Q(\sqrt{-p})$ to the characters of the class group of $Q(\sqrt{-p})$ valued in $\bar{F_p}$ we get the following: -PROPOSITION: Let $A$ be the class group of $Q(\sqrt{-p})$. Assume that for every $a\in A$ there exists a pair of non trivial characters $\chi_i:A\rightarrow\bar{F_p}^*$, with -$\chi_1\neq\chi_2$ and $\chi_1\neq \chi_2^{-1}$, such that $\chi_1(a)+\chi_1(a^{-1}) =\chi_2(a)+\chi_2(a^{-1})$. Then the integral Hecke ring $\mathbb{T}$ in weight $(p+1)/2$ and level one cannot be generated by a single $T_\ell$, for $\ell$ prime. -Few remarks: 1) The above does not say anything about the possibility of having a $T_n$ (with $n$ not prime) generating $\mathbb{T}$. In fact I think we cannot rule this out a priori. -2) Probably a little more group theoretic work can be done to reformulate the condition on the mod $p$ characters of $A$, and turn it into something nicer. I think if $A$ is not cyclic and $3$ does not divide its order then the assumption is satisfied. -3) Notice that we do not need to worry about $T_p$ generating $\mathbb{T}$ when $n>1$. In fact, since $P$ splits in $H/Q({\sqrt{-p}})$, where $H$ is the Hilbert Class Field of $Q(\sqrt{-p})$, all the $\rho_i$'s are the same locally at $p$ and one can show that the eigenvalue $a_p$ is $1$ for all of them. -An example. The cuspidal, integral Hecke ring of level one and weight $k=246=(491+1)/2$ is so that $p=491$ does not divide its discriminant. However every Hecke operators $T_n$, for $2\leq n\leq 153$ generates a subring of $\mathbb{T}$ of index divisible by $p$. -May be it could be nice to see whether the class group of $Q(\sqrt{491})$ satisfies the assumption of the proposition. I suspect it does! -[EDIT: The assumption of the proposition is NOT satisfied for $p=491$, unless I am wrong to an infinite amount. Therefore my suspicion is not confirmed. The proposition above, as far as I can tell, remains valid, although I have no example of a prime $p\equiv 3$ mod $4$ such that the class group $A$ of $Q(\sqrt{-p})$ satisfies the assumption of the proposition. -There still remains to give an explanation of the fact that $p=491$ seems to divide the index of the subring generated by $T_\ell$ in the integral, cuspidal Hecke ring of level $1$ and weight $246$. The arguments above explain such divisibility for primes $\ell$ that are non-quadratic residue mod $p$. For what concerns the other primes $\ell$, I am tempted to say that there should be a mod $491$ Galois representation arising from weight $246$ and level $1$ that is tamely ramified, reducible at $p$, and non-dihedral (may be with small image?): this in fact would cause the existence of a system of mod $p$ eigenvalues $(a_\ell)$, with $\ell\neq p$, arising from $S_{246} (\Gamma(1))$ such that its quadratic twist $(a_\ell \ell^{(p-1)/2})$ also arises from the same space. -Summarising -PROPOSITION: Let $p\equiv 3$ mod $4$. Assume that -1) the class number of $Q(\sqrt{-p})$ is > 3; -2) there exists a mod $p$ representation $\rho$ of $G_Q$ arising from $S_{(p+1)/2}(\Gamma(1))$ that is tamely ramified and reducible at $p$ and it is non-dihedral; -then for every prime $\ell$, $p$ divides the index of the subring generated by $T_\ell$ in the integral, cuspidal Hecke ring of weight $(p+1)/2$ and level $1$. -When $p=2083$ I learnt from some tables that there exists an odd, $A_5$ extension of $Q$ unramified outside $p$ that gives rise to a representation of the type we want in 2). Since condition 1) is also satisfied we see that the Hecke ring in weight $1044$ and level $1$ cannot be generated by a single $T_\ell$, for $\ell$ prime.]<|endoftext|> -TITLE: Fundamental group of R^2-Q^2 -QUESTION [8 upvotes]: After learning about the fundamental group, and proving that $\mathbb{R}^n$ minus any countable set is path-connected, I started wondering if the fundamental group of $\mathbb{R}^2-\mathbb{Q}^2$ is known. Does anyone know whether or not it is? Or how one might go about determining it? - -REPLY [21 votes]: The space $X = \mathbb{R}^2 - \mathbb{Q}^2$ is not semilocally simply connected, and so in a sense the fundamental group is a poor measure of the homotopy 1-type of $X$. It is an exercise in Hatcher's book Algebraic Topology that this group is uncountable. -Here is an attempt to motivate a different view of the fundamental group: Consider the space $X_{n,m}$ which is the complement in $\mathbb{R}^2$ of all points with coordinates $(p/n,q/m)$ where $n,m$ are positive integers and $q,p$ are integers. This forms a cofiltered diagram $\mathbb{N}^2 \to Top$ whose limit is $X$, and where each space in the diagram is semilocally simply-connected. There is a map $X_{n,m} \to X_{n',m'}$ when $n'|n$ and $m'|m$. The fundamental group of $X_{n,m}$ is free on a countable number of generators (a generator is given by a small circle around each deleted point, and connect this circle to the basepoint, $(\pi,\pi)$, say, by a path), and the maps between them kill off certain generators. Already, you can see how complicated $\pi_1(X)$ is going to be, since it has a countable number of quotients, each of which is free on a countable number of generators (actually, each of them is isomorphic to the free group on generators $\mathbb{Z}\times\mathbb{Z}$). -This system of groups is known as a (strict) progroup, and can be thought of as a formal limit This is usually the 'right' algebraic object to consider when one has a badly behaved space such as this one. It is this object which 'controls' what the covering spaces of $X$ look like, as $X$ doesn't have a universal covering space (which if it did exist, would have fibre $\pi_1(X)$, and other connected covering spaces would be quotients of this one). This gets into the realm of shape theory (wikipedia, nLab), developed to consider spaces with bad local homotopy properties.<|endoftext|> -TITLE: A question about large denumerable ordinal numbers -QUESTION [6 upvotes]: Let S be the set of all ordinal-definable real numbers and let A(S) be the statement that S is -denumerably infinite. If ZFC is consistent, it has been proved (by A. Levy) that it remains -consistent if we adjoin A(S) to it as a new axiom-so let us assume that this has been done. -Now S has a natural well-ordering inherited from the well-ordering of the class of all -ordinal-definable sets. Let O(S) be the ordinal number of this natural well-ordering of S. -My question is, how does the size of O(S)-which must be very large indeed-compare with other -large denumerable ordinal numbers, such as the smallest non-recursive ordinal number (in the -sense of CHURCH/KLEENE) or some of the so-called "admissible" ordinal numbers? - -REPLY [8 votes]: The class HOD consisting of the hereditarily ordinal definable sets is transitive proper class inner model of ZFC, and your assumption A(S) amounts to the assertion that $\mathbb{R}^{\text{HOD}}$, the set of reals of this model, is a countable set in the ambient set-theoretic universe V. The shortest enumeration of this set in HOD, however, has the order type of the continuum of HOD, which we may denote $\frak{c}^{\rm \text{HOD}}$, and this would seem to be (one natural interpretation of) the ordinal to which you are referring. -Although your assumption means that this ordinal is countable in the ambient universe $V$, we may also consider it inside the inner model HOD, where it is simply the continuum. Thus, from the perspective of HOD, the ordinal is very large indeed by the standards that you have mentioned. In particular, HOD thinks that it is much larger than $\omega_1^{ck}$, the first non-computable (non-recursive) ordinal, and also that it is admisssible, and a limit of admissibles and a limit of limits of admissibles, and so on. In particular, it is a cardinal in the constructible universe L, since it is a cardinal in HOD, which contains L. -And once we know that HOD thinks those things, then it follows that our universe V also agrees, since those properties about this ordinal are absolute from HOD to V, since HOD computes the L hierarchy the same as we do in V. -Finally, let me point out that the forcing method allows us enormous flexibility in controlling the precise value of this particular ordinal. Namely, suppose that V is any model of ZFC and that $\delta$ is any cardinal of $V$ with uncountable cofinality and above the continuum. First, by a theorem of Solovay, there is a forcing extension $V[G]$ in which the continuum becomes exactly $\delta$; one simply adds $\delta$ many Cohen reals. Next, by a theorem of MacAloon, there is a further extension $V[G][H]$ not adding any reals in which the reals of $V[G]$ become ordinal-definable in $V[G][H]$; one may do this by forcing to code the individual reals into the GCH pattern high above $\delta$. Finally, we may perform collapse forcing to build a model $V[G][H][K]$ in which $\delta$ becomes countable. If the reals of $V[G]$ were coded into the GCH pattern up high, this last step will preserve those reals as ordinal definable, and it will create no new ordinal definable sets since it is almost homogeneous forcing. Thus, the $\mathbb{R}^{\text{HOD}}$ of $V[G][H][K]$ is the same as $\mathbb{R}^{V[G]}$ and it will have cardinality $\delta$ there. Thus, we have arranged your ordinal to be exactly $\delta$ in $V[G][H][K]$, a model satisfying your assertion $A(S)$. But $\delta$ was an arbitrary cardinal with uncountable cofinality above the continuum in $V$.<|endoftext|> -TITLE: Can we bound the minimal degree of a field extension required to obtain semi-stable reduction -QUESTION [7 upvotes]: Let $K$ be a number field and let $X$ be a smooth projective geometrically connected curve over $K$. -There exists a finite field extension $L/K$ such that $X_L=X\otimes_K L$ has semi-stable reduction, i.e., there exists a semi-stable arithmetic surface $\mathcal{X}$ over the ring of integers $O_L$ with generic fibre $L$-isomorphic to $X_L$. Let $L_m$ be such an extension of minimal degree over $K$. -Question 1. Can we bound $[L_m:K]$ in terms of data depending only on $X$? - -REPLY [7 votes]: For an abelian variety $A/K$, Serre-Tate says that you get semistable reduction if you adjoin enough torsion to $K$. For example, adjoining all of the 3-torsion will suffice. It seems plausible (but I don't know for certain) that if the Jacobian $J$ of your curve $X$ has semistable reduction, then so does $X$. If that's the case, then you can take $L$ to be $K(J[3])$, whose degree is bounded by a function of $\dim(J)=\text{genus}(X)$. - -REPLY [7 votes]: I think the answer to Question 1 is yes. One may use the fact that a curve has semistable reduction iff its Jacobian does and apply Grothendieck's theorem which says that an abelian variety has semistable reduction (over a local field) iff the representation of the intertia group on the Tate module is unipotent (it is always quasi-unipotent). One may ensure the unipotence by requiring the Jacobian variety have level $n$ structure for some $n> 2$; to do this one it suffices to make an extension of the base field of degree at most $|Sp(2g,\mathbb{Z}/n)|$, where $g$ is the dimension of the abelian variety, so the genus of the curve in your case. -(Note that a regular scheme over the ring of integers of a number field remains regular when it is base changed to the ring of integers of its completion at any finite prime, so semistability of a curve is not affected by completion.)<|endoftext|> -TITLE: Which countable linear orders are $\aleph_0$-categorical? -QUESTION [10 upvotes]: The question is: Which countable linear orders are $\aleph_0$-categorical? -I have a bit of progress on this: -Define a discrete tuple to be a set of elements, ordered discretely, such that if $a$ and $b$ are in the tuple, and $c$ is between $a$ and $b$ in the structure, then $c$ is part of the tuple. -Then if there are only finitely many discrete tuples (which may contain beginning or ending points of the structure), each of which has finite length, then it's fairly clear that the structure is $\aleph_0$-categorical (and finitely axiomatizable) with a back-and-forth argument. -If there is a discrete tuple of infinite length, then the structure is not $\aleph_0$-categorical, as there are infinitely many distinct formulas with two free variables (saying there are precisely $n$ elements between $x$ and $y$, for example, are all mutually exclusive and satisfied on the structure). More generally, if there are discrete tuples of arbitarily large finite length, it's still not $\aleph_0$-categorical. -I know there are models which fit neither characterization; for example $\mathbb{Q}\times 2$, with the dictionary ordering, has infinitely many discrete tuples, all with length 2. By a quick back-and-forth argument this is still $\aleph_0$-categorical, so the characterizations aren't inclusive enough. - -REPLY [12 votes]: In the paper below Rosenstein gives a complete characterization of $\aleph_0$-categorcial theories of linear order. -Rosenstein, Joseph G. -$\aleph _0$-categoricity of linear orderings. -Fund. Math. 64 1969 1–5. -You can also find the result, and many other gems, in Rosenstein majestic text on linear orders.<|endoftext|> -TITLE: covering by spherical caps -QUESTION [16 upvotes]: Consider the unit sphere $\mathbb{S}^d.$ Pick now some $\alpha$ (I am thinking of $\alpha \ll 1,$ but I don't know how germane this is). The question is: how many spherical caps of angular radius $\alpha$ are needed to cover $\mathbb{S}^d$ completely? There is an obvious bound coming from dividing the volume of the sphere by the volume of the cap, but I am assuming this is far from sharp. I assume that the coding theorists among us have considered this sort of question at great length... One can consider either fixed $d$ or asymptotic results for large $d.$ - -REPLY [2 votes]: I think that when $d$ is fixed, the first order term when $\alpha\to0$ is given by the case of a suitably renormalized cube of the same dimension $d$. This is what happens when you compute the minimal $L^p$ average distance from a uniform point on a manifold to a set of $N$ points, see Approximation by finitely supported measures. -One important lesson I learned while writing this article is that for this kind of questions, it is worth asking statisticians (see e.g. S. Graf & H. Luschgy Foundations of quantization for probability distributions, Lecture Notes in Mathematics, vol. 1730). Often they work on domains of $\mathbb{R}^d$, but then a cutting method as in the above paper can extend the result to manifolds. -You should get an estimate on the number $N$ of $\alpha$-caps needed to cover $S^d$ of the form $N=\theta(d)\alpha^{-d} + o(\alpha^{-d})$ where $\theta(d)$ is explicit in terms of the volume of $S^d$ and the covering constant of $\mathbb{R}^d$, which is unknown when $d>3$.<|endoftext|> -TITLE: representing tensor-C*-categories in BIM -QUESTION [5 upvotes]: Given a factor M (=von Neumann alg. with center ℂ), let us write BIM for the ⊗-C*-category of M-M-bimodules. - -Which ⊗-C*-categories can one faithfully embed into BIM? - -⓵ Are there necessary conditions for a ⊗-C*-category to be representable in BIM? -⓶ Are there sufficient conditions for a ⊗-C*-category to be representable in BIM? -Comment: -I suspect that a lot is known about ⓶ in relation with the theory of planar algebras... -I specially care about ⓵: are there examples of ⊗-C*-categories that don't embed? - -REPLY [5 votes]: I don't know about necessary conditions, but here are some results concerning sufficient conditions: - -In MR1749868, Hayashi and Yamagami realize amenable $C^*$-tensor categories in the category of bifinite (Jones index) bimodules of the hyperfinite $II_1$-factor.- -In arXiv:0811.1764v4, Stefaan Vaes and Sébastien Falguières showed that "the representation category of any compact group is the [bifinite] bimodule category of a $II_1$-factor," i.e., given a compact group $G$ with representation category $C$, there is a $II_1$-factor $M$ whose category of bifinite bimodules is exactly $C$.- -Recently, Sven Raum and Sébastien Falguières showed that "all finite $C^*$-tensor categories are [bifinite] bimodule categories of a $II_1$-factor," i.e., given a finite $C^*$-tensor category $C$, there is a $II_1$-factor $M$ whose category of bifinite bimodules is exactly $C$. This paper has yet to appear on the arXiv. Here is the link to the conference at which the talk was given.- - -I don't know of such results for type $III$ factors.<|endoftext|> -TITLE: Why was it reasonable to ask what the higher K-groups are? -QUESTION [12 upvotes]: To say I am a novice in $K$-theory is to overstate my experience with the field. I've been reading the various wiki articles so as to have some preparation before jumping in, and I couldn't answer the following question to myself: -I understand that $K$-theory had started with the Grothendieck-Riemann-Roch in mind, and that the only thing that was needed for that purpose from $K$-theory was just to define the Grothendieck Group ($K_0$). Once the idea of the Grothendieck group was established, this was generalized to topological spaces, as well as for other kinds of modules. Then comes the step I don't understand -- it seems that people were then trying to find the higher $K$-groups that would make $K$-theory into a cohomological theory. Milnor came up with Milnor $K$-theory, which I understand from wiki is different from later notions of higher $K$-theory. But why would one leap from the concept of the Grothendieck group to thinking that this construction is the $0^{th}$ step in a cohomological theory? What was the context/motivation for that? - -REPLY [11 votes]: The idea of considering higher K-groups comes from topology, and is due to Atiyah, Bott, and Hirzebruch. Atiyah and Hirzebruch defined topological K theory and observed that Bott periodicity says that $K(X)$ is more or less the same as $K(S^2X)$. This suggested to them defining a generalized cohomology theory of period 2 by using all the groups $K(S^nX)$ (this was the first example of a generalized cohomology theory). Once one realizes that topological $K^0$ can be extended to topological $K^n$, it does not take much imagination to suggest that algebraic $K^0$ also has an extension to algebraic $K^n$. (Of course, finding this extension was much harder than guessing it existed.)<|endoftext|> -TITLE: Sum of products of p-th powers of roots of 1 and monomial symmetric functions -QUESTION [6 upvotes]: Hello mathematicians, -i'm looking for explicit computations of expressions like -$$ - \sum_{\substack{0\leq i,j,k -TITLE: Computing the dimensions of representations in a reducible induced representation -QUESTION [6 upvotes]: This is a question on math.se that got no answers. -1) Is there a relatively general method of computing the dimensions of representations in a reducible induced representation? -An explicit specific example is: $G=Sp_4(\mathbb{F}_q)$, $J$ the antidiagonal $(1,1,-1,-1)$, $P$ the parabolic corresponding to $4=2+2$, $\tau$ a representation of $P$ that comes from a representation of $GL_2(\mathbb{F}_q)$, which will also be denoted $\tau_0$ (the matrix $\[\[g,\*\],[0,\*\]\]$ acts as $\tau_0(g)$). Mackey's irreducibility criterion shows that $Ind_P^G \tau$ is reducible iff the central character $\omega_{\tau_0} =1$. How does one compute the dimensions of the (two) subrepresentations in this case? Note that this specific case is solved: "The characters of the finite symplectic group $Sp(4,q)$", B. Srinivasan, but the computations there are quite elaborate and not in the spirit of the question. -In the above example a short attempt to use Deligne-Lusztig theory seems to fail since the irreducible subrepresentations seem to be geometrically conjugate. Maybe the short attempt is ridiculous. -2) Is there a way to compute the dimensions, or even the characters themselves, using Deligne-Lusztig theory? (i.e. same question only restricted to reductive groups over finite fields) - -REPLY [5 votes]: Certainly for the finite groups of Lie type there is a general technology for treating induced characters from parabolics, based on Deligne-Lusztig theory. Even without getting fully into that story, Lusztig's methods yield (recursively) the degrees of irreducible characters. The 1985 book by Roger Carter Finite Groups of Lie Type develops most of the theory up to that time, though later papers by Lusztig and others refine the results in many directions. See for example Chapter 10 of Carter's book for the decomposition of induced characters in the Hecke algebra spirit. (Some of this was worked out earlier by people including Curtis, Iwahori, Kilmoyer in the important special case of the character induced from the trivial character of a Borel subgroup in the finite group, or parabolic analogues.) -Small rank groups had been treated earlier in some detail, as noted in the question for $Sp_4$, but the general methods lead much farther than the ad hoc methods used in early papers. -It should be emphasized that general methods like Mackey theory for finite groups stop well short of solving these kinds of problems for groups of Lie type. Even the original Deligne-Lusztig paper has to be supplemented by further work on Hecke algebra methods for decomposition of induced characters. But eventually this all becomes a unified theory for these particular families of groups. (A short treatment by Digne-Michel is also given in their LMS Student Text softcover book, but with less concrete detail than Carter gives.)<|endoftext|> -TITLE: First eigenvalue of the Laplacian on Berger spheres -QUESTION [12 upvotes]: Consider the Hopf fibrations $S^1\to S^{2n+1}\to CP^n$ and $S^3\to S^{4n+3}\to HP^n$. These are Riemannian submersions with totally geodesic fibers. Consider now their canonical variations (the so-called Berger spheres), i.e., for each $t>0$, let $S^{2n+1}_t$ (resp $S^{4n+3}_t$) be the Riemannian manifold $S^{2n+1}$ (resp $S^{4n+3}$) endowed with the metric obtained by scaling the round metric by a factor of $t^2$ in the direction of the fibers $S^1$ (resp $S^3$). For each $t>0$, they are the total space of a Riemannian submersion with totally geodesic fibers isometric to $tS^1$ (resp $tS^3$). In this context, the Laplacian $\Delta_t$ of the canonical variation can be related to the original Laplacian $\Delta_1$ in terms of the vertical Laplacian $\Delta_v$, by the formula $$\Delta_t=\Delta_1+(\tfrac{1}{t^2}-1)\Delta_v,$$ see Berard-Bergery and Bourguignon [Illinois Journ of Math, 1982]. The operator $\Delta_v$ acts on functions of the total space and is defined by $$(\Delta_v f)(p)=\Delta_{F_p}(f\vert_{F_p})$$ where $\Delta_{F_p}$ is the Laplacian of the fiber $F_p$. -Since the fibers are totally geodesic, the spectrum of $\Delta_v$ coincides with the spectrum of the Laplacian of the fiber, and all the above operators commute. In particular, $L^2$ of the total space admits a decomposition in simultaneous eigenspaces of $\Delta_t$ and $\Delta_v$. As a consequence, we have the following inclusion of spectra: $$Spec(\Delta_t) \subset Spec(\Delta_1)+(\tfrac{1}{t^2}-1) Spec(\Delta_F).$$ -I am interested in computing $\lambda_1(t)$, the first non-zero eigenvalue of $\Delta_t$, for all $t$. By the above, for all $t>0$, we know $\lambda_1(t)$ is of the form $\mu_k+(\tfrac{1}{t^2}-1)\phi_j$ where $\mu_k\in Spec(\Delta_1)$ is an eigenvalue of the Laplacian on the original total space and $\phi_j\in Spec(\Delta_F)$ is an eigenvalue of the Laplacian of the fiber. The problem is that not all combinations of $\mu_k$'s and $\phi_j$'s give an eigenvalue of $\Delta_t$ and the first non-zero eigenvalue of $\Delta_t$ might be given by different combinations of $\mu_k$'s and $\phi_j$'s for each $t$. Recall that since total space and fibers are spheres, both $\mu_k$'s and $\phi_j$'s are easy to compute, namely the $k$th eigenvalue of the $m$-sphere is $k(k+m-1)$. -In the case of the first family with 1-dim fibers, this computation follows from a paper of Tanno [Tohoku Math Journ, 1979]. The trick is to look at a trajectory of the vector tangent to the $S^1$ fiber (a great circle) and solve the eigenvalue equation there (which becomes an ODE). Using this he proves that the only combinations of $\mu_k$'s and $\phi_j$'s permitted are when $0\leq j\leq k$ and $k-j$ is even. He then also finds explicit eigenfunctions when $k=j$; $j=1$ and $k$ odd; $j=2$ or $j=0$ and $k$ even. With these, it is very easy to compute $\lambda_1(t)$ for this family where the fibers are 1-dim. -However, I do not know any way of extending the result to the case of 3-dim fibers, or also the case of 7-dim fibers, $S^7\to S^{15}\to S^8$. It seems that Tanno's technique uses strongly the fact that the fibers are 1-dim. Any suggestion on how to compute this other first eigenvalues would be greatly appreciated. - -REPLY [5 votes]: Tanno, the same author of the paper above mentioned in the case of 1-dim fibers, has another paper one year later [Tanno, Shûkichi. Some metrics on a $(4r+3)$-sphere and spectra. -Tsukuba J. Math. 4 (1980), no. 1, 99–105. ] in which he addresses exactly my original question, regarding the case of 3-dim fibers. As a consequence, one has an explicit formula for the first eigenvalue of the Laplacian $\Delta_t$ of $S^{4n+3}_t$ for all $t>0$. The technique is indeed related to Claudio's suggestion of using representation theory, since the Laplacian acts as the Casimir element. The case of $S^7\to S^{15}\to S^8$ can also be treated similarly, obtaining the first eigenvalue explicitly.<|endoftext|> -TITLE: Reference for the odd dihedral case of Artin's conjecture -QUESTION [10 upvotes]: The example that Matt Emerton cited here prompted me to become interested in how one proves that odd two dimensional dihedral Galois representation are modular. This is the first case of the strong Artin conjecture for two dimensional representations and I feel like understanding it would be helpful in getting some sense for why Galois representations are modular. Emerton mentioned that the theorem was proved by Hecke; according to Gelbart's review of the Serre/Deligne paper on Galois representations attached to weight 1 modular forms; the dihedral case follows from Hecke's work on theta series attached to binary quadratic forms. -Chandan Singh Dalawat give some more detail on the example that Emerton gives on pp. 5-6 of his article titled Splitting Primes, citing an article of Serre for still more detail. I have some glimmerings of how this goes in the case under discussion; in that case one needs to show that the Artin L-function is 1/2 of the difference of two theta series; presumably one uses class field theory for the splitting field viewed as a cubic extension of the quadratic subfield. The two quadratic forms used to define the relevant theta series correspond to the nonprincipal ideal classes of the quadratic subfield. But I don't see exactly how it should go. -I've seen references to -J.-P. Serre, Modular forms of weight 1 and Galois representations. In: Algebraic -Number Fields (1977), pp. 193–268 = Œuvres/Collected Papers III, Springer- -Verlag, Berlin, 1986, pp. 292–367. - -but given that the result goes back to Hecke it seems like there should be expositions along classical lines from an earlier time (1930's-1960's). and I haven't been able to find them. Does anyone know such a reference? - -REPLY [6 votes]: There's a beautiful history behind this. -Basically, Artin and Hecke were working on different sides of this "dihedral modularity conjecture" at the same time (the 20s) and at the same place (Hamburg), but apparently they never discussed this aspect of their research. -So they had the tools to prove this instance of what would become the Langlands program by 1927, but they didn't know it! -There is a brief account of this in Tate's paper "The general reciprocity law" (note that he was Artin's doctoral student), and a more extended historical survey of Artin and Hecke's work during that time on Cogdell's article "On Artin L-functions". -I think this is how the proof would have looked like back in 1927 (although in modern notation, and not in german!) -Arithmetic side (Artin) -Let $\rho:\mathrm{Gal}(L/K)\to \mathrm{GL}_2(\mathbb{C})$ be 2-dimensional dihedral complex representation. From representation theory we know that $\rho$ is monomial, that is, induced from a 1-dimensional representation. -Artin had proved in 1923 that his L-functions behave well under representation theoretic operations and, in particular, induction. Therefore, there is an L-function $L(\varrho,s)=L(\rho,s)$, with $\varrho$ one-dimensional (abelian). -From Artin reciprocity (1927) we have that $L(\varrho,s)=L(\chi,s)$, with $L(\chi,s)$ a Hecke L-function. -The last step is Hecke's proof from 1917 that abelian L-functions are meromorphic for non-trivial characters. Since $\varrho \neq 1$, the original L-function $L(\rho,s)$ is meromorphic on the complex plane, and we have proved the Artin conjecture for dihedral representations. -Automorphic side (Hecke) -Hecke had been studying theta series, and in particular in 1927 he constructed a cusp form $f_\theta$ of weight $1$ as a linear combination of $\theta$-series of binary quadratic forms attached to $K$. -He had alredy proved the basic properties of the L-functions of arbitrary modular forms and Hecke characters, so he knew their functional equation. -In the case of his $f_\theta$ the gamma-factor was very simple, just $\Gamma(s)$. -So, according to Tate, he listed all the Hecke L-functions that shared that same gamma-factor. After weeding out the one coming from Eisenstein series (which in turn correspond to cylic (reducible) two-dimensional representations), he was left with a correspondence $L(\chi,s)=L(f_\theta,s)$. -Arithmetic side revisited -This would have been an easy step for either one of them, if they had known what the other one was up to. -A quick inspection of the gamma-factor of the Artin L-function shows that the only representations for which it equals $\Gamma(s)$ are the ones odd and two-dimensional. Since the other two-dimensional odd representations are irreducible (except the cyclic, which we have alredy mentioned correspond to Eisenstein series), we have showed: -$$L(\rho,s)=L(f_\theta,s)$$ -Jacquet-Langlands proof -The first actual proof of the result follows from the converse theorem for $\mathrm{GL}_2$ in "Automorphic forms on GL(2)" (1971). But I don't think they mention the dihedral case in particular. Langlands does, saying that it is implicit in the works of Hecke and Maass, in his 1975 book "Base change for GL(2)". -A different proof follows from the results by Deligne and Serre in "Formes modulaires de poids 1" (1974). -I'm not sure of what relevance Maass' work has in this case. The same goes for some attributions to Brauer, since his induction theorem isn't really needed here. -To answer the actual question, no, there's no direct reference for this result before 1971. That said, technically Artin's 1927 paper implies this case of the (weak) Artin conjecture, and we now know (by a result of Booker, 2003) that this "weak" case implies the strong Artin conjecture.<|endoftext|> -TITLE: Literature on the Springer resolution -QUESTION [10 upvotes]: Could you suggest me a basic reading list on the Springer resolution? Is there a textbook I can refer to? Or do I need to start with the original paper? -Unfortunately googling for "Springer" and "resolution" was not very helpful so far, due to the existence of a certain publisher called Springer which, from some strange reasons, produces many books discussing resolution of singularities ... :p - -REPLY [7 votes]: Springer developed his resolution of singularities for the unipotent (or nilpotent) variety of a semsimple algebraic group (or its Lie algebra) as part of a bigger program to construct Weyl group representations in a new way by making the Weyl group act on cohomology groups of fibers of the resolution. This is a multifaceted subject by now and can be approached in somewhat different ways. As Mike indicates in his answer, Ginzburg and others provide at least a partial pathway (mainly in characteristic 0) using modern algebraic geometry and emphasizing the best-behaved example $SL_n$. Some of the machinery used is necessarily rather sophisticated, as in Springer's original papers. -Papers by Steinberg and Spaltenstein developed a lot of detail about the Springer resolution and its fibers, including dimension formulas. Steinberg's 1974 Springer Lecture Notes 366 volume is based on Vinay Deodhar's write-up of Steinberg's Tata lectures covering some of this material. In my 1995 AMS monograph Conjugacy Classes in Semisimple Algebraic Groups I gave a more comprehensive treatment in the framework of the Borel-Chevalley structure theory of semisimple (or reductive) groups over arbitrary algebraically closed fields. -See especially Chapter 6 for the Springer resolution in this setting, as well as the brief survey in Chapter 9 of Springer's construction of Weyl group representations (which Ginzburg outlines in his Chapter 3). -The Springer resolution taken in isolation is well enough presented in these sources, though in different styles and with different assumptions on the field. But there is no standard textbook treatment of the related Weyl group story, which is much harder to approach straightforwardly. Carter's 1985 book Finite Groups of Lie Type does contain a lot of concrete details in special cases, however. Which source to consult depends a lot on what you want to do next.<|endoftext|> -TITLE: Convergence of the series $\sum_p p^{-s}$ ($p$ prime and $s>1$) -QUESTION [10 upvotes]: I know that $\sum_p p^{-s}$, $s>1$, converges. Now, I define $J(s) = \sum_p p^{-s}$. Are there any "well known" values for $J(2)$, $J(3)$, $J(4)$, etc? We all know that $\zeta(2)= \frac{\pi^2}{6}$, $\zeta(4)=\frac{\pi^4}{90}$, etc. - -REPLY [5 votes]: No, in the sense that there are (for all I know) no identies along the lines of those for $\zeta(s)$, that you recalled, known. -Your function $J$ is sometimes called the prime zeta function. -You can find some information, some approciamte numerical values, plots, and pointers to the literature, e.g., at -http://mathworld.wolfram.com/PrimeZetaFunction.html -and -http://en.wikipedia.org/wiki/Prime_zeta_function -Two related hand-waving/heuristic arguments for the difficulty (not sure how good/convincing they are): - -The values would 'encode' quite precise information on the set of primes. -The arithmetic function you are summing, that is, $f(n) = n^{-s}$ if $n$ is prime, and $f(n)=0$ if $n$ is not prime, is not a 'nice' arithmetic function; for example it is not multiplicative. - -A related note that might interest you, in case you are not aware of it: -As you say $\sum_p p^{-1}$ diverges. However, the rate of divergence is fairly precisely known. -Namely, by Mertens's Second Theorem -$$\lim_{n \to \infty} \left ( \sum_{p\le n} p^{-1}\right ) - \log \log n $$ -exists, and is equal to (or perhaps, rather defines) the Meissel--Mertens constant, -which is approxiamtely $0.2614972$.<|endoftext|> -TITLE: Inverse function theorem for manifolds with boundary as the domain -QUESTION [7 upvotes]: I wonder that whether there exists a version of the inverse function theorem for smooth maps from a smooth manifolds with boundary to a smooth manifold without boundary? More precisely, whether the following assertion is true? -Let $M$ be a smooth manifold with boundary $\partial M$ and $N$ be a smooth manifold without boundary whose dimension $d$ is equal to the dimension of $M$. Let $f$ be a smooth map from $M$ to $N$. Assume that there exists a point $x\in \partial M$ such that the rank of $f_M$ -at $x$ is $d$ and the rank of $f_{\partial M}$ on $x$ is $d-1$. Then there exists a open neignborhood $U$ of $x$ on $M$ such that $f_{U}$ is a diffeomorphism from $U$ onto $f(U)$. -Any references or comments are well appreciated. Thanks a lot! - -REPLY [3 votes]: It seems to me that what you ask is a straightforward consequence of the usual local inversion theorem. More precisely: up to taking a suitable chart at $x$, we can assume $M$ is a half space and $\partial M$ a hyperplane, in $\mathbb{R}^d$. You then just have to apply the inversion theorem in a neighborhood of $X$ in $\mathbb{R}^n$ and restrict to a neighborhood of $x$ in $M$. In other words, manifold with boundary are dealt with just the same as manifold without boundary, but with an embedded hypersurface. - -REPLY [3 votes]: Note that, the question being local you can work in local charts. Also, recall that, by definition of manifold with boundary, and by definition of smooth maps between manifolds with boundary, you can assume w.l.o.g. that $f$ is the restriction to $U:=V\cap H$ of a $C^1$ map $\tilde f$ defined on a nbd $V$ of $x:=0\in\mathbb{R}^d$, where $H$ is a closed half-space. So $\tilde f$ is locally invertible by the usual inverse mapping theorem on open sets of $\mathbb{R}^d$, and such is f by restriction. Note that you don't have to assume anything on the invertibility of $f_{|\partial M}$. The same argument works for Banach manifolds.<|endoftext|> -TITLE: Finitely presented groups which are not residually amenable -QUESTION [7 upvotes]: What are examples of finitely presented but not residually amenable groups? -Well, the examples that I want to have are simple f.p. groups as well as examples of non residually amenable groups arise from other reasoning then simplicity. -Thank you for all your references! - -REPLY [5 votes]: Cornulier has a finitely presented sofic group which is not the limit of amenable groups: -http://arxiv.org/pdf/0906.3374<|endoftext|> -TITLE: Cohomology of Hypersurface complement -QUESTION [9 upvotes]: Suppose $Y$ is a smooth hypersurface in projective space $\mathbb{P}^n$, $X = \mathbb{P}^n - Y$ is the hypersurface complement. Is there a general method to compute cohomology of $X$? In particular, for small n, is there any examples or references? - -REPLY [3 votes]: Griffiths (On the periods of certain rational integrals. I, II. Ann. of Math. 90 (1969), 460-495 & 90 (1969), 496–541.) gave a procedure to calculate the Hodge numbers of Y. Let $S=\mathbf{C}[x_0,\dots,x_n]$, let $f$ be an equation for $Y$ and $d$ be its degree. Let $J_f$ be the jacobian ideal of $f$, i.e., the ideal generated by the partial derivatives of $f$. Then -$$ h^{i,n-1-i}(Y)=\delta_{i,n-1-i}+\dim (S/J_f)_{id-n-1}.$$ -The other Hodge numbers can be obtained from by hyperplane theorems: i.e., if $p+q\neq n-1$ then $h^{p,q}(Y)=1$ if and only if $0\leq p=q\leq n-1$ holds. All other Hodge numbers are zero. -Here, $\delta_{i,j}$ is the Kronecker delta, $S$ has a natural grading, $J_f$ is generated by homogeneous elements and therefore $S/J_f$ is a graded ring. $(S/J_f)_k$ means all elements of degree $k$. -The above formula can be generalized, see e.g., Steenbrink (Intersection form for quasi-homogeneous singularities. Compositio Math. 34 (1977), 211–223) for weighted projective spaces of Dimca (Betti numbers of hypersurfaces and defects of linear systems. -Duke Math. J. 60 (1990), 285–298) for hypersurfaces with isolated weighted homogeneous singularities.<|endoftext|> -TITLE: Cartier divisors on the moduli space of two pointed elliptic curves -QUESTION [5 upvotes]: The moduli space of stable $2$-pointed genus $1$ curves has cyclic quotient singularities, so any Weil divisor on $\overline{M}_{1,2}$ is $\mathbb{Q}$-Cartier. -Is one of the two boundary divisors $\Delta_{irr}$ and $\Delta_{0,2}$ of $\overline{M}_{1,2}$ Cartier ? - -REPLY [2 votes]: The singularities of $\overline{M}_{1,2}$ are located as follows: -a singularity of type $\frac{1}{4}(2,3)$ representing an elliptic curve of Weierstrass representation $C_{4}$ with marked points $[0:1:0]$ and $[0:0:1]$; -a singularity of type $\frac{1}{3}(2,4)$ representing an elliptic curve of Weierstrass representation $C_{6}$ with marked points $[0:1:0]$ and $[0:1:1]$. -a singularity of type $\frac{1}{6}(2,4)$ representing a reducible curve whose irreducible components are an elliptic curve of type $C_{6}$ and a smooth rational curve connected by a node; -a singularity of type $\frac{1}{4}(2,6)$ representing a reducible curve whose irreducible components are an elliptic curve of type $C_{4}$ and a smooth rational curve connected by a node. -The divisor $\Delta_{irr}$ is contained in the smooth locus of $\overline{M}_{1,2}$ and it is Cartier, while the other boundary divisor contains two singular points, it is $\mathbb{Q}$-Cartier but not Cartier.<|endoftext|> -TITLE: Why does one consider the dual of the Steenrod algebra? -QUESTION [10 upvotes]: Why does one consider the dual of the Steenrod algebra? - -REPLY [3 votes]: Decided to move a link in one of my comments to an answer - just for convenience. -Some detailed information about close connection between the Milnor dual of the Steenrod algebra and the automorphism group scheme of the additive group scheme is given in the paper "On Realizations of the Steenrod Algebras" by Lebedev and Leites (Journal of Prime Research in Mathematics 2 (2007) 1–13) -Also see Is there a high-concept explanation of the dual Steenrod algebra as the automorphism group scheme of the formal additive group? here on MO<|endoftext|> -TITLE: What is a "best" transcendence basis for R/Q ? -QUESTION [7 upvotes]: It is easy to show, using the axiom of Zorn, that there exists a transcendence basis for $\mathbb{R}/\mathbb{Q}$, i.e. a set $S$, algebraically independent over $\mathbb{Q}$, such that $\mathbb{R}/\mathbb{Q}(S)$ is an algebraic extension. -What can we say about $T=\mathbb{R} - \mathbb{Q}(S)$? It is easy to show that $\mathbb{R}/\mathbb{Q}$ is not purely transcendental, so that $T \neq \emptyset$. (Indeed, no element of $\mathbb{R}-\mathbb{Q}$ algebraic over $\mathbb{Q}$ may be contained in $\mathbb{Q}(S)$ - an algebraic relation for such an element over $\mathbb{Q}$ would immediately translate into a relation of algebraic dependence in $S$ over $\mathbb{Q}$.) -Thus, all real, non-rational algebraic numbers are contained in $T$. Is it possible, for example, to choose $S$ so that this inclusion is an equality? If not, how "small" can we make $T$? -I'm sorry if this turns out to be trivial, or if the answer to my question can be easily found in the literature. I tried! -Thank you very much and have a pleasing day. - -REPLY [7 votes]: No, you cannot choose $T$ so nicely. For cardinality reasons, we can choose some $s\in S$ with $s\notin \overline{\mathbb{Q}}$. Now, consider the element $\sqrt{s}$. I'll leave it to you to consider how large this makes $T$.<|endoftext|> -TITLE: A potential resolution of $R/r$ -QUESTION [6 upvotes]: The DGA -For $k$ some field, let $R$ be a $k$-algebra, and let $r\in R$. -Define a differential graded algebra $\mathbf{R}_r$ as follows. As a graded algebra, it is isomorphic to $R\langle t\rangle$, the free $k$-algebra over $R$ with non-central variable $t$ added. The algebra $R$ has degree zero, and $t$ has degree one. The differential $d$ is defined by $d(R)=0$ and $d(t)=r$, and extend by the Leibniz rule. -As a chain complex, the DGA $\mathbf{R}_r$ then looks like -$$ R\leftarrow RtR\leftarrow RtRtR \leftarrow ...$$ -Note that each $t$ may be replaced by $\otimes_k$ without changing the $R$-bimodule structure. So what are the homology groups? Clearly, $d(RtR)=RrR\subset R$, and so $H_0(\mathbf{R}_r)=R/r$ (the quotient by the two-sided ideal generated by $r$). -The Question -My question is, for what $R$ and $r$ do all higher homology groups of $\mathbf{R}_r$ vanish? Equivalently, when is $\mathbf{R}_r$ quasi-isomorphic to $R/r$? -Examples and Counterexamples -There are some immediate answers. When $r=1$, the complex $\mathbf{R}_1$ is the (augmented) bar resolution of $R$ as a bimodule over itself. As the name implies, the homology groups vanish. The standard argument for the exactness of the bar resolution (constructing a chain homotopy between the identity map and the zero map on $\mathbf{R}_1$) can be generalized to $\mathbf{R}_r$ for any $r$ which is a central unit. -If $R=k[x]$ and $r=x$, then a direct computation shows that the higher homology groups vanish, and so $\mathbf{R}_x$ is a resolution of $R/x\simeq k$. -As a counterexample, consider $R=k[x]$, and $r=x^2$. For $xt-tx\in (\mathbf{R}_{x^2})_1$, we have -$$ d(xt-tx)=x(x^2)-(x^2)x=0$$ -However, any 1-boundary $d(atbtc)$ must have $x$-degree at least 2, and so $xt-tx$ is not exact. - -REPLY [6 votes]: This complex (in simultaneously a more general setting, where you have several elements $t_1,\ldots,t_p$, and a more special setting, because only the case of $R$ being a free algebra was studied then) was introduced by Shafarevich [E. S. Golod, I. R. Shafarevich, “On the class field tower”, Izv. Akad. Nauk SSSR Ser. Mat., 28:2 (1964), 261–272], and was studied somewhat thoroughly recently, see, e.g. [E. S. Golod, “Homology of the Shafarevich complex and noncommutative complete intersections”, Fundam. Prikl. Mat., 5:1 (1999), 85–95] - this paper again dealing with the case of free $R$. As the latter paper suggests, this complex is a noncommutative generalisation of the Koszul complex in the commutative case, detecting the "noncommutative complete intersections" usually known as "strongly free sets" from a paper of Anick [D. J. Anick, Non-commutative graded algebras and their Hilbert series, J. Algebra, Volume 78, Issue 1, September 1982, Pages 120-140]. -Finally, in the case of arbitrary (graded) $R$ this complex is discussed in a paper of Piontkovski [http://arxiv.org/abs/math/0606279] for the purpose of studying "relative noncommutative complete intersections". You would be especially interested in "Theorem-Definition 2.3" from that paper. -(For those who cannot be bothered to check the references, one condition of Piontkovski's paper is completely analogous to the complete intersection result stating that $A=B[f_1,\ldots,f_k]$ for a regular sequence $f_1,\ldots,f_k$: the complex in question is acyclic iff $R=(R/r)\langle r\rangle$. Here one has to make some assumptions, e.g. assume $R$ graded and $r$ being homogeneous of positive degree, so this does not cover central units mentioned in the question, whereas Tyler Lawson's example is the simplest instance of this result.)<|endoftext|> -TITLE: Sheaves of $\mathbb Z$-modules = sheaves of abelian groups -QUESTION [8 upvotes]: In his "Algebraic Geometry", Hartshorne proves that for any ringed spaces $(X,\mathcal O_X)$, category $Mod(X)$ of sheaves of $\mathcal O_X$-modules has enough injectives. If we take $\mathcal O_X$ to be constant sheaf of rings $\underline{\mathbb Z}_X^{\natural}$ (i.e. sheaf associated to a constant presheaf $\underline{\mathbb Z}_X$), Hartshorne claims that $Mod(X)=Ab(X):=$ category of sheaves of abelian groups on X. -Why does this last claim hold? Of course, if one takes open subset $U \subseteq X$ that is connected, then section $\mathcal O_X(U)$ actually is ring isomorphic to $\mathbb Z$. But, for more 'complicated' open subsets $U \subseteq X$, sections $\mathcal O_X(U)$ become more complicated rings than $\mathbb Z$. So generally, many sections of a sheaf of $\mathcal O_X$-modules have more complicated structure than merely abelian group (i.e. $\mathbb Z$-module). -So, why is it obvious, according to Hartshorne, that $Mod(X)=Ab(X)$? I can only see that $Mod(X)$ is subcategory of $Ab(X)$. - -REPLY [10 votes]: $\def\sh#1{\mathcal{#1}}\def\csheaf#1{\underline{#1}}\def\on#1{\operatorname{#1}}$First of all, if $X$ is an irreducible scheme (or any such topological space), then all of its open subsets are connected and there are no complications such as you describe. However, the complications dry up under close examination no matter what, because we are talking about sheaves of abelian groups versus sheaves of $\sh{O}$-modules, and thus the same gluing property that alters the sections of the constant sheaf on $\mathbb{Z}$ alters the sections of its modules in such a way that nothing goes wrong. Here are the details: -Given a sheaf of abelian groups $\sh{F}$ and a sheaf of rings $\sh{O}$, the structure of an $\sh{O}$-module on $\sh{F}$ is the same as a homomorphism of sheaves of rings $\sh{O} \to \sh{Hom}(\sh{F}, \sh{F})$, where the last object is the sheaf of group homomorphisms from $\sh{F}$ to itself. If $\sh{O}$ is the constant sheaf on a ring $R$, then it is the sheafification of the constant presheaf $\csheaf{R}$ on $R$, and by the universal property of sheafification, it suffices to give a map $\csheaf{R} \to \sh{Hom}(\sh{F}, \sh{F})$ (this extends uniquely to $\sh{O}$). But the sections of $\csheaf{R}$ are just $R$ for any open set, even disconnected ones, and so that's the same as a map $R \to \on{Hom}(\sh{F}|_U, \sh{F}|_U)$ for any open $U$, compatible with restrictions. This gives in particular a collection of maps $R \to \on{Hom}(\sh{F}(U), \sh{F}(U))$ compatible with restrictions, and these two data are equivalent since a map of sheaves is just a collection of maps of sections compatible with restrictions. Of course, the last kind of data is the same thing as an $R$-module structure on $\sh{F}$, as you want.<|endoftext|> -TITLE: How many definitions are there of the Jones polynomial? -QUESTION [19 upvotes]: Even with the connection to quantum groups being made clearer (I believe it was not known when the Jones polynomial was first introduced), it seems to me that still we don't have the "right" definition of the Jones polynomial. It is certainly true though that we know a lot of different definitions, some more useful than others. I'm thinking of: -1) Kauffman bracket (i.e. the skein relation). This defines the Jones polynomial by giving a straightforward algorithm to compute it, but leaves any other significance a mystery. -2) Quantum groups. Take $U_q(\mathfrak s\mathfrak l_2)$, and observe via universal $R$-matrices that "its category of representations is a braided monoidal tensor category", so that in particular, $\overbrace{V\otimes\cdots\otimes V}^{n\text{ times}}$ gives a representation of $B_n$, for any given representation $V$ of $U_q(\mathfrak s\mathfrak l_2)$. The Jones polynomial is easily derived from this representation of $B_n$. -3) KZ equations (closely related to (2)). Let $X_n$ be the configuration space of $n$ points $(z_1,\ldots,z_n)$ in $\mathbb C$. Now write down the one-form $A=\hbar\cdot\sum_{i -TITLE: Grothendieck's question on the Brauer group for groups -QUESTION [18 upvotes]: Let $G$ be a group, and let $M(G)=H^2(G,\mathbb{C}^*)$ be the Schur multiplier of $G$. There is a group $Br(G)$ of complex projective representations of $G$ modulo those that can be lifted to linear representations. Projective representations of degree $n$ are classified by cohomology classes $H^1(G,PGL_n(\mathbb{C}))$. There are maps of pointed sets $$H^1(G,PGL_n(\mathbb{C}))\rightarrow M(G)$$ arising from the exact sequences of (trivial) $G$-modules -$$1\rightarrow \mathbb{C}^*\rightarrow GL_n(\mathbb{C})\rightarrow PGL_n(\mathbb{C})\rightarrow 1.$$ -Let $Br(G)$ be the subgroup of $M(G)$ generated by the images of these boundary maps for all $n$. (Of course, $Br(G)$ can be defined directly without using group cohomology.) -My question is: when is $Br(G)\rightarrow M(G)$ surjective? This is analogous to the question of Grothendieck on the difference between $Br(X)$ and $H^2(X_{et},\mathbb{G}_m)$ of schemes. If one works with the classifying space of $G$, then everything may be interpreted in terms of topological Azumaya algebras, and the question is precisely the same as Grothendieck's. -I know of three broad cases where this is true. First, when $G$ is a finite group, this follows fairly easily from group cohomology. Roughly speaking, one takes a cocycle representative $\alpha\in Z^2(G,\mathbb{C}^*)$. Then, one uses the cocycle to create a twisted group algebra $\mathbb{C}^{\alpha}[G]$. This algebra is a semi-simple algebra over the complex numbers, so it has various finite-dimensional representations. Any one of them gives a projective representation for $G$ with cohomology class $\alpha$. See for instance Karpilovsky's book Projective representations of finite groups. -Second, when $G$ is a group such that $BG$, the classifying space, has the homotopy type of a finite CW-complex, it follows from Serre's theorem that $Br=Br'$ for finite CW-complexes. See Grothendieck's Groupe de Brauer I. -Third, when $G$ is a good group in the sense of Serre, then $Br(G)=M(G)$. A group $G$ is good if it has the same cohomology as its finite completion. This result is proven in Schroer's paper Topological methods for complex analytic Brauer groups. -So, my real question is: does anyone know any more general results than these? For instance, is $Br(G)=M(G)$ for any finitely presented group? - -REPLY [3 votes]: Note that it is not hard to construct examples where $Br(G)$ is not equal to $M(G)$. To see this, let $H_1=\mathbb{Z}/p\times\mathbb{Z}/p$, and let $\alpha_1\in H^2(H_1,\mathbb{C}^*)=\mathbb{Z}/p$ be a generator. The class $\alpha_1$ is represented by a projective representation $p_1:H_1\rightarrow PGL_p(\mathbb{C})$. Now, let $H_i=(H_1)^i$, and let $\alpha_i$ be the Brauer class of the projective representation $$p_i:H_i=(H_1)^i\xrightarrow{p_1\times\cdots\times p_1}(PGL_p(\mathbb{C}))^i\rightarrow PGL_{p^i},$$ where the final arrow is given by embedding $PGL_p(\mathbb{C})^i$ into $PGL_{p^i}$ via block matrices down the diagonal. Then, it is not hard to show that the index of $\alpha_i$ is $p^i$. Recall that the index of a class of $Br(G)$ is the least common divisor of all projective representations having that class. -Now, define $G=*_i H_i$, the free product of the $H_i$ for all positive integers $i$. The corresponding classifying space $BG$ is the wedge sum of the $BH_i$. Therefore, the collection of continuous pointed maps -$$\alpha_i:BH_i\rightarrow K(\mathbb{C}^*,2)$$ -induces a continuous pointed map -$$\alpha:BG\rightarrow K(\mathbb{C}^*,2) -$$such that the composition $$BH_i\rightarrow BG\rightarrow K(\mathbb{C}^*,2)$$ is $\alpha_i$. -It is easy to see that $ind(\alpha_i)$ divides $ind(\alpha)$ for all $i$. But, since $ind(\alpha_i)=p^i$, it follows that $ind(\alpha)$ is infinite, and hence that $\alpha\notin Br(G)$, as desired. -Obviously, $G$ is not finitely presented. In relation to André's comment above, it seems likely that it is not linear either.<|endoftext|> -TITLE: Lie algebras with trivial second cohomology group -QUESTION [8 upvotes]: Let $\mathfrak{L}_n(\mathbf{C})$ be the set of all $n$-dimensional complex Lie algebras. One know that in $\mathfrak{L}_n(\mathbf{C})$ there exist only a finite number of Lie algebras with trivial second cohomology group with coefficients in the adjoint representation (by using a rigidity argument of Nijenhuis and Richardson). -Question: Fixing $n$, are Lie algebras $\mathfrak{g} \in \mathfrak{L}_n(\mathbf{C})$ such that $H^2(\mathfrak{g},\mathfrak{g})=0$ classified? -References: - -Nijenhuis, Richardson - Deformations of Lie algebra structures -Nijenhuis, Richardson - Cohomology and deformations in Graded Lie algebras -Nijenhuis, Richardson - Cohomology and deformations of algebraic structure - -REPLY [4 votes]: Even if we consider more generally cohomologically rigid Lie algebras $L$, which satisfy $H^i(L,L)$ for all $i\ge 0$, a classification is not possible. -Roger Carles proved in 1985 in his paper "Sur certain classes d'algebres de Lie rigides", that every complete Lie algebra $L$ with abelian nilradical ${\rm nil}(L)$ is cohomologically rigid. He proves a lower bound of the number $s_n$ of non-isomorphic solvable cohomologically rigid Lie algebras of dimension $n$, which is indeed -$$ -\Gamma(\sqrt{n})\le s_n, -$$ -for all $n\ge 81$, with the Gamma function. This shows that a classification is more or less hopeless.<|endoftext|> -TITLE: How to prove this Poincare Inequality -QUESTION [6 upvotes]: I want to ask a question about a statement that I found on the paper: Principal Eigenvalues for Problems With indefinite Weight Function in $R^N$. -The statement is the following: -Suppose that $g:\mathbb{R}^2\to\mathbb{R}$ is a $C^{\infty}$ function which changes sign on $\mathbb{R}^2$ and there exist constants $K,R>0$ such that $g(x)\leq-K$ for $|x|>R$. Let $B$ a ball such that -$$ -\int_{B} g(x) dx <0 -$$ -and $g(x)<0$ if $x\in \mathbb{R}^2\setminus B$. - I would like to know why is true that there exist a positive constant $C_1>0$ such that -$$ -\int_{B} u^2 dx \leq C_1\int_{B}|\nabla u|^2 dx -$$ -for all $u\in H^1(B)$ with $\int_{B}g u^2 dx>0$ ? - -The authors give the following reference for this result: -K.J. Brown, S.S. Lin and A. Terkitas, Existence and noexistence of steady-state solutions for a selection-migration model in popular genetics. J. Math. Biol. 27 (1989), 91-104. -I can't have access to this paper that's why I am posting this question here. -I am trying to prove this inequality, but the results I know are based on arguments that requires compact support for $u$ or mean zero. I appreciate if some specialist can give me a hint on how to prove this inequality for the general case (neither compact support or mean zero) or provide an alternative reference where it is proved. Thanks. - -REPLY [8 votes]: Just assume that $g$ is a bounded function with $\int_B g < 0$, and positive somewhere in th ball $B$ . Then the set -$$\Big \{u\in H^1(B)\, : \|u\|_2= 1\, ,\, \int_B g u^2\ge 0 \Big \}$$ -is not empty and does not contain constant functions. By weak compactness $\|\nabla u\|_2$ attains a non-zero minimum on it, and by homogeneity we get the thesis.<|endoftext|> -TITLE: Special values of Artin L-function -QUESTION [9 upvotes]: Ok, so this might be a really naive question (and clearly related to Special values of Artin L-functions). -The Stark conjecture postulates that all Artin L-functions has a transcendental (over $\mathbb{Q}$) leading coefficient at $s=1$ (or $s=0$) in the Taylor expansion (well, it implies this). -My question: are all such leading coefficients, at various values of $s$, transcendental over $\mathbb{Q}$, or can there be algebraic ones? -Edit: I exclude the Dirichlet L-functions for which there is a positive answer to my question. -So the question becomes: given an Artin L-function (corresponding to some representation of some Galois group), are there always $s\in\mathbb{C}$ such that the leading term is algebraic? -My gut feeling tells me that there can indeed be algebraic ones. -As I said, this is in all probability an extremely naive question, but a fleeting google search led to nothing (at least nothing I could understand). - -REPLY [8 votes]: Ok so I think I have worked out which integers are critical. Let $F$ be a finite Galois extension of $\mathbf{Q}$ and $\rho:\operatorname{Gal}(F/\mathbf{Q}) \to GL(V)$ be an irreducible Artin representation, with $\rho \neq 1$. -First I should mention that there is a functional equation relating $L(\rho,s)$ and $L(\overline{\rho},1-s)$ but it involves some power of $\pi$, so one should be careful when formulating an algebraicity conjecture. So let me restrict to the case $s=1-m$ where $m$ is an integer $\geq 1$. -As proved in Deligne's article Valeurs de fonctions L et périodes d'intégrales, if $1-m$ is critical then the associated period is trivial and Deligne's conjecture is true, so that $L(\rho,1-m)$ is a nonzero algebraic number. One can even prove, using a theorem of Siegel, that $L(\rho,1-m)$ belongs to the number field generated by the values of the character of $\rho$, and that $L(\rho,1-m)^{\sigma}=L(\rho^\sigma,1-m)$ for any $\sigma \in \operatorname{Aut}(\mathbf{C})$, for a detailed proof see Thm 1.2 in Coates-Lichtenbaum, On $\ell$-adic zeta functions, Annals of Math. 98 n°3 (1973) (I'm grateful to Junkie for pointing out to me this reference). If $1-m$ is not critical then $L(\rho,1-m)=0$ but algebraicity of the leading term is not expected (regulators are expected to be transcendental, but even their irrationality is very difficult to prove). -Thus it remains to find the critical integers. By definition $1-m$ is critical iff the gamma factor $L_\infty(\rho,s)$ has no pole at $s=1-m$. By definition $L_\infty(\rho,s)=\Gamma_{\mathbf{R}}(s)^{\dim V^+} \Gamma_{\mathbf{R}}(s+1)^{\dim V^-}$ where $\Gamma_{\mathbf{R}}(s)=\pi^{-s} \Gamma(s/2)$ and $V^{\pm}$ is the $\pm$-eigenspace for the action of $\rho(c)$, where $c \in \operatorname{Gal}(F/\mathbf{Q})$ is a choice of complex conjugation. A small computation then gives : -$$s=1-m \textrm{ is critical if and only if } V=V^{(-1)^m}.$$ -Examples and remarks. - -If $\rho$ is $1$-dimensional, this is consistent with the situation for Dirichlet characters. -If $F$ is totally real then $V=V^+$ so the result is consistent with the Klingen-Siegel theorem. -If $\rho$ is an odd irreducible $2$-dimensional Artin representation (correponding to a weight 1 newform thanks to the proof of Serre's conjectures) then $\dim V^+ = \dim V^{-}=1$ so there is no critical integer for $L(\rho,s)$. -On the other hand if $\rho$ is $2$-dimensional and even (corresponding conjecturally to a non-holomorphic Maass cusp form), then half of the integers are critical for $L(\rho,s)$. -It is possible to extend the above analysis to Artin representations associated to arbitrary Galois extensions of number fields.<|endoftext|> -TITLE: Rationale for Hadamard's finite part of a divergent integral -QUESTION [10 upvotes]: (Note: I asked this question a few days ago on math.stackexchange but didn't get any responses. I've therefore decided to post it here instead.) -I have a problem justifying throwing away the divergent term in order to obtain Hadamard's finite part. I find this step to be highly unusual, and it is not obvious to me how the resulting expression can be valid. I'd appreciate, if possible, intuitive arguments -- I assume there are some. Its not the mathematics that I have problems with at this stage. Its the intuition. -For example (taken from the Wikipedia page) the finite part of the following integral -$$\int_a^b \frac{f(t)}{(t-x)^2}\, dt = \lim_{\varepsilon \to 0} \left[ \int_a^{x-\varepsilon}\frac{f(t)}{(t-x)^2}\,dt + \int_{x+\varepsilon}^b\frac{f(t)}{(t-x)^2}\,dt -\frac{2f(x)}{\varepsilon} \right]$$ -involves throwing away the term $\frac{2f(x)}{\varepsilon}$. I find it hard to justify this step especially when the term is neither finite nor negligible. - -REPLY [3 votes]: Looking at Fourier transforms can provide an intuitive context for the Hadamard finite part (F.P.) regularization. -Monkey around with this ladder of expressions (understood as F.P.s): -$$A)\int_{-\infty }^{\infty }\frac{exp(i2\pi fx)}{(i2\pi f)^2}df=\frac{sgn(x)}{2}x = \int_{0}^{x}\frac{sgn(u)}{2}du$$ -$$B) \int_{-\infty }^{\infty }\frac{exp(i2\pi fx)}{i2\pi f}df=\frac{sgn(x)}{2}$$ -$$C)\int_{-\infty }^{\infty }exp(i2\pi fx)df=\delta(x) = \frac{d}{dx}\frac{sgn(x)}{2}$$ -To descend the ladder, formally take the derivative of both sides above or of the explicit F.P. expressions below (second equalities), which is equivalent to multiplying the integrands above by $i2\pi f$. To climb, integrate from $0$ to $x$ both sides below, using the explicit expressions for the integrands for the F.P. given below in the second equalities, or simply divide the integrands on the L.H.S. above by $i2\pi f$. (Note that $x$ can be negative or positive and that the Dirac delta function contributes only a value of $1/2$ when evaluated on the boundary of the integral.) So, the explicit F.P. integrals below commute with differentiation and integration w.r.t. $x$ and can be naturally defined in terms of the two ops, and the implicit symbolic formulas above allow us to formally retain the representation of the two ops as multiplication and division operations in the Fourier transform integrands. -For finite limits for the integrals, you'll end up with the expressions on the right above being convolved with a sinc function with some phase, that should agree with the L.H.S. if the Hadamard finite finesse is applied. -The OP's example is closely related to A) with $x=0$ and is more palatable within this context. In detail, in the limits $\varepsilon \to 0^+$ and $L \to \infty,$ -$C)\displaystyle\delta(x)=\int_{-L }^{L }exp(i2\pi fx)df$ -$B)\displaystyle\frac{sgn(x)}{2}=F.P.\int_{-\infty }^{\infty }\frac{exp(i2\pi fx)}{i2\pi f}df=\left [ \int_{\varepsilon}^{L }+\int_{-L }^{-\varepsilon} \right ]\frac{exp(i2\pi fx)-1}{i2\pi f}df$ -$=\displaystyle\left [ \int_{\varepsilon}^{L }+\int_{-L }^{-\varepsilon} \right ]\frac{exp(i2\pi fx)}{i2\pi f}df-\frac{ln(L/\varepsilon)}{i2\pi}-\frac{ln(\varepsilon/L)}{i2\pi}$ -$=\displaystyle\left [ \int_{\varepsilon}^{L }+\int_{-L }^{-\varepsilon} \right ]\frac{exp(i2\pi fx)}{i2\pi f}df=C.P.V\int_{-\infty }^{\infty }\frac{exp(i2\pi fx)}{i2\pi f}df$ -where $F.P.$ denotes the Hadamard finite part and $C.P.V.$, the Cauchy principle value. (Of course, the $\frac{1}{f}$ terms pose no serious problems since $\frac{1}{f}$ is an odd function and we are integrating symmetrically about $0$.) -Similarly, -$A)\displaystyle\frac{sgn(x)}{2}x=F.P.\int_{-\infty }^{\infty }\frac{exp(i2\pi fx)}{(i2\pi f)^2}df=\left [ \int_{\varepsilon}^{L }+\int_{-L }^{-\varepsilon} \right ]\frac{exp(i2\pi fx)-(1+i2\pi fx)}{(i2\pi f)^2}df$ -$=\displaystyle\left [ \int_{\varepsilon}^{L }+\int_{-L }^{-\varepsilon} \right ]\frac{exp(i2\pi fx)}{(i2\pi f)^2}df-\frac{2}{(i2\pi)^2 \varepsilon}=\frac{|x|}{2}.$ -It's even more convincing when you plot the integrals (including C) and observe how they evolve as $L$ increases for small $\varepsilon.$ -Another context for the Hadamard finite limit is given in MSE-Q13956. -For a comparison of different methods of regularization for integrals of this type see http://arxiv.org/abs/hep-th/0202023 "Improved Epstein-Glaser renormalization in coordinate space I. Euclidean framework" by Gracia-Bondia (pg. 14-). -Edit 2/11/21: Another example, in fractional calculus, of where the Hadamard finite part could obviously be invoked where it is equivalent to another route of analytic continuation is in my reply to this MO-Q.<|endoftext|> -TITLE: Why does the parameterization (F:F':1) happen? -QUESTION [12 upvotes]: 1) To parameterize the conic $x^2+y^2=1$, we can use $(x,y)=(\sin t,\sin't)$ ($\sin'$ meaning the derivative of $\sin$, namely $\cos$). -2) To parameterize an elliptic curve $y^2=4x^3-g_2x-g_3$, we can use $(x,y)=(\wp(t),\wp'(t))$. -(I know how to prove (1) and (2), and that it is possible to parameterize curves of higher genus using the unit disc.) -Question: Is there a common explanation for why both parameterizations are of the form $(F(t),F'(t))$? (as opposed to $(F(t),G(t)$) Does this phenomenon generalize to curves of higher genus? - -REPLY [6 votes]: There is a very deep reason for this. The general principle goes back to Andre Bloch. -If you have a non-constant holomorphic map from the plane to a compact complex manifold, -then there is another such map satisfying a differential equation. -A nice popular explanation of this can be found in the survey of Demailly, -Variétés hyperboliques et équations différentielles algébriques. (French) [Hyperbolic manifolds and algebraic differential equations] Gaz. Math. No. 73 (1997), 3–23.<|endoftext|> -TITLE: structure theorem for modules -QUESTION [7 upvotes]: Can structure theorem for modules be extended to modules over UFDS , to modules over Neotherian rings ? if yes then can one get the statement and reference? -Since operations on matrices with coefficients as polynomials in several variables some extension seems possible . - -REPLY [6 votes]: Let me start by something classical: extending the classical result for PIDs, by Steinitz's Theorem (1912) all finitely generated modules over Dedekind domains are characterized, see http://en.wikipedia.org/wiki/Dedekind_domains and scroll down. -Beyond Dedekind domains things get complicated, but there is considerable recent work going on. The following paper by Levy and Klingler gives an overview on their expansive (recent) work on this. Very roughly, there is a tame/wild dichotomy (very informally 'wild' is more or less 'hopeless') and they give an (essentially) complete answer for those noetherian rings were the answer is not 'wild'; the key-word here is Dedekind-like. -The 'essential' refers to the fact that, as they point out, there are or at least might be some exceptions in some characteristic two cases. -I am convinced that I once saw some subsequent work on this particular cases (though I do not remember whether it was partial or complete), yet unfortunately I am unable to find it right now.<|endoftext|> -TITLE: The double of a smooth manifold with boundary? -QUESTION [7 upvotes]: $\def\mc#1{\mathcal#1}\def\seq#1{\langle#1\rangle}\def\bbR{\mathbb R}\def\gt{>}\def\dom{{\rm dom\ }}$In some instances, I have seen an appeal to the concept of "the double" of a smooth manifold with non-empty boundary. Wikipedia gives a pure nonsense for this: "Precisely, the double is $M \times \{0,1\} / \sim $ where $ (x,0) \sim (x,1) $ for all $ x \in \partial M $ ." The essential problem here is how to construct pairwise smoothly compatible charts covering the boundary. I ask whether anyone reading this knows how to do this. -My natural idea of constructing "the" double would be the following. Let $\mc A$ be an atlas for the given $n-$dimensional smooth compact manifold with boundary. Let $M_0$ be the "interior" and $M_1$ the "boundary". Let $\mc A_0$ contain the charts $\phi:\dom\phi\to\bbR^n$ belonging to $\mc A$ with $\dom\phi\cap M_1=\emptyset$ , and let $\mc A_3$ contain the rest. So the functions belonging to $\mc A_3$ are bijections from some subset of $M_0\cup M_1$ onto some set of points $x=\seq{x_0,x_1,\ldots x_{n-1}}\in\bbR^n$ where $x_0\ge 0$ , and mapping points $m\in M_1$ to $x$ with $x_0=0$ . -Then one would take as a generating atlas of a "doubled manifold" the set $\bar{\mc A}=\mc A_0\cup\mc A_1\cup\mc A_2$ where $\mc A_2$ contains the functions $\tilde\phi:(m,{\rm w})\mapsto\phi(m)$ where $\phi\in\mc A_0$ and the fixed ${\rm w}$ is chosen so that $(m,{\rm w})\not\in M_0\cup M_1$ for $m\in M_0$ . As the elements of $\mc A_1$ one would take the functions $\bar\phi$ for $\phi\in\mc A_3$ constructed as follows. Let $P:\bbR^n\to\bbR^n$ be the bijection $\seq{x_0,x_1,\ldots x_{n-1}}\mapsto\seq{-x_0,x_1,\ldots x_{n-1}}$ . Then define $\bar\phi$ so that $\dom\phi\owns m\mapsto\phi(m)$ and $(M_0 \cap \dom \phi ) \times \{ {\rm w} \} \owns(m,{\rm w})\mapsto P(\phi(m))$ . -Then $\bar{\mc A}$ defines a "continuous atlas for the double", i.e. the chart changes $\psi\circ\phi^{-1}$ are homeomorphisms between some open subsets of $\bbR^n$ for $\phi,\psi\in\bar{\mc A}$ , but they need not be differentiable at points $\seq{0,x_1,\ldots x_{n-1}}$ . - -REPLY [17 votes]: I believe that the usual remedy is a collar. That is, for any smooth manifold there is a suitable diffeomorphism from a neighborhood of $\partial M$ to $[0,1)\times \partial M$, or in other words a smooth embedding $[0,1)\times \partial M\to M$ that is "the identity" on the boundary. This allows you to glue along the boundary and get a smooth manifold. To see that the result is independent of the choice of an embedding you use the fact that any two such embeddings are smoothly isotopic. - -REPLY [4 votes]: The doubled manifold is only a (piecewise smooth) $C^0$-manifold, unless you put more structure on the initial manifold with boundary. - -In dimension one, then you get a little bit more: you get a $C^1$-structure on the double. But still, you do not get a $C^2$-structure. -Here's how it goes: -Take $\mathbb R_+$ with its standard smooth structure. -Its double is $\mathbb R$. -Now let's analyze this further: -If you want that construction to be functorial (w.r.t diffeomorphisms), then you would like the diffeomorphism group of $\mathbb R_+$ to act on $\mathbb R$. -In other words, you want a group homomorphisms -$$ -Di\!f\!f(\mathbb R_+) \longrightarrow Di\!f\!f(\mathbb R),\qquad \varphi\mapsto\bar\varphi, -$$ -where $\bar\varphi$ is defined by $\bar\varphi(x):=\varphi(x)$ for positive $x$, and -$\bar\varphi(x):=-\varphi(-x)$ for negative $x$. -Now take $\varphi(x):=x+x^2$. -One easily checks that $\bar\varphi$ is not $C^2$! -$\qquad$ Conclusion:$\qquad$ the double of $\mathbb R_+$ is only equipped with a canonical $C^1$ structure.$\qquad$ It does NOT have a canonical $C^2$ structure. -Note: The same argument as above with $\mathbb R\times \mathbb R_+$, and the map -$\varphi(x,y):=(x+y,y)$ shows that the double of $\mathbb R\times \mathbb R_+$ is not $C^1$. - -On the positive side, here are two situations where it is possible to equip the double with a canonical smooth strucutre: - -if your manifold is Riemannian structure, and the boundary totally is geodesic. -in two dimensions, a complex structure induces a smooth structure on the double. -(no compatibility required between the cx structure and the boundary)<|endoftext|> -TITLE: Is the Brauer group of a surface an elliptic curve? -QUESTION [28 upvotes]: Of course not. -But after reading a bit, some points make me believe it should be: -Let $S$ be a nice$^{\*}$ surface defined over $Spec\ \mathbb{Z}$. - -The Brauer group $Br(S\otimes \bar{\mathbb{Q}})$ is an abelian divisible group, -It is also a $Gal(\bar{\mathbb{Q}}/\mathbb{Q})$ module, -For good primes there are reductions $Br(S)\rightarrow Br(S\otimes \mathbb{F}_q)$, -These $Br(S\otimes \mathbb{F}_q)$ are finite, -There is a formal Brauer group $\hat{Br}(S)$ of dimension 1, -The coefficients of $\hat{Br}$, in suitable natural coordinates, relate to $|Br(S\otimes \mathbb{F}_q)|$. -There are some examples where the associated L-function comes from a modular form (of weight 3). I'm not sure if this is conjectured (let alone known) in general. - -Since the Brauer group observes many characteristics of an abelian variety (all properties) of dimension 1 (properties 5 and 7 [weight isn't two, but it's the right space]), my vague question is: how far is it from actually being a variety? -There are some easy examples of $S$ with $|Br(S\otimes \mathbb{F}_q)|$ varying between $1$ and $4(q-4)$, as $q$ varies over the primes. This is a clear point of departure from elliptic curves and varieties in general. -Maybe there's a family of natural galois-module homomorphisms into certain abelian varieties defined over $\mathbb{Q}$, commuting with the reduction maps and restriction (or some other appropriate term) to formal groups? -What's going on with these Brauer groups? -$^\*$ say a K3 surface. Something that (1) is true for (so not a rational surface) and (4) is proven for. - -REPLY [16 votes]: This is not a general answer to your question, but evidence of the intriguing connection between Brauer groups of surfaces and elliptic curves. Let $X$ be a K3 surface over the complex numbers $\mathbb{C}$. Then, the rank of $H^2(X,\mathbb{Z})$ is $22$, and the Hodge numbers are $h^{0,2}=h^{2,0}=1$ and $h^{1,1}=20$. If the Neron-Severi group of $X$ is as big as possible, namely if it has rank $20$, then the image of $H^2(X,\mathbb{Z})\rightarrow H^2(X,\mathcal{O}_X)$ has rank $2$ and so is a lattice. Therefore, the cokernel of this map is an elliptic curve. But, since $H^3(X,\mathbb{Z})=0$, there is an exact sequence $$H^2(X,\mathbb{Z})\rightarrow H^2(X,\mathcal{O}_X)\rightarrow H^2(X,\mathcal{O}_X^*)\rightarrow 0.$$ Therefore, $H^2(X,\mathcal{O}_X^*)$ is an elliptic curve $E$. The torsion of $H^2(X,\mathcal{O}_X^*)$ is therefore precisely the Brauer group of $X$, and these are precisely the torsion points of $E$. -One can produce similar examples using abelian surfaces. For instance, if $E$ is a CM elliptic curve, then $E\times E$ is an abelian surface such that $H^2(E\times E,\mathcal{O}^*)$ is isomorphic to $E$.<|endoftext|> -TITLE: Rigid nilpotent Lie algebras -QUESTION [6 upvotes]: I am very interested in the Nicola Ciccoli's answer to a question formulated in mathoverflow:67717 -My questions are: - -Can a nilpotent Lie bracket be rigid in the scheme of Lie brackets on $\mathbb{C}^n$? Is it easy to prove a rigid nilpotent Lie algebra exists? -Do there exist rigid Lie brackets in the scheme of nilpotent Lie brackets on $\mathbb{C}^n$ for $n \geq 7$? - -REPLY [9 votes]: Vergne's conjecture is still open. It says that there is no complex $n$-dimensional nilpotent Lie algebra which is rigid in the variety $\mathcal{L}_n(\mathbb{C})$ of all $n$-dimensional complex Lie algebras. The Grunewald and O’Halloran conjecture, which implies Vergne's conjecture, states that every complex nilpotent Lie algebra is the proper degeneration of another Lie algebra of the same dimension. -The Heisenberg Lie algebra, with $[x,y]=z$ is rigid in $\mathcal{N}_3(\mathbb{C})$, but -not in $\mathcal{L}_3(\mathbb{C})$, since it is a proper degeneration of $sl_2(\mathbb{C})$. -R. Carles gives a necessary condition for a nilpotent Lie algebra $L$ to be rigid in $\mathcal{L}_n(\mathbb{C})$: it must be characteristically nilpotent, such that every ideal -of codimension $1$ in $L$ is again characteristically nilpotent. It is not clear, whether -Carles condition is also sufficient. Certainly the condition to be characteristically nilpotent is not sufficient. It is easy to construct characteristically nilpotent Lie algebras, which are not rigid. But even for the stronger condition of Carles, I found -filiform nilpotent Lie algebras of dimension $n\ge 13$, which are characteristically nilpotent, and every ideal of codimension $1$ also being characteristically nilpotent. -Unfortunately I do not know whether these algebras are rigid or not -(there is a claim in the literature that no filiform nilpotent Lie algebra can be rigid in $\mathcal{L}_n(\mathbb{C})$, but I could not verify this, and did not find a valid proof).<|endoftext|> -TITLE: Why does Hom need an identity in the definition of the category? -QUESTION [7 upvotes]: I was studying the axioms of a category, and noted that one axiom says there is an element $1_X\in Hom(X,X)$ for any object $X$ which serves as the identity. Why is this axiom necessary? What happens if I drop this axiom? - -Background: I can define the category of affine holomorphic symplectic varieties, by saying - -The objects are semisimple algebraic groups -The morphisms $Hom(G,G')$ are affine holomorphic symplectic varieties with Hamiltonian $G\times G'$ action -The composition of two morphisms $X\in Hom(G,G')$ and $Y\in Hom(G',G'')$ is given by the holomorphic symplectic quotient $X\times Y//G'$. - -This becomes a nice symmetric monoidal category; the identity in $Hom(G,G)$ is $T^*G$. -Suppose I want to consider the category of hyperkähler manifolds instead. I can try the following - -The objects are semisimple compact groups -The morphisms $Hom(G,G')$ are hyperkähler manifolds with Hamiltonian $G\times G'$ action -The composition of two morphisms $X\in Hom(G,G')$ and $Y\in Hom(G',G'')$ is given by the hyperkähler quotient $X\times Y///G'$. - -Now, the problem is that $T^*G_\mathbb{C}$ has a hyperkähler metric (constructed by Kronheimer) and almost acts like an identity, but not quite: given a hyperkähler manifold $X$ with $G$ action, $T^*G_\mathbb{C} \times X /// G$ is equivalent to $G$ as holomorphic symplectic varieties but not equivalent as hyperkähler manifolds. -What should I do? - -For my purpose, I guess using the terminology semigroupoid would suffice (I just want to define the target "category" of a TQFT precisely.) But I'm curious what kind of hell will break loose if I drop this axiom, why the people who originally defined categories included this into the axiom, etc. - -REPLY [13 votes]: As Fernando points out, you can't talk about isomorphisms in a semicategory, which means that they won't be as much use as categories in describing universes of mathematical objects. But the category of semicategories has a surprisingly interesting relationship to that of categories. There is of course a forgetful functor $\mathrm{Cat} \to \mathrm{Semicat}$, and as Scott says it has a left adjoint that does what you expect. But it also has a right adjoint, which takes a semicategory S to the category of idempotents in S: the objects are idempotents $e \colon a \to a$ and a morphism $e \to e'$ is a morphism $f \colon a \to a'$ in S such that $fe = f = e'f$. So we get a monad on Cat whose unit is the canonical functor from a category to its idempotent-splitting completion, or Cauchy completion, or Karoubi envelope. -Böhm, Lack and Street use this framework here to talk about weak Hopf algebras. They show that 'weak monoids' fall naturally out of the formal theory of monads if instead of working directly in a bicategory you Cauchy-complete the hom-categories first. -Another application of semicategories and semifunctors is in computer science: Hayashi, Adjunction of semifunctors: categorical structures in nonextensional $\lambda$-calculus, TCS 41, shows how to describe $\lambda$-calculus without the $\eta$-law quite elegantly. I haven't worked it out, but it seems to me that this framework should also give a way of talking about 'weak limits' (the kind with not-necessarily-unique mediating morphisms) in terms of adjunctions. - -REPLY [6 votes]: I suppose the extent to which hell breaks loose depends entirely on the purpose you're using categories for. Dropping identities presumably invalidates the Yoneda lemma, and therefore all the results in category theory that depend on it. But if you just want to use (monoidal) functors as bookkeeping devices without expecting "deeper" category theory to predict things for you, nothing much happens. To follow up on Scott's answer, there is for example a perfectly good theory about adjunctions when one ignores identities, and there is in fact a relation to formally adding identities. See Hayashi "Adjunction of semifunctors" in Theoretical Computer Science 41:95--104, 1985, and Hoofman and Moerdijk "A remark on the theory of semi-functors" in Mathematical Structures in Computer Science 5:1--8, 1995.<|endoftext|> -TITLE: Morita invariance of Drinfeld centre -QUESTION [5 upvotes]: Given a monoidal category $M$ one can consider its Drinfeld centre $Z(M)$. Objects of the Drinfeld centre are pairs $(m, \alpha)$ where $m$ is an object and $\alpha$ is an isomorphism $\alpha: - \otimes m \to m \otimes -$ satisfying some "obvious" conditions. -A simple and important example of a monoidal category is the category of $G$-equivariant sheaves of $k$-vector spaces on $Y \times Y$ where $Y$ is a finite $G$-set. The monoidal structure is given by -$V\otimes W = p_{13*} (p_{12}^*V \otimes p_{23}^* W)$ -where $p_{ij} : Y \times Y \times Y \to Y \times Y$ denotes the projection map (in a hopefully obvious notation). -For example, if $Y$ is a point then one recovers the (tensor) category of representations of $G$. If $Y = G$ then one recovers a monoidal category equivalent vectors spaces graded by $G$. If $G$ is the trivial group then one obtains the tensor category of ``matrices of vector spaces'' over $Y$. -Now there is a result, which I have heard (by Ostrik) called ``Muerger's Morita invariance of Drinfeld centre''. It should have the consequence that, with $G$ and $Y$ as above: - -The Drinfeld centre $Z(Sh_G(Y \times Y))$ does not depend on $Y$ up to equivalence. - -(I guess the baby example of $G$ the trivial group explains the term ``Morita invariance''.) -My question is: - -Where can I read about these results in the literature? - -REPLY [9 votes]: To expand on Noah's answer, Müger shows that if two fusion categories are Morita equivalent, in the sense that there is an invertible bimodule between them, then their Drinfeld centers are equivalent. In fact, the reverse implication is also true: this is Theorem 3.1 in a different Etingof-Nikshych-Ostrik paper. So two fusion categories having equivalent centers is the same as Morita equivalence. -In your case, if $Y$ and $Y'$ are two $G$-sets, then $Sh_G(Y \times Y')$ is a bimodule from $Sh_G(Y \times Y)$ to $Sh_G(Y' \times Y')$ with the obvious left and right actions. This bimodule is invertible, with inverse $Sh_G(Y' \times Y)$. In other words, $$Sh_G(Y \times Y') \boxtimes_{Sh_G(Y' \times Y')} Sh_G(Y' \times Y) \cong Sh_G(Y \times Y)$$ as $Sh_G(Y \times Y)$-bimodules, and similarly with $Y$ and $Y'$ switched.<|endoftext|> -TITLE: Average squared distance vs diameter in vertex-transitive graphs -QUESTION [6 upvotes]: Let $X=(V,E)$ be a finite, connected graph on $n$ vertices, endowed with its graph metric $d$. The average squared distance of $X$ is $avg(d^2)=\frac{1}{n(n-1)}\sum_{x,y\in V,x\neq y} d(x,y)^2$; it satisfies the obvious bound $avg(d^2)\leq diam(X)^2$, where $diam(X)$ is the diameter of $X$. -Now assume that $X$ is vertex-transitive. My intuition is that, in this case, for ``many'' pairs of vertices, the distance is much smaller than the diameter, which should entail an inequality $avg(d^2)\leq \lambda(diam(X))^2$, where $\lambda<1$ is some constant, maybe depending only on the common degree of the vertices. Is this intuition correct? If yes, can $\lambda$ be estimated? -EDIT: Thanks to all for your input. The example of the complete graph is somewhat embarrassing, meaning that the OP was poorly formulated. As Aaron sort of guessed, I'm interested in families of $k$-regular graphs ($k$ fixed) with number of vertices increasing to infinity. So the new question would be: does a bound $avg(d^2)\leq \lambda(diam(X))^2$ hold for $|V|$ large enough? Observe that, for vertex-transitive graphs, the lower bound $avg(d^2)\geq\frac{(diam X)^2}{8}$ holds: see proposition 3.4 in -http://toctest.cs.uchicago.edu/articles/v005a006/v005a006.pdf - -REPLY [2 votes]: For Random 3-regular graphs \lambda is asymptotically 1. I suspect that for Ramanujan graphs it will be 1 as well? -Regarding random graphs papers to look are papers citing Bollobas and de la Vega, Combinatorica 2 -(1982), 125-134. -http://www.stanford.edu/class/msande337/notes/the%20diameter%20of%20random%20regular%20graphs.pdf -which does not contain it but a related result. -A reference which contains the claim is Remco van der Hofsted's book: see Theorem -10.15 and Theorm 10.16 (in the latest version on his webpage).<|endoftext|> -TITLE: obstruction theories in algebraic geometry -QUESTION [20 upvotes]: I'd like to know about the history of obstruction theories in algebraic geometry, as well as the relationship with concepts of the same name in topology. I would also like to know where obstruction theories have been used in algebraic geometry. The applications I know of are in Artin's criteria for showing that stacks are algebraic, and in the construction of virtual cycle classes in Gromov-Witten, Donaldson-Thomas, and similar theories. -Morally, an obstruction theory is supposed to control infinitesimal liftings: suppose $X \rightarrow Y$ is a map of algebraic geometry objects (schemes or stacks or whatever) and $S \subset S'$ is a square-zero extension of $Y$-schemes. An obstruction theory for $X$ over $Y$ is, vaguely speaking, a way of associating to any $Y$-morphism $S \rightarrow X$ an obstruction to the existence of an extension of that map to a $Y$-morphism $S' \rightarrow X$. In any individual lifting problem, an obstruction can usually be found in some cohomology group or other, but it is useful for some purposes (like the ones noted above) to have an abstract definition. -I know of four attempts to axiomatize the notion of an obstruction theory in algebraic geometry already: -1) Artin, M. Versal deformations and algebraic stacks -2) Fantechi, B. and Manetti, M. Obstruction calculus of functors of Artin rings, I -3) Li, J. and Tian, G. Virtual moduli cycles and Gromov--Witten invariants of algebraic varieties -4) Behrend, K. and Fantechi, B. The intrinsic normal cone - -REPLY [4 votes]: Fabien Morel has developed a version in $\mathbb{A}^1$-homotopy theory of the obstruction theory which is familiar in algebraic topology. See for example his paper http://www.mathematik.uni-muenchen.de/~morel/bgln.pdf . He uses it to prove some results about vector bundles on smooth affine varieties which are analogues of classical results about vector bundles on compact (real) manifolds.<|endoftext|> -TITLE: Images of Borel subsets of non-metric compact spaces -QUESTION [11 upvotes]: The following question was noted by Jan Pachl in connection with the -study of Arens products and he has not received a satisfactory -answer from the various experts he has asked. Let $X$ and $Y$ be -compact Hausdorff spaces and let $F$ be a continuous mapping from $X$ -onto $Y$. Let -$A\subseteq Y$ and suppose that $F^{-1}A$ is Borel. Does it follow -that $A$ is also Borel. -Certainly if $X$ and $Y$ are metric then the answer is positive; in -this case both $Y\setminus A = F(X\setminus F^{-1}A)$ and $A = -F(F^{-1}A)$ are analytic and hence Borel. But even for $X=Y=2^{\omega_1}$ -the argument that disjoint analytic sets can be separated by Borel -sets does not seem to be available. - -REPLY [6 votes]: It is true that if $f:K \to L$ is a continuous mapping from a compact space $K$ onto $L$ and $A \subseteq L$ has Borel preimage in $K$ then $A$ is Borel in $L$. Jan Pachl points out that this is a very special case of a general theorem (Theorem 10) by -P. Holický and J. Spurný in "Perfect images of absolute Souslin and absolute Borel Tychonoff spaces", Topology Appl. 131 (2003), 281--294.<|endoftext|> -TITLE: What are examples of theorems which were once "valid", then became "invalid" as standard definitions shifted? -QUESTION [25 upvotes]: That is, results established by correct proofs within some framework, yet the manner in which their author or the general mathematical community at the time would describe these results would, in later times, be interpreted as constituting a false claim, due to changing fashions as to how to standardly formalize some of the relevant concepts. -I imagine this sort of thing has happened often (e.g., with shifting accounts of "polyhedra" a la Lakatos' "Proofs and Refutations", or a motley of different definitions of "continuity" before standardization on the one we use now), but I do not have enough awareness of history to be able to provide solid examples (e.g., it seems plausible to me that Darboux may have considered himself to have proven that every derivative is continuous, taking the intermediate value property to be defining for continuity, but I do not know if this is an accurate account of what he claimed). - -REPLY [8 votes]: I am still not 100% sure I understand what the question is asking for, but it occurred to me today that one example might be the following statement: - -$(*)$ If every S is P then some S is P. - -Today, we would say that $(*)$ is not valid, because if S is vacuous then "every S is P" is true but "some S is P" is false. However, for most of the history of Western civilization, $(*)$ was considered valid. This is usually explained by saying that the "classical" statement that "every S is P" really means, in modern language, "there exists some S and every S is P." -This point is discussed in detail in the article on The Traditional Square of Opposition in the Stanford Encyclopedia of Philosophy, where it is also pointed out that if we additionally translate "some S is not P" into modern language as "if there exists some S then some S is not P" then we can recover the entire traditional square of opposition. -This seems to meet Sridhar Ramesh's request for an example not of changing standards of rigor but of "shifting standard formalizations of preformal concepts" (in this case, the concepts of "every" and "some").<|endoftext|> -TITLE: The classical Krein-Rutman theorem -QUESTION [8 upvotes]: The classical Krein-Rutman theorem states that any positive compact linear endomorphism $T:X \to X$ on a Banach space $X$ with positive spectral radius $r(T)$ has an eigenvalue $r(T)$ with a positive eigenvector. Papers and textbooks seem to write off the theorem as "standard" and "well-known", but I have not been able to locate any exposition with a proof of the theorem. -Is there a reference (preferably a textbook) including the statement and the proof of the theorem? (The original paper of Krein and Rutman appears to be in Russian, which I cannot read.) - -REPLY [6 votes]: For Banach lattices, a statement and a sketch of the proof can be found in Abramovich and Aliprantis's article "Positive operators" in the Handbook of the Geometry of Banach Spaces: see Google books excerpt<|endoftext|> -TITLE: Growth of Laplacian eigenvalues on a compact domain? -QUESTION [8 upvotes]: Let $\mathcal{M}$ be a compact Riemannian manifold and let $\Delta$ be the (scalar) Laplace-Beltrami operator on $\mathcal{M}$. Then $\Delta$ has a discrete spectrum and if we order its distinct eigenvalues $\lambda_i$ by magnitude then some very simple examples suggest that the magnitude of $\lambda_i$ might be roughly quadratic in $i$. For instance, on the circle $S^1$ eigenfunctions have the form $\cos(nx)$ or $\sin(nx)$ for $n \in \mathbb{N}_0$; hitting these functions with $-\frac{\partial}{\partial x^2}$ yields $n^2\cos(nx)$ and $n^2\sin(nx)$, respectively. Similar analysis can be done for the geometrically flat torus $T^2$. On the 2-sphere, we have $\lambda_i=i(i+1)$. This rough idea of "differentiating twice leads to a square" makes me suspect that a similar relationship might hold for other domains -- what can be said in general? I'm particularly interested in smooth surfaces embedded in $\mathbb{R}^3$. -Update: Weyl's formula provides some valuable information about the Laplace spectrum, but does not determine the asymptotic growth of $\lambda_i$. For instance, suppose we have a manifold such that $N(R) \approx R$, i.e., the number of eigenvalues with value no greater than $R$ is roughly equal to $R$ itself. Letting $n_i$ be the multiplicity of $\lambda_i$, this relationship holds for, say, $\lambda_i = i(i+1)$ and $n_i = 2i+1$ (which is the situation on the sphere), since -$$ N(\lambda_i) = \sum_{j=0}^i n_j = \sum_{j=0}^i 2i+1 = i^2 + 2i + 1 \approx i(i+1) = \lambda_i. $$ -But it also holds for $\lambda_i = i$ and $n_i = 1$ since then -$$ N(\lambda_i) = \sum_{j=0}^i n_j = \sum_{j=0}^i 1 = i + 1 \approx i = \lambda_i. $$ - -REPLY [13 votes]: Weyl's formula: -$$N(R)=\frac{1}{(4{\cdot}\pi)^{d/2}{\cdot}\Gamma\left(\frac d2+1\right)}{\cdot}V{\cdot}R^{d/2}+o(R^{d/2}).$$ -where $d$ --- dimension, $V$ --- volume, $N(R)$ --- number of eigenvalues $\le R$. -It works for any compact Riemannian manifold.<|endoftext|> -TITLE: Invariants and orbits of $n$-tensors -QUESTION [14 upvotes]: My question may be absolutely elementary and is probably answered in 19th century. A reference or a short clear argument would be highly appreciated. -Let $V_1, \ldots V_n$ be finite dimensional vector spaces over the same field (may assume complex numbers). What are $GL(V_1)\times \ldots \times GL(V_n)$-orbits on $V_1 \otimes \ldots \otimes V_n$? -The only invariant of an orbit I can see is "a multirank" $(k_1, \ldots k_n)$ where $k_i$ is the dimension of support of an element in $V_i$. The multirank satisfies inequalities $k_i \leq \prod_{j\neq i} k_j$. Would it be too naive to suggest that orbits are in 1-1 correspondence with legal multiranks? - -REPLY [8 votes]: Here is a start, suppose that $V_i$ is $\mathbb C^{k_i}$ (and restricting to $k_1,k_2,\dots,k_n, n\geq 2$). The tuples $(k_1,k_2,\dots,k_n)$ for which the action of $GL_{k_1}\times\cdots\times GL_{k_n}$ on $\mathbb{C}^{k_1}\otimes \cdots\otimes \mathbb{C}^{k_n}$ has only finitely many orbits are $(k,l),(2,2,k),(2,3,k)$, for positive integers $k,l$. This was proven in - -V. G. Kac, "Some remarks on nilpotent orbits", J. Algebra 64 (1980), 190–213. - -These orbits are classified in "Orbits and their closures in the spaces $\mathbb{C}^{k_1}\otimes \cdots\otimes \mathbb{C}^{k_r}$" by P.G. Parfenov (MR). This paper doesn't seem to be freely online in English, but Russian version is here, and I believe you can find a summary in section 5 here. - -REPLY [4 votes]: I don't think that's right, Bugs. Say that $V_1$ and $V_2$ are $d$-dimensional and $V_3=\mathbb C^2$. Then an element of $V_1\otimes V_2$ is a linear map $V_1^\star\to V_2$, generically an isomorphism; an element of $V_1\otimes V_2\otimes V_3$ is an ordered pair $(A,B)$ of these; the unordered $d$-tuple of eigenvalues of $B\circ A^{-1}$ is an invariant of the $GL(V_1)\times GL(V_2)$-action; and an element of $GL(V_3)$ will just perform some fractional linear transformation on all of these numbers, so that if $d\ge 4$ then there is a complex invariant here.<|endoftext|> -TITLE: Complex manifolds in which the exponential map is holomorphic -QUESTION [28 upvotes]: Let $X$ be a complex manifold and $g$ a hermitian metric on $X$. Consider the Riemannian exponential $\exp_p: T_p X \to X$. -If $\exp_p$ is holomorphic for every $p \in X$, then $(\exp_p)^{-1}$, suitably restricted, provide holomorphic normal coordinates near $p$, with respect to which the metric osculates to order 2 to the standard metric at the origin. This shows that $g$ is a Kähler metric. -However, Kähler is not sufficient to ensure that $\exp_p$ is holomorphic: take $X$ a curve of genus $g \geq 2$. If $\exp_p:T_pX \to X$ is holomorphic, then it lifts to a holomorphic map from $T_pX$ to the universal cover $\widetilde{X} = \Delta$, giving a holomorphic map $T_pX \simeq \mathbb{C} \to \Delta$, which must be constant by Liouville's theorem. In fact, one can see that $\exp$ cannot be holomorphic if $X$ is Kobayashi hyperbolic. -This leaves the question: What are the hermitian manifolds/metrics whose exponential map is holomorphic? - -REPLY [22 votes]: NB: I've had a little time to think about this and can now improve my answer, in particular, removing the real-analytic assumption, which, as I suspected, was not necessary. Here is the improved answer: -If the metric $g$ is Kähler, then having the exponential map from a point $p\in M$ be holomorphic makes it flat in a neighborhood of $p$. -Suppose that $\exp_p:T_pM\to M$ is holomorphic near $0_p\in T_pM$ (where we use the natural holomorphic structure on the complex vector space $T_pM$). Let $z:T_pM\to\mathbb{C}^n$ be a complex linear isometry, so that the hermitian metric on $T_pM$ is just $|z|^2$ in the usual sense. Let $Z$ be the holomorphic 'radial' vector field on $\mathbb{C}^n$, whose real part is the standard radial vector field on $\mathbb{C}^n$. -Then -$$ -{\exp_p}^*g = g_{i\bar j}(z)\ dz^i\ d\overline{z}^j -$$ -for some functions $g_{i\bar j}$ on a neighborhood of $0\in\mathbb{C}^n$. Since $g$ is Kähler, there is a function $f$ defined on a neighborhood of $0\in\mathbb{C}^n$ such that -$$ -g_{i\bar j} = \frac{\partial^2f}{\partial z^i\ \partial\overline{z}^j}. -$$ -Now, the condition that $z$ furnish Gauss normal coordinates for ${\exp_p}^*g$ is easily seen to be that -$$ -\mathcal{L}_Z\bigl(\bar\partial f\bigr) = \bar\partial\bigl(|z|^2\bigr). -$$ -In particular, $ \bar\partial\bigl(\mathcal{L}_Z(f - |z|^2)\bigr) = 0$, -so $\mathcal{L}_Z(f - |z|^2) = h$ for some holomorphic function $h$ on a neighborhood of $0$. This $h$ must vanish at $0$, so it is easy, by adding the real part of the appropriate holomorphic function to $f$ (which won't change $g$) to arrange that $h\equiv0$ and, moreover, that $f(0) = 0$. But this now implies that the real-valued function $f-|z|^2$ vanishes at the origin and also is constant along the radial vector field. Thus, $f = |z|^2$, and the metric $g$ is flat in these coordinates.<|endoftext|> -TITLE: What extra conditions are necessary for the following version of Koszul duality? -QUESTION [7 upvotes]: Conventions: So that I don't have to worry about, fix a field $k$ of characteristic zero, and always work over it. Categories of modules, etc., are always $\infty$-categories of dg modules. Algebras are "associative" in the coherent-homotopy sense. All tensor products are left-derived. Etc. I am of course always interested in hearing about subtleties and generalizations of such conventions, but for the sake of this question you may pretend that I understand this part of the story. -A construction: Let $A$ be an algebra and $V$ a left $A$-module. Write $V^\ast$ for the linear dual to $V$; then $V^\ast$ is a right $A$-module. Set $C = V^\ast \otimes_A V$. I claim that $C$ is a coalgebra, at least when $V$ satisfies some finiteness condition. Indeed, if $V$ satisfies a finiteness condition, then $\operatorname{End}_k(V) = V\otimes_k V^\ast$, and the action of $A$ on $V$ is encoded in a map $A \to \operatorname{End}_k(V)$, which by associativity is a map of $A$-$A$-bimodules. Then the comultiplication is: -$$ C = V^\ast \otimes_A V = V^\ast \otimes_A A \otimes_A V \to V^\ast \otimes_A (V \otimes_k V^\ast) \otimes_A V = C \otimes_k C $$ -This is coassociative on account of the coassociativity of tensor products. -For $X$ any left $A$-module, there is a corresponding left $C$-comodule defined by $V^\ast \otimes_A X$; the comodule structure is -$$ V^\ast \otimes_A X = V^\ast \otimes_A A \otimes_A X \to V^\ast \otimes_A (V \otimes_k V^\ast) \otimes_A X = C \otimes_k (V^\ast \otimes_A X) $$ -Aside: Let $C$ be any (coassociative) coalgebra, $Y$ a left $C$-comodule, and $Z$ a right $C$-comodule. Recall that the underived cotensor product is the 1-categorical equalizer of the two maps $Z \otimes Y \rightrightarrows Z \otimes C \otimes Y$. The (derived) cotensor product $Z \Box_C Y$ is the right-derived version thereof (cotensor is left-exact, if I haven't made an error); it should also be the $\infty$-categorical equalizer. -At least in the 1-categorical non-dg setting cotensor products of bi-comodules and so on are not always associative. But they are associative when all coalgebras are flat over whatever ground ring you're working over, and we're working over a field, so this is not an issue. I don't know what the correct statement is in the dg $\infty$-categorical level. -The construction, continued: In a similar way, $V$ is a right $C = V^\ast \otimes_A V$ comodule, and so for any left $C$-comodule $Y$, I can define a left $A$-module $V \Box_C Y$. -All together, I've constructed functors -$$V^\ast \otimes_A: A\text{-mod} \leftrightarrow C\text{-comod} :V \Box_C.$$ -One composition of these functors (the one from $C$-comod to $C$-comod) is the identity: -$$ V^\ast \otimes_A V \Box_C = C \Box_C $$ -The other composition cannot always be the identity — just imagine what would happen if I took $V$ to be the zero module! -Question: A statement that perhaps deserves to be called Koszul duality is that these two functors are an equivalence of $\infty$-categories. Of course, that seems almost nonsense to me, because in the non-dg-$\infty$ world categories of modules and categories of comodules seem quite different. So part of my question is to clarify the statement of the statement. But the bulk of my question is: What conditions do I need to add in the above exposition to have the statement I'd like? For example, I would like to identify $\operatorname{End}_k(V) = V \otimes_k V^\ast$, and so I would expect to need some "finiteness" condition on $V$, or I would expect to need some topology. -Note that what many people call "Koszul duality" consists also of taking the dual $C^\ast$ to the coalgebra $C$, so as to get an algebra. Every left comodule of $C$ is a left module of $C^\ast$, but there are generically more of the latter, at least in the 1-categorical version of the story. Given that many discussions of Koszul duality use the (derived) $\operatorname{End}_A(V,V)$, which is an algebra that should be essentially the same as $C^\ast$, I worry that maybe the algebra-to-algebra version is more robust. But I'm not sure. -Bonus question and further reading: Sometime soon I will be writing up a construction related to the one above, as part of a larger project. I picked up the above ideas from discussions with various people. But I would like to give credit where it's due, so I would like to hear about any particular papers I should be sure to cite. -For further reading, you might check out this question from the first month of MathOverflow, and also perhaps Jacob Lurie's ICM address. - -REPLY [8 votes]: This version of Koszul duality (as well as many others) can be deduced from the Barr-Beck-Lurie theorem (cf. Lurie's book Higher Algebra). You can consider the functor -from k-mod to A-mod of tensoring by V and ask for it to have a left or right adjoint, giving rise to a comonad or monad on k-mod (ie the coalgebra or algebra form). These are two finiteness conditions on V (k-dualizability and A-dualizability - ie dualizability as a vector space or compactness as A-module - though maybe I'm mixing up the order). If you ask for both simultaneously then you guarantee that the functor from A-mod to (co)modules over the (co)monad is both limit and colimit preserving -- this is overkill but an easy way to ensure the hypotheses of the Barr-Beck-Lurie theorem are satisfied. You now get an equivalence between your category of (co)modules and a COMPLETION of A-mod --- i.e., the part of A-mod that your particular chosen module V sees (the category it generates). (This is how you satisfy the other requirement of the theorem, i.e., that the corresponding functor is conservative). -The classical version of this is A=S, symmetric algebra in one variable, ie A-mod = quasicoherent sheaves on the line, and V is the augmentation module (skyscraper at the origin). We then have a descent theorem, saying that the completion of A at the augmentation is derived equivalent to sheaves on a point with descent data for inclusion of point to line -- which is exactly modules for the Koszul dual exterior coalgebra (or dually exterior algebra). -EDIT: In response to the comment about "seeing" modules: technically the condition is whether there are any Exts between your given object V and some other object W. -I think of this geometrically: if V stands for a skyscraper on a variety, it will see the entire formal neighborhood of the point, i.e. intuitively you can manufacture Exts with the skyscraper for anything that has a stalk at this point. So for instance for enveloping algebras a representation will see only representations that have all invariants the same as V (i.e. the center must act with the same generalized character). So I'm not sure I understand the question regarding Lie algebras (take the example of $\mathfrak g$ the trivial one-dimensional Lie algebra and we're back in the original Koszul duality discussed in the paragraph above).<|endoftext|> -TITLE: Zeros of holomorphic one-forms on Riemann surface -QUESTION [7 upvotes]: Is it true that for any point on any compact Riemann surface there exists a global holomorphic one-form, which does NOT have a zero at that point. - -REPLY [13 votes]: On $\mathbb P^1$, there is no non zero holomorphic $1$-form, on any elliptic curves, the holomorphic forms are "constant" (the canonical bundle is trivial), so never vanish if they are not identically zero. -As for the other surfaces, namely if $g(X) \geqslant 2$, then $|K_X|$ has no base point (cf Hartshorne, IV, lemma 5.1), which amouts to saying that for all point $x\in X$, there exists a holomorphic form non-vanishing at $x$. -Moreover, if $X$ has genus $g\geqslant 2$ as previously and $X$ is not hyperelliptic, then $K_X$ is very ample (cf Hartshorne, IV, proposition 5.2), which means that the linear system given by the (global) holomorphic $1$-forms induces an embedding into $\mathbb P H^0(X, K_X)^* \simeq \mathbb P^{g-1}$.<|endoftext|> -TITLE: Eigenvalues of the sum of a diagonal and a unit matrix -QUESTION [10 upvotes]: I'm trying to find information on the eigenvalues of an $n \times n$ matrix A such that -$A = D + J$ -Where $D$ is some complex valued diagonal matrix, and $J$ is an matrix consisting of all $1$'s. -When $D$ has identical values, the problem is equivalent to finding the eigenvalues of $J$. -So my question is this: -If $D$ has non-identical values (specifically, non-identical imaginary components), -is there an elementary way to compute the eigenvalues of $A$ ? -The problem comes from linearising about the origin of a system of $n$ near identical coupled resonators. $D$ relates to the behaviour of each resonator, $J$ relates to the coupling process. - -REPLY [13 votes]: There is a formula. Recall that $\det(I-AB)=\det(I-BA)$ for any matrices -$A$ and $B$ such that both products $AB$ and $BA$ are defined. Now -$$ - \det(tI-D-J) = \det(tI-D) \det(I-(tI-D)^{-1}J). -$$ -If $u$ is the column vector with each entry equal to 1 then $J=uu^T$ and -$$ - \det(I-(tI-D)^{-1}J) = \det(I-(tI-D)^{-1}uu^T) = \det(1- u^T(tI-D)^{-1}u) = 1-u^T(tI-D)^{-1}u. -$$ -If we write $\phi(M,t)$ for the characteristic polynomial off $M$, this yields -$$ - \phi(D+J,t) = \phi(D,t) \left(1-\sum_i \frac1{t-D_{i,i}}\right). -$$ -The sum is equal to $\phi'(D,t)/\phi(D,t)$ and therefore the right side equals $\phi(D,t)-\phi'(D,t)$.<|endoftext|> -TITLE: Zero sets of harmonic fucntions -QUESTION [6 upvotes]: Can a two variable Harmonic function f(x,y) be zero on a curve with a cusp? - -REPLY [3 votes]: There exists an honor thesis dedicated to exactly this problem, and its generalizations: Problems in Harmonic Function Theory Ronald A. Walker, see http://scholarship.richmond.edu/cgi/viewcontent.cgi?article=1488&context=honors-theses<|endoftext|> -TITLE: Models of computation with decidable halting problem? -QUESTION [7 upvotes]: There are numerous examples of models of computation in which all programs halt, for example primitive recursion. -Are there (non-trivial) examples of models in which only some programs halt, but the halting problem is still decidable? -Does the decision procedure need to lie outside of the original model itself? -EDIT: -Carl Mummert gives very good example of a model of computation that has this property. But the example of polynomially clocked Turing machines is weaker than primitive recursion. -Are there examples which are stronger than primitive recursion? - -REPLY [5 votes]: Joel Hamkins points out that the decision procedure for any reasonable notion of "computability" is not going to be solvable by a function that is "computable" within that notion. -Here is a contrasting example of a nontrivial model of computation in which the halting problem is solvable in the usual sense of computation. An index $e$ in our new system is a pair $(e_1, p)$ where $e_1$ is an index for a Turing machine and $p$ is a code for a polynomial over $\mathbb{N}$. In our new system, program $e$ is said to compute output $o$ on input $n$ (write $P_e(n) = o$) if and only if Turing machine $e_1$ computes $o$ on input $n$ in less than $p(|n|)$ steps. If $e_1$ runs for more then $p(|n|)$ steps then we say the computation of $P_e(n)$ is undefined (i.e. does not halt). Here we assume the Turing machine uses binary coding for numbers and we let $|n|$ be the number of bits required to express $n$ in binary notation. -This restricted model of computation is relatively common in the study of polynomial-time computability, where an index of the form $e = (e_1, p)$ is called a "polynomially clocked Turing machine". It's immediate from the definitions that a function is computable in the restricted model if and only if it is computable in polynomial time. Thus the model includes a very wide class of functions. However, because the time bound for index $e$ is already included in $e$, we can solve the halting problem for this model of computation with a normal Turing machine. (We cannot solve it with any polynomially clocked machine, of course.)<|endoftext|> -TITLE: Is there a Levi decomposition for Lie group and algebraic group? -QUESTION [6 upvotes]: Let $G$ be a Lie group and $R$ be the largest connected solvable -normal subgroup of $G$. -Question 1 -Is there a Lie subgroup $S$ such that: (1) $G=SR$; (2) -every real representation of $S$ is semisimple? -Question 2 -Is there a Lie subgroup $S$ such that: (1) $G=SR$; (2) -every complex representation of $S$ is semisimple? -Let $G$ be an algebraic group and $R$ be the largest connected -solvable normal subgroup of $G$. Is there a algebraic subgroup $S$ -such that: (1) $G=SR$; (2) every representation of $S$ is -semisimple? -I want to know the formal statement and references. - -REPLY [5 votes]: I was unsure if I should post this because the question has not been active for a year. I decided to go ahead anyway, because people might have the same question in the future. -The following paragraph is taken from Structure of Lie Groups and Lie Algebras by -V. V. Gorbatsevich, A. L. Onishchik and E. B. Vinberg. -Let $A$ be a simply-connected covering group for $SL_2(\mathbb{R})$. As is -known, $Z(A) \simeq \mathbb{Z}$. Let $z_0$ be a generator of the group $Z(A)$, and let $t_0 \in \mathbb{T}=SO_2$ be a rotation through an angle incommensurable with $\pi$. Then $\Gamma = \langle(t_0, z_0)\rangle$ is a discrete normal subgroupof $\mathbb{T} \times A$. Let $G = (\mathbb{T} \times A)/\Gamma$. Then the image $L$ of the subgroup A under the natural homomorphism $\mathbb{T} \times A \rightarrow G$ is a Levi subgroup of G. The fact -that the subgroup $\langle t_0 \rangle$ is dense in $\mathbb{T}$ implies that $L$ is dense in $G$, i.e. it is not a Lie subgroup. -This shows that question (1), which is true for algebraic groups in characteristic 0 as explained in the other answer, is not generally true for Lie groups, as the Levi factor can fail to be closed.<|endoftext|> -TITLE: Groupoids vs Pseudogroups -QUESTION [9 upvotes]: (Warning: I'm not an expert in the topic) Let's work in a "geometric" category, for example the category $\mathfrak{Diff}$ of "manifolds" (without the requirements of connectedness and second countability and even of finite dimensionality), and consider also the subcategory of "usual" finite dimensional (Edit: possibly disconnected, Hausdorff, second countable) manifolds $\mathrm{Diff}$. We can than consider two kinds of objects: - -Lie groupoids $\mathcal{G}=(\mathcal{G}_1 \rightrightarrows \mathcal{G}_0)$ where $\mathcal{G}_0$ and $\mathcal{G}_1$ are smooth manifolds (in $\mathfrak{Diff}$) and the source $\mathrm{src}$ and target $\mathrm{trg}$ maps are submersions. -Pseudogroups $\Gamma$ of diffeomorphisms of some manifold $M$ in $\mathrm{Diff}$. - -Remark 1. From a pseudogroup $\Gamma$ on $M$ (for example the local symplectomorphisms if $M$ is equipped with a symplectic form, or the local diffeomorphisms preserving a foliation...) an (abstract) groupoid $\mathcal{G}=\mathrm{Germ(\Gamma)}$ can be obtained via the construction of germs, and I believe it's also a (usually infinite dimensional) Lie groupoid. In this case $\mathcal{G}_0$ is some space of germs of manifolds, and $\mathcal{G}_1$ consists of germs of local diffeomorphisms belonging to $\Gamma$. -(Edit: for the benefit of the readers let me say D.Carchedi in his answer below has pointed out that this $\mathcal{G}$ is actually finite dimensional). -Remark 2. Given a Lie group action $\varphi\colon K\times M \to M$, the "action groupoid" $G_\varphi=(K\times M \rightrightarrows M)$ in $\mathrm{Diff}$ can be constructed. If I'm not mistaken, also an "action pseudogroup" $\Gamma_\varphi$ can be constructed, which consists of all the restrictions of the diffeomorphisms of the form $\varphi_g$, $g\in K$, to open subsets of $M$. -Remark 3. We can apply the germ construction to $\Gamma_{\varphi}$, obtaining a new groupoid $\mathcal{G}_{\varphi} = \mathrm{Germ} (\Gamma_{\varphi})$. Given $\gamma\in \mathcal{G}$ and $x,y \in M$ such that $\mathrm{src} (\gamma)=\mathrm{germ}_M (x)$ and $\mathrm{trg} (\gamma)=\mathrm{germ}_M (y)$, we can think of the element $\gamma$ both as a $\mapsto$ arrow between points (with no internal structure) $x,y$ (since $\varphi_g \colon x\mapsto y$) and as a $\to$ arrow between objects $x$ and $y$ (once we identify $x$ and $y$ with their germs of neighbourhoods in $M$). -Remark 4. Groupoids can be viewed as a generalization of group actions (and, more generally, of equivalence relations). In the theory of orbifolds (either in the differentiable category or in others) the morphisms $G_1$ of a groupoid $G_1\rightrightarrows G_0$ in $\mathrm{Diff}$ representing the orbifold $X=[G_1\rightrightarrows G_0]$ are viewed as abstract "glueing data" to obtain a "space" $X$ by patching the pieces of the (possibly very disconnected) space $G_0$ (please correct me if this view is mistaken). So, ideally, a morphism $(g\colon x\to y) \in G_1$ corresponds to the assertion that $\varphi_g\colon x \mapsto y$ for some map $\varphi_g\colon U \to V$ between local patches around $x$ and $y$ (producing identifications in an "étale" way). -Remark 5. On the contrary to remark 4, in the theory of moduli (we leave for a moment the differentiable setting) when a groupoid $G_1\rightrightarrows G_0$ represents a moduli stack $\mathcal{M}$, elements (closed points) of $G_1$ are seen as maps (isomorphisms) between the "structured things" that $\mathcal{M}$ parametrizes. - - -Q1. Concerning Remark 3, which is the relation btween $G_{\varphi}$ and $\mathcal{G}_{\varphi}$? Are they equivalent? - - - -Q2. Given a Lie groupoid $G=(G_1\rightrightarrows G_0)$, say with finite dimentional object space $M:=G_0$, which conditions (if any) on $G$ ensure that there is a pseudogroup $\Gamma$ on $M$ so that $G$ is equivalent to $\mathrm{Germ}(\Gamma)$ ? So, for example, can Remark 4 be made rigorous? - In other words, can the $\mapsto$ viewpoint (examplified by Remark 4) be interchanged with the $\to$ viewpoint (examplified by Remark 5) ? Is étaleness of $G$ sufficient or necessary? - - - -Q3. More generally, given a groupoid $\mathcal{G}=(\mathcal{G}_1 \rightrightarrows \mathcal{G}_0)$, is there a manifold $\mathcal{S}$ in $\mathfrak{Diff}$ such that $\mathcal{G}$ is equivalent to the groupoid $\mathrm{Germ} (\Gamma)$ of germs of some pseudogroup $\Gamma$ of diffeomerphisms of $\mathcal{S}$ ? - - - -Q4. What about other "geometries" (i.e. other sites in which to consider internal groupoids and -assuming the notion can be generalized- pseudogroups)? - -REPLY [3 votes]: You could look at -arXiv:1107.5511 - Pseudogroups and their etale groupoids - Mark V. Lawson, Daniel H. Lenz -and othjer works by Lawson in this area, for example the relation between inverse semigroups and ordered groupoids, originally studied by C. Ehresmann.<|endoftext|> -TITLE: A survey on positive mass theorem? -QUESTION [9 upvotes]: Could you suggest a good survey paper on positive mass theorem? - -REPLY [3 votes]: Dan Lee's book Geometric Relativity available here: https://bookstore.ams.org/gsm-201/<|endoftext|> -TITLE: Is the simplicial completion of a localizer always a bousfield localization of the injective model structure? -QUESTION [12 upvotes]: Background -Recall (from Cisinski's Astérisque volume 308) that given a small category $A$, we define an $A$-localizer to be a class $W$ of morphisms of $\mathrm{Psh}(A)$ satisfying the following axioms: - -The class $W$ satisfies 2-for-3 -The class $W$ contains $\mathrm{rlp}(\mathrm{Mono}(A))$, where $\mathrm{Mono}(A)$ denotes the class of all monomorphisms in $Psh(A)$. -The class $W\cap \mathrm{Mono}(A)$ is closed under transfinite composition and pushouts. - -Given a class $C$ of morphisms in $\mathrm{Psh}(A)$, define $W(C)$, the localizer generated by $C$, to be the intersection of all $A$-localizers containing $C$. We say that a localizer $W$ is accessible if there exists a small set $S$ such that $W=W(S)$. -Let $(A,W)$ be a pair comprising a small category $A$ and an $A$-localizer $W$. We define the simplicial completion of $(A,W)$ to be the pair $(A\times \Delta, W_\Delta)$, where we define the $A\times \Delta$-localizer $W_\Delta$ as follows: The class $W_\Delta$ is the localizer generated by $W^{\Delta^{op}}$, the class of levelwise $W$-equivalences (viewing $\mathrm{Psh}(A\times \Delta)$ as $\mathrm{Psh}(A)^{\Delta^{op}}$, so the levels are indexed by the objects of $\Delta$) together with the class of all maps $T\otimes \Delta^1\to T$ where $T$ is an object of $\mathrm{Psh}(A\times \Delta)$ (given an object $T$ and a simplicial set $X$, we define $(T\otimes X)(a,s)=T(a,s)\times X(s)$ for a pair of objects $a$ in $A$ and $s$ in $\Delta$). -It is a theorem of Cisinski that the functor $pr_1^\ast:\mathrm{Psh}(A)\to \mathrm{Psh}(A\times \Delta)$ obtained from the projection onto the first factor induces an equivalence of categories $W^{-1}Psh(A) \to W_\Delta^{-1}\mathrm{Psh}(A\times \Delta)$. Further, if the $A$-localizer $W$ is accessible, the functor $pr_1^\ast$ is the left Quillen functor of a Quillen equivalence between the model categories $(\mathrm{Psh}(A), \mathrm{Mono}(A), W)$ and $(\mathrm{Psh}(A\times \Delta),\mathrm{Mono}(A\times \Delta), W_\Delta)$. -Recall that the injective (with respect to $\Delta$) model structure on $\mathrm{Psh}(A\times\Delta)\cong \mathrm{Psh}(\Delta)^{A^{op}}=\mathrm{sSet}^{A^{op}}$ is the model structure for which the cofibrations are exactly the monomorphisms and the class of weak equivalences $W_{\mathrm{inj}}$ comprises the levelwise (indexed by the objects of $A$) weak homotopy equivalences. -We will say that an $A$-localizer $W$ is a Bousfield localization of another $A$-localizer $W^\prime$ if $W^\prime \subseteq W$ (because if $W$ and $W^\prime$ are accessible, the model structure associated with $W$ is a left Bousfield localization of the model structure associated with $W'$. -Question: -Given a small category $A$ and an $A$-localizer $W$, is the $A\times \Delta$-localizer $W_\Delta$ a Bousfield localization of the class of levelwise weak homotopy equivalences $W_{\mathrm{inj}}$? -Remarks -I can prove that this is true in the special case that $A$ has the following property: -The category $A$ is a Reedy category such that the Reedy model structure on $sSet^{A^{\mathrm{op}}}$ coincides with the injective model structure. Somewhat surprisingly, this special case happens to include some of the most interesting examples. Cisinski has given an axiomatic characterization of a wide swath of these categories using his formalism of "catégories squelletiques" (Astérisque 308 ch. 8). Also see this MO Question. - -REPLY [19 votes]: Let $A$ be a small category, and $W$ an $A$-localizer. Then we say that $W$ is regular if any presheaf $X$ over $A$ is canonically the homotopy colimit of the representable presheaves above $X$; see Definition 3.4.13 (all references are in Astérisque 308). Except stated otherwise, all the assertions below about regular $A$-localizers are proved in section 3.4. -Given any $A$-localizer $W$, there is a minimal regular $A$-localizer containing $W$. This process of regularization preserves all the interesting properties of $W$ you might expect (namely, being accessible, proper, cartesian, being closed under small filtered colimits, respectively). What is nice is that there is the following -Theorem 3.4.36. -An $A$-localizer $W$ is regular if and only if its simplicial completion $W_\Delta$ contains the class $W_{inj}$ of termwise simplicial weak equivalences. -This has useful consequences. For instance, any regular $A$-localizer is closed under small filtered colimits. There is also a kind of internal relative version of Quillen's theorem A. See Corollaries 3.4.41 and 3.4.47. -There is a whole class of squelettique categories $A$ for which any $A$-localizer is regular; see Proposition 8.2.8 (the proof I give relies on the general results of regular localizers, but there is a more elementary proof). However, for this property to hold, you need all the automorphisms of objects of $A$ to be trivial. Indeed, you may find in Proposition 3.4.57 that equivariant homotopy theory (à la Peter May) gives a nice example of a squelettique category $A$ with a nice explicit complete homotopical structure whose associated $A$-localizer $W$ is proper, cartesian, closed under filtered colimits, but not regular (in particular, its simplicial completion does not contain the class of termwise simplicial weak equivalences). This $W$ also has the property that any map between representable presheaves is a weak equivalence, and that, for any representable $a$, the associated $A/a$-localizer is regular (in this example, the objects of $A/a$ have not any non trivial automorphisms). In particular, this shows that the characterization of the $A$-localizer of $\infty$-equivalences for a (local) test category given by Proposition 6.4.26 is the optimal one (at least in this language). -N.B. This kind of question has also applications in comparing the various notions of $\infty$-categories. For instance, the Quillen equivalence between Joyal's model category for quasi-categories and Rezk's model category for complete Segal spaces can be explained rather immediately from the correspondance $W\mapsto W_\Delta$ and from the fact that any $\Delta$-localizer is regular (and using Theorem 3.4.36 above). The same thing will happen for the analogs for $(\infty,n)$-categories by the same argument.<|endoftext|> -TITLE: Cone of effective divisors! -QUESTION [12 upvotes]: Let $X$ be a smooth simply connected projective variety of dimension $n$ (over complex numbers of course). For such $X$ we have two famous cones which are cone of effective curves and ample cone and are dual to each other. -Question: Is there any thing as Cone of effective divisors? Is there any problem to define such a thing? Has any body studied that? -For surfaces, it is just cone of effective curves. So the smallest dimension at which we would get some thing new is three. - -REPLY [22 votes]: As mentioned in the comments, the (pseudo)effective cone $\overline{\mathrm{Eff}}(X)$, defined as the closure of the cone of all effective divisors on $X$, is certainly an object of study, and Lazarsfeld's book is a good reference. Your complaint that he doesn't say much about its structure is surely related to the fact that so little is known! Here are a few general things I'm aware of: - -The interior of the effective cone is the big cone, i.e., the cone of line bundles with positive volume. -The dual of the effective cone is the cone of moveable curves, see Boucksom-Demailly-Paun-Peternell. -As part of their work on the minimal model program, Birkar-Cascini-Hacon-McKernan prove that log Fano varieties have finitely generated effective cones. - -And here are a couple specific instances where one knows more: - -When $X$ admits an action by a solvable group with a dense orbit, the effective cone is generated by the components of the complement of the orbit. (This works when $X$ is, e.g., a toric variety or a Schubert variety.) -There's been a lot of recent work on the case $X=\overline{M}_{0,n}$, see e.g., Hu-Keel, Hassett-Tschinkel, Castravet-Tevelev.<|endoftext|> -TITLE: Cardinality of connected manifolds -QUESTION [11 upvotes]: Consider the assertion: -Every connected, but not necessarily paracompact, n-manifold is of cardinality -$2^{\aleph_0}$ (at least assuming the axiom of choice). -For n=1 this may be proved via enumeration of the short list of examples. The essential point is that while there is a Long Line, there is no Extra Long Line. -What is the situation for n>1? - -REPLY [5 votes]: Stephen's argument can also be phrased in the language of model theory. Given a connected manifold $X$, consider an elementary submodel of a large fragment of set theory $\mathfrak{M}$ that (1) contains all the reals, -(2) is closed under countable subsets, (3) contains $X$ as an element -and (4) is of size $2^{\aleph_0}$. -It suffices to show that $X\subseteq \mathfrak{M}$. But $X\cap \mathfrak{M}$ is open since each point of $X$ has a neighbourhood of size $2^{\aleph_0}$ and, by elementarity there is a bijection from the reals to this neighbourhood and, since the $\mathfrak M$ contains the reals it must also contain the image of this bijection, namely the neighbourhood. But $X\cap \mathfrak{M}$ is also closed since $\mathfrak M$ contains all sequences from $X\cap \mathfrak{M}$ and hence their unique (by Hausdorffness) limits. By connectedness $X\cap \mathfrak{M} = X$.<|endoftext|> -TITLE: Does intersection pairing on `$IH^*(X)$` agree with cup-product on `$H^*(X)$`? -QUESTION [7 upvotes]: Let $X$ be a proper singular variety over $k=\overline{\mathbb F}_p,$ irreducible of dimension $d.$ Let $H^*(X)$ and $IH^*(X)$ be the $l$-adic cohomology groups and $l$-adic intersection cohomology groups of $X,$ resp. Then, is the natural map $H^*(X)\to IH^*(X)$ compatible with the cup-product on $H^*(X)$ and the intersection product on $IH^*(X)?$ -Background and Motivation: Given $X_0/\mathbb F_q$ an $\mathbb F_q$-structure of $X,$ one deduces a Galois action on $H^*(X)$ and $IH^*(X),$ with respect to which they are "mixed" (the 2nd one being pure), and the weight filtrations $W$ on both of them are independent of the choice of $X_0.$ One has a natural morphism -$$ -H^n(X) \to IH^n(X) -$$ -which factors as -$$ -Gr^W_n H^n(X) \hookrightarrow IH^n(X), -$$ -and this turns out to be injective. -As $X$ is singular, Poincaré duality might fail, i.e. the cup-product -$$ -H^i(X)\otimes H^{2d-i}(X) \to H^{2d}(X), -$$ -which is Galois equivariant, may be degenerate. This is the case when $H^i(X)$ (resp. $H^{2d-i}(X)$) is not pure of weight $i$ (resp. $2d-i$) for the reason of Galois, and $W_{i-1}H^i(X)$ (resp. $W_{2d-i-1}H^{2d-i}(X)$) is contained in the left kernel (resp. right kernel) of the cup-product pairing. I would like to know if this is the only obstruction for Poincaré duality to hold, namely they are exactly the left/right kernel. -Since the intersection pairing -$$ -IH^i(X)\otimes IH^{2d-i}(X)\to\mathbb Q_l(-d) -$$ -is perfect (I don't know if the pairing has a geometric definition in char. $p$ --- $D_XIC_X\simeq IC_X$ is the only reason I know), if my question in the beginning has a positive answer, then the cup-product pairing on $Gr^W_*H^*(X)$ will be perfect. -Correction: The argument above for non-degeneracy on $Gr^W_*H^*(X)$ is wrong. Here's a counter-example. Let $X$ be the projective cone of a projective smooth curve of genus $g,$ either over char. 0 or $p.$ Then $H^1(X)=0$ but $H^3(X)$ is of dimension $2g$ and pure of weight 3. - -REPLY [3 votes]: I think the answer to your question is yes, at least if $X$ is irreducible. -One may think of cup product in cohomology arising from the obvious pairing in the derived category $\mathbb{Q}_{l,X} \otimes^L \mathbb{Q}_{l,X} \to \mathbb{Q}_{l,X}$. The pairing on intersection cohomology comes from a pairing $IC_X \otimes^L IC_X \to \omega_X$, where $\omega_X$ is the dualising complex on $X$. There are natural maps $\mathbb{Q}_{l,X}[d] \to IC_X$ and $\mathbb{Q}_{l,X}[2d] \to \omega_X$; the first comes from the construction of $IC_X$ and the second, by adjunction, from the natural surjective map $H^{2d}(X) \to \mathbb{Q}_l(-d)$. This gives rise to a commutative diagram involving the two pairings which gives the commutativity at the level of cohomology that you wanted. (Note that one can check the commutativity of the diagram after restricting to the smooth locus of $X$.)<|endoftext|> -TITLE: Kolmogorov Complexity and Proof Techniques -QUESTION [14 upvotes]: I'm interested in examples of theorems that employ the proof techniques that are utilized in the proof of the undecidability of Kolmogorov Complexity. -Definition:(Sipser) Let x be a binary string. We say that the minimal description of x, written as d(x), is the shortest string $\langle$M,w$\rangle$ where TM M on input w halts with x on its tape. So, the Kolmogorov Complexity K(x) is written as, K(x)=|d(x)|. K(x) is defined to be the length of minimal description of x. -Theorem: K(x) is not a computable function. -Proof/Sketch of Proof (attributed to Chor): -Proof of negation. $\forall$n, let $y_{n}$ be the lexicographical first string y that satisfies n < K(y). -Consider the following TM M: -On input n (encoded in binary), M generates one by one all binary strings $x_{0}$, $x_{1}$, $x_{2}$, $x_{3}$... in lexicographic order. -For each $x_{i}$ it produces, M computes K($x_{i}$). -If K($x_{i}$) > n, then the TM M, outputs $x_{i}$ and halts. -Else, the TM M, continues to examine the next lexicographical string $x_{i+1}$. -Since the function K is unbounded, it is guaranteed that M will eventually come across a string x satisfying K(x) $>$ n. -Question: what will the TM M output on input n? -By definition on input n TM M outputs $y_{n}$ (the lexicographical first string whose Kolmogorov complexity exceeds n, K(x) > n), but the length of n is $log_{2}$(n). -So we have $K_{M}$($y_{n}$) $\leq$ $log_{2}$(n). There is a constant $c_{M}$ such that $\forall$y, K(y) $\leq$ $K_{M}$(y) + $c_{M}$, so $\forall$n K($y_{n}$) $\leq$ $log_{2}$(n) + $c_{M}$. -By definition of $y_{n}$ for all n, n < K($y_{n}$). By combining the two inequalities we get: n < $log_{2}$(n) + $c_{M}$, but for large enough n this is false. Thus a contradiction. -Question: What other theorems utilize a similar proof technique in their proofs? -For example: -The proof that the set of incompressible strings is undecidable is very similar with some slight modifications. - -REPLY [4 votes]: Using the same technique, one can construct infinitely many statements which are true with probability arbitrarily close to 1, but are nonetheless unprovable. See lemma 4 in http://theory.stanford.edu/~trevisan/cs172/notek.pdf<|endoftext|> -TITLE: Counting non-isomorphic graphs with prescribed number of edges and vertices -QUESTION [10 upvotes]: I'd love your help with this question. -Let $n\geq3$ be a fixed integer. How many non-isomorphic graphs with $p$ vertices and $q$ edges are there where $p+q=n$? -Thank you very much. -Crossposted at MSE. - -REPLY [23 votes]: Using the Combinatorica package in Mathematica, the command NumberOfGraphs$[p,q]$ returns the number of non-isomorphic graphs with $p$ vertices and $q$ edges. If you want to implement this yourself, you may want to proceed here first. -There is an explicit (but rather complicated) formula which you can find here. The formula is obtained via Pólya's Enumeration Theorem. -Edit: Indeed it is a standard application of Pólya theory to obtain formulas for the number of nonisomorphic graphs with $p$ vertices and $q$ edges. (Counting the number where the total number of vertices and edges is $n$ can be obtained from this.) The standard book on graph enumeration is "Graphical enumeration" by Harary and Palmer. There is a web site with many sequences arising from results discussed in the book.<|endoftext|> -TITLE: Cubic hypersurfaces of complex projective space -QUESTION [7 upvotes]: Given the equation of a cubic hypersurface $C\subset\mathbb{P}^{N}_{\mathbb{C}}$ ($N\geq 4$), -there is an algorithm (or better a software) that allows to determine if $C$ is factorial (i.e., all of whose local rings are unique factorization domains, and -hence there is no distinction between Cartier divisors and Weil divisors), -and if $\mathrm{Pic}(C)=\mathbb{Z}\langle\mathcal{O}_C(1)\rangle$ ? -Of course this is trivial if $C$ is smooth. -Thanks. - -REPLY [6 votes]: First concerning your question: most people use $\operatorname{Pic}(C)$ for Cartier divisors. I interpret your question as follows: you want to determine whether every divisor on $C$ is linear equivalent with a Cartier divisor (this means factorial). -Now if $C$ is smooth this is true and I think this is also true if $\dim C-\dim C_{sing}>3$. -If $\dim C_{sing}\geq \dim C-3$ things are much more complicated. A necessary condition for being factorial is (roughly said) that the rank of $H^{N-2,N-2}(C,\mathbb{C}) \cap H^{2N-4}(C,\mathbb{Z})$ equals one. (If the MHS on H^{2N-2}$ does not have pure weight you have to be a bit more careful here.) -If $\Sigma=C_{sing}$ then you have an exact sequence -$$H^{2N-5}(C)\to H^{2N-5}(C\setminus \Sigma)\to H^{2N-4}_\Sigma(C)\to H^{2N-4}(C).$$ -If I remember correctly there should be a copy of $H^2(\Sigma) $ inside -$H^{2N-4}_{\Sigma} (C)$. -If this is all of $H^{2N-4}_\Sigma$ then you can relatively easily show that $H^{2N-4}(C)$ is one-dimensional and hence each divisor on $C$ is homologically equivalent to a Cartier divisor. -If $H^{2N-4}_\Sigma(C)$ is bigger then $H^2(\Sigma)$ things are getting complicated. -In the case that $\dim \Sigma=0$, i.e., $C$ has isolated singularities then the only interesting case is $N=4$. Now $H^4_\Sigma$ is the part of the cohomology of the Milnor fiber that is invariant under the monodromy. This can be calculated using Singular. -In some case you can actually calculate the cokernel $K$ of -$H^3(C\setminus \Sigma)\to H^4_\Sigma(C)$. -For this see e.g., Dimca's paper on Betti numbers and defects of linear systems. -It turns out that $K$ is the primitive cohomology group $H^4(C,\mathbb{C})$ -The formula Francesco mentioned is a special case of Dimca's approach. -Grooten-Steenbrink and Hulek-K. gave similar formula as Dimca for certain classes of nonisolated singularities.<|endoftext|> -TITLE: connected component of the identity section in the special fiber of the Neron model under isogenies -QUESTION [5 upvotes]: Let $A$ and $B$ be two isogenous abelian varieties over a number field $K$ and let $\mathcal{A}$ and $\mathcal{B}$ denote their Neron models over $\mathcal{O}_K$. Let $v \in M_K^0$ denote a finite prime of $K$, $k_v$ its residue field, $\mathcal{A}_v = \mathcal{A} \times _{\mathcal{O}_K} k_v$ the special fiber of the reduction of $A$ at $v$, and $\mathcal{A}_v^0$ the connected component of the identity section in the special fiber $\mathcal{A}_v$. -Is it true that the cardinalities of the $k_v$-rational points of $\mathcal{A}_v^0$ and $\mathcal{B}_v^0$ are the same, i.e. -$|\mathcal{A}_v^0(k_v)| = |\mathcal{B}_v^0(k_v)| ,\ \forall v \in M_K^0$? - -REPLY [6 votes]: I think the answer is yes. This can be deduced for example from the results of SGA7, Expose IX (p.14-15) as follows: -Firstly, the dimension of the unipotent part, the toric part and the abelian part of the connected component of the special fibre are the same for $\mathcal{A}_v^0$ and $\mathcal{B}_v^0$. The toric and abelian parts of $\mathcal{A}_v^0$ and $\mathcal{B}_v^0$ are moreover isogenous and so have the same number of points in the residue field. The unipotent parts also have the same number of points since that only depends on the dimension. Since the connected component of the special fibre is an iterated extension of these group schemes the claim follows follows from Lang's theorem i.e., that torsors for connected groups over finite fields are trivial.<|endoftext|> -TITLE: Basic Question about Kodaira Dimension -QUESTION [5 upvotes]: Greetings friends, -Let $X$ be a smooth complex projective variety with canonical divisor $K$. -For $n \in \mathbb{N}$, let $\lambda_n$ denote the rational map $X \to \mathbb{P}^M$ induced from $H^0(X, nK)$, and let $d_n = dim(im(\lambda_n))$. Let $\kappa_1$ be the maximum value $d_n$ attains as $n$ ranges over $\mathbb{N}$, where we say that $\kappa_1=-1$ if $H^0(X,nK)=\emptyset$ for all $n$. -Let $\kappa_2$ be the smallest integer so that the sequence $\frac{dim(H^0(X,nK))}{n^b}$ is bounded, where we make the same convention of $\kappa_2=-1$ for above. -I'd like to understand why $\kappa_1 = \kappa_2$; could someone please provide me with a reference that discusses this, or provide a brief sketch as to the proof? -Thank you, -Robert - -REPLY [4 votes]: Dear Robert, a reference is Lazarsfeld, Positivity in Algebraic Geometry I, Corollary 2.1.38. Note that the statement there concerns the Iitaka dimension of any line bundle, not just the canonical bundle. -The basic idea of the proof is as follows: given a line bundle $L$ on $X$, one can find a birational morphism $f: X' \rightarrow X$ such that $X'$ fibres over another variety $Y$ of dimension $\kappa_1(L)$. One then compares the numbers $h^0(X,nL)=h^0(X',f^\ast (nL))$ with the number of sections of powers of some ample line bundles on $Y$, which one can compute with Riemann--Roch, to get the desired estimates.<|endoftext|> -TITLE: References for holomorphic foliations -QUESTION [14 upvotes]: I'm looking for an introduction to holomorphic foliations and foliations of complex manifolds. -Any little helps, but I'm particularily interested in problems of the type where we have a hermitian manifold $(X,h)$ (not necessarily compact) and a foliation $\mathcal F$ of $X$, such that the restriction of $h$ to any leaf of the foliation is Kahler. Anything that could help to describe existence of such foliations, or consequences of their existence, would be greatly appreciated. But, to begin with: -Is there any general introduction to the theory of foliations on complex manifolds? - -REPLY [10 votes]: Let $M$ complex manifolds admitting a smooth, positive, proper plurisubharmonic exhaustion, $\rho:M\to[0,\infty)$, whose we have complex Monge-Ampere foliation $(\partial\bar\partial \rho)^n=0$. Patrizio, Giorgio;and Wong, Pit Mann in -Stability of the Monge-Ampère foliation,Mathematische Annalen,March 1983, Volume 263, Issue 1, pp 13–29 -showed that if the volume function is continuous on the level sets of $\rho$, then the leaf space, $\mathcal L$, admits a Kähler form $\omega$, so that, if $\pi:M\to\mathcal L $ is the projection, we have $$\partial\bar \partial \log \rho=\pi^*\omega.$$ -We have always the following theorem: -Theorem: If $\omega$ is non-negative, $\omega^{n-1}\neq 0$, $\omega^n=0$, and $d\omega=0$, then -$$\mathcal F=\text{ann}(\omega)=\{W\in TX|\omega(W,\bar W)=0, \forall W\in TX\}$$ -define a foliation $\mathcal F$ on $X$ and each leaf of $\mathcal F$ being a Riemann surface(which is Kaehler always) -You can see my short not about fiberwise Calabi-Yau foliation and semi-Ricci flat metric introduced by Greene,Shapire,Vafa, and Yau -https://hal.archives-ouvertes.fr/hal-01551080 -Take a holomorphic foliation map $\pi:X\to Y$ such that the leaves of the foliation coincide with the fiber of $\pi$, then the pull back of any Kahler metric on $Y$ to $X$ gives rise to a homogeneous holomorphic Monge-Ampère foliation and the degenerate Kahler form can be the pull back of a Kahler metric on $Y$. See Proposition 6.4 of the following paper of Ruan. -In fact by Theorem 1.3 in the following reference, when the homogeneous Monge-Ampère equation comes from a collapsing, the foliation is holomorphic. -In fact holomorphic foliation correspond to Cheeger-Fukaya-Gromov theory about collapsing Riemannian manifolds. -If you want to study the collapsing part of degeneration of K\"ahler-Einstein metrics, then you are in deal with holomorphic foliation(see Wei-Dong Ruan's paper in bellow) and also fiberwise Kahler-Einstein foliation (which is a foliation in fiber direction and may not be foliation in horizontal direction. See this preprint) -Wei-Dong Ruan, On the convergence and collapsing of Kähler metrics, J. Differential Geom.Volume 52, Number 1 (1999), 1-40.<|endoftext|> -TITLE: Are there any very hard unlinks? -QUESTION [8 upvotes]: This question is closely related to a question of Gowers: Are there any very hard unknots? . -I'm thinking about how to create interesting knots from small numbers of local moves on unlinks. The "standard embedded n-component unlink" (let's call it the untangled unlink) is defined as a union of n circles in n disjoint 3-balls; and an unlink is a link which is ambient isotopic to an untangled unlink. In order to get some feel for what is possible and to have a non-trivial example to play with, I'm looking for an unlink which is difficult to untangle. Paraphrasing Greg Kuperberg's formulation: - -Can you untangle any unlink with relatively little work, say a polynomial number of geometric moves of some kind? I'm interested in untangling the components one from another, as opposed to to untangling individual components. - -I mean this very much in the same sense that Tim Gowers meant his question: Is there a geometric algorithm to untangle components of any unlink, which "makes the unlink simpler" (this is intentionally vague) at each stage? Conversely, is there an unlink which, if it were given to me as a physical object (some tangled loops of rope), I would not be able to disentangle one component from the other without considerable ingenuity? -At the moment, I am most interested in the question for 2-component links and for 3-component links. It's a bit embarassing that I have no intuition at all for what the answer to this question might look like- the "link case" seems completely different from the "knot case". -Edit: Just an aside: experimentally, it's known that fluids made of long closed molecules flow much faster than fluids made of open molecules, although I don't think that the mathematics behind this is understood. But intuitively, it's clear what's going on- closed molecules don't get tangled up in one another as easily as open ones do. So if hard unlinks exist (and in rheology we're talking unlinks with thousands or millions of components), experimentally we can argue that they must at least be rare. Maybe. - -REPLY [7 votes]: I conjecture that the answer is yes, that you may undo a split link with polynomially many moves. This would yield another proof that unlinking is in NP, but it would be somewhat more satisfying, since the certificate would say that you could actually show someone how to tease apart the two components in polynomial time, rather than some abstract normal surface coordinate certificate. -I have a vague idea how one might use Dynnikov's argument to prove this. He analyzes a sphere in the complement of an arc presentation for the link, by considering an induced singular foliation from an open book, following a method that Birman and Menasco applied to braids. One ought to be able to bound the complexity of this sphere using normal surface theory. There is a nice triangulation of $S^3$ with $n^2$ tetrahedra which has an arc presentation of the link in the 1-skeleton (think of $S^3$ as the join of the two components of the Hopf link with each component having a cell structure with $n$ 1-cells). I think one ought to be able to relate the number of singularities of the foliation with the complexity of the normal 2-sphere with respect to this triangulation, which is at most exponential in $n^2$ (Joel Hass, Jeffrey C. Lagarias, Nicholas Pippenger, The Computational Complexity of Knot and Link Problems, J.A.C.M. 46 (1999) 185–211). -Now Dynnikov shows that one may perform exchange moves to decrease the number of singularities of the foliated sphere. In order to get a polynomial number of moves, one would have to show that the number of singularities could be decreased by a definite factor by each move. I suspect this should be true by analyzing things from the normal surface perspective: in a normal surface with exponentially many cells, there are large swaths of the surface which are parallel. If one could eliminate a singularity in the foliation in one part of the surface, then one ought to be able to eliminate singularities in the parallel sheets, and thus eliminate a definite fraction of the singularities. Hopefully this would be similar to the type of algorithm for counting components of normal curves (see Ian Agol, Joel Hass, William P. Thurston, The Computational Complexity of Knot Genus and Spanning Area). -Incidentally, Dynnikov's argument is a bit easier to understand in the split link case, since there are no "boundary conditions" to consider for a sphere versus a disk. An algorithm to detect the unlink gives an algorithm to detect the unknot: just push off a parallel unlinked copy of the knot, and see if the resulting two-component link is split.<|endoftext|> -TITLE: When are representation rings special lambda-rings? (variations of an old question) -QUESTION [11 upvotes]: Status: Questions 2 and 4 answered in the negative. Questions 1 and 3 ARE STILL UNANSWERED, despite previous claims. - -On the third page of Wolfang K. Seiler's paper "lambda-rings and Adams operations in algebraic K-Theory" (pp. 93-102 in Beilinson's Conjectures), it is claimed that the Grothendieck ring of representations of a given group $G$ over a given commutative ring $A$ modulo exact sequences is a special $\lambda$-ring. Here, "representations" means finitely-generated projective $A$-modules with a $G$-module structure. ("Special $\lambda$-ring" is what most people just call "$\lambda$-ring", as opposed to "pre-$\lambda$-ring".) -1. Does anybody know a proof of this? I don't understand the sketched proof in 1, and I don't have the reference (R. G. Swan, A splitting principle in algebraic $K$-theory, in: Representation theory of finite groups and related topics, Proc. Symp. Pure Math. 21 (1971), 155-159) either. I have been told that Atiyah-Tall has a proof, but I am very skeptical about it (and if it has a proof, then I am unable to find it amid all the geometry). -I know how this is proven for the case of a characteristic $0$ field (yes, this is in Atiyah-Tall, but it is trivial using characters). My problem is to show this over arbitrary commutative rings. -2. What if we take the Grothendieck ring modulo split exact sequences (i. e., modulo direct sums) rather than modulo all exact sequences? Is this still a special $\lambda$-ring? What if we work over some field? What if the group is finite? -Update on 2: This is easily seen to be wrong, even if the field is $\mathbb F_2$ and the group is the $2$-element group. In fact, applying Krull-Schmidt and considering the Jordan decomposition of the matrix $\left(\begin{array}{cc} 1&1\\ 0&1\end{array}\right)$, we easily see that the $\lambda^2\left(uv\right)=\left(\lambda^1\left(u\right)\right)^2\lambda^2\left(v\right)+\left(\lambda^1\left(v\right)\right)^2\lambda^2\left(u\right)-2\lambda^2\left(u\right)\lambda^2\left(v\right)$ equation is not satisfied if both $u$ and $v$ are the representation of the $2$-element group on $\mathbb F_2^2$ in which the generator acts as $\left(\begin{array}{cc} 1&1\\ 0&1\end{array}\right)$. -3. Can we hope for any reasonable results if we replace the group by a cocommutative bialgebra? Note that cocommutativity is required to define the exterior powers of an $H$-module (where $H$ is our bialgebra). What if the bialgebra is Hopf? -4. What if we remove the projectivity condition on representations of $G$? I know that removing the finite-generatedness condition is a very bad idea (in fact, it makes the Grothendieck group collapse because of the Eilenberg swindle), but I have seen nobody talking about the Grothendieck group of finitely generated but not-necessarily-projective modules. Is it that embarrassingly stupid? -Update on 4: Question 4 has now been answered by Ben Wieland in the comments to this post. -Oh, and I have already asked this question over the "field" $\mathbb F_1$. The answer is "no" to all three questions in this case. But I am interested in "real" rings and fields this time. -EDIT: I might be understanding what Seiler is doing in 1, but in this case, he is doing it wrong, so I think I am not understanding him. -If I am understanding him right, Seiler's somewhat cryptic formulation "where the module induced by $M$ itself has a one-dimensional quotient, given by the linear functions on $S\left(M\right)$" means that we consider the surjection $M\otimes S\left(M\right) \to S\left(M\right)$ which sends $m\otimes m_1m_2...m_k$ to $mm_1m_2...m_k$, and its kernel is then a projective $S\left(M\right)$-module with $\lambda$-dimension one less than that of $M\otimes S\left(M\right)$ (because for every exact sequence $0\to U\to V\to W\to 0$ and every $k\geq 0$, we have the equality $\left[\wedge^k V\right] = \sum\limits_{i=0}^k \left[\wedge^i U \otimes \wedge^{k-i} W\right]$ in $K_0$). Unfortunately $M\otimes S\left(M\right)$ and $S\left(M\right)$ are not quite $S\left(M\right)$-modules with $G$-representations, because $G$ does not act $S\left(M\right)$-linearly on them (unless I did something wrong). Also we need a device to conclude that two projective $R$-modules $V$ and $W$ with $G$-module structure are equal in $K_0$ if and only if the corresponding $S\left(T\right)$-modules $V\otimes S\left(T\right)$ and $W\otimes S\left(T\right)$ with $G$-module structure are equal in $K_0$, where $T$ is some projective $R$-module; this looks like it should follow from the gradedness of symmetric algebras, but I don't see how (mostly because I have no idea what equality in $K_0$ actually means in terms of modules). - -REPLY [3 votes]: After wondering about this question, I went and looked up Swan's paper. His argument is basically the one Seiler is trying to say; it also uses a paper by J. Burroughs: http://www.ams.org/mathscinet-getitem?mr=269719, which I have not looked at. -According to Swan, his result only applies to the case when $A$ is noetherian for technical reasons, but that the general case would be discussed more generally elsewhere. -In the case of interest, suppose $C$ is the category of $G$ representations over $A$. Given an object $M$ of $C$, you are supposed to consider a category $C'$. In this case, this is a category of graded $S(M)$-modules with compatible $G$-action; that is the objects are graded $S(M)$-modules $Q$, together with a $G$-action on $Q$ such that the action map $S(M)\otimes Q\to Q$ is $G$-equivariant; here $G$ is acting in the evident non-trivial way on $S(M)$. -Seiler's (and Swan's) map $M\otimes S(M)\to S(M)$ is a morphism in $C'$, so you get a splitting relation of the form $[K]+[S(M)]=[M\otimes S(M)]$ in $K_0(C')$. -Swan then has an argument that the map $K_0(C)\to K_0(C')$, induced by tensoring up, is injective. I find it hard to parse, and don't really understand it. It involves a map of the form -$$\theta: K_0(C')\to K_0(C)[[t]]/K_0(C)[t],$$ -defined by sending $[Q]$ to $\sum [Q_k]t^k$. (It's necessary to quotient by $K_0(C)[t]$, in order to get something that takes exact sequences to sums.) The composite of $\theta$ with $K_0(C)\to K_0(C')$ is then shown to be injective, somehow using the fact that $\sum [S^n(M)]t^n$ has a multiplicative inverse in $K_0(C')$.<|endoftext|> -TITLE: Intersection theory over non algebraically closed fields -QUESTION [5 upvotes]: Let $k$ be a field, consider intersecting schemes over $k$. - -Is there a version of intersection theory which keeps track of the extension of $k$ over which the intersections are happening? - -Ideally I would imagine that $A_0(pt)$ was the representation ring of $Gal(\overline{k}/k)$, so that I could take various traces to extract information about who is living where. - -REPLY [4 votes]: I don't see how such a thing could exist. Rationally equivalent classes should define the same element in an intersection theory. But, on $\mathbb{P}^1$, you can have two different divisors of the same degree, which are both $\mathrm{Gal}(\overline{k}/k)$ stable but have different Galois actions, and they will be rationally equivalent. -If you try to abandon the condition that rationally equivalent classes define the same element, you will run into trouble with self intersections. For example, consider a conic $C$ in $\mathbb{P}^2$. Its self intersection has degree $4$, representing that, for a generic perturbation $C'$ of $C$, the intersection $C' \cap C$ has four points. However, the Galois action on those $4$ points will depend on which perturbation you choose. -It is possible that your question is really the more general one of ``how do people study the sort of questions studied in intersection theory, while keeping track of Galois action"? There is tons of research on this front. Some papers I like are -J. Harris, Galois groups of enumerative problems, Duke Math. J. 46 (1979), 685–724 -R. Vakil, Schubert Induction -A. Leykin and F. Sottile, Galois groups of Schubert problems via homotopy computation -However, as you will see, these papers do not use the kind of generalized intersection theory you are visualizing.<|endoftext|> -TITLE: History of Gauss' Law -QUESTION [18 upvotes]: Does any one know actual references for the discovery of Gauss' Law (a corollary of the Divergence Theorem)? -The entry in Wikipedia for Divergence Theorem says it was discovered by - Lagrange 1762, Gauss 1813, Green, 1825 - -while the entry in Wikipedia for Gauss' law says it was discovered by - Gauss in 1835, - -but not published until 1867. -In the case of the divergence theorem, only dates are given, no references. I couldn't find anything in the collected works of Lagrange around 1762 that seemed to be close to any form of the divergence theorem. -In the case of Gauss' law, a reference is given to the book by Balone, "A Word on Paper: Studies on the Second Scientific Revolution", but no page reference is given; the only -index entry for Gauss is a brief mention of his name in connection with some one else; and perusing through the subject matter makes it seem very unlikely that anything about Gauss' law is in this book. - -REPLY [9 votes]: I found the following excerpt on this web site: http://jeff560.tripod.com/d.html. It gives references to specific papers of both Lagrange and Gauss. The inverse square law for the electric force is usually associated with Coulomb, but was apparently first inferred by Priestley on the basis of Franklin's observation that there is no electric field inside a hollow conductor and analogy to the known analogous property for gravity. -The history of the theorem is bewildering with many re-discoveries. -O. D. Kellogg Foundations of Potential Theory (1929, p. 38) has the following note on the result “known as the Divergence Theorem, or as Gauss’ Theorem or Green’s Theorem”: -A similar reduction of triple integrals to double integrals was employed by Lagrange: Nouvelles recherches sur la nature et la propagation du son Miscellanea Taurinensis, t. II, 1760-61; Oeuvres t. I, p. 263. The double integrals are given in more definite form by Gauss Theoria attractionis corporum sphaeroidicorum ellipticorum homogeneorum methodo novo tractate, Commentationes societas scientiarum Gottingensis recentiores, Vol III, 1813, Werke Bd. V pp. 5-7. A systematic use of integral identities equivalent to the divergence theorem was made by George Green in his Essay on the Mathematical Theory of Electricity and Magnetism; Nottingham, 1828 [Green Papers, pp. 1-115]. -Kline (pp. 789-90) writes that Mikhail Ostrogradski obtained the theorem when solving the partial differential equation of heat. He published the result in 1831 in Mem. Ac. Sci. St. Peters., 6, (1831) p. 39. J. C. Maxwell had made the same attribution in the 2nd edition of the Treatise on Electricity and Magnetism (1881). See also the Encyclopaedia of Mathematics entry Ostrogradski formula -For Gauss’s theorem Hermann Rothe “Systeme Geometrischen Analyse, Erster Teil” Encyklopädie der mathematischen Wissenschaften mit Einschluss ihrer Anwendungen Volume: 3, T.1, H.2 p. 1345 refers to Gauss’s Allgemeine Lehrsätze in Beziehung auf die im verkehrten Verhältnisse des Quadrats der Entfernung wirkenden Anziehungs- und Abstossung-Kräfte 1839 in Werke Bd. V (especially pp. 226-8.)<|endoftext|> -TITLE: Combinatorial interpretations of integral transforms -QUESTION [14 upvotes]: It is well known that the ordinary and exponential generating functions of a sequence of numbers are related by an integral transform (the Borel transformation). -Does there exist a combinatorial theory of integral transforms? The example above indicates that something might be going on "behind the scenes". Has anyone been able to formulate a precise combinatorial explanation of this phenomenon? If not this one, might other types of integral transforms have combinatorial interpretations? -Thanks for your interest! - -REPLY [7 votes]: I claim that there is a construction here similar to the one in "Lagrange inversion for species" by Gessel and Labelle, which can explain the general picture. -To every species $S:\mathcal{B}\to \mathcal{B}$ there should correspond a "labelled" version $L(S):\mathcal B\to \mathcal B$ whose exponential generating function is the ordinary generating function of $F$. The construction should start by looking at the span -$$\mathcal B \leftarrow \mathcal B\times\mathcal B\rightarrow \mathcal B$$ with maps $p,q$ and introducing a "kernel" on $\mathcal B\times \mathcal B$ in a manner that $L$ can be defined from -$$F(\mathcal B)\leftarrow F(\mathcal B\times \mathcal B)\rightarrow F(\mathcal B)$$ by a pull-push formalism. -This kernel $\kappa (X,Y)$ on $F(\mathcal B^2)$ should be a species of 2 sorts, but since labelling is the inverse of $\sum \frac{x^ny^n}{n!}$ it will actually be a virtual species (formal difference of species, so $\kappa$ is a categorification of $e^{-xy}$). Then one has to check that $$L=p _{\ast}(\kappa q^{\ast})$$ or something analogous to it holds. I.e. I suspect that there is a categorification of the Borel (also called Laplace-Carson?) transform which is a natural operation on species, and I believe that the appearances of these integral transforms arise from decategorifications of properties of spans of groupoids. -Unfortunately, I don't know where to find a treatment of this. I was looking a while back for a treatment of the (closely related) Laplace transform along similar lines but couldn't find any references. Since I can't work out details right now, perhaps someone who knows this can elaborate? -Also, there is a section on differential-integral equations with species in "Combinatorial species and tree-like structures" by F. Bergeron, G. Labelle and P. Leroux, which is worth looking at.<|endoftext|> -TITLE: Mukai-Umemura 3-fold and Kaehler-Einstein metrics -QUESTION [8 upvotes]: The "most symmetric" Mukai-Umemura 3-fold with automorphism group $PGL(2,\mathbb{C})$ admits a Kaehler-Einstein metric according to Donaldson's result. -On the contrary, there are some arbitrarily small complex deformations of the above $3$-fold which do not admit Kaehler-Einstein metrics, as shown by Tian. All examples considered by Tian seem to have no symmetries at all. Is it possible to find similarly arbitrarily small complex deformations with $\mathbb{C}^*$-action and which do not admit any Kaehler-Einstein metric ? - -REPLY [5 votes]: [Edited] Such a manifold cannot exist. -Indeed the small deformations of the "symmetric" Mukai-Umemura $3$-fold $X$ are described explicitly by Donaldson in this paper, pages 43-44. There he describes 5 classes of deformations. Classes 1,2,3 correspond to points in $H^1(X,TX)$ whose $PSL(2,C)$ orbits are closed, and the corresponding small deformations admit Kahler-Einstein metrics thanks to a theorem of Székelyhidi, see Propositions 7,8 and page 12 of this paper. -Cases 4,5 have orbits that are not closed, and do not admit Kahler-Einstein metrics thanks to Tian's argument (case 5 contains the manifold considered by Tian). These are the only possible cases where you could hope to find an example. Any such manifold has a nontrivial C* action precisely if the $PSL(2,C)$-stabilizer of the corresponding point in $H^1(X,TX)$ contains a C*. -However one can check directly that in both cases the stabilizer cannot contain any C*<|endoftext|> -TITLE: How does one compute induced representations for modular representations? -QUESTION [7 upvotes]: The set-up is this: Let $G$ be a finite group, and $H$ a subgroup. We are given an irreducible representation of $H, \rho: H\rightarrow GL_n(K)$ (I will notationally identify $\rho$ with its character). I want to decompose $Ind^G_H(\rho)$ into irreducibles. I am given character tables of both $G$ and $H$. -If $K$ were $\mathbb{C}$, Frobenius reciprocity (https://planetmath.org/FrobeniusReciprocity) will do the trick. However, I am in the modular case; meaning: $char(K)||H|$. I still have all the character tables, except now they are Brauer character tables for the correct characteristic. -Is there a method for decomposing $Ind^G_H(\rho)$ into irreducible (Brauer) characters? -Edit: I wanted to make clear that since in the modular case we don't have Maschke's theorem, the ``decomposition'' into irreducibles would be in the Grothendieck group of Brauer characters of $G$. (representations of $G$ wouldn't nec. be direct sums of irreducible representations) - -REPLY [4 votes]: Frobenius reciprocity for Brauer characters is a little more complicated (a lot more complicated if you don't have complete tables). You need the projective characters to compute the multiplicities. I don't use anything special about the character being induced, though sometimes you can leverage that information (especially if you don't have complete tables). -In GAP, this is easily done: -g:=CharacterTable("ON"); -# Let g be the O'Nan simple group -chi:=Sum([1..10],i->Random(Irr(g mod 2)));; -# chi is a random reducible Brauer character -ipr:=Irr(g)*DecompositionMatrix(g mod 2);; -# ipr is the list of projective Brauer characters -vec:=MatScalarProducts(ipr,[InducedClassFunction(chi,g)])[1]; -# vec is the multiplicity of each Brauer character in chi -chi = vec * Irr( g mod 2 ); -# should be true: we have decomposed it correctly. - -To get the decomposition matrix, you must not only have the Brauer table but also the ordinary table. You don't need much from the subgroup H, just a character to induce and the element fusion from H into G. -This is theorem 2.13 on page 25 of Navarro's textbook on Characters and Blocks of Finite Groups. -By projective, I mean in the module theory sense, an indecomposable direct summand of the regular representation, usually these are denoted Φi corresponding to a Brauer character φi. I don't mean representation into projective groups, like PGL.<|endoftext|> -TITLE: Orders of automorphism groups of p-groups -QUESTION [10 upvotes]: There is a theorem that says that if $p$ is a prime and $G$ is a $p$-group with $|G| = p^{n}$, $|Aut(G)|$ divides $\Pi_{k=0}^{n-1} (p^{n}-p^{k})$. -This theorem is sharp, since $\Pi_{k=0}^{n-1} (p^{n}-p^{k}) = |GL(n,p)| = |Aut(E)|$, where $E$ is an elementary abelian group of order $p^{n}$. -The proof I know is by proving the $p$-part and the $p^{'}$-part of the divisibility separately. -The $p^{'}$-part of the divisibility boils down to the integrality of the binomial coefficient analogues which count decompositions of a vector space over $\mathbb{Z}/(p)$ into two subspaces whose sum is the space and whose intersection is $0$. -The $p$-part of the divisibility is the divisibility statement for the order of a Sylow $p$-subgroup of $Aut(G)$, and it uses induction and the fact that a $p$-group will have fixed points whenever it acts on a set whose cardinality is a nonmultiple of $p$. -Is there a book that covers this theorem? If so, how far (and where) does the book run with it? - -REPLY [5 votes]: This is just Burnside's basis theorem. See for instance theorem 12.2.2 on page 178 of M. Hall, Jr.'s textbook on the Theory of Groups. The original reference for the phrasing in terms of automorphisms is from P. Hall (1933). -As far as where to go from it, this is roughly how automorphism groups of p-groups are calcuated in O'Brien (1992), Eick–Leedham-Green–O'Brien (2002) and the AutPGrp package of GAP. This is often useful in understand fusion systems, where the p-core of automorphism groups is under good control, and so the GL(n, p) part is the primary interest. -Another application (known to the OP, but interesting enough to describe clearly) is a result of Burnside (1905) classifying for which powers b = b(p, a, q) there is a group of order paqb with no non-identity normal p-subgroup: the classification is based simply on the orders of the automorphism groups of the p-subgroups. Burnside had an error in his analysis of the associated arithmetical condition that was corrected in Coates–Dwan–Rose (1976). Burnside's result was generalized in Glauberman (1975) and Bialostocki (1975, 1987). Many of these and further results are based on analyzing nilpotent p′-subgroups of GL(n, p), resting ultimately on the fact that that every p′-subgroup of the automorphism group of a p-group of rank n is isomorphic (including in some sense, its action) to a subgroup of GL(n, p). - -Burnside, W. -On groups of order pαqβ -Lond. M. S. Proc. (2) 1, 388-392 (1904). -JFM35.0162.01 -DOI:10.1112/plms/s2-1.1.388 -Burnside, W. -On groups of order pαqβ (second paper). -Lond. M. S. Proc. (ser 2) 2, (1905) 432-437. -JFM36.0198.02 -DOI:10.1112/plms/s2-2.1.432 -Hall, P. -A contribution to the theory of groups of prime-power order. -Proc. Lond. Math. Soc., Ser. 2, 36, (1933) 29-95. -Zbl0007.29102 -DOI:10.1112/plms/s2-36.1.29 -Glauberman, G. -On Burnside's other paqb theorem. -Pacific J. Math. 56 (1975), no. 2, 469–476. -MR412269 -URL:euclid.pjm/1102906371 -Bialostocki, Arie. -On products of two nilpotent subgroups of a finite group. -Israel J. Math. 20 (1975), no. 2, 178–188. -MR407148 -DOI:10.1007/BF02757885 -Coates, Martin; Dwan, Michael; Rose, John S. -A note on Burnside's other pαqβ theorem. -J. London Math. Soc. (2) 14 (1976), no. 1, 160–166. -MR419594 -DOI:10.1112/jlms/s2-14.1.160 -Bialostocki, Arie. -On the other pαqβ theorem of Burnside. -Groups–St. Andrews 1985. -Proc. Edinburgh Math. Soc. (2) 30 (1987), no. 1, 41–49. -MR879428 -DOI:10.1017/S0013091500017946 -O'Brien, E. A. -Computing automorphism groups of p-groups. -Computational algebra and number theory (Sydney, 1992), 83–90, -Math. Appl., 325, Kluwer Acad. Publ., Dordrecht, 1995. -MR1344923 -Eick, Bettina; Leedham-Green, C. R.; O'Brien, E. A. -Constructing automorphism groups of p-groups. -Comm. Algebra 30 (2002), no. 5, 2271–2295. -MR1904637 -DOI10.1081/AGB-120003468<|endoftext|> -TITLE: When is a sheaf on a scheme extendable to a representable functor? -QUESTION [5 upvotes]: I'll start with example: -Let $X$ be a scheme, and $O_X$ be its structure sheaf. It is defined at the moment on open sets of $X$, and it takes them to $Sets$. However, it is extendable to a sheaf on the Zariski site of $Sch$ by: Take a scheme $S$ to $\mathbb{G}_a(S)$. Now that it is a functor $Sch \rightarrow Sets$, it makes sense to ask whether it is representable, which in this case it is (by $\mathbb{Z}[X]$), and it is even a group scheme. -My, somewhat vague, question is: how prevalent is the phenomenon? For example, are all coherent sheaves on any scheme extendable to representable functors? To group schemes? Is there an iff condition for this to happen? - -REPLY [12 votes]: If $F$ is a coherent sheaf on a noetherian scheme $X$, there is a natural extension of $F$ to the large Zariski site of $X$: with an object $f\colon T \to X$, you associate the group of global sections of the pullback $f^*F$. According to a result of Nitin Nitsure, this is representable if and only if $F$ is locally free (see http://arxiv.org/abs/math/0308036). What Keerthi says is not quite correct: the functor represented by the spectrum of the symmetric algebra of $F$ is that sending $f\colon T \to X$ to the group of global sections of the dual of $f^*F$, which does not coincide with the group of global sections of $f^*(F^\vee)$. -On the other hand there are many ways of extending a sheaf on the small Zariski site; for example, one can extend it to the small étale site, where it is always represented by an algebraic space with an étale map to $X$ (the analogue of the "espace étalé" for the usual topology), which then you can extend to the large étale site. This would have a somewhat better chance of being representable by a scheme; however, this construction is very different in spirit, and the resulting scheme would be enormous, and probably not very useful.<|endoftext|> -TITLE: How much linear algebra can be done with graphs? -QUESTION [32 upvotes]: Let G be a finite directed acyclic graph, with sources $A=\{a_1,\ldots,a_n\}$ and sinks $B=\{b_1,\ldots,b_n\}$, with edge weights $w_{ij}$. The weight of a directed path P is the product of weights of edges in P. Set -$e(a,b)= \sum\limits_{P\colon\, a\to b}w(P).$ -Then we can form a matrix $M=\left(e(a_i,b_j)\right)_{i,j}$, and the Lindström-Gessel-Viennot(-Karlin-MacGregor) Lemma tells us that the determinant of M is the signed sum of all n-tuples of non-intersecting paths from A to B. -$\det(M)=\sum\limits_{(P_1,\ldots,P_n)\colon\,A\to B} \mathrm{sign}(\sigma(P))\prod\limits_{i=1}^n w(P_i)$. -This is an extremely beautiful result, and indeed, it makes sense to take it as the definition of the determinant (maybe Kasteleyn-Percus is even better). In the context of quantum topology, graphs naturally occur, and it seems to me that determinants appear because we are (or should be) counting n-tuples of weighted non-intersecting paths. -If this is the case, then my question, in some sense, is why we need matrices at all. Finite directed acyclic weighted graphs carry more information than mere matrices, they appear naturally, and it seems a shame to trade them in for puny matrices just in order to do some linear algebra. So I'd like to ask whether I can do all of the linear algebra which I need directly off the graph: - -Question 1: The above lemma gives us a graph-theoretical definition of the determinant. Is there a corresponding purely graph-theoretical definition of eigenvalues? Is anything at all known beyond the determinant? Edit: What I really want is the signature and the rank- the rest is just fishing for what is possible. - -In the context in which I am most interested, the weights are valued in a non-commutative (skew-power-series) ring. Is the Lindström-Gessel-Viennot Lemma valid in this context? Are there any references? (My naive search on Zentralblatt and MathSciNet didn't turn anything up). - -Question 2: Over a skew-power-series ring (or over some more general nice class of noncommutative rings), does the signed sum of all n-tuples of non-intersecting paths from A to B recover the K1-class of M? - -Finally, to understand the big picture a little bit better, is there a reference for what Qiaochu said about the meaning of the (Lindström-Gessel-Viennot) determinant, as coming from some sort of quantum mechanics picture, where "the entries of the matrix describe transition amplitudes and that the determinant is an alternating sum over transition amplitudes in which "histories" of n particles can constructively or destructively interfere."? Does it make (physical?) sense to say something like "Graph G is a Feynman diagram. Shine light through the sources. The determinant is the amount of light you see at the sinks."? - -REPLY [14 votes]: I will try to contribute a partial answer. First I want to comment on the Lindstrom-Gessel-Viennot determinant coming from quantum mechanics stuff, in physics this is known as the Slater determinant, giving the formula for the wavefunction of a multi-fermionic system. This gives a nice picture to think of LGV as yet another instance of the Boson-Fermion correspondence. In the boson case, one allows all paths and obtains the total number as the permanent of the LGV matrix (this is obvious from the definition of the permanent), and in the fermion case one gets a system with states counted by the determinant of the LGV matrix. Of course the non-intersecting part comes into play because fermions in addition satisfy the Pauli exclusion principle, therefore they cannot occupy the same site at the same time. -Now, LGV and related results have an interesting history even in fields other than combinatorics. Fisher, in "Walks Walls, Wetting and Melting" 1984, considered the vicious walkers model in statistical mechanics, which considers mutually avoiding directed lattice paths. From this perspective it is interesting to look at some configurations which aren't solved by the usual LGV theorem, for instance when the paths are allowed to intersect at vertices but not edges, or when two paths are allowed to intersect in at most 2 consecutive vertices (the terminology for this classification is $n$-friendly walkers, see here). Viennot and others considered such variants after the relation between the combinatorics of lattice paths and statistical mechanics was established, it turns out that some of these models also have determinantal formulas associated to them. One main article is "From the Bethe Ansatz to the Gessel-Viennot Theorem" by R. Brak, J. W. Essam, and A. L. Owczarek, the point here is that LGV related results can be proven using transfer matrix methods as well, which is a powerful point of view in light of the models I mentioned above where the usual LGV fails (i.e. outside of the six vertex model). -Now if you need something more rigorous relating LGV matrices to fermion models, this can be done, but it doesn't seem to have been written nicely anywhere. Sometimes this is mentioned in the literature in the case of graphs like $\mathbb Z^2$, see for example "Domino tilings and the six-vertex model at its free fermion point" by P.L. Ferrari and H. Spohn, but I believe there should be a more general setting to talk about this. If you take the point of view that Greg Kuperberg mentioned in his answer to this previous MO question, that Kasteleyn-Percus matrices are essentially equivalent to LGV matrices, then I believe there is more literature on interpreting these as models of Majorana fermions living on the graph. The article I'm thinking of is "Dimer Models, Free Fermions and Super Quantum Mechanics" by Dijkgraaf, Orlando and Reffert. -As a last note, I wanted to say that I don't fully understand your motivation to want to identify every occurrence of the determinant with a LGV (or Kasteleyn-Percus) context, given that even within graph theory there are families of objects (even paths or random walks, as mentioned above) which are counted by determinants of a different sort of flavor. As to the question about non-commutative weights to LGV, I can't offer any insight, except perhaps to suggest looking at previous work on non-commutative extensions of the LGV theorem, such as the extension proved in "Noncommutative Schur Functions and their Applications" by Fomin and Greene (available from Fomin's website). But this is probably not very useful since even in their case the ring is almost commutative. - -With regard to your question on rank and signature, there is a lot one could explore. As a start, if you look at the Laplacian matrix associated to a graph, then the rank has a clear combinatorial meaning, it measures the number of connected components of the graph (well, the maximum number of edges in an acyclic spanning subgraph, to be more precise), but there is not much to say about the signature since all eigenvalues of the Laplacian are nonnegative. When it comes to the adjacency matrix of the graph, the connection to combinatorial quantities gets even weaker (but it is an area of research), for example the rank of the adjacency matrix was conjectured to have something to do with the chromatic number of the graph, but the relation can not be so nice as was shown by Alon and Seymour. There has been some work on interpreting the signature of the adjacency matrix as well, with some early articles by Torgasev (for example here), but it is of similar nature. One can compute rank and signature easily by linear algebra methods and therefore hopes to use these to bound graph-theoretic quatities, but apparently not the other way around.<|endoftext|> -TITLE: Gently falling functions -QUESTION [32 upvotes]: I wonder if it is possible to characterize the class of -gently falling functions, which I would like to define -as follows. -Let $g(x)$ be a $C^2$ function defined on an interval -$R \subseteq \mathbb{R}$ of the real line. -It will be no loss of generality for my purposes to -assume that $g(x) \ge 0$ on $R$, -and that $\max_{x \in R} \; g(x) = 1$. -Imagine placing a particle on the curve at any point -and letting it frictionlessly slide down the curve -under the influence of gravity -(with nominal gravitation constant 1). -Say that $g(x)$ is gently falling if, -for every start point, -the particle never "ski-jumps off" the curve in its -downward descent. -Example 1. -$g(x) = \sqrt{1-x^2}$ for $R=[0,1]$ is not gently falling. -If I've calculated correctly, a particle starting near -the max $(0,1)$ will separate from the curve after falling -height $1/3$. - -  -  -  -  -  - - -Example 2. -$g(x) = \cos^2(x)$ for $R=[0,\pi/2]$. -Under the assumptions above, I believe this is gently falling: -a particle released at any point remains on the curve throughout -its descent. -Example 3. (Added later.) -$g(x)$ is a scaled and translated piece of $\cos^2(x)$ for $x>2$, preceded by a linear ramp -tangent to $\cos^2(x)$: - -  -  -  -  -  - - -Now a particle released at the max gathers enough momentum to lift off at the $x=2$ transition. -So $g(x)$ is not gently falling. -I've worked out a condition that (I think!) -needs to be satisfied for $g(x)$ -to be gently falling, under my assumptions. -Let $g=g(x)$, $\dot{g} = dg/dx$, and $\ddot{g} = d^2g /dx^2$. -Then, for $g$ to be gently falling, -$$2 (1-g) \ddot{g} + (1 + \dot{g}^2) \ge 0$$ -should be satisfied for all $x$. -For the quarter circle, the left side of this inequality becomes -$$ -\frac{x^2}{1-x^2} - 2 - \left(\frac{x^2}{\left(1-x^2\right)^{3/2}}+\frac{1}{\sqrt{1-x^2}} - \right) - \left(1-\sqrt{1-x^2}\right)+1 -$$ -which simplifies to -$$ -\frac{3}{1-x^2} -\frac{2}{\left(1-x^2\right)^{3/2}} -$$ -which is positive for $x$ less than the root $\sqrt{5}/3 \approx 0.745$, in accord with the picture -above. When I work out the calculation for the second example, -there are no roots within $R$. -I have two questions. -First, has anyone seen a curve definition like this before? -If so, even for tangentially related ideas, I'd appreciate a pointer. -Second, if my gently-falling inequality on $g$, or some -analogous inequality, is correct, are there general methods to characterize -all the functions $g$ that satisfy it? -Right now I can determine if any given function is gently falling, -but I don't have a sense for the contours of the set of all gently falling -functions. Thanks for any ideas! -(The motivation for these questions is a bit far from their -form above, and it might distract from the mathematics to -explain.) - -REPLY [9 votes]: Let $p(t) = (x(t), y(t))$ be a parametrization of a smooth curve $(a,b) \to [0,\infty) \times (-\infty,0)$, and assume it satisfies the "zero total energy parametrization" $(x')^2 + (y')^2 = -2y$. We want the apparent force (given by combining acceleration and gravity) at any time to be in a counterclockwise direction from the velocity vector. Using cross products, we find that this is equivalent to -$$x'(t)(y''(t) + 1) \geq x''(t)y'(t).$$ -If you choose initial conditions $x'(0) > 0$ and $y'(0) = 0$, and set the difference between the sides to a non-negative constant $c$, numerical integration will yield one of three types of trajectory, depending on the sign of $x'(0)-c$. If it is negative, you get a periodic sequence of narrow loops, like a frictionless roller coaster. If it is zero, you get a constant speed horizontal trajectory. If it is positive, the trajectory will fall gently but increasingly steeply. -Now, we consider the special case where the path satisfies the vertical line test, so $y(t) = f(x(t))$ for some function $f$. By the chain rule, we see that $y'(t) = f'(x(t)) x'(t)$ and $y''(t) = f''(x(t))x'(t)^2 + f'(x(t))x''(t)$. Feeding this into the inequality and simplifying, we obtain the condition -$$f''(x(t))x'(t)^2 + 1 \geq 0.$$ -In order to express this purely in terms of $f$, we substitute our variables into the zero energy condition to get $x'(t)^2 + f'(x(t))^2 x'(t)^2 = -2f(x(t))$, which simplifies to $x'(t)^2 = -2\frac{f(x(t))}{1 + f'(x(t))^2}$. The inequality then becomes: -$${f'}^2 + 1 \geq 2f'' f$$ -If we apply the substitution $g = f+1$, we get the equation in the question. -You can find similar equations by searching the web with terms like "roller coaster differential equation". For example the second version of equation 4.17 in this document looks like the second displayed equation, but with "real-world" terms like mass included.<|endoftext|> -TITLE: Infinitely Divisible Distributions and Maximal Entropy -QUESTION [5 upvotes]: The normal distribution on $\mathbb{R}$, the exponential distribution on $\mathbb{R}_{\geq 0}$, and the geometric distribution on $\mathbb{N}$ are examples of distributions that are both infinitely divisible and entropy maximizers. On the other hand, the Poisson distribution is an infinitely divisible distribution on $\mathbb{N}$ without maximizing entropy, while the uniform distribution on the interval $[a,b]$ maximizes entropy but is not infinitely divisible. -Can anything be said about the relationship between these two classes of distributions? - -REPLY [5 votes]: The Poisson distribution has maximum entropy under the condition of having a specific mean and being a sum of Bernoulli random variables. The condition of being a sum of Bernoulli random variables can be weakened to being ultra log-concave.<|endoftext|> -TITLE: Qcoh(-) algebraic stack? -QUESTION [15 upvotes]: The $2$-functor $\text{Qcoh} : \text{Sch}^{op} \to \text{Cat}$, which sends a scheme to its category of quasi-coherent modules, is a stack by Descent Theory. Is it actually an algebraic stack? If not, how far is it from being algebraic? For example, which fragment of Artin's criteria is satisfied? I expect that this should be well-known, but currently I don't know where this example is treated in the literature. -For example I can show that for every pair of surjective ring homomorphisms $A \to B, C \to B$, the canonical functor $\text{Mod}(A \times_B C) \to \text{Mod}(A) \times_{\text{Mod}(B)} \text{Mod}(C)$ is an equivalence of categories. In particular, the functor has a deformation theory (in the sense of Artin), sending $(A_0,M,P \in \text{Mod}(A_0))$ to the category of $A_0[M]$-modules $Q$ with an isomorphism $Q/MQ \cong P$. - -REPLY [19 votes]: Artin's axioms do not apply in this case, because the stack is not limit-preserving. They only work with stacks that are locally finitely presented. -In any case, it is easy to give examples of quasi-coherent sheaves whose functor of automorphisms is not representable (for example, an infinite dimensional vector space), and this of course prevents the stack from being algebraic. -As Matthieu says, one should consider coherent sheaves. -[Edit] The question is: what about the stack of coherent sheaves, without flatness hypothesis? First of all, one should interpret "coherent" as meaning "quasi-coherent of finite presentation". The notion of coherent sheaf, as defined in EGA, is not functorial, that is, pullbacks of coherent sheaves are not necessarily coherent. Hartshorne's book defines "coherent" as "quasi-coherent and finitely generated", but this is a useless notion when working with non-noetherian schemes. -The stack of quasi-coherent finitely presented sheaves is not algebraic either. For example, let $k$ be an algebraically closed field, $k[\epsilon] = k[t]/(t^2)$ the ring of dual numbers. If $S = \mathop{\rm Spec} k[\epsilon]$, consider the coherent sheaf $F$ corresponding to the $k[\epsilon]$-module $k = k[\epsilon]/(\epsilon)$. Then I claim that the functor of automorphisms of $F$ over $S$ is not represented by an algebraic space. -Suppose it is represented by an algebraic space $G \to S$. Denote by $p$ the unique rational point of $S$ over $k$; the tangent space of $S$ at $p$ has a canonical generator $v_0$. Furthermore, if $X$ is a $k$-scheme with a rational point $x_0$ and $v$ is a tangent vector of $X$ at $x_0$, then there exists a unique $k$-morphism $S \to X$ sending $p$ to $x_0$ and $v_0$ to $v$. The inverse image of $S_{\rm red} = \mathop{\rm Spec} k$ in $G$ is isomorphic to $\mathbb G_{\mathrm m, k}$; so $G_{\rm red}$ is an affine scheme, hence $G$ is an affine scheme. The differential of the projection $G \to S$ at the origin of $G$ has a $1$-dimensional kernel, the tangent space of $\mathbb G_{\mathrm m, k}$ at the origin. On the other hand there is a unique section $S \to G$ sending $p$ to the origin, corresponding to $1 \in k^* = \mathrm{Aut}_{k[\epsilon]}k$; this means that there is a unique tangent vector of $G$ at the origin mapping to $v_0$. These two facts give a contradiction. -Here is another way to look at this. Given a contravariant functor $F$ on $k$-schemes and an element $p$ of $F(\mathop{\rm Spec}k)$, one can define the tangent space of $F$ at $p$ as the set of element of $F(\mathop{\rm Spec}k[\epsilon])$ that restrict to $p$. However, in order for this tangent space to be a $k$-vector space, one needs a Schlessinger-like gluing condition on $F$ (this is standard in deformation theory). The analysis above shows that this condition is not satisfied for the functor of automorphisms of $k$ over $k[\epsilon]$.<|endoftext|> -TITLE: fundamental groups of surfaces -QUESTION [8 upvotes]: What are the properties that hold for a fundamental group of a surface and does not hold necessary for the fundamental groups of manifolds of higer dimensions ?? - -REPLY [3 votes]: Poincar\'e duality groups of dimension 2 are surface groups.<|endoftext|> -TITLE: Regular, Gorenstein and Cohen-Macaulay -QUESTION [29 upvotes]: All the statements below are considered over local rings, so by regular, I mean a regular local ring and so on; -It is well-known that every regular ring is Gorenstein and every Gorenstein ring is Cohen-Macaulay. There are some examples to demonstrate that the converse of the above statements do not hold. For example, $A=k[[x,y,z]]/(x^2-y^2, y^2-z^2, xy, yz, xz)$ where $k$ is a field, is Gorenstein but not regular, or $k[[x^3, x^5, x^7]]$ is C.M. but not Gorenstein. -Now, here is my question: -I want to know where these examples have come from, I mean, have they been created by the existence of some logical translations to the Algebraic combinatorics (like Stanley did), or even algebraic geometry, or they are as they are and they are some kind of lights that have been descended from heaven to their creators by any reason! - -REPLY [41 votes]: I will argue that the examples you gave are "simplest" in some strong sense, so although they look unnatural, if Martians study commutative algebra they will have to come up with them at some point. -Let's look at the first one $A=k[[x,y,z]]/(x^2-y^2, y^2-z^2, xy,yz,zx)$. Suppose you want - -a $0$-dimensional Gorenstein ring which is not a complete intersection (complete intersections are the cheapest way to get Gorenstein but non-regular, for example $k[[x]]/(x^2)$). - -Then it would look like $A=R/I$, where $R$ is regular and $I$ is of height equals to $\dim R$. If $\dim R=1$ or $2$, then $I$ would have to be a complete intersection, no good (now, the poor Martian may not know this at the begining, but after trying for so long she will have to give up and move on to higher dimension, or prove that result for herself). Thus $\dim R=3$ at least, and we may assume $R=k[[x,y,z]]$ (let's suppose everything are complete and contains a fied). -Since $I$ is not a complete intersection, it must have more than $3$ minimal generators. If it has $4$ then it would be an almost complete intersection, and by an amazing result by Kunz (see this answer), those are never Gorenstein. -So, in summary, a simplest $0$-dimensional Gorenstein but not complete intersection would have to be $k[[x,y,z]]/I$, where $I$ is generated by at least $5$ generators. At this point our Martian would just play with the simplest non-degenerate generators: quadrics, and got lucky! -(You can look at this from other point of view, Macaulay inverse system or Pfaffians of alternating matrices, see Bruns-Herzog, but because of the above reasons all the simplest examples would be more or less the same, up to some linear change of variables) -On to your second example, $k[[t^3,t^5,t^7]]$. Now very reasonably, our Martian wants - -a one dimensional Cohen-Macaulay ring B that is not Gorenstein. - -Since $\dim B=1$, being a domain would automatically make it Cohen-Macaulay. The most natural way to make one dimensional domains is to use monomial curves, so $B=k[[t^{a_1},...,t^{a_n}]]$. But if $n=2$ or $n=3$ and $a_1=2$ you will run into complete intersections, so $a_1$ would have to be $3$ and $n=3$ at least, and on our Martian goes... -(Again, you can arrive at this example by looking at things like non-symmetric numerical semi-groups, but again you would end up at the same simplest thing). -Reference: I would recommend this survey for a nice reading on Gorenstein rings. - -REPLY [17 votes]: All of these conditions are very important in algebraic geometry. I don't know much about the algebraic combinatorics aspect of these notions, but my feeling is that came from geometry and not vice versa. -The reason we care about these notions is that even though it would be nice to always work with non-singular varieties (a.k.a. regular) it can't always be done. For instance, families of non-singular varieties may degenerate to singular ones and most of the time there is no way to resolve these singularities in the families. -For instance, any families of hypersurfaces (e.g., plane curves) degenerate to singular ones. However, hypersurfaces are Gorenstein so if we can handle those we are fine. So, in particular, to give an example of a Gorenstein but not regular ring you only need to find a singular hypersurface. For example $k[[x,y]]/(x^2-y^3)$ is such an example. -Now if you study more general varieties than hypersurfaces you might not always be able to guarantee that they degenerate to Gorenstein varieties. On the other hand, if you consider stable families, then if the general fiber is smooth, then all fibers are Cohen-Macaulay. This is a non-trivial result. You can find it here. -As Kevin mentioned, the Gorenstein and Cohen-Macaulay properties can be measured by the dualizing complex. $X$ is Cohen-Macaulay if and only if its dualizing complex is a sheaf and it is Gorenstein if and only if it is Cohen-Macaulay and its dualizing sheaf is a line bundle. I am not totally sure what he means by the last statement, but $X$ is regular if the sheaf of differentials is locally free. If it isn't regular one needs to think about what "top differentials" mean. For a discussion of that see this MO answer. -Anyway, this gives us an easy way to construct Cohen-Macaulay but not Gorenstein varieties. You "only" need a Cohen-Macaulay variety whose canonical bundle is not a line bundle. -An easy way to do that is to use the fact that rational singularities are always Cohen-Macaulay. For surface singularities you can ensure that they are rational from their resolution graph (see Artin's paper) and it is easy to cook up a resolution graph that makes sure that the canonical sheaf of the singularity is not a line bundle. -Another way to make sure that a singularity is Cohen-Macaulay is to compute its local cohomology. See Lemma 4.1 of this paper of Patakfalvi for a condition. That tells you when a cone is Cohen-Macaulay and then just pick a variety with the right embedding and it will give you something non-Gorenstein. For instance, take a cone over $\mathbb P^1\times \mathbb P^1$ embedded by the $(2,1)$ line bundle.<|endoftext|> -TITLE: Periods and L-values of modular forms -QUESTION [10 upvotes]: Suppose $f$ is a modular form of weight $k \ge 2$. -It's "well-known" that there are "periods" $\Omega_-$ and $\Omega_+ \in \mathbb{C}$, such that the $L$-values $L(f, \chi, j)$, for $\chi$ a Dirichlet character and $1 \le j \le k-1$, are $(2\pi i)^j$ times an algebraic multiple of one of $\Omega_{\pm}$. -What is the correct statement of this "well known" result? Colmez claims in his Bourbaki seminar on $p$-adic BSD (here, Section 3.1.3) that -$$L(f, \chi, j) \in \begin{cases} \overline{\mathbb{Q}} \cdot (2\pi i)^j \Omega_+ & \text{if $\chi(-1) = (-1)^j$} \\ \overline{\mathbb{Q}} \cdot (2 \pi i)^j\Omega_- & \text{if $\chi(-1) =-(-1)^{j}$.}\end{cases} $$ -On the other hand, Vatsal claims in Theorem 0.1 of this paper (in which he defines canonical choices for the periods $\Omega_{\pm}$ up to $p$-adic units for a given prime $p$) that the conditions should be -$$L(f, \chi, j) \in \begin{cases} \overline{\mathbb{Q}} \cdot (2\pi i)^j \Omega_+ & \text{if $\chi(-1) = 1$} \\ \overline{\mathbb{Q}} \cdot (2 \pi i)^j\Omega_- & \text{if $\chi(-1) =-1$.}\end{cases} $$ -Vatsal's statement is repeated verbatim in the MathSciNet review of the paper (here if you have an institutional subscription). -These can't both be right, surely? If $k = 2$ then the only possibility for $j$ is $1$, so the two claims can be reconciled by simply switching the labelling of the two periods; but for $k \ge 3$ then the only way they can both hold is if $\Omega_+ / \Omega_-$ is algebraic, which I gather isn't expected to be the case unless $f$ is CM. -Colmez doesn't give a reference, while Vatsal cites a 1976 paper of Shimura, which I haven't been able to get hold of to check the exact statement. -Which of these two statements are correct? Or are they both correct but I've misunderstood them? - -REPLY [13 votes]: Check out Manin's paper Periods of parabolic forms and $p$-adic Hecke series. He only deals with level 1 and even weight, but this is enough to know that Vatsal's statement is not the right one (though not enough to fully confirm Colmez's statement, though it's true as given in Theorem 1 of Shimura's paper you can't get your hands on; you should also be able to dig it out of Mazur–Tate–Teitelbaum or Emerton–Pollack–Weston, which generalizes Vatsal). In particular, see section 1.2: The Periods Theorem of Manin's paper. -Update: In fact, the statement in the form you're looking for is written down in Theorem 1(ii) of Shimura's On the periods of modular forms (doi:10.1007/BF01391466), where he even tells you what you can take as $\Omega_\pm$ (and the result in the weight 2 case in Shimura's aforementioned 1976 paper was still conditional on a result proved in this 1977 paper).<|endoftext|> -TITLE: Identification of conformal classes of pos def quadratic forms on R^2 with unit ball -QUESTION [5 upvotes]: One of the lemmas at the foundation of Teichmuller theory is as follows. Let $Q(x,y)$ be a positive definite quadratic form. Then there exists unique $\lambda \in \mathbb{R}$ and $\mu \in \mathbb{C}$ with $\lambda > 0$ and $\|\mu\| < 1$ such that the following hold. Let $Q_{\mu}(x,y)$ be the quadratic form -$$Q_{\mu}(x,y) = \|z + \mu \overline{z}\|^2 \quad \quad (z=x+i y)$$ -Then -$$Q(x,y) = \lambda Q_{\mu}(x,y).$$ -I can prove this using brute force (it comes down to solving a system of 3 equations in 3 unknowns). However, this shed no light on why it should be true. Does anyone know a conceptual reason for it? - -REPLY [2 votes]: This is a special case of a general fact, that the space of positive definite quadratic forms on $\mathbb{R}^n$ with determinant $=1$ is a symmetric space for the Lie group $PSL(n,\mathbb{R})$. A matrix acts on a quadratic form by change of basis, and a stabilizer is a conjugate of $SO(n,\mathbb{R})$. When $n=2$, the symmetric space for $PSL(2,\mathbb{R})$ is $\mathbb{H}^2$, hyperbolic 2-space, and you are considering the identification with the unit disk model. From this perspective, your question reduces to: why is upper half-space the symmetric space for $PSL(2,\mathbb{R})$? Of course, this is a special fact related to the existence of $\mathbb{C}$, a commutative field containing $\mathbb{R}$. The action here is the restriction of the action of $PSL(2,\mathbb{C})$ on $\mathbb{CP}^1$ to the subgroup $PSL(2,\mathbb{R})$, so that the point stabilizers of $\mathbb{CP}^1-\mathbb{RP}^1$ are compact. I don't know a good explanation for this, other than the fact that the eigenvalues of the two complex conjugate eigenvectors must be unit complex conjugates, which forms a compact group $U(1)\cong PSO(2)$.<|endoftext|> -TITLE: Is there a closed form expression for the inverse of the matrix with elements $A_{i,j}=x_i$ for $i=j$ and $A_{i,j}=1$ for $i\neq j$? -QUESTION [5 upvotes]: Hello All - -Consider a matrix with elements: -$A_{i,j}=x_i$ for $i=j$ -$A_{i,j}=1$ for $i\neq j$ -Is there a closed form expression for the elements of $A^{-1}$? -Will be glad to know of any reference. -Thanks -HC - -REPLY [14 votes]: You can use the Sherman-Morrison formula. -In the notation of the Wikipedia article, let $u=v=(1,\ldots,1)'$ and $A$ (not the same as your $A$) be the diagonal matrix with $(x_{1}-1, \ldots, x_{n}-1)$ on the diagonal. -Then, if I haven't made a mistake, the entry of the inverse matrix you're looking for is -$\frac{1}{x_{i}-1} - \frac{ \frac{1}{(x_{i}-1)^{2}} }{ 1 + \sum_{k} \frac{1}{x_{k}-1}}$ if $i=j$, and $ \frac{ -\frac{1}{ (x_{i}-1)(x_{j}-1) }}{ 1 + \sum_{k} \frac{1}{x_{k}-1}}$ otherwise.<|endoftext|> -TITLE: Self-injective basic algebras -QUESTION [5 upvotes]: Do you know of any self-injective basic algebra $A$ over a field $k$ such that its enveloping algebra $A^{\mathrm{op}}\otimes_k A$ is not self-injective? -The algebra $A$ cannot be finite-dimensional, since then $A$ is Frobenius and so is $A^{\mathrm{op}}\otimes_k A$. - -REPLY [5 votes]: [Nagata, Masayoshi. A conjecture of O'Carroll and Qureshi on tensor products of fields. Japan. J. Math. (N.S.) 10 (1984), no. 2, 375--377. MR0884425] proved that the Krull dimension of the tensor product $\mathbb C(x,y)\otimes_{\mathbb C}\mathbb C(x,y)$ is $2$. If this tensor product is reduced (I think it is, but it is late...) then it cannot be self-injective. In that case, $\mathbb C(x,y)$ is an example.<|endoftext|> -TITLE: The asymptotic growth of global sections of powers of a complex line bundle -QUESTION [7 upvotes]: L is a holomorphic line bundle on a compact complex manifold X. The Kodaira dimension of L is defined as the maximal dimension of the image of the map associated to the powers $ mL(m \in N)$. I want to prove the asymptotic estimate -$$ h^0 (X,mL) \leq O(m^{k(L)})$$ -I heard that it is an easy consequence of the Schwarz lemma. Maybe it is a similar argument used by Siegel to prove the theorem "the transcendental degree of the meromorphic function field of a compact complex manifold is not bigger than the dimension of the manifold". But I'm afraid to deal with meromorphic mappings and singularities in the image.So I cannot complete the argument myself. -Can somebody tell me how the argument goes? - -REPLY [3 votes]: Hi, -An enlightening and very elementary proof of this fact can be found in the very complete book of X. Ma and G. Marinescu "Holomorphic Morse inequalities and Bergman kernels". -You will find this in Chapter 2. Their approach is exactly what you are looking for (only elementary complex analysis in several variables and a slightly modified Schwarz inequality).<|endoftext|> -TITLE: Volume of Minkowski sum of a ball and an ellipsoid -QUESTION [6 upvotes]: Is there a simple way to calculate/estimate the volume of Minkowski sum of an n-dimensional unit ball and an n-dimensional ellipsoid? Even a simple ellipsoid like $\frac{x_1^2}{a^2} + x_2^2 + \ldots + x_n^2 = 1$ will do. - -REPLY [3 votes]: For more on this, check out "Surface area and other measures of ellipsoids", by one I. Rivin: -[math/0403375] Surface area and other measures of ellipsoids -by I Rivin -► -Surface area and other measures of ellipsoids - Elsevier -by I Rivin - 2007 -Advances in Applied Mathematics 39 (2007)<|endoftext|> -TITLE: Families of curves for which the Belyi degree can be easily bounded -QUESTION [13 upvotes]: I know (edit: three) families of smooth projective connected curves over $\bar{\mathbf{Q}}$ for which the Belyi degree is not hard to bound from above. - -The modular curves $X(n)$. They are constructed by compactifying the quotient $Y(n) = \Gamma(n)\backslash \mathbf{H}$. The natural morphism $X(n) \longrightarrow X(1)$ is Belyi of degree $n^2$ (up to a constant factor). This also bounds the Belyi degree of a modular curve given by a congruence subgroup $\Gamma$. In general, Zograf proves that the Belyi degree of a (classical congruence) modular curve is bounded by $128(g+1)$. -The Fermat curves $F(n)$. They are given by the equation $x^n+y^n+z^n =0$ in $\mathbf{P}^2$. The morphism $(x:y:z)\mapsto (x^n:z^n)$ is Belyi of degree $n^2$. It is known that $F(n)$ is not a modular curve for $n$ big enough. So this example is really different than the one above. (Also note that $n^2\leq 10g+10$ by the Plucker formula.) -Wolfart curves are curves $X$ over $\overline{\mathbf{Q}}$ with a Galois Belyi morphism $X\to \mathbf{P}^1$; I took this terminology from a preprint by Pete L. Clark. Such curves are also called Galois Belyi covers or Galois three-point covers in the literature. The Belyi degree of a Wolfart curve is bounded by $84(g-1)$. (In particular, the latter implies that there are only finitely many Wolfart curves of given genus.) - -The following family of curves is not so easily dealt with. - -For an elliptic curve $E$ over the rational numbers, the Belyi degree can be bounded in the height of the $j$-invariant of $E$ following Belyi's proof of his theorem. This was written down explicitly by Khadjavi and Scharaschkin. - -I'm looking for families of curves for which the Belyi degree is ``easy to read off''. That is, a collection (finite or infinite) of smooth projective connected curves $X_i$ over $\bar{\mathbf{Q}}$ for which the Belyi degree can be bounded easily. -Are there any other nice examples? - -REPLY [12 votes]: Another example, like JSE's, that comes already equipped with a Belyi map but is not as familiar as modular curves and Fermat curves: For any relatively prime integers $m,n$ with $0 -TITLE: Is there an algebraic construction of the Quillen (determinant) Line Bundle? -QUESTION [17 upvotes]: Let's consider the moduli space of representations of $\pi=\pi_1(\Sigma)$ (a surface group) into $G$ (a lie group). Call this $X=\operatorname{Hom}(\pi,G)$, and let $Y=\operatorname{Hom}(\pi,G)/\\!/G$, where $G$ acts by conjugation on $X$ (and we take the GIT quotient). Let's denote the quotient map by $f:X\to Y$. Goldman has constructed a natural symplectic form $\omega$ on $Y$. By a construction of Quillen (the so-called determinant line bundle) there exists a line bundle $\mathcal L$ over $Y$ with curvature form equal to $\omega$ (or perhaps $\omega$ times some constant). -To construct this line bundle, though, one apparently needs to think of $\pi$ as the fundamental group of some specific Riemann surface (i.e. pick a holomorphic structure on $\Sigma$), and consider the entire (infinite-dimensional) moduli space of flat $G$-connections on $\Sigma$ (of which $Y$ is the quotient modulo the gauge group). We then need some infinite-dimensional analysis, and the notion of a Fredholm operator. I'm looking for a construction of $\mathcal L$ which stays in the algebraic world (and in particular avoids picking a holomorphic structure on $\Sigma$). -Some questions along those lines: -1) Is $f^\ast\omega$ exact? If so, then is there a natural choice of $1$-form $\gamma$ on $X$ such that $d\gamma=f^\ast\omega$? This would essentially answer my question, as taking the trivial bundle with connection form $\gamma$ (some might say $\exp\gamma$) would give $f^\ast\mathcal L$. -2) Is there another way to construct $\mathcal L$ staying in the algebraic category? -Perhaps a comment is relevant: Goldman proves that $\omega$ is closed by appealing to the infinite-dimensional picture of the moduli space of flat connections. This is apparently the standard proof, though I have found in the literature an entirely algebraic proof that $\omega$ is closed, via some direct (but coordinate independent, and thus not all that messy) calculations, see http://www.ams.org/journals/proc/1992-116-03/S0002-9939-1992-1112494-2/. - -REPLY [14 votes]: For $G={\rm PGL}_n({\mathbb C})$ there is no complex algebraic Quillen line bundle on the complex character variety $Y$ as the cohomology class $[\omega]=\alpha_2\in H^2(Y;{\mathbb Q})$ is not pure by Proposition 4.1.8 in http://arxiv.org/abs/math.AG/0612668. -EDIT: It is instructive to think about the case $G={ GL}_1({\mathbb C})$. Then the character variety is $M_B=GL_1^{2g}$ the complex torus while $M_{DR}$ is an affine bundle over $Jac(C)$. Analytically $M_B\cong M_{DR}$ but the universal bundle $L$ on $M_{DR}\times C$ which is just the pull back of the Poincare bundle from $Jac(C)\times C$ is not algebraic on $M_B$, for example because the cohomology class $[\omega]$ of the symplectic form (which shows up in $c_1(L)^2$) has weight $2$ in $H^2(M_{DR};{\mathbb Q})$ but has weight $4$ (homogeneous weight $2$) in $H^2(M_B;{\mathbb Q})$. -Thus in particular there is no complex algebraic Quillen bundle on $M_B$ with first Chern class $[\omega]$. -EDIT 2: To take David's example let $g=1$, then $M_B={\mathbb C}^* \times {\mathbb C}^*$ which does not have non-trivial pure cohomology in $H^2(M_B;{\mathbb Q})$ and even the Picard group is trivial (see an argument here); thus $M_B$ does not support any non-trivial algebraic line bundle.<|endoftext|> -TITLE: Simple Ore extensions -QUESTION [6 upvotes]: Let $R[x;\sigma,\delta]$ be an Ore extension, where $R$ is an associative and unital ring and $\sigma : R\to R$ is a (not necessarily injective!) ring endomorphism. (In the literature it is often assumed to be injective). -My question is the following: -If $R[x;\sigma,\delta]$ is a simple ring, is $\sigma$ necessarily an injective map? -Note that the answer is affirmative in the case when the maps $\sigma$ and $\delta$ commute. Does anyone know of an example of a simple Ore extension where $\sigma$ is NOT injective? - -REPLY [2 votes]: Actually, the argument in my earlier answer can be refined to show the following. -Proposition: Suppose $R$ is a reduced abelian ring, $\sigma:R\rightarrow R$ is a ring homomorphism and $\delta:R\rightarrow R$ is a $\sigma$-derivation. If the Ore extension $A=R[x,\sigma,\delta]$ is simple, then $\sigma$ has to be injective. -Proof: Suppose $\sigma$ is not injective. Then we find a non-zero element $b\in\ker\sigma$. Because $R$ is reduced we know that all powers of $b$ are non-zero. Define $I=\{P\in A\mid \exists k\in \mathbb{N}:P\;b^k=0\}$. Then it is clear that $I$ is a left ideal of $A$, that it is a right $R$-submodule of $A$, that it does not contain $1$ and that $bx-\delta(b)\in I$. To show that $I$ is a non-trivial ideal, we only have to show that $Ix\subset I$. Take any $P\in I$, so $P\;b^k=0$ for some $k$. Now we see that $(P\; x)b^{k+1}=P\;b^k\delta(b)=0$ and hence $P\;x\in I$. QED<|endoftext|> -TITLE: What is higher dimensional algebra? -QUESTION [7 upvotes]: Could anyone explain what higher dimensional algebra is? -I tried to look on the web but I couldn't find a satisfactory definition, the ones that I found are too vague. What I'm looking for is a good definition of higher dimensional algebra, what it deals with and some references about the subject. - -REPLY [14 votes]: My definition of higher dimensional algebra is that it is the study and application of algebraic theories with partial algebraic structures whose domains are defined by geometric conditions. I think the first study of such partial algebraic structures was in -Higgins, P.~J. "Algebras with a scheme of operators". Math. Nachr. 27 (1963) 115--132. -The intuitive idea is also: do mathematical formulae have to be confined to a line? Can we have and use formulae in 2 or 3 or many dimensions? and can this be useful? If so we can expect to need rewriting of such formulae, and indeed this exists and is useful. You can find many 2-dimensional rewrites in the new book Nonabelian Algebraic Topology (European Mathematical Society, 2011), and with a pdf available. See for example the rotations in section 6.4, used in this mathoverflow answer on computing homotopies. -Those who have read Flatland will remember that the Linelanders had a rather limited existence! Can we really think that the brain works in one dimension only? -I mention that the idea of all this occurred to me in 1965 when I thought that the proof I had written out many times for the van Kampen Theorem for the fundamental groupoid should generalise to higher dimensions, if one had the right gadgets, seen as higher homotopy groupoids. 9 years later: whoopee! (work with Chris Spencer and with Philip Higgins). The first joint papers with Philip were published in 1978 and 1981. -Note that group objects in the category of groups are just abelian groups. But double groupoids, which are groupoid objects in the category of groupoids, are much more complicated than groups, and can model $2$-dimensional homotopical notions. -You can find presentations on some of these ideas on my preprint page, and including a presentation Theoretical Neuroscience in Delhi in 2003, which went down well! arXiv:math/0306223. -For my purposes I have needed a cubical context, which is not commonly followed, but in which it was easier to conjecture and prove the theorems I was thinking of. But the definition above seems quite general, and allows for a melding of algebra and geometry. -I also believe that Henry Whitehead's term `Combinatorial homotopy' and his actual work show he was thinking of generalising some methods of combinatorial group theory to higher dimensions. His notion of crossed module (1946) has proved an important feature of such ideas, and crossed modules later appeared in a key relations with $2$-groups and more generally with double groupoids. -In another direction the following diagram - -occurs in this paper on double semigroups.<|endoftext|> -TITLE: What should philosophers know about math? -QUESTION [14 upvotes]: I know, I know... this is not a technical question. Nevertheless, I believe this is the right place to ask such question. -I am sure many of you read about philosophy, including philosophy of science, ethics and so on. If you read works from the ancient times, especially those written by non mathematicians, you may find mistakes simply because they used to think on the science the knew at the time. Actually, until a couple of hundred years ago, most natural philosophers (if not all of them) were scientists as well. -My question is: - -Does the lack of knowledge about math or science in general, have an impact on a philosopher's ability to reason about the world and figure the right path to follow? - -I know many philosophers or at least, people who have a degree in philosophy (either master or phd), and I find that they have no basic understanding of very simple facts, from physics all the way down to mathematics. For example, many of them would guess that, even in absence of friction, an object heavier than another would reach the ground faster. But, again this is only one of the many examples. I'm not even going to mention absence of both knowledge and intuition about markov-based processes, probability in general and so forth. -What is your opinion about this. Do you think philosophers should have a deeper basis in natural sciences in order to better understand the world and develop better ways of thinking? - -REPLY [2 votes]: I do not see math or physics as a requisite for a philosopher. For example, one can look to the ancient eastern philosophers and see that they possessed a profound perception of the world without any knowledge of modern physics or mathematics. And for those not as philosphically inclined (as well as those suitably apt), knowledge of modern mathematics and physics can only be a good thing, as they are languages of nature. However, I do not see one's philosophy being hampered by lack of knowledge of the fact that rocks of a different size dropped from the top of the tower of Pisa should hit the ground roughly at the same time.<|endoftext|> -TITLE: Cohen-Macaulay domain with non-Cohen-Macaulay normalization? -QUESTION [12 upvotes]: Is the normalization of a Cohen-Macaulay domain necessarily Cohen-Macaulay? I suspect that the answer is no, but I don't have a counterexample. I am most interested in "geometric" situations, so one can place assumptions like excellence on the domain if it's relevant. - -REPLY [13 votes]: This is an algebraic and affine version of what Karl wrote. I could not produce a concrete example, but here is how one can try to do it. -Take a normal non-CM domain $B$ over some field of characteristic $0$ (see this question for some concrete examples). Find a Noether normalization $A \subset B$ (Macaulay 2 can do it for you). -Next, use the primitive element theorem to find $z \in B$ such that $R=A[z]$ has the same quotient field as $B$. Clearly then $\bar R = B$, and $R$ is a hypersurface since $A$ is a polynomial ring. The equation is likely to be messy, though.<|endoftext|> -TITLE: Existence of certain identities involving characteristic 2 "thetas" -QUESTION [5 upvotes]: Let l=2m+1 be prime. In my previous MO question, "What are the polynomial relations between these characteristic 2 thetas?", I defined a subring of Z/2[[x]] as follows: -The subring, S, is generated by [1],...,[m] where [i] is the sum of the x^(n^2), n running over all integers congruent to i mod l. -QUESTION...... Let F=x+x^9+x^25+x^49+...,G=F(x^l), and H=G(x^l). Are G and H in S? -The answer is yes when l=3,5 or 7. When l=7, if we set a=[1],b=[2] and c=[3], we have the curious identities H=(abc)^3*(abc+ba^3+cb^3+ac^3), and G=(abc)^2+a^7+b^7+c^7+H. -Remark 1... Kevin Buzzard explained to me that one can decide whether an explicitly given identity such as the ones we've displayed holds by using the theory of characteristic 2 modular forms and computer calculation. But how does one produce these putative identities? -Remark 2... For all l one can show in an elementary way that H is in the field of fractions of S. In fact if a=[i], b=[2i] and c=[4i], then H is the quotient of a^8(a^8+b^2) by b^4+c. Furthermore for l at most 13, H is in S. (One shows that the quotient lies in S, by combining the "quintic relations" of my MO question cited earlier with Groebner basis computer calculations.) -I'll sketch an argument giving the l=7 identities. Let C be the curve in affine 3-space defined by the ideal of quintic relations. C has 3 linear branches at the origin and 3 -linear branches at each of the seven points (r,r^4,r^9) with r^7=1. Passing to projective 3-space we find that (the Zariski closure of) C has 14 simple points at infinity. The formula for H as a quotient shows that H has zeros of order 49 at the branches at the origin, simple zeros at the branches at the other singular points, and poles of order 12 at infinity. This -leads to the identity for H. To get the identity for G one notes that (GH)+(GH)^2+(G+H)^8=0--see my MO question, "What's known about the reduction...?" It follows from this that if G is in the field of fractions of S then G+H has zeros of order 7 at the branches at the origin, of order 3 at the branches at the other singular points, and poles of order 6 at infinity. This suggests that G+H=(abc)^2+a^7+b^7+c^7. To verify this we set J=(abc)^2+a^7+b^7+c^7+H, and use Groebner basis computer calculations to show that JH+(JH)^2+(J+H)^8=0; it then follows that J=G. -EDIT: I think I can now show that when l=11, G is NOT in the field of fractions of S, -even though H is in S. I'll make this an answer once I'm surer of it. -EDIT #2: My supposed counterexample when l=11 is incorrect; G like H is in S. I had -the wrong modular equation of degree 11 relating G and H. Once I found the correct equation, -in Cayley's article, I was able to argue as in the case l=7. -FINAL(?) EDIT: As I've shown in my answer, G and H are indeed always in S. And I've produced a simple conjectural explicit formula for G+H that holds for l<1500. Whether there is anything comparably simple for H isn't clear. At any rate here are formulas for H when l<24. I write C(a,b,c) for the sum of the [ra][rb][rc] where r runs from 1 through (l-1)/2; -more generally (a,b,c) can be replaced by any multi-set. P is the product of the [r] where -r runs from 1 through (l-1)/2. The identity when l=17 is striking. -l=3.......... [1]^9 +[1]^12 -l=5.......... P^5 +P^6 -l=7.......... (P^3)(P+C(1,1,1,2)) -l=11.........(P^2)(C(1,1,3)+C(1,1,2,4)) -l=13.........P*(P+[1][2][3][5]+[1][4][5][6]+[2][3][4][6]+C(1,1,2,2,2,5)) -l=17.........P*([1][2][4][8]+[3][5][6][7]) -l=19.........P*([2][3][5]+[4][6][9]+[1][7][8]+C(3,3,2,4)) -l=23.........P*(C(1,2,3,3)+C(1,2,4,5)+C(1,4,4,6)+C(1,2,2,5,9)) - -REPLY [3 votes]: I suppose it's bad form to answer one's own MO question, but I now have an almost complete solution to this one. I can prove: -1.----H is always in the ring S generated by the [j]. -2.----The same holds for G except perhaps when l=15 mod 16. (In "More questions involving characteristic 2 theta series identities" I provide some experimental evidence when l=15 mod 16.) -To prove 1. note that I gave a formula in my question expressing H as a quotient of elements of S. Now I have made a study of the variety V consisting of the zeros of the polynomial relations between the various [j]. V is a curve; when l>3 it has exactly l+1 singular points, each of which is an ordinary multiple point of multiplicity (l-1)/2. Using my formula for H, I can show that it has ord at least 0 at every non-singular point of V, and ord> 0 at every branch centered at every singular point. So it lies in all the local rings of S. -EDIT:NOT SO--the condition of being in the local ring at a singular point is more stringent. -For a correct argument see the FINAL EDIT below. -To prove 2. let C be the sum of the x^(ln) where n runs over all (non-zero) integers of the form (square) or 2(square) or l(square) or 2l(square). Note that C^2+C is G+H. So in view of 1. it suffices to show that C is in S. In my previous answer I indicated why this is true when l=1 mod 4 or l=3 mod 8, writing C explicitly as a polynomial in the [j]. I will edit this answer shortly to handle the more difficult case l=7 mod 16. -EDIT: Suppose now l=7 mod 16. Here's a proof that C lies in S. Let T, contained in a product of 4 copies of Z/l, consist of all (r1,r2,r3,r4) other than (0,0,0,0) with (r1^2)+(r2^2)+(r3^2)+(r4^2)=0. There is a group of order (24)(16)=384 acting on T by permutation of co-ordinates and sign changes of co-ordinates. Using the fact that l=7 mod 8, we find that every orbit has size 384 or 192 or 64. Call an orbit "small" if it has size 192 or 64. We shall show that C is a sum of terms attached to the small orbits. To each small orbit attach the power series [r1][r2][r3][r4] where (r1,r2,r3,r4) is an orbit representative; this is independent of the representative. I'll show that C is the sum of these contributions. Clearly every exponent appearing in the sum of these power series is -divisible by l. It remains to show that x^ln appears in the sum if and only if n is the product of a non-zero square by 1,2,l or 2l. -Now the coefficient of x^ln in [r1][r2][r3][r4] is the mod 2 reduction of M where M is the number of integer 4-tuples (a,b,c,d) satisfying: -(1)---(a^2)+(b^2)+(c^2)+(d^2)=0 -(2)---(a,b,c,d) reduces to (r1,r2,r3,r4) mod l -Modulo 2, M is (1/64)(the number of (a,b,c,d) satisfying (1) and reducing to an element in the orbit of (r1,r2,r3,r4)). So the sum of the M, modulo 2, is the number of (a,b,c,d) satisfying (1) and reducing to a point in some small orbit. Also the number of (a,b,c,d) -satisfying (1) and reducing to a point in an orbit of size 384 is clearly a multiple of 384. -So the coefficient of x^ln in our sum is the mod 2 reduction of (1/64)(the number of (a,b,c,d) that satisfy (1) and do not reduce to (0,0,0,0)). -Let R(n) be the number of representations of n as a sum of 4 squares. We have just shown that the coefficient of x^ln in our sum is the mod 2 reduction of (1/64)(R(ln)-R(n/l)). It remains to show that (1/64)(R(ln)-R(n/l)) is odd precisely when n is the product of a square by 1,2,l or 2l. Jacobi proved that R(n) is 8(the sum of the divisors of n) when n is odd, and 24(the sum of the odd divisors) when n is even. So (1/8)(R(ln)-R(n/l)) is a product of -local factors, one from each prime. The factor attached to 2 is 1 or 3. That attached to l is l^t(1+l) where t=(ord_l)(n). Since l=7 mod 16, this is 8(an odd number). Finally that attached to an odd prime p other than l is the sum of the divisors of p^s where s is (ord_p)(n). This factor is odd just when s is even, and the result follows. -A couple of remarks. When l=15 mod 16 the same argument shows that the sum we've constructed is not C, but 0. Also an orbit is small precisely when it has a representative with r1=r2 or a representative with r4=0. -FINAL EDIT: I now have an answer I'm prepared to accept, unless some spoilsport finds a flaw; it shows that G,H (and F) all lie in the subring S of Z/2[[x]] generated by the [j] -irrespective of l. Unlike the approach taken in the last edit which exhibited G+H explicitly as a polynomial in the [j], (except when l is 15 mod 16), this one doesn't seem to give nice explicit formulas. I'll be using results from other MO questions of mine, and some further results in manuscript. Let K be an algebraic closure of Z/2, and S' be the subring of K[[x]] generated over K by the [j]. It,s enough to show that G,H and F lie in S'. -First I show that they're all in the field of fractions, L, of S'. In another MO post I wrote H as a quotient of 2 elements of S. To handle F I use the following: -(1)___For l>3, Spec(S') is a curve with l+1 singular points, among them the maximal ideal m generated by [1],...,[l-1]. These are ordinary singular points of multiplicity (l-1)/2. -(2)___There is a group of automorphisms of S'/K isomorphic to PSL_2(Z/l). These automorphisms stabilize the space spanned by [0],...,[l-1] and act transitively on the (l+1)(l-1)/2 valuation rings in L/K containing the local rings at the singular points. The group is generated by the maps [j]-->[rj], r prime to l, [j]-->a^(j^2) [j] where a is an l'th -root of unity in L, and a sort of characteristic 2 "Fourier transform". -Now the maps [j]-->[rj] and [j]-->a^(j^2) [j] generate a subgroup B of PSL_2 of order l(l-1)/2, and my "quotient formula for H" shows that B fixes H. So the orbit of H under PSL_2 has size at most l+1. A rather formal calculation with the "Fourier transform" shows that the orbit consists of H and the F(ax) where a^l=1. I claim that each of these elements lies in the local ring of m on S'. For H this is easy; H has ord l^2 at each valuation ring containing m. Taking E to be the sum of [1],...[(l-1)/2] we find that E+E^4=F+H. So F is in this local ring as well, and the result follows easily for each F(ax). The fact that PSL_2 -acts transitively on the singular points now shows that H and the F(ax) lie in the local ring at every singular point. Also the quotient formula for H shows that H has ord 0 at every non-singular point, and the same then holds for the F(ax). Thus H and the F(ax) are in S'; this corrects the argument I gave earlier. -I now turn to G. There is a degree l+1 2-variable symmetric polynomial P over Z/2 with P(F,G)=0. Furthermore P(z,G) is monic of degree l+1, and has H and the F(ax) as roots. Also the constant term of P(z,G) is G^(l+1), while the coefficient of z is G+ higher degree terms. Since the product of H and the F(ax), as well as the l'th symmetric function of H and the F(ax), are in S', both G^(l-1) and G+... are in S'. Now over K these 2 elements generate a field between K(G^(l+1)) and K(G); since G+... is in this field it is all of K(G), and G is in L. Also G^(l+1), as the product of H and the F(ax), is fixed by PSL_2. -Since every homomorphism from PSL_2 to the l+1 th roots of unity is trivial, G is fixed by -PSL_2. -At the valuation rings lying over m, G has ord l. So G is in the local ring of m, and consequently in the local ring at every singular point. Furthermore, like H and the F(ax), G -has ord 0 at the non-singular points. So it is in S'. (Note also that like H and the F(ax), G has poles of order 12 at every valuation ring in L/K that doesn't contain S').<|endoftext|> -TITLE: Bound on the von Neumann entropy of a positive, positive semi-definite, and unit-trace density matrix? -QUESTION [8 upvotes]: Can the von Neumann entropy of a positive, positive semi-definite, and unit-trace density matrix with equal on-diagonal terms be bounded by equalizing all off-diagonal elements to their highest/lowest value? -Statement of problem -Consider the density matrix $M = (m_{i,j})$ in $d$-dimensions with all positive elements: $m_{i,j} > 0$. From physics, a density matrix is Hermitian, positive semi-definite, and has unit trace: -$\quad M^\dagger = M, \quad 0 \le M \le 1, \quad \mathrm{tr} \, \, \, M = 1 .$ -For now, we also assume that $M$ has equal on-diagonal elements: $m_{i,i} = 1/d$. -Now, consider the two density matrices $M^\uparrow$ and $M^\downarrow$ formed by setting all off-diagonal elements of $M$ to their maximum and minimum value, respectively: -$\quad m_{i,j}^{\uparrow} = \max_{k \neq l} (m_{k,l}) , \qquad i \neq j$ -$\quad m_{i,j}^{\downarrow} = \min_{k \neq l} (m_{k,l}) , \qquad i \neq j$ -$\quad m_{i,i}^{\uparrow} = m_{i,i}^{\downarrow} = m_{i,i}$ -(Above, the extremizations are taken over all off-diagonal elements, i.e. all $k$ and all $l$ such that $k \neq l$.) With the von Neumann entropy of a matrix $A$ defined as -$\quad S[A]= \mathrm{tr} (-A \ln A),$ -can we bound -$\quad S[M^\uparrow] \le S[M] \le S[M^\downarrow] \quad ?$ -Physics Intuition -Because $M$ is a positive matrix, it can be expressed as a Gram matrix for some set of $d$ vectors $\{v_i\}$. That is, the elements of $M$ are just the inner products of the $v_i$: -$\quad m_{i,j} = (v_i , v_j )$ -These $v_i$ are unique up to a global unitary. If we want, we can work with the normalized set $\{e_i = v_i/ \sqrt{p_i} \}$ with probability distribution $\{p_i = (v_i , v_i ) \}$. -The density matrix we describe above can be obtained by starting with a system $\mathcal{S}$ in a pure state $\vert \psi \rangle= \sum_{i = 1}^d \sqrt{p_i} \vert s_i \rangle$ (where $\{{\vert s_i \rangle\}}$ is an orthonormal basis for $\mathcal{S}$) and environment $\mathcal{E}$ in pure state $\vert e_0 \rangle$, and evolving forward under the unitary $U$ which sends -$\quad \vert s_i \rangle \otimes \vert e_0 \rangle \to \vert s_i \rangle \otimes \vert e_i \rangle$ -where $\langle e_i \vert e_j \rangle \equiv (e_i,e_j) = m_{i,j} \sqrt{p_i p_j}$. Then -$\quad \mathrm{tr}_\mathcal{E} [ U ( \vert \psi \rangle \langle \psi \vert \otimes \vert e_0 \rangle \langle e_0 \vert ) U^\dagger ] = \rho_{\mathcal{S}} \equiv M $ -Having equal on-diagonal elements of $M$ is equivalent to $p_i = 1/d$. $M$ being positive means that all the vectors $\vert e_0 \rangle$ can be fit in the first "orthant", i.e. there is a single basis in which all the $\vert e_0 \rangle$ have all positive components. -The physics intuition is that by decreasing the off-diagonal elements of this density matrix to their min value (i.e. we are making more distinguishable the environment states corresponding to distinct system states) we are just increasing the decoherence. Therefore, the entropy should go up as we pass from $M$ to $M^{\downarrow}$. Likewise, $M^{\uparrow}$ has less decoherence, and should have a lower entropy. -Numerical Evidence -I've sampled many millions of density matrices of the described form, and have never found a violation of the inequality in question. However, it's also always been true (numerically) -that -$\quad M \succ M^{\downarrow}$ -but -$\quad M^{\uparrow} \nsucc M \quad \mathrm{and} \quad M^{\uparrow} \nprec M$ -where $\succ$ is the majorization partial order on density matrices. (Entropy is a Shur-concave function, and therefore preserves the majorization order.) It was a surprise to me that $M^{\uparrow} \nsucc M$, and this lowers confidence in the physics intuition described above. If the inequality concerning entropies is true, it must make use of the specific properties of the entropy function, not just that it's Shur-concave. -Generalizing to unequal diagonal elements -When the diagonal elements of $M$ are unequal, we go back to the physics motivation to define $M^{\uparrow}$ and $M^{\downarrow}$. This situation corresponds to unequal $p_i$: -$\quad m_{i,i} = (v_i,v_i) = p_i$ -$\quad m_{i,j} = (v_i,v_j) = \sqrt{p_i p_j} (e_i,e_j), \quad i \neq j .$ -Additional decoherence will correspond to less overlap (more distinguishability) between the environmental states $e_i$, which remain normalized. This means we define -$\quad \gamma^{\uparrow} = \max_{k \neq l} (e_k,e_l) = \max_{k \neq l} [m_{k,l} / \sqrt{m_{k,k} m_{l,l}}]$ -$\quad \gamma^{\downarrow} = \min_{k \neq l} (e_k,e_l) = \min_{k \neq l} [m_{k,l} / \sqrt{m_{k,k} m_{l,l}}]$ -$\quad m_{i,j}^{\uparrow} = \gamma^{\uparrow} \sqrt{m_{i,i} m_{j,j}} , \qquad i \neq j$ -$\quad m_{i,j}^{\downarrow} = \gamma^{\downarrow} \sqrt{m_{i,i} m_{j,j}} , \qquad i \neq j$ -$\quad m_{i,i}^{\uparrow} = m_{i,i}^{\downarrow} = m_{i,i} = p_i$ -(Here, $\gamma^{\uparrow}$ and $\gamma^{\downarrow}$ are the largest and smallest "decoherence factors".) We can then ask whether the above inequality is true in this more general case. - -REPLY [5 votes]: Indefiniteness is not an issue. If M is positive semidefinite, then M-uparrow and M-downarrow are positive semidefinite too; that's easy to prove. -However, I'm afraid one can find counterexamples, at least for the lower bound. Take two random numbers between 0 and 1/d, and allow all off-diagonal elements of M to be one or the other (while preserving symmetry and checking for positive semidefiniteness of M). The entropy of M can then be below the entropy of M-uparrow. -The following matrix, for example: -M=[1 a b b; a 1 b b;b b 1 b;b b b 1]/4 -with a=0.01 and b=0.6 -has entropy 0.934, while its M-uparrow has entropy 0.940.<|endoftext|> -TITLE: Mathematicians working on social choice theory -QUESTION [8 upvotes]: Can someone tell me which mathematicians are actively working on social choice theory, or point to a place where they may be listed? - -REPLY [2 votes]: Graciela Chichilnisky is one of them (as well as her co-authors). Late Beno Eckmann has a nice chapter in one of his books on social choice theory as a direct application of an average rule in arbitrary spaces. Other people include Luc Lauwers or Shmuel Weinberger. These are examples of mainly algebraic topology at work. -On the other hand, social choice theory is a wide area. It ranges from aggregating preferences to voting and even mechanism design to some extent where tools applied are different.<|endoftext|> -TITLE: (Dis)similarity between groups and Lie algebras -QUESTION [18 upvotes]: There are many questions which sound similar or the same for groups and Lie algebras. Some (very few, actually) of those questions have identical solutions and answers. Some have identical answers but very different solutions. Some have different answers and different solutions. -I am not talking about situations where one have a strictly defined (functorial) correspondence of some sort which allows to reduce almost automatically some questions from groups to Lie algebras, like Lie groups-Lie algebras, or Malcev(-like) correspondence. -For simplicity, let us confine to finite groups and finite-dimensional Lie algebras over a field. A few examples of what I have in mind: - -Suppose a group (resp. Lie algebra) is represented as a (not necessarily direct) product (resp. sum) of two nilpotent subgroups (resp. subalgebras). Is it true that the group (resp. Lie algebra) is solvable? -Suppose a group (resp. Lie algebra) is commutative-transitive: $[x,y] = 1$ (resp. 0), $[y,z] = 1$ (resp. 0), and $y \ne 1$ (resp. 0) implies $[x,z] = 1$ (resp. 0). What can be said about its structural properties? -(edit: added a couple of hours later) (Lyndon-)Hochschild-Serre spectral sequence connecting group (resp. Lie algebra) (co)homology with (co)homology of its normal subgroup (resp. ideal) and the respective quotient. - -As far as I know, both questions 1. and 2. have very similar answers for groups and Lie algebras, in some particular situations admit (almost) identical proofs or fragments of proofs, but in the whole generality the proofs are very different, with group-theoretic proof not utilizing in any way a Lie-algebraic one, and vice versa. As for 3., initially the group- and Lie-algebraic cases were given very similar, but disjoint proofs, and later were interpreted as particular instances of the Grothendieck (?) spectral sequence. -Question. -Is it possible, looking on a property of groups/Lie algebras expressed as a formula in the appropriate first- or second-order theory, to predict solely on syntaxical/logical ground, whether this property would convey similar or different results in respective categories? On the other hand, is there some (metamathematical, I dare to say) principle which would make this impossible? -I realize that the question is probably too vague, and I would appreciate any help in making it more precise and interesting. - -REPLY [3 votes]: This isn't getting directly at your question, but you might consider looking at Ellis' notion of a multiplicative Lie algebra. This all started in: -G.J. Ellis, On five well-known commutator identities, J. Aust. Math. Soc. Ser. A 54 (1993), 1–19. -Then just search for "multiplicative Lie algebra" or "multiplicative Lie ring" for more recent papers. Multiplicative Lie rings provide an interesting framework in which to think about some group and Lie algebra results in a unified way.<|endoftext|> -TITLE: Tomita-Takesaki versus Frobenius: where is the similarity? -QUESTION [29 upvotes]: I've often heard Alain Connes say that the modular flow of Tomita-Takesaki theory should be thought of as a characteristic zero analog of the Frobenius endomorphism. ... can anyone justify this claim? - -Given a von Neumann algebra $M$, its modular flow - is a canonically defined homomorphism - $$ -\mathbf{\Phi}: i\mathbb R\quad\longrightarrow\quad \text{BIM}^\times(M) -$$ - that, in the presence of a state (or weight), lifts to a homomorphism $i\mathbb R\to Aut(M)$. - Here, $\text{BIM}^\times(M)$ denotes the 2-group of invertible $M$-$M$-bimodules. - The bimodule $\mathbf{\Phi}(it)$ is the non-commutative $L^p$-space for the value $\frac 1 p=it$. - - Given a ring $R$ of characteristic $p$, its Frobenius is a canonically defined homomorphism - $$ -\mathbf{F}:\mathbb N\quad\longrightarrow\quad End(R) -$$ - such that $\mathbf{F}(1)$ sends $x$ to $x^p$. - More generally, $\mathbf{F}(n)$ sends $x$ to $x^{p^n}$. - -So far, the only analogy I can see is that both $\mathbf{F}$ and $\mathbf{\Phi}$ are canonically defined actions... - -REPLY [13 votes]: A low tech (naive?) piece of intuition comes straight from the definition of the modular operator and what happens if one tries to carry it over to finite fields. -The nontrivial automorphism $z\mapsto\overline{z}$ in $Gal(\mathbb{C}/\mathbb{R})$ is encoded in a von Neumann algebra via the existence of a *-operation $(zX)^\ast=\overline{z}X^\ast$. When $M$ is faithfully represented in a Hilbert space $H$ with cyclic and separating vector $\Omega$ we construct $SX\Omega:=X^\ast\Omega$, then $\Delta:=|S|^2$, then $\sigma_{t}(X)=\Delta^{it}X\Delta^{-it}$, so the modular group encodes $z\mapsto\overline{z}$ in some sense. -Let $n\in\mathbb{N}$ and note that the Frobenius automorphism of $F_{p^n}$ generates $Gal(F_{p^n}/F_p)$. Take an associative, unital algebra $R$ over $F_{p^n}$ equipped with a bijection $Q:R\to R$, satisfying $Q(zx)=Fr(z)Q(x)$ for all $z\in F_{p^n}$ and $x\in R$. If $R$ is faithfully represented on an $F_{p^n}$ vector space $V$ with cyclic and separating vector $\Omega$ then we obtain a map $S:V\to V$, $S x\Omega=Q(x)\Omega$, which has the property $$S z\xi=Fr(z)S\xi\quad\hbox{for all}\quad z\in F_{p^n},~\xi\in V.$$ Given such a map $T$ we can extract an $F_{p^n}$-linear map $T^n$ (the analogue of $T\rightarrow |T|^{2i}$ for an antilinear operator). Set $\Delta:=S^n$. As $\Omega$ is cyclic and separating and $Q$ is a bijection, $\Delta$ is invertible and we can form the maps $\sigma_{m}(x)=\Delta^{m} x\Delta^{-m}$ for each $m\in\mathbb{Z}$. Then we find that $\sigma_m$ is an $F_{p^n}$ algebra homomorphism and $\sigma_{m_1}\circ\sigma_{m_2}=\sigma_{m_1+m_2}$. If $R$ is a field over $F_{p^n}$ then we have the canoncial map $Q(x)=x^p$ and $\sigma_m(x)=Fr^{nm}(x)$. This extends easily to the case $R=M_n(K)$ for a field $K$. I'm not sure if a `nice' $Q$ exists for a general division algebra.<|endoftext|> -TITLE: fundamental group of the complement of lines in C^2 -QUESTION [6 upvotes]: Here is an interesting problem, which I could not solve and would appreciate any comment in solving it; -Assume that we have omitted infinitely many lines from $\mathbb{C}P^2$ to obtain $\mathbb{C}^2$ and now, consider the following lines in which $(z,w)$ is the coordinate of $\mathbb{C}^2$: $z=0, z=1, w=0, w=1, z=w.$ -What is the fundamental group of the complement of these lines in $\mathbb{C}^2$? - -REPLY [5 votes]: Following up on JSE's answer, this arrangement is, up to linear change of coordinates, the projectivization of the braid arrangement of rank three, so the fundamental group of the complement is the quotient of the pure braid group (of the plane) on four strands by the infinite cyclic subgroup generated by the full twist (= the product of the usual generators), which is central. You can see this as follows: the four points are labelled x,y,z,w; the requirement that they are distinct means you throw away the hyperplanes x=y, x=z, x=w, etc. The complement of these six hyperplanes in C^4 is homotopy equivalent to the x=0 slice: the homotopy is given by sliding points parallel to the line x=y=z=w, which lies in all the hyperplanes. This gives the arrangement of six planes in C^3 with equations y=0,z=0,w=0,y=z,y=w,z=w. projectively, once you throw away y=0 you can set y=1 in the other five equations, and you get your arrangement.<|endoftext|> -TITLE: Laminations as a limit of ideal triangulations -QUESTION [7 upvotes]: I am wondering about the following: - -Suppose that $S$ is a non-compact - hyperbolic surface of finite area. - Suppose that $\lambda \subset S$ is a - non-trivial, geodesic, - measured lamination. Forget the transverse measure. Is there a - (non-compact) geodesic lamination $\lambda'$ - containing $\lambda$, and a sequence of - ideal geodesic triangulations $T_i$, so - that $T_i \to \lambda'$ in the - Chabauty topology? - -A little bit of context: - -The Chabauty topology is a generalization of the Hausdorff topology to non-compact sets. It is characterized by the condition that every point of $\lambda'$ is the limit of a sequence of points $x_i \in T_i$, and conversely every convergent sequence $x_{i_n} \in T_{i_n}$ limits to a point of $\lambda'$. See, e.g. Notes on Notes of Thurston. -I am mainly interested in the answer in the setting where $\lambda$ is the pleating lamination on the boundary of the convex core of a quasifuchsian $3$-manifold. This places some additional hypotheses on $\lambda$: for example, it would have to be compact. But the question seems to be intrinsically $2$--dimensional, and it's not clear to me how to use compactness of $\lambda$ as a hypothesis. -If the ideal triangulations $T_i$ are replaced by simple closed curves, the result is well-known. So one approach would be to take a sequence of closed curves $C_i$, limiting to $\lambda' \supset \lambda$, approach each $C_i$ by triangulations (twisting more and more), and then take a diagonal sequence of triangulations. But it's not clear that this diagonal sequence even converges. - -Anyway, either a reference or a way to argue would be much appreciated! - -REPLY [5 votes]: It turns out that the answer to the question is "yes". Saul Schleimer and I needed this result for a paper that we just finished writing, so we ended up sorting it out. The full argument is written down in Lemma A.6 in the Appendix of this paper, so what follows below is an outline. -Take a sequence of closed curves $C_i$, which limits to $\lambda$ in the measure topology. This sequence has a subsequence (which I will still call $C_i$) limiting to $\lambda' \supset \lambda$ in the Chabauty topology. Now, each $C_i$ is contained in the Chabauty limit of a sequence of triangulations $T_{i,j}$. This means that one can take a representative triangulation $T_{i,j(i)}$ that is very close to $C_i$, where "very close" can be quantified (say, closer than distance $1/i$) because the Chabauty topology is metrizable. Now, the sequence $T_{i,j(i)}$ will converge to a lamination $\lambda'' \supset \lambda' \supset \lambda$.<|endoftext|> -TITLE: Grothendieck connections and jets -QUESTION [9 upvotes]: The following question is based on some remarks in section I.2 of Deligne's book Equations Différentielles à Points Singuliers Réguliers. -Let $X$ be a smooth complex variety and $X_1$ the first infinitesimal neighborhood of the diagonal in $X \times X$, so there is a natural morphism $X \to X_1$. If we write $p_1,p_2 : X_1 \to X$ for the two projections, then the first-order jet bundle of a vector bundle $V$ on $X$ is defined to be $J^1(V) = p_{1*} p_2^*V$. Here "upper star" is used in the sense of $\mathcal{O}$-modules. This allows for a convenient way of expressing the notion of a connection: this is just an isomorphism $p_1^*V \to p_2^*V$ which restricts to the identity over $X$, which is the same as an $\mathcal{O}$-linear map $V \to J^1(V)$ such that the composition $V \to J^1(V) \to V$ is the identity. -Here Deligne says something I don't understand: he refers to a first-order differential operator $j^1 : V \to J^1(V)$ "which associates with any section a first-order jet" (I'm translating from the French). What is he talking about, and what makes the map he's talking about a differential operator? I don't have much intuition for jets, although I know they have something to do with taking Taylor expansions, so any general explanation of that would be greatly appreciated as well. - -REPLY [12 votes]: Here's one way to say what's going on. -For a vector bundle $V$ on $X$ and a point $x\in X$, two local sections of $V$ are said to have the same $1$-jet if their Taylor series have the same constant and linear terms. To make sense of "Taylor series" you might need a coordinate chart in $X$ and a local basis for $V$, but this equivalence relation is independent of those choices. The equivalence classes form a vector space. Call it $J^1_x(V)$ and write $j^1_x(v)$ for the class of the section $v$. Varying $x$ now, this yields a vector bundle $J^1(V)$ on $X$. By remembering only the constant term of the Taylor series we get a surjective map of vector bundles $J^1(V)\to V$. The kernel is canonically isomorphic to $T^\star X\otimes V$. -A connection may be defined as a splitting of this, in other words a vector bundle map $V\to J^1(V)$ such that the composed map $V\to V$ is the identity. In the case of the trivial line bundle, where section means function, there is such a splitting, defined using $1$-jets of constant functions. In the general case there is no canonical splitting. A connection may be informally thought of as a choice of which ($1$-jets of) sections of $V$ are being considered locally constant. (Locally connections exist but are not canonical. Globally they need not exist at all in algebraic geometry, although they do in smooth topology.) -There is in any case a canonical map $j^1:\Gamma(V)\to \Gamma(J^1(V))$ of section spaces (or sheaves) that splits that surjection. It takes $v$ to $x\mapsto j^1_x(v)$. But it is not $\mathcal O$-linear, i.e. it is not an order zero operator. The quantity $j^1(fv)-fj^1(v)$, which is a section of that kernel, may be written as $df\otimes v$. -A connection yields an $\mathcal O$-linear map $\Gamma(V)\to \Gamma(J^1(V))$ that splits the surjection. Subtracting it from $j^1$ we get a map $\nabla:\Gamma(V)\to \Gamma(T^\star X\otimes V)$, a first order operator satisfying $\nabla (fv)=f\nabla (v)+df\otimes v$, namely the covariant derivative associated with the connection. Of course the connection can be recovered from $\nabla$.<|endoftext|> -TITLE: "Modular forms from Feynman integrals "? -QUESTION [13 upvotes]: I would like to learn more about the background of this talk, but found no text on that theme. Do you know more? Edit: An interesting talk by Miranda Cheng (slides). -Edit: A talk today on the theme, has anyone a text or slides?: http://www.mpim-bonn.mpg.de/de/node/4590 - -REPLY [18 votes]: I can say a little about the work of Brown and Schretz, since Brown gave a talk at BIRS last month. -If you take a graph with $N$ edges and some restriction on valence (called a Feynman graph), there is a certain integral on an $N-1$-dimensional domain with boundary that in small cases appeared to yield linear combinations of multiple zeta values. This was a discovery of Broadhurst and Kreimer, by numerical brute force. -This raised the question of which graphs yield combinations of multiple zeta values. This is interesting in part because when multiple zeta values show up in an integral, it is a sign that mixed Tate motives are lurking somewhere. In fact, Brown showed last year that $MT(\mathbb{Z}) = \pi_1^{mot}(\mathbb{P}^1 \setminus \{0,1,\infty \})$, i.e., every mixed Tate motive over $\mathbb{Z}$ has periods in $\mathbb{Q}[(2 i \pi)^{-1}, MZV]$. The connection between graphs and motives comes through work of Bloch, Esnault and Kreimer, where they defined motives of graphs $G$ by making a variety $X_G$ by some blow-up construction and taking $H^{N-1}$. You can ask if these motives are mixed Tate, but this is apparently way too hard. An easier question is computing cohomology, and easier still is counting points over finite fields. -Kontsevich conjectured that the function that takes a prime power $q$ and returns the number of $\mathbb{F}_q$-points in $X_G$ is a polynomial, and it was true for small $G$, but this recent work gives a construction of counterexamples. Some of the functions are (or appear to be) $q$-expansions of modular forms of small level and weight, but that seems to be a transition zone between polynomials and wilder functions that no one has been able to identify. - -REPLY [5 votes]: You may have a look here : http://people.math.jussieu.fr/~brown/K3inphi4.pdf<|endoftext|> -TITLE: Image of the Coproduct of a Hopf Algebra -QUESTION [5 upvotes]: Let $H$ ba a Hopf algebra with coaction $\Delta: H \to H \otimes H$. Denote the action of $\Delta$ by $\Delta (h) = h_{(1)} \otimes h_{(2)}$. I was wondering if every element of $H$ can arise as a $h_{(2)}$, for some $h$. To be more precise, for any $g \in H$, does there exist a $h \in H$, such that $\Delta(h) = f \otimes g + \sum_i h_i \otimes h'_i$, for some set $h'_i$ which is linearily indpt with $g$? Or, alternatively, does there exist $H' \subseteq H$, such that $\Delta(H) \subseteq H \otimes H'$? - -REPLY [8 votes]: Counitarity implies that $$(\varepsilon\otimes\mathrm{id}_H\circ\Delta)(H)=H.$$ If $\Delta(H)\subseteq H\otimes H'$ for some subspace $H'\subseteq H$, then this tells us ---since $\varepsilon$ takes scalar values!--- that $H\subseteq H'$.<|endoftext|> -TITLE: Posets of cosets and contractibility -QUESTION [7 upvotes]: For this question let $G$ be a group, perhaps infinite, and let $H_i$ for $i\in I$ be a (finite) family of subgroups closed under taking intersections. I am interested in the coset poset $\mathcal{C}(G,\{H_i\})$ which is defined as the set of cosets $g H_i$ with ordering by inclusion. Note that $g H_i\subseteq g'H_j$ implies that $H_i\subseteq H_j$ and $g^{-1} g'\in H_j$. -The family of subgroups $H_i$ defines a diagram of groups with maps given by inclusion inside $G$. Taking classifying spaces $B(H_i)$ gives a diagram of simplicial sets and I am interested in the colimit of this, denote it $B(G,H_i)$. -Question 1 Where is there a reference for: - -The realisation $N(\mathcal{C}(G,H_i))$ is contractible if and only if $B(G,H_i)$ is a classifying space for $G$. - -I've got a sketch proof which takes the coset poset and applies the Borel construction using the category $\mathcal{E}G$ with object set $G$ and singleton homsets: -$\mathcal{C}(G,H_i)\times\mathcal{E}(G)/G$ -This quotient category can be seen to be equivalent to a category with objects the family $H_i$ and homsets $Hom(H_i,H_j)\cong H_j$. The nerve of this is then seen to be both equivalent to the colimit $B(G,H_i)$ and the Borel construction applied to $N(\mathcal{C}(G,H_i))$. -Question 2 What are some nice examples? I can think of right-angled Artin groups, with subgroups indexed by the simplices of the flag complex; I guess that the coset complex is CAT(0). Also if one of the subgroups is $G$ itself then the result holds. -Of course a valid answer to this whole question could be that I've got everything wrong. - -REPLY [5 votes]: Here are some references that should be of use. -(i) H. Abels and S. Holz, Higher generation by subgroups , J. Alg, 160, (1993), 311– 341. -(ii) S. Holz, 1985, Endliche Identifizier zwishen Relationen , Ph.D. thesis, Univerist\"{a}t Bielefeld. -(iii) A. Bak, R. Brown, G. Minian and T.Porter, Global Actions, Groupoid Atlases and Applications, Journal of Homotopy and Related Structures, 1(1), 2006, pp.101 - 167. -Abels and Holz consisder a slightly different construction. They take the nerve of the covering given by all the cosets but discuss some of the other equivalent notions including yours. Several of the things you find, they prove in that paper (and some were in Holz's thesis.) They discuss some of the Artin group examples and give references to Soulé and others (I won't list their references here.) The paper on global actions includes some pretty (I think!) examples of the interpretation of these nerves in very elementary examples. (e.g. $S_4$, with its usual presentation by transpositions, with subgroups generated by pairs of those. The nerve, if I remember rightly, has realisation $S^2$, the 2-sphere.) -Abels and Holz's paper is very neat indeed and deserves to be more widely known. Hope this helps.<|endoftext|> -TITLE: Extremal Fano with non constant scalar curvature vs Kaehler-Einstein Fano manifolds -QUESTION [6 upvotes]: The first Hirzebruch surface $F_1$ does not admit any Kaehler metric of constant scalar curvature. Yet it admits an extremal metric representing each Kaehler class as shown by Calabi. The two points blow-up of $\mathbb{CP}^2$ behaves similarly. -Do you know any other example of Fano manifold such that: - - The anti-canonical class does not contain any Kaehler-Einstein metric. - The anti-canonical class contains an extremal metric. - -REPLY [2 votes]: In fact, I am asking the question because I just proved that there are certain deformations of the Mukai-Umemura $3$-fold that belong to Tian's family and admit extremal metrics in the anti-canonical Kaehler classes. However they do not admit any Kaehler-Einstein metric according to the celebrated result of Tian. The remarkable thing is that these examples are arbitrarily small complex deformations of the Mukai-Umemura $3$-fold which does admit a Kaehler-Einstein metric by Donaldson's computation of the $\alpha$-invariant. -I would like to know how much this example of phenomenon is original, and if there are others of this type.<|endoftext|> -TITLE: Base Change for Eigenvarieties -QUESTION [5 upvotes]: Let $E/F$ be a Galois extension of number fields, and $G$ a reductive group over $F$. If Langlands Base Change is known for $G/F$ and $G/E$, and moreover the eigenvarieties for $G/F$ and $G/E$ have been constructed, is there a rigid map between the eigenvarieties which interpolates the base change? -Assuming the answer to the previous question is yes, will the map be a closed immersion? What is known about the image with respect to the subvariety of Gal($E/F$)-invariants? -Finally, are there any instances where we have such a map between eigenvarieties but base change is not known? -Sorry for the barrage of questions; I would be happy with even a partial answer to the first. - -REPLY [4 votes]: This is all a bit complicated -- the theory is still in its infancy and some arguments aren't quite as smooth as they should be. -If all you know is that "the eigenvarieties have been constructed" then you're in a hopeless situation -- in fact in some sense I don't even know what this statement means. You need to know some facts about the eigenvarieties before you can prove anything. Let's say for example that you're in the happy situation where classical points are dense (for example, $G$ compact mod centre, or perhaps you've defined an eigenvariety to have that property by taking the closure of the classical points in an a priori bigger object). Then you might be able to get somewhere (see next para). However there are examples of Hida families over non-totally real bases where there appear to be natural $p$-adic objects where classical points do not seem to be dense, so then things might be really tough. -If classical points are dense, then there is still another big problem: as far as I know Langlands Base Change (and functoriality in general) only transfers packets to packets. However, eigenvarieties don't parametrise packets, they parametrise systems of Hecke eigenvalues. If you are dealing with inner forms of $GL(2)$ then packets have size 1 so Chenevier doesn't see this problem and he can prove the best possible theorem (this is one of the papers you cite). If however there are packets then one runs into combinatorial issues -- Flicker finds himself with such problems in the other paper you cite. So, even if classical points are dense, things might not be so easy. -I have a student, Judith Ludwig, who is making some headway with issues of this nature. The questions can be quite delicate.<|endoftext|> -TITLE: Minimize trace of inverse of convex combination of matrices. -QUESTION [5 upvotes]: Hello! (First question--please forgive me if its unclear.) -I am interested in efficient/approximate optimization techniques for minimizing a norm of a convex combination of symmetric, positive semi-definite matrices. In particular I am wondering what techniques exist to efficiently solve the following problem: -$\alpha^{*}=\arg\min\big\{\operatorname{trace}((\sum_i\alpha_i \mathcal{I}_i)^{-1}):\sum_i\alpha_i=1, \alpha_i\ge0 \big\}$ -where $\mathcal{I}_i$ are positive semi-definite matrices and $\operatorname{trace}$ is the sum of the eigenvalues. Edit: Furthermore, assume $\mathcal{I}_i\succ0$ for at least one $i$. -Approximations are acceptable--I am interested in any "standard" approaches, should they exist. Should some other scalar function be more convenient this would be interesting to me too (ie determinant, nuclear norm, spectral radius, etc). The matrices are often sparse and could be extremely large. -The simplest idea I had was to lower bound the sum of matrices using Weyl's theorem, ie, $\lambda_i(A+B)\ge \lambda_j(A)+\lambda_{n+i-j}(B)$ for $j\ge i$ giving me an upper bound to the trace of the inverse of the sum. I am wondering if another bound would be tighter or better still some clever variational approach exists. -The motivation (shouldn't be relevant to the answer) for this problem comes from statistical machine learning. I would like to develop techniques for characterizing maximum composite likelihood estimators (a generalization of the maximum likelihood estimator). Classical statistical analysis yields a convex combination of the expected Hessian matrices and is as above. (The $\mathcal{I}$ denotes the Fisher information matrix.) -There is some similarity to the Eigenvalues of Matrix Sums question although I believe this question is sufficiently different to ask anew. -Thanks, -Josh - -REPLY [3 votes]: Since the functional is convex, I would try modified gradient descent. The main problem with this approach is that you'll really have to compute the inverse matrix a few times, which may be out of question due to the "extremely large size" condition. If so, just disregard this answer. -The algorithm is fairly simple: if you have $m$ matrices $A_j$ start with $a_j=1/m$ for all $j$. Compute $A(a)=\sum_j a_j A_j$. Compute $b_j=\operatorname{trace} A_jA^{-2}$. Now trace the path in the simplex $a_j\ge 0,\sum_j a_j=1$ that goes from $a$ in the direction of $b$ until it reaches some vertex (so you start moving in the direction of the projection of $b$ to the plane $\sum_j x_j=0$ until you reach a face, then move along the face in the direction that is the projection of $b$ to that face and so on). Let $a(t)$ be the point on that path at distance $t$ (along the path) and $T$ be the path length (you can use the $\ell^2$ or the $\ell^1$ distance, it doesn't seem to matter). Now start computing $\operatorname{trace}A(a(2^{-k}T))^{-1}$ ($k\ge 0$) until $2^{-k}T<\varepsilon$. If the minimal one you attained is less than $\operatorname{trace}A(a)^{-1}$, move $a$ to the corresponding point and repeat. -To keep track of how close you are to the optimal value, just compare $\max_j b_j$ and $\operatorname{trace}A(a)^{-1}$. Their difference dominates the error. -I tried it with thirty 100 by 100 random matrices and it converged pretty fast but, as I said, if you are in no position to attempt a single matrix inversion or multiplication, this is pretty useless.<|endoftext|> -TITLE: A heuristic for the density of solutions to Diophantine equations -QUESTION [13 upvotes]: Let $f\in\mathbb{Z}[X_1,\ldots,X_n]$ be a Diophantine equation which, for the purposes of this question, I will assume is homogeneous and nonsingular on $\mathbb{R}^n\setminus\{0\}$ (so that $\nabla f\not=0$). Supposing that it has infinitely many primitive integer zeros, we can posit that they are smoothly distributed in an asymptotic sense. Writing $V(R)\subseteq R^n$ for the set of primitive solutions to $f(x)=0$ in a ring $R$, the integer solutions $V(\mathbb{Z})$ clearly lie on the manifold $V(\mathbb{R})$. So, I am looking for a density $\rho\colon V(\mathbb{R})\to\mathbb{R}$ with -$$ -\vert V(\mathbb{Z})\cap U\vert\sim\int_{V(\mathbb{R})\cap U}\rho(x)\,d\sigma(x),\qquad\qquad{\rm(1)} -$$ -for subsets $U\subseteq\mathbb{R}^n$, where $d\sigma$ is the standard surface integral on $V(\mathbb{R})$. This should hold asymptotically as $U$ is scaled up, and for reasonably regular regions $U$. -My question is regarding a simple (but incorrect -- see below) heuristic argument for calculating $\rho$. Choosing positive integer $N$ and real $a\gg N$ then, for large regions $U$, the set of $x\in U$ with $\vert f(x)\vert < 2a$ has volume about $2a\int_{V(\mathbb{R})\cap U}\Vert\nabla f\Vert^{-1}\,d\sigma$, so should contain about that number of integer points. The probability of a random $x\in\mathbb{Z}^n$ being relatively prime to $N$ and satisfying $f(x)=0$ (mod $N$) is $N^{-n}\vert V(\mathbb{Z}/N\mathbb{Z})\vert$. Conditional on $\vert f(x)\vert < 2a$ and $f(x)=0$ (mod $N$), it seems reasonable to suppose that $f(x)=0$ with probability $N/(2a)$. Multiplying these terms together and taking the limit as $N$ increases to include all prime-powers as factors, we get the following expression for $\rho$. -$$ -\begin{align} -&\rho(x)=\Vert\nabla f(x)\Vert^{-1}\prod_p c_p,\qquad\qquad{\rm(2)}\\ -&c_p=\lim_{r\to\infty}p^{-r(n-1)}\left\lvert V(\mathbb{Z}/p^r\mathbb{Z})\right\rvert. -\end{align} -$$ -The product is taken over all primes $p$. This seems like a very neat expression, and can be seen that it gives the correct result for linear equations. However, it is not correct in general. Just looking at quadratic forms for $f$, the expression given by (2) is wrong. I do not have any good feeling as to where exactly this heuristic goes astray, and if it is possible to fix it. Maybe this approach and the reason that it does not quite work is well known. This is not an area in which I am any kind of expert, so maybe others on MathOverflow would be able to help? -For example, consider $f=x^2+y^2-z^2$, so that we are looking for primitive Pythagorean triples. Euclid's parameterization $(x,y,z)=(a^2-b^2,2ab,a^2+b^2)$ can be used to show that $\rho=\sqrt{2}\pi^{-2}\vert z\vert^{-1}$. However, on $V(\mathbb{R})$ we have $\Vert\nabla f\Vert = 2\sqrt{2}\vert z\vert$ and you can calculate $c_2=1$ and $c_p=1-p^{-2}$ for odd prime $p$. Using (2) would lead to $\rho=2\sqrt{2}\pi^{-2}\vert z\vert^{-1}$, which is out by exactly a factor of 2. If we look at Pythagorean quadruples $f=w^2+x^2+y^2-z^2$ instead, then we can calculate $c_p=(1-p^{-1})(1+2p^{-1}1_{\{p\equiv1{\rm\ mod\ }4\}}+p^{-2})$ for odd primes $p$, so the product in (2) is not unconditionally convergent. - -Is there a known or, even, just conjectural expression for the asymptotic density $\rho$? And, is it possible to explain precisely how the heuristic used to derive (2) fails? - -It would be great if my expression (2) above could be fixed. Heuristics like the one used here are often very useful to understand what the integer solutions to Diophantine equations look like, and it is a bit worrying that it gives the wrong answer in this case. It is also consistent with the idea that find rational solutions to an equation, you should first check for solutions in the completions of $\mathbb{Q}$, according to the Hasse principle. Also, it so nearly works (only being a factor of 2 out for Pythagorean triples) and gives perfectly sensible looking results in many cases, that I am loath to give up and just accept that it doesn't work without a good reason as to why. For example, it does seem perfectly consistent with Falting's theorem (as given in my answer to a previous MO question) and with the Birch and Swinnerton-Dyer conjecture. In the case where $f$ is a a cubic describing an elliptic curve, then $c_p=(1-p^{-1})N_p/p$ for all but finitely many primes $p$, where $N_p$ is the number of $\mathbb{F}_p$-points on the elliptic curve reduced modulo $p$. Then, up to finitely many terms, $\prod_pc_p$ coincides with the Euler product at $s=1$ of $(L(s)\zeta(s))^{-1}$, where $L$ is the L-function of the curve. According to the Birch and Swinnerton-Dyer conjecture, I would expect this to be zero, finite, or infinite when the curve has rank $r=0$, $r=1$ and $r>1$ respectively. Putting this back into (2) is consistent with $\vert V(\mathbb{Z})\cap B_R\vert$ growing at rate $(\log R)^r$, which you would expect for an elliptic curve of rank $r$. - -REPLY [4 votes]: Searching including the key phrases "Hardy-Littlewood circle method" and "singular series", as suggested by the other answers, turned up some interesting references which shed light on the question and why I obtained the results I did for the cases mentioned. As the question is already quite long and would become rather unmanageable to add this as a large update, I'm adding it as an answer here. -The product $\prod_pc_p$ is indeed called the singular series and $\int\Vert\nabla f\Vert^{-1}\,d\sigma$ is the singular integral. The product does asymptotically give the number of solutions subject to hypotheses and/or conditions on $f$, generally seeming to work better when the number of variables is large relative to the degree. For quadratic forms, the paper A new form of the circle method, and its application to quadratic forms by D.R. Heath-Brown (Journal für die reine und angewandte Mathematik, 1996. Preprint available here) gives expressions for the asymptotic density which show why I obtained the results I did for Pythagorean triples and quadruples mentioned in the question. For a quadratic form $f$ in $n$ variables, they show the following. - -For $n\ge5$, expression (2) for the asymptotic density given in the question is correct, so the heuristic works! (Theorem 5 of the D.R. Heath-Brown paper). -For $n=4$ and the determinant of $f$ not a perfect square, you should multiply the terms $c_p$ by $1-\chi(p)p^{-1}$ and the overall expression by $L(1,\chi)$ (Theorem 6 of the paper). Here, the character $\chi$ is the Jacobi symbol $\left({\rm det}(f)\over\ast\right)$. Leaving out these terms will give a product which is not unconditionally convergent, as happened for Pythagorean quadruples. -For $n=4$ and the determinant of $f$ a perfect square, the product $\prod_pc_p$ will diverge to infinity (assuming that there are solutions in every $p$-adic field, so $c_p\not=0$). Instead, $c_p$ should be multiplied by $1-p^{-1}$ and the overall density by $\log\Vert x\Vert$ (Theorem 7 of the paper). -For $n=3$ then expression (2) given in the question works after multiplying by a factor of $\frac12$ (Theorem 8 and Corollary 2 of the paper)! This is why I was out by a factor of 2 for Pythagorean triples. The D.R. Heath-Brown paper has the following to say on this. - - -...It therefore remains to understand the appearance of the factor $\frac12$ in the case $n=3$, which can be thought of as corresponding to a Tamagawa number of 2. In the proof of Theorem 8 this factor arises from the residue at s = 0 of -$$\zeta(2s+1)\frac{P^s}{s}.$$ - -I'm not very familiar with Tamagawa numbers and am not yet sure whether this is the same as the factor $\alpha=\frac12$ mentioned in Daniel's answer or how it comes into the heuristic derivation.<|endoftext|> -TITLE: Question on an example about flatness in Hartshorne -QUESTION [5 upvotes]: I have been having trouble understanding some statements regarding flatness in Hartshorne - in particular relating to some of the examples in the text. Any help would be appreciated! -Here is the issue: - -In example III.9.8.4 Hartshorne discusses an example of a family of twisted cubics arising from the projection of $\mathbb{P}^3$ to $\mathbb{P}^2$ from a point. The result is that the flat limit of the twisted cubic is not only singular, but it has an embedded point at the singular point. -In example III.9.10.1 he explains why if one takes the reduced induced structure on the fibers, then the family is not flat. - -My question is: If one takes the flat family $Y\to\mathbb{A}^1$ from the first bullet, and uses the canonical map $Y_{red}\to Y$, then by composing one gets a family $Y_{red}\to \mathbb{A}^1$, which should be flat by proposition III.9.7 (which states that you have flatness over a smooth curve if every associated point of $Y$ maps to the generic point of the curve). Now, I expected $Y_{red}$ to be the family in the second bullet, but since it is not flat, this cannot be the case. What is going on? -Thanks! - -REPLY [3 votes]: Sandor is right. -Indeed, in Hartshorne's example the total family $Y$ is defined by the ideal -$$I=(a^2(x+1)-z^2, ax(x+1)-yz, xz-ay, y^2-x^2(x+1)) \subset k[a, x,y,z],$$ -whereas the central fibre (corresponding to $a=0$) is defined by -$$I_0 = ( z^2, yz, xz, y^2-x^2(x+1)) \subset k[x,y,z].$$ -The following Macauley2 script shows that - -$I$ is a radical ideal, hence $Y$ is a reduced affine scheme, that is $Y=Y_{red}$. -$I_0$ is not radical. In fact, the central fiber $Y_0$ has an embedded point at the node $(0,0,0)$ corresponding to the nilpotent element $z$. -i1 : k=ZZ/32003; -i2 : S=k[a,x,y,z]; -i3 : I=ideal (a^2*(x+1)-z^2, a*x*(x+1)-y*z, x*z-a*y, y^2-x^2*(x+1)); -o3 : Ideal of S -i4 : I==radical I -o4 = true -i5 : T=k[x,y,z]; -i6 : I0=ideal (z^2, y*z, x*z, y^2-x^2*(x+1)); -o6 : Ideal of T -i7 : I0 == radical I0 -o7 = false<|endoftext|> -TITLE: What do coherent topoi have to do with completeness? -QUESTION [27 upvotes]: There is a theorem of Deligne in SGA4 that a "coherent" topos (e.g. one on a site where all objects are quasi-compact and quasi-separated) has enough points (i.e. isomorphisms can be detected via geometric morphisms to the topos of sets). I've heard it said that this is a form of Goedel's completeness theorem for first-order logic. -Why is that? I get the intuition that having a model for a formula is supposed to be analogous to a point of a suitable topos, but this is very vague. -I'm sorry for not providing more motivation, but I don't know enough about this connection to do so! -(This was first posted on math.SE here, where it did not (yet) receive a response.) - -REPLY [8 votes]: A good explanation of this topic can now be found here: - -Benjamin Frot, Gödel's Completeness Theorem and Deligne's Theorem, arXiv:1309.0389,<|endoftext|> -TITLE: Motivating the category of chain complexes -QUESTION [48 upvotes]: Let $R$ be a commutative ring. For awhile I have been trying to motivate to myself more fully the definition of and various structures on the category $\text{Ch}(R)$ of chain complexes of $R$-modules (and various subcategories thereof). One significant piece of motivation is the Dold-Kan correspondence, at least when $R = \mathbb{Z}$, which tells us that studying connective chain complexes is like studying linearized homotopy theory (or linearized higher category theory). This is a great idea, but I don't have much intuition for what's going on in the proof of Dold-Kan, and I don't see how one could have predicted in advance that something like Dold-Kan might be true just by looking at all the definitions in the right way. I like the idea of linearized homotopy but I don't know what the conceptual path is from linearized homotopy to, for example, the braiding $a \otimes b \mapsto (-1)^{|a| |b|} b \otimes a$. -Consider also the differential. I can think of various ways to motivate $d^2 = 0$, and I don't quite know how they fit together. For example, one can talk about boundaries of manifolds with boundary, the exterior derivative, and Stokes' theorem. If one starts from the simplicial / higher-categorical perspective, the differential encodes something like the generalized source / target of a higher morphism, and somehow the fact that this generalized source / target ought to satisfy a natural "gluing law" (for example if $a \to b$ is a $1$-morphism then $d(a \to b + b \to c)$ ought to equal $d(a \to c)$) is equivalent to it squaring to zero. I can sort of picture how this works in low dimensions but I don't completely grasp what the exact relationship between these two ideas is. -Keeping in mind the symmetric monoidal structure, the differential behaves like an element of a super Lie algebra, concentrated in degree $-1$, acting on a representation (see for example Theo Johnson-Freyd's MO answer here). The action of a super Lie algebra should be related to infinitesimal symmetry coming from a super Lie group, but I don't have a clear idea of what this super Lie group is or what it has to do with homotopy theory. This seems to have something to do with the supergeometric definition of differential forms, but I don't really know anything about this. -Algebraically, the relation $d^2 = 0$ seems to come from at least two different ideas: first-order approximation, and odd things anti-commuting with themselves. Both of these ideas seem relevant to what I'm confused about, but I can't put them together into a cohesive story. - -So what is that cohesive story? - -Edit: If the bulk of the question seems sort of silly to you, feel free to focus on that last bit about super Lie algebras. I remember hearing that this has something to do with the action of the automorphism group of the odd real line; I would appreciate if someone could clarify that for me. - -REPLY [5 votes]: I think the picture is clearer if you think globes rather than simplices. -The starting neat observation is that a homomorphism $d : A_1 \to A_0$ is the same thing as an internal category; the $A_0$ gives the objects and $A_0 \times A_1$ gives the arrows; $(x,f)$ is construed as an arrow from $x$ to $x + df$, and composition is given by $(x+df, g) \circ (x,f) = (x, f+g)$. -Conversely, every internal category is isomorphic to one constructed in this way; addition gives us a unique way to translate any arrow to be one whose source is zero, and composition is given by addition by using linearity of composable pairs: -$$ ((x+df, g), (x,f)) = ((x+df, 0), (x,f)) + ((0,g), (0,0)) $$ -$$ (x+df, g) \circ (x,f) = (x+df, 0) \circ (x,f) + (0,g) \circ (0,0) -= (x,f) + (0,g) = (x, f+g) $$ -It's easy to imagine this extended to higher degree; we can translate a $2$-morphism so its source is the identity morphism of zero. Letting $A_2$ be the group of such things, and write $2$-morphisms as elements of $A_0 \times A_1 \times A_2$. All of the compositions should (?) again be addition, and if we have a $2$-morphism -$$ (0,0,n) : (0,0) \to (0,f) $$ -then we set $dn = f$ to give the target of the $2$-morphism, and we require $df = 0$ so that the two arrows are parallel. This gives $d^2 = 0$, and it's clear how this continues into higher dimension. -This formulation makes it, in my opinion, nearly obvious (at least on the level of algebra) that the category of globular objects should be equivalent to the category of connective chain complexes. -If we're imagining $n$-morphisms as being $n$-balls, then we can even see $\ker d$ as being those balls whose boundary is identified with zero — i.e. as spheres meeting the basepoint.<|endoftext|> -TITLE: Orthogonal Complements of Root Lattices in E_8 -QUESTION [6 upvotes]: I have a rather stupid lattice theory question. Suppose $L$ is a root lattice that can be primitively embedded in the $ E_8 $ lattice. Is the orthogonal complement of $ L$ in $E_8$ unique up to isomorphism, or for different primitive embeddings could I get non-isomorphic complements? - -REPLY [7 votes]: You can get different orthogonal complements for different embeddings. There are two different embeddings of $A_{7}$ in $E_{8}$ so that for the first embedding the orthogonal complement is the lattice $A_{1}$, and for the second embedding the orthogonal complement is the lattice $\langle 8 \rangle$.<|endoftext|> -TITLE: Does the action of an affine group scheme preserve the nilradical of an algebra? -QUESTION [7 upvotes]: Let $k$ be a commutative ring and let $G$ be a flat affine algebraic group scheme over $k$. -Let $G$ act by algebra automorphisms on the commutative $k$-algebra $A$. So $G(R)$ acts by $R$-algebra -automorphisms on $A\otimes_k R$ for any commutative $k$-algebra $R$. -Let $N$ be the nilradical of $A$. Is $N$ always a $G$ submodule? So is the image of $N\otimes_k R$ in $A\otimes_k R$ -invariant under $G(R)$? I only need it when $G$ is a Chevalley group scheme, in which case it is true. -But my question is if this is a general fact. - -REPLY [14 votes]: This is true if you assume that $G$ is smooth. Consider the coaction $A \to A \otimes_k k[G]$; since $k[G]$ is a smooth $k$-algebra, the nilradical of $A \otimes_k k[G]$ is $N \otimes_k k[G]$; since $N$ is sent to the nilradical of $A \otimes_k k[G]$, this implies the thesis. -Otherwise it is false in general, even over algebraically closed fields. For example, take $G = \alpha_p = \mathop{\rm Spec}k[x]/(x^p)$ in characteristic $p$, and consider the action of $G$ over itself by translation. This corresponds to the coaction $k[x]/(x^p) \to \bigl(k[x]/(x^p)\bigr) \otimes_k \bigl(k[x]/(x^p)\bigr)$ sending $x$ into $x\otimes 1 + 1 \otimes x$. The nilradical of $k[x]/(x^p)$ is $(x)$; but $x\otimes 1 + 1 \otimes x$ is not in $(x) \otimes \bigl(k[x]/(x^p)\bigr)$.<|endoftext|> -TITLE: Are almost commuting hermitian matrices close to commuting matrices (in the 2-norm)? -QUESTION [19 upvotes]: I consider on $M_n(\mathbb C)$ the normalized $2$-norm, i.e. the norm given by $\|A\|_2 = \sqrt{\mathrm{Tr}(A^* A)/n}$. -My question is whether a $k$-uple of hermitian matrices that are almost commuting (with respect to the $2$-norm) is close to a $k$-uple of commuting matrices (again with respect to the $2$-norm). More precisely, for an integer $k$, is the following statement true? - -For every $\varepsilon>0$, there exists $\delta>0$ such that for any $n$ and any matrices $A_1,\dots, A_k\in M_n(\mathbb C)$ satisfying $0\leq A_i\leq 1$ and $\|A_iA_j - A_j A_i\|_2 \leq \delta$, there are commuting matrices $\tilde A_1,\dots,\tilde A_k$ satisfying $0\leq \tilde A_i\leq 1$ and such that $\|A_i - \tilde A_i\|_2 \leq \varepsilon$. - -The important point is that $\delta$ does not depend on $n$. -I could not find a reference to this problem in the litterature. However, this question with the $2$-norm replaced by the operator norm is well-studied. And the answer is known to be true if $k=2$ (a result due to Lin) and false for $k=3$, and hence $k\geq 3$ (a result of Voiculescu). - -REPLY [3 votes]: As part of my dissertation, "Almost Commuting Operators on von Neumann Algebras," -I have extended Glebsky's result to the normalized Schatten class for $1 \leq p < \infty$. Moreover, for $p=2$ we recover Filonov and Kachkovskiy's theorem with the same estimate. In our work, however, we use different techniques as the normalized Schatten p-norm does not arise from an inner product for $p \neq 2$.<|endoftext|> -TITLE: Laplace-Beltrami Operator on Surfaces -QUESTION [11 upvotes]: I would like to know what is known about the spectrum of the Laplace-Beltrami operator on 2-dimensional negatively curved surfaces of constant curvature. -For instance, - -What is the spectrum of the Hyperbolic plane of constant curvature $-k$? -What is the Laplace-Beltrami operator and its spectrum for a compact surface of constant curvature $-1$ and genus $g\geq 2$? -If the surface is compact then the spectrum of the L-B operator is discrete and their eigenvalues are -$$ -0=\lambda_1<\lambda_2<\lambda_3<\ldots<\lambda_n<\ldots -$$ -In the case of constant curvature $-1$, is there a lower bound for its second eigenvalue? upper bound? what else is known in this case? - -Can anyone point me to the right reference? Thanks! - -REPLY [6 votes]: If the surface is closed of genus $g$, there is a universal lower bound on $\lambda_{2g-2}$.<|endoftext|> -TITLE: Clifford algebra non-zero -QUESTION [14 upvotes]: This should be a very easy question, but the proof in Lawson/Michelson (Spin geometry) is wrong and I do not find a really correct and complete argument: -Let V be a nonzero real vector space with scalar product: Why is the Clifford-algebra (constructed from the tensor algebra by quotiening out an ideal) non-zero? - -REPLY [5 votes]: Here is a braindead way to generate PBW theorems like this one: -One is given a presentation of an algebra, which allows one to put words in the generators $x_1,x_2,\dots,x_n$ into a "normal form" (in the case of the Clifford algebra, if $x_1,x_2,\dots,x_n$ is a basis of $V$, then a normal form might be words $x_{i_1} x_{i_2} \cdots x_{i_p}$ with $1 \leq i_1 < i_2 < \cdots < i_p \leq n$). One suspects that the set of words in normal form is actually a basis for the algebra. So one constructs the regular representation of the algebra in question. If the theorem is to be true, there is no choice about this: it is the vector space spanned by words in normal form, and the left (and right) multiplication operators are determined by the relations. Usually the easiest way to write down the formulas is recursively. One is then left to check that these operators satisfy the defining relations. This is "just linear algebra", but the computations in any particular example (or class of examples) may get messy. When it works, it usually works in arbitrary characteristic (and even integrally). -In the case of the Clifford algebra and similar algebras (e.g. enveloping algebras of Lie algebras, symplectic reflection algebras and their generalizations) this all works without too much difficulty. It also works for the Hecke algebra attached to a Coxeter system, though to make the calculations manageable in a case-free fashion it's good to use the Bourbaki trick of employing both the right and left representations simultaneously. There is even a general theorem here, which usually goes by the name "Bergman diamond lemma" (but in the cases I'd care most about, checking that its conditions are satisfied is just about the same level of difficulty as doing the work directly).<|endoftext|> -TITLE: Can we say anything about the Krull dimension of a localization? -QUESTION [12 upvotes]: I'm looking for a theorem of the form - -If $R$ is a nice ring and $v$ is a reasonable element in $R$ then Kr.Dim$(R[\frac{1}{v}])$ must be either Kr.Dim$(R)$ or Kr.Dim$(R)-1$. - -My attempts to do this purely algebraically are not working, so I started looking into methods from algebraic geometry. I thought that Grothendieck's Vanishing Theorem might help (i.e. if dim$(X)=n$ then $H^i(X,\mathcal{F})=0$ for any sheaf of abelian groups $\mathcal{F}$ and any $i>n$) but the problem is that the converse for this theorem fails, so I can't conclude anything about dimension. Perhaps this theorem could give some sort of test for when dimension drops, but I'm hoping for a better answer. -We'll definitely need some hypotheses. For the application I have in mind we can assume $R$ is commutative and is finitely generated over some base ring (e.g. $\mathbb{Z}_{(2)}$), but we should not assume it's an integral domain. If necessary we can assume it's Noetherian and local, but I'd rather avoid this. As for $v$, it's not in the base ring and it has only a few relations with other elements in $R$, none of which are in the base ring. If we can't get the theorem above, perhaps we can figure out something to help me get closer: - -Are there any conditions on $v$ such that the dimension would drop by more than 1 after inverting $v$? - -One thing I know: to have any hope of dimension dropping by $1$ I need to be inverting a maximal irreducible component. I'm curious as to the algebraic condition this puts on $v$. - -REPLY [10 votes]: The dimension of $R[1/v]$ is the biggest height of some prime ideal $P$ such that $v\notin P$. So, let $I_{d-1}$ be the intersection of all primes of height at least $d-1$ ($d= \dim R$), then - -$\dim R[1/v] \geq d-1$ if and only if $v\notin I_{d-1}$. - -Under a mild condition (all maximal ideals has height at least $d-1$), then $I_{d-1}$ is inside the Jacobson radical of $R$. So in this case $v$ is not inside the Jacobson radical would suffice. -EDIT (in response to the comment by the OP): Now let $I_d$ be the intersection of all primes with height $d$. Then $\dim R[1/v] \leq d-1$ iff $v\in I_d$. So - -$\dim R[1/v] =d-1$ iff $v \in I_d- I_{d-1}$. - -Here is a short calculation to show that in Fernando's example above, $I_1=(x)$. Any prime that does not contain $x$ would have to contain $(x-1, y, z)$, but this is a maximal ideal of height $0$. So $x \in I_1$. On the other hand $R/(x)$ is $k[y,z]$ so the intersection of all height one there is $0$. Thus $I_1=(x)$. This gives you precisely what elements satisfies your condition.<|endoftext|> -TITLE: Simplest examples of nonisomorphic complex algebraic varieties with isomorphic analytifications -QUESTION [60 upvotes]: If they are not proper, two complex algebraic varieties can be nonisomorphic yet have isomorphic analytifications. I've heard informal examples (often involving moduli spaces), but am not sure of the references. - -What are the simplest examples of nonisomorphic complex algebraic varieties with isomorphic analytificaitons? - -By "simplest", I mean by one of the following measures. - -(best) An example whose proof is as elementary as possible, and ideally short. This of course requires proof that the complex algebraic varieties are nonisomorphic, and that the analytifications are isomorphic. -A known example that is simple to state, but may have a complicated proof. (Ideally there should be a reference.) -An expected, folklore, or conjectured example. - -REPLY [3 votes]: If you look at local rings, it is easy to construct such examples. For example, if $X,Y$ are smooth of same dimension, then for any point $x\in X,y\in Y$, the completions of $O_{X,x}, O_{Y,y}$ are isomorphic, but of course the algebraic local rings are not necessarily isomorphic (for example, if $X,Y$ are not birational, then even their fraction fields are not isomorphic.)<|endoftext|> -TITLE: Growth zeta-functions of regular languages -QUESTION [6 upvotes]: Dear All, -my following question may be known and ought to be known, so in case it is folklore please could you give me the references. -To start, it is obvious that growth of rational languages are always either polynomial or exponential. That is, if $L$ is a regular language, then the sequence $a_n$, where $a_n$ is the number of all words $w\in L$ of length $\leq n$, grows polynomially or exponentially. -But what I am really curious about is -$\textbf{Question}:$ If $L$ is a regular language, is then the correspondent growth zeta-function $\sum\limits_{n=1}^{\infty}a_nx^n$ rational? - -REPLY [9 votes]: Yes (though the standard term for these is generating functions rather than zeta functions); in fact, there's a relatively straightforward explicit construction for finding the generating function for a regular language given an unambiguous regular expression for it; replace null with $0$, any symbol with $x$, concatenation by multiplication, union with addition, and if $f(x)$ is the function for some expression $E$, then the function for $E^*$ is just $1\over 1-f(x)$. You should be able to convince yourself that this works by examining terms. On the other hand, not every rational function occurs as the generating function of a regular language. For details, you might have a look at http://www.morris.umn.edu/academic/math/Ma4901/Sp2011/Final/BrianGoslinga-final.pdf .<|endoftext|> -TITLE: What the heck is the Continuum Hypothesis doing in Weibel's Homological Algebra? -QUESTION [49 upvotes]: On page 98 of Weibel's An Introduction to Homological Algebra he mentions that the ring $R = \prod_{i=1}^\infty \mathbb{C}$ has global dimension $\geq 2$ with equality iff the continuum hypothesis holds. He doesn't give any clue as to the proof of this fact or why the continuum hypothesis got involved. On page 92 he mentions some examples of Osofsky and says the continuum hypothesis gets involved there because of non-constructible ideals over uncountable rings. I think this explains at least the "why" of the appearance of the continuum hypothesis (though I would welcome more details on this!), but it leaves me with some other questions: - -How is the continuum hypothesis used in this proof? -Why wouldn't the proof work without the continuum hypothesis? - -I will understand if the above have to do with some work of Osofsky that is not widely known. If I can't get answers for those questions, perhaps I can still get help on the below. I got involved with this because I wanted to understand an example of a ring that is von Neumann regular but not semisimple (and an infinite product of fields is such an example). I had hoped all such examples would have weak dimension zero (to be VNR) and right global dimension 1. In particular, I wanted to know that the global dimension of $A = \prod_{i=1}^\infty \mathbb{F}_2$ was $1$. According to this MO answer and its comments, $Spec A$ is the Stone-Cech compactification of $\mathbb{N}$. Now I'm concerned that things from set theory which I try to avoid thinking about will come into play in this example as well as in the above ring $R$. - -What is the global dimension of $\prod_{i=1}^\infty \mathbb{F}_2$? Do we need to assume the continuum hypothesis at any point? What about an uncountable product of $\mathbb{F}_2$? - -REPLY [35 votes]: Since Tilemachos Vassias asked (in a comment to Mariano Suárez-Alvarez's answer) about "the appearance of set theory in seemingly random and completely unexpected places in mathematics," I'd like to give a more general answer than the one given by Juris Steprans, who concentrated on dimension. I've come to expect set theory to appear in any area of mathematics that gets beyond the consideration of countable or "essentially countable" (e.g. separable, in topology) structures. This expectation is not so much based on the foundational role of set theory, mentioned in Fernando Muro's comment, but on its role as the study of combinatorial structures on infinite sets. It is not surprising (to me) that when one analyzes problems or structures in depth, combinatorial issues arise. Indeed, it happens surprisingly often that analysis of a problem reduces it entirely to a combinatorial question, and then one expects to use set-theoretic tools. In some cases, that leads to independence results; in other cases, the tools produce solutions in ZFC. One difference between these two sorts of outcomes is that we know how to prove independence results only when uncountability is involved in an essential way (unless you count applications of Gödel's incompleteness theorems, but they're not really set theory), whereas the use of infinite combinatorics to prove results in ZFC can also arise in essentially countable situations.<|endoftext|> -TITLE: The divisor bound in number fields -QUESTION [23 upvotes]: The divisor bound asserts that for a large (rational) integer $n \in {\bf Z}$, the number of divisors of $n$ is at most $n^{o(1)}$ as $n \to \infty$. It is not difficult to prove this bound using the fundamental theorem of arithmetic and some elementary analysis. -My question regards what happens if ${\bf Z}$ is replaced by the ring of integers in some other number field. For sake of concreteness let us work with the simple extension ${\bf Z}[\alpha]$, where $\alpha$ is some fixed algebraic integer. Of course, one may now have infinitely many units in this ring, but if we restrict the height then it appears that we have a meaningful question, namely: -Question: Let $n \in {\bf Z}[\alpha]$ be of height $O(H)$ (by which I mean that $n$ is a polynomial in $\alpha$ with rational integer coefficients of size $O(H)$ and degree $O(1)$). Is it true that the number of elements of ${\bf Z}[\alpha]$ of height $O(H)$ that divide $n$ is at most $H^{o(1)}$? -Here $o(1)$ denotes a quantity that goes to zero as $H \to \infty$, holding $\alpha$ fixed. (Actually, for my applications I would like $\alpha$ to not be fixed, but to have a minimal polynomial of bounded degree and coefficients of polynomial size in $H$, but for simplicity let me stick to the fixed $\alpha$ question first.) -It is tempting to take norms and apply the divisor bound to the norm, but then I end up needing to bound the number of elements in ${\bf Z}[\alpha]$ with a given norm and of controlled height, and I don't know how to do that except for quadratic extensions. A related problem comes up if one tries to exploit unique factorization of ideals to answer this problem. (On the other hand, it appears to me from the Dirichlet unit theorem that the number of units of height $O(H)$ is at most polylogarithmic in H, so the unit problem at least should go away.) - -REPLY [23 votes]: As long as you allow a fixed number field $F = {\bf Q}(\alpha)$ you can prove $H^{o(1)}$ as you more-or-less suggest towards the end, by first showing that the number of ideals of $F$ that divide $n$ is $H^{o(1)}$ and then proving that any ideal has $O(\log^r H)$ generators of height at most $H$, where $r = r_1 + r_2 - 1$ is the rank of the unit group $U_F$ of $F$. -The first part is basically the same as the argument over $\bf Z$. If the ideal $(n)$ factors into prime powers as $\prod_i \wp_i^{e_i}$ then there are $\prod_i (e_i + 1)$ ideals that divide $n$. Given $\epsilon > 0$ there are finitely many choices of rational prime $p$ and integer $e>0$ such that $e+1 > (p^e)^\epsilon$, and therefore only finitely many choices of a prime $\wp$ of $F$ and $e>0$ such that $e+1$ exceeds the $\epsilon$ power of the norm of $\wp^e$. Therefore $\prod_i \wp_i^{e_i} \ll_\epsilon N^\epsilon$ where $N$ is the absolute value of the norm of $n$. But $N \ll H^{[F : {\bf Q}]}$, and $\epsilon$ was arbitrary, so we've proved the $H^{o(1)}$ bound. -For the second part, Dirichlet gives a logarithm map $U_F \rightarrow {\bf R}^r$ whose kernel is finite (the roots of unity in $F$) and whose image is a lattice $L$. A unit of height at most $H$ has all its conjugates of size $O(H)$, so is mapped to a ball of radius $\log(H) + O(1)$. Therefore there are $O(\log^r(H))$ units of height at most $H$. Much the same argument (involving a translate of $L$) shows that he same bound applies also to generators of any ideal $I$, since the ratio of any two generators of $I$ is a unit. -[I see that "Pitt the Elder" just gave much the same answer.]<|endoftext|> -TITLE: Einstein metrics and conformal geometry -QUESTION [10 upvotes]: I recall reading somewhere that if a conformal class contains an Einstein metric then that metric is the unique metric with constant scalar curvature in its conformal class, with the exception of the case of the round sphere. Does this sound right? If it is true: where can I find the proof of this result? - -REPLY [5 votes]: The first proof of the statement "Einstein metrics are the unique metrics with constant scalar curvature in their conformal class, except for round spheres" is due to Obata in 1971, see MR0303464 M. Obata, The conjectures on conformal transformations of Riemannian manifolds. -J. Differential Geometry 6 (1971/72), 247–258. In the beginning of the paper he lists many previous works with partial results in this direction as well.<|endoftext|> -TITLE: What could be some potentially useful mathematical databases? -QUESTION [75 upvotes]: This is a soft question but it's not meant as a big-list question. I have recently been asked whether I want to provide feedback at the pre-beta stage on a forthcoming website that will provide a platform for data sharing, and rather than giving just my personal opinion I'd rather consult other mathematicians first. I was going to write a blog post but then I thought that Mathoverflow was a more suitable place since I have a question and I'm looking for answers of a certain type rather than general comments. The website seems to be aimed mostly at scientists who want to share raw data, so at first I thought it probably wouldn't be much use to mathematicians since our data is (or are if you prefer) mostly highly interlinked -- the connections are often more interesting than what they connect. -But on further reflection, it seems to me that a good data sharing site could be a valuable resource, even if it doesn't do absolutely everything any mathematician would ever want. For instance, Sloane's database is fantastically useful. A rather different sort of database that is also useful is Scott Aaronson's Complexity Zoo. So useful databases exist already. Is this an aspect of mathematical life that could be greatly expanded given the right platform? And if so, what should the platform be like? -I don't know anything about the design of the site, but if I'm going to comment intelligently on what features it would need to have to be useful to mathematicians, I'd like to be armed with some examples of the kind of data sharing we might actually go in for. Here are a few ideas off the top of my head. - -Diophantine equations: one could have a list of what is known about various different ones. -Mathematical problems: listed in some nice categorized way, each problem accompanied by a description, complete with reading list, of what you really ought to know before thinking about the problem. (As an example, if you are thinking about the P versus NP problem, then you really ought to know about the Razborov/Rudich natural proofs paper.) -Key examples in various different areas and subareas of mathematics. -Sometimes you have a whole lot of related mathematical properties with a complicated pattern of implications between them. Under such circumstances, it could be nice to have this information presented in a nice graphical way (something I think this site may be able to do well -- they seem to be keen on visualization) with links to proofs of the implications or counterexamples that demonstrate when the implications do not hold. (The example I'm thinking of while writing this is different forms of the approximation property for Banach spaces, but there are presumably several others.) -List of special functions and the facts about each one that are the main facts one uses to prove things about them. -List of integrals that can be evaluated, with descriptions of how they can be evaluated. -List of important irrational numbers with their decimal expansions to vast numbers of places. (I'm not sure why this would be useful but it might be amusing.) - -These are supposed to be examples where people could usefully pool the background knowledge that they pick up while doing research. I'm not particularly pleased with them: they should be thought of as a challenge to come up with better ones, which almost certainly exist. If you've ever thought, "Wouldn't it be nice if there's somewhere where I could look up X," then X would make a great answer. I think the most interesting answers would be research-level answers (unlike some of the suggestions above). -If there were a site with a lot of databases, it would make a great place to browse: it would be much easier to find useful data there than if it was scattered all round the internet. -One constraint on answers: there should be something about a suggested database that makes it unsuitable for Wikipedia, since otherwise putting it on Wikipedia would appear to be more sensible. - -REPLY [7 votes]: I would love to see a database of theorems published online with links to the theorems they depend on. For example, if I looked up the Hahn-Banach theorem it'd point me to the Riesz extension theorem and any other major theorems required in its proof. If there are multiple published proofs, then more than one could be listed, each with its own set of links to previous theorems. -An application might be something like this: if I wanted to learn something like the Atiyah-Singer index theorem, it'd look it up, and the database could generate a complete dependency graph (or multiple such graphs) of major theorems I need to learn before I can get there. The database could even be personalised so it knows that it knows what theorems I'm familiar with and can use that to find an optimal path for me to learn something new, -In principle, any published book or paper would be representable as a subgraph of the full database graph. So it'd be nice to be able to look up a paper in this database and see that, in essence, this paper proves X, Y and Z using results A, B and C from papers P, Q and R. -Ultimately such a thing could morph into something where the proofs and dependencies are represented formally, but we don't have the technology for that yet.<|endoftext|> -TITLE: Picturing Riemannian geometry on a surface as a flow -QUESTION [6 upvotes]: This question asks pictures or intuitions of Riemannian geometry on a compact surface. -Pull the connection 1 form on the tangent circle bundle down to the tangent bundle with a vector field with isolated singularities. Take its dual vector field under the Riemannian metric. Now there is a vector field that is dual to the connection. In fact there are many such forms, one for each vector field with isolated singularities. -My question is: What picture of the connection does this vector field or its flow give us? -Is there a particularly clear picture if the vector field is dual to a meromorphic 1 form? - -REPLY [4 votes]: Most of your question is about local Riemannian geometry on a surface $(M,g)$, and that part can be answered without too much trouble, so I'll do that. Whether you can get anything global out of it is another matter, and I'm somewhat doubtful, but maybe the following will be of some use. -First, let's avoid the singular points for a moment and work on the part of the surface where the vector field isn't zero. Let's start with a unit vector field $\mathbf{e}_1$ on an open set $U\subset M$. Let $\mathbf{e}_2$ be such that $(\mathbf{e}_1,\mathbf{e}_2)$ is an oriented $g$-orthonormal frame field on $U$. -Direct computation shows that the vector field your process associates to $\mathbf{e}_1$ is the vector field $[\mathbf{e}_1,\mathbf{e}_2]$, i.e., the Lie bracket of the two unit vector fields. -Let $\mathbf{v}$ be another vector field on $U$. One can write $\mathbf{v} = r\cos\theta\ \mathbf{e}_1+r\sin\theta\ \mathbf{e}_2$ and, on the open set where $r>0$ (so that $\theta$ is well-defined), and your process associates to $\mathbf{v}$ the vector field -$$ -\bigl[\cos\theta\ \mathbf{e}_1+\sin\theta\ \mathbf{e}_2, -\ -\sin\theta\ \mathbf{e}_1+\cos\theta\ \mathbf{e}_2\bigr] -= [\mathbf{e}_1,\mathbf{e}_2] - \nabla^g\theta, -$$ -where $\nabla^g\theta$ is the gradient of $\theta$ with respect to the metric $g$. -It is not clear to me that you can get much local geometric information from this formula, and I'm not sure what you are hoping to see globally, but maybe you have something in mind. Locally, what you see is that the vector fields you get this way all differ by gradient vector fields. If you think of this as a first-order differential operator from vector fields to vector fields, it's nonlinear, but if you think of the map from $C^\infty(M)$ to vector fields given by $\theta\mapsto [\mathbf{e}_1,\mathbf{e}_2] - \nabla^g\theta$, then this is obviously an affine operator, so, you can almost think of it as a linear operator. -Note that, if $\mathbf{v}$ has isolated zeros of nonzero index, then $\nabla^g\theta$ blows up at those points, so your process yields vector fields with unbounded lengths. -As for the case of a meromorphic $1$-form, I guess you mean a vector field $\mathbf{v}$ that is $g$-dual to the real part of a meromorphic $1$-form $\omega$. Otherwise, I'm not sure what you mean. In this case, you can see, without too much trouble, that, away from the poles and zeros (where it isn't defined) the flow of the resulting vector field, say $\mathbf{w}$, is $g$-area preserving. The converse is true also (at least locally); i.e., if the flow of the resulting vector field is $g$-area preserving, then, up to a (real, nonzero) multiple, it is dual to a $g$-harmonic $1$-form (i.e., one that is closed and co-closed), so that (again, locally) it is dual to the real part of a holomorphic $1$-form. -Locally, at least, there is not anything more to say.<|endoftext|> -TITLE: Katz Modular Functions and Emerton's Completed Cohomology -QUESTION [18 upvotes]: Early this year, I started to learn about p-adic modular forms. Very recently, a mathematician tells me Emerton constructed an object called completed cohomology group with very rich structure, and the author could use it to prove fantastic results about Galois representations. (see Emerton's paper "Local-Global Compatibility In The p-adic Langlands Programme For $GL_{2/\Bbb{Q}}$". In page 45, he defines the huge $\hat{H}^{1}_{A}$) -With little understanding of it, can I ask if we can have a "natural" embedding of the space of Katz generalized p-adic modular functions (as defined in Section I.3 of Gouvea's Arithmetic of p-adic Modular Forms) into Emerton's completed cohomology group, assuming the base ring is $\Bbb {Z}_p$? If so, can we describe the image? This would be some type of Eichler-Shimura theorem for Katz generalized p-adic modular functions. Could this be one of the motivations when he defined such completed cohomology group? - -REPLY [29 votes]: You already have two helpful answers related to general aspects of Eichler--Shimura isomorphisms in a $p$-adic context. Here is an answer that more directly addresses your original question. - -I will begin by recalling/stating some facts on the $p$-adic modular form side: -Fix a tame (i.e. prime-to-$p$) level $N$, and let $\mathbb T(N)$ be the completed Hecke algebra generated by the $S_{\ell}$ and $T_{\ell}$ for $\ell \nmid N p$ acting on Katz's space $V(N)$ of generalized $p$-adic modular fuctions of level $N$. Fix a maximal ideal $\mathfrak m$ in $\mathbb T(N)$, and let $V(N)_{\mathfrak m}$ be the localization of Katz's space at the maximal ideal $\mathfrak m$. -In fact, in order to deal sensibly with oldforms at $N$, I find it helpful to do the -following: fix the finite set of primes $\ell_1,\ldots,\ell_n$ dividing $N$, and take -a direct limit of $V(N)_{\mathfrak m}$ as $N$ ranges over all levels divisible by -just $\ell_1,\ldots,\ell_n$. Note that $\mathbb T_{\mathfrak m}$ may grow as $N$ increases (if $N$ divides $N'$ then $\mathbb T(N)\_{\mathfrak m}$ is a quotient of $\mathbb -T(N')\_{\mathfrak m}$), but eventually stabilizes (even thought the $V(N)_{\mathfrak m}$ don't stabilize), because if $\rho$ is any lift of -the Galois representation $\overline{\rho}$ attached to $\mathfrak m$ then the difference between the prime-to-$p$ conductor of $\rho$ and $\overline{\rho}$ is bounded. -So now let $V_{\mathfrak m}$ be this direct limit of $V(N)\_{\mathfrak m}$, and let $\mathbb T_{\mathfrak m}$ be the Katz Hecke algebra acting on it. Note that $V_{\mathfrak m}$ is a smooth representation of the product $\prod_{i = 1}^n GL_2(\mathbb Q_{\ell_i}).$ -Note also that there is an action of $U_p$ on $V_{\mathfrak m}$. (Let me reiterate that I did not include $U_p$ in my Hecke algebra $\mathbb T_{\mathfrak m}$!) - -Now completed cohomology: -If we take completed cohomology at tame level $N$, we get a $p$-adically complete -$\mathbb Z_p$-module $\widetilde{H}^1(N)$, with an action of $\mathbb T_{\mathfrak m}$, -as well as of $G_{\mathbb Q}$ and $GL_2(\mathbb Q_p)$. We can complete it at $\mathfrak m$ -to get $\widetilde{H}^1(N)\_{\mathfrak m}$. -If we then take the direct limit over all $N$ -which are divisible exactly by $\ell_1,\ldots,\ell_n$, we get a module I'll denote -$\widetilde{H}^1_{\mathfrak m}$, which has an action of $\mathbb T_{\mathfrak m}$, of $G_{\mathbb Q}$ -and $GL_2(\mathbb Q_p)$, and also of the product $\prod_{i=1}^n GL_2(\mathbb Q_{\ell_i})$. - -Now suppose that $\overline{\rho}$ satisfies some technical conditions, irreducibility -being the most significant one. (The precise conditions are in the local-global compatibility paper that you mention. Note also that the irreducibility assumption eliminates the distinction between cohomology and cohomology with compact support, and --- more or -less equivalently --- the distinction between working on closed vs. open modular curves.) -Then you can show that there is an isomorphism of $\mathbb T_{\mathfrak m}[G_{\mathbb Q}\times GL_2(\mathbb Q_p) \times \prod_{i = 1}^n GL_2(\mathbb Q_{\ell_i})]$-modules -$$\widetilde{H}^1\_{\mathfrak m} = \rho^u \otimes_{\mathbb T\_{\mathfrak m}} \pi^u -\hat{\otimes}_{\mathbb T\_{\mathfrak m}} V\_{\mathfrak m}^{U_p =0}. -$$ -Here $\rho^u$ is the univeral modular deformation of $\overline{\rho}$ over $\mathbb T_{\mathfrak m}$, $\pi^u$ is a representation of $GL_2(\mathbb Q_p)$ on an orthonormalizable $\mathbb T_{\mathfrak m}$-Banach module constructed from $\rho^u_{| G_{\mathbb Q_p}}$ via the $p$-adic local Langlands, and $V_{\mathfrak m}^{U_p = 0}$ is, as indicated, the kernel -of $U_p$ on $V\_{\mathfrak m}$. Also, the completed tensor product $\hat{\otimes}$ has to be suitably interpreted. (One should cut back to a fixed tame level $N$, then form the completed tensor product, and then take a direct limit over all $N$.) -So this is a kind of $p$-adic Eichler--Shimura isomorphism relating $p$-adically completed cohomology and $p$-adic modular forms. -It brings out the difference between the two sides quite clearly: on the completed cohomology side, we have a Galois action, which is encoded in the appearance of $\rho^u$. -(This reflects the classical fact that every cuspform appears "twice" in cohomology.) -Also, completed cohomology has an action of $GL_2(\mathbb Q_p)$, while $p$-adic modular forms just have an action of $U_p$. -So to compare the two, we have to first get rid of the $U_p$-action on $p$-adic modular forms (which we do by passing to $U_p = 0$), and then add in a $GL_2(\mathbb Q_p)$-action, which we do by tensoring with $\pi^u$. -Note also that this isomorphism is not canonical. In this sense, it is analogous -to looking at classical cohomology of modular forms with say $\mathbb Q$-coefficients, -and modular forms with $\mathbb Q$-coefficients. These will be isomorphic as Hecke modules ---- up to the issue of cuspforms appearing twice in cohomology --- but not canonically so. In order to make the Eichler--Shimura isomorphism canonical, one has to extend scalars to an appropriate period ring. Whether this is possible with completed cohomology I'm not sure about at the moment. - -One more remark: trading in a $U_p$-action for a $GL_2(\mathbb Q_p)$-action is a fairly significant upgrading of structure, and this is why completed cohomology provides a useful tool for proving modularity theorems for Galois representations, over and above the already-existing theories of $p$-adic modular forms and $p$-adic modular symbols. - -Added: You asked about motivation. The original motivation for defining completed cohomology was to construct eigenvarieties. Later it became clear that it was an important object in its own right, providing a global counterpart to the representations of $p$-adic groups that were beginning to appear as part of $p$-adic local Langlands. E.g. the theorem that locally algebraic vectors in cohomology are classical was first proved as an ingredient in the proof of an analogue of Coleman's "small slope implies classical" result for the eigenvariety constructed from completed cohomology. Only later was it realized that this could be combined with a local-global compatiblity result to prove modularity theorems for Galois representations. -Note that the rough relation with $p$-adic modular forms, namely that one gets the same Hecke algebra via either approach, was clear from the beginning, even though the more precise Eichler--Shimura-like statement above was not. Since eigenvarieties (as their name indicates) only care about Hecke eigenvalues, this meant that completed cohomology was good enough for constructing them.<|endoftext|> -TITLE: How to calculate the Witten-Reshetikhin-Turaev invariants from a triangulation? -QUESTION [17 upvotes]: I'm interested in the Witten-Reshetikhin-Turaev invariants for 3-manifolds, and in particular, how to calculate them from a triangulation of the 3-manifold (recall that as they were first introduced, their calculation is based on a surgery diagram for the manifold). Is there any some sort of "state sum" model giving the WRT invariant for a $3$-manifold, which is based on the triangulation of the $3$-manifold? -I am well aware of the Turaev-Viro construction, and it is exactly the type of definition I'm hoping for, but of course the TV invariant is $\left|\cdot\right|^2$ of the WRT invariant, not the WRT invariant itself. -The simpler the construction, the better. Certainly one can go algorithmically from a triangulation to a surgery diagram, but I want something which is conceptually based on the triangulation. -I guess any answer I get will also include the case for a pair $(Y^3,L)$, and even perhaps allow for $M^3$ to have nonempty boundary (i.e. the full tqft). However, I will be satisfied even if I just get a construction for a closed $3$-manifold $M$. - -REPLY [3 votes]: Hi, we (Ricardo Machado and Sostenes Lins) put in arxiv, 3 articles which shows how to get a framed link presentation from a triangulation. We show an O(n^2) algorithm which I implemented and it is really working. So with the framed link we can calculate the WRT-invariants -The link to the articles: -http://arxiv.org/find/all/1/all:+AND+lins+AND+sostenes+AND+ricardo+Machado/0/1/0/all/0/1<|endoftext|> -TITLE: Weyl modules and reduction modulo $p$. -QUESTION [5 upvotes]: Representation theory of Lie groups and Lie algebras are quite close and in a suitable sense they become equivalent. However, some subjects are typically found on a Lie group theory language. For example, Weyl modules almost always start from a (Chevalley) group context and not from a Lie algebra context. Another examples are about hyperalgebras. -Let's just focus on Weyl modules. Starting with a finite-dimensional complex Lie algebra $g$ one constructs Weyl modules as a module given by generators and relations, i.e as a quotient of the traditional $U(g)$, and then one can proceed with an algebraic approach. -QUESTION: What is necessary to start from a purely algebraic setting the following ideas? - -Let $G$ be a simply connected Chevalley group over an algebraically closed field of characteristic $p$. Let $G_\mathbb{C}$ be the Chevalley group over the complex numbers which corresponds to $G$. -A Weyl module is the reduction modulo $p$ of an irreducible module in characteristic $0$. The Weyl module $\overline{W}(\lambda)$, the reduction modulo p of the $G_\mathbb{C}$-module of highest weight $\lambda$, has a unique maximal submodule $M(\lambda)$ and $V(\lambda) = \overline{W}(\lambda)/M(\lambda)$ is the irreducible $G$-module of highest weight $\lambda$. -It was found in http://www.jstor.org/pss/2374222. - -Particularly, what is the reference for the affirmation "A Weyl module is the reduction modulo $p$ of an irreducible module in characteristic $0$" ? - -REPLY [9 votes]: It's not difficult to answer this question, but for this it's useful to sketch briefly the origins of the term Weyl module. As usual in mathematics, the history and attributions are somewhat convoluted, but the main developments can be reconstructed in outline form: -1) Following decades of classical development in Lie theory, a major breakthrough came with Chevalley's 1955 Tohoku paper in which he showed uniformly how to construct a $\mathbb{Z}$-basis (unique up to sign choices) for each simple Lie algebra over $\mathbb{C}$ which permits a good "reduction mod $p$ for arbitary primes. In effect he incorporated factorial denominators into his basis construction, so certain "exponentials" can be constructed as polyunomials even in characteristic $p$. After change of basis to an arbitrary field $k$, these elements generate a Chevalley group, essentially a $k$-form of an associated simple algebraic group of adjoint type. Over finite fields he gets related simple groups including some never constructed before. -2) In his 1956-58 Paris seminar on classification of semisimple algebraic groups, Chevalley observed that groups of all possible types exist based on his $\mathbb{Z}$-form approach. His Bourbaki seminar talk (exp. 219, 1960-61) explained how to convert this into something like scheme language in terms of a $\mathbb{Z}$-form of the function algebra. But not all details are given (and some are out of focus). A more definitive version was given by Lusztig in J. Amer. Math. Soc. 2009. -3) Steinberg's Yale lectures on Chevalley groups (1967-68) streamlined the construction of all types of groups, including the various twisted groups, and began the study of representations in characteristic $p$ in terms of reduction mod $p$ of simple finite dimensional modules with dominant integral highest weights. A crucial role is played by Kostant's construction of a suitable integral form in the universal enveloping algebra, which he presented at the 1965 Boulder summer conference (and where Steinberg also lectured, while Verma and I were among the graduate students attending). The basic constructions here were included in my 1972 Springer graduate text. Meanwhile Warren Wong had extended some of the ideas in terms of "contravariant" forms. -4) In my 1971 J. Algebra paper, I didn't yet give a name to the modules obtained from a minimal "admissible" $\mathbb{Z}$-form, but I was able to extend the theory of these highest weight modules and their possible composition factors. This work got taken up by Verma in his lecture notes from the 1971 Budapest summer school organized by Gelfand (published some years later); he emphasized the hidden but essential role of an affine Weyl group. In a footnote he also called attention to the 1974 paper by Carter and Lusztig (Proc. London Math. Soc.), where the interplay of representations for special linear and symmetric groups was treated -in the Chevalley group setting mod $p$. They made my "linkage" result more precise and dubbed the unnamed modules of highest weight Weyl modules (in part at least because their formal characters are still given by Weyl's formula even though they tend to have complicated composition series in characteristic $p$). -5) Independently, Jantzen was making rapid progress from a different angle on the same problems; he quickly adopted the label "Weyl module" for arbitrary semisimple groups, using as I did the module obtained from a minimal admissible lattice. For the special linear case, he was also able to show by ad hoc methods that these Weyl modules have the natural universal property among finite dimensional highest weight modules. Taking advantage of Kempf's 1976 vanishing theorem for dominant line bundles in characteristic $p$, I observed that this would easily prove the universal property of Weyl modules in general (written down by Jantzen in Satz 1 of his 1980 Crelle paper). In effect Weyl modules are dual to global section modules of line bundles (with an obvious dualization of the highest weight). This became in fact the preferred definition of "Weyl module" in Jantzen's book Representations of Algebraic Groups. -6) The 1979 Invent. Math. paper by Kazhdan and Lusztig led not only to their conjecture on composition factor multiplicities for Verma modules in characteristic 0 but also to Lusztig's analogous conjecture for Weyl modules. This is still not quite proved in the correct generality (and fails to cover primes smaller than the Coxeter number). The story is not complete, but the literature by now is extensive.<|endoftext|> -TITLE: Asymptotics for Ramsey Theory -QUESTION [9 upvotes]: Ramsey Theory says that every sufficently large (but finite) complete graph having $d-$coloured edges contains a monochromatic complete subgraph with $k$ vertices. -One could ask for asymptotics: Let $A(n,d,k)$ be the minimal number of -monochromatic complete subgraphs with $k$ vertices contained in any complete graph with $n$ -vertices whose edges are coloured with $d$ colours. -One has the obvious bounds: $n+1-N\leq A(n,d,k)\leq {n\choose k}\sim \frac{n^k}{k!}$ -where $N$ is the corresponding Ramsey number. -There exists thus a critical exponent $\alpha$ -such that $\lim_{n\rightarrow\infty}\frac{A(n,d,k)}{n^{\beta}}=\infty$ -for all $\beta<\alpha$ and -$\liminf_{n\rightarrow\infty}\frac{A(n,d,k)}{n^{\beta}}=0$ -for all $\beta>\alpha$. -What is $\alpha$? Perhaps the function $\frac{A(n,d,k)}{n^{\alpha}}$ has a nice behaviour? -Are there any instance where one can say anything interesting? (One has -obviously $\alpha\geq 1$ a strict inequality would probably already be interesting.) -The same type of question can be asked for Erd\"os-Szekeres, for van der Waerden -(with a different, much smaller but still affine lower bound) etc. -Added after Gowers solution: Gowers gave the following easy proof that $\alpha=k$. -Denote by $m=\inf\lbrace n\ \vert\ A(n,d,k)>0\rbrace$ the Ramsey number -corresponding to our problem. Since a fixed monochromatic $k-$clique is contained -in exactly ${n-k\choose m-k}$ subsets of size $m$ we have -$$A(n,d,k)\geq\frac{{n\choose m}}{{n-k\choose m-k}}\sim \frac{n^k}{m^k}\ .$$ -The only interesting question is thus the exact value of $\lim_{n\rightarrow\infty}\frac{A(n,d,k)}{n^k}$ which exists by Gowers first argument. - -REPLY [4 votes]: OK, this is an attempt to amplify the very brief sketch of what could be done with the regularity lemma. Of course, I'll still give a sketch, but I'll try to say more clearly what I mean. -First of all, the regularity lemma itself says that for every $\epsilon>0$ there exists a constant $K$ such that for every graph $G$ you can partition the vertex set into $k\leq K$ sets $V_1,\dots,V_k$ of roughly equal size such that for all but $\epsilon k^2$ of the pairs $(i,j)$ the induced bipartite subgraph (or just subgraph if $i=j$) is $\epsilon$-regular. Here, $\epsilon$-regular means something like "behaves to within $\epsilon$ like a typical random graph of the same density". I won't say precisely what that means, but I hope it will be plausible that this quasirandomness property will have the consequences I claim it has. -Now if we $r$-colour a complete graph, that's the same as partitioning the edges into $r$ graphs. It turns out that a simple modification of the proof of the regularity lemma allows us to apply it to all these $r$ graphs at once, obtaining a single partition that works for all of them. (Of course, now the constant $K$ will depend on $r$ as well as on $\epsilon$.) What I mean here is that for all but $\epsilon k^2$ pairs, all $r$ of the induced bipartite subgraphs are $\epsilon$-quasirandom. -For many purposes, all random graphs of a given density behave in the same way. That is the case here, so once we have our partition the only information we really care about is the densities of the various colours in the various bipartite graphs. That is, we care about numbers like $\alpha_{ijs}$ where that is the density of edges in $V_i\times V_j$ that are coloured with colour $s$. -Let's call a pair $(i,j)$ good if all the colour classes of edges in $V_i\times V_j$ are quasirandom. The main consequence we need of quasirandomness is that if you have $d$ of the sets $V_1,\dots,V_k$ such that all the pairs from among those $d$ sets are good, then the number of cliques of colour $s$ will be approximately $|V_{i_1}|\dots|V_{i_d}|\prod_{j,h\leq d}\alpha_{i_ji_hs}$, where $V_{i_1}\dots V_{i_d}$ are the sets in question. Since almost all pairs are good, almost all $d$-tuples generate $\binom d2$ pairs that are all good. It follows that up to some error that tends to zero with $\epsilon$, the number of monochromatic cliques, as a fraction of $n^k$, depends just on the numbers $\alpha_{ijs}$. (To see this, you just add up over all possible $d$-tuples.) It follows that the minimum number of monochromatic cliques can be found by calculating a minimum over all choices of $\alpha_{ijs}$. Since $k$ is bounded above by a number that depends on $\epsilon$ and $r$ only, this is in principle a finite problem for any given $\epsilon$, $d$ and $r$. (However, the bound on $k$ is absolutely vast, so this problem is absolutely not feasible in practice.) More to the point, it proves that $A(n,d,r)/n^d$ tends to a limit as $n$ tends to infinity with fixed $d$ and $r$. (Apologies -- I've changed your $k$ to a $d$ and your $d$ to an $r$. That's because I chose $k$ in the regularity lemma, which I now regret.) -It remains to establish that this limit isn't zero. A quick and dirty way of doing that is to make $\epsilon$ so small that there must exist $R(r,d)$ of the sets $V_i$ such that all pairs are good, where $R(d,r)$ is the number needed to get a monochromatic clique of size $d$ when you have $r$ colours. For each pair we can now choose the colour that occurs most frequently and colour that pair with that colour. Applying Ramsey's theorem, we obtain a monochromatic clique of size $d$, which translates into $d$ sets $V_{i_1}\dots V_{i_d}$ that are all quasirandomly joined in that colour with density at least $1/r$. By quasirandomness, that gives us (to within a small error) at least $r^{-\binom d2}|V_{i_1}|\dots|V_{i_d}|$. Each $V_i$ has at least $n/K$ elements, and $K$ is independent of $n$, which completes the proof.<|endoftext|> -TITLE: Cohomology of the quotient of a Lie group by a finite subgroup -QUESTION [11 upvotes]: Let $G$ denote the $\operatorname{Spin}(n)$ group with $n>4$ and let $\Gamma$ be a cyclic subgroup $G$ of a prime order $p >2$. When does the projection $G \to G/\Gamma$ induce a surjection -between cohomology groups $H^3$ with integral coefficients? - -REPLY [7 votes]: Here's another approach using the Leray-Serre spectral sequence. Using the fact that $\Gamma$ acts freely on $G$, the map $G \to G/\Gamma$ is a covering space, and so the cohomology of $G/\Gamma$ may be computed by the spectral sequence -$$E_2^{s, t} = H^s(\Gamma, H^t(G)) \implies H^{s+t}(G/\Gamma).$$ -Here $H^s(\Gamma, M)$ is the s'th group cohomology of $\Gamma$ with coefficients in the $\Gamma$-representation $M$. We record three facts to start: - -If $\Gamma \cong \mathbb{Z} / p$, then $H^s(\Gamma, \mathbb{Z})$ is 0 for $s$ odd, and $\mathbb{Z} / p$ for $s$ even. -For $G=Spin(n)$, $H^t(G) = \mathbb{Z}, 0, 0, \mathbb{Z}$, for $t=0,1,2,3$. -For $p$ odd, $\mathbb{Z} / p$ cannot act nontrivially on $\mathbb{Z}$, since $Aut(\mathbb{Z}) = \mathbb{Z} / 2$. - -Let's compute part of the $E_2$-term of the spectral sequence. Fact 2 implies that the $t=1$ and $2$ rows vanish entirely. Fact 3 implies that $H^0(G) = \mathbb{Z}$ and $H^3(G) = \mathbb{Z}$ are trivial $\Gamma$-modules, so the $t=0$ and $3$ rows are the group cohomology described in Fact 1. -Thus the only possible term that can contribute to $H^3(G/\Gamma)$ is $E_2^{0, 3} = H^3(G) = \mathbb{Z}$. There is, however, a possiblity of a single differential in this region of the spectral sequence, namely -$$d_4: E_2^{0, 3} = \mathbb{Z} \longrightarrow E_2^{4, 0} = H^4(\Gamma, \mathbb{Z}) = \mathbb{Z} / p.$$ -Therefore, $H^3(G/\Gamma)$ surjects onto $H^3(G)$ precisely when this differential $d_4= 0$. We note that if it's not 0, it is surjective; thus at worst $H^3(G/\Gamma)$ may be identified with an index $p$ subgroup of $H^3(G)$. -So, how do we compute $d_4$? I claim that it's given as: -$$H^3(G) \cong H^4(BG) \to H^4(B\Gamma) = H^4(\Gamma)$$ -where the map is the restriction in cohomology, induced by the (inclusion) homomorphism $\Gamma \subseteq G$. This can be seen, for instance, by comparing this with the spectral sequence for the (rather dumb) fibration $G \to G/G=pt$. -So a long winded answer to your question is: The map is surjective if and only if none of the $H^4(BG) = \mathbb{Z}$ is supported on $H^4(\Gamma) = \mathbb{Z} /p$. I would imagine that determining when that is the case is highly dependent upon the subgroup in question.<|endoftext|> -TITLE: Has anyone studied the Prym map for double covers with two ramification points? -QUESTION [9 upvotes]: If $f \colon C \to C'$ is a dominant morphism of smooth projective curves, there is a norm map $f_\ast = \mathrm{Nm} \colon JC \to JC'$ between their Jacobians, and we can consider the abelian subvariety $Z = (\ker \mathrm{Nm})^0$ and the polarization on $Z$ induced from $JC$. In two particularly interesting cases this polarization is twice a principal polarization $\Xi$, namely when $f$ has degree two and is either unramified or has two ramification points. In this case, one calls the ppav $(Z,\Xi)$ the Prym variety of $f$. -This construction works just as well in families, so it defines morphisms between some moduli spaces. If $R_g$ denotes the moduli space of smooth genus $g$ curves $C'$ and an unramified connected double cover $C \to C'$, one gets a Prym map $R_g \to A_{g-1}$. Here $A_g$ is the moduli space of dimension $g$ principally polarized abelian varieties. This map is rather well studied (e.g. the papers of Beauville, Donagi-Smith, Donagi and too many others to mention here). -On the other hand, one could also let $R_{g,2}$ be the moduli space of smooth genus $g$ curves with a double cover with two branch points. One gets in the same way a Prym map $R_{g,2} \to A_g$. But I have never seen any paper dealing with the properties of this map; maybe this is because of my incomplete knowledge of the literature. (One reason it may be less interesting is that unlike the ordinary Prym map in genus six, this one should never be generically finite for any $g$.) -Here are for instance some natural questions about this map: Is it dominant when $g \leq 4$? If yes, what is the structure of the generic fiber? Is it generically injective when $g \geq 5$? Does it extend to a compactification $\overline R_{g,2}$ using admissible covers, either by mapping to the Satake compactification or to a toroidal compactification like the 2nd Voronoi? -Is any of this known? - -REPLY [3 votes]: This is not as up to the minute as the beautifully detailed thesis linked in the answer above, but just a couple of historical comments. -Such a 2 point ramified double cover occurs from an unramified one when a loop shrinks to a point, which... let's see now, does intersect [I believe] the loop on which the double cover is based. Then I believe the induced double cover of the normalized curve is ramified at the 2 points over the singular point. -Vice versa, 2 point ramified double covers are a subcase of Beauville's "star - double covers" of curves with one node. E.g. 2 point ramified covers of genus 5 curves are a special case of Beauville's "star - double covers" of genus 6 curves with one node. -Anyway such ramified double covers have thus been studied by Wirtinger in the 19th century, and also explicitly by Fay in his book on theta functions, as well as more generally by Beauville.<|endoftext|> -TITLE: Unsolved Problem from AmMathMonthly -QUESTION [6 upvotes]: Here is a simply described but fiendishly diophanterrorizing problem -I asked on AMM eons ago. Maybe you can shed some light upon it. -0.2 (base 4) = 0.2 (continued fraction) -0.24 (base 6) = 0.24 (continued fraction) -Find all examples of -0.$xyz$... (base B) = 0.$xyz$... (continued fraction). -First of all, both choices of notation define a rapidly closing interval nesting, -and already on post-comma digit 2, you're down to one number by a -simple > / < argument. But you may not use 0 for CF and $\ge B$ for base $B$, -and thus almost any base $B$ will run into a dead end sooner or later. -(It's fun to experiment with low $B$.) -Obvious Thing 1: 1-digit solution 0.$n$ for $B=n^2$. -Educated guess 2: There are only two solutions with two digits. -(The second was listed in the MAA Answer Column; juggling with -Chebyshev polynomials I had a sort of proof for that case, but -it probably had more holes than a Menger sponge and so it wasn't -printed there). -Wild guess 3: There is no solution with more than two digits, -for the reasons above. -Can you at least prove case 2? (The MAA discussion splits it into -two subcases; $239^2+1=2\times 13^4$ killed one of them.) -This is Monthly problem 10507. The problem appeared in February 1996 (volume 103, issue 2, page 173), and the discussion (and solution to other parts) in March 1998 (volume 105, issue 3, page 276). Thanks to Gerald Edgar for the pointer. - -REPLY [6 votes]: Here is a solution for the case you ask. But first let me say that given the nature of the question it would probably get better answers at artofproblemsolving. What follows is a lot of very elementary number theory, and an appeal to a result of Ljunggren from 1942. -So we have $B\geq 2$ and $x,y\in \{0,1,\dots,B-1\}$ satisfying -$$\frac{x}{B}+\frac{y}{B^2}=\frac{1}{x+\frac{1}{y}}$$ or in other words $B^2y=(xy+1)(Bx+y)$. Let $a=\gcd(x,y)$ and $x=am, y=an$. We have $$B^2n=(Bm+n)(a^2mn+1)$$ where $\gcd(m,n)=1$. Since $\gcd(n,a^2mn+1)=1$ we have that $n$ is a factor of $Bm+n$ therefore there is an integer $k$ so that $B=kn$. The equation simplifies to $$n^2k^2=(km+1)(a^2mn+1)$$ -We see that $\gcd(km+1,k^2)=1$ so $km+1$ divides $n^2$, but also $\gcd(n^2,a^2mn+1)=1$ so $n^2$ divides $km+1$. We conclude that $km+1=n^2$ and $a^2mn+1=k^2$. In particular $k^2-1$ is divisible by $n$, and $n^2-1$ is divisible by $k$, so that $$\frac{k^2+n^2-1}{kn}=t\in \mathbb Z.$$ Now some Vieta jumping shows that $k,n$ are consecutive terms in the sequence $a_0=0,a_1=1$ and $a_{n+1}+a_{n-1}=ta_n$. Let $k=a_{p+1}$ and $n=a_p$, the equations reduce to $$m=nt-k=a_{p-1}, a^2m=kt-n=a_{p+2}.$$ -Now, it is not hard to prove that our sequence is a strong divisibility sequence so that $$a^2=\frac{a_{p+2}}{a_{p-1}}$$ implies that $p+2$ is divisible by $p-1$ which only happens if $p\in \{2,4\}$. So in particular we either have $a^2=t^3-2t$ or $a^2=t^3-3t$. The second equation doesn't have non-trivial solutions because if $\gcd(t,t^2-3)=1$ then $t^2-3$ is a square which is not possible, and if $\gcd(t,t^2-3)=3$ then $3(t/3)^2-1$ is a square $-1\pmod{3}$ which is also a contradiction. For the first equation, similarly we conclude that $t=2r$ must be even and that $r(2r^2-1)$ is a perfect square, so $r=s^2$ is also a perfect square. This finally brings us to the equation $2s^4-1=l^2$, which has solutions only for $s=1$ and $s=13$ as was proved by W. Ljunggren in "Zur Theorie der Gleichung x^2+1=Dy^4" (Avh. Norske Vid. Akad. Oslo I. 5, 27pp.). This proves that the two solutions you had are the only ones.<|endoftext|> -TITLE: Are there two Raynauds? -QUESTION [8 upvotes]: The Wikipedia page on Michel Raynaud says that he worked with Grothendieck on the SGA volumes. However, in SGA 1 and 2 the name Mme Michèle Raynaud is given. According to the Genealogy Project, the two are different persons, but I cannot find any further information on Michèle Raynaud. Are there really two Raynauds, or did Michèle Raynaud just become Michel Raynaud at some point? - -REPLY [2 votes]: While Kevin Buzzard's comment is more informative than what I could come up with (not knowing the involved persons) here is some generally accessible information showing that there is a Michèle and a Michel. -First, the MathSciNet author database lists both (and MathSciNet is very good at 'merging' different names even pseudonyms under which one and the same person published). -Second, and more significantly, during a period of several years, sixties/seventies, one can find a variety of papers published by Michèle Raynaud as well as papers published by Michel Raynaud. -For example: - -Raynaud, Michèle Théorèmes de Lefschetz en cohomologie étale des faisceaux en groupes non nécessairement commutatifs. (French) C. R. Acad. Sci. Paris Sér. A-B 270 1970 -Raynaud, Michel Anneaux locaux henséliens. (French) Lecture Notes in Mathematics, Vol. 169 Springer-Verlag, Berlin-New York 1970<|endoftext|> -TITLE: On the independence of the Kurepa Hypothesis -QUESTION [5 upvotes]: Kurepa Hypothesis says there is a Kurepa tree, which is a $\omega_1$-tree has at least $\omega_2$ many branches. -It is known that beginning from a model with an inaccessible cardinal $\kappa$, after collapes $\kappa$ to $\omega_2$ using the Levy collape, then in the generic extension, Kurepa Hypothesis fails. -In above generic extension, $\omega_2$ is equal to $\kappa$ and by a counting argument for nice names, $2^{\omega_1}=\omega_2$. My question is that "is it consistent that Kurepa Hypothesis fails and $2^{\omega_1}>\omega_2$?" -(The reason I think this question: The biggest possible value of the number of branches is $2^{\omega_1}$, so in the environment of $2^{\omega_1}=\omega_2$, it is most difficult for the living of a Kurepa tree. So I want to know whether this requiement is necessary.) - -REPLY [2 votes]: Another solution can be obtained by adding many reals and using a lemma of Spencer Unger that generalizes the lemma used by Silver. - -Lemma (Unger) Suppose $\mu \leq \kappa$ are regular, there is $\tau \leq \mu$ such that $2^\tau \geq \kappa$, and $\mathbb{P}$ is $\mu$-c.c. and $\mathbb{Q}$ is $\mu$-closed. Then $\Vdash_\mathbb{P} \mathbb{Q}$ does not add branches to $\kappa$-trees. - -MR2945572 Unger, Spencer. Fragility and indestructibility of the tree property. Arch. Math. Logic 51 (2012), no. 5-6, 635–645. -Start with Silver's model of $\neg KH$, in which an inaccessible $\theta$ is Levy collapsed to $\omega_2$ by $\mathbb{Q} = Col(\omega_1,<\kappa)$. Then add at least $\omega_3$ many Cohen or Random reals, call that $\mathbb{P}$. Any $\omega_1$-tree $T$ in $V^{\mathbb{Q} \times \mathbb{P}}$ is captured in some intermediate extension by a subforcing $\mathbb{Q}_0 \times \mathbb{P}_0$ of size $<\kappa$, where $\mathbb{Q}_0$ is rank initial segment of $\mathbb{Q}$. By the well known fact about Knaster posets $\mathbb{P} / \mathbb{P}_0$ adds no branches to $T$, and by Unger's lemma, $\mathbb{Q} / \mathbb{Q}_0$ does not add branches over $V^{\mathbb{P} \times \mathbb{Q}_0}$. Finally, since $\kappa$ is inaccessible in $V$, the number of branches of $T$ in $V^{\mathbb{Q}_0 \times \mathbb{P}_0}$ is $<\kappa$.<|endoftext|> -TITLE: Teaching a pedagogy course -QUESTION [18 upvotes]: At my institution incoming graduate students must take a semester long course on pedagogy taught by current grad students. I may soon be in the position of having to teach this course and I'm looking for advice for readings to give the students. The problem is that our grad students don't teach till after they pass their quals, so effectively the pedagogy course is teaching people how to teach when they are more focused on their qual courses and know that they won't have to teach for at least a year. Consequently, most students view it as a complete waste of time and gain little from it. When I took this course I really tried to get something out of it and some of the lessons stuck with me. Still, at that time we mostly had readings taken out of guides for high school teachers and I'd rather have more applicable readings. -Here's what I've got so far: - -A small booklet the university produces with rules, resources for getting help when you teach, online resources the university uses, and honor code -Old syllabi we can discuss -All students will go observe other grad students teaching and write a report which we can discuss. -Discuss some teaching mechanics, e.g. blackboard use, dealing with student questions, etc. Good resource seems to be the AMS blog and a document from Williams -All students will need to prepare one lecture and deliver it. We can then give them feedback. As recommended by Gerhard, I think I'll try to video tape this and give them the tapes. Below are some further readings... -Some notes by V.I. Arnold, some by John Baez, and some (recommended below) by Bruce Reznick. I'm starting to think a good exercise would be to assign these as readings and ask the students to find 3 things in a given reading which they disagree with. We can then discuss. -Silly, but how to curve grades -Perhaps some discussions of issues of pedagogy, e.g. why teach calculus the way we do, and pros/cons of giving handouts -Discuss evaluations, teaching statements, and teaching letters with an eye towards applying for jobs -Discuss giving talks in general - - -Does anyone have any other ideas for topics worth discussing or readings worth giving? I'm particularly interested in books on this subject which I could draw readings from. - -Also (and this is probably too general), since these students will have at least a year before they actually teach, I'm keeping my eyes open for lessons that will stick with them and come in handy when they finally get in front of a class. If you have any nuggets that fit the bill, I'd be interested to hear them (all mine right now come from John Baez's notes or the notes from Williams) - -REPLY [2 votes]: Thanks everyone for your advice. The course is over now and it seems to have been a success. I created a webpage which contains all the materials I used as well as the syllabus. Please feel free to make use of this material if you should ever need to -http://dwhite03.web.wesleyan.edu/pedagogyLinks.html<|endoftext|> -TITLE: A subcategory of top where subspaces and subobjects coincide? -QUESTION [5 upvotes]: I think that my question is easily answerable. The question is: What is a nice subcategory of topological spaces where the subobjects are subspaces. I would like the category of compactly generated Haussdorf spaces to be such a category, since this category is convienent in many other ways. -Some backround definitions: - -A subspace (of X) of a topoogical space is a subset of X with the induced topology. -The notion of a subobject (of X) is a categorical notion. It is an equivalence class of monomorphisms of the form, $m:Y\rightarrow X$. This $m$ is equivalent to another subobject, $m':Y'\rightarrow X$ if their exists an isomorphism, $i:Y\rightarrow Y'$ making the evident triangle commute. - -REPLY [11 votes]: A supplement to Todd's answer with the observation that the requested property fails in all the usual cartesian closed full subcategories of $\mathbf{Top}$. Indeed, there are even regular subobjects (see Todd's comment for the definition) that are not subspaces. (Todd's own counterexample in compactly generated spaces describes a non-regular subobject that is not a subspace.) -By the "usual" subcategories, I mean those that fall under the general treatment of cartesian closed full subcategories of $\mathbf{Top}$ in: -M. Escardó, J.Lawson, A. Simpson. -Comparing Cartesian closed categories of (core) compactly generated spaces. -Topology and its Applications, 143 (2004) 105–145. -This considers categories of $\mathcal{C}$-generated spaces for suitable collections $\mathcal{C}$ of topological spaces. (Compactly generated spaces arise by taking $\mathcal{C}$ to be the collection of compact Hausdorff spaces.) -Every such category of $\mathcal{C}$-generated spaces includes the integers $\mathbb{Z}$ with discrete topology. Using cartesian closedness, define the iterated function space -$F_2 := \mathbb{Z}^{\mathbb{Z}^\mathbb{Z}}$. I observe below that $F_2$ has a (countable) subspace $Y$ that is not $\mathcal{C}$-generated. Then $Y$ endowed with its $\mathcal{C}$-generated topology gives the promised regular subobject of $F_2$ in the category of $\mathcal{C}$-generated spaces that is not a subspace. -For the observation, first, by Corollary 7.3 of op. cit., the topology of $F_2$ is -independent of the choice of $\mathcal{C}$. For convenience, we consider $F_2$ with its -topology as a sequential space. -It is known that $F_2$ is not a Fréchet–Urysohn space. That is, there is a countable subset $X \subseteq F_2$ and element $x_\infty \in F_2$ such that $x_\infty$ is in the closure of $X$ but is not the limit of any sequence in $X$. (This known fact is fairly easily verified once one has a concrete grasp of $F_2$ as the sequential space of continuous functions from Baire space $\mathbb{Z}^\omega$ to $\mathbb{Z}$.) Define $Y := X \cup \{x_\infty\}$. Then the subspace $Y$ of $F_2$ is (obviously) not sequential, but (by Lemma 6.3 of op. cit.) has a countable pseudobase in (all) the sense(s) of op. cit. Thus, by Theorem 6.10 of op. cit., $Y$ is not $\mathcal{C}$-generated. -Edit. I should have said, Todd's counterexample to the original question also applies to $\mathcal{C}$-generated spaces. The main point of my answer is to show the perhaps more surprising fact that "convenience" (i.e. cartesian closedness) is not even compatible with regular subobjects being subspaces. (This does not conflict with the characterisation of subspaces as regular subobjects in $\mathbf{Top}$, stated by Todd, because equalizers in the category of $\mathcal{C}$-generated spaces are computed differently from in $\mathbf{Top}$.)<|endoftext|> -TITLE: Is $Sym^g$ of a Riemann Surface of genus $g$ Calabi-Yau? -QUESTION [11 upvotes]: The $g$-fold symmetric product of a Riemann surface of genus $g$ naturally has both a symplectic structure as well as a complex structure. Is it in fact Calabi-Yau? If so, is anything known about a mirror for it (in the sense of mirror symmetry)? -My motivation for this comes from Mirror Symmetry and Heegaard-Floer Homology (neither of which I've done any serious work in so the following might be complete garbage). Heegaard-Floer homology is essentially the study of the Floer homology of various special Lagrangian tori in $Sym^g S$ for $S$ a surface of genus $g$. If $Sym^g S$ admits any sort of sensible mirror, under HMS one should be able to extract a lot of information about the Heegaard-Floer homology of 3-manifolds admitting a genus $g$ Heegaard splitting by looking at morphisms of sheaves on the mirror side. -So modulo the CYness of $Sym^g S$ a related question is what work has been done in this direction? - -REPLY [16 votes]: This is not a Calabi-Yau if $g\ne 1$ (for any definition of Calabi-Yau). Indeed, there is a degree one map from the symmetric power of the curve to a torus of dimension $g$. Pull-back of the volume form on the torus to $Sym^gS$ will have zeros at the set where the differential of the map is degenerate, this set is non-empty if $g\ne 1$.<|endoftext|> -TITLE: How many normal subgroups a primitive group can have? -QUESTION [5 upvotes]: Let $G$ be a primitive permutation group of degree $n$, that is $G$ acts transitively and faithfully on a set consisting of $n$ elements and $G$ preserves no nontrivial partition of $X$. In a sense primitive groups are the 'simple' permutation groups. -For example one can show that a primitive group has at most 2 minimal normal subgroups. this can be regarded as a 'width' result. -My question is about the 'depth', namely: -Let $G$ be a primitive permutation group of degree $n$, Let $1\lneq G_m\lneq \cdots\lneq G_0=G$ be a strictly decreasing sequence of normal subgroups of $G$. Is there a good bound for $m$ in terms of $n$? -A stupid bound is $m\leq \log_2(n!)$. My guts feeling is that $m$ should be at most $n-1$, because it feels that every subgroup in the sequence should contribute at least one element to the orbit of a point. -Edit: I ran a computer check in the meanwhile up to degree $n=50$, and it turns out that $m$ is bounded by $n-1$ in this region. Moreover $m$ grows much slower than $n$, e.g., it never exceeded $7$. - -REPLY [6 votes]: I have been trying to find results in the literature on this topic, but finding it frustrating. The difficulty is that estimates on composition and chief series lengths of various types of groups are typically studied not as end in themselves, but as a means to proving other results, such as bounding the minimal generator numbers of groups. -But it is certainly possible to prove a bound that is linear in $n$ for general permutation groups of degree $n$ and logarithmic in $n$ for primitive permutation groups. It is harder to get sensible estimates of the constants involved. -It is proved in: -Cameron, P. J.; Solomon, R. G.; Turull, A. -Chains of subgroups in symmetric groups. -J. Algebra 127 (1989) 340--352. -that any chain of subgroups in $S_n$ is less than $3n/2$. -To study primitive permutation groups, you can use the O'Nan-Scott Theorem. For the affine case, the group $G$ is a semidirect product of elementary abelian $n=p^d$ with an irreducible subgroup of ${\rm GL}(d,p)$. In Theorem 2.3 of -Lucchini, A.; Menegazzo, F.; Morigi, M. -On the number of generators and composition length of finite linear groups. -J. Algebra 243 (2001) 427--447. -a bound is established on the composition length of a completely reducible subgroup of ${\rm GL}(d,p)$ that is linear in $d$ and logarithmic in $p$, so we get a bound on the composition length of $G$ that is logarithmic in $n$. -For the other cases of the O'Nan-Scott Theorem, the socle of $G$ is a direct product of isomorphic nonabelian simple groups, and the number of direct factors is logarithmic in $n$, and we can use the Cameron et al result to bound the composition length of the section of $G$ that permutes these factors. You still have to estimate the composition factors involving the automorphsim groups of these simple factors, but they cannot be very large compared with $n$, so it is possible to get the logarithmic bound.<|endoftext|> -TITLE: Is there always a toric isomorphism between isomorphic toric varieties? -QUESTION [25 upvotes]: Suppose two toric varieties are isomorphic as abstract varieties. Does it follow that there exists a toric isomorphism between them? - -Edit: the comments below lead me to believe that I'm using the terms "toric variety" and "toric morphism" in a non-standard way, so let me clarify. Here are the definitions I have in mind. -Definition: A toric variety is a normal variety $X$ together with (1) a dense open subvariety $T\subseteq X$, and (2) a group structure on $T$ making it a torus, such that the action of $T$ on itself extends to an action on $X$. (Note: I do not include the data of an isomorphism between $T$ and $\mathbb G_m^{\dim X}$, but only require that such an isomorphism exists.) -Definition: If $(X,T_X)$ and $(Y,T_Y)$ are toric varieties (group structures on $T_X$ and $T_Y$ implicit), a toric morphism between them is a morphism $f:X\to Y$ which restricts to a group homomorphism $f|_{T_X}:T_X\to T_Y$. - -REPLY [10 votes]: I don't know if you are still interested in a reference for this, but I recently found that this is proved in section 4 of "Lifting of morphisms to quotient presentations" by F. Berchtold. The proof uses the Cox construction and it seems to me that the argument is the one that David suggested above, but I am not done reading the paper. I hope this helps.<|endoftext|> -TITLE: Groups of order $n$ with a character whose degree is at least $0.8\sqrt{n}$ (say) -QUESTION [20 upvotes]: [edited in response to some corrections by Geoff Robinson and F. Ladisch] -Throughout, all my groups are finite, and all my representations are over the complex numbers. -If $G$ is a group and $\chi$ is an irrreducible character on it, of degree $d$, say that $\chi$ has quite large degree if $d\geq (4/5)|G|^{1/2}$. -This condition has arisen in some work I am doing concerning (Banach algebras built on) restricted direct products of finite groups, and while our main concern was to find some examples with this property, I am curious to know if anything more can be said. -(I've had a quick look at the papers of Snyder (Proc AMS, 2008) and Durfee & Jensen (J. Alg, 2011), but these don't seem to quite give what I'm after. However, I may well have missed something in their remarks.) -Examples. -The affine group over a finite field with $q$ elements has order $q(q-1)$ and a character of degree $q-1$, which has "quite large degree" for $q\geq 3$. The affine group of the ring ${\mathbb Z}/p^n{\mathbb Z}$ has order $p^{2n-1}(p-1)$ and a character of degree $p^{n-1}(p-1)$, which has quite large degree for $p\geq 3$. -Being fairly inexperienced in the world of finite groups, I can't think of other examples (except by taking finite products of some of these examples). -Some simple-minded observations from a bear of little brain. -A group can have at most one character of quite large degree (this is immediate from $|G|=\sum_\pi d_\pi^2$). If such a character $\chi$ exists, it is real (hence rational) valued (edit: as pointed out by Geoff Robinson in comments, $\chi$ must be equal to its Galois conjugates, and hence rational valued), and its centre is trivial (consider its $\ell^2$-norm). -In particular $G$ can't be nilpotent. Moreover, since $\chi\phi=\chi$ for any linear character $\phi$, a bit of thought shows that $\chi$ vanishes outside the derived subgroup of $G$. -Now, let ${\mathcal C}$ be the class of all groups $G$ that possess a character of quite large degree clearly this is closed under taking finite products. -Let ${\mathcal S}$ denote the class of all solvable groups. -Question 1. -Does there exist a finite collection ${\mathcal F}$ of groups such that every group in ${\mathcal C}$ is the product of groups in ${\mathcal F}\cup{\mathcal S}$? -Question 2. Can we bound the derived length of groups in ${\mathcal C}\cap{\mathcal S}$? -Question 3. If the answers to Q1 and Q2 are negative, or beyond current technology, would anything improve if we replaced $4/5$ with $1-\epsilon$ for one's favourite small $\epsilon$? -Question 4. It may well be the case that hoping for a description of the class ${\mathcal C}$ in terms of familiar kinds of group is far too naive and optimistic. If so, could someone please give me some indications as to why? Perhaps it reduces to a set of known open problems? - -REPLY [3 votes]: The following is a list of all groups whose order is $\leq 1000$ and not equal to -one of $384, 768, 864, 896$ or $960$ which satisfy the condition in the title -of the question -- for each group we give its GAP ID number, the largest -character degree, its quotient by the square root of the group order and a -structure description of the group: - -$[ 1, 1 ]$, $1$, $1.00000$, $1$ -$[ 6, 1 ]$, $2$, $0.81650$, ${\rm S}_3$ -$[ 12, 3 ]$, $3$, $0.86603$, ${\rm A}_4$ -$[ 20, 3 ]$, $4$, $0.89443$, ${\rm C}_5 \rtimes {\rm C}_4$ -$[ 42, 1 ]$, $6$, $0.92582$, $({\rm C}_7 \rtimes {\rm C}_3) \rtimes {\rm C}_2$ -$[ 54, 5 ]$, $6$, $0.81650$, $(({\rm C}_3 \times {\rm C}_3) \rtimes {\rm C}_3) \rtimes {\rm C}_2$ -$[ 54, 6 ]$, $6$, $0.81650$, $({\rm C}_9 \rtimes {\rm C}_3) \rtimes {\rm C}_2$ -$[ 56, 11 ]$, $7$, $0.93541$, $({\rm C}_2 \times {\rm C}_2 \times {\rm C}_2) \rtimes {\rm C}_7$ -$[ 72, 39 ]$, $8$, $0.94281$, $({\rm C}_3 \times {\rm C}_3) \rtimes {\rm C}_8$ -$[ 72, 41 ]$, $8$, $0.94281$, $({\rm C}_3 \times {\rm C}_3) \rtimes {\rm Q}_8$ -$[ 110, 1 ]$, $10$, $0.95346$, $({\rm C}_{11} \rtimes {\rm C}_5) \rtimes {\rm C}_2$ -$[ 156, 7 ]$, $12$, $0.96077$, $({\rm C}_{13} \rtimes {\rm C}_4) \rtimes {\rm C}_3$ -$[ 192, 184 ]$, $12$, $0.86603$, $(({\rm C}_2 \times {\rm C}_2 \times {\rm C}_2 \times {\rm C}_2) \rtimes {\rm C}_3) \rtimes {\rm C}_4$ -$[ 192, 185 ]$, $12$, $0.86603$, $(({\rm C}_4 \times {\rm C}_4) \rtimes {\rm C}_3) \rtimes {\rm C}_4$ -$[ 192, 1008 ]$, $12$, $0.86603$, $((({\rm C}_4 \times {\rm C}_4) \rtimes {\rm C}_3) \rtimes {\rm C}_2) \rtimes {\rm C}_2$ -$[ 192, 1009 ]$, $12$, $0.86603$, $((({\rm C}_2 \times {\rm C}_2 \times {\rm C}_2 \times {\rm C}_2) \rtimes {\rm C}_3) \rtimes {\rm C}_2) \rtimes {\rm C}_2$ -$[ 192, 1023 ]$, $12$, $0.86603$, $((({\rm C}_4 \times {\rm C}_4) \rtimes {\rm C}_2) \rtimes {\rm C}_2) \rtimes {\rm C}_3$ -$[ 192, 1025 ]$, $12$, $0.86603$, $(({\rm C}_2 \times {\rm C}_2) . ({\rm C}_2 \times {\rm C}_2 \times {\rm C}_2 \times {\rm C}_2)) \rtimes {\rm C}_3$ -$[ 240, 191 ]$, $15$, $0.96824$, $(({\rm C}_2 \times {\rm C}_2 \times {\rm C}_2 \times {\rm C}_2) \rtimes {\rm C}_5) \rtimes {\rm C}_3$ -$[ 272, 50 ]$, $16$, $0.97014$, ${\rm C}_{17} \rtimes {\rm C}_{16}$ -$[ 342, 7 ]$, $18$, $0.97333$, $({\rm C}_{19} \rtimes {\rm C}_9) \rtimes {\rm C}_2$ -$[ 486, 31 ]$, $18$, $0.81650$, $({\rm C}_{27} \rtimes {\rm C}_9) \rtimes {\rm C}_2$ -$[ 486, 39 ]$, $18$, $0.81650$, $(({\rm C}_9 \rtimes {\rm C}_9) \rtimes {\rm C}_3) \rtimes {\rm C}_2$ -$[ 486, 40 ]$, $18$, $0.81650$, $({\rm C}_3 . (({\rm C}_9 \times {\rm C}_3) \rtimes {\rm C}_3)) \rtimes {\rm C}_2$ -$[ 486, 41 ]$, $18$, $0.81650$, $(({\rm C}_9 \rtimes {\rm C}_9) \rtimes {\rm C}_3) \rtimes {\rm C}_2$ -$[ 486, 127 ]$, $18$, $0.81650$, $(({\rm C}_3 \times ({\rm C}_9 \rtimes {\rm C}_3)) \rtimes {\rm C}_3) \rtimes {\rm C}_2$ -$[ 486, 129 ]$, $18$, $0.81650$, $(({\rm C}_3 \times ({\rm C}_9 \rtimes {\rm C}_3)) \rtimes {\rm C}_3) \rtimes {\rm C}_2$ -$[ 486, 131 ]$, $18$, $0.81650$, $(({\rm C}_3 \times (({\rm C}_3 \times {\rm C}_3) \rtimes {\rm C}_3)) \rtimes {\rm C}_3) \rtimes {\rm C}_2$ -$[ 486, 176 ]$, $18$, $0.81650$, $(({\rm C}_3 \times ({\rm C}_9 \rtimes {\rm C}_3)) \rtimes {\rm C}_3) \rtimes {\rm C}_2$ -$[ 486, 177 ]$, $18$, $0.81650$, $(({\rm C}_3 \times ({\rm C}_9 \rtimes {\rm C}_3)) \rtimes {\rm C}_3) \rtimes {\rm C}_2$ -$[ 486, 178 ]$, $18$, $0.81650$, $(({\rm C}_3 \times (({\rm C}_3 \times {\rm C}_3) \rtimes {\rm C}_3)) \rtimes {\rm C}_3) \rtimes {\rm C}_2$ -$[ 486, 179 ]$, $18$, $0.81650$, $(({\rm C}_3 . (({\rm C}_3 \times {\rm C}_3) \rtimes {\rm C}_3)) \rtimes {\rm C}_3) \rtimes {\rm C}_2$ -$[ 486, 236 ]$, $18$, $0.81650$, $(({\rm C}_3 \times (({\rm C}_3 \times {\rm C}_3) \rtimes {\rm C}_3)) \rtimes {\rm C}_3) \rtimes {\rm C}_2$ -$[ 486, 238 ]$, $18$, $0.81650$, $(({\rm C}_3 \times ({\rm C}_9 \rtimes {\rm C}_3)) \rtimes {\rm C}_3) \rtimes {\rm C}_2$ -$[ 500, 17 ]$, $20$, $0.89443$, $(({\rm C}_5 \times {\rm C}_5) \rtimes {\rm C}_5) \rtimes {\rm C}_4$ -$[ 500, 21 ]$, $20$, $0.89443$, $(({\rm C}_5 \times {\rm C}_5) \rtimes {\rm C}_5) \rtimes {\rm C}_4$ -$[ 500, 18 ]$, $20$, $0.89443$, $({\rm C}_{25} \rtimes {\rm C}_5) \rtimes {\rm C}_4$ -$[ 504, 160 ]$, $18$, $0.80178$, $(({\rm C}_7 \rtimes {\rm C}_3) \times {\rm A}_4) \rtimes {\rm C}_2$ -$[ 504, 161 ]$, $18$, $0.80178$, ${\rm A}_4 \times (({\rm C}_7 \rtimes {\rm C}_3) \rtimes {\rm C}_2)$ -$[ 506, 1 ]$, $22$, $0.97802$, $({\rm C}_{23} \rtimes {\rm C}_{11}) \rtimes {\rm C}_2$ -$[ 600, 148 ]$, $24$, $0.97980$, $(({\rm C}_5 \times {\rm C}_5) \rtimes {\rm C}_3) \rtimes {\rm C}_8$ -$[ 600, 149 ]$, $24$, $0.97980$, $(({\rm C}_5 \times {\rm C}_5) \rtimes {\rm C}_8) \rtimes {\rm C}_3$ -$[ 600, 150 ]$, $24$, $0.97980$, $(({\rm C}_5 \times {\rm C}_5) \rtimes {\rm Q}_8) \rtimes {\rm C}_3$ -$[ 672, 1257 ]$, $21$, $0.81009$, $({\rm C}_2 \times {\rm C}_2 \times (({\rm C}_2 \times {\rm C}_2 \times {\rm C}_2) \rtimes {\rm C}_7)) \rtimes {\rm C}_3$ -$[ 672, 1258 ]$, $21$, $0.81009$, $(({\rm C}_2 \times {\rm C}_2 \times {\rm C}_2) \rtimes {\rm C}_7) \times {\rm A}_4$ -$[ 702, 47 ]$, $26$, $0.98131$, $(({\rm C}_3 \times {\rm C}_3 \times {\rm C}_3) \rtimes {\rm C}_{13}) \rtimes {\rm C}_2$ -$[ 812, 7 ]$, $28$, $0.98261$, $({\rm C}_{29} \rtimes {\rm C}_7) \rtimes {\rm C}_4$ -$[ 840, 139 ]$, $24$, $0.82808$, $({\rm C}_5 \rtimes {\rm C}_4) \times (({\rm C}_7 \rtimes {\rm C}_3) \rtimes {\rm C}_2)$ -$[ 930, 1 ]$, $30$, $0.98374$, $(({\rm C}_{31} \rtimes {\rm C}_5) \rtimes {\rm C}_3) \rtimes {\rm C}_2$ -$[ 992, 194 ]$, $31$, $0.98425$, $({\rm C}_2 \times {\rm C}_2 \times {\rm C}_2 \times {\rm C}_2 \times {\rm C}_2) \rtimes {\rm C}_{31}$<|endoftext|> -TITLE: Magma "actions" (or alternatively, "What is the Yoneda lemma for magmas?") -QUESTION [9 upvotes]: Arguably the most import thing about groups, semigroups and more generally categories, is that they can act on sets (or even collections of sets in the case of a category). This is the basis for all the major applications of these structures that I can think of (for example, interpreting groups as symmetries, using Galois groups to classify extensions, semigroups to classify automata, etc.), and it is also the most basic hook one can use to get intuition about these structures. Of course the reason that these structures have non-trivial actions follows from the associativity axiom which leads to Cayley's theorem/Yoneda's lemma. -As a result, associativity is a necessary and sufficient condition for a binary algebraic structure to `act' like a collection of functional relations on some set (or between many sets in the case of categories). If one believes that functions are the end-all-be-all of mathematics, then the story is over and the situation is hopeless. However, I believe (perhaps naively) that functions are just one kind of relation -- granted a very important one! -- and that there are many other useful things out there. -One alternate picture of magmas that I find quite amusing, is to think of them as forests of binary trees with labelled leaves, together with some rewriting rules that can be applied to subtrees. This type of structure shows up quite frequently in programming languages (for example when implementing an interpreter). As a result, I wonder if there is some canonical way to think of a magma as something like a Turing complete language. -As an example, it is easy to translate the lambda calculus into a magma; let $\Sigma = ${$ x_0, x_1, ... x_n $} be a collection of variables and let $\lambda \Sigma = ${$ \lambda x_0 , .... \lambda x_n $} be a second collection of distinct symbols denoting name binding. Then a lambda expression is determined by the free magma generated by $\Sigma$, $\lambda \Sigma$ subject to the relations: -$(\lambda x_i M) N = M [ x_i \mapsto N ]$ -Where the $[ x_i \mapsto N ]$ denotes the capture avoiding subsitution rule. -Now this tells something about how a magma interacts with itself (specifically that it can do pretty much anything it wants), but what I am searching for is some way to interpret a magma interacting with some external set. Is there a general mechanism that would allow one to think of a magma as a relation with some extra structure in the same sense that a semigroup is like a transitive relation, a monoid like a preorder and a group like an equivalence relation? - -REPLY [5 votes]: An action $S$ of a magma $M$ is a function $M \times S \to S$ satisfying no extra conditions. (If you imposed any conditions then $M$ wouldn't act on itself.) This lets you write down trees where some of the nodes are labeled by elements of $S$ rather than elements of $M$.<|endoftext|> -TITLE: Algorithm to calculate $Wh(G)$ for finitely presented group $G$? -QUESTION [8 upvotes]: Let $G$ be a finitely presented group. -Are there any algorithm to calculate whitehead group $G$, $Wh(G)$ in terms of presentation of $G$? - -REPLY [6 votes]: There is no algorithm to decide from a finite presentation of a group $G$ whether $\mathrm{Wh}(G)$ is zero. See Corollary 5.7 of this paper which studies various computability issues in higher-dimensional topology. -I think the original question that asks whether one can compute $\mathrm{Wh}(G)$ from a presentation of $G$ does not quite make sense because $\mathrm{Wh}(G)$ need not be finitely generated, e.g. Example on page 2 of this paper gives a virtually $\mathbb Z^3$-group whose Whitehead group is infinitely generated. -For finite $G$, the original question does make sense because the group $\mathrm{Wh}(G)$ is finitely generated (thanks to a theorem of Bass; see again Oliver's book mentioned in comments), and abelian (Whitehead groups are abelian by definition), and isomorphism problem is solvable for finitely generated ableian groups. However, the paper mentioned in the first paragraph above shows that even for finite groups there is no algorithm. On the other hand, a lot is known about Whitehead groups of finite groups, see again Oliver's book.<|endoftext|> -TITLE: What is the replacement for a "sufficiently small disc" in characteristic p? -QUESTION [7 upvotes]: How do I make the following sort of argument work in characteristic p? - -Let $f:X \to Y$ be a proper equidimensional map of smooth algebraic varieties, assume all fibres are reduced. Say at some point $y \in Y$, I have computed the differential and know that $df(T_x X) \supset V$ for all $x \in X_y$ and some vector subspace $V \subset T_y Y$. Then over a sufficiently small polydisc $D$ such that $T_y D + V = T_y Y$, the total space of $X_D$ is smooth. Thus the same is true of a (say) one-dimensional thickening $\tilde{D}$ of $D$, and so moving $D$ a little bit in $\tilde{D}$ produces many deformation equivalent smooth manifolds (with boundary) $X_{D'}$. -In particular if I want to know something about $H^*(X_y)$, I can first thicken to the smooth $X_D$ which is homotopy equivalent to it, then move $X_D$ to the diffeomorphic $X_{D'}$, which I prefer because say $D'$ avoids some bad points in $Y$. - -REPLY [10 votes]: I think it would be difficult to give a general result that covers everything you want and can do but there is a collection of techniques (maybe better described as a dictionary) that works in many cases. - -A polydisc should be replaced by the strict Henselisation of some (smooth) subvariety $T$ passing through $y$. The fact that $X_y$ is homotopy equivalent to $X_D$ is then replaced by the proper base change theorem. -The transversality property has a direct analogue in that one may arrange that $T$ is transversal to $f$ and hence that $X_T$ is smooth. -Deformation of $D$ is replaced by choosing, locally on $Y$, a smooth map $Y\to Z$ such that $T$ is one of the fibres. To get closer to the topological situation one should probably also assume that a section has been chosen. Deformations correspond to fibres over other other points of $Z$ (or rather the Henselisation at the corresponding point of the section). -To compare (the analogues of) $X_D$ and $X_{D'}$ one probably also would need the smooth base change theorem together with the proper base change theorem. - -As I said to see how this works in any particular situation one would need to have more details on it, no a priori success is guaranteed.<|endoftext|> -TITLE: Basic questions about stacks 2 -QUESTION [7 upvotes]: I have again three basic questions about stacks. -1) When we consider categories fibered in groupoids, do we always mean small or essentially small groupoids? Especially I want to know if algebraic geometers always impose this condition when they talk about categories fibered in groupoids, especially stacks. -2) In the proof of Artin's criterion for algebraic spaces/stacks $X/S$ for every point $p \in X$ of finite type over $S$ a "local approximation" $X_p$ is constructed. Then $X = \coprod_p X_p$ does the job. But in order to show that this is actually a scheme in the given universe, we need that the points of finite type constitute a set. Perhaps I'm overlooking something trivial here, but I cannot see how we can use Artin's criterions to deduce this. -3) What is the current status of the book "Algebraic Stacks" by Kai Behrend, Brian Conrad, Dan Edidin, William Fulton, Barbara Fantechi, Lothar Göttsche und Andrew Kresch? I would love to read it as soon as it is completed. - -REPLY [9 votes]: Regarding 3), Andrew Kresch just told me that they gave up on the project.<|endoftext|> -TITLE: number of galois extensions of local fields of fixed degree -QUESTION [5 upvotes]: Let $K$ be a local field (of characteristic 0) with (finite) residue field of characteristic $l$ and let $p$ be a prime. -Considering the cases, whether the $p$-th roots of unity are in $K$ and whether $l$ equals $p$ (and maybe whether $p=2$) or not, my question is: -How many Galois extensions of $K$ of degree $p$ exist? - -REPLY [10 votes]: I am sorry if I see this question only now, but since no one gave the following answer, it seems worth posting it. -There is a general formula for the number of extensions of degree $d$ of a $p$-adic field $K$ contained inside a fixed algebraic closure $\overline{K}$, which is given by -$$ -\# \{ L \mid K \subseteq L \subseteq \overline{K}, \, [L \colon K] = d \} = \sigma(h) \cdot \sum_{j = 0}^m \frac{p^{m+j+1} - p^{2j}}{p - 1} \cdot (p^{\varepsilon_p(j) \cdot d \cdot d_0} - p^{\varepsilon_p(j - 1) \cdot d \cdot d_0}) -$$ -where: - -$\sigma$ denotes the sum of divisors function; -$h, m \in \mathbb{N}$ are the unique natural numbers such that $p \nmid h$ and $d = h \cdot p^m$; -$\varepsilon_p(j) := \sum_{k=1}^j p^{-k}$ if $j \geq 1$, $\varepsilon_p(0) := 0$ and $\varepsilon_p(-1) := -\infty$, i.e. $p^{\varepsilon_p(-1) \cdot d} = 0$. In particular, $p^{\varepsilon_p(j) \cdot d} \in \mathbb{N}$ if $-1 \leq j \leq m$; -$d_0 := [K \colon \mathbb{Q}_p]$. - -This formula is due to Krasner, and has been proved in the paper "Nombre des extensions d'un degré donné d'un corps $\mathfrak{p}$-adique". The proof uses the same analytic techniques that go into the proof of the (much more famous) Krasner lemma. -Observe that this number is different from the number of $K$-isomorphism classes of extensions of $K$ having a given degree. This is of course due to the presence of non-Galois extensions. Here are two examples of this phenomenon: - -if $q \neq p$ is a prime then there are $q + 1$ fields $K \subseteq L \subseteq \overline{K}$ having degree $[L \colon K] = q$, but there are only two isomorphism classes of these fields: one containing the only unramified extension, and the other containing the tamely and totally ramified extensions; -if $p \geq 3$ there are $1 + p + (p^2 - p) \cdot p$ extensions $\mathbb{Q}_p \subseteq L \subseteq \overline{\mathbb{Q}_p}$ such that $[L \colon \mathbb{Q}_p] = p$, but they form $1 + p + p^2 - p = p^2 + 1$ isomorphism classes. $p + 1$ of these contain a unique extension (which is Galois over $\mathbb{Q}_p$) and every other isomorphism class contains $p$ extensions (see for example Proposition 2.3.1 in the paper "A database of local fields" by Jones and Roberts). - -Finally, let me remark that this formula is related to Serre's "mass formula", which is valid in any characteristic. This formula says that a certain "count" of totally ramified extensions of a local, non-Archimedean field $K$ of degree $d$ is equal to $d$. More precisely, -$$ -\sum_{L \in \Sigma_d} (\# \kappa)^{d - 1 - \mathrm{v}_K(\mathrm{disc}(L/K))} = d -$$ -where $\Sigma_d$ denotes the set of totally ramified extensions of $K$ which have degree $d$, and $\kappa$ is the residue field of $K$. Observe that if $p \nmid d$ then the formula can be written simply as $\# \Sigma_d = d$. -Two useful references for this are: - -Serre's original paper "Une 'formule de masse' pour les extensions totalement ramifiées de degré donné d'un corps local"; -Krasner's paper "Remarques au sujet d'une note de J.-P. Serre...", in which he reproves the formula using his methods.<|endoftext|> -TITLE: Some questions on moduli of stable maps -QUESTION [7 upvotes]: Let $\overline{M}_{0,k}(\mathbb{P}^n,d)$ -denote the moduli space of genus zero degree $d$ stable maps with $k$ marked points. This is an orbifold of expected dimension. Let $\overline{U}_{0,k}(\mathbb{P}^n,d)$ be the corresponding universal curve. Then we have two morphisms -$$\pi_1: \overline{U}_{0,k}(\mathbb{P}^n,d) \rightarrow \overline{M}_{0,k}(\mathbb{P}^n,d)$$ -and -$$\pi_2: \overline{U}_{0,k}(\mathbb{P}^n,d) \rightarrow \mathbb{P}^n$$ -So if we have a coherent sheaf $F$ on $\mathbb{P}^n$ we can pull it back to universal curve and push it forward to moduli space, i.e. we can consider $F_i=(R^i\pi_{1*})(\pi_2^*F)$ on moduli space. -Here are my questions: -1) Are $F_i$ zero for $i>0$ ? (is $\pi_{1*}$ exact ) -2) If $F$ is equal to ideal sheaf of some smooth projective subvariety $X$, and $\beta \in H_2(X)$ is of degree $d$, then is there in general any relation between some component of support of $F_0$ and the moduli space $\overline{M}_{0,k}(X,\beta)$ ? -3) if the answer to question one is No, what is the interpretation of $F_i$, $i>0$, for the case when $F$ is the ideal sheaf of $X$ as in question two? - -REPLY [2 votes]: 1) It depends on $F$. Roughly speaking fibers of $\pi_1$ are the curves parametrized by $\overline{M}_{0,k}(P^n,d)$. So if you want $R^{>0}\pi_{1*}\pi_2^*F$ to vanish, you have to check that $H^{>0}$ of the restriction of $F$ to any such curve vanishes. Since all the curves are rational it suffices (but not necessary) to assume that $F$ is generated by global section. -2,3) It seems that you want something else. The usual approach is the following. Assume that $X$ is the zero locus of a regular section $s$ of a vector bundle $F$. Then $s$ gives a global section of $\pi_2^*F$, and hence of $\pi_{1*}\pi_2^*F$. The claim is that (the virtual fundamental class of) $\overline{M}_{0,k}(X,\beta)$ is the zero locus of this section of $\pi_{1*}\pi_2^*F$ on $\overline{M}_{0,k}(P^n,d)$.<|endoftext|> -TITLE: Hodge diamond of a Calabi-Yau fourfold -QUESTION [6 upvotes]: I am trying to compute the Hodge diamond of a Calabi-Yau fourfold which is a complete intersection inside of a projective bundle over a threefold base. I have computed the arithmetic genera $\chi_1$ and $\chi_2$ which give me two linear relations between the four independent Hodge numbers $h^{(1,1)}$, $h^{(1,2)}$, $h^{(1,3)}$ and $h^{(2,2)}$. So if I can compute two of them the arithmetic genera will give me the other two. Since my variety is a Calabi-Yau fourfold, $h^{(1,1)}$ is the second Betti number and $h^{(1,2)}$ is twice the third Betti number, so really all I need are the second and third Betti numbers to finish the calculation. As such, I wanted to use the Lefschetz hyperplane theorem to relate the cohomology of my fourfold to the cohomology of the ambient projective bundle, but it turns out that the divisors for which my fourfold is a complete intersection of are not ample (they are only relatively ample). So does anyone have any ideas on how to relate the cohomology of my variety to the cohomology of the projective bundle if the divisors for which my variety is a complete intersection of are only relatively ample? - -REPLY [10 votes]: I encountered a similar problem a few years ago, but then in dimension 3. In that case a master student wanted to calculate the hodge diamond of a threefold which was a hypersurface $W$ in a $P^2$-bundle over a del Pezzo surface. This threefold was birational to a singular hypersurface $Y$ in some weighted projective 4-space. -Now you can apply Lefschetz' hyperplane theorem for $Y$, but it only works for the lower cohomology groups, since Poincar\'e duality might fail. There are methods to calculate the higher cohomology groups if the singularities of $Y$ are nice enough. -Then you consider a factorization of the birational map $W\dashrightarrow Y$, calculate exceptional divisors etc, in order to figure out the difference between the Hodge numbers of $W$ and of $Y$. -In the end this turned out to be too hard for the student in question, so I worked out some of the details. This particular example is worked out at the end of the first version (on arXiv) of my paper with Klaus Hulek on MW-groups of elliptic threefolds. The referee considered this example superfluous, therefore we omitted this example in the later versions and the published version of this paper.<|endoftext|> -TITLE: Algebra Counterexample Request: Linear Quotients -QUESTION [8 upvotes]: A result of Herzog, Hibi, and Zheng in "Monomial ideals whose powers have a linear resolution" states that: -Theorem: Let $I\subseteq\Bbbk[x_1,\ldots,x_n]$ be a monomial ideal generated in degree 2. The following conditions are equivalent: - -I has a linear resolution. -I has linear quotients. -Each power of I has a linear resolution. - -This, when combined with (a slightly extended version of) Froberg's characterization of degree 2 (squarefree) ideals with a linear resolution, gives a complete description of all degree 2 (monomial) ideals with linear quotients. -However, this still leaves open a characterization of monomial ideals generated in higher degree with linear resolutions/quotients/linear resolutions of their powers. Sturmfels [in "Four Counterexamples in Combinatorial Algebraic Geometry"] and Conca [in "Regularity jumps for powers of ideals"] have a number of examples of ideals $I$ with linear quotients and linear quotients of their powers $I, I^2,\ldots,I^{k-1}$, which fail to have a linear resolution for $I^k$. -The counterexample request then is this: Are there any (monomial) homogeneous ideals which have characteristic independent linear resolutions but no linear quotients under any order of the generators? The theorem above implies that such an example would have to be generated (at least partially) in degree 3 or higher. -Edit: As Professor Conca pointed out, triangulations of the projective plane provide such an example. I was hoping to find a characteristic independent counterexample, but forgot to include this in the description. The request was largely because my research partner A. Hoefel and I were searching for such characteristic independent examples to test a conjecture - but were unable to locate any in the literature (or on a few targeted searches with M2 and GAP.) - -REPLY [7 votes]: If you take square-free monomials ideals and consider the Alexander duality, then linear resolution is dual to Cohen-Macaulay and linear quotient is dual to shellable. Hence you are looking for a non-shellable simplicial complex that is CM in all the characteristic. -Volkmar Welker mentioned to me that the EG models web page (http://www.eg-models.de/) lists many examples of simplicial complexes with unusual or difficult to find proprieties. -There you find what you are looking for. For example, EG Model No. 2003.05.003 is a nonshellable but constructible 2-dimensional simplicial complex.<|endoftext|> -TITLE: Understanding (the wiki page on) Verdier duality -QUESTION [6 upvotes]: My familiarity with concepts related to derived categories is only tangential, and little by little I intend to get more comfortable with them. I was playing around with Caldararu's introduction to the topic, and looking up various things on the web. -Here is my question (that I have every confidence is trivial for experts): -On the wiki page on Verdier duality http://en.wikipedia.org/wiki/Verdier_duality it says the following: Let $F$ be a field, and $X$ a finite dimensional (dimension is defined here cohomologically, but for our purposes a finite dimensional manifold will do) locally compact space. -In the part about Poincare duality, it says: -$H^k(X,F)=[F,X[k]]$. -What is the interpretation of this notation? As I see it, $[F,X[k]]$ means $Hom(F,X[k])$ in the derived category. But this means that $X$ is seen as a complex. How? And why would $Hom(F,X[k])$ equal $H^k(X,F)$? - -REPLY [2 votes]: I just revamped what was written. Perhaps now it is more understandable:New Wiki Entry on Verdier duality.<|endoftext|> -TITLE: A character-free proof that a permutation group is doubly transitive iff the associated permutation module over $\mathbb C$ has irreducible augmentation submodule? -QUESTION [5 upvotes]: Let $G$ be a group of permutations of a finite set $X$. By the augmentation submodule of $\mathbb CX$, I mean the set of vectors whose coefficients sum to $0$. It is easy to show via character theory that the augmentation submodule is irreducible iff $G$ is doubly transitive. Does anybody know a proof using only the definitions of doubly transitive and irreducible representation and avoiding character theory and the orthogonality relations? It is known (I learned this from Peter Cameron, but don't know a good reference) that replacing $\mathbb C$ by $\mathbb R$ characterizes double homogeneity. So the field being algebraically closed is somehow important. -The motivation for this question comes from trying to understand the relationship between double transitivity for transformation monoids and irreducibility of the augmentation submodule. Character theory for monoids is harder to apply and so an answer to my question may provide some insight. - -REPLY [3 votes]: Maybe this is the sort of direct proof you're looking for? -Suppose the action of $G$ is not $2$-transitive. Then $G$ preserves some nontrivial directed graph $\mathcal{G}$ with vertex set $X$ (the $G$-orbit of your favourite edge). Therefore $G$ preserves the eigenspaces of this graph, i.e., the eigenspaces of the adjacency operator defined by -$$\mathcal{A} f (x) = \sum_{y:(x,y)\in\mathcal{G}} f(y).$$ -If the whole augmentation submodule were an eigenspace then $\mathcal{A}$ would act as some scalar $\lambda$, and then it's easy to see that $\lambda$ must be $-1$ and all edges must be in $\mathcal{G}$, a contradiction. Therefore each eigenspace is a proper invariant subspace. -(To be fair, I feel sure that on inspection this proof can be little more than the usual proof, but for dummies. If we insist that $\mathcal{G}$ is the orbit of a single directed edge then $\mathcal{A}$ is closely related to the averaging operator often seen in proofs of complete reducibility, we're using the fact that $G$ and $\mathcal{A}$ commute, and then we're essentially copying the proof of Schur's lemma.)<|endoftext|> -TITLE: Is there an analog of Clifford Theorem in the setting of Lie algebras? -QUESTION [11 upvotes]: A classical theorem of Clifford states that if G is a finite group and K a field, then every irreducible right KG-module is a completely reducible right KN-module, where N is any normal subgroup of G. -Is there a Lie theoretic analog of this result? That is, if L is a finite-dimensional Lie algebra, I an ideal of L, and M an irreducible L-module, is M a completely reducible I-module? I expect the answer is negative, but what about a counterexample? - -REPLY [3 votes]: The positive results seem at most to be tied to finite dimensional representations in positive characteristic: -Let $\frak H$ be the Heisenberg algebra with basis $x,y,$ where $c$ is central and $[y,x]=c$. It acts on polynomials $k[x]$ with $x$ acting by multiplication by $x$ and $y$ acting as $d/dx$ and $c$ acting as the identity. If $k$ is a field of characteristic $0$ this is an irreducible representation. On the other hand, $y$ and $c$ span a normal subalgebra and the restriction of this representation to it is not irreducible, polynomials of degree $\le n$ is a subrepresentation for all $n$. -Similarly, in characteristic $p$ the action passes to $k[x]/(x^p)$ which is a finite dimensional irreducible representation whose restriction to $\langle y,c\rangle$ is not irreducible (for the same reason as in characteristic $0$). -The problem as compared with the group case is that automorphisms must be replaced by derivations. If $\phi$ is an automorphism of a group $N$ and $V$ is a $N$-representation, then we may define a new $N$-action on $V$ given by $n\cdot v:=\phi(n)v$. It is clear that $V$ is irreducible precisely when this new action is and that if $V$ is a subrepresentation of a $G$-representation $W$ and $\phi\in G$, then $\phi(V)$ is isomorphic to this twisted action. The fact that $\phi(V)$ is semi-simple when $V$ is irreducible is the key fact used in order to prove the group case. -In the Lie algebra case we instead have a derivation $D$ of a Lie algebra $\frak n$ and if $V$ as before lies in $W$ we have $xD(v)=[x,D]v+Dxv=D(x)v+Dxv$. Hence, instead of, in the abstract case, defining a twisted action on $V$ we have to define an action on $V\times V$ given by $x(u,v)=(xu+D(x)v,xv)$. This instead is an extension of $V$ by $V$ and may not be even semi-simple when $V$ is irreducible. This is exactly what happens in the two examples above. -Hence, I am not totally convinced that the statement is true even in the finite dimensional characteristic zero case when say the Lie algebra is not algebraic (i.e., not the Lie algebra of an algebraic group). I do not however have a counterexample.<|endoftext|> -TITLE: $N$-step simplicial complexes -QUESTION [8 upvotes]: Recently, answering a question here, Dror Bar-Natan observed that «way too often two-step complexes have a natural extension to become many-step complexes». By such a thing I mean (and I think Dror means too!) a "complex" in which $d^N=0$ and not $d^2=0$, as usual. I like to think of $N$-complexes are modules over the quiver $$\cdots\to\bullet\to\bullet\to\bullet\to\bullet\to\cdots$$ bound by the relations that say that the product of $N$ consecutive arrows is zero. -Many-step complexes have appeared in recent work of Mikhail Kapranov (arXiv:q-alg/9611005, which appears not to be published, according no MathSciNet?) and [Kassel, C.; Wambst, M. Algèbre homologique des N-complexes et homologie de Hochschild aux racines de l’unité. Publ. Res. Inst. Math. Sci. 34, (1998), 91–114], for example, but they certainly predate that: they appear already in [Mayer, W. A new homology theory. I, II. Ann. of Math. 43, (1942). 370–380, 594–605] and in [Spanier, E.H. The Mayer homology theory, Bull. Amer. Math. Soc. 55 (1949), 102- 112], and maybe earlier than that. - -Is there a reasonable analogue for what an $N$-step simplicial complex might be? - -REPLY [3 votes]: Oh, there are many papers recently devoted to $N$-complexes, in particular in connection with theoretical physics. A lot of these papers have simplicial stuff in it. I don't have access to MathSciNet now so some of my suggestions you have to look up yourself as I sometimes don't know the exact titles. -Here goes: - -Tikaradze: "Homological cobstructions on $N$-complexes" Journal of pure and applied algebra (I'm pretty sure that there are some simplical stuff in this.) -Michel Dubois-Violette has a lot of papers that are very relevant. Especially a paper called $d^N=0$ or something to that effect in $K$-theory. -Angel and Diaz: "Differential graded algebras" Journal Pure and Applied Algebra I think. -Connes (Alain, that is) et al has some papers on so called Homogenous algebras and Yang-Mills algebras, you can look up. -Berger and Marconnet: "Koszul and Gorenstein properties of Homogenous algebras" Algebras and representation theory. -Same goes for Richard Kerner and Victor Abramov -And, if I may bang my own drum, I ever so briefly dabbled a bit in this area a few years back: Larsson and Silvestrov: "On $q$-differential graded algebras and $N$-complexes". Nothing very deep though ;) - -Finally, and maybe most importantly, my friend Goro Kato tried to construct a derived category of $N$-complexes but managed to show instead that in reality one gains almost nothing with $N$-complexes instead of ordinary $2$-complexes, at least not from a homological perspective: From an $N$-complex one can construct, essentially in a unique manner, an ordinary $2$-complex. I don't remember the title but MathSciNet should solve that easily.<|endoftext|> -TITLE: Tensor products of permutation representations of symmetric groups. -QUESTION [7 upvotes]: I am looking for a reference for the following fact which must be classical (which makes it harder, for me, to track a reference down). I am interested because there are similar (more complicated) statements about the cohomology of symmetric groups. -If $P$ is a partition, namely $p_{1} + \cdots + p_{k} = n$, we let $\rho_{P}$ denote -the permutation representation of $S_{n}$, induced up from the trivial representation of $S_{P}$. -If $P$ and $Q$ are partitions of $n$ then consider any matrix $\hat{A}$ with nonnegative integer entries such that the entries -of $i$th row of $A$ add up to $p_{i}$ and those of the $j$th column of $A$ add up to $q_{j}$. Then the entries of $\hat{A}$ -form another partition of $n$, which we call $A$ and say that $A$ is a product-refinement of $P$ and $Q$. For example if $P = Q = 1 + 2$ then two possibilities for $\hat{A}$ are $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and $\begin{pmatrix} 0 & 1 \\ 1 & 1 \end{pmatrix}$. -Proposition: If $\rho_{P}$ and $\rho_{Q}$ are permutation representations of $S_n$ then $\rho_{P} \otimes \rho_{Q} \cong \bigoplus_{A} \rho_{A},$ -where the sum is over $A$ which are product-refinements of $P$ and $Q$. -Questions: 1) what is the reference for this fact? and 2) what is standard terminology (for product-refinement in particular)? - -REPLY [8 votes]: Hi Dev, -It looks to me like a proof of this fact is given in the answer to Exercise 7.84(b) of Richard Stanley's Enumerative Combinatorics, volume 2, along with a reference to Example I.7.23(e), page 131, of I. G. Macdonald's Symmetric Functions and Hall Polynomials (2nd edition).<|endoftext|> -TITLE: The odd power of copositive matrix -QUESTION [7 upvotes]: If $A$ is copositive, what about $A^3$? Is it also copositive? More generally, -my question is whether the odd power of a copositive matrix is still copositive. -Any reference is appreciated - -REPLY [11 votes]: My preliminary experiments show that the answer is no. Here is why. -In the paper Constructing copositive matrices from interior matrices, the following matrix (from Horn's quadratic form) is mentioned to be copositive: -$$ -A=\begin{bmatrix} - 1 & -1 & 1 & 1 & -1\\\\ - -1 & 1 & -1 & 1 & 1\\\\ - 1 & -1 & 1 & -1 & 1\\\\ - 1 & 1 & -1 & 1 & -1\\\\ - -1 & 1 & 1 & -1 & 1\\\\ -\end{bmatrix} -$$ -Now plug this matrix into Matlab, and search to see if $x^TA^3x < 0$ occurs for a nonnegative $x$. Here is one particular value: -$$x = \begin{pmatrix} -4\\\\ -6\\\\ -3\\\\ -0\\\\ -0\\\\ -\end{pmatrix} -$$ -because with this choice of $x$, we obtain $x^TA^3x = -11$<|endoftext|> -TITLE: Counterexamples in PDE -QUESTION [71 upvotes]: Let us compile a list of counterexamples in PDE, similar in spirit to the books Counterexamples in topology and Counterexamples in analysis. Eventually I plan to type up the examples with their detailed derivations. -Please give one example per answer, preferably with clear descriptions and pointers to literature. -A related question: - -REPLY [2 votes]: In this paper of Caicedo and Castro https://www.aimsciences.org/journals/displayArticles.jsp?paperID=4083 they prove that for the seminilinear wave equation -$$\square u + \lambda u + h(u) = c\sin(x+t)$$ subject to double periodic conditions $u(x,t)=u(x,t+2\pi)=u(x+2\pi)$ there is no continuous solutions for |c| large enough. Here $h$ could be any continuous function with compact support and $-\lambda\notin\sigma(\square)$. It is not very hard to prove the existence of weak solutions. The thing here is that the data $c\sin(x+t)$ is smooth (actually analytic) but there is no regularity at all.<|endoftext|> -TITLE: More questions about Verdier duality (and related math) -QUESTION [5 upvotes]: The first set of questions can be found here: Understanding (the wiki page on) Verdier duality -I'm fairly confident that I understand something wrong, so I'll write down here clearly what my set of beliefs is about what is right, and you feel free to shoot down any falsehood: -Let $X$ be our geometric object, be it a topological space, variety, scheme, or what have you. I will do two cases, one Poincare duality, and the other Serre duality. -Serre duality -I will assume $X$ is nice (a variety, projective, smooth,... of dimension $n$). Here I will look at the (abelian) category of coherent $O_X$-modules. In this case the "dualizing module" is $\omega_X[n]$. I take that to mean that $[\mathcal{E},\mathcal{F}]$ is dual to $[\mathcal{F},\mathcal{E}\otimes \omega_X[n]]$. -I also believe $H^k(X,\mathcal{F})\cong [O_X,\mathcal{F}[k]]$. Going from there, the rest is easy: $H^k(X,\mathcal{F})\cong [O_X,\mathcal{F}[k]]\cong [$dual of $\mathcal{F},O_X[k]]$, which is dual to $[O_X[k],\omega_X[n]\otimes$ dual of $\mathcal{F}]\cong [O_X, \omega_X \otimes$ dual of $\mathcal{F}[n-k]] \cong H^{n-k}(X,\omega_X \otimes $ dual of $\mathcal{F})$. -Great! However... -Poincare duality -Again assume $X$ is nice (orientable compact smooth manifold of dimension $n$). Let $K$ be field. Here I will look at the (abelian) category of $K$-vector spaces. Here the "dualizing sheaf" is $K[-n]$. I will continue somewhat similarly to the Serre duality case. I interpret the dualizing sheaf as meaning that $[\mathcal{E},\mathcal{F}]$ is dual to $[\mathcal{F},\mathcal{E}\otimes K[-n]]$ (here $\mathcal{E}$ and $\mathcal{F}$ are $K$-vector spaces). -For whatever reason, I believe that $H_k(X,\mathcal{E})\cong[\mathcal{E}[-k],D_X]$ ($D_X$ is the dualizing sheaf in general. But what does this even mean in general? For example in the Serre duality case, what would $H_k(X,\mathcal{F})$ even mean?). So: -$H_k(X,K) \cong [K[-k],K[-n]] \cong [K,K[k-n]] \cong H^{k-n}(X,K)$. Wait. What? Makes no sense! -If we use the "duality" it still makes no sense: -$[K,K[k-n]] \cong [K[k-n],K[-n]] \cong [K,K[-k]]\cong H^{-k}(X,K)$. What? Huh? -So in conclusion, I desperately want to have a handle on this, but I clearly don't. Hopefully just a nudge in the right direction would lead me to a better understanding of this yoga. - -REPLY [2 votes]: I'm not sure if this will answer your question or not, but let $\mathbb{D}$ be the Verdier dualizing sheaf on the locally compact space $X$. If $M$ is a manifold, then $\mathbb{D}[-n]$, where $n$ is the dimension of $M$, is isomorphic in the derived category to the orientation sheaf $\omega$ on $M$ (see, for example, Borel's book on interesection cohomology, I think around section V.7). The Borel-Moore homology of $X$, which is equivalent to the ordinary homology of $X$ if $X$ is compact (otherwise it's the homology theory built from locally-finite chains) is defined to be $H^{BM}_{k}(X)=H^{-k}(X; \mathbb{D})$. If you unwind all that indexing business, it says that, for a compact manifold, $H_k(M)=H^{n-k}(M; \omega)$, which is probably the duality statement you're looking for. -Of course showing that this has anything to do with classical Poincare duality (say via the cap product) is pretty far from obvious. This is part of the content of a paper I currently have in preparation with Jim McClure.<|endoftext|> -TITLE: Groups with triple system of self-normalizing subgroups -QUESTION [13 upvotes]: Does there exist a group $G$ (finite or infinite) with three subgroups $A, B, C \leq G$ satisfying the following three conditions? - - -$A = N_G(A)$, $B = N_G(B)$, $C = N_G(C)$; - -$AB = BC = CA = G$; - -$A \cap B = B \cap C = C \cap A = 1$. - - - -(This question turned up in a more specific setting, but a negative answer to the existence of such a group would settle our case.) - -REPLY [16 votes]: I revise earlier edits to give a coherent account of the construction which shows that such subgroups can exist. -The underlying idea of the strategy is as follows: Let $X$ be a non-trivial finite group with trivial center which admits an automorphism $\alpha$ fixing only the identity (informally, and by a slight abuse, known as fixed-point free automorphism). Note that $\alpha$ can't have prime order, for otherwise $X$ would be nilpotent by Thompson's theorem, and hence would have non-trivial center. Then the direct product $X \times X$ is a product of two self normalizing subgroups of order $|X|$ which intersect trivially. One is $\Delta(X) = \{ (x,x): x \in X \}$. The other is -$\Delta^{\alpha}(X) = \{ (x,x\alpha): x \in X\}$. That $\Delta(X)$ is self-normalizing is clear, -since $Z(X) = 1$. For notice that if $(x,x)^{(a,b)} \in \Delta(X)$ for each $x \in X$, then -$ab^{-1} \in Z(X)$. The argument for the other subgroup is similar. Now we seek such a group $X$ with trivial center admitting a fixed point free automorphism $\alpha$ of composite odd order. If such a group exists, we may set $\beta = \alpha^{-1}$. Then the group $G = X \times X$ has the desired three self-normalizing subgroups $\Delta(X)$, $\Delta^{\alpha}(X)$ and $\Delta^{\beta}(X)$. For notice that if $x \alpha = x \beta$, then $\alpha^{2}$ fixes $x$, so that $\alpha$ fixes $x$ as $\alpha$ has odd order. But then $x$ is the identity by hypothesis. Hence any two of the three subgroups have trivial intersection, and have product $X \times X$ by order considerations. -There does exist a group $X$ of order $7^{6} .2^{4}$ which admits a fixed point free automorphism $\alpha$ of order $9$, and which has trivial center. Hence the construction above does work in this case, and $G = X \times X$ has three self-normalizing subgroups of the required form. We find a subgroup $Y$ of order $144$ of ${\rm GL}(6,7)$ which has an elementary Abelian normal subgroup $U$ of order $16$, acted on by an element $a$ of order $9$ whose cube centralizes $U$ but such that $a$ itself acts fixed-point freely on $U$, and such that, furthermore, $a$ does not have the eigenvalue $1$ in the given representation. We then take $X$ to be the semidirect product $VU$, where $V$ is a $6$-dimensional vector space over ${\rm GF}(7)$ and $U$ acts as the given elementary Abelian subgroup of ${\rm GL}(6,7)$. Then allowing $a$ to act on $U$ as it does within ${\rm GL}(6,7)$, and to act on $V$ as the given matrix yields an action of $a$ on $VU$ as a fixed-point free automorphism of order $9$. In terms of matrices, $a$ is the matrix -$ \left( \begin{array}{clcrc} 0 & 1 & 0 & 0 & 0 & 0\\ 0&0&1&0&0&0\\ 2 &0 &0 &0&0&0\\ -0&0&0&0&1&0\\ 0&0&0&0&0 &1\\0 &0 &0 &4&0& 0 \end{array}\right)$. The group $U$ -is the set of diagonal matrices with diagonal entries $\pm 1$ such that the product of the -first three diagonal entries is $1$ and the product of the last three diagonal entries is $1$. -The subgroup $U$ is $a$-invariant, $a^3$ centralizes $U$, but $C_{U}(a) = I$. The group $VU\langle a \rangle$ is the semidirect product $VY$. -A similar construction works for other odd primes $p$ by considering ${\rm GL}(2p,q)$, where $q$ -is a prime congruent to $1$ (mod p).<|endoftext|> -TITLE: Dualizable objects are flat? -QUESTION [9 upvotes]: Let me first recall some basic well-known definitions: Let $R$ be a ring (as always commutative). A (cocomplete) abelian tensor category is defined to be a symmetric monoidal category, whose underlying category is also a $R$-linear (cocomplete) abelian category, such that the tensor product is right exact (cocontinuous) and $R$-linear in each variable. An object $X$ of such a category $C$ is called flat if $X \otimes -$ is an exact functor. Also recall that an object $X$ of a symmetric monoidal category is called dualizable with dual $X^\*$ if it comes equipped with morphisms $1_C \to X \otimes X^\*$ and $X^\* \otimes X \to 1_C$ satisfying the two triangular identities. -Now these notions are connected in the case of $C = \text{Mod}(R)$: The dualizable objects are precisely the finitely generated projective $R$-modules (equivalently, locally free of finite rank). Thus, every dualizable object is flat, and Lazard's Theorem implies that every flat module is a directed colimit of locally free modules. Now I wonder if these statements remain true in the general case. I've already this concerning Lazard's theorem here for the category of quasi-coherent modules on a nice scheme (without answer). Thus my question is: -Question: Is every dualizable object in a (cocomplete) abelian tensor category flat? -Remark that we cannot insert "projective" as an intermediate step. It is easy to see that $1$ is projective iff every dualizable object is projective. But $1$ is not projective in $\text{Qcoh}(S)$ forr some non-affine concentrated scheme $S$, whereas the question in this case is answered with "yes" since every locally free quasi-coherent sheaf is locally flat and thus (globally) flat. - -REPLY [16 votes]: A dual pair as you describe above induces an adjunction $(X^*\otimes-)\dashv(X\otimes-)$, and since right adjoints preserve limits, $X$ is flat. - -REPLY [15 votes]: In a symmetric monoidal category, $X$ and $X^\ast$ are mutually dual. (We can compose with the symmetry to obtain a unit $1_C \to X^\ast \otimes X$ and counit $X \otimes X^\ast \to 1_C$ which realize $X$ as the dual of $X^\ast$.) This means we have both $X \otimes - \dashv X^\ast \otimes -$ and $X^\ast \otimes - \dashv X \otimes -$. Since $X \otimes -$ is both a left and right adjoint, it is exact.<|endoftext|> -TITLE: Name of "slice" category with 2-cells as morphisms ? -QUESTION [5 upvotes]: Hi, -I would like to know whether there is a standard name for the following "slice" category: -Let $\mathcal{C}$ be a 2-category and $c \in \mathcal{C}$ an object of $\mathcal{C}$. -We can form the category where an object $(d,f)$ is a pair of an object $d\in\mathcal{C}$ and an -arrow $f : d\to c$. A morphism $(h, \alpha)$ from $(d,f)$ to $(e,g)$ is given by -$h : d \to e$ and a 2-cell $\alpha : f \Rightarrow g \circ h$. -Thanks a lot, -ben -[Edit: fixed typo mentioned by Martin] - -REPLY [2 votes]: See this answer to much the same question. I would call this the 'lax' slice category, although it's not so common a notion that everyone would know what you meant, so maybe you should keep the scare quotes around 'lax'. -A propos of Martin's comment, the correct notion of slice 2-category depends on what you're doing -- you might want the strict version, with strictly commuting triangles, or the pseudo version, with invertible 2-cells (this is the strictest one that makes sense for non-strict 2-categories), or this lax version. Or you might want to restrict to (discrete) (op)fibrations as objects.<|endoftext|> -TITLE: A totally categorical structure with trivial geometry which is not interpretable in the trivial structure -QUESTION [10 upvotes]: Among the theorems of early geometric model theory there is one by Lachlan stating that every totally categorical structure with a trivial pregeometry is intrepretable in a dense linear order. -That suggests in particular that there are totally categorical structures with a trivial pregeometry that are not interpretable in the trivial structure $(M,=)$. Could someone kindly give an example of such a structure? - -REPLY [4 votes]: An example of such a structure is given here: ω-categorical, ω-stable structure with trivial geometry not definable in the pure set. Essentially, it is a non-split cover of the theory of the pure set. As in the paper of Hrushovski on Totally Categorical Structures.<|endoftext|> -TITLE: Larger cycle than 4, 2, 1 in Collatz iteration? -QUESTION [9 upvotes]: (Here I discuss the Collatz problem only for positive integers.) -It is possible, by computation, to find all cycles in the Collatz iteration of a fixed length. -It is clear that an increase must be followed by a decrease (for if $n$ is odd, then $3n+1$ is even) and a decrease can be followed by either an increase or decrease (for if $n$ is even, $\frac{n}{2}$ may be odd or even). -Using this, it is easy, for example, to show $4, 2, 1$ is the only cycle of length $3$: -Over the course of a cycle, we must have both increases and decreases. Position an increase at the beginning of the cycle. Then the changes of the cycle must be $IDD$ (where $I$ stands for increase and $D$ stands for decrease), for there is no other way to include an increase and avoid two consecutive increases. Then the cycle consists of $a, 3a+1, \frac{3a+1}{2}$, so that $a = \frac{3a+1}{4}$ and $a = 1$, leading to the familiar $1, 4, 2 = 4, 2, 1$ cycle. -For length $4$, there are two possibilities for a pattern of increase and decrease along a cycle which start with an increase: $IDID$ and $IDDD$. -The $IDID$ possibility leads to $a = \frac{9a+5}{4}$ so $a = -1$. -The $IDDD$ possibility leads to $a = \frac{3a+1}{8}$ so $a = \frac{1}{5}$. Neither of these values of $a$ is a positive integer, so there are no cycles of length $4$. -Likewise, for length $5$, there are only $3$ possibilities: $IDDDD$, $IDIDD$, and $IDDID$. Since last two are equivalent, this leads to $IDDDD$ and $IDIDD$. -$IDDDD$ leads to $a = \frac{3a+1}{16}$ so $a = \frac{1}{13}$. -$IDIDD$ leads to $a = \frac{9a+5}{8}$ so $a = -5$. So there are no cycles of length $5$. -There is more to be said for this way of considering cycles in the Collatz iteration. If a sequence of increases and decreases leads to the equation $a = Ta + U$, then $T$ is positive and $T \neq 1$, since $T = \frac{3^{m}}{2^{n}}$ for some positive integers $m, n$. The positivity of $T$ and the fact that $T \neq 1$ ensure that the solution of $a = Ta + U$ also solves (and therefore is the only solution to) $a = T(Ta + U) + U$ (or the equation for fixed points of higher-multiplicity iterates of the function $f(a) = Ta + U$), so that only cycles that form primitive circular words need to be considered (and $IDID$ was unnecessary, given how it reduces to $ID$). With this noted, it becomes very easy to handle cycles of length $6$: -The number of increases in a cycle is $1$, $2$, or $3$, since no two of them can be consecutive (and since there must be at least one increase). $3$ increases in a cycle can only be realized by $IDIDID$, which is not a primitive circular word. $2$ increases in a cycle can be realized by $IDIDDD$, $IDDIDD$, or $IDDDID$. $IDIDDD$ and $IDDDID$ are equivalent, while $IDDIDD$ is not a primitive circular word. So the only cases to consider are $IDDDDD$ and $IDIDDD$. -$IDDDDD$ leads to $a = \frac{3a+1}{32}$ so $a = \frac{1}{29}$. -$IDIDDD$ leads to $a = \frac{9a+5}{16}$ so $a = \frac{5}{7}$. -So there are no cycles of length $6$. -This is natural and simple enough that someone must have considered it before. Who has, and in what paper(s)? Also, has this been used to settle the question of whether or not there is a cycle in the positive integers larger (both in length, and in member-wise comparison) than the $4, 2, 1$ cycle? - -REPLY [4 votes]: Edit: Thanks to all for your comments. Because of the phrasing of the review of Eliahou's paper on MathSciNet, I was confused between the $3n+1$ and the $3n-1$ problems. But Eliahou's paper is indeed about the $3n+1$, hence is relevant to the OP. I edited my answer so that it should now make sense. -It is known that any non-trivial cycle must start after $2^{40}$, and have length $17 087 915b + kc$, where $b,c\in\mathbb{N}$ and $k\in\{301 994 , 85 137 581\}$: see Shalom Eliahou, {\it The $3x + 1$ problem: new lower bounds on nontrivial cycle lengths}. Discrete Math. 118 (1993), 45–56.<|endoftext|> -TITLE: Trace identities and the Kauffman Bracket skein module -QUESTION [7 upvotes]: Let's consider $K_t(M)$, the Kauffman bracket skein module (see this and this papers) of a three-manifold $M$. When $t=-1$, $K_t(M)$ is easily seen to be isomorphic to the ring of functions on the character variety $X=\operatorname{Hom}(\pi_1(M),\operatorname{SL}(2))/\\!/\operatorname{SL}(2)$, and the isomorphism sends a loop $\gamma$ in $K_t(M)$ to the function on $X$ given by $\rho\mapsto\operatorname{tr}(\rho(\gamma))$. The proof of this amounts to checking the following three identities: -1) $\operatorname{tr}(A)=\operatorname{tr}(A^{-1})$ -2) $\operatorname{tr}(AB)+\operatorname{tr}(AB^{-1})=\operatorname{tr}(A)\operatorname{tr}(B)$ -3) $\operatorname{tr}(\operatorname{id})=2$ -for $A,B\in\operatorname{SL}(2)$. (1) corresponds to the loops in the Kauffman bracket skein module being unoriented. (2) and (3) correspond to the skein relations. Note I'm being a bit cavalier and forgetting that actually everything is framed, but it isn't particularly important for this question. -Question: Is there a natural class of objects to allow $A$ and $B$ to range over, and a natural "function" $\operatorname{tr}$ on this class of objects, such that the following more general identities are satisfied? -1) $\operatorname{tr}(A)=\operatorname{tr}(A^{-1})$ -2) $-t\operatorname{tr}(AB)-t^{-1}\operatorname{tr}(AB^{-1})=\operatorname{tr}(A)\operatorname{tr}(B)$ -3) $\operatorname{tr}(\operatorname{id})=-(t+t^{-1})$ -Notice that here $\operatorname{tr}$ can't be a genuine function, since (2) (along with (1)) implies that product operation on trace values is noncommutative. -Of course, the motivation of this question is to obtain a description of $K_t(M)$ just in terms of $\pi_1(M)$, and it is natural to try something involving $U_q(\mathfrak s\mathfrak l_2)$. - -REPLY [10 votes]: The old answer is that the trace identity you give in 2) is not quite right, -Let $R=\sum_i a_i\otimes b_i$ be the $R$ matrix for $U_q(sl_2)$ and let $t$ be the -$4$th root of $q$, then -$$t tr(XY)+ t^{-1}tr(S(X)Y)=\sum_itr(a_iX)tr(b_iY),$$ -where $S$ is the antipode and $tr$ is the ordinary trace in the fundamental representation -of $U_q(sl_2)$. That discovery really cost me, as I spent a long time trying to find -the identity using the quantum trace, and it wasn't true. The appearence of the $R$-matrix -on the right hand side of the equation is because diagrammatically, that is where the crossing -is. -You can find it in a paper of Bullock, Frohman and Bartoszynska in Communications -in Mathematical Physics in the late 90's where we proved that the space of observables -for $U_q(sl_2)$ lattice gauge field theory based on a fat graph is the Kauffman bracket skein algebra -of the surface which is a regular neighborhood of your graph. -The skeins are functions on the space of connections on the lattice. We based our construction of lattice gauge field theory in the work of Alexeev, Schomerus and Grosse. -Also we were inspired somewhat by the work of Buffenoir and Roche on lattice gauge field theory. In both their approaches, the algebra of observables were defined via generators -and relations. They used a Wick ordering to get functions. We saw that the connections were -actually a coalgebra, and the dual product on the observables satisfied the relators given -by the authors in the physical literature. This allowed us to give a coordinate free construction of lattice gauge field theory based on a quantum group, that led to structural -control over the algebra of observables. Falling back on an observation of Fock we were -able to show that the algebra of observables quantizes the characters of the underlying surface group with respect to a Poisson structure constructed by Goldman. -To get the signs to work like you want you need to work with $-tr$. The minus sign -has been explained nicely by Bonahon and Wong. It comes from the fact that you are looking -at $PSL_2(\mathbb{C})$ representations and lifting them to to $SL_{2}(\mathbb{C})$ representations. -Our ultimate step in this direction was "The Yang-Mills measure in the Kauffman bracket -skein module". The lattice models were equipped with a "path" integral which is -a trace on the algebra of observables that is topologically invariant. Once we had -identified the algebra of observables as the Kauffman bracket skein algebra, we carried the -path integral over to the purely topological picture to get a trace which deforms either -integration against Haar measure in the case where the surface has boundary, or integration -against the symplectic measure if the surface is closed. Although the physicists treated -the path integral as a formal power series, we were able to show that as long as Planck's constant -does not lie on the unit circle, or if it does, it is a root of unity, that the integral actually -converges. -In more modern terms, quantum Teichm\"{u}ller theory as developed by Fock, Checkov, Bonahon -and Kashaev, constructs a dual lattice gauge field theory, whose representation theory -has been worked out by Bonahon and his collaborators. What is nice about this is you -can emulate steps of the proof of the geometrization conjecture in the quantum setting -and find fixed representations. Bonahon and Wong recently proved that the space of -observables contains a large subalgebra which is the Kauffman bracket skein algebra of the -underlying surface. -The shortcoming of all of this is that it doesn't address characters of closed surfaces. -Generally the way you deal with that is by defining a projector. Ultimately more -geometric approaches via a line bundle over the character variety will probably give more -satisfactory answers in that case. -The quantum hyperbolic invariants of Baseilhac and Bennedetti end up assigning quantum invariants to knots and links in manifolds with a $PSL_2(\mathbb{C})$ representation. When -the underlying representation is trivial these are the invariants of Kashaev which is have been equated with evaluations of the colored Jones polynomial by Murakami and Murakami. -Some citations. -Bullock, Doug; Frohman, Charles; Kania-Bartoszyńska, Joanna Topological interpretations of lattice gauge field theory. Comm. Math. Phys. 198 (1998), no. 1, 47–81. -Bullock, Doug; Frohman, Charles; Kania-Bartoszynska, Joanna The Yang-Mills measure in the Kauffman bracket skein module. Comment. Math. Helv. 78 (2003), no. 1, 1–17. -Bonahon, Francis; Liu, Xiaobo Representations of the quantum Teichmüller space and invariants of surface diffeomorphisms. Geom. Topol. 11 (2007), 889–937. -Kashaev, Rinat M. On quantum moduli space of flat PSL2(R)-connections on a punctured surface. Handbook of Teichmüller theory. Vol. I, 761–782, IRMA Lect. Math. Theor. Phys., 11, Eur. Math. Soc., Zürich, 2007 -Baseilhac, Stephane; Benedetti, Riccardo Quantum hyperbolic geometry. Algebr. Geom. Topol. 7 (2007), 845–917. -An alternate approach to quantum hyperbolic invariants: -Kashaev, R.; Reshetikhin, N. Invariants of tangles with flat connections in their complements. Graphs and patterns in mathematics and theoretical physics, 151–172, Proc. Sympos. Pure Math., 73, Amer. Math. Soc., Providence, RI, 2005. -Kauffman brackets, character varieties, and triangulations of surfaces -Francis Bonahon (USC), Helen Wong (Carleton College) arXiv:1009.0084v1 [math.GT]<|endoftext|> -TITLE: Almost Flat Connections On Principal G-Bundles -QUESTION [6 upvotes]: Let $E \rightarrow F$ be a principal $G$-bundle. Let $\alpha$ be a connection 1-form with values in the Lie algebra of $G$. Let $\omega$ denote the curvature 2-form of connection $\alpha$. -We know that $\omega =0$ everywhere i.e., $\alpha$ is a flat connection if and only if the distribution $ker (\alpha)= \{v_x \in T_x E \ | \ \alpha (v_x)=0 \ ; x \in E \} $ on $E$ is completely integrable. -Now, suppose we have a norm on the space of 2-forms. We start with an "almost flat connection -$\alpha$ " i.e., the curvature form satisfies inequality, $|\omega| < \epsilon$ everywhere, for sufficiently small $\epsilon$. Is it true that the distribution $\ker(\alpha)$ on $E$ is "close" to a completely integrable distribution? and "close" in what sense? We may assume that the base $F$ is compact. -I have a feeling that with appropriate notion of "closeness" of distributions the above question has an affirmative answer. While I am trying to show that it is so, I have a difficulty deducing any useful information about connection 1-form from bounds on curvature. Thanks in advance for bringing in any new insight. -I have posted this question on Math SE also. https://math.stackexchange.com/questions/47571/curvature-and-connections-in-principal-g-bundles - -REPLY [3 votes]: The answer depends on the setting. If the group $G$ is compact, $M$ is compact and you are working modulo the gauge group, then the answer is positive and follows from Uhlenbeck's compactness theorem (first proven in her paper "Connections with $L^p$-bounds on curvature"). This theorem was extended by Katrin Wehrheim to the case when the base-manifold $M$ noncompact, see her book "Uhlenbeck compactness." In the case when $G$ is noncompact, I do not think there is a meaningful way to get a positive answer. For instance, $S^3$ admits "almost flat" symmetric affine connections, see Agaoka's example in -http://www.jstor.org/stable/2161101 On the other hand, the frame bundle of $S^3$ admits unique flat connection, namely, the trivial one (up to gauge transformations), but it is highly nonsymmetric. Thus, there is no way you can approximate Agaoka's connections (lifted to the frame bundle) by flat ones. I am sure there are other examples.<|endoftext|> -TITLE: How to relate equivariant symplectic cohomology, Contact Homology, Cyclic Homology and String Topology? -QUESTION [13 upvotes]: I am trying to understand how all the players in the title relate, but with all the grading shifts,and difficult isomorphisms involved in the subject I am having a hard time being sure that I have the picture right. I am going to write what I think is true, and if someone would confirm or deny it, that would be really nice. -The basic jumping off point is that if N is a simply connected manifold, symplectic cohomology of the cotangent disk bundle $D^*(N)$, symplectic cohomology $SH^*(D^*(N))$ as defined in Seidel's "A Biased View of Symplectic Cohomology" is naturally identified with isomorphic to $H_{n-*}(LN)$. And symplectic homology is isomorphic to $H^*(LN)$, these are the Hochschild cohomologies and homologies respectively of the algebra $C^*(N)$. -The reason is that the zero section generates the compact Fukaya category $Fuk^{cpt}(D^*(N))$ and $End(N_0,N_0) \cong C^*(N)$. There is expected to be a geometric Seidel map from $SH^*(D^*(N)) \to HH^*(Fuk^{cpt}(D^*(N))$ basically one considers cylinders in with a puncture which satisfy a deformed d-bar equation and which are asymptotic to periodic orbits of the Hamiltonian vector field and whose boundary lies on the zero section and is able to deform the compositions in the category to first order. -Question 1) Has anybody checked in this example that Seidel's map is an isomorphism? -Now we take equivariant versions of this, $SH^*_{eq}$ which should be identified with cyclic cohomology $CC^*(C^*(N))$ and we need to consider. Now we move to (linearized) contact homology, which Bourgeois and Oancea claim can be identified with $H_{eq}(LN,N)$(I give up on the gradings at this point :)) where N is included as the constant loops. Reasonable enough, since those are somehow the generators missing from contact homology. However, they also seem to be making an identification that $SH_* \cong H_*(LN)$ and not the cohomology... I get that it's not really a huge deal as vector spaces to identify a vector space and it's dual, but it adds to the confusion below. -With contact homology one can try a similar map to the Seidel map, namely work in the symplectic completion $T^*(N)$ and consider holomorphic disks with a puncture, boundary on the zero section, and as asymptotic now to a Reeb orbit as $|\rho|\to \infty$ `($|\rho|$ is some norm on $T^*(N)$). This gives a map from $CH_{*}\to CC^{*} $ which is mysterious because...(or maybe this is a map from "contact cohomology", I'm getting confused). Edit: A reference for this map is in a paper by Xiaojun Chen called "Lie Bialgebras and cyclic homology of A(\infty structures) in topology" -Question 2) We are supposed to somehow map $H_{*,eq}(LN,N) \to H_{*,eq}(LN)$. I'm not great with topology but this is a strange map, since it seems to go the wrong way. On the other hand one should expect it to be interesting by analogy with the Seidel map. What is this map at least conjecturally supposed to be? The only thing I could think of is the kernel of the map: $C_*(LN) \to C_*(N)$ induced by the map $LN \to N$ but this map doesn't even exist equivariantly. - -REPLY [10 votes]: Some blah on symplectic homology vs. cohomology. There's an invariant $SH(M)$ of Liouville domains $M$ which some people call symplectic homology and some symplectic cohomology. This is the direct limit of Hamiltonian Floer groups associated with functions of increasing eventual slope. The dual theory has two rather unpleasant features: it involves inverse limits, hence one must worry about $\varprojlim^1$-terms; and in general it's not countably generated. It's not often used. -Why the confusion about terminology? Well, depending on your convention for the sign of the symplectic action functional, you may regard this as Morse homology or compactly supported Morse cohomology of this function. From the perspective of Lagrangian Floer cohomology, consistency demands that one calls it symplectic cohomology; I do. However, symplectic field theorists (including Bourgeois-Oancea, I think) prefer the contrary convention. -Blah about grading. The integer grading on $SH^\ast(M)$ is defined when $c_1=0$, and is canonical when $H^1(M)=0$. One has Viterbo's map $H^\ast(M)\to SH^\ast(M)$, and one convention makes this preserve degree (I'll take that option), while another makes it shift degree by the complex dimension $n$ of $M$. -Seidel's map for cotangent bundles. -(Edited: my first version was not correct.) There are actually two versions of Seidel's open-closed string map, derived from the same moduli spaces: -$$ \kappa: SH^\ast(M) \to HH^\ast(F(M),F(M))$$ -and -$$ \lambda : HH_\ast(F(M),F(M)) \to SH^{n+\ast}(M). $$ -Here $F(M)$ is the Fukaya category of exact, compact Lagrangians. Moreover, there are extensions of these maps to the wrapped Fukaya category $W(M)$. -The absence of a dualisation - hence the connection of symplectic cohomology to both Hochschild homology and cohomology - looks strange. Mohammed Abouzaid points out in his comment below that this is a manifestation of a self-duality property for the wrapped category. He shows in http://arxiv.org/abs/1003.4449 that for $M=T^\ast L$ and the wrapped category, the map $\lambda$ is an isomorphism. -Cyclic version. My expectations are slightly different from those stated in the question. I'd guess that $\lambda$ extends to a map from cyclic homology to circle-equivariant symplectic cohomology, -$$ HC_\ast(F(M)) \to SH^{n+\ast}_{S^1}(M) $$ -and that this should be an isomorphism when $\lambda$ is. -For cotangent bundles $T^\ast L$ of simply connected, spin manifolds, one has $SH^\ast_{S^1}(M) \cong H_{n-*}^{S^1}(\mathcal{L}L)$. -Linearised contact (co)homology is, according to Bourgeois-Oancea (if I have it right), the mapping cone of the (cochain level) Viterbo map $H^\ast(M; H^\ast_{S^1}(pt.)) \to SH^\ast_{S^1}(M)$. For cotangent bundles as before, Viterbo's map should be identifiable with the map induced by the equivariant inclusion of constant loops: $H^\ast(T^\ast L)[u] = H^\ast (L)[u] = H_{n-\ast}(L)[u] \to H_{n-\ast}^{S^1}(\mathcal{L}L)$.<|endoftext|> -TITLE: When does the homological dimension of a tensor product equal the sum of dimensions? -QUESTION [9 upvotes]: The notion of dimension I prefer most is right global dimension, but the question can also be asked for other notions (e.g. weak dimension, injective dimension, Krull dimension). Letting $d$ be whichever dimension you pick, the question then becomes: - -Let $A$ and $B$ be $R$-algebras. When is $d(A\otimes_R B) = d(A)+d(B)$? - -This question was addressed in the '50s by Eilenberg et al and Auslander. The first proved this result for $R$ commutative and $B$ is a ring of matrices, polynomials, or rational functions. The second proved it for $R$ a field and $A,B,A\otimes B$ semiprimary algebras of finite global dimension. In 1996 Vladimir Bavula came up with some different sufficient conditions on $A$ and $B$ when $R$ is a field but they seem much more complicated. I can't find anything on the problem since then, so any newer references would be much appreciated. Generally I turn to Weibel or Lam for questions like this, but neither book mentions anything. At this moment I'm mostly asking out of pure curiosity, so I'm fine with any assumptions on $R$. Depending on the answers I get, I might try to say something about ring spectra and then I'd have to be much more careful about hypotheses. The case where $R$ is a field has been done a lot, so perhaps $R$ could just be a commutative ring or commutative noetherian. Whatever gives a nice answer. -Here's what seems to be known. For $d =$ weak dimension or Krull dimension, $d(A\otimes B) \geq d(A)+d(B)$ and this also holds for left-Noetherian algebras $A$ and $B$ in the case where $d$ is left global dimension. One case where equality fails is $A = B =$ Division ring of $K[x_1,\dots,x_n]$ and $d =$ left global dimension. Then $d(A\otimes_K B) = n$ but $d(A)=d(B)=0$. -If $K$ is algebraically closed and $A$ and $B$ are finite dimensional $K$-algebras then gl.dim$(A\otimes B)=$ gl.dim$(A)+$ gl.dim$(B)$ -Eilenberg et. al give the following as Proposition 10: If $K$ is a field then -l.gl.dim$(A)+$ weak dim$(B) \leq$ gl.dim$(A\otimes B) \leq$ l.gl.dim$(A) + $ dim$_K(B)$, and -l.proj.dim$(A) +$ weak dim$(B)\leq$ dim$_K(A\otimes B) \leq$ dim$_K(A) +$ dim$_K(B)$. -So this theorem reduces the problem to finding when weak dimension (as a ring) equals dimension as a $K$-algebra. It's not true in general that dim$_K(A\otimes_R B) =$ dim$_K(A)+$ dim$_K(B)$, e.g. if $A$ and $B$ are locally separable algebras over $K$ with $[A:K]=[B:K]=\infty$ then $dim_K(A)=dim_K(B)=dim_K(A\otimes B)=1$ because $A\otimes B$ satisfies the same properties just listed for $A$ and $B$. - -REPLY [3 votes]: I sometime back investigated this question for the n dimensional quantum torus, that is, $A$ and $B$ are of type $F \ast \mathbb Z^n$, where $F$ is a field. For such algebras the Krull and the global dimensions coincide. It was found that dimension of $A \otimes_F B$ is superadditive in general but when $B$ has dimension $n$ or $n - 1$ the dimension is additive with tensoring.<|endoftext|> -TITLE: Jacquet Langlands correspondence -QUESTION [6 upvotes]: I have one issue with the Jacquet Langlands correspondence. The Weyl law for $H$ modulo a congruence subgroup and the Weyl law for cocompact groups are different. So why does this not contradict this functoriality? What am I missing? -I have not yet studied the Jacquet Langlands correspondence explicitly yet. How explicit are the lifts, about the level etc.? I know that there is not an expansion formula for cocompact groups available as we have it for groups with a parabolic element. -Update: After a reading a little bit, I found a paper which focuses exactly on the first part of the question and also gives references for the second part of the question: -Risager, Morten S. Asymptotic densities of Maass newforms. J. Number Theory 109 (2004), no. 1, 96–119. - -REPLY [4 votes]: In what sense is the Weyl law different for congruence subgroups and cocompact groups? -At any rate, the Jacquet-Langlands correspondence is not a bijection between the two cuspidal spectra. More precisely, let $D$ be a quaternion algebra over a number field $F$, and consider the groups $G=PD^\times$ and $G'=PGL_2$. Then the Jacquet-Langlands correspondence injects the automorphic representations of $G(\mathbb{A}_F)$ into those of $G'(\mathbb{A}_F)$. A cuspidal representation $\pi$ of $G'(\mathbb{A}_F)$ lies in the image of this map if and only if $\pi_v$ is a discrete series representation of $G'(F_v)$ at all places $v$ where $D$ ramifies. So unless $G'=G$, the image will miss several cuspidal representations of $G'(\mathbb{A}_F)$. -I think the lifts are not explicit in the sense that they are not given by an explicit construction.<|endoftext|> -TITLE: Quantum field theory in Solovay-land -QUESTION [15 upvotes]: Constructing quantum field theories is a well-known problem. In Euclidean space, you want to define a certain measure on the space of distributions on R^n. The trickiness is that the detailed properties of the distributions that you get is sensitive to the dimension of the theory and the precise form of the action. -In classical mathematics, measures are hard to define, because one has to worry about somebody well-ordering your space of distributions, or finding a Hamel basis for it, or some other AC idiocy. I want to sidestep these issues, because they are stupid, they are annoying, and they are irrelevant. -Physicists know how to define these measures algorithmically in many cases, so that there is a computer program which will generate a random distribution with the right probability to be a pick from the measure (were it well defined for mathematicians). I find it galling that there is a construction which can be carried out on a computer, which will asymptotically converge to a uniquely defined random object, which then defines a random-picking notion of measure which is good enough to compute any correlation function or any other property of the measure, but which is not sufficient by itself to define a measure within the field of mathematics, only because of infantile Axiom of Choice absurdities. -So is the following physics construction mathematically rigorous? -Question: Given a randomized algorithm P which with certainty generates a distribution $\rho$, does P define a measure on any space of distributions which includes all possible outputs with certain probability? -This is a no-brainer in the Solovay universe, where every subset S of the unit interval [0,1] has a well defined Lebesgue measure. Given a randomized computation in Solovay-land which will produce an element of some arbitrary set U with certainty, there is the associated map from the infinite sequence of random bits, which can be thought of as a random element of [0,1], into U, and one can then define the measure of any subset S of U to be the Lebesgue measure of the inverse image of S under this map. Any randomized algorithm which converges to a unique element of U defines a measure on U. -Question: Is it trivial to de-Solovay this construction? Is there is a standard way of converting an arbitrary convergent random computation into a measure, that doesn't involve a detour into logic or forcing? -The same procedure should work for any random algorithm, or for any map, random or not. -EDIT: (in response to Andreas Blass) The question is how to translate the theorems one can prove when every subset of U gets an induced measure into the same theorems in standard set theory. You get stuck precisely in showing that the set of measurable subsets of U is sufficiently rich (even though we know from Solovay's construction that they might as well be assumed to be everything!) -The most boring standard example is the free scalar fields in a periodic box with all side length L. To generate a random field configuration, you pick every Fourier mode $\phi(k_1,...k_n)$ as a Gaussian with inverse variance $k^2/L^d$, then take the Fourier transform to define a distribution on the box. This defines a distribution, since the convolution with any smooth test function gives a sum in Fourier space which is convergent with certain probability. So in Solovay land, we are free to conclude that it defines a measure on the space of all distributions dual to smooth test functions. -But the random free field is constructed in recent papers of Sheffield and coworkers by a much more laborious route, using the exact same idea, but with a serious detour into functional analysis to show that the measure exists (see for instance theorem 2.3 in http://arxiv.org/PS_cache/math/pdf/0312/0312099v3.pdf). This kind of thing drives me up the wall, because in a Solovay universe, there is nothing to do--- the maps defined are automatically measurable. I want to know if there is a meta-theorem which guarantees that Sheffield stuff had to come out right without any work, just by knowing that the Solovay world is consistent. -In other words, is the construction: pick a random Gaussian free field by choosing each Fourier component as a random gaussian of appropriate width and fourier transforming considered a rigorous construction of measure without any further rigamarole? -EDIT IN RESPONSE TO COMMENTS: I realize that I did not specify what is required from a measure to define a quantum field theory, but this is well known in mathematical physics, and also explicitly spelled out in Sheffield's paper. I realize now that it was never clearly stated in the question I asked (and I apologize to Andreas Blass and others who made thoughtful comments below). -For a measure to define a quantum field theory (or a statistical field theory), you have to be able to compute reasonably arbitrary correlation functions over the space of random distributions. These correlation functions are averages of certain real valued functions on a randomly chosen distribution--- not necessarily polynomials, but for the usual examples, they always are. By "reasonably arbitrary" I actually mean "any real valued function except for some specially constructed axiom of choice nonsense counterexample". I don't know what these distribtions look like a-priory, so honestly, I don't know how to say anything at all about them. You only know what distributions you get out after you define the measure, generate some samples, and seeing what properties they have. -But in Solovay-land (a universe where every subset S of [0,1] is forced to have Lebesgue measure equal to the probability that a randomly chosen real number happens to be an element of S) you don't have to know anything. The moment you have a randomized algorithm that converges to an element of some set of distributions U, you can immediately define a measure, and the expectation value of any real valued function on U is equal to the integral of this function over U against that measure. This works for any function and any distribution space, without any topology or Borel Sets, without knowing anything at all, because there are no measurability issues--- all the subsets of [0,1] are measurable. Then once you have the measure, you can prove that the distributions are continuous functions, or have this or that singularity structure, or whatever, just by studying different correlation functions. For Sheffield, the goal was to show that the level sets of the distributions are well defined and given by a particular SLE in 2d, but whatever. I am not hung up on 2d, or SLE. -If one were to suggest that this is the proper way to do field theory, and by "one" I mean "me", then one would get laughed out of town. So one must make sure that there isn't some simple way to de-Solovay such a construction for a general picking algorithm. This is my question. -EDIT (in response to a comment by Qiaochu Yuan): In my view, operator algebras are not a good substitute for measure theory for defining general Euclidean quantum fields. For Euclidean fields, statistical fields really, you are interested any question one can ask about typical picks from a statistical distribution, for example "What is the SLE structure of the level sets in 2d"(Sheffield's problem), "What is the structure of the discontinuity set"? "Which nonlinear functions of a given smeared-by-a-test-function-field are certainly bounded?" etc, etc. The answer to all these questions (probably even just the solution to all the moment problems) contains all the interesting information in the measure, so if you have some non-measure substitute, you should be able to reconstruct the measure from it, and vice-versa. Why hide the measure? The only reason would be to prevent someone from bring up set-theoretic AC constructions. -For the quantities which can be computed by a stochastic computation, it is traditional to ignore all issues of measurability. This is completely justified in a Solovay universe where there are no issues of measurability. I think that any reluctance to use the language of measure theory is due solely to the old paradoxes. - -REPLY [4 votes]: I don't know anything about the Solovay land, but I can say a little about the random functions and this may be related to what you're going for. -People have for a while been considering random functions which are generated in this way in the context of nonlinear dispersive equations. Probably the most interesting examples are the Gibbs measures associated to infinite dimensional Hamiltonian systems like nonlinear Schrödinger or wave equations. See for example these slides of Jim Colliander: -http://blog.math.toronto.edu/colliand/files/2010/08/2010_08_Colliander_Istanbul_Final.pdf -which has an outline of the technicalities to determine on which Banach space your measure will be supported. More can be found from this lecture of Gigliola Staffilani: -http://www-math.mit.edu/~gigliola/Milan-lecture3-4.pdf -Now, if you want to put a measure on an infinite dimensional space (like the space of distributions), its support will necessarily be extremely thin even if you do get a dense subset. So for instance if you start with some $f \in L^2$ and randomize its Fourier coefficients to make a random function $f^\omega$, you get an increased integrability $f^\omega \in L^p$ for all $p < \infty$ almost surely. There are other measures you can use besides Gaussian measures where the same phenomenon will occur (random $\pm 1$'s will also do the trick by the well-known Khintchine's inequality). -Using Gibbs measures (or just ad hoc randomizations like randomized Fourier coefficients), people have been able to establish almost surely globally defined flows for random data which is "supercritical" when measured in a Sobolev space. The Gibbs measures just come with the nice feature of being invariant under the flow provided you can construct the flow. In physical space, the random data looks much better than a typical element in the Banach space. For this reason, you can show solutions to nonlinear evolution equations exist almost surely and even establish some kind of almost sure well-posedness when you deterministically would not have such a result. But there is a limit to what can be achieved with this freedom. In particular, if your variances not only fail to decay but even grow with the frequency, then it can be very difficult or simply impossible to construct a solution to the equation. The example you gave of variance like $k^2$, for instance, is likely to be too large at frequency infinity -- i.e. too irregular -- for any sensible solution to a familiar nonlinear evolution equation to exist, even though, of course, it will make sense as a measure on the space of distributions and have support in some negative Sobolev space you can explicitly compute. Since Fourier multipliers applied to the random Fourier series will also be random Fourier series, you will also be able to solve linear PDE with the random data with no difficulty, and these solutions will also possess higher integrability than you would ever sensibly ask for. But if the equation you have in mind is the cubic nonlinear Schrödinger and you're insistent about very large data, what you're asking for could likely be hopeless for reasons much more serious than this axiom of choice stuff. You have to pay attention to the space because even cubing -- let alone solving the equation -- may be impossible.<|endoftext|> -TITLE: Suggestions for teaching advanced high school students -QUESTION [5 upvotes]: Hi all, -I'm a grad student and just joined a mentoring program in which I will visit a group of advanced year ten high school students (around 16 years old) from a group of schools in the area. I don't quite know where they're at just yet. I assume they've done Euclidean geometry, analytic algebra are just getting used to differential calculus. I'm fairly free to do what I want with them short of cruel and unusual punishment (such as making them read Gilbarg and Trudinger). They meet weekly and I had in mind meeting them every two weeks and going through a shotgun summary of what maths is all about, with an eye for a historical development. Of course, I want it to be fun! I have a semseter, so that makes about 6 or 7 two hour meetings. -To make the question precise: What do the potential answerers believe should not be left out of such an introduction? -Thanks, -Mat - -REPLY [4 votes]: The most valuable thing that you have to offer these students is your own experience as a college student and graduate student. This might seem surprising, but is based on my own experience of working with high school students in the U.S. Many of these students aren't sure about going to college, don't know what will be expected of them when they do go to college, and aren't sure that they're "smart enough" to succeed once they get to college. Mostly they need mentoring and encouragement more than they need instruction in any particular aspect of mathematics. -There's a good chance that you know more about what's going on in college than even their high school teachers (who may have been out of college for many years.) You're closer to their own age and not an authority figure, so they're more likely to believe what you have to say. You should talk to them to find out what would be most useful for them to know.<|endoftext|> -TITLE: Is it possible to define a closure operator in terms of partial ordering? -QUESTION [6 upvotes]: For boolean algebra, let's take Roman Sikorski's Boolean Algebras as our reference. After giving a set of axioms, he proves (p.9) that the join of A and B is the least element of the algebra such that A and B are its subelements. He also asserts that since that's so, the join of A and B can be defined in terms of the ordering relation. -In the Appendix (p. 198), Sikorski defines a closure algebra as a boolean algebra with an additional operator, a closure operator, and gives axioms for it. He also defines an interior operator in terms of the closure operator. -The assertion about the definability of join in terms of ordering leads one to ask whether the same might be true of the closure and interior operators. Given sufficiently strong axioms for them, one might find that: - -The closure of A is the least element -of which A is a subelement. -The interior of A is the greatest -element which is a subelement of A. - -This idea just occurred to me this evening. I imagine the answer depends on the axioms for the closure operator. Has this topic already been explored? Thanks for any help or references. - -REPLY [8 votes]: Oh sure, this is quite well-known. The closure of an element is the smallest closed element which is greater than or equal to the given element. Dually for the interior operator. -In general, a closure operator $\phi: P \to P$ on a poset $P$ is an order-preserving, inflationary ($x \leq \phi(x)$), idempotent ($\phi(\phi(x)) = \phi(x)$) operation, and if you want a topological closure operator, then you demand $\phi(x \vee y) = \phi(x) \vee \phi(y)$ as well. Alternatively, a closure operator can be specified by an inclusion $i: C \hookrightarrow P$ such that every $p \in P$ has a least upper bound $c \in C$. The assignment $p \mapsto c$ gives an order-preserving mapping $j: P \to C$ with the property -$$j(q) \leq d \Leftrightarrow q \leq i(d)$$ -(where the left side is to be read in $C$ and the right in $P$), and the closure operator $\phi$ is the composite $i \circ j$; it is easy to check the order-preserving inflationary idempotent properties. In the context of Boolean algebras $P$, as in Sikorski, you demand that $C$ be closed under joins as well (and that $i$ preserve them), to make $\phi$ a topological closure. -I am writing this answer to bear out a far wider connection with category theory. The posets here are special cases of categories, where there is at most one arrow in any hom-set $\hom(x, y)$, which we write as $x \leq y$. Functors between poset categories amount to order-preserving maps. A closure operator amounts to a monad on a poset. The subcollection $i: C \subseteq P$ of closed elements for that operator is the category of algebras for that monad. The corresponding mapping $j: P \to C$ is the left adjoint to $i$. Any adjoint pair (known under another name as a Galois connection) between posets always induces a closure operator. And so on. -There is an embarrassment of riches of references (i.e., so many that it's hard to think of one that stands out). But a trade secret among category theorists is to understand what general concepts mean in the simplified case of poset categories, and this point of view is explicitly declared in Paul Taylor's Practical Foundations of Mathematics. I think just about any introductory book on category theory will refer to this, though, and you will also find it used heavily in more specialized treatises like Johnstone's Stone Spaces that involve looking at lattices from a categorical point of view.<|endoftext|> -TITLE: Can cotangent bundles see exotic smooth structures? -QUESTION [40 upvotes]: I have two questions that are inspired by a couple of questions here on MO (referenced below), as well as by a conversation with some other grad students at a summer school. -Caveat: I'm not a symplectic geometer, nor a differential topologist in the 'classical' sense, so my questions might have a well-known answer, or be open. - -It's known that exotic $\mathbb{R}^4$'s have diffeomorphic cotangent bundles, as it's known (see here) that the Milnor spheres have isomorphic (topological) cotangent bundles. - -1. Can the smooth structure of cotangent bundles distinguish exotic smooth structures on the base? - -Phrased differently, does it exist a pair of smooth manifolds $M,M'$ that are homeomorphic, such that $T^*M$ and $T^*M'$ are isomorphic (see footnote) as vector bundles, but not diffeomorphic as smooth manifolds? - -As said before, exotic smooth structures on $\mathbb{R}^4$ are not detected by the cotangent bundle as a smooth manifold. But the cotangent bundle admits a canonical symplectic structure, so... - -2. Can the symplectic structure of cotangent bundles see the base? - -I.e. Does it exist a pair $M,M'$ of smooth manifolds that have diffeomorphic tangent bundles, but such that their symplectic cotangent bundles are/are not symplectomorphic? In this phrasing, I'm including also pairs like $\mathbb{R}^3$ and the Whitehead manifold (see this question), but we can impose some further restrictions: what if $M$ and $M'$ are homeomorphic? -UPDATE: on this page, Igor Belegradek gives a reference to a paper where it's proved that homeomorphic $n$-spheres have diffeomorphic cotangent bundles. As Andy Putman points out in his answer, Mohammed Abouzaid found examples of spheres with non-symplectomorphic cotangent bundles, so the answer is YES: cotangent bundles can (at least sometimes) see the base. -UPDATE (16/07/2012): regarding question 2, Tobias Ekholm and Ivan Smith have an interesting preprint: Corollary 1.4 says that the symplectic topology on $T^*(S^1\times S^{8k-1})$ detects the smooth topology of the underlying manifold. Their paper is about double points of Lagrangian immersions (mentioned in Tim Perutz's answer, below). - -As Igor Belegradek remarked, the word "isomorphic" might be a bit confusing, in this context. What I mean is that there is a homeomorphism $M\to M'$ that pulls back $T^*M'$ to $T^*M$. This can be strengthened/weakened by asking that this homeomorphism is topologically isotopic to the identity (since $M$ and $M'$ are homeomorphic, this makes sense). For example, I might want $M$ and $M'$ to be both parallelisable. - -REPLY [3 votes]: The questions at hand are much easier to answer for open manifolds than for closed manifolds. Here is a large family of simple examples: There are uncountably many diffeomorphism types of exotic $\mathbb{R}^4$s. As mentioned already in the question, the total space of the cotangent bundle of each exotic $\mathbb{R}^4$ is diffeomorphic to $\mathbb{R}^8$. -Moreover, the bundle is trivial, because its base is contractible. Hence, for any homeomorphism $f:M_0\to M_1$ between exotic $\mathbb{R}^4$s, the $f$-pullback of the cotangent bundle of $M_1$ is base-preservingly isomorphic (as a topological vector bundle) to the cotangent bundle of $M_0$. This yields uncountably many (parallelisable) examples where the answer to question 1 is NO. -All these total spaces are also symplectomorphic; this yields uncountably many examples where the answer to question 2 is NO. That the total spaces of the cotangent bundles of any two exotic $\mathbb{R}^4$s $M_0,M_1$ are symplectomorphic follows from Corollary 1.12.D in Eliashberg/Gromov: Convex symplectic manifolds, because $M_0,M_1$ are tangentially equivalent in the Eliashberg/Gromov sense: There is (again by triviality of the bundles) a vector bundle morphism $T^*M_0\to T^*M_1$ which is fibrewise bijective and whose underlying map $M_0\to M_1$ is a homotopy equivalence.<|endoftext|> -TITLE: Kähler metric on projectivised bundle -QUESTION [7 upvotes]: Let $E\rightarrow M$ be a holomorphic bundle over a Kähler manifold. Does its projectivisation $\mathbb{P}(E)$ always admit a Kähler metric? If yes, how to see that? - -REPLY [9 votes]: Let me recall you briefly how to obtain such a Kähler metric on the total space of the projectivized bundle. -Start with any given hermitian metric $h$ on $E$ and consider on the projectivized bundle $\mathbb P(E)$ of hyperplanes of $E$ the tautological line bundle $\mathcal O_E(1)\to\mathbb P(E)$ of rank one quotients of $E$. Then, on $\mathcal O_E(1)$ you have a natural induced quotient hermitian metric, which I shall call again $h$. -If you compute the Chern curvature $\Theta(\mathcal O_E(1))$ of $\mathcal O_E(1)$ with respect to $h$, you will find a closed $(1,1)$-form on $\mathbb P(E)$ which is positive along the relative tangent bundle (after all, the restricion of $\mathcal O_E(1)$ to fibers $\simeq\mathbb P^{\operatorname{rk}E-1}$is just the usual $\mathcal O(1)$, so that its curvature restricted to fibers is just the usual Fubini-Study metric). No more can be said along "horizontal" directions. -Now, suppose that $M$ is compact Kähler, with Kähler form $\omega$ and call $\pi\colon\mathbb P(E)\to M$ the natural projection. Since $E$ is holomorphic (see Francesco's answer), $\pi$ is holomorphic as well. Thus, $\pi^*\omega$ is again a closed $(1,1)$-form on $\mathbb P(E)$ which is zero on vertical directions (as a pull-back) and strictly positive on horizontal ones. -Finally, by compactness of $M$, and so of $\mathbb P(E)$, you can find a large constant $C$ such that the closed $(1,1)$-form -$$ -C\,\pi^*\omega+i\,\Theta(\mathcal O_E(1)) -$$ -is positive definite everywhere (such a large multiple is chosen in such a way that $\omega$ compensates the possible lack of positivity of $\Theta(\mathcal O_E(1))$ along horizontal directions, and observe that $\omega$ does not interfere with the vertical ones). -This gives you the desired Kähler metric on $\mathbb P(E)$.<|endoftext|> -TITLE: How to invert the matrix [n choose 2j - i] ? -QUESTION [13 upvotes]: In a certain model of a stat-physics type, one encounters a matrix -$$ -A_n:=\left[\binom{n}{2j-i}\right]_{i,j=1}^{n-1}. -$$ -The determinant of this matrix (equal to $2^{\binom n2}$) counts the number of all possible configurations, and our understanding of the model would greatly increase if we would know the inverse of this matrix. So the question is, - -is there a closed-form expression for the inverse of the matrix $A_n$? - -A little more information about the matrix: -its eigenvalues are $2^i\colon i=1,\dots,n-1$. The eigenvectors are not orthogonal to each other (and $A_n$ is not a symmetric matrix, for sure), but the vectors corresponding to even $i$'s are orthogonal to the ones which correspond to the odd $i$'s (i.e., the whole space orthogonally decomposes into the "even" and the "odd" parts). -PS I would appreciate any help or reference. I've looked through Krattenthaler's seminal "Advanced determinant calculus", but didn't find such a matrix there. - -REPLY [9 votes]: You probably know the following already. Oh, well. -Anyway, Eric Nordenstam and I answered exactly this problem. -It is a special case of Theorem 1 in our preprint, http://arxiv.org/abs/1201.4138 and it also appears in the proceedings of FPSAC 2012. Our method was inspired (and suggested) by Krattenthaler: We guessed the answer from empirical data, and then proved our guess was right by multiplying the matrices out and confirming that the answer was the identity matrix.<|endoftext|> -TITLE: The ring generated by all functions from a set to itself -QUESTION [8 upvotes]: Let $S$ be a finite set. Now $\mathop{End}(S)$ is a monoid, and we may build a ring $R$ by allowing formal sums of functions. -Preliminary questions, since $R$ is surely well-known: What is it called? In a general category, what is the name of the construction that builds a ring out of the endomorphisms of an object? -My real question is about $A = R \otimes \mathbb{C}$. Restricting to the subalgebra generated by the automorphisms of $S$, it is clear that understanding the representation theory of symmetric groups is a prerequisite for understanding $A$. Are the symmetric groups on sets of size smaller than $| S |$ responsible for all nontrivial properties of $A$? If so, what suitably strong formulation of inclusion/exclusion is at work? - -REPLY [13 votes]: The monoid of all maps on $n$ letters is denoted $T_n$ and called the full transformation monoid. Your intuition is both right and wrong. The irreducible representations of $T_n$ are in bijection with irreducible representations of all symmetric groups of degree at most $n$. The character table is block upper triangular with diagonal blocks character tables of symmetric groups of degree at most $n$. Thus inverting the table, which is how you decompose characters into irreducibles, is a sort of generalized inclusion-exclusion. Putcha computed the character table, although Hewitt and Zuckermann stated a portion of the result without a complete proof in 1957. But this is only the semisimple part of the story. -The algebra of $T_n$ is not semisimple. For example, nobody knows what the projective indecomposables look like in general. The algebra is quasihereditary and so has finite global dimension. Putcha computed a lot of information about the quiver. Ponizovsky showed $T_1,T_2,T_3$ has finite representation type, Ringel showed $T_4$ has finite representation type and Putcha showed the representation type is infinite in all other cases. I think this part of the representation theory is not completely controlled by symmetric groups. -Update 1/15. -In case you are still interested I just put up a paper -http://arxiv.org/abs/1502.00959 computing the global dimension of this algebra. From the paper you can get a very precise idea of how the representations of the symmetric groups of different degrees interact to control the Ext spaces between simple modules in a non trivial way. Also the paper gives a more accurate survey of the representation theory of this algebra than my answer.<|endoftext|> -TITLE: How does one view the De Rham spectral sequence as a Grothendieck spectral sequence? -QUESTION [6 upvotes]: I was rereading basic results on de Rham cohomology, and this led me inevitably to the fact that $H^q(X,\Omega^p)$ converges to $H^*(X)$ for any smooth proper variety (over any field). How does one view this spectral sequence "maturely" as a Grothendieck spectral sequence? - -REPLY [7 votes]: You probably want one (the second one...) of the two hypercohomology spectral sequences which compute $\mathbb H^\bullet(X,\Omega^\bullet)$, the hypercohomology of the de Rham complex. A reference for this is Weibel's book. -I doubt you can view this as a Grothendieck spectral sequence, but it has sufficiently much hyper in it to be considered mature, I guess. - -REPLY [6 votes]: If by "Grothendieck spectral sequence" you mean the spectral sequence associated to the composite of functors (fulfilling the Grothendieck condition) then I am skeptical as to whether this is possible. Also I do not see that there would be any particular point in being able to view it in that light (unless the functors involved would be interesting in themselves). As you seem to be looking for some general principle that would give the dRss I would like to a mention one such which seem to give most spectral sequences in use very quickly. Usually one constructs the dRss by taking a Cartan-Eilenberg resolution of the de Rham complex and then consider the spectral sequence associated to a double complex. However, one can instead use the Massey exact couple construction; once one has an exact couple one automatically gets a spectral sequence. One systematic way of constructing such exact couples is to start with a triangulated category $T$, a sequence of morphisms $\cdots\to X_{i-1}\to X_i\to X_{i+1}\to\cdots$ and a homological functor $H$ on $T$. This gives a spectral sequence starting with $H^i(Y_j)$ and going towards $\varinjlim_iH^\ast(Y_i)$. (Convergence is not assured but is OK for instance of $Y_i$ is $0$ for small $i$ and equal if $i$ is large.) Starting with the naive truncations of de Rham complex gives the dRss (and starting with canonical truncations in the algebraic case gives the conjugate spectral sequence). -This setup works very generally, for instance it is how Adams first constructed the Adams spectral sequence.<|endoftext|> -TITLE: On which space does $GL_n(F_p[X])$ act nicely? -QUESTION [9 upvotes]: The group $GL_n(\mathbb{Z})$ acts properly and isometrically on the space of homothety classes of scalar products on $\mathbb{R}^n$. This is a Riemannian manifold of nonpositive sectional curvature. -Is there a similar space for the case of $GL_n(F_p[x])$. Maybe one can construct a building or something like this. I guess if the space I am looking for exists it should be rather well known. Otherwise I apologize for the vagueness of this question. - -REPLY [10 votes]: The group $GL_n(\mathbb{F}_p[x])$ acts on the Bruhat-Tits building for $GL_n$. The vertex set is $GL_n(\mathbb{F}_p((x^{-1})))/GL_n(\mathbb{F}_p[[x^{-1}]])$, and the higher simplices form sets of the form $GL_n(\mathbb{F}_p((x^{-1})))/I$ for various parahoric groups $I$. The action of $GL_n(\mathbb{F}_p((x^{-1})))$ on the left restricts to an action of $GL_n(\mathbb{F}_p[x])$. -When $n=2$ there is a nice exposition of the action in chapter 2 of Serre's Trees (as long as you're okay with $PGL_2$ instead of $GL_2$). You get an infinite sequence of 1-simplices that vaguely resembles the $SL_2(\mathbb{Z})$-quotient of the complex upper half-plane. In higher rank, I don't think things can be quite so explicit. -You can also find actions of $GL_n(\mathbb{F}_p[x])$ on many other spaces, namely those that are defined over rings containing $\mathbb{F}_p[x]$ and that come equipped with a natural algebraic or analytic action of $GL_n$. For example, the analytification of the flag variety for $GL_n$ over the completion of an algebraic closure of $\mathbb{F}_p((x^{-1}))$ admits such an action.<|endoftext|> -TITLE: Source on functorial algebraic geometry -QUESTION [21 upvotes]: I want to understand algebraic geometry from the functorial viewpoint. I've found a set of notes (linked below) that develop algebraic geometry from the elementary beginnings in this framework. They go under the name "Introduction to Functorial Algebraic Geometry" (following a summer course held by Grothendieck), and are in parts in an almost unreadable shape. Is there an available, elementary, and readable source which takes the functorial viewpoint? -http://www.math.jussieu.fr/~leila/grothendieckcircle/FuncAlg.pdf -EDIT: Great answers so far, thanks! I would like to add that if anyone were to have a better scan of the document linked to above, or if anyone has vol. 2 of the notes I would be very interested in aquiring a copy. - -REPLY [4 votes]: Arun Debray has written up (mostly complete) lecture notes for a course on algebraic geometry taught by Sam Raskin which takes a completely functorial perspective, and doesn't take the perspective that $\mathrm{Spec}(R)$ is a topological space. -https://www.math.purdue.edu/~adebray/lecture_notes/m392c_Raskin_AG_notes.pdf<|endoftext|> -TITLE: Is the Lie algebra-valued curvature two-form on a principal bundle P the curvature of a vector bundle over P? -QUESTION [8 upvotes]: I am an analyst struggling through some geometry used in physics. -Some background: For some Lie group $G$, let $P$ be a principal $G$-bundle over the smooth manifold $M$. Let $\omega$ be a connection 1-form on $P$ (a "principal connection"). This is a Lie algebra-valued 1-form. -As for the curvature two-form, either you see definitions with no explanation at all, e.g. "The curvature is given by $\Omega = d\omega + \frac{1}{2}[\omega,\omega]$". This is obviously less than ideal for improving one's intuitive appreciation. Or one defines something called the exterior covariant derivative $D$ (see wiki) and then the curvature is simply the exterior covariant derivative of the connection one-form. -The issue: I can't get round the following observation though: From the point of view of the manifold $P$, $\omega$ is just a one-form with values in some vector space which happens to be $\mathfrak{g}$. Usually when you need to covariantly differentiate such an object, you would need a connection in a bundle $E \to P$ with fibre $\mathfrak{g}$, no? Then $\omega$ would be an $E$-valued one-form on $P$, i.e. in $\Gamma(E) \otimes\Omega^1(P)$, and you can differentiate covariantly in the normal way using the connection. Why is this scenario different? -The exterior covariant derivative $D$ satisfies $D^2\phi = \Omega\wedge\phi$... So you have some sort of covariant differentiation $D$ which differentiates forms $\eta$ taking values in $\mathfrak{g}$ and for which $D^2$ is some sort of curvature... but of $P \to M$ and not of the bundle in which $\eta$ is taking values. Isn't this strange? Or is this indeed just how things are? This prompts my more precise question: - -Is the Lie algebra-valued curvature - two-form on a principal bundle P the - curvature of some vector bundle over P with fibre $\mathfrak{g}$? - -REPLY [7 votes]: Your confusion is revealed in this sentence "Or one defines something called the exterior covariant derivative D (see wiki) and then the curvature is simply the exterior covariant derivative of the connection one-form." This is just not true; one does not take the `exterior covariant derivative of the connection $1$-form' to get the curvature. -Let's be precise: Let $P\to M$ be a principal right $G$-bundle, and let $\omega$ be a $\frak{g}$-valued $1$-form on $P$ that defines a connection on $P$ (I won't repeat the well-known requirements on $\omega$). The curvature $2$-form $\Omega = d\omega +\frac12[\omega,\omega]$ on $P$ is the $2$-form that vanishes if and only if it is possible to find local trivializations $\tau: P_U \to U\times G$ such that $\omega = (\pi_2\circ\tau)^*(\gamma)$ where $\gamma$ is the canonical left-invariant $1$-form on $G$. -Now, given a representation $\rho:G\to \text{GL}(V)$, where $V$ is a vector space, one can define an associated vector bundle $E = P\times_\rho V$. Using $\omega$, it is possible to define an 'exterior covariant derivative operator' $D_\omega:\Gamma(E\otimes A^p)\to \Gamma(E\otimes A^{p+1})$ (where $A^p\to M$ is the bundle of alternating (i.e., 'exterior') $p$-forms on $M$). The operator ${D_\omega}^2:\Gamma(E\otimes A^p)\to \Gamma(E\otimes A^{p+2})$ then turns out to be linear over the $C^\infty$ alternating forms, so it is determined by its value when $p=0$, i.e., by ${D_\omega}^2:\Gamma(E)\to \Gamma(E\otimes A^{2})$, which can be regarded as an section of $\text{End}(E)\otimes A^2$, i.e., a $2$-form with values in $\text{End}(E)$. The formula for this section, when pulled back to $P$, can now be expressed in terms of $\Omega$ in the usual way. In particular, ${D_\omega}^2$ vanishes identically if $\Omega$ does. -Note that one does not take the 'exterior covariant derivative' of the $1$-form $\omega$ anywhere. Instead, one takes the exterior covariant derivative of the exterior covariant derivative of a section of $E$. -I suspect that what you may be trying to do is interpret $\omega$ as a $1$-form on $P$ with values in the trivial bundle $P\times\frak{g}$ and then say that the curvature is the 'exterior covariant derivative' of $\omega$. However, to make this work, you have to specify a connection on the trivial bundle $P\times\frak{g}$, which amounts to choosing a $1$-form $\eta$ on $P$ that takes values in $\text{End}(\frak{g})$ and setting $D_\eta(s) = ds + \eta\ s$. By setting $\eta = \frac12\text{ad}(\omega)$, one gets $D_\eta\omega = \Omega$ (just by definition), so it is possible to do this, but I don't think that this is that useful an observation, since, after all, you could have taken the connection on the trivial bundle $P\times\frak{g}$ to be $\eta = \frac13\text{ad}(\omega)$ (for example) or even $\eta=0$. What justifies the $\frac12$, other than the desire to get the `right' answer?<|endoftext|> -TITLE: Axiom of choice and bases of vector spaces over a fixed field -QUESTION [18 upvotes]: Let $k$ be a field. In 1984 Andreas Blass proved that the axiom "for every extension $K|k$, every vector space over $K$ has a basis" implies the axiom of choice. He also raised the question - -Does the axiom "every vector space over $k$ has a basis" imply the axiom of choice ? - -What's the current status of the question ? Has there been progress ? - -REPLY [12 votes]: It has been shown for $K=\mathbb F_2$ (the field with two elements) by Keremedis (Available here) -In the dictionary of AC equivalences it shows that not a lot is known on the connection between the existence of a basis over a fixed field and the axiom of choice.<|endoftext|> -TITLE: Geometry of complex elliptic curves -QUESTION [23 upvotes]: Is there an elliptic curve in CP^2 whose induced Remannian metric ( induced from the Fubini-Sudy metric on CP^2) is Euclidian flat? - -REPLY [6 votes]: A much more general result holds. If $M$ is a compact complex manifold and $f:M\to\mathbb{CP}^n$ is a holomorphic embedding such that $f^*g_{FS}$ is an Einstein metric, then the Einstein constant must be strictly positive. This is a theorem of D. Hulin. -One can also wonder what manifolds $M$ one can obtain in general (i.e. what compact complex submanifolds of $\mathbb{CP}^n$ are Einstein for the induced Fubini-Study metric). It is believed that these must all be complex homogeneous spaces, and a complete classification is known in the case of (complex) codimension at most $2$, and for complete intersections. See this other paper of Hulin and the references there.<|endoftext|> -TITLE: An example where finitistic dimension does not equal right global dimension? -QUESTION [5 upvotes]: The (right) big finitistic dimension of a ring is Findim$(R) =$ sup{proj.dim(M) | $M$ a right $R$-module of finite projective dimension}. The (right) little finitistic dimension findim$(R)$ is the sup over f.g. right modules of finite projective dimension. -The right global dimension of a ring is r.gl.dim$(R) =$ sup{proj.dim(M) | $M\in R$-mod} -It is clear that Findim$(R)\leq$ r.gl.dim$(R)$. According to T.Y. Lam's book Lectures on Modules and Rings, the right global dimension of $R$ is infinite iff there is some right $R$-module $M$ of infinite projective dimension. So if r.gl.dim$(R)<\infty$ then all $R$-modules have finite projective dimension and Findim$(R)=$r.gl.dim$(R)$. If r.gl.dim$(R)=\infty$ then there must be a module of infinite projective dimension but there could also be a chain of modules of finite but increasing projective dimension, i.e. big finitistic dimension can be finite or infinite. Clearly the only way to have Findim$(R)\neq$ r.gl.dim$(R)$ is to have infinite right global dim but finite big finitistic dimension. - -What is an example of a ring with Findim$(R)\neq$ r.gl.dim$(R)$? - -The classical example of a ring of infinite global dimension is $k[t]/(t^2)$ or really any ring with a nilpotent element. I suspect this ring has finite Findim, but I don't have a good enough sense of what $k[t]/(t^2)$-modules look like to easily pick out the ones of finite projective dimension. - -If Findim$(R)=n$, can we find such an example for any $n$? - -I have also seen the term "finitistic global dimension" running around. Here it is defined exactly as little finitistic dimension above. - -Are these two terms always interchangeable? - -REPLY [8 votes]: A non-semisimple self-injective algebra like $k[t]/(t^2)$ has the property that its modules are either of infinite projective dimension or projective. So its Findim is actually zero, while its gldim is infinite. -For your second question, pick a ring $R$ with global dimension $n$ and consider the direct product ring $R\times k[t]/(t^2)$. Its global dimension is infinite, and its Findim is $n$.<|endoftext|> -TITLE: Maximal number of connected components of complement to an affine plane real algebraic curve -QUESTION [7 upvotes]: Let $X$ be a (singular, reducible) affine plane real algebraic curve of degree $d$. -How we can estimate maximal number of connected components of it's complement in $R^2$ in terms of degree? - -REPLY [2 votes]: We can also prove that the maximal number of components for reducible nonsingular curve is $\frac{n^2+n+2}{2}$ by the following computation. -Let $m$ be a partition of $n$, where $m_i$ is the degree of $i$-th irreducible component of our curve. -Denote $s=\{i|m_i=1\}$, $t=\{i|m_i>1\}$. -Then, using Harnack inequality and Besout's theorem(for intersections with infinite line and for intersections between irreducible components) we can write that the number of connected components is lesser than -$$ -\max_{m\vdash n}(\sum_{i}(\frac{(m_i-1)(m_i-2)}{2}+1)+1+n-s+\sum_{i\neq j}m_im_j)=$$ -$$=\frac{n^2}{2}-n+1+\max_{m\vdash n}(2t+s))\leq $$ -$$\leq \frac{n^2}{2}-n+1+ \max_{0\leq 2x+y\leq n,0\leq x,0\leq y}(2x+y)=\frac{n^2+n+2}{2} -$$ -Therefore the only question we have is the following: -why do Harnack bound (in terms of degree) holds for singular curves as well as for nonsingular? -The proof could be obtained using First Harnack Theorem.<|endoftext|> -TITLE: Symmetric Ricci Tensor -QUESTION [6 upvotes]: Let $M$ be a pseudo-Riemannian manifold. Assume that $M$ comes with a zero-torsion affine connexion $\nabla.$ There is no need for $\nabla$ to be the Levi-Civita connexion. Recall that the curvature tensor of $\nabla$ is given by $R(X,Y)Z = \nabla_X\nabla_YZ - \nabla_Y\nabla_XZ - \nabla_{[X,Y]}Z,$ while the Ricci tensor is given by the trace of the curvature tensor: $\mbox{Ric}(Y,Z) = \mbox{trace}\left[ X \mapsto R(X,Y)Z \right].$ -The connexion $\nabla$ is called locally equi-affine if around each point $x \in M$ there is a parallel volume form, i.e. a non-vanishing, skew-symmetric, $n$-form such that $\nabla\omega = 0.$ -In Nomizu & Sasaki's "Affine Differential Geometry", in Prop 3.1 on page 14, they say that a zero-torsion connexion has symmetric Ricci tensor if and only if it is locally equi-affine. -In fact, there is a differential one-form $\tau$ such that $\nabla_X\omega = \tau(X)\omega$ for any vector field $X \in \mathfrak{X}(M).$ Thus, we see that $\nabla$ has symmetric Ricci tensor if and only if $\tau \equiv 0.$ -However, in Section 4 of page 237, they say that $\nabla$ has symmetric Ricci tensor if and only if the exterior derivative $\mbox{d}\tau \equiv 0.$ -Clearly only one of them is correct, but the proof to Prop 3.1. is so sparing with details that I can't follow it. It's one of those proofs that if you can't prove the statement thenyou can't understand the proof. -So, is it $\tau \equiv 0$ or $\mbox{d}\tau \equiv 0$? - -REPLY [8 votes]: They are both correct. :-) -I gave a somewhat detailed write-up last year on my blog, but the gist of the argument is that if $\tau = 0$, then $d\tau = 0$. On the flip side, if $d\tau = 0$, locally you can lift $\tau = du$ for some function $u$, and by replacing $\omega$ with $e^{-u}\omega$ you construct the desired equi-affine volume form. -In other words, the second statement applies to arbitrary volume forms, while the first statement that $\tau$ vanishes applies to a special volume form, the one that gives the equi-affine condition. - -Just to make clear that there's no contradiction. We first note that for any volume form (non-vanishing differential form of top degree), the expression $\nabla_X \omega = \tau(X)\omega$ holds for some one-form $\tau$ depending on $\omega$. Since we are interested in local statements, assume now our manifold is diffeomorphic to the unit ball. -The first statement says that "$\exists \omega$ a volume form such that $\tau = 0$." -The second statement say that "$\forall \omega$ a volume form, the corresponding $d\tau = 0$." -The two statements are equivalent using what I sketched in the second paragraph above.<|endoftext|> -TITLE: Strict Transform under Blow-Up along nonsingular Subvariety -QUESTION [7 upvotes]: Let $\beta:\widetilde{X}\mathrel{\mathop:}=\mathop{\mathrm{Bl}}_Z(X)\to X$ be the blow-up of a nonsingular algebraic variety $X$ along a nonsingular subvariety $Z$. Let $E\mathrel{\mathop:}=\beta^{-1}(Z)$ be the exceptional divisor. Now, let us assume I have a divisor $D$ on $X$. Then I was told that $\beta^\ast D \sim \widetilde{D} + \alpha E$, where $\widetilde{D}$ denotes the strict transform of $D$ and "$\sim$" is linear equivalence. I was also told that $\alpha$ is the "multiplicity of $D$ along $Z$". First, what does multiplicity mean here and second, does anyone know (possibly by reference) a proof? -Note: This bears some relation to Hartshorne Exercise II.8.5. For fibred surfaces, there is also Proposition 9.2.23 in Liu's Book. However, it seemed very specific to the twodimensional case. - -REPLY [7 votes]: First of all, let $X$ be a smooth variety and $D$ an effective divisor on $X$. Denote $\operatorname{mult}_x(D)$ the multiplicity of $D$ at a point $x\in X$. The function $x\mapsto\operatorname{mult}_x(D)$ is known to be upper-semicontinuous on $X$. Therefore, if $Z\subset X$ is any irreducible subvariety, one can define the multiplicity of $D$ along $Z$, denoted $\operatorname{mult}_Z(D)$, to be the multiplicity $\operatorname{mult}_x(D)$ at a general point $x\in Z$. -Now, the second part of your question is just a matter of local computation. I will sketch it in the case where $X$ is a smooth surface and $Z$ is a point $x_0\in X$, leaving to you to work out the details in more general situations. -So, fix local coordinates $(x,y)$ centered at $x_0$ and consider the blow-up map $\mu\colon\widetilde X\to X$ given by $\mu(u,v)=(uv,v)$. In this chart, the exceptional divisor $E$ is given by the single equation $\{v=0\}$. Now, take a divisor $D\subset X$ whose local equation near $x_0$ is given by $\{f(x,y)=0\}$. Let -$$ -f(x,y)=\sum_{j,k\ge 0}a_{jk}x^jy^k. -$$ -Then, $\operatorname{mult}_{x_0}(D)=m$, where $m=\inf\{j+k\mid a_{jk}\ne 0\}$. -Finally, a local equation for $\mu^*D$ is given by -$$ -f\circ\mu=f(uv,v)=\sum_{j,k\ge 0}a_{jk}u^jv^{j+k}=v^m\underbrace{\sum_{j,k\ge 0}a_{jk}u^jv^{j+k-m}}_{\text{holomorphic and does not vanish along $E$}}. -$$ -Thus, you see that $\mu^*D$ consist of the sum of one irreducible component given by $mE$ and the remaining part is just the proper transform of $D$.<|endoftext|> -TITLE: Is there a way to define a prime ideal object via diagrams in the category of rings? -QUESTION [9 upvotes]: I like to think in terms of commutative diagrams rather than referring to elements. So to me a group is really a group object, i.e. an object with some maps satisfying certain commutative diagrams. You can define rings and modules similarly. You can define a field object as a commutative ring object with the inverse axiom from a group object applied to the multiplication (alternately as a commutative ring object over which every module is free). Assuming your category is abelian, one would hope that you could define a maximal ideal $M$ by an exact sequence $0 \rightarrow M \rightarrow R \rightarrow k \rightarrow 0$ where $k$ is a field object. I'm thinking ahead to generalizing this to a triangulated category where $M$ could be uniquely defined via a fiber sequence. -I’d like to define an ideal $I$ as a subobject of $R$ which is also an $R$-bimodule. Both of these can be defined by diagrams, without reference to elements. However, I'm a bit afraid that this is completely wrong since I've never heard of an "ideal object", and it seems Google has not either. Is there something I’m missing here that makes my definition fail? -The next object I’d like to define is a prime ideal. I have no idea how to do this without referring to elements. The trick with maximal ideals above doesn’t seem to work unless someone has a way to classify “integral domain objects” via diagrams. - -Has anyone ever heard of a way to define prime ideals purely diagrammatically? - -Alternately, assuming that I totally botched that attempt to define an "ideal object," one could still ask about prime modules, i.e. nonzero modules $M$ such that $IN = (0)$ implies $N = (0)$ or $IM = (0)$ for any ideal $I$ of $R$ and any submodule $N$ of $M$. Has anyone heard of a diagrammatic way to define these? -I'm tagging this Topoi because an answer of Peter Arndt to a totally different MO question gives me a small amount of hope that some of the mysteries of Topos Theory could help me. - -REPLY [13 votes]: The lattice of ideals of $R$ is isomorphic to the lattice of regular quotients of $R$. Here, a regular quotient is an equivalence class of regular epimorphisms $R \to S$ in the category of rings (which are precisely the surjective ring homomorphisms). So this also serves as a categorical definition of an ideal. Also the ideals of $R$ are exactly the subobjects of $R$ in $\text{Mod}(R)$, but I think that you want to keep in the category of rings, right? Now, a prime ideal is characterized by the fact it is a proper ideal $\mathfrak{p}$ and for all ideals $\mathfrak{a}, \mathfrak{b}$ we have that $\mathfrak{a} * \mathfrak{b} \subseteq \mathfrak{p}$ implies $\mathfrak{a} \subseteq \mathfrak{p}$ or $\mathfrak{b} \subseteq \mathfrak{p}$. Unfortunately, there is no lattice-theoetic definition of the product of ideals, see this previous MO question, but there I give a definition which comes from ideas of Rosenberg's noncommutative algebraic geometry and uses subcategories of $\text{Mod}(R)$. -However, you can also give the following characterization within the category of rings: A prime ideal corresponds to a regular quotient $R \to S$ where $S$ is an integral domain. Now an integral domain is characterized by the property that it embeds into a field, where embeddings are given by monomorphisms (which actually coincide with the injective ring homomorphisms as in every algebraic category). A field $K$ is characterized by the following property: $K \neq 0$, where $0$ is the terminal object, and the lattice of ideals has exactly two elements, namely $0$ and $K$. -You may also consult the overlooked book "Categories of commutative algebras" by Yves Diers. Here the author developes all the basic commutative algebra for so-called Zariski categories, which are special locally presentable categories which resemble categories of commutative rings. In the chapter "Classical objects" we defines integral objects etc. in every Zariski category. Namely: First a simple object is defined to have exactly two congruences, and an integral object is defined to be one which embeds into a simple one.<|endoftext|> -TITLE: The Galois representation of a p-divisible group is crystalline -QUESTION [8 upvotes]: Can someone explain (or give a reference) why the Galois representation attached to a p-divisible group over the ring of integers of a p-adic ring is Crystalline? - -REPLY [8 votes]: This is shown in §6 of Fontaine's paper "Sur certains types de représentations p-adiques du groupe de Galois d'un corps local; construction d'un anneau de Barsotti-Tate", see the point i) after Theorem 6.2. -The paper is available on jstor: http://www.jstor.org/pss/2007012<|endoftext|> -TITLE: The category of l-adic sheaves -QUESTION [25 upvotes]: I'm currently trying to understand the construction of the category of l-adic constructible sheaves as in SGA5, and it seems that quite a lot of machinery (the MLAR condition, localization of the category of projective systems, etc.) has to be gone through before one can even construct this category and show that it's abelian, for instance. On the other hand it is not even true that the derived category of l-adic sheaves is defined in the obvious manner, since it is defined as a 2-limit of the derived categories of $\mathbb{Z}/l^n$ constructible sheaves. -I understand that the categorical machinery in existence today is a lot more powerful than it was in the 1970's, which makes me curious: is there a cleaner and more transparent way of doing this, and a more modern presentation than SGA or Frietag-Kiehl? - -REPLY [4 votes]: Zheng and Liu are using $\infty$-categories to study constructible sheaves on stacks, and they have a $\ell$-adic version too. (Though most of the details for the $\ell$-Adic version should appear in a second paper that is still in preparation, and I would not call their first paper easy to read. But it is certainly a modern presentation...) Reference : -http://math.columbia.edu/~zheng/bc1.pdf -By the way, they use Gabber's finiteness results, and there is now a nice reference for these too ! (This is really cool.) -http://www.math.polytechnique.fr/~orgogozo/travaux_de_Gabber/<|endoftext|> -TITLE: Categorical definition of the ideal product within the category of rings -QUESTION [44 upvotes]: This is an extension of this question. Let $I,J$ be ideals of a ring $R$; every ring is commutative and unital here. Is it possible to define $R \to R/(I*J)$ out of $R \to R/I$ and $R \to R/J$ in categorical terms within the category of rings? To be more precise: Is there a formula in the language of category theory $\phi$ with a parameter of type "category" and three parameters of type "morphism", such that $\phi(\text{Ring},R \to R/I,R \to R/J,R \to R/K)$ is true if and only if $K = I*J$? -It is easy to do so for the ideal sum and the ideal intersection. Namely $K=I + J$ is characterized by $R/K = R/I \otimes_R R/J$ via the natural maps, the tensor product being the coproduct of $R$-algebras. And $K=I \cap J$ is characterized by the fact that $R/K$ is the universal regular quotient of $R$ such that $R \to R/I \times R/J$ factors through it; this is a fancy way of saying that $I \cap J$ is the kernel of $R \to R/I \times R/J$. But somehow it is quite difficult for the ideal product. Note that the linked question above shows that there will be no characterization just using regular quotients of $R$. -Although $I*J$ is the image of the natural morphism $I \otimes J \to R$, this takes place in the category of $R$-modules and thus leaves the given category of rings. Perhaps unitalizations are useful, but I cannot get rid of the factors $I,J$ in the canonical morphism $\tilde{I} \otimes \tilde{J} \to R$. Another idea is the following: $I*J \subseteq I \cap J$ and it suffices to characterize (the quotient of) $I \cap J / I*J$. This is an $R$-module isomorphic to $\text{Tor}_1(R/I,R/J)$. But again a priori this leaves the category of rings. -There is a categorical definition of prime ideals (see here) and thus also of radical ideals. Since we can also define intersections and inclusions, we also have $\text{rad}(I*J) = \text{rad}(I \cap J)$ as a categorical information, but of course this does not suffice to recover $I*J$. -EDIT: There is a somewhat nonsense positive answer: The ring $\mathbb{Z}[x]$ can be defined categorically, see here. Actually we also get the coring structure, including the multiplication resp. addition $\mathbb{Z}[x] \to \mathbb{Z}[x,y]$ and zero $\mathbb{Z}[x] \to \mathbb{Z}$. Then you can define the underlying set $|R|=\hom(\mathbb{Z}[x],R)$ categorically and also $|I|$ as the equalizer of two maps $|R| \to |R/I|$, the one being twisted by the zero morphism $\mathbb{Z}[x] \to \mathbb{Z} \to \mathbb{Z}[x]$. Now $|I*J| \subseteq |R|$ is the subset defined as the union of the images of the maps $(|R|^2)^n \mapsto |R|$ induced by $\mathbb{Z}[x] \to \mathbb{Z}[x_1,y_1,...,x_n,y_n], x \mapsto x_1 y_1 + ... + x_n y_n$, which comes from the coring structure. But now $R \to R/I$ is characterized by $|I|$, so we win. -Thus I want to change my question: Is there a more direct categorical definition of the ideal product, not going through $\mathbb{Z}[x]$ and thereby just imitating the element definition? -More generally (and here the above element definition does not work): If $X$ is a scheme and $Z \to X, Z' \to X$ are two closed subschemes, what is a categorical characterization of the closed immersion $Z'' \to X$ corresponding to the ideal product? What is the geometric meaning of it (this was partially discussed here)? The underlying space is just the union, but the structure sheaf not. For $Z'=Z$ this process is called thickening. - -REPLY [4 votes]: The unitalization approach can be made to work. -Let $C_K = \{ (r,s) \in R \times R \mid r-s \in K \}$ be the congruence defined by an ideal $K$. -Then, we have three maps defined on $S = C_I \otimes_R C_J$ : - -$\pi_0 : S \to C_I$ induced by the first projection $C_J \to R$ and the identity on $C_I$ -$\pi_1 : S \to C_J$ induced by the first projection $C_I \to R$ and the identity on $C_J$ -$\mu : S \to R \times R$ induced by the inclusions $C_I \to R\times R$ and $C_J \to R \times R$. - -Letting $\Delta \subseteq R \times R$ be the image of the diagonal, define $T = \{ x \in S \mid \pi_0(x) \in \Delta \wedge \pi_1(x) \in \Delta \}$. I claim that $\mu(T) = C_{IJ}$. -On the level of $R$ modules, we have isomorphisms $C_K \cong R \oplus K$, such as $(r,s) \mapsto (r, s-r)$, and so we have -$$ S \cong R \oplus I \oplus J \oplus (I \otimes_R J) $$ -In this form, the maps $\pi_i$ become projections onto the relevant summands, so $T$ is precisely the submodule $R \oplus (I \otimes_R J)$, so we've eliminated the $I$ and $J$ summands you were having trouble with. -By splitting the $R$-module maps, $T$ is genenerated as an $R$-module by elements of the form -$$ (r, r+i) \otimes (s, s+j) - (0,i) \otimes (s,s) - (r,r) \otimes (j,0) $$ -and applying $\mu$ to such a thing gives the element $(rs, rs + ij)$, and now it's easy to see that $\mu(T) = C_{IJ}$ as claimed.<|endoftext|> -TITLE: Are gaps in $n^2\sqrt{2}$ poissonian? -QUESTION [7 upvotes]: I would like to know gaps about the sequence $n^2 \sqrt{2} \mod 1$. Van der Corput's trick shows that $n^2 \sqrt{2}$ is equidistributed on the circle. For large $N$, the fraction -$$ \frac{\# \{ 1 \leq i \leq N: a < n^2 \sqrt{2} \mod 1 < b \} }{N} \approx b-a.$$ -Do the spacings between these values approach the Poisson distribution (as they would for uniformly random numbers)? Is there a proof of this involving Homogenous spaces? This is different from Elkies and McMullen's paper where they consider $\sqrt{n}\mod 1$ and relate it to the 5D space of Euclidean lattices. - -REPLY [13 votes]: Rudnick and Sarnak (MR1628282) conjecture that any not-too-well-approximable $\alpha$ has this property in place of $\sqrt{2}$, in particular any algebraic number should have this property. They prove that almost every $\alpha$ (in the Lebesgue sense) have this property, but no specific $\alpha$ has been exhibited so far. Rudnick-Sarnak-Zaharescu (MR1839285) and Zaharescu (MR1957276) have examined what happens for too-well-approximable $\alpha$'s. The above mentioned papers and their AMS reviews should serve as good starting points.<|endoftext|> -TITLE: Survey of Algebraic K-Theory Since 1980? -QUESTION [5 upvotes]: I just came across Charles Weibel's Development of Algebraic K-Theory until 1980, and found it really helpful. Is there been anything analogous which surveys the developments in the last 30 years? I'd be particularly interested in understanding links (if they exist) to motivic theory, geometric Langlands and higher class field theory. - -REPLY [3 votes]: I would suggest the lectures of Friedlander and Weibel: "An overview of algebraic K-theory" in Algebraic K-theory and its applications (Trieste 1997), 1999; MR. The later lectures include the modern point of view in terms of motivic cohomology and so forth together with connections to various theorems like the Milnor conjecture.<|endoftext|> -TITLE: reference for Noether's theorem -QUESTION [18 upvotes]: What is a good reference for a geometric version of Noether's theorem about Lagrangians, symmetries and conserved currents? - -REPLY [5 votes]: The simplest, most elegant and strongest version I know is, by far, the one in Aderson's book (see page 106 and ss.) He deals directly with the variational equation, with no explicit mention to the lagrangian. -(By the way, it is surprising why this statement is not mentioned in Schwarzbachs' book!)<|endoftext|> -TITLE: A simple proof of the Weyl algebra's rigidity. -QUESTION [10 upvotes]: I am wondering if there is a nice presentation of the Hochschild cohomology of $A_n$ the Weyl algebra. It is known that $H^m(A_n,A_n)=0$ for $m>0$, and thus it is rigid. A proof can be found in Sridharan, but this proof seems to be doing a lot more and is fairly complicated. -I was wondering if there was a simpler way to see this fact specifically. Essentially, I am being a bit lazy. -Thanks! - -REPLY [11 votes]: Since the Weyl algebra is a deformation of the polynomial ring in two variables, there is a short transparent proof using deformation theory, cf. M. Gerstenhaber and A. Giaquinto, On the cohomology of the Weyl algebra, the quantum plane, and the q-Weyl algebra, arXiv:1208.0346.<|endoftext|> -TITLE: Basic Question on action of Galois group on Tate module -QUESTION [5 upvotes]: I am trying to show that for an elliptic curve $E/K$ with complex multiplication the action of $G_{\overline{K}/ K}$ on the $T_{l}(E)$, the Tate module is abelian. -An approach: Let $\rho$ denote the Galois representation on the (rational) Tate module $(T_{l} \otimes \mathbb{Q}_{l})$. -Complex Multiplication implies that $dim(End(E) \otimes \mathbb{Q}_{l}) \geq 2 $. Since, each element in $End(E)$ acts as a intertwiner of $\rho$. -By Schur's lemma, the representation $\rho$ is evidently reducible. So the irreducible components of $\rho$ are one dimensional and thus the action of $G_{\overline{K}/ K}$ is abelian. -While this is a essentially representation theoretic, is there a more arithmetic proof of this -result? - -REPLY [6 votes]: Since $E$ has CM over $K$, the ring $F:= \operatorname{End}_K(E) \otimes \mathbb{Q}$ is an imaginary quadratic field. Suppose $\ell$ is a prime integer unramified in $F$. Now $F_\ell:= F \otimes_\mathbb{Q} \mathbb{Q}_{\ell}$ is either two copies of $\mathbb{Q}_{\ell}$ or a quadratic extension of $\mathbb{Q}_{\ell}$. -The two dimensional $\mathbb{Q}_{\ell}$-vector space $V_{\ell}:= T_{\ell} \otimes_{\mathbb{Z}_{\ell}} \mathbb{Q}_{\ell}$ is a rank one free module over $F_{\ell}$. The Galois action on $T_{\ell} \otimes \mathbb{Q}_{\ell}$ respects the action of $F_{\ell}$ and hence factorizes as $\operatorname{Gal}(\bar{K}/K) \to F^{\times}_{\ell} \hookrightarrow \operatorname{Aut} (V_{\ell}) = GL_2(\mathbb{Q}_{\ell})$. Here $F^{\times}_{\ell} $ is the units of $F_{\ell}$. -(This is a standard argument, see for example Silverman AEC or Milne EC)<|endoftext|> -TITLE: Binomial coefficients and derivatives of modular forms -QUESTION [10 upvotes]: Let $E_2$, $E_4$, and $E_6$ denote the standard Eisenstein series. -The usual variables $q=e^{2\pi i\tau}$ allow us to regard the -$E_n$'s as functions on either the upper half plane or the unit -disk and we can define $E_n'=\frac{1}{2\pi -i}\frac{d}{d\tau}E_n(\tau)=q\frac{d}{dq}E_n(q)$. I had cause to -calculate a few of these and saw -$$ E_4'=\frac{1}{3}(-E_6+E_4E_2) $$ -$$ E_4''=\frac{5}{36}(E_8-2E_6E_2+E_4E_2^2)$$ -$$E_4^{(3)}=\frac{5}{72}(-E_{10}+3E_8E_2-3E_6E_2^2+E_4E_2^3) $$ -$$E_4^{(4)}=\frac{35}{864}(E_4^3-4E_{10}E_2+6E_8E_2^2-4E_6E_2^3+E_4E_2^4)-40\Delta $$ -and -$$E_6'=\frac{1}{2}(-E_8+E_6E_2) $$ -$$E_6''=\frac{7}{24}(E_{10}-2E_8E_2+E_6E_2^2) $$ -$$E_6^{(3)}=\frac{7}{36}(-E_4^3+3E_{10}E_2-3E_8E_2^2+E_6E_2^3)+168\Delta $$ -It's a standard fact that the derivative of a modular form is -quasimodular, so it's not surprising that we have polynomials in -$E_2$. I am surprised about the appearance of the binomial -coefficients though. Is there a deeper reason for their -appearance? Also, I wonder if the/a pattern continues. For -instance, it would be interesting if it happens that there always -is some $\alpha \in \mathbb{Q}$ so that -$$E_4^{(n)}-\alpha \sum_{k=0}^{n} (-1)^{k+n}\binom {n}{k}E_{4+2n-2k}E_2^{k}$$ -is modular (and similarly for $E_6$). The other direction you -could ask if the pattern extends is for other modular forms -besides $E_4$ and $E_6$. I've taken a handful of derivatives of -other Eisenstein series and saw similar results. You don't get -the binomial coefficients though when you take derivatives of -$\Delta$, so maybe at most something general can be said is for -non-cusp forms. - -REPLY [16 votes]: The constant $\alpha$ in your question can be in fact written explicitly as $(k)_n/12^n$, where $(a)_n=\Gamma(a+n)/\Gamma(n)$ is the Pochhammer symbol (shifted factorial) and $k$ denotes the (even) weight of the corresponding Eisenstein series. -Your observation is indeed related to the Rankin--Cohen brakets; see Section 5.2 in [D. Zagier, Elliptic modular forms and their applications, The 1-2-3 of modular forms, Universitext (Springer, Berlin, 2008), pp. 1–-103]. Preserving the notation $D$ of Zagier's lectures for your differential operator and picking a modular form $f$ of weight $k$, one can show that $D^nf$ transforms under the modular group as -$$ -D^nf\biggl(\frac{a\tau+b}{c\tau+d}\biggr) -=\sum_{r=0}^n\binom{n}{r}\frac{(k+r)_{n-r}}{(2\pi i)^{n-r}} -c^{n-r}(c\tau+d)^{k+n+r}D^rf(\tau), -$$ -by the induction on $n\ge 0$. In addition, the function $E_2(\tau)$ transforms as -$$ -E_2\biggl(\frac{a\tau+b}{c\tau+d}\biggr) -=\frac{12c(c\tau+d)}{2\pi i}+(c\tau+d)^2E_2(\tau). -$$ -Therefore, it remains to verify that the difference -$$ -g_n=D^nE_k-\frac{(k) _ n}{12^n}\sum_{r=0}^n(-1)^{n-r}\binom{n}{r}E_{k+2n-2r}E_2^r -$$ -satisfies -$$ -g_n\biggl(\frac{a\tau+b}{c\tau+d}\biggr)=(c\tau+d)^{k+2n}g_n(\tau). -$$ -Technicalities. -Indeed, I found the remaining details quite boring, but going through my yesterday -writing I have realised that your expectation fails already for $D^5E_4$ and $D^4E_6$. -Here is my explanation why. -Because $g_n(\tau)$ is a $q$-series, so it is invariant under $\tau\mapsto\tau+1$, -we can restrict to verifying the claim under the transformation $\tau\mapsto-1/\tau$ -(that is, $a=d=0$, $b=-1$, and $c=1$). Then (we take $s=n-r$ in the above formula) -$$ -D^nE_k(-1/\tau) -=\sum_{s=0}^n\binom ns\frac{(k+n-s) _ s}{(2\pi i)^s}\tau^{k+2n-s}D^{n-s}E_k(\tau) -$$ -and -$$ -\begin{aligned} -& -\frac{(k) _ n}{12^n}\sum_{r=0}^n(-1)^{n-r}\binom nrE_{k+2n-2r}E_2^r\bigg|_{\tau\mapsto-1/\tau} -\cr &\qquad -=\frac{(k) _ n}{12^n}\sum_{r=0}^n(-1)^{n-r}\binom nr\tau^{k+2n-r}E_{k+2n-2r}\biggl(\tau E_2+\frac{12}{2\pi i}\biggr)^r -\cr &\qquad -=\frac{(k) _ n}{12^n}\sum_{r=0}^n(-1)^{n-r}\binom nr\tau^{k+2n-r}E_{k+2n-2r}\sum_{s=0}^r\binom rs\frac{12^s}{(2\pi i)^s}\tau^{r-s}E_2^{r-s} -\cr &\qquad -=\frac{(k) _ n}{12^n}\sum_{s=0}^n\binom ns\tau^{r-s}\frac{12^s}{(2\pi i)^s}\tau^{k+2n-s} -\sum_{r=s}^n(-1)^{n-r}\binom{n-s}{r-s}E_{k+2n-2r}E_2^{r-s}. -\end{aligned} -$$ -Subtracting the latter from the former we obtain -$$ -\begin{aligned} -g_n(-1/\tau) -&=\sum_{s=0}^n\binom ns\frac{(k+n-s) _ s}{(2\pi i)^s}\tau^{k+2n-s}g_{n-s}(\tau) -\cr -&=\tau^{k+2n}g_n(\tau)+\sum_{s=1}^n\binom ns\frac{(k+n-s) _ s}{(2\pi i)^s}\tau^{k+2n-s}g_{n-s}(\tau). -\end{aligned} -$$ -Therefore, $g_n(-1/\tau)=\tau^{k+2n}g_n(\tau)$, hence $g_n(\tau)$ is a modular form (of weight $k+2n$), -if and only if the additional sum over $s$ vanishes, that is, $g_{n-s}=0$ for $s=1,\dots,n$. -The latter however does not happen when $k+2n>12$.<|endoftext|> -TITLE: Do coarse moduli spaces respect Galois actions? -QUESTION [10 upvotes]: To explain, I will use the following concrete example: Let $\mathcal{M}_g$ be the functor for the moduli problem of classifying genus $g$ smooth projective curves (taking a scheme $S$ to the set of ways that $S$ parametrizes genus $g$ curves). This, as is well known, has a coarse moduli space: $M_g$. -Let $\sigma \in Aut(\mathbb{C})$, and say that we begin with a specific $\mathbb{C}$-curve of genus $g$. This $\sigma$ may act in two ways on this curve (by act I mean turn it into a different curve, not act as meaning an automorphism): -1. By taking a fiber product over the automorphism $Spec(\mathbb{C}) \rightarrow Spec(\mathbb{C})$ (this is the usual action people use, and has nothing to do with the moduli space) -2. By letting $\sigma$ act on $M_g(\mathbb{C})$, the point corresponding to our curve, will be taken to another $\mathbb{C}$-point, which in turn corresponds to another curve. -The question is: do these two actions agree in general? Are there conditions that need to be fulfilled for this to be true? -As an aside, this is true if the moduli space is fine. For example, if the moduli of genus $g$ curves were fine (which it isn't), then the curve corresponding to $Spec(\mathbb{C})\rightarrow M_g$ will be given simply by taking the fiber product with the universal family $E_g\rightarrow M_g$. Since fiber products commute, we get that the two actions above agree. This reasoning breaks down, however, if the moduli space is not fine. - -REPLY [5 votes]: There are curves (with automorphisms) for which the field of definition is not the field of moduli. Concretely, there are curves $C$ with corresponding point $P \in M_g$, such that $C$ cannot be defined over $K =\mathbb{Q}(P)$, only over larger fields. That means that there are automorphisms of $\mathbb{C}$ (fixing $K$) whose action don't change $P$ but change $C$.<|endoftext|> -TITLE: On the field of invariants of a finite group -QUESTION [9 upvotes]: So let $G$ be a finite group and let $\iota:G\rightarrow S_n$ be an embedding of $G$ in a symmetric group of degree $n$ for some fixed integer $n$. Let $K$ be a fixed field of characteristic $0$. The group $S_n$ permutes the variables $\{x_1,\ldots,x_n\}$ and therefore -acts on the field -$$ -L:=K(x_1,\ldots,x_n). -$$ -One may look at the invariant subfield $L^{G}\subseteq L$. From Galois theory one has that -$L/L^G$ is a Galois extension with Galois group $G$. In particular, the transcendence degree -of $L^G$ over $K$ is equal to $n$. In general, the field $L^G$ is not purely transcendental -so the following question makes sense: -Q: Does the isomorphism class of $L^G$ depend on the embedding $\iota$ ? -Intuitively I would say no, but this is really just a guess! - -REPLY [6 votes]: It seems likely (and may be known) that it does depend on the permutation representation. However, it is a subtle question as the stable isomorphism class (of any faithful permutation, or in fact arbitrary, representation) does not. Here two extensions $K$ and $K'$ of a field $k$ are stably isomorphic if for some $m$ and $n$ $K(x_1,\ldots,x_m)$ and $K'(x'_1,\ldots,x'_n)$ are isomorphic as $k$-extensions. -Addendum: The result is well-known but I cannot at the moment come up with a reference so instead I give the proof: Let $V$ be a faithful $G$-representation and $U$ the non-empty Zariski open subset where $G$ acts freely. Then $k(V)^G$ is the fraction field of $U/G$. If now, $V'$ is another faithful representation with open subset $U'$ we have that $U\times V'$ has a free $G$-action with a linear action on the second factor. Hence $U\times V'/G$ is a vector bundle over $U/G$ (by descent theory) and in particular its fraction field is stably isomorphic to that of $U/G$. However, $U\times V'/G$ is birational to $U\times U'/G$ which in turn is birational to $V\times U'/G$. The fraction field of the latter is for the same reason stably isomorphic to that of $U'/G$.<|endoftext|> -TITLE: The Origin of the Musical Isomorphisms -QUESTION [26 upvotes]: In Riemannian geometry, the "lowering indices" operator is denoted by $\flat:TM \to T^*M$ and the "raising indices" operator by $\sharp:T^*M \to TM$. These isomorphisms are sometimes referred to as musical isomorphisms, as stated on Wikipedia and in several other sources. Surely, the motivation for such terminology is clear. I would nevertheless like to know who decided to adpot these (rather amusing) notations, so here is a question: - -What was the first paper / textbook that made use of the notations $\flat$ and $\sharp$? - -and a possible follow-up question: - -If such notations were not adopted widely after the first appearance, who popularized them? - -REPLY [13 votes]: The musical isomorphisms already appear in 1971 on page 21 of BERGER M. et al, Le spectre d'une variete riemannienne, Lecture Notes in Math. 194, 1971, Springer, see http://ci.nii.ac.jp/naid/10003477917/ -However, I am not convinced that this is the first place where they appear, or that Berger was the inventor. One would have to trace German textbooks in Riemannian geometry from the 1960s or perhaps earlier. Interesting question!<|endoftext|> -TITLE: Effective Chebotarev density results for arbitrary number fields -QUESTION [5 upvotes]: So let $f(x)\in\mathbf{Z}[x]$ be a monic polynomial of degree $d$ and let $K$ be the splitting field of $f$. Let us define -the "heigt of $f$" $:=||f||$ to be the maximum of the abolute values of -the coefficients of $f$. (Instead of the height it might be better to work with the abolute value of the discriminant of $K$). -Let us denote the Galois group of $f$ over $\mathbf{Q}$ by -$G$. For each prime number for which the roots of $f$ modulo $p$ are distinct we denote by -$G_p$ the Galois group of $f\pmod{p}$. A cute result that may be found for example in Van der Waerden first algebra book says that there exists an (non-canonical but well defined up to conjugation in $G$) injection of $G_p$ in $G$. By elementary group theory, if we take the group generated by a set of representatives of the conjugacy classes of $G$ then it generates $G$. Thus by Chebotarev density theorem, we know that there exists a finite set of prime numbers -$S$ of $\mathbf{Q}$ such that -$$ -G_S:=\langle G_p| p\in S\rangle=G. -$$ -In particular, we may always choose a set $S$ with $G_S=G$ and -$|S|\leq r$ where $r$ is the number of conjugacy classes of $G$. -Q: So let $S_x$ be the set of all prime numbers less than $x$. Is it possible to find explicitly a lower bound for $x$ in terms $||f||$ (or $|disc(K)|$) and $d$ such that -$$ -G_{S_x}=G? -$$ -added: So basically, I'm just asking for an effective version of the Chebotarev density theorem for the splitting field of $f$, this is probably well known to the expert. -So probably one should consider $|disc(K)|$ rather than $||f||$ which can be arbitrary large for a fixed $K$ (even though $f$ may have bad reduction for many primes $p$ which do not divide $|disc(K)|$) - -REPLY [10 votes]: You are looking for the very useful paper -Effective versions of the Chebotarev density theorem, J. C. Lagarias and A. M. Odlyzko, pp. 409-464 in Algebraic Number Fields, A. Frohlich (ed.), Academic Press, 1977. -The bounds there are quite large, as I recall, especially if you don't want to assume GRH. There is very nice recent work of Jouve, Kowalski, and Zywina that give a lot of insight into how to do this in practice and how many prime numbers you shoudl expect to have to use: see Kowalski's blog post about this and work cited there.<|endoftext|> -TITLE: Limits of p-adic Representations -QUESTION [5 upvotes]: I'm thinking about properties of "limits" of p-adic representations, in the following sense. -Notations: $p$ denotes a prime. For a field $F$, let $G_F$ be the absolute Galois group of $F$. Representations are always continuous. -Definition: Let $\rho:\ G_{\Bbb{Q}_p}\rightarrow GL_d(\Bbb{Q}_p)$ be a representation, - $(\rho_n:\ G_{\Bbb{Q}_p}\rightarrow GL_d(\Bbb{Q}_p))_{n\geq 1}$ a sequence of representations. We say that the sequence converges uniformly to $\rho$ if it converges uniformly as a sequence of continuous functions. More explicitly, for every $\epsilon>0$, -for all sufficiently large $n$, and for every $\sigma$ in $G_{\Bbb{Q}_p}$, the (sup)norm of the matrix $\rho_n(\sigma)-\rho(\sigma)$ is less than $\epsilon$. From now on, simly say $(\rho_n)$ converges to $\rho$ (ie: drop the word "uniform"). -Remark+ Questions: -(0) Obviously, we can define convergence for representations from topological groups over topological rings. But just stick with the case above. -(1) Do we have an existing definition of limits of p-adic representations that are different (better?) than the naive one above? -(2) Assume the sequence $(\rho_n)$ converges to $\rho$, and all (actually infinitely many) of them are de Rham/ potentially semistable / semistable / crystalline, can we say that $\rho$ has the same property? -(3) Conversely, if $\rho$ is DR/ pst/ st/ cryst, can we say that almost all of the $\rho_n$ are too? -More remarks: -(1) A person gave me the following counterexample of (2) in the de Rham case: look at a nonclassical point $x$ on the eigencurve, it is the limit of points $x_n$ that have big integer weights hence classical (Coleman's result). The restriction to $G_{\Bbb{Q}_p}$ of the representation induced by $x$ is not de Rham (because it is not classical, and by the Fontaine-Mazur conjecture (?)). But the restrictions of representations given by $x_n$ are! -(2) I realize there is a paper of Laurent Berger "Limites De Representations Cristallines". Thm IV.2.1 in page 25, he proved that under extra conditions on the Hodge-Tate weights of the $\rho_n$, then (2) holds for the crystalline case. It seems to me his word "limites" have the same meaning as in the definition above. - -REPLY [2 votes]: Regarding $(1)$, the paper Converging Sequences of p-adic Galois Representations and Density Theorems by Bellaiche-Chenevier-Khare-Larsen studies four different notions of convergence: (pointwise) trace convergence, (pointwise) physical convergence, uniform trace convergence, and uniform physical convergence. The definition you give is that of uniform physical convergence. The results in the first section of the paper are of the form "trace convergence implies physical convergence." -http://arxiv.org/abs/math/0406102<|endoftext|> -TITLE: Can one use Atiyah-Singer to prove that the Chern-Weil definition of Chern classes are $\mathbb{Z}$-cohomology classes? -QUESTION [18 upvotes]: In Chern-Weil theory, we choose an arbitrary connection $\nabla$ on a complex vector bundle $E\rightarrow X$, obtain its curvature $F_\nabla$, and then we get Chern classes of $E$ from the curvature form. A priori it looks like these live in $H^*(X;\mathbb{C})$, but by an argument that I don't really understand they're actually in the image of $H^*(X;\mathbb{Z})$ (which is where they're usually considered to live). Meanwhile, I've heard people say that whenever I see a arbitrary real constants that end up having to be integers I should wonder whether the Atiyah-Singer index theorem is lurking in there somewhere. Is there anything to this wild guess? - -REPLY [5 votes]: My first remark about this question is a little bit pithy - the standard cohomological index formula works only for even dimensional manifolds while Chern-Weil theory is of course more general. -For a more substantive answer, let's look at the index formula itself: -$Index(D) = \int_{T^*M} ch(\sigma_D) Todd(TM \otimes \mathbb{C})$ -Here $ch(\sigma_D)$ refers to the chern character of the "symbol class" in K-theory of the elliptic operator $D$. My first observation is that this formula places special emphasis on the Todd class of the tangent bundle, and it seems to me that one would have to place similar emphasis on the Todd class in Chern - Weil theory in order to answer your question in the affirmative. I certainly do not claim that this is impossible - the Todd polynomial seems to take a privileged position in many places where it does not seem to belong - but I do not know how to do it. -Additionally, we are forced to ask what characteristic classes can be written in the form $ch(\sigma_D)$. As Tom Goodwillie points out, we will get an integrality result if we plug in any K-theory class $x$ rather than a K-theory class of the form $\sigma_D$ (although it may be that all classes in $K(T^*M)$ are symbol classes), but there are still constraints on which characteristic classes can appear. For example, we will only get classes in cohomology of even degree. -So there are a number of hints that the right hand side of the index formula is too specialized for Chern-Weil theory. But the left hand side also helps answer your question. Here are two general principles in index theory that will help me explain: - -The index theorem for any operator can be deduced from the index theorem for a special class of operators on a special class of manifolds - the "spin$^c$ Dirac operators". These operators are completely determined by the geometry of the manifolds on which they live. -The index depends continuously on the operator in an appropriate sense. In particular, if $D_t$ is a continuous family of operators then $Index(D_t)$ is constant in $t$. - -Combining these two facts, we see that the integers which we can obtain from the index theorem depend only on homotopy theoretic information about the manifold and the coefficient bundle (if applicable). This is not true of all integers that can be obtained via Chern-Weil theory, and it demonstrates that index theoretic invariants are much too crude for ordinary cohomology theory.<|endoftext|> -TITLE: Lawvere theories versus classical universal algebra -QUESTION [28 upvotes]: A Lawvere theory is a small category with finite products such that every object is isomorphic to a finite product of copies of a distinguished object x. A model of the theory in a category with finite products is a product preserving functor from the theory to that category. -This notion is supposed to be the right categorical viewpoint on universal algebra. Given a classical variety of universal algebra, one gets a theory by considering the opposite of the category of finitely generated free algebras. -My question is, what are the advantages/disadvantages of theories versus classical universal algebra? -I believe that theories are more general. What are natural examples of theories that are not covered by classical universal algebra? -Does the classical equational theory of universal algebra have a nice analog for theories? - -REPLY [3 votes]: One reason that Lawvere theories might be useful is in homotopy theory: Badzioch has apparently done work on formalizing the notion of a "homotopy algebra" (over a given Lawvere theory): - -Algebraic theories in homotopy theory, Annals of Mathematics 155 Issue 3 (2002) 895-913, https://doi.org/10.2307/3062135 - -The idea seems to be that an algebra over a Lawvere theory is something you can make homotopyish without recourse to things like operads: given $T$, a homotopy $T$-algebra is a functor $T \to \mathrm{Spaces}$ which preserves products up to homotopy equivalence (rather than on the nose). -In general, it seems that having a categorical language rather than explicit operations is much better for making algebraic structures homotopy invariant. An earlier example is given by the $\Gamma$-spaces of Segal: one writes the axioms of an abelian monoid in terms of a suitable product-preserving functor $\mathrm{Fin}_* \to \mathbf{Sets}$ and replaces those then by weakly product-preserving functors to spaces. It turns out grouplike $\Gamma$-spaces are essentially equivalent to infinite loop spaces. -I don't understand too much of this yet, but what's a little puzzling to me is that homotopy $T$-algebras (according to Badzioch) turn out to be essentially the same as ordinary $T$-algebras. (By contrast, the homotopy-theoretic analog of an abelian group—an infinite loop space—is generally very different from an abelian group.)<|endoftext|> -TITLE: Why is the degree:rank ratio of a vector bundle called its "slope"? -QUESTION [23 upvotes]: Whenever one studies moduli spaces of vector bundles on curves, one of the first things to be introduced is the "slope" of a vector bundle, i.e., its degree:rank ratio. Is there a nice (preferably geometric) intuition behind the use of the word "slope" for this? - -REPLY [6 votes]: Mumford studied the quasi-projectivity of moduli space of vector bundle. But such big family is unbounded. So he needed to consider moduli space of vector bundles with some sort of GIT. He arrived to the notion of slope stability. One advantage to restricting to semistable bundles of fixed rank and degree is that -the moduli problem is bounded. In fact Mumford extended Giesecker-stability notion. See section 2.5 -Suppose $\mathcal F$ is a coherent sheaf on a projective $X$. Define the Euler characteristic -$$\chi(\mathcal F)=\sum_{i=0}^{dim X}(-1)^i\dim H^i(X,\mathcal F)$$ -Let $L$ be an ample line bundle with $c_1(L) =\beta$ on a smooth projective variety $X$ of -dimension $g$. A torsion-free sheaf $E$ is Gieseker stable (respectively Gieseker -semistable) with respect to $\beta$ if for each subsheaf $F$ we have -$$P(F) < P(E) \;\; \;\; (P(F) \leq P(E))$$ -where $P(E) = \frac{\chi(E \otimes L^n)}{r(E)} $, is the reduced Hilbert polynomial. -Using Riemann-Roch theorem we have -$$P(E\otimes \mathcal O_X(1)^{\otimes m})=rk E\frac{m^n}{n!}+(deg E-rk E\frac{deg K}{2})\frac{m^{n-1}}{(n-1)!}+\cdots$$ -where $K$ is the canonical divisor. -We have nice interpretation between the notion of semi-stability for vector bundles and the notion of -semistability coming from an associated GIT problem -We can interpret the slope of vector bundle over a curve of genus $g$ using Riemann-Roch formula. Let me explain it -From of differential geometric point of view the degree of a holomorphic vector bundle can be computed by Chern–Weil formula in terms of curvature, and the fact that curvature decreases in sub-bundles. We explain Chern-Weil formula which gives an effective way for degree of holomorphic vector bundle. -A reflexive sheaf (i.e its double dual is equal itself) is locally free (i.e., a holomorphic vector bundle) outside a subvariety of codimension greater than or equal to two. Let $\mathcal F$ be a coherent subsheaf of holomorphic vector bundle $E$, then there is an analytic subset $S \subset M$ of codimension bigger than two and a holomorphic vector bundle $F$ on $X \setminus S$ such that -$$\mathcal F|_{X\setminus S}=\mathcal O(F)$$ -and $F$ is a sub-bundle of $E|_{X\setminus S}$ and there is an orthogonal projection $\pi:E|_{X\setminus S}\to F$ -which $\pi\in L_1^2(End(E))$ lying in the Sobolev space of $L^2$ sections of $End(E)$ with $L^2$ first-order weak derivatives and satisfying $\pi=\pi^*=\pi^2$ where $\pi^*$ denotes the adjoint of $\pi$. The Chern-Weil formula is $$deg_\omega \mathcal F=\frac{\sqrt[]{-1}}{2\pi n}\int_X tr(\pi\Lambda_\omega F_h)\omega^n-\frac{1}{2\pi n}\int_X|\nabla''\pi|^2\omega^n$$ -where $\nabla''\pi$ is computed in the sense of currents using the $(0,1)$ part of the Chern connection of $E$. -We define the slope of $\mathcal E$, to be $$\mu(\mathcal E)=\frac{deg \mathcal E}{rk \mathcal E}$$ -For any non-trivial vector bundle $E$ on curve $X$ by using the Riemann-Roch formula we can compute the slope of a vector bundle over a curve as follows, -$$\mu(E)=\frac{dim H^0(X,E)-dim H^1(X,E)}{rank E}+g_X-1$$ -where $g_X$ is the genus of curve $X$. -Another interpretation about semi-stability via slope -It is known that a holomorphic -line bundles on a compact connected Riemann surface $\Sigma_g$ -do not admit non-zero global holomorphic sections if their degree is negative. -A non-zero homomorphism from line bundles $L_1$ and $L_2$(which is a section of the line bundle $L^∗_1 ⊗ L_2$) exist if $deg (L^∗_1 ⊗ L_2) ≥ 0$, which is equivalent to $deg L_1 ≤ deg L_2$ -Note that, for higher rank vector bundles, the degree of $E^∗_1 ⊗ E_2$ is -$$deg (E^∗_1 ⊗ E_2) = rk (E_1)deg (E_2) − deg (E_1)rk E_2$$ -so the semi-positivity condition is equivalent to -$$\frac{deg E_1}{rk E_1} -≤ -\frac{deg E_2}{rk E_2} -$$<|endoftext|> -TITLE: normal crossing divisor v.s. strict normal crossing divisor -QUESTION [9 upvotes]: On wikipedia, the normal crossing divisor is defined to be (by my understanding): -(Assume $X/k$ be a smooth geometrically integral scheme of finite type over a field $k$). -Let $D = \sum_{i=1}^n C_i$ be a Weil Divisor, here $C_i$ are irreducible closed subsets of codimension 1 of $X$. Endow $C_i$ with the reduced scheme structure (hence they are integral closed $k$-scheme of $X$ of codimension 1.) We call $D$ is a normal crossing divisor if each $D_i$ is smooth over $k$ and $D_i$'s intersect transversely. -But in somewhere, I saw the notion "strict normal crossing divisor". For example, -Definition 1.5.1, p.8, in http://math.arizona.edu/~swc/aws/07/KedlayaNotes10Mar.pdf -or -5.1, p.16, in http://www.uni-due.de/~bm0032/publ/TubNbdDocumenta.pdf -The definition looks the same as "normal crossing divisor". -I would like to know what's the difference between these two definitions. It -would be great with examples. Also, if the base scheme $S$ is not a field, then is the definition the same as above with the modification "each $D_i$ is smooth over $S$"? - -REPLY [3 votes]: The definition you gave is currently the accepted definition of snc. As a friend of mine is fond to say: "...and of course, everything you read on wikipedia is correct..." -Anyway, -this definition essentially means that $D$ is supposed to look like the intersection of coordinate hyperplanes locally at any point. The key is what locally means. In the definition of snc it means Zariski locally, but if you replace that by étale locally, then you get the notion of nc. -As Donu remarked the difference between nc and snc is often negligable. For instance if you want to talk about a resolution of singularities where the preimage of the singular set is nc, then you can usually assume snc, because it just means replacing your resolution with a further blow-up. So, probably people working with those kind of things tend not to worry about the difference. -On the other hand there is at least one place where the difference is important. -This is when you want to understand the singularities of a pair $(X,D)$. For simplicity let's say that $X$ is smooth and $D$ is a reduced integral divisor on $X$. Then $(X,D)$ has dlt singularities if and only if $D$ is snc. In particular, an nc pair has "worse" singularities than an "snc" pair. There's a lot more one could say about this, but since this kind of goes out of the scope of the original question I'll stop here. -Remark I realize that you might not know what dlt singularities mean, but in some sense it is not important at the moment. It is an important class of singularities. You can read about them in Kollár-Mori: Birational Geometry of Algebraic Varieties and/or Kollár -Singularities of the Minimal Model Program.<|endoftext|> -TITLE: Reference for openness of stable locus of holomorphic family of vector bundles on a compact riemann surface. -QUESTION [6 upvotes]: I am looking for a reference (or an easy explanation) for the openness of the stable locus of a holomorphic family of (holomorphic) vector bundles on a compact Riemann surface parametrized by a (compact) complex manifold. For me, a holomorphic family of vector bundles on a compact Riemann surface $X$ parametrized by a (compact) complex manifold is just a holomorphic vector $E$ bundle over $T \times X$. -There is a proof in Narasimhan and Seshadri's paper "Stable and Unitary Vector Bundles on a Compact Riemann Surface" but the proof there depends on their proof of the unrelated theorem that describes stable vector bundles on a compact riemann surface via certain unitary representations of suitably defined fuchsian groups. -If there is a simpler proof in the case that all parametrized bundles are of degree zero I would like to know it. -EDIT: Note that I do not want to assume that $T$ and the vector bundle $E$ over $T \times X$ are algebraic. So e.g. Huybrechts-Lehn or Le Potier's "Lectures on Vector Bundles" aren't of any help to me, I think. - -REPLY [3 votes]: I think this question is discussed in Kobayashi's "Differential geometry of complex vector bundles", at least the result follows implicitly. -One way to see is as follows: For simplicity, consider holomorphic rank 2 bundles $V$ of degree 0. They are not stable if there exists a holomorphic $f\colon L\to V$ of a holomorphic line bundle $L$ of degree $0.$ If you think of a holomorphic bundle as given by a holomorphic structure $\bar\partial,$ then the result follows from the observation, that, for a family of Fredholm operators (like the $\bar\partial$ on $L^*\otimes V,$ where you vary the holomorphic structures on $L$ and $V$) the minimal kernel dimension is attained on an open subset.<|endoftext|> -TITLE: The different types of stacks -QUESTION [17 upvotes]: This question is very naive, but it will help me a lot in getting in to the vast literature about stacks. -The question is this: there are many kinds of stacks (algebraic spaces, DM, algebraic stacks, Artin stacks, geometric stacks...). What would be a catch-phrase you would use to describe them, and more importantly to differentiate their uses one from the other? -I'll give you an example (which may be off because of my lack of familiarity with subject): it seems to me that the catchphrase for algebraic spaces is that they are the result of looking at the "orbit space" of the action of a finite group on a scheme. -What catchphrases would you give for the rest of them (and how would you change the catchphrase I gave for algebraic spaces, if at all) to best describe their utility? - -REPLY [9 votes]: Algebraic spaces are non-stacky algebraic (Artin) stacks, and DM-stacks, although stacky, have only finite stabilizer groups (or étale stab. groups; I'm sloppy here) for all points on the "underlying space". -To get to a scheme from an algebraic space, one can either pass to an étale cover or just restrict oneself to an open dense subspace. To get to a scheme from a DM-stack, one can still pass to an étale cover, or, if one prefers finite group actions, pass to a Zariski open, which supports a $G$-bundle with total space an affine scheme, and $G$ is a finite group. The DM-stack is covered by such Zariski opens (though with different groups $G$). But for a general Artin stack there is really a long way to get to a scheme (or alg. space): if schemes are floating on the surface of the ocean, then Artin stacks rest deep in the ocean --- they are "covered" by schemes but the relative dimension are usually positive. -This somehow explains (at least to me) why for algebraic spaces (or more generally, DM-stacks) it suffices to use the étale topology to define and compute cohomology of sheaves, whereas for general Artin stacks, lisse-étale topology is necessary. -Edit: This is actually a comment for 4) in unknowngoogle's answer. It is possible to have a continuously varying family of algebraic groups parameterized by a space (i.e. not iso-trivial, even if one passes to an algebraic stratification of the base space). Therefore, I don't think all algebraic stacks (say over $k$) are locally quotients of $k$-schemes by $k$-algebraic groups: one needs group schemes over a larger base. There exists a stratification of any algebraic stack such that the inertia is flat over each stratum, but one cannot make this flat family a constant family.<|endoftext|> -TITLE: Atkin-Lehner involution and class number -QUESTION [6 upvotes]: I was told of a relation between the number of fixed points of the Atkin-Lehner involution and the class number of certain number fields. -Can someone point me to a reference where I could learn about it? (I am not very familiar with CM theory so I don't want a one-line proof without further explanation). -Thanks. - -REPLY [7 votes]: Let me try to explain the CM connection, as I think it really is the most intuitive way to understand this. Fear not, this is not a one line answer! -I'll be working with $\mathbb{C}$ throughout. Every elliptic curve over $\mathbb{C}$ is of the form $\mathbb{C}/\Lambda$, where $\Lambda$ is a discrete rank two sublattice of $\mathbb{C}$. This description is not unique: If $\alpha$ is any nonzero complex number, then $\mathbb{C}/\Lambda$ and $\mathbb{C}/\alpha \Lambda$ define the same elliptic curve. -If we have a two elliptic curves, $\mathbb{C}/\Lambda_1$ and $\mathbb{C}/\Lambda_2$, and a map $\phi$ between them, then there is a unique complex number $\beta$ such that $\beta \Lambda_1 \subseteq \Lambda_2$ and $\phi$ arises as the map which takes the coset $z+\Lambda_1$ to the coset $\beta z + \Lambda_2$. - -I'll first explain the basic idea of complex multiplication, and then talk about the Atkin-Lehner case. Complex multiplication is all about describing the possible self maps of an elliptic curve. Consider a complex number $\beta$ and a lattice $\Lambda$. When does multiplication by $\beta$, as a map from $\mathbb{C}$ to $\mathbb{C}$, descend to a map from $\mathbb{C}/\Lambda$ to $\mathbb{C}/\Lambda$? This happens, if and only if $\beta \Lambda \subseteq \Lambda$. -Now, let's fix $\beta$ and consider which $\Lambda$ have this property. Notice that, if $\lambda$ is a nonzero element of $\Lambda$, and $\theta$ is any element in the ring $\mathbb{Z}[\beta]$, then $\theta \lambda$ is in $\lambda$. This means that $\mathbb{Z}[\beta] \cdot \lambda$ must form a discrete sublattice of $\Lambda$, so the ring $\mathbb{Z}[\beta]$ must be a discrete sublattice of $\mathbb{C}$. -Case 1: $\mathbb{Z}[\beta]$ is a rank $1$ sublattice of $\mathbb{C}$. In this case, $\beta$ is in $\mathbb{Z}$ and $\beta \Lambda \subseteq \Lambda$ for every $\Lambda$. -Case 2: $\mathbb{Z}[\beta]$ is a rank $2$ sublattice of $\mathbb{C}$. In this case (this is not obvious) $\mathbb{Z}[\beta]$ must either be of the form $\mathbb{Z}[\sqrt{-d}]$ or $\mathbb{Z}[(1+\sqrt{-d})/2]$, where $d>0$ and, in the latter case, $d$ must be $3 \mod 4$. In this case, there are finitely many lattices $\Lambda$ such that $\beta \Lambda \subseteq \Lambda$ (up to treating $\Lambda$ and $\alpha \Lambda$ as equivalent, as mentioned in the second paragraph.). The number of these lattices is more or less the class number of $\mathbb{Q}[\sqrt{-d}]$. (It is exactly this if $d$ is square free and $1$ or $2$ mod $4$; otherwise there are some details to fix up.) -Case 3: $\mathbb{Z}[\beta]$ is not a discrete sublattice of $\mathbb{C}$. As discussed above, in this case there are no $\Lambda$'s for which $\beta \Lambda \subseteq \Lambda$. - -Now, for the Atkin-Lehner connection. The modular curve $Y_0(p)$ (the one without the cusps) parameterizes ordered pairs $(\Lambda_1, \Lambda_2)$, where $\Lambda_2$ is an index $p$ sublattice of $\Lambda_1$, and where $(\Lambda_1, \Lambda_2)$ is identified with $(\alpha \Lambda_1, \alpha \Lambda_2)$ for any nonzero complex number $\alpha$. -The Atkin-Lehner involution sends $(\Lambda_1, \Lambda_2)$ to $(\Lambda_2, p \Lambda_1)$. So a fixed point of Atkin-Lehner must correspond to a pair $(\Lambda_1, \Lambda_2)$ such that $(\Lambda_2, p \Lambda_1) = (\alpha \Lambda_1, \alpha \Lambda_2)$ for some $\alpha$. In particular, -$$p \Lambda_1 = \alpha (\alpha \Lambda_1) = \alpha^2 \Lambda_1.$$ -Set $\gamma = \alpha^2/p$. Then both $\gamma$ and $\gamma^{-1}$ take $\Lambda_1$ to itself, so both $\mathbb{Z}[\gamma]$ and $\mathbb{Z}[\gamma^{-1}]$ are discrete lattices. Looking at the case by case analysis above, one works out that $\gamma$ is one of $1$, $-1$, $\pm i$, $\pm e^{2 \pi i/6}$ and $\pm e^{4 \pi i/6}$. Now, $\alpha = \sqrt{p \gamma}$ and we have $\alpha \Lambda_1 = \Lambda_2 \subset \Lambda_1$. So $\sqrt{p \gamma}$ must also generate a discrete sublattice of $\mathbb{C}$. Looking at the previous list of cases, the only one that survives is $\gamma = -1$ and $\alpha = \sqrt{-p}$. -So the fixed points of Atkin-Lehner come from lattices $\Lambda_1$ such that $\sqrt{-p} \Lambda_1 \subset \Lambda_1$; for each such lattice $\Lambda_1$ we get the fixed point $(\Lambda_1, \sqrt{-p} \Lambda_1)$. Using the previous discussion, the number of such $\Lambda_1$'s is essentially the class number of $\mathbb{Q}(\sqrt{-p})$.<|endoftext|> -TITLE: Is every algebraic space the quotient of a scheme by a finite group? -QUESTION [16 upvotes]: In this MO question it is claimed that a catchphrase for "algebraic spaces" could be that they are "the result of looking at the orbit space of the action of a finite group on a scheme". -Hence my question: - -Given an algebraic space $X$, is it true that there exists a scheme $S$ and an action of a finite group $G$ on $S$ such that $X=S/G$ ? - -If I remember correctly, every algebraic space is the quotient of an affine scheme by an étale equivalence relation, so I tend to think that there could exist such equivalence relations that are not "implemented" by a group action... - -REPLY [17 votes]: As I just remembered, the answer is yes if $X$ is quasiseparated, noetherian and normal: see Laumon and Moret-Bailly, Champs algébriques, (16.6.2). The quesiseparated assumption is needed (see Scott Carnahan's answer). Without the normality condition, I don't have a counterexample but my guess is that there is one. -EDIT after Chris' and Jason's comments: in fact the proof in the case of normal algebraic spaces can be made substantially simpler than in the book (which proves a more general result about noetherian Deligne-Mumford stacks). It goes like this: -Assume $X$ noetherian, integral, normal and, to simplify, separated (I am not sure how much this helps). Cover $X$ by étale maps $X_i\to X$ with each $X_i$ integral and affine. There is a dense open subspace $U$ of $X$ which is a scheme and such that each induced map $U_i:=X_i\times_X U\to U$ is finite. Let $V\to U$ be an étale Galois cover, with Galois group $G$, dominating all the $U_i$'s. Now let $\overline{V}\to X$ (resp. $\overline{X_i}\to X$) be the normalization of $X$ in $V$ (resp. in $X_i$); we have dense open immersions $V\subset \overline{V}$ and $X_i\subset \overline{X_i}$. By functoriality, $G$ acts on $\overline{V}$, with quotient $X$. -I claim that $\overline{V}$ is a scheme. Indeed, for each $i$, $\overline{X_i}$ is also the normalization of $X$ in $U_i$. In particular there is an $X$-morphism $f_i:\overline{V}\to\overline{X_i}$ (deduced from $V\to U_i$) which must be finite surjective (everyone is integral, finite and surjective over $X$). Put $V_i:=f_i^{-1}(X_i)$: this is an open subspace of $\overline{V}$ which is finite over $X_i$, hence an affine scheme. So, the union $W$ of the $V_i$'s is an open subspace of $\overline{V}$ which is a scheme and maps surjectively to $X$ (since $V_i\to X_i$ is surjective), hence $\overline{V}$ is covered by $\{gW\}_{g\in G}$.<|endoftext|> -TITLE: Lifting of a ucp map with values in a von Neumann algebra ultraproduct of matrix algebras -QUESTION [8 upvotes]: Let $u:A \to \prod_{\mathcal U} M_n$ be a unital completely positive map (ucp) from a unital separable $C^*$algebra into the von Neumann algebra ultraprodut $\prod_{\mathcal U} M_n$. -Here $\mathcal U$ is an ultrafilter on $\mathbb N$ and $\prod_{\mathcal U} M_n$ is the quotient of $B=\{(x_n)_{n \in \mathbb N}, x_n \in M_n(\mathbb C) , \sup_n \|x_n\|<\infty\}$ by the ideal $I_{\mathcal U}=\{(x_n)_n, \lim_{\mathcal U} Tr(x_n^* x_n)/n = 0\}$. - -Does there exist a ucp lifting of $u$, i.e. a sequence $u_n:A \to M_n(\mathbb C)$ of ucp maps such that $u=q \circ (u_n)$, where $q:B\to B/I_{\mathcal U}$ is the quotient map? - -If not, is $u$ locally liftable? That is: given a finite dimensional operator system $E\subset A$ (= a subspace $E$ of $A$ containing $1$ and stable under $a \mapsto a^*$), does there exist a ucp lifting of the restriction of $u$ to $E$? -Some comments: as in my related question, I know that the answer is no in general if one replaces the von Neumann algebra ultraproduct by the $C^*$-algebra ultraproduct. But I hope that again, the situation might be much simpler in the von Neumann algebra setting. (I even have the feeling that I have known the answer to this question, but that I have forgotten it). - -REPLY [4 votes]: The answer is no, in general there is no lifting. A lifting exists if the $C^{\ast}$-algebra has the so-called lifting property (LP), and local liftings exist if it has the local lifting property (LLP). -I constructed in -Andreas Thom, Examples of hyperlinear groups without factorization property, Groups Geom. Dyn. 4, no. 1 (2010) 195-208. -an example of a group $G$, such that the universal group $C^{\ast}$-algebra of $G$ does not have the LLP. The idea is that $G$ is hyperlinear, but cannot have Kirchberg's factorization property. The hyperlinearity is shown by a concrete construction of micro-states, Kirchberg's factorization property has to fail since $G$ has property (T), but is not residually finite. Note that Kirchberg showed that Kazhdan groups with factorization property are residually finite. See also -Narutaka Ozawa, About the QWEP conjecture, Internat. J. Math. 15 (2004), no. 5, 501– -530. -where all these concepts are explained.<|endoftext|> -TITLE: A very basic question about Abel-Jacobi map -QUESTION [5 upvotes]: Let $C$ be a compact Riemann surface, let $C^2$ be the cartesian square of $C$, let $J(C)$ be the degree zero Jacobian of $C$, and let $\delta : C^2 \to J(C)$ be the map $(x,y) \mapsto [\mathcal{O}(x-y)]$. -In this paper http://arxiv.org/abs/math/9810054 of Hain and Reed, page 9, they say that it is an elementary exercise in algebraic topology to show that $\delta^\ast(\phi) = \Delta + (\psi_1 + \psi_2)/2$. -Explanation of notation: - -$\phi$ denotes the symplectic form $\sum dx_i \wedge dy_i$ on $J(C)$, where $x_i, y_i$ are coordinates on the torus $J(C) = H^1(C;\mathcal{O}) / H^1(C;\mathbb{Z})$ corresponding to a symplectic basis of $H_1(C;\mathbb{Z}) \cong H^1(C;\mathbb{Z})$. This is also the class of the theta divisor $\Theta$. -$\psi_i$ is the first Chern class of the relative cotangent bundle of the $i$th projection $C^{2} \to C$. -$\Delta$ is the class of the diagonal $C \to C^{2}$, $x \mapsto (x,x)$. - -My question is: How to do this "elementary exercise"? It ought to be easy but I'm just not seeing it... - -REPLY [4 votes]: Let me give another proof in the spirit of algebraic geometry. -By Riemann's theorem, for any symmetric theta divisor $\Theta$ there exists a theta characteristic $\kappa$ (i.e., a line bundle such that $\kappa^{\otimes 2}=\omega_C$) on $C$ such that $$W_{g-1}=\Theta + \kappa,$$ -where $W_{g-1}$ is the image of the abelian sum mapping -$$u \colon \textrm{Sym}^{g-1}(C) \to J(C).$$ -$\kappa$ is called the Riemann's constant, and one has -$$\delta^* \Theta = q_1^* (\kappa) \otimes q_2^*(\kappa) \otimes \mathcal{O}_{C^2}(\Delta), \quad \quad (\star)$$ -where $q_i \colon C^2 \to C$ are the natural projections. -The proof of such a formula is easy, and it is based on the Seesaw Principle: in other words, one shows that the restrictions to $C \times \{ p \}$ and $\{p \} \times C$ of both sides of $(\star)$ coincide for all $p \in C$. -For the details, see [Birkenhake-Lange, Complex Abelian Varieties, Proposition 11.10.2 $(a)$], putting $\eta=\kappa$ into the statement. -Since $\kappa$ is a theta characteristic it follows that the cohomology class of -$q_1^* (\kappa) \otimes q_2^*(\kappa)$ is exactly $\frac{1}{2}(\psi_1 + \psi_2)$. On the other hand, as you noticed, the cohomology class of $\Theta$ is $\phi$, so we are done.<|endoftext|> -TITLE: Is a profinite group with a finite number of simple quotients and Jordan-Hölder factors finitely generated? -QUESTION [5 upvotes]: Assume $G$ is a profinite group such that the Jordan-Hölder factors appearing in the finite quotients vary in a finite number of isomorphism classes of simple groups. Assume also $G$ to have a finite number of subgroups whose corresponding quotient is simple. Does this imply that $G$ is (topologically) finitely generated? -I'm asking here after some attempt to make work a modification of the principle the for a $p$-group $P$ each set of elements generating $P/\Phi(P)$ is a generating set for $P$. For $P$ groups the question is clearly much simpler, and i have been thinking that elements generating each simple quotient had to be be enough (this is not true, as shown by the simple example $Sym_n$. But the different symmetric groups have bigger and bigger Jordan-Holder factors). However the issue is not totally trivial, because maximal (non-normal) subgroups are in generally not contained in a proper normal subgroup, so it is not possible to replicate a similar proof smoothly. Note that the hypothesis of having a finite number of factors rules out silly couterexamples like $\prod_{i=4}^\infty Alt_i$. -Is anything know about this question? Thanks for the attention! - -REPLY [4 votes]: Yves has answered your first question, so I'll address your more general aim. -One can regard the Frattini subgroup of a profinite group $G$ as the set of 'non-generators': a non-generator is an element $x$ such that for any subset $X$ of $G$ for which $\overline{\langle X,x \rangle}=G$, then $\overline{\langle X \rangle} = G$. It is a general fact about profinite groups that the Frattini subgroup is pronilpotent, so if the groups you are interested in are not virtually pronilpotent, there is no easy reduction of generation problems to a fixed finite quotient. So the situation in pro-$p$ groups really is very different to the general case. -One can also define the Mel'nikov subgroup (it goes by other names) $M(G)$ to be the intersection of all normal subgroups $N$ such that $G/N$ is simple. This plays a similar role to the Frattini subgroup, but you have to consider normal closures: we have $x \in M(G)$ if and only if, whenever the subset $X$ of $G$ is contained in a proper closed normal subgroup, then $X \cup \{x\}$ is also contained in a proper closed normal subgroup. A consequence is that if any subset $X$ of $G$ satisfies $XM(G) = G$, then $G$ is the closure of the group generated by the conjugates of $X$. -The natural question at this point is: when is $G/M(G)$ finite? In the pro-$p$ case, this condition precisely corresponds to finite generation (as a topological group); indeed $M(G) = \Phi(G)$ if $G$ is pronilpotent. In the general case, the property '$G/M(G)$ is finite' is not comparable with finite generation, and is not even stable under passing to subgroups of finite index. However, if you additionally assume $G$ is a pro-$\pi$ group, where $\pi$ is a finite set of primes, then $G$ finitely generated implies $G/M(G)$ is finite, although this appears to need the fact that there are only finitely many finite simple $\pi$-groups up to isomorphism (this follows from the classification of finite simple groups - I don't know any elementary proof). In the other direction, $G$ may be infinitely generated even if $G/M(G)$ is finite and $G$ is prosoluble, involving only two primes - there is an example in a recent paper of J. Wilson, for instance: my internet is bad here but I think the paper is called 'Large hereditarily just infinite groups'.<|endoftext|> -TITLE: Distributions on product spaces -QUESTION [10 upvotes]: I hope this is suitable to MO. -Question. Let $X$ and $Y$ be two open sets in $\mathbb R^n$ and $\mathbb R^m$, respectively. In what sense can we consider $\mathcal{D}^{\prime} \left(X\times Y\right)$ as a tensor product of $\mathcal{D}^{\prime} \left(X\right)$ and $\mathcal{D}^{\prime} \left(Y\right)$ (where $\mathcal{D}^{\prime}$ means distributions of the standard kind, i. e., those acting on $\mathcal{C}^{\infty}$ functions of compact support)? What if $\mathcal{D}^{\prime}$ is replaced by $\mathcal{E}^{\prime}$ (distributions with compact support) or $\mathcal{S}^{\prime}$ (tempered distributions)? -Remarks. I am trying to understand in how far distributions form a coalgebra, and what can be derived from this viewpoint. The applicability of coalgebras to distribution theory seems to be one of the selling points of coalgebra and Hopf algebra theory, but I have yet to see a place where this is actually elaborated upon and applied to yield nontrivial results. "I have yet to see" does not mean much, though, as I am a complete greenhorn at analysis, and there is not much literature avaliable on the coalgebra side. - -REPLY [10 votes]: According to the Schwartz Kernel Theorem and its variants, there are the canonical isomorphisms - $$\mathcal{D}^{\prime} \left(X\right)\tilde\otimes \mathcal{D}^{\prime} \left(Y\right)\simeq\mathcal{D}^{\prime}\left(X\times Y\right),$$ -$$\mathcal{E}^{\prime} \left(X\right)\tilde\otimes \mathcal{E}^{\prime} \left(Y\right)\simeq\mathcal{E}^{\prime} \left(X\times Y\right),$$ -$$\mathcal{S}^{\prime} \left(\mathbb R^n\right)\tilde\otimes \mathcal{S}^{\prime} \left(\mathbb R^m\right)\simeq\mathcal{S}^{\prime} \left(\mathbb R^{n+m}\right),$$ -where $E\tilde\otimes F$ is the completion of the space $E\otimes F$. -Roughly speaking, this follows from the fact that the corresponding spaces of test functions $\mathcal{D}$, $\mathcal{C}^{\infty}$, and $\mathcal{S}$ are nuclear Fréchet spaces, and one has the canonical isomorphisms -$$E^{\prime}\tilde\otimes F^{\prime}\simeq \left(E\tilde\otimes F\right)^{\prime}\simeq L(E; F'),$$ -provided that $E$ and $F$ are nuclear Fréchet spaces. (Here the duals carry the strong dual topology and the space $L(E;F ')$ of continuous linear mappings is endowed with the topology of bounded convergence.) -As Johannes mentioned in his comment, a detailed presentation of the Schwartz Kernel Theorem and its versions for various spaces of distributions can be found in Topological Vector Spaces, Distributions and Kernels by Trèves. (More specifically, take a look at Chapt. 51, "Examples of Nuclear Spaces. The Kernels Theorem".)<|endoftext|> -TITLE: P-adic Arithmetic Software -QUESTION [8 upvotes]: I would like to automate a huge amount of computation that involve basic arithmetic operations with $p$-adic numbers. I have found a Mathematica packages for it, but it is old and acts quite erratically. Do you know of any computational software that does it reliably? - -REPLY [11 votes]: SAGE has p-adic arithmetic (for example, see http://www.math.utah.edu/~carlson/cimat/python-sage.pdf), and has the added benefit of being completely free and open-source! - -REPLY [3 votes]: This is just a very partial answer that is based on my experience trying to do some work with extension of p-adic numbers. -There is p-adic arithmetic in the free software programs SAGE, PARI and GAP, but their main limitation (i don't know if things have changed recently) was their inability to with relative extension, that is extensions of another field which is itself a proper extension of $\mathbb{Q}_p$. However if you have limited need for extensions they (mostly SAGE and PARI, because GAP is more group-theory oriented) have a very good interface to work with p-adic numbers. -For my thesis work i absolutely needed relative extensions, so i had to use Magma which is not free, but for small computation it can be used online: http://magma.maths.usyd.edu.au/calc/. It has a very good library, which requires some time to learn, like the magma language which requires some learning too, but however the functionality provided is worth the effort.<|endoftext|> -TITLE: Semantic definition of sentence -QUESTION [5 upvotes]: This is a follow-up to question Completeness vs Compactness in logic 68788. One common theme was that compactness in logic is a purely semantic notion, so should have no need of completeness. -The definition of compactness seems to depend in an irreducible way on the concept of a sentence, which appears to be a syntactic notion. So my question is: Is there any purely semantic definition of a sentence in first-order logic? - -REPLY [7 votes]: I suspect you've misunderstood what you quoted from the earlier question, since the semantics discussed there was the semantics of sentences, presupposing a notion of sentence. Nevertheless, your present question makes sense, and the (probably rather uninformative) answer is yes. The idea is to mimic semantically the definitions of formulas and their satisfaction, without ever mentioning the strings (or trees, etc.) of symbols that syntax deals with. Fix a vocabulary $L$ and (to avoid set-versus-class problems) consider only $L$-structures included in some fixed large set. "Formulas" will, in this approach, be certain collections of pairs $(A,s)$ where $A$ is a structure and $s$ is a finite tuple of elements of $A$. (The length of the tuple $s$ is to be the same for all pairs in a specific formula, but it may be different for different formulas). To define which collections of pairs count as formulas, one uses an inductive definition whose main clauses look like - -If $R$ is an $n$-ary predicate symbol of $L$ then there is a formula consisting of exactly those pairs $(A,s)$ where $s$ is an $n$-tuple satisfying $R$ in $A$. -If $\phi$ and $\psi$ are two formulas in which the $n$-tuples have the same length, then $\phi\cup\psi$ is a formula. -The complement of any formula $\phi$, within the collection of all pairs $(A,s)$ whose tuples $s$ have the same length, is again a formula. -If $\phi$ is a formula whose tuples have non-zero length, say $n+1$, then there is a formula consisting of those pairs $(A,s)$ in which $s$ is an $n$-tuple and $(A,(a)^\frown s)\in\phi$ for some $a\in A$. -If $\phi$ is a formula whose tuples have length $n$ and if $f:\{1,2,\dots,m\}\to\{1,2,\dots,n\}$ then there is a formula consisting of exactly the pairs $(A,s\circ f)$ with $(A,s)\in\phi$. - -The first clause says that atomic formulas are formulas; the next three say that the class of formulas is closed under the usual first-order constructors (disjunction, negation, and existential quantification), and the last allows trivial manipulations of the components in tuples (permutation, identifications, and adding dummy components). - -REPLY [7 votes]: One way to completely avoid syntax is to use Ehrenfeucht–Fraïssé games. If $\mathfrak{A}$ and $\mathfrak{B}$ are two structures with the same signature, then two $k$-tuples $\bar{a}$ and $\bar{b}$ from these respective structures satisfy the same type if and only if duplicator wins every finite length EF game between $(\mathfrak{A},\bar{a})$ and $(\mathfrak{B},\bar{b})$. This is a perfectly semantical way of defining the space $S_k$ of $k$-types for a given signature. One can show directly that the resulting space $S_k$ is a compact zero-dimensional space. The $k$-ary formulas of the language basically correspond to clopen sets in $S_k$. The compactness of $S_0$ is basically a restatement of the Compactness Theorem. Similarly, the Omitting Types Theorem is basically a restatement of the Baire Category Theorem for these spaces $S_k$. -This is the approach used by the Fraïssé school of model theory. One recent book which tries to promote this approach is Bruno Poizat's A Course in Model Theory.<|endoftext|> -TITLE: Name of a metric space concept -QUESTION [8 upvotes]: I have a metric space with the following property (a bit like having unique geodesics): for any points $a,b,x,y$ with $d(a,b)=d(a,x)+d(x,b)=d(a,y)+d(y,b)$ and $d(a,x)=d(a,y)$, we have $x=y$. Is there an established name for this? -(UPDATE: the condition $d(a,x)=d(a,y)$ was omitted by mistake in the original question.) - -REPLY [3 votes]: It is easy to prove that the completion of your space can be any separable metric space -where metric spheres are nowhere dense. - -Does not it scare you? - -Answer to the comment: -Not all of your spaces can be embedded into a metric tree. -BTW, there is a nice characterization of subsets in a metric tree: -$$ | x - y | + | a - b | \le \max\{|x-a| + |y-b|,|x-b|+|y-a|\}$$ -for all points $x,y,a,b$ -(here $|x-y|$ denotes the distance from $x$ to $y$). -In other words, the values -$$X=|x-a|+|y-b|,$$ -$$Y=|x-b|+|y-a|,$$ -$$Z=|x-y|+|a-b|$$ satisfy ulrtatriangle inequality. -This inequality can be also thought as a discrete analog of CAT[−∞] inequality.<|endoftext|> -TITLE: 3-orbifolds with a Seifert geometry that are not actually Seifert fibered -QUESTION [8 upvotes]: It is well-known that Seifert fibered $3$--manifolds are geometric: they admit one of the Thurston geometries $S^2 \times R$, $R^3$, $H^2 \times R$, $S^3$, $Nil$, and $PSL(2,R)$. Furthermore, the converse is also true: every $3$--manifold admitting one of these 6 geometries has a Seifert fibration over a $2$--orbifold. -In the realm of $3$--orbifolds, it is still true that every Seifert fibered $3$--orbifold is geometric, with the same 6 geometries. However, the converse is false: there exist geometric $3$--orbifolds with one of the 6 "Seifert" geometries, which do not Seifert fiber. My question is: - -What are the orientable $3$--orbifolds that have a Seifert geometry, but are not Seifert fibered? Do we know a complete list? - -One example of this weird phenomenon is the figure-8 knot, labeled 3. This orbifold is Euclidean. But it is not Seifert fibered, because any order-3 singular locus must be vertical in a Seifert fibration. Thus, if there was a Seifert fibering of the orbifold, drilling out the singular locus would produce a Seifert fibration on the knot complement, which is absurd because the knot complement is hyperbolic. -More generally, Dunbar has classified the spherical $3$--orbifolds that are not Seifert fibered. There are 21 such examples in total: http://www.ams.org/mathscinet-getitem?mr=1118824 -What is known about the other 5 Seifert geometries? - -REPLY [8 votes]: From Three-dimensional orbifolds and their geometric structures, by Boileau, Maillot and Porti (p. 15): - -Remark. There are orbifolds with geometry $E^3$ and $S^3$ which are not orbifold-Seifert fibered [52, 54]. - -[54] is this article of Dunbar's you quoted. [52] is Geometric orbifolds, of the same author.<|endoftext|> -TITLE: $H^4$ of the Monster -QUESTION [43 upvotes]: The Monster group $M$ acts on the moonshine vertex algebra $V^\natural$. -Because $V^\natural$ is a holomorphic vertex algebra (i.e., it has a unique irreducible module), there is a corresponding cohomology class $c\in H^3(M;S^1)=H^4(M;\mathbb Z)$ -associated to this action. - -Roughly speaking, the construction of that class goes as follows: - -For every $g\in M$, pick an irreducible twisted module $V_g$ (there is only one up to isomorphism). -For every pair $g,h\in M$, pick an isomorphism $V_g\boxtimes V_h \to V_{gh}$, -where $\boxtimes$ denotes the fusion of twisted reps. -Given three elements $g,h,k\in M$, the cocycle $c(g,h,k)\in S^1$ is the discrepancy between - $$ -(V_g\boxtimes V_h)\boxtimes V_k \to V_{gh}\boxtimes V_k \to V_{ghk}\qquad\text{and}\qquad -V_g\boxtimes (V_h\boxtimes V_k) \to V_g\boxtimes V_{hk} \to V_{ghk} -$$ - - -I think that not much known about $H^4(M,\mathbb Z)$. But is anything maybe known about that cohomology class? Is it non-zero? Assuming it is non-zero, would that have any implications? -More importantly: what is the meaning of that class? - -REPLY [25 votes]: In arXiv:1707.08388, I calculate that the cohomology class you described has order 24 and that it is not a characteristic class in the ordinary sense.<|endoftext|> -TITLE: Proof strength of Calculus of (Inductive) Constructions -QUESTION [16 upvotes]: This is a follow-on from this question, where I pondered the consistency strength of Coq. This was too broad a question, so here is one more focussed. Rather, two more focussed questions: - -I've read that CIC (the Calculus of Inductive Constructions) is interpretable in set theory (IZFU - intuitionistic ZF with universes I believe). Is there a tighter result? - -And - -What is the general consensus of the relative consistency of constructive logics anyway? - -I am familiar, in a rough-and-ready way, with the concept of consistency strength in set theory, but more so of the 'logical strength' one has in category theory, where one considers models of theories in various categories. Famously, intuitionistic logic turns up as the internal logic of a topos, but perhaps this is an entirely different dimension of logical strength. - -I guess one reason for bringing this up is the recent discussion on the fom mailing list about consistency of PA - Harvey Friedman tells us that $Con(PA)$ is equivalent to 15 (or so) completely innocuous combinatorial statements (none of which were detailed - if someone could point me to them, I'd be grateful), together with a version of Bolzano-Weierstrass for $\mathbb{Q}\_{[0,1]} = \mathbb{Q} \cap [0,1]$ every sequence in $\mathbb{Q}_{[0,1]}$ has a Cauchy subsequence with a specified sequence of 'epsilons', namely $1/n$). A constructive proof of this result would be IMHO very strong evidence for the consistency of PA, if people are worried about that. - -REPLY [2 votes]: I just stumbled upon this old question by chance, and I thought maybe you should have a look at Alexandre Miquel's thesis if you haven't already done so (and if you can read French). Conjecture 9.7.12 on page 329 (331 of PDF) suggests that the Calculus of Constructions with universes should be equiconsistent with Zermelo set theory with universes (assuming I'm not misreading—I'm easily confused between all these theories), which at least gives some lower bound.<|endoftext|> -TITLE: Projective modules over free groups -QUESTION [6 upvotes]: Consider the ring of Laurent polynomials $R := \mathbb{Z}[s,s^{-1}]$ with integer coefficients. Are all projective $R$-modules free? (Let's say left modules by convention.) -More generally, let $G$ be the free group on a finite set $S$ of generators, and consider the integral group ring $R := \mathbb{Z}G$. Are all projective $R$-modules free? Note that in part (1), $S$ was the singleton $\{ s \}$. -Is the answer the same if we allow an infinite set $S$ of generators? Note that the group $G$ itself is infinite either way. - -I feel like work of Serre, Swan, Bass, Kaplansky, Suslin, Quillen, and others may be relevant, but I couldn't find the answers. -Thank you for your help. - -REPLY [8 votes]: The answer to your questions is 'yes' and, as you feel, it is a result of Bass: -MR0178032 (31 #2290) -Bass, Hyman -Projective modules over free groups are free. -J. Algebra 1 1964 367–373.<|endoftext|> -TITLE: Is every algebraic smooth hypersurface of affine space parallelizable? -QUESTION [25 upvotes]: Consider a hypersurface $X=V(f) \subset \mathbb A^n_{\mathbb C}$, where $f(T_1, T_2,\ldots,T_n)\in \mathbb C[T_1,Y_2,\ldots, T_n]$ is a polynomial . -Assume that $X$ is smooth, i.e. that $df(x)\neq 0 \;$ for all $x\in X$ . My question is simply whether $X $ is parallelizable i.e. whether its tangent bundle $T_X$ is algebraically trivial. -I've asked a few friends and their answer was unanimously "no, why should it be?", but they couldn't provide a counter-example. Here are a some considerations which might show that the question is not so ridiculous as it looks. -We have the exact sequence of vector bundles on $X$ -$$0 \to T_X\to T_{A^n_{\mathbb C}}|X\to N(X/A^n_{\mathbb C})\to 0$$ -Now, the normal bundle $N(X/A^n_{\mathbb C})$ is trivial (trivialized by $df$) and the restricted bundle $T_{A^n_{\mathbb C}}|X$ is trivial because already $T_{A^n_{\mathbb C}}$ is trivial. Moreover the displayed exact sequence of vector bundles splits, like all exact sequences of vector bundles, because we are on an affine variety. So we deduce (writing $\theta$ for the trivial bundle of rank one on $X$) -$$\theta^n=T_X\oplus \theta $$ -In other words the tangent bundle is stably trivial, and this is already sufficient to deduce (by taking wedge product) that $\Lambda ^{n-1}T_X=\theta$ (hence the canonical bundle $K_X=\Lambda ^{n-1}T_X^\ast$ is also trivial). This suffices to prove that indeed for $n=2$ the question has an affirmative answer: every smooth curve in $A^2_{\mathbb C}$ is parallelizable. -Another argument in favor of parallelizability is that there are no analytic obstructions: O. Forster has proved a result for complex analytic manifolds which implies that analytically (and of course differentiably) our hypersurface is parallelizable: $T_{X_{an}}=\theta _{an}^{n-1}$.This is why I choose $\mathbb C$ as the ground field: the question makes perfect sense over an arbitrary algebraically closed field but I wanted to be able to quote the related analytic result.[ As ulrich remarks, parallelizability can't be deduced over the non-algebraically closed field $\mathbb R$, as shown by a 2-sphere] -Edit ulrich's great reference not only answers my question but seems to yield more results in the same direction. For example consider a smooth complete intersection: $X=\{ x\in \mathbb C^n|f_1(x)=f_2(x)=\ldots=f_k(x)=0 \} $ with the $f_i$'s polynomials and the $df_i(x)$'s linearly independent at each $x\in X$ . Then, just as above, the normal bundle is trivial and the tangent bundle is stably trivial: $\theta^n=T_X\oplus \theta ^{k} $ -So Suslin's incredible theorem again allows us to conclude that $X$ is parallelizable. -However not all affine smooth algebraic varieties are parallelizable: for example the complement of a smooth conic in $\mathbb P^2(\mathbb C)$ is a smooth affine variety (Veronese embedding !) but is not even differentiably parallelizable. I wonder if these differentiable obstructions are the only ones preventing algebraic parallelizability of smooth algebraic subvarieties of $\mathbb C^n$. Any thoughts, dear friends? - -REPLY [3 votes]: There is a huge amount of work on these kinds of questions (I am aware of them since one of my colleagues, Satya Mandal, works on related topics). For example, this recent paper: -http://128.84.158.119/abs/0911.3495 -shows that stably free modules on smooth affine threefolds over alg. closed fields of char. not $2,3$ are free. -Another interesting relevant issue for your new question is when you can split a rank $1$ free off a projective module of rank $d=\dim R$ (Serre showed you can always do if the rank $>d$). Nori outlined a program to find the obstruction for this, something now called the Euler class group. You can find some relevant information and references at the second paper here: -(Local Coefficients and Euler Class Groups) -http://www.math.ku.edu/~mandal/publ.html -Cheers,<|endoftext|> -TITLE: Diffusion convergence -QUESTION [5 upvotes]: Consider diffusion:$$d\eta^x_{t}=\sigma(\eta^x_{t})dW_{t}+\mu(\eta^x_{t})dt,\quad \eta_0^x =x,$$ where $W$ is a Wiener process. We assume that $\sigma,\mu$ are such that the diffusion is well-defined and that it converges to the invariant measure $\eta_{\infty}$ (e.g. the Ornstein-Uhlenbeck process with $\sigma(\eta)=\sigma,$ $\mu(\eta)=-\mu\eta$ ). -By these assumptions we have -$$\mathbb{E} f(\eta^x_{t})-\mathbb{E}f(\eta_{\infty})\rightarrow0,$$ for “nice functions” $f$ (compactly-supported, $C^{2}$ should be enough). What is the speed of this convergence? E.g. in the case of the OU process we have $$e^{\mu t}(\mathbb{E}f(\eta^x_{t})-\mathbb{E}f(\eta_{\infty}))\rightarrow x\mathbb{E}f'(\eta_{\infty}),$$ which is easy to check by direct calculation using the transion density. What about other diffusions. I hope for something like $$l(t)(\mathbb{E} f(\eta^x_{t})-\mathbb{E}f(\eta_{\infty}))\rightarrow h(x)\mathbb{E}f'(\eta_{\infty}),$$ where $l(t)$ is some function (probably such that $\frac{\log l(t)}{t}$ is a slowly varying function) and $h$ is some other function (hopefully continuous and integrable with respect to the law of $\eta_{\infty}$ ). -The problem is probably very hard in general (some spectral gap theory) but I would be happy to know that the convergence holds for some “large class” of diffusions. -I looked into books of Revuz, Yor and Protter but have not found anything. Do you know any other book? - -REPLY [3 votes]: Preliminaries and Notation -I can proof your result without using the explicit transition density. The approach could generalize to the case, where the drift $\mu$ is a potential, so set $\mu(x)= -\nabla H(x)$. Then in the case of the OU-Process $H(x)=\frac{1}{2}x^2$. So we consider a process of the form -$$ \mathrm{d} X_t = - \nabla H(X_t) \mathrm{d}t + \sqrt{2} \mathrm{d} W_t . $$ -Then this process under some growth assumptions on $H$ has a unique invariant measure absolute continuous wrt. the Lebesgue measure $\mathrm{d} x$ given by $\mu(\mathrm{d} x) = \frac{1}{Z} \exp(- H(x))\mathrm{d}x $. -The evolution of the density $p_t$ is given by the Fokker-Planck-Equation -$$ \partial_t p_t = \nabla\cdot ( \nabla p_t + p_t \nabla H) . $$ -If we start the process not with a Dirac measure, but with a measure $p_0$ absolute continuous wrt. to $\mu$. We find a $\rho_t$ with $p_t = \rho_t \mu$. This relative density satisfies the equation -$$ \partial_t \rho_t = \Delta \rho_t - \nabla H \cdot \nabla \rho_t =: L \rho_t . $$ -The operator $-L$ is symmetric in $L^2(\mu)$ and under some growth condition on $H$ also elliptic. -Formal solution -Assume that the first eigenvalues of $-L$ are non-degenerated, i.e. $\lambda_0=0 < \lambda_1 < \lambda_2 \leq \lambda_3 \dots$. Then for a initial density $p_0$ absolute continuous wrt. to $\mu$ we have the expansion for $\rho_t$ -$$ \rho_t = 1 + \langle \rho_0, v_1 \rangle_{L^2(\mu)} v_1 e^{-\lambda_1 t} + O(e^{-\lambda_2 t}) , $$ -where $v_1$ is the eigenfunction of $L$ to the eigenvalue $\lambda_1$. Now we find -$$ \begin{split} - \mathbb{E}_{\rho_t \mu} f- \mathbb{E}_{\mu} f & = \int f(x) (\rho_t - 1)\mathrm{d}\mu \\ - &= \int f(x) \langle \rho_0,v_1\rangle_{L^2(\mu)} e^{-\lambda_1 t} v_1 \mathrm{d}\mu + O(e^{-\lambda_2 t}) -\end{split}$$ -Multiplying by $e^{\lambda_1 t}$ we find -$$ e^{\lambda_1 t} (\mathbb{E}_{\rho_t \mu} f- \mathbb{E}_{\mu} f) = \langle \rho_0 , v_1 \rangle_{L^2(\mu)} \int f(x) v_1 \mathrm{d} \mu + O(e^{-(\lambda_2-\lambda_1)t}). \tag{1}$$ -OU-semigroup -Now setting $H(x) = \frac{1}{2} x^2$ we have in 1d -$$ L = \partial_{xx} - x \partial_x .$$ -The invariant measure is Gaussian $\mu = \frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} x^2}$ and the eigenfunctions are the probabilistic Hermite polynomials with eigenvalues $0,-1,-2,\dots$. Hence $\lambda_1=1$ and $v_1(x)=x$ and $(1)$ takes the form -$$ e^{t} (\mathbb{E}_{\rho_t \mu} f- \mathbb{E}_{\mu}f) = \langle \rho_0, x\rangle_{L^2(\mu)} \int f(x) x \mathrm{d}\mu + O(e^{-t})$$ -Integrating by parts and using the fact $\mu'(x) = -x \mu(x)$ yields -$$ e^{t} (\mathbb{E}_{\rho_t \mu} f- \mathbb{E}_{\mu}f) = \langle \rho_0, x\rangle_{L^2(\mu)} \int f'(x)\mathrm{d}{\mu} + O(e^{-t}).$$ -Now with an approximation argument we take a series of inital distribution $\rho_0^n$ such that $\rho_0^n \mu \to \delta_{\xi}$ and find that $\langle \rho_0^n , x\rangle_{L^2(\mu)}\to \xi$, which shows your formula. -Problems in the general case -We used the spectral gap of the operator $L$, which holds in a wide class of potential $H$, but is hard to find explicit. There are estimates to bound it from below. Also sharp asymptotics for small noise are known. -Further we need the first eigenfunction, which corresponds to the metastable state of the system. Again for general $H$ explicit formulas are not known, but there exists again small noise expansions. -For your formula in the case of the OU-semigroup we heavily used the fact, that $-v_1(x) \mu(x) = \mu'(x)$, which is also not true in general. So the general formula would be $(1)$, which maybe can again be simplified, depending on the form of $v_1$.<|endoftext|> -TITLE: Is Mac Lane still the best place to learn category theory? -QUESTION [90 upvotes]: For a student embarking on a study of algebraic topology, requiring a knowledge of basic category theory, with a long-term view toward higher/stable/derived category theory, ... - -Is Mac Lane still the best place to start? - -Or has the day arrived when it is possible to directly learn ($\infty$,n)-categories, without first learning ordinary category theory? (So the next generation will be, so to speak, natively derived.) If so, via what route? If not, what's the most efficient path through the classical core material to a modern perspective? - -REPLY [4 votes]: I recommend for a first reading on category theory: -Martin Brandenburg, Einführung in die Kategorientheorie -It is really an excellent exposition with some nice perspectives on the concepts, -supported by plenty of examples. :) -(Note that it is provided in German only.) -A lot of thanks to Konrad S. for suggesting to me the reference! -(Such a pity, that I just missed the chance to meet the author in person..)<|endoftext|> -TITLE: Are there any solutions to $2^n-3^m=1$? -QUESTION [13 upvotes]: Are there any positive integer solutions to $2^n-3^m=1$ with $n,m>2$ ? -By way of justifying the question, I've found lots of info on what happens when $m=n$ (mostly FLT variations, Darmon + Merel,...), but I don't really know where to look for $m\not=n$. -Also it's pretty obvious that you can't have solutions to similar equations, e.g., $2^n-3^m=2$. There are no solutions for $n,m<1000$ aside from $n=2, m=3$. It seems pretty likely to me that it should happen for some large numbers at some point though. -Are there any theorems I don't know about regarding primes $p,q$ and $p^n-q^m=k$, $k \in \mathbb{N}$ that might rule out a solution or help me find one? - -REPLY [7 votes]: There is another method that allows one to handle $a^{n}-b^{m}=k$ (here I call the bases $a$ and $b$ because primality is not important to how the method works). Specifically, if one has a solution, it allows a larger solution to be found, or proven to not exist. -As an example of this method, it is easy to outline a proof that $2^{n} - 5^{m} = 3$ has no solutions larger than $(m,n)=(3,7)$: -Suppose $m>3$, $n>7$, and $2^{n}-5^{m}=3$. -Rewrite the equation as $2^{n}=5^{m}+3$. -Now, to use the largest solution we know, subtract $2^{7}=5^{3}+3$ from both sides to obtain $2^{7}(2^{n-7}-1)=5^{3}(5^{m-3}-1)$. -Since $m$ and $n$ give a solution larger than the one we know, both sides are positive integers. Since $n>7$, the highest power of $2$ dividing the right side is $2^{7}$. -Since the order of $5$ in $(\mathbb{Z}/(128))^{\times}$ is $32$, we know that $32$ divides $m-3$, and $5^{32}-1$ divides both sides. Then $29423041$, as a prime factor of $5^{32}-1$, divides both sides. Then $29423041$ divides $2^{n-7}-1$, so since the order of $2$ in $(\mathbb{Z}/(29423041))^{\times}$ is $122596$, $2^{122596}-1$ divides both sides. (This is probably not a profitable direction to take, but it can work as an illustration of the method.) -The contradiction would be obtained by concluding that $5^{4}$ divides the right side, or $2^{8}$ divides the left side. -In the case where a larger solution exists, the ability to bounce back and forth between the two sides of the equation only goes as far as concluding that the larger solution (that is, the common value of both sides when the larger solution is plugged in) minus the common value from the known solution divides both sides.<|endoftext|> -TITLE: Ratio of triangular numbers -QUESTION [18 upvotes]: Which natural numbers can be written as the ratio of two triangular numbers? -That is, which natural numbers can be written as -$$\frac{n(n+1)}{m(m+1)}$$ for natural numbers $m,n$. - -REPLY [3 votes]: A small complement to the nice answers so far: there are infinitely many squares that are not the ratio of two triangular numbers. -Indeed, with Gjergji Zaini's notation, let $k=p+1$, where $p\equiv 3\pmod{4}$ is a prime. Then the equation -$$k^2-1=(ky-x)(ky+x)$$ -becomes -$$p(p+2)=((p+1)y-x)((p+1)y+x).$$ -The first factor on the right is at most $p$, hence if $p$ divides this factor, then the factors on the right are $p$ and $p+2$, so that $y=1$, which corresponds to $n=0$. If $p$ divides the second factor on the right, then the factors on the right are $q$ and $pr$ such that $qr=p+2$. Now -$$q-r\equiv q+pr=2(p+1)y\equiv 0\pmod{p+1},$$ -hence $|q-r|\geq p+1$, which forces $q=1$ and $r=p+2$ (since $1\leq q\leq p$). In other words, the factors on the right are $1$ and $p(p+2)$, so that $y=(p+1)/2$. The last expression is even, hence it does not correspond to any $n$.<|endoftext|> -TITLE: Why should I believe the Singular Cardinal Hypothesis? -QUESTION [17 upvotes]: The Singular Cardinal Hypothesis (SCH) is the statement that $\kappa^{cf(\kappa)} = \kappa^+ \cdot 2^{cf(\kappa)}$ for every singular cardinal $\kappa$ (or various equivalent statements). -It is obviously implied by the Generalized Continuum Hypothesis. It is also implied by the Proper Forcing Axiom (under which $2^{\aleph_0} = \aleph_2$). Nonetheless it doesn't seem terribly compelling to me. But I am trying to learn to appreciate it! -Why should I believe SCH? (Now when I say "believe," it isn't clear exactly what I mean by that. There are obviously plenty of models of set theory in which SCH holds, and they are certainly worth studying, but somehow they aren't the models that feel most "realistic" in my little head.) What I want is an answer in the style of Maddy's Believing the axioms, explaining why I should "like" (or not) this hypothesis. -Some thoughts on why SCH isn't so unreasonable: - -Somehow I find very compelling the result that SCH holds above the first strongly compact cardinal. This is what makes it seem most reasonable to me that SCH should hold everywhere. -SCH is implied by various contradictory axioms. Its negation is equiconsistent with the existence of a fairly large cardinal. -It is obvious that $\kappa^{cf(\kappa)} \geq \kappa^+ \cdot 2^{cf(\kappa)}$, so SCH is just saying that the cardinality on the left shouldn't be any larger than strictly necessary. -It makes cardinal arithmetic much easier! (But maybe this is an argument against SCH too...) - -Why else should SCH seem reasonable? - -REPLY [10 votes]: $SCH$ follows from $MM$ (Foreman-Magidor-Shelah), in fact from $PFA$ (Viale). One reason why one might not want to believe $MM$ is that there is another ``maximality principle'' which also implies $2^{\aleph_0}=\aleph_2$, namely $(*)$, and the logical relation between $MM$ and $(*)$ is somewhat of a mystery. I proposed an axiom, though, called $MM^{*,++}$, which implies both $MM$ and $(*)$, see my "Woodin's axiom (*), or Martin's Maximum, or both?", in: Foundations of mathematics, essays in honor of W. Hugh Woodin's 60th birthday, Harvard University (Caicedo et al., eds.), pp. 177-204, available at https://ivv5hpp.uni-muenster.de/u/rds/paper_for_woodin_volume_new_version.pdf . It's a really cool axiom to believe in! The mathematical challenge is to prove the consistency of $MM^{*,++}$ relative to anything which is believed to be consistent. In short, you should believe $SCH$ because you should believe $MM^{*,++}$.<|endoftext|> -TITLE: Does the extension to the pro-completion of a left exact (finite) colimit preserving functor preserve (finite) colimits -QUESTION [6 upvotes]: Let $\mathbf C$ be a category with finite limits. Then a left exact functor $F\colon \mathbf C\to \mathbf{Set}$ is pro-representable and hence extends to the pro-completion $\mathbf C$. My question is whether it is true that the extension of $F$ preserves finite colimits whenever $F$ does, and if so what is a reference? I'm also curious about arbitrary colimits but for the application I care about, finite colimits (or even coequalizers) are enough. - -REPLY [2 votes]: I am not certain if he provides the answer to your question, but Dan Isaksen's paper: -Calculating limits and colimits in pro-categories, Fundamenta Mathematicae 175 (2002) 175--194. -is relevant I think. The point is that the reindexing lemmas in procategories provide a powerful tool for calculating limits and colimits.<|endoftext|> -TITLE: Why are noetherian rings such natural objects in algebraic geometry? -QUESTION [19 upvotes]: I assume it is partially because they are good generalizations of polynomial rings, but what makes this generalization better than graded algebras or other generalizations of polynomial rings? - -REPLY [2 votes]: I study mainly modules and abelian groups, but even I find the Noetherian property useful. For me, probably the main feature lurking in the background is that submodules of a finitely generated $R$-module, $R$ Noetherian, are finitely generated. This fact also allows us to conclude that a finitely generated module over a Noetherian ring $R$ is finitely presented. For a finitely generated module over an arbitrary ring, we always have the exact sequence $$0 \rightarrow K \rightarrow R^n \rightarrow M \rightarrow 0$$ where $K$ is the kernel of the map $R^n \rightarrow M$ that sends the standard basis elements of $R^n$ to the generators of $M$. We say $M$ is finitely presented if $K$ is finitely generated. If $R$ is Noetherian, this is automatic. -It is also nice that when $R$ is Noetherian, any finitely generated $R$-module has a primary decomposition. -Facts like this are standard fare in commutative algebra. Now it has been a good while since I've done much algebraic geometry, but if memory serves, when decomposing an affine variety into irreducible subvarieties, one uses a primary decomposition of the ideal$I$ of the variety (in the polynomial ring in $n$ variables over the field underlying the affine space; this field is usually assumed algebraically closed). But the ideal $I$ turns out to be equal to its own radical, so the primary decomposition involves only prime ideals, and each of these corresponds to an irreducible (affine) variety, each being a component of the original variety.<|endoftext|> -TITLE: Confusion about D-modules and functors -QUESTION [9 upvotes]: Given a morphism $f:X\rightarrow Y$ between smooth complex varieties, one can define functors from the bounded derived category with holonomic cohomology on $Y$ to the same category on $X$. The easiest one is $Lf^{*}$ which can be obtained by putting a $D$-module structure on the inverse image of $\mathcal O$-modules and deriving it. From this one can get two more functors: - -$f^!:=D \circ Lf^{*} \circ D$, where $D$ is the duality functor and -$f^{\dagger}:=Lf^{*}[dim X-dim Y] $ - -Now my question is, under what conditions are these two isomorphic? -Edit: These notations are bad/confusing/wrong, since they are not compatible with the formalism of six functors. See answers below for better notations. - -REPLY [12 votes]: Just to clear up some notational confusion (there doesn't seem to be any completely standard notation): -In Bernstein's notes: -The "easy" pullback (i.e. the one that coincides with the pullback of the underlying $\mathcal O$-module) is denoted $Lf^\Delta$. -$f^! = Lf^\Delta [dim X - dim Y]$ (right adjoint to $f_!$) -$f^\ast = \mathbb D f^! \mathbb D$ (left adjoint to $f_\ast$) -Note that $f^\ast$ and $f^!$ agree with the corresponding functors for constructable sheaves (not for the underlying $\mathcal O$-modules). -However... In Hotta, Takeuchi, Tanisaki -The easy pullback is denoted $Lf^\ast$ to agree with the $\mathcal O$-module functor. -Berstein's $f^!$ is now called $f^\dagger$. -Bernstein's $f^\ast$ is now called $f^\star$ -Ok, now in David Ben-Zvi's answer above (and in many other places) -The easy pullback is $f^\dagger$, and the rest agrees with Berstein's notation. -In my opinion, the most important notational feature to be preserved is that $(f^\ast , f_\ast)$ and $(f_! , f^!)$ form adjoint pairs. -To answer your question -When $f$ is smooth, $f^\ast = \mathbb D f^! \mathbb D = f^! [2(dim Y - dim X)]$, and the easy inverse image functor is self dual (and preserves the t-structure). -One way to think about these different pullbacks is that the easy inverse image preserves the structure sheaf $\mathcal O_Y$. This corresponds to the constant sheaf shifted in perverse degree under the RH correspondence. On the other hand $f^\ast$ preserves the "constant sheaf" whereas $f^!$ preserves the dualizing sheaf (for a smooth complex variety, these correspond to the D-modules $\mathcal O[-n]$ and $\mathcal O[n]$). - -REPLY [4 votes]: The two standard D-module pullbacks agree (up to a shift) for smooth morphisms. Bernstein, Borel or Kashiwara are standard references for this. -A couple of comments: First you should restrict to coherent D-modules to get the duality functor (on the derived level). Next I think the notation is a little off from the standard. -The easy D-module pullback functor is usually the one denoted with a dagger, while its -shifted version is $f^!$, not $f^\ast$ - which is its dual. These conventions are set up so as to have the "formalism of six operations" -- i.e. we have adjoint pairs $(f^\ast,f_\ast)$ and $(f_!,f^!)$ for general morphisms (on holonomic D-modules), related by duality. The two pushforwards agree for proper maps and the two pullbacks agree (after dimension shift) for smooth maps.<|endoftext|> -TITLE: Deformations of Hirzebruch surfaces and toric action -QUESTION [7 upvotes]: Hi, -the Hirzebruch surface $F_n$ admits a deformation for $0\leq m\leq n$ defined by the equation -$$ -\mathcal{M}=\{ ([x_0:x_1],[y_0:y_1:y_2],t) \in \mathbb{P}^1 \times \mathbb{P}^2 \times \mathbb{C}| x_0^ny_1-x_1^ny_0 + t x_0^{n-m}x_1^m y_2 =0\} -$$ -The central fiber is $F_n$ while a noncentral fiber is isomorphic to $F_{n-2m}$. -Each fiber $\mathcal{M}_t$ is toric. However it seems almost always impossible to find a differentiable action of the torus $T^2$ on $\mathcal M$ that would preserve and act holomorphically and torically on the fibers. -I would like to know whether this is correct and why. -Thanks a lot. - -REPLY [3 votes]: This is more of a comment than an alternative solution to Dmitri's solution (unfortunately I am again unable to comment; I still don't understand the rules here). Yann can also prove his assertion by infinitesimal deformation theory. This has the feature that it works over an arbitrary base field (even in positive characteristic). The first-order deformations of a smooth, toric variety $X$ are the same as vectors $v$ in $H^1(X,T_X)$. For the torus $T$ acting on $X$, the tangent sheaf $T_X$ is $T$-linearized. For an algebraic subgroup $S$ of $T$, the action of $S$ on $X$ extends to the first-order deformation if and only if $v$ is $S$-invariant. -By the description of the tangent sheaf $T_X$ in Section 4.3 of Fulton's book, $H^1(X,T_X)$ equals the direct sum over all irreducible $T$-divisors $D$ of $H^1(D,N_{D/X})$, where $N_{D/X}$ is the normal sheaf of $D$ in $X$, i.e., $\mathcal{O}_X(D)|_D$. For such a $D$, there is a one-parameter subgroup $S_D$ of $T$ which fixes $D$ pointwise, not just setwise (this can be checked on a $T$-invariant open affine which intersects $D$, hence follows from the special case that $X$ equals $\mathbb{A}^m$ cross a torus). The invertible sheaf $N_{D/X}$ is $T$-linearized, hence $S_D$-linearized. But $S_D$ acts trivially on $D$, hence this $S_D$-linearization is just scaling by a character of $S_D$. Again by looking at a $T$-invariant open affine, one can see that this character is an embedding, i.e., it has trivial kernel. This $S_D$-linearization gives an action of $S_D$ on $H^1(D,N_{D/X})$ which is scaling by the same character (or perhaps the dual character, depending on your sign conventions). In particular, for a first-order deformation of $X$ corresponding to a vector $v$ in $\oplus_D H^1( D, N_{D/X} )$, if the projection of $v$ into the factor $H^1(D,N_{D/X})$ is nonzero, then $v$ is not $S_D$-invariant. Thus -the action of $S_D$ on $X$ does not extend to the first-order deformation, in fact the action of no nontrivial subgroup of $S_D$ extends.<|endoftext|> -TITLE: Preduals of B(E) -QUESTION [7 upvotes]: For a Hilbert space $H$ it is well known that the algebra $B(H)$ has a unique predual; the Banach space of trace class operators. -If $E$ is a Banach space then is it known whether - -$B(E)$ is always a dual Banach algebra? -The predual is always unique? - -I'm aware that 2 can fail in the case of a general `dual Banach algebra', so if the answer is "No!" can we place appropriate conditions on $E$ to ensure that 1 and 2 hold? If this is well-known then appropriate references would be useful. - -REPLY [4 votes]: As Yemon mentioned ages ago (sorry!) I explored this a bit in http://arxiv.org/abs/math.FA/0604372 -We say that a Banach algebra $A$ is a dual Banach algebra if $A$ is isomorphic to $E^*$ for some Banach space $E$, such that the multiplication in $A$ becomes separately weak$^*$-continuous. If $X$ is a dual space, then $B(X)$ is the dual of $N(X_*)$ but a little calculation shows that the multiplication is only weak$^*$-continuous on one side. To get a dual Banach algebra, you need $X$ to be reflexive. -My little result is that if $X$ is reflexive, and also has the approximation property, then $N(X)$ is the unique dual Banach algebra predual of $B(X)$. To be precise, if $B$ is another dual Banach algebra, and $\theta:B(X)\rightarrow B$ is a linear bijection, is bounded, and is an algebra homomorphism, then $\theta$ is necessarily weak$^*$-continuous (so I don't need to assume that $\theta$ is an isometry).<|endoftext|> -TITLE: A four-dimensional counterexample? -QUESTION [13 upvotes]: Does anyone know an example of a smooth hyperbolic surface bundle over a hyperbolic surface (surface = compact two-manifold) which does not have a complex structure? Is there any decision procedure to tell, given such a bundle, whether it has a complex structure, or is it more of a black art? - -REPLY [8 votes]: Igor, suppose the total space of a bundle is complex. Then the fibration can be realized -holomorphically. Such fibrations are classified by holomorphic maps of the base Riemann surface to the moduli space of the fiber. Thus, your question about a "procedure" essentially reduces to the problem of what homomorphisms from surface groups to the mapping class group are realized holomorphically. There are obstructions of Hodge-theoretic nature (since the total space of the bundle is Kahler), for instance, Betti numbers of the 4-manifold should be even, etc. If the fiber has genus 2 (and, thus, is hyperelliptic), then its moduli space admits a finite cover which is a domain in ${\mathbb C}^3$, so no non-constant holomorphic maps in this case. However, for higher genus of the fiber, there is nothing like a complete set of invariants, even conjecturally. -There are interesting examples of complex structures on fibrations which fall into two classes: -(a) Using the period map (or a similar construction) one can show that the moduli space $M_g$ (except for $g\le 2$) contains a lot of compact Riemann surfaces. For instance, S.Diaz ["Complete subvarieties of the moduli space of smooth curves", 1987] used this to show that one can construct a projective curve passing through any finite subset of $M_g$ for $g\ge 3$. -(b) Kodaira's direct construction of complex surfaces which fiber: They are obtained as ramified covers over products of two Riemann surfaces, where the ramification locus is a (possibly disconnected) smooth Riemann surface in the product (graph of a, possibly multivalued, locally invertible holomorphic map between the surfaces).<|endoftext|> -TITLE: Some unpublished notes of Hofstadter -QUESTION [16 upvotes]: I'm looking for some unpublished notes called "Eta Lore," which are apparently related to a talk Douglas Hofstadter first gave at the Stanford Math Club in 1963. I know these notes exist because they're cited in the following articles: - -Hendel, R.J.; Monteferrante, S. "Hofstadter's Extraction Conjecture." The Fibonacci Quarterly 32(2), 1994. -Chuan, W. "Extraction Property of the Golden Sequence." The Fibonacci Quarterly 33(2), 1995. -Nillsen, R.; Tognetti, K.; Winley, G. "Bernoulli (Beta) and integer part sequences." A teaching module developed by the Australian Mathematical Society. - -I've e-mailed the authors of the articles above, but I also thought I'd try asking around here, since none of the authors who've responded so far have been able to help. Any information on alternative references would also be appreciated. -Motivation and Background -I'm specifically interested in the definition of the function INT mentioned in Section I.33 of Gödel, Escher, Bach. The description there, in case it rings any bells, is: "The basic idea behind INT is that plus and minus signs are interchanged in a certain kind of continued fraction." -As far as I can tell, the notes I'm looking for are mostly about continued fractions and integer sequences---in particular, things called $\eta$-sequences and $\beta$-sequences (unfortunately, I have no idea what those are, and I haven't found any leads on OEIS). - -REPLY [10 votes]: A copy of Hofstadter's paper is on OEIS here: Link to Eta-Lore paper -It is also referenced on OEIS at A001468.<|endoftext|> -TITLE: Topology of SU(3) -QUESTION [24 upvotes]: $U(1)$ is diffeomorphic to $S^1$ and $SU(2)$ is to $S^3$, but apparently it is not true that $SU(3)$ is diffeomorphic to $S^8$ (more bellow). Since $SU(3)$ appears in the standard model I would like to understand its topology. -By one of the tables here $SU(3)$ is a compact, connected and simply connected 8-dimensional manifold. This MO post says that its $\pi_5$ is $\mathbb{Z}$ thus it can not be homeomorphic to $S^8$(e.g.: see this wiki article). Even if it was a homotopy sphere Poincaré conjecture would not be helpful (at least in the smooth category: there exists exotic 8-spheres, right?). -I guess that this is what the author of this question was trying to know... -Anyway, is it known any manifold diffeomorphic to $SU(3)$? - -REPLY [5 votes]: As explained in the survey paper The Geometry of Compact Lie Groups by L.J. Boya, and by several other people here, $SU(3)$ is an $SU(2)\cong S^{3}$ fiber bundle over $S^{5}$, but in this case in particular, you can say a little more: The $SU(2)$-bundles over $S^{5}$ are classified by $\pi_{4}(S^{3})=\mathbb{Z}/2\mathbb{Z}$. We know that $SU(3)$ is not the trivial bundle, hence, it should be the unique square root of the trivial principal bundle.<|endoftext|> -TITLE: $p$-adic continuity for exponents in product decomposition of the $j$-invariant -QUESTION [7 upvotes]: Let j be the usual j-invariant -$j(z) = 1/q + 744 + 196884q + ...$ -where $ q = $exp$(2 \pi i z),$ -and let $x_n$ denote the exponent of $1 - q^n$ in the product decomposition -$j(z) = (1/q) (1 - q)^{x_1} (1-q^2)^{x_2} .... .$ -What, if anything, is known about p-adic continuity of the function $n \mapsto x_n$, especially for p = 2, 3, and 5? - -REPLY [11 votes]: The function $x_n$ will not be continuous. -Note that $j$ is a function of $\tau$, and -$$\frac{1}{2 \pi i} \frac{d}{d \tau} = -q \frac{d}{d q}.$$ -If $f$ is a meromorphic modular function, then $dlog(f)$ is a meromorphic modular form of weight two (easy exercise). Applying this with $f = j$, we find that -$d log(j)$ is such a function, which one easily computes to be $-E_6/E_4$. -Hence -$$\frac{-E_6}{E_4} = \frac{j'}{j} = q \frac{d}{dq} -\left(- \log(q) + \sum_{n=1}^{\infty} x_n \log(1 - q^n) \right) -= - 1 + \sum_{n=1}^{\infty} \frac{n x_n q^n}{1 - q^n}.$$ -Expanding the RHS in the usual way, we find that -$$\frac{-E_6}{E_4} = - 1 + \sum_{n=1}^{\infty} - q^n \sum_{d|n} n x_n.$$ -Suppose the form on the left is overconvergent, which it is whenever $i$ is a supersingular $j$-invariant (so in particular for $p=2$, $p = 3$, and $p = 5$). Then one (roughly) expects a decomposion into (generalized) overconvergent eigenforms of weight two, so -$$\frac{E_6}{E_4} = \lambda E^{*}_2 + \sum \lambda_i f_i,$$ -where $E^{*}_2$ is the Eisenstein series of weight $2$ and level $\Gamma_0(p)$, -and $f_i$ are (non-classical) cuspidal generalized eigenforms (when $p = 2$, this is actually a theorem in this case of David Loeffler). -Write -$E_6/E_4 = \sum a_n q^n$. -If $x_n$ is $p$-adically continuous, then - $a_l \equiv a_{l'}$ for $l \equiv l'$ modulo a high power of $p$. -This is easily seen to be true for $E^*_2$. Is it also true for $f_i$? -What would it mean if $a_l$ was continuous as a function of the primes -$l$? Remember that associated to $f_i$ is a Galois representation -$\rho_{i}$ such that the trace of Frobenius at $l$ is $a_l(f_i)$. If this was a continuous function for primes $l$, then by Cebotarev density, it would follow that the corresponding -Galois representation $\rho_{i}$ would factor (modulo $p^n$) -through an abelian extension. In particular, $\rho_{f_i}$ itself would -have to be reducible, contradicting known facts. So the chances that $x_n$ -are continuous are zero. (With more effort I could produce a rigorous proof of this fact, but it is not worth it.) -Numerical computation will be misleading in this case, for a reason first noted by Serre and Swinnerton-Dyer. Take Ramanujan's function $\Delta = q \prod_{n=1}^{\infty} (1-q^n)^{24}$. -Then for primes $l$ it appears that $\tau(l)$ is $2$-adically continuous. -This is related to the fact that $\rho_{\Delta}$ is essentially abelian -modulo quite a large power of $2$, something like $2^{11}$. But it cannot be so in chararacteristic zero because $\rho_{\Delta}$ is irreducible, even though showing this by naive computation is -surprisingly hard. The forms $f_i$ will have a similar property, which is why your computations falsely suggest that the $x_n$ are continuous.<|endoftext|> -TITLE: Geometric Interpretation of $Q$-curvature -QUESTION [12 upvotes]: Let $(M,g)$ be a Riemannian manifold of dimension $n>2$. Thanks to the late T.Branson we have the following definition of the so-called $Q$-curvature: -$Q= \Delta R + \frac{n^3-4n^2+16n-16}{4(n-1)(n-2)^2} R^2 - \frac{8(n-1)}{(n-2)^2}|Ric|^2.$ -Here $\Delta = -div\nabla$, $R$ is the scalar curvature, and $|Ric|$ is the norm of the Ricci tensor. There has been much research on $Q$ curvature since its discovery in the eighties, motivated in large part by the conformal transformation properties that it possesses. A question that appears to be open, though, is whether or not there is a (relatively) concise geometric interpretation of this scalar curvature invariant. For $R$ we have the nice interpretation that it determines the rate at which the growth of a ball around a point differs from the flat case. Similarily the Ricci tensor measures the deviation of a solid angle from the Euclidean case. Can you think of a geometric interpretation of $Q$-curvature that is similarily elegant? - -REPLY [4 votes]: I think this is more like a remark than an answer. I gave seminar on Q-curvature (more precisely, Q-curvature flow) twice. In both seminars, I was asked, "What is the geometric meaning of Q-curvture? For example, if Q-curvature is zero, what can we conclude about the manifold $M$?" I was surprised that same question has been raised, and how little we know about Q-curvature. But my answer to their questions is this: "If Q-curvature is zero, then we are solving a 4th order partial differential equation. But we know very little about 4th order PDF, even though it is elliptic. However, if we impose more condition on the manifold $M$, we may be able to get something. For example, if $M$ is locally conformally flat with zero Q-curvature, then the Euler-characteristic of $M$ must be zero, which follows from the Chern-Gauss-Bonnet Theorem in 4 dimension: -$$\chi(M)=\frac{1}{8\pi^2}\int_M(|W|^2+Q),$$ -where $W$ is the Weyl tensor."<|endoftext|> -TITLE: Is there an algebraic description of Yang-Mills measure? -QUESTION [14 upvotes]: Consider $X=\operatorname{Hom}(\pi,\mathrm{SL}(2,\mathbb C))/\! \!/\mathrm{SL}(2,\mathbb C)$ and $Y=\operatorname{Hom}(\pi,\mathrm{SU}(2))/\mathrm{SU}(2)$, where $\pi$ is a surface group. Note that if we use the right coordinates, then $Y$ is exactly the real valued points of $X$ (for the purposes of this question, though, this is irrelevant). -For those reading the comments below, I used to use $X_{\mathbb R}$ instead of $Y$. -Since $\pi$ is a surface group, the cup product (intersection pairing) gives rise to a symplectic form $\omega$ on $Y$ (which can also be naturally viewed as a holomorphic symplectic form on $X$). Now for $f\in\Gamma(X,\mathcal O_X)$, we have a natural linear map: -$$f\mapsto \int_{Y}f\cdot\wedge^{\operatorname{top}}\omega$$ -(I believe it is more or less correct to call this the Yang-Mills measure, e.g. see this paper by Bullock, Frohman, and Kania-Bartoszynska) -Question: Is there a completely algebraic description of this integral? i.e. one that just works with input $\pi$ and the algebraic group $\mathrm{SL}(2)$, and avoids the notion of real and complex points? -Note: there is certainly a simple algebraic description of $\omega$, so perhaps we just need to algebraically define the homology class of $Y$ in $X$ . . . -A paper of Witten's is perhaps related (and great to read even if it isn't). - -REPLY [5 votes]: I am a little reticent to do this. This is just an outline that is based heavily on the BFK paper -you cited in the question, and what seems to be your observation. Maybe I am just repeating back -to you what you are thinking. -To start with $SL_2$ is the restricted dual of $U(sl_2)$ that is the linear functionals that factor through a finite dimensional representation. You can also think of $SL_2$ as the coordinate ring -of $SL_2\mathbb{C}$. There is a linear functional $\mu:SL_2\rightarrow \mathbb{C}$ given by restricting the function to $SU(2)$ and integrating against Haar measure. Even though its analytic -its really algebraic, as Weyl orthogonality gives a complete set of rules for evaluating it. -$SL_2$ is a Hopf algebra, which you will need for the next construction. -Taking the tensor product of $\mu$ with itself $k$-times you get $\mu^{\otimes k}:SL_2^{\otimes k}\rightarrow \mathbb{C}$. -There is an action of $U(sl_2)$ on $SL_2^{\otimes k}$ that is the analogy of the diagonal action -by conjugacy of $SL_2\mathbb{C}$ on the coordinate ring of the cartesian product of $SL_2\mathbb{C}$ with itself $k$-times, $C[SL_2\mathbb{C}^k]=SL_2^{\otimes k}$, so that the ring -of invariants can be identified with the characters of $SL_2\mathbb{C}$ representations of -the free group on $k$ letters. -From the Bullock paper in Commentari, or if you wish the Przytycki-Sikora paper about the same time, you can identify this ring with Kauffman bracket skein algebra with -$A=-1$ of a cylinder over any orientable surface $F$ whose fundamental group is the free group -on $k$ letters. That means we can use the algebra of Jones-Wenzl idempotents to construct things. -Let $F$ be a closed oriented surface of genus $g$ and let $F'$ be the result of removing an open disk from $F$. -Let $\sum_c (-1)^c(c+1)s_c(\partial F')$ be series coming from coloring $\partial F'$ with the $c$th Jones Wenzl idempotent. We can define a linear functional on the characters ring of $F'$ by letting -$$YM(\alpha)=\lim_{N\rightarrow \infty} \mu^{\otimes 2g}(\sum_{c=0}^N\alpha(-1)^c(c+1)s_c(\partial F'))$$ The linear functional cannot see handleslides across the boundary of $F'$. That was how -we defined the Yang-Mills measure in the paper. -Lets see if we can see it more algebraically. -Complete $ SL_2^{\otimes 2g}$ so that -$$ \zeta=\sum_{c=0}^{\infty}(-1)^c(c+1)s_c(\partial F') $$ -is -in the completion and annihilates handleslides. -Here is how to complete. The Yang-Mills measure defines a symmetric pairing -$$<\alpha,\beta>=YM(\alpha\beta)$$ We say a sequence $\alpha_n$ is Cauchy if for every -character $\beta$, the sequence of complex numbers $<\alpha_n,\beta>$ is Cauchy. We complete -by including the characters of the free group into the equivalence classes of Cauchy sequences. -The completion is no longer an algebra, but it is a module over the the characters of the free group. -Notice for any character $\alpha$, the sequence of partial sums of $\alpha \zeta$ is Cauchy, -and hence defines an element of the completion. The span of all $\alpha\zeta$, the cyclic module -generated by $\zeta$ is isomorphic to the $SL_2\mathbb{C}$ characters of the fundamental group -of $F$. Furthermore the restriction of the extension of the $\mu^{\otimes 2g}$ to that cyclic -module is the Yang-Mills measure. -Finally, instead of using characters, you can use the Kauffman bracket skein algebras of cylinders over surfaces as long as $A$ doesn't lie on the unit circle, or it it does, it needs to be a $2p$th root of unity for some counting number $p$. -There are other ways of getting there. I always thought that the Yang-Mills measure had something -to do with type $II_1$ factors, at least in the quantized case, but I could never get there. No doubt -you need $A$ to be real, and then you need to do some some sort of transform as initially skeins act like unbounded operators.<|endoftext|> -TITLE: Probabilistic Solution of the Porous Medium Equation -QUESTION [8 upvotes]: It is well known that the transition density for standard Brownian motion $B_t$ in $\mathbb{R}^d$ yields a solution to the global Cauchy problem for the heat equation $$u_t = \Delta u$$ with initial condition given by the Dirac distribution $\delta_0$. -Unlike the heat equation, the porous medium equation $$u_t = \Delta(u^m)$$ with exponent $m>1$ has finite speed of propagation. - -When is the infinitesimal generator of a stochastic process linear? -Is there a probabilistic solution of this non-linear diffusion equation? - -REPLY [2 votes]: The processes described by Andre work by having the interaction act at the level of the mobility of the particles. -There are other ways, too. One is in the work of Philipowski (see also Figalli & Philipowski). Here the idea is to take interactions of potential type, i.e. for instance -$ -dX^i = -\sum_{j\neq i} \nabla W_\epsilon(X^i-X^j) \, dt + \delta \, dB^i. -$ -The parameter $\epsilon$ is the spatial range of $W$, and in the limit $\epsilon\to0$ the interaction becomes purely local, and leads to a nonlinear diffusion term. If one also lets $\delta\to0$, then the purely Brownian contribution also vanishes. Only the nonlinear diffusion is then left.<|endoftext|> -TITLE: Riemann hypothesis via absolute geometry -QUESTION [95 upvotes]: Several leading mathematicians (e.g. Yuri Manin) have written or said publicly that there is a known outline of a likely natural proof of the Riemann hypothesis using absolute algebraic geometry over the field of one element; some like Mochizuki and Durov are thinking of a possible application of $\mathbf{F}_1$-geometry to an even stronger abc conjecture. It seems that this is one of the driving forces for studying algebraic geometry over $\mathbf{F}_1$ and that the main obstacle to materializing this proof is that the geometry over $\mathbf{F}_1$ (cf. MO what is the field with one element, applications of algebaric geometry over a field with one element) is still not satisfactorily developed. Even a longer-term attacker of the Riemann hypothesis from outside the algebraic geometry community, Alain Connes, has concentrated recently in his collaboration with Katia Consani on the development of a version of geometry over $\mathbf{F}_1$. -Could somebody outline for us the ideas in the folklore sketch of the proof of the Riemann hypothesis via absolute geometry ? Is the proof analogous to the Deligne's proof (article) of the Riemann-Weil conjecture (see wikipedia and MathOverflow question equivalent-statements-of-riemann-hypothesis-in-the-weil-conjectures) ? -Grothendieck was not happy with Deligne's proof, since he expected that the proof would/should be based on substantial progress on motives and the standard conjectures on algebraic cycles. Is there any envisioned progress in the motivic picture based on $\mathbf{F}_1$-geometry, or even envisioned extensions of the motivic picture ? - -REPLY [27 votes]: Last fall, there was a conference in Nagoya about precisely this question (oddly enough, funded by a "Riemann Hypothesis" DARPA grant). Since I was attending a different conference at the same university at the same time, I didn't get to see all of the talks. However, Kedlaya's overview talk, which is listed among others on the schedule page, is rather informative. -Essentially, one hopes to get the completed $L$-function of an $\mathbb{F}_1$-scheme $X$ by cohomological means, by choosing a holomorphic family of operators (analogous to $1-q^{-s}\text{Frob}_q$ in the function field setting), and taking the determinant of the action on the cohomology of $X$ (which is expected to be infinite dimensional). This is basically a generalization of the Grothendieck-Lefschetz trace formula to a cohomology theory that is not yet known. There is some algebraic evidence that some form of the de Rham-Witt complex with a suitable alteration at infinity is such a cohomology theory, but I don't know what the appropriate family of operators ought to be. I am told that there are promising hints coming from the world of dynamical systems and foliated spaces, and this is where non-commutative geometry seems to enter the picture.<|endoftext|> -TITLE: Stylistic question -QUESTION [7 upvotes]: I'm writing up a paper now where I'm the only author and have a stylistic question. -Should I write ''I'' or ''we'' as in ''I/we recall the definition...'' etc. I think this simple example will make everyone understand what I'm talking about. -Or should/can I mix? Is this too confusing. Or simply bad? Or ok? -I feel that ''we'' is maybe a bit too formal and ''I'' gives the paper a certain personal cosy touch. But in certain respects ''we'' is more ''we go through this ordeal together you and I, dear reader'', where ordeal can mean anything in a whole spectrum of things. You know what I'm saying. -So? - -REPLY [2 votes]: In the interest of having an answer and since this is CW anyway: -English is not my native language and (thus) I read several articles/books/chapeters on mathematical writing in English, among others by Halmos, Krantz, and Knuth (et al.) -I do not have the references handy and, since it's been a while, do not recall what precisely they said (and in particular not who said what exactly), but the consesus is definitely 'we' (with exceptions for things like thanks and so on; or, somebody said roughly: one of the rare situations where 'I' feels appropriate is to say 'I could not prove this conjecture' where both 'we' and 'one' seem odd; and also for the 'thanks') -So, I strongly recommend using 'we', in the sense you mentioned yourself (the author and the reader) and using 'I' only if it is a really somehow personal statement. -There is the additional question of using 'the author' instead of 'I' (in the above exceptional cases). Here, I think the situation is slightly less clear. Typically, if I am the sole author I write 'I' as writing myself 'the author' feels somehow strange for me. But, I had at least once a journal that changed this (in the Acknowledgment something like 'I thank..' was change to 'The author thanks...'). -In case of multiple authors I am a bit more likely to write 'the authors' (in the exceptional cases) in order not to have two different 'we' (authors and reader as well as just the authors). -Personally, I think an additional advantage of the 'we' (instead of 'I') is that thus articles are written in the same way no matter whether it is single authored or coauthored. -To have single authored ones with 'I' and coauthored ones with 'we' seems confusing (well, maybe confusing is a too strong word, but I'd find it at least mildly distracting). -And, finally, how incovenient would it be, one starts a project alone and starts to write a bit of it up, and latter one finds a coauthor...then one would have to change all the 'I' to 'we'.<|endoftext|> -TITLE: Is Lebesgue/Borel non-measurability actually caused by non-uniqueness? -QUESTION [7 upvotes]: In ZFC, every construction of a Lebesgue or Borel non-measurable set uses the axiom of choice. None of them that I've seen use choice to define a unique set, even though it's entirely possible to do so (e.g. under the AoC, if $\kappa = |A|$ is the cardinality of set $A$, then $\kappa$ is unique). So I've been wondering lately whether the strength of the AoC is enough by itself to construct a non-measurable set. -Here's an attempt at a specific question. In the language of set theory (first-order logic with equality extended with the ZFC axioms), is there a formula without parameters that identifies a unique, Lebesgue/Borel non-measurable set? - -REPLY [4 votes]: Concerning Borel measurability, it was already pointed out that there is an explicit formula s(x) such that ZFC proves "The set { x in R: s(x) } is not Borel". -(This is not true for ZF, as was pointed out elsewhere.) -Concerning Lebesgue measurability, ZFC neither proves nor refutes the following: - -There is an OD-definition (or: OD(R)-definition) of a subset of the reals which is non-measurable. - -There is a slight fuzziness here, because there are many non-equivalent notions of definability; OD(R)-definability is perhaps the most prominent and useful. -But an explicit formula can be given: -There is a formula phi(x) in the language of set theory (without parameters) such that ZFC neither proves nor refutes - -"The set { x in R : phi(x) } is non-measurable". - -In fact, phi(x) can be of a rather simple form ($\Delta^1_2$, as you remarked above). Here is an abbreviated version of phi: For each real number x, let $x_1$ be the number obtained from x by deleting all even decimal places, $x_2$ by deleting all odd decimal places (do what you want for the countably many reals where this is not well-defined). This defines a measure-preserving Borel map from $\mathbb R$ to $\mathbb R\times \mathbb R$. Now consider the set M of all reals x for which there is some $\alpha$ such that $x_1\in L_\alpha$, but $x_2\notin L_\alpha$. ZFC does neither prove nor refute that M is Lebesgue-measurable. -(I think that the fact that ZFC does not prove that M is measurable is already due to Gödel.)<|endoftext|> -TITLE: Spectra of a Symmetric Toeplitz Operator -QUESTION [8 upvotes]: For a physics application, I would like to be able to compute the eigenvalues of the linear operator (acting on the Hilbert space $\ell^2$) given by an infinite matrix of the form -$\begin{bmatrix} -a_0 & a_1 & a_2 & a_3 & \dots\cr -a_1 & a_0 & a_1 & a_2 & \dots\cr -a_2 & a_1 & a_0 & a_1 & \dots\cr -a_3 & a_2 & a_1 & a_0 & \dots\cr -\vdots & \vdots & \vdots &\vdots & \ddots \end{bmatrix}$ -(where the $a_i \in \mathbb{R}$). Some Googling tells me these (or at least, the finite dimensional analogue) are known as symmetric Toeplitz matrices, but I'm having trouble finding answers to the following questions: - -What conditions must there be on the $a_i$ for the above operator to be compact (so we can apply the spectral theorem). Is it enough for $\sum_{i=0}^{\infty} a_i^2$ to be finite? (This question might be easy, it's just been a long time since I've done any functional analysis). -In the case that the above operator is compact (and so the spectral theorem applies), is there any good way to approximate its eigenvalues/eigenvectors? In particular, do the eigenvalues and eigenvectors of the upper-leftmost $N$ by $N$ finite submatrix (which is also symmetric Toeplitz) "approach" the eigenvalues/eigenvectors of the operator in any sense. -Is there a known closed form for the eigenvalues/eigenvectors? I really doubt this, but since these operators "kind of" look like circulant matrices (whose eigenvectors do have a fairly nice closed form), perhaps there is some really subtle roots-of-unity trick? - -EDIT: I can't seem to get the LaTeX for the above matrix to display properly on my computer. If other people are having the same problem, it is supposed to look like this: -a_0 a_1 a_2 a_3 ... -a_1 a_0 a_1 a_2 ... -a_2 a_1 a_0 a_1 ... -a_3 a_2 a_1 a_0 ... - . . . . . - . . . . . - -REPLY [2 votes]: I came across this question today (a month after you posted) so not sure if you have had the answers. Anyway here is what I know. -(1) and (2): There are no non-zero compact Toeplitz operators (I'm talking about the infinite case). In fact, for an operator represented by a matrix $(a_{ij})_{i,j=0}^{\infty}$ to be compact (on $l^2$), the necessary condition is that the limit of the entries on each diagonal (parallel to the main diagonal) must be zero: $\lim_{m\to\infty}a_{i+m, j+m}=0$ for each fixed $i$ and $j$. -(3): unless the operator is a multiple of the identity, a symmetric Toeplitz operator does not have an eigenvalue. On the other hand, the spectrum of such an operator can be described using the symbol (as mentioned above by Romanov). See Theorem 7.20 and Proposition 7.24 in Chapter 7 in "Banach algebra techniques in operator theory" by R. Douglas. - TL<|endoftext|> -TITLE: The rank of a not necessarily finitely generated module. -QUESTION [7 upvotes]: This question is motivated by this one. The main point of the question (was) to try to weaken the notion of rank. After the answers and comments, it seems this is not a good way to do it, but perhaps it is still useful for anyone trying to do the same. -Remark: Conditions added after Hailong's comment and Tom's answer: $R$ reduced and $M$ indecomposable. -Let $R$ be a reduced noetherian ring and $M$ an indecomposable $R$-module. -For any prime $p\in\mathrm{Spec} R$ with residue field $\kappa(p)$ define -$$\delta_M(p)=\dim (M\otimes_R \kappa(p))$$ -the local rank of $M$ at $p$. - -Question 1: If $\delta_M(p)=1$ for all $p$, does it follow that $M_p\simeq R_p$? Or more generally, if $\delta_M$ is constant, does it imply that $M_p$ is a free $R_p$-module for all $p$? - -Remarks -1 If $M$ is finitely generated, then by Nakayama lemma these questions are easy. (See Exercise II.5.8 on page 125 in [Hartshorne]). -2 Tom Goodwillie points out that if $R=\mathbb Z$ and $M\subset \mathbb Q$ consists of all rational numbers $a/b$ such that $b$ is square-free, then $M_p\simeq R_p$, but it is not invertible, so that would be too much to ask. -3 Yves Cornulier shows here that if $M$ is projective and $M_p\simeq R_p$, then $M$ is finitely generated. In other words, if the answer to (the first part of) Question 1 is "YES" then $M$ cannot be projective. So this suggests a subquestion... - -Question 1a: - Does there exist an example of an $R$ and a non-finitely generated projective $M$ for which $\delta_M$ is constant? - -And let me include also a somewhat vague, but related question: - -Question 2: Is the class of modules $M$ for which $\delta_M$ is finite for all $p$ interesting? Is there some kind of a finiteness condition they satisfy? (Other than the one that this means directly). Maybe with some additional hypetheses? (projective?) - -REPLY [5 votes]: Here are a few comments, too long to fit in the comments box. -1) One can modify Tom Goodwillie's example as follows: for simplicity lets pick $R$ to be a local domain of dimension $1$ (so there are only $2$ prime ideals, $0$ and $\mathfrak m$, with residues $K$, the quotient field and $k$, the residue field of $R$ respectively). Then any module $M$ that fits into a short exact sequence: -$$0 \to k \to M \to K \to 0 $$ would satisfy: $\delta_M(0)= \delta_M(\mathfrak m) =1$. But $M_{\mathfrak m} =k$ is not a free $R_{\mathfrak m}=R$-module. -Note that one can not try the exact sequence with $k, K$ swapped, since $Ext^1_R(k,K)=0$. On the other hand $Ext^1(K,k)$ is complicated, since projective resolutions of $K$ depend on the continuum hypothesis! -2) To weed out examples like above, perhaps more relevant than indecomposability or projectivity is to require $M$ to be torsion-free, so $M$ injects into $M\otimes Q(R)$ ($Q(R)$ is the total ring of quotients). -Say we assume this and the "rank" is $1$. Then immediately we know that $M$ is a submodule of $Q(R)$, and this more or less describes $M$: at every prime $p$, once we tensor with $\kappa (p)$, basically only one denominator of $M$ (that is not in $R$) is left. When $R=\mathbb Z$, one can see the example in your 2) very clear from this point of view. -3) Finally, I am not sure we should call $\delta_M(p)$ the local "rank". When $M$ is finitely generated, $\delta_M(p)$ is the minimal number of generators of $M$ locally at $p$. Of course, if $M$ is locally free, it will be the rank, but we definitely do not want to assume that. -This perhaps explains why the easy counter-examples by me and Tom Goodwillie exist: $M$ can locally have a constant number of generators, but it generally does not mean $M$ is locally free.<|endoftext|> -TITLE: locally constant constructible sheaves and finite etale coverings -QUESTION [10 upvotes]: Maybe it is well known to experts or maybe it is just a stupid idea, but I will ask any way. -We know that if $X$ is a topological space, then there is an equivalence of categories between the category of locally constant sheaves (of sets) on $X$ and the category of covers (sous-entendu local homoemorphism) of $X$. -The equivalence is given by "$\Rightarrow$" using the "espace \'etal\'e" of sheaves, "$\Leftarrow$" taking the sheaf of sections. -Now I replace $X$ by a scheme (locally noetherien or something?), and I think there is an equivalence of categories between the category of locally constant constructible sheaves of sets (By constructible I mean the constant values should be finite) on the \'etale site of $X$ and the category of finite \'etale coverings of $X$. -I tried to construct the functor "$\Leftarrow$": Given $Y\rightarrow X$ finite \'etale, we associate to any $T\to X$, the set of sections $T\to Y\times_XT$. This is a locally constant constructible sheaf if $X$ is locally noetherian: You decompose $X$ into connect components and by SGA1 corollary 5.3 then you can see easily that on each connected component the association is a constant sheaf with a finite constant value. -Is this an equivalence? If it is how one constructs the quasi-inverse? -Furthermore, do we have any formulation like the finite representations of $\pi_1^{\text{et}}(X,x)$ is equivalent to the category of locally constant constructible \'etale sheaves (of vector spaces) on $X$. If this is true it should be a direct consequence of Grothendieck's main theorem on $\pi_1^{\text{et}}$ and the above statement. - -REPLY [3 votes]: To construct a quasi-inverse, you may use the equivalence of categories between affine morphisms and sheaves of quasicoherent algebras described in EGA2 Chapter 1. Given a locally constant constructible sheaf $F$ of sets, you can take the étale sheafification of the presheaf of algebras $U \mapsto \mathscr{O}_U^{F(U)}$ on the small étale site of $X$. The relative spectrum of this sheaf is the finite étale cover you want. It looks like you need some descent to prove this, so this construction is more or less a disguised version of Angelo's. -If you have a pointed connected scheme $(X,x)$, then there is an equivalence between finite representations of the fundamental group on a vector space, and locally constant constructible étale sheaves of vector spaces. In one direction, a sheaf $F$ is taken to the fiber over $x$. In the other direction, you pass to a trivializing cover, take a constant sheaf of appropriate dimension, and apply the associated sheaf construction (which is just descent). If your scheme is not connected, you replace the fundamental group with the fundamental groupoid.<|endoftext|> -TITLE: Obstructions to Einstein metrics in high dimensions -QUESTION [12 upvotes]: It is well known that there exists three and four manifolds that do not admit an Einstein metric, but I wonder if this question is still open for manifolds of dimension higher than four. That is, does anyone know of a compact n-manifold, $n>4$, that does not admit an Einstein metric? - -REPLY [4 votes]: In general, for dimensions $n>5$ we don't have topological obstructions to the existence of Einstein metrics (like as Hitchin-Thorpe inequality when $n=4$). -However, there exist compact homogeneous Riemannian manifolds with no $G$-invariant -Einstein metrics. In this case, one works with cosets $M=G/K$ of a compact Lie group $G\subset {\rm Isom}(M)$ and the Einstein equation $Ric = c \cdot g$ is written with respect to a $G$-invariant Riemannian metric $g$ on $M=G/K$. -For such a metric the Einstein equation reduces to a polynomial system and positive real solutions correspond to invariant Einstein metrics. -In particular, let $\frak{g}=\frak{k}\oplus\frak{m}$ be a reductive decomposition for $M=G/K$ (such an orthogonal decomposition always exists for a homogeneous Riemannian manifold $(M=G/K, g)$). For simplicity, let us assume that the tangent space ${\frak{m}}\cong T_{o}G/K$ decomposes into $q$ pairwise inequivalent -isotropy summands (i.e., irreducible $Ad(K)$-submodules) -$ -\frak{m}= \frak{m}_1 \oplus \frak{m}_2 +\oplus ..... \oplus \frak{m}_q. -$ -Then any $G$-invariant Riemannian metric is given by -$ -g = < , > = x_1\cdot (-B)|_{\frak{m}_1}+\cdots+x_s\cdot (-B)|_{\frak{m}_q} -$ -for some positive real numbers $x_1, x_2, ... x_q$, where $-B$ denotes the negative of the Killing form of $\frak{g}$ (restricted on $\frak{m}$). -Homogeneous Einstein metrics are positive real solutions of the system $\{r_1-r_2=0, ...., r_{q-1}-r_{q}=0\}$ -where $r_{i}$ are the components of the invariant Ricci tensor on any isotropy summand. These are -given by -\begin{equation}\label{ricc} - r_{k}=\frac{1}{2x_{k}}+\frac{1}{4d_{k}}\sum_{i, j}\frac{x_{k}}{x_{i}x_{j}}[ijk]-\frac{1}{2d_{k}}\sum_{i, j}\frac{x_{j}}{x_{k}x_{i}}[kij], \qquad (k=1, \ldots, q). - \end{equation} -Here $[ijk]$ are the structure constants of $G/K$. Their computation is usually non-trivial. For particular classes of homogeneous spaces (e.g. isotropy irreducible), they are related with the Casimir operator of $\frak{g}$. They are non-zero only for these triples $\frak{m}_i$, $\frak{m}_j$, $\frak{m}_k$, for which -$B([\frak{m}_i, \frak{m}_j], \frak{m}_k)\neq 0$. -Notice that the study of homogeneous Einstein metrics becomes more complicated if some of the isotropy summands are equivalent each other. -Finally, it turns out that there are several homogeneous spaces with no invariant Einstein metrics. For example, we know by M. Wang and W. Ziller that the 12-dimensional space $SU(4)/SU(2)$ does not admit any homogeneous Einstein metric. -see: M. Wang and W. Ziller: Existence and non-existence of homogeneous Einstein metrics, - Invent.~Math.~84 (1986) 177--194. -This space seems to be the lowest dimensional example of a compact homogeneous manifold with no invariant Einstein metrics. -Other interesting examples of compact homogeneous spaces with no invariant Einstein metrics have been presented for instance by -J-S. Park and Y. Sakane: - Invariant Einstein metrics on certain homogeneous spaces, - Tokyo J. Math. 20 (1) (1997) 51--61.<|endoftext|> -TITLE: Is the square of the covering radius of an integral lattice/quadratic form always rational? -QUESTION [6 upvotes]: This is one of many observations from Pete L. Clark's questions on "Euclidean" quadratic forms. I sent Pete many positive integral forms that obeyed his condition. In turn, his condition turns out to be what Conway, Sloane, and in particular Gabriele Nebe refer to as "covering radius less than $\sqrt 2,$" see -http://www.math.rwth-aachen.de/~nebe/pl.html -and -http://www.math.rwth-aachen.de/~nebe/papers/CR.pdf -One of Pete's relevant questions is -Must a ring which admits a Euclidean quadratic form be Euclidean? -My own observations, confirmed by Pete with the Magma command CoveringRadius, are that the square of the covering radius is rational, and the denominator can always be taken to be a small power of 2 times the determinant of the lattice. Also the power of 2 seems to depend merely on the dimension. -So, that is the question, is the squared covering radius always rational with denominator a (dimension-dependent) power of 2 times the determinant of the lattice? -(note that it may be necessary to have the fraction not be in lowest terms to see the denominator as requested.) -(Also, Pete considers indefinite forms, other rings, etc. The covering radius stuff is for positive forms over the rational integers, which, except for the occasional annoying power of 2, are often called lattices. Not my fault). - -REPLY [8 votes]: Yes, the square $R^2$ of the covering radius is always rational; and in small dimensions its denominator is always a factor of $2^{n+1} \Delta$ where $\Delta$ is the lattice discriminant, but possibly not for all $n$. -[I see that David Speyer just posted a very similar answer...] -A point $P$ at maximal distance $R$ from a lattice $L \subset {\bf R}^n$ is at distance $R$ from some $n+1$ lattice points $P_1, P_2,\ldots,P_{n+1}$ whose convex hull contains $P$ (proof: find $P_i$ by induction on $i$, using the fact that $P$ is a local maximum of the distance from a point to $L$). Fix some choice of the $P_i$. Then -the simplex spanned by them has circumcenter $P$ and circumradius $r$. There's a formula (see below) for the circumradius of a simplex, generalizing the familiar $R=abc/4K$ of triangle geometry, of the form $R^2 = N/D$, where $N$ is a determinant formed from the squares of the side-lengths, and $D = 2^n (n! V)^2$ with $V$ being the volume of the simplex. Since $(n!V)^2$ is the covolume of the lattice $L' \subset L$ generated by the $n$ vectors $P_i-P_{n+1}$ ($i=1,\ldots,n$), this gives what you want — provided $L' = L$. -Now in small dimensions $L'$ is always all of $L$, but in higher dimensions I suspect there may be counterexamples — and if $[L:L']$ is odd then $L$, or a lattice sufficiently close to some large scaling of $L$, will be a counterexample to your conjecture. Note that in such a counterexample the covering radius $R'$ of $L'$ must exceed $R$, because the complement of $L$ in $L'$ is outside the union of translates of $L'$ by a radius-$R$ ball whereas $R'$ is the smallest radius of a ball whose translates by $L'$ cover ${\bf R}^n$. -As for the formula $R^2 = N/D$: it follows from the formula $(n! V)^2 = \pm \det M$ where $M$ is the symmetric matrix of order $n+2$ with 0's on the diagonal, 1's on the last row and column, and $(i,j)$ entry $|P_i-P_j|^2$ for $i,j \leq n+1$. Now apply this formula to the degenerate simplex in ${\bf R}^{n+1}$ formed by the $P_i$ together with $P$. This volume is zero. Let $M_+$ be the resulting matrix, of order $n+3$. Subtract $R^2$ times the bottom row from the $(n+2)$nd row, and then subtract $R^2$ times the last column from the next-to-ast column. This gives $\det M_+ = -2R^2 \det M - \det M_0$, where $M_0$ is the minor of $M$ obtained by deleting the last row and column. Now solve for $R^2$ and we're done. -For $n=2$, the determinant of $M_0$ has only two nonzero terms, both equal $(abc)^2$, and we recover $4KR=abc$ where $K$ is the area. For $n=3$ there's still a nice description of $\det M_0$: it's proportional to the square of the area of the tetrahedron's Ptolemy triangle! That is, of the triangle of side lengths $|P_1-P_2| |P_3 - P_4|$, -$|P_1-P_3| |P_4-P_2|$, $|P_1-P_4| |P_2-P_3|$ that exists thanks to the Ptolemy inequality. -It follows that $6VR$ is the area of that triangle. For $n\geq 4$ there doesn't seem to be such a simple description of $\det M_0$ except for its appearance in the formula for $R$. - -REPLY [7 votes]: It's rational. However, I am not sure whether or not the denominator is what you think it is. -Let $\Lambda \subset \mathbb{R}^n$ be your lattice. -The covering radius is the smallest $r$ such that every point of $\mathbb{R}^n$ is within $r$ from some lattice point. Let $w$ be a point whose closest distance to $\Lambda$ is exactly $r$. -Let $S$ be the set of points of $\Lambda$ which are $r$ away from $w$. -Lemma: $S$ is not contained in any hyperplane. -Proof: Suppose, to the contrary, that $S$ is contained in the hyperplane $H$. Let $u$ be a normal vector to $H$. If $w$ is not in $H$, let $u$ point to the side of $H$ on which $w$ lies; if $w$ is in $H$, choose the sign of $u$ arbitrarily. Look at the point $w+\epsilon u$ for small positive $\epsilon$. It is more than $r$ away from every point of $S$. However, if $\epsilon$ is small enough, then it is also more than $r$ away from every point in $\Lambda \setminus S$. So $w+\epsilon u$ is not within $r$ of any point of $\Lambda$, contradicting the definition of $r$. QED -Since $S$ is not contained in a hyperplane, we can choose $n+1$ points of $S$ which do not lie in a hyperplane. Without loss of generality, let one of these points be $0$, and call the others $v_1$, $v_2$, ..., $v_n$. I will show that $r^2$ is rational, and its denominator divides $2^{n+1} \Delta$ where $\Delta$ is the determinant of the lattice generated by the $v_i$'s. However, it is not obvious to me that the lattice generated by the $v_i$ is always $\Lambda$. Moreover, I could imagine that it might happen that the $v_i$'s usually generate $\Lambda$, but every once in a very rare while they don't, which would explain why a numerical search wouldn't find this phenomenon. So I am not sure whether the denominator is exactly what you think it is. -Let's finish the proof. Since the $v_i$ form a basis for $\mathbb{R}^n$, let's write $w = \sum a_i v_i$. -Now, $w$ is equidistant from $0$ and from $v_i$, so $w$ lies on the hyperplane $\{ x: \langle v_i, x \rangle = |v_i|^2/2 \}$. In other words, -$$\sum_j a_j \langle v_i, v_j \rangle = |v_i|^2/2 \quad (*)$$ -For every $i$, $(*)$ gives a linear equation in the $a_i$. The right hand side is a half integer. The matrix $\left( \langle v_i, v_j \rangle \right)$ has determinant $\Delta$, and each entry of this matrix is a half integer. So the inverse matrix has entries whose denominators divide $2^{n-1} \Delta$, and we see that the denominators of the $a_i$ divide $2^n \Delta$. -Then -$$r^2 = \langle w,w \rangle = \sum_{i,j} a_{i} a_{j} \langle v_i, v_j \rangle. \quad (**)$$ -It is immediately obviously that this is rational. -To get the denominator bound, use $(*)$ to turn $(**)$ into -$$\sum_i a_i |v_i|^2/2.$$ -As we saw above, $a_i$ has denominator dividing $2^n \Delta$, and $|v_i|^2/2$ is a half integer. So the denominator of $r^2$ divides $2^{n+1} \Delta$, as I promised.<|endoftext|> -TITLE: More four-dimensional counterexamples -QUESTION [6 upvotes]: To follow up on A four-dimensional counterexample?, I am probably being dense, but are there examples of spaces which are homotopy equivalent to bundles of surfaces over surfaces (or three-manifolds over the circle, or circle over three-manifold) and yet are not such. Same question if you change "bundle" to "product". In Hillman's book -he seems very careful to sidestep this question and talk about homotopy equivalence only... -I am interested primarily in spaces where the fiber and the base are $K(\pi, 1)$ spaces (so not spheres) and are oriented (if that makes any difference). - -REPLY [8 votes]: Borel's conjecture predicts that any homotopy equivalence of closed aspherical manifolds is homotopic to a homeomorphism. The conjecture has been proved for many fundamental groups, see e.g. -The Borel Conjecture for hyperbolic and CAT(0)-groups by Bartels-Lueck. -Basic ingredients are topological surgery, and computations of $L$-groups and $K$-groups. -Surgery works perfectly in dimensions $\ge 5$, but in dimension $4$ one is limited to fundamental groups of subexponential growth, see -Subexponential groups in 4-manifold topology by Krushkal-Quinn. -I am not up to date with L-theory computations but for polycyclic groups the reference is -[Farrell-Jones, The surgery L-groups of poly-(finite or cyclic) groups, Invent. Math. 91 (1988), no. 3, 559–586, EuDML]. -Combining the two ingredients gives Borel's conjecture for closed 4-manifolds homotopy equivalent to an infranil 4-manifold. This includes direct products of two closed surfaces with zero Euler characteristic. -Looking at recent papers by Bartels, Farrell, and Lueck at arxiv will bring you to state of the art on the L-theory computations, in particular, see -Survey on aspherical manifolds by Lueck, -but I am not aware of the result that covers surface bundles specifically, and in any case one is limited to groups of subexponential growth.<|endoftext|> -TITLE: Looking for ideas concerning the teaching of lower-division differential equation courses... -QUESTION [5 upvotes]: I'm looking for problems/lessons plans that could be used in a lower-division differential equations course that involve discerning properties of solutions of an equation, IVP, or BVP, without looking for an explicit/implicit solution (general or particular, given the context). Flow lines are an example of this, but I'm looking for something more advanced. One idea I've used is using first-order autonomous equations to figure out the dynamical properties of solutions for different initial conditions. I'm looking for similar ideas. Also: recommendations on how to present existence/uniqueness issues, besides showing a lot of examples, would be appreciated. (Boyce and DiPrima try to give a sketch of a proof of the basic existence/uniqueness result for first order IVPs. I wonder if this can be done without a course on analysis under your belt.) -Bonus: What about introducing group theoretic concepts at an early level? There are textbooks that claim to do this, but I wonder if this is as untenable as trying to teach measure theory in a calculus course. - -REPLY [3 votes]: I am not sure this answer would be helpful since the question is two years old. Anyway, since the question has popped up again, I am happy to introduce this lovely short book that hopefully will give you some ideas: Ordinary Differential Equations: A Brief Eclectic Tour, written by David A. Sánchez.<|endoftext|> -TITLE: How to construct log-canonical (or Calabi-Yau), non-Cohen-Macaulay singularities of low codimensions? -QUESTION [7 upvotes]: (EDIT 07/06/11: although the question has not been settled definitely, Sándor's excellent answer and the comments by Angelo and ulrich have highlighted many potential obstructions to the constructions I wanted. Thank you all! I am still very interested in any leads, so please keep them coming if you have some more.) -I would like to know examples of log-canonical singularities of low codimension which is normal but non-Cohen-Macaulay. A silly way to do it may be just adjoining variables, so here is the precise question: - -Fix a number $c$, for what $n$ can one construct an affine variety $X \subseteq \mathbb A^n_{\mathbb C}$ such that: $X$ is indecomposable and normal of codimension $c$, $X$ has at worst log-canonical singularity but $X$ is not Cohen-Macaulay? Given $c$, can we construct such $X$ for all $n$ big enough? - -The case $c=1$ is easy, there are no example since hypersurfaces are Cohen-Macaulay, so let's begin with $c=2$. -Motivation/Comments: I am actually looking for $F$-pure rings (i.e., the Frobenius is a pure morphism), but conjecturally my question above is virtually the same. Karl Schwede told me one can also try to look for (projective) Calabi-Yau varieties with some non-vanishing middle cohomolgy and low codimension embedding, then take their cones. But not being a geometer, I do not know how to construct such things. - -REPLY [6 votes]: 1 Examples, smallest $n$ -Let $A$ be an abelian variety and $X$ a cone over $A$. Then $X$ is log canonical, but as soon as $\dim A\geq 2$, then $X$ is not Cohen-Macaulay. (This you can see by computing the local cohomology at the vertex). -For the $c=1$ case: those are obviously CM. -As ulrich points out, there exist abelian surfaces in $\mathbb P^4$, so those give you $c=2$ with $3$-dimensional singularities, that is, $n=5=2c+1$. For $c=2$ this is the smallest $n$ you can get. Here is why: For $n\leq 4$ anything of codimension $c=2$ would be of dimension at most $2$ and hence if it is normal, it is $S_2$ and in particular CM. -So, let's assume that $c>2$. -As any quasi-projective variety $A$ of dimension $d$ maybe embedded in $\mathbb P^{2d+1}$ (Embed in some $\mathbb P^N$ and notice that the closure of the secant variety of $A$ is of dimension $2d+1$, so as long as $N>2d+1$, one may find a point and a projection that gives an embedding of $X$ into $\mathbb P^{N-1}$. Repeat.), you can do that with abelian varieties as well. This gives you $c=d+1$, so turning it around, for any $c>2$ you can find an $X$ -that you're looking for in $\mathbb A^n$ with $n=2c-1$. -2 Non-indecomposables -(Note: this is here, because I originally did not read the requirements carefully. Then I didn't feel like erasing it. ) -If you did not required an indecomposable singularity, then you could take the product of this $X\subset \mathbb A^{2c-1}$ and an arbitrary $\mathbb A^r$. Of course, this is why you made that requirement. Anyway, you'd get -$$ -X_{c,r}:=X\times \mathbb A^r\subset \mathbb A^{2c-1+r} -$$ -of codimension $c$ with $n=2c-1+r$. In other words, yes, you can construct such an $X$ for all $n$ big enough. I am not sure whether the bound $2c-1$ is optimal, but I have a feeling that you can't get much better than that. Note that this construction works for $c=2$ as well, so you can get examples in all dimensions $n\geq 5=2\cdot 2+1$. -If you wanted indecomposable singularities, I would expect that the codimension is actually increasing with the dimension. In other words, I would expect low codimensional examples in low dimension and not in (arbitrarily) high dimension. -3 Indecomposable vs. isolated -The previous point shows why Hailong assumed indecomposable. I claim that one may as well also assume isolated. At least to start. -First, let $X$ be an example as required, of dimension $d=\dim X$ and of codimension $c$. Let $s=\dim\mathrm{Sing} X$ the dimension of the singular set of $X$ and assume that $\mathrm{Sing} X$ is irreducible. Take a general complete intersection of codimension $s$. This will have dimension $d-s$, codimension $c$ and it is an isolated non-CM log canonical singularity. -Next, let $X$ be a codimension $c$ isolated non-CM log canonical singularity and consider a $\mathbb Q$-Gorenstein deformation of $X$ over a base of dimension $s$. -(A $\mathbb Q$-Gorenstein deformation means that the relative dualizing sheaf of the family is a $\mathbb Q$-line bundle and its line bundle powers restrict to the appropriate power of the dualizing sheaf of the members of the family). -The total space of the deformation will be an example of the kind you want. (For an indecomposable example you need a deformation without a trivial component). -The fact that this gives you an example that you want is non-trivial. -It is log canonical by inversion of adjunction see the main result of Kawakita's paper and it is non-CM at every point of the singular set by Corollary 1.3 of this paper. -From this you can conclude that any example you get will be a deformation of an isolated example. A priori you get that at the generic points of the singular set. At non-generic points those are still degenerations (i.e., non-small deformations) of the general ones. -The problem you run into is that you need non-trivial deformations of these singularities and they have finite dimensional versal deformation spaces, so you can't get too far with this idea. (Angelo will correct me if this is wrong, since he is one of the ultimate experts on this. See also Artin's extended work on this topic.) -This suggests that given your restriction of being indecomposable, regarding your "every big enough $n$" question, it seems that in order for that to happen you really need low codimensional isolated examples, which will be hard to construct since you can't get too far with cones (cf. ulrich's and Angelo's comments). -On the other hand we do get new examples out of this: A flat family of polarized abelian/CY/etc varieties will give you a flat family of the cones over them. As long as the family has maximal variation (i.e., the moduli map of the base is generically finite) the resulting singularity will be what you want. So, this way you can get higher dimensional examples, but it seems that all constructions are limited in dimension. -All of this suggests that the answer to your big enough question will be "no". -4 Better examples (manageability over low codimension) -To get an example that you want, you do not need to have an abelian variety for the cone construction. If $A$ is a smooth projective variety of dimension $d$ such that $\omega_A\simeq \mathscr O_A$ and there exist two integers $i,m\in\mathbb Z$ such that $0 -TITLE: A Priori proof that Covering Radius strictly less than $\sqrt 2$ implies class number one -QUESTION [18 upvotes]: It turns out that each of Pete L. Clark's "euclidean" quadratic forms, as long as it has coefficients in the rational integers $\mathbb Z$ and is positive, is in a genus containing only one equivalence class of forms. In the language of (positive) integral lattices, the condition is that the covering radius be strictly smaller than $\sqrt 2.$ - Please see these for background: -Intuition for the last step in Serre's proof of the three-squares theorem -Is the square of the covering radius of an integral lattice/quadratic form always rational? -Must a ring which admits a Euclidean quadratic form be Euclidean? -http://www.math.rwth-aachen.de/~nebe/pl.html -http://www.math.rwth-aachen.de/~nebe/papers/CR.pdf -I have been trying, for some months, to find an a priori proof that Euclidean implies class number one. I suspect, without much ability to check, that any such Euclidean form has a stronger property, if it represents any integral form (of the same dimension or lower) over the rationals $\mathbb Q$ then it also represents it over $\mathbb Z.$ This is the natural extension of Pete's ADC property to full dimension. Note that a form does rationally represent any form in its genus, with Siegel's additional restriction of "no essential denominator." If the ADC property holds in the same dimension, lots of complicated genus theory becomes irrelevant. -EDIT: Pete suggests people look at §4.4 of http://math.uga.edu/~pete/ADCFormsI.pdf . -EDIT 2: It is necessary to require Pete's strict inequality, otherwise the Leech lattice appears. -So that is my question, can anyone prove a priori that a positive Euclidean form over $\mathbb Z$ has class number one? -EDIT 3: I wrote to R. Borcherds who gave me a rough idea, based on taking the sum of a given lattice with a 2-dimensional Lorentzian lattice. From page 378 in SPLAG first edition, two lattices are in the same genus if and only if their sums with the same 2-dimensional Lorentzian lattice are integrally equivalent. I hope someone posts a fuller answer, otherwise I'll be spending the next six months trying to complete the sketch myself. The references: Lattices like the Leech lattice, J.Alg. Vol 130, No. 1, April 1990, p.219-234, then earlier The Leech lattice, Proc. Royal Soc. London A398 (1985) 365-376. Quoted: - -But there is a good way to - show that some genus of lattices L has class number 1 if its - covering - radius is small. What you do is look at the sum of L and a - 2-dimensional - Lorentzian lattice. Then other lattices in the genus correspond - to some - norm 0 vectors in the Lorentzian lattice. On the other hand, - if the - covering radius is small enough one can use this to show that a - fundamental domain of the reflection group of the Lorentzian - lattice has - only one cusp, so all primitive norm 0 vectors are conjugate - and therefore - there is only 1 lattice in the genus of L. This works nicely when - L is the - E8 lattice for example. - There are some variations of this. When the covering radius is - exactly - sqrt 2 the other lattices in the genus correspond to deep holes, - as in the - Leech lattice (Conway's theorem). The covering radius sqrt 2 - condition - corresponds to norm 2 reflections, and more generally one - can consider - reflections corresponding to vectors of other norms; see - http://dx.doi.org/10.1016/0021-8693(90)90110-A - for details - -REPLY [5 votes]: Alright, make this a separate answer... there are tables of positive forms available up to dimension 5. So I revised some C++ programs and found the "odd" lattices that satisfy Allcock's rule, the ambient vector space is covered when one places an open ball of radius 1 around each lattice point of odd norm, then an open ball of radius $\sqrt 2$ around each lattice point of even norm (such as the origin). A lattice is called "borderline" if those open balls do not suffice but closed balls of the same radii do the job. -So I am posting the table, so far (dimension 1,2,3,4,5), at this LINK. One can see that these are integral lattices as the matrix entries listed (one matrix per line) gives inner products always integral, meaning all f_ij are even when i,j are distinct, but the lattices are "odd" in that at least one diagonal entry f_ii is odd. -The computer that hosts my websites is down for the next month, so here is the table of "odd" lattice successes up to five variables. It is probably complete up to dimension 4, in dimension 5 I was getting near the upper bound of Nipp's tables so I am less certain about completeness. There are surely some of these in dimensions 6,7,8, probably not 9,10. -EDIT 24 November 2011: It is my educated guess that the sum of eight squares is one of the "odd" borderline lattices. I'm trying to figure out how to check this. -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= - f11 - 1 - 3 - 5 -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= - f11 f12 f22 - 1 0 1 - 1 0 2 - 1 0 3 - 1 0 5 - 2 0 3 - 2 2 3 - 3 0 3 -------------------------- - 1 0 4 borderline - 2 2 5 borderline - 3 2 3 borderline -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= - disc: f11 f22 f33 f23 f13 f12 -succeed 4: 1 1 1 0 0 0 -succeed 8: 1 1 2 0 0 0 -succeed 12: 1 1 3 0 0 0 -succeed 12: 1 2 2 2 0 0 -succeed 20: 1 1 5 0 0 0 -succeed 20: 1 2 3 2 0 0 -succeed 24: 1 2 3 0 0 0 -succeed 28: 2 2 3 2 2 2 -succeed 36: 1 2 5 2 0 0 -succeed 36: 1 3 3 0 0 0 -succeed 36: 2 2 3 0 0 2 -succeed 60: 2 3 3 0 0 2 ------------------------------------------------------------ -succeed 16: 1 1 4 0 0 0 borderline -succeed 16: 1 2 2 0 0 0 borderline -succeed 32: 1 3 3 2 0 0 borderline -succeed 32: 2 2 3 2 2 0 borderline -succeed 48: 2 3 3 2 2 2 borderline -succeed 64: 3 3 3 -2 2 2 borderline -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= - disc: f11 f22 f33 f44 f12 f13 f23 f14 f24 f34 -succeed 16: 1 1 1 1 0 0 0 0 0 0 -succeed 32: 1 1 1 2 0 0 0 0 0 0 -succeed 48: 1 1 1 3 0 0 0 0 0 0 -succeed 48: 1 1 2 2 0 0 0 0 0 2 -succeed 80: 1 1 2 3 0 0 0 0 0 2 -succeed 96: 1 1 2 3 0 0 0 0 0 0 -succeed 112: 1 2 2 3 0 0 2 0 2 0 -succeed 144: 1 1 2 5 0 0 0 0 0 2 -succeed 144: 1 2 2 3 0 0 2 0 0 0 -succeed 240: 2 2 3 3 2 2 0 2 0 0 -succeed 240: 1 2 3 3 0 0 0 0 2 0 ------------------------------------------------------------ -succeed 64: 1 1 1 4 0 0 0 0 0 0 borderline -succeed 64: 1 1 2 2 0 0 0 0 0 0 borderline -succeed 64: 1 2 2 2 0 0 2 0 2 0 borderline -succeed 80: 1 1 1 5 0 0 0 0 0 0 borderline -succeed 128: 1 1 3 3 0 0 0 0 0 2 borderline -succeed 128: 1 2 2 3 0 0 0 0 2 2 borderline -succeed 128: 2 2 2 3 2 2 0 2 0 0 borderline -succeed 144: 1 1 3 3 0 0 0 0 0 0 borderline -succeed 192: 1 2 3 3 0 0 2 0 2 0 borderline -succeed 192: 2 2 2 3 0 0 0 2 2 2 borderline -succeed 256: 1 3 3 3 0 0 2 0 2 -2 borderline -succeed 256: 2 2 3 3 0 2 2 2 2 2 borderline -succeed 256: 2 2 3 3 2 2 0 0 2 0 borderline -succeed 304: 2 2 3 3 0 2 2 0 2 0 borderline -succeed 512: 3 3 3 3 2 2 -2 -2 -2 2 borderline -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= - - - - - disc: f11 f22 f33 f44 f55 f12 f13 f23 f14 f24 f34 f15 f25 f35 f45 -succeed 16: 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 -succeed 32: 1 1 1 1 2 0 0 0 0 0 0 0 0 0 0 -succeed 48: 1 1 1 1 3 0 0 0 0 0 0 0 0 0 0 -succeed 48: 1 1 1 2 2 0 0 0 0 0 0 0 0 0 2 -succeed 80: 1 1 1 2 3 0 0 0 0 0 0 0 0 0 2 -succeed 112: 1 1 2 2 3 0 0 0 0 0 2 0 0 2 0 -succeed 144: 1 1 2 2 3 0 0 0 0 0 2 0 0 0 0 ------------------------------------------------------------ -succeed 64: 1 1 1 1 4 0 0 0 0 0 0 0 0 0 0 borderline -succeed 64: 1 1 1 2 2 0 0 0 0 0 0 0 0 0 0 borderline -succeed 64: 1 1 2 2 2 0 0 0 0 0 2 0 0 0 2 borderline -succeed 64: 1 2 2 2 2 0 0 0 0 0 0 0 2 2 2 borderline -succeed 96: 1 1 1 2 3 0 0 0 0 0 0 0 0 0 0 borderline -succeed 112: 1 1 1 2 4 0 0 0 0 0 0 0 0 0 2 borderline -succeed 128: 1 1 1 3 3 0 0 0 0 0 0 0 0 0 2 borderline -succeed 128: 1 1 2 2 3 0 0 0 0 0 0 0 0 2 2 borderline -succeed 128: 1 2 2 2 3 0 0 2 0 2 0 0 2 0 0 borderline -succeed 144: 1 2 2 2 3 0 0 2 0 2 0 0 0 0 2 borderline -succeed 144: 1 1 1 2 5 0 0 0 0 0 0 0 0 0 2 borderline -succeed 160: 1 1 2 2 3 0 0 0 0 0 0 0 0 0 2 borderline -succeed 192: 1 1 2 3 3 0 0 0 0 0 2 0 0 2 0 borderline -succeed 192: 1 2 2 2 3 0 0 0 0 0 0 0 2 2 2 borderline -succeed 208: 1 1 2 3 3 0 0 0 0 0 0 0 0 2 2 borderline -succeed 240: 1 1 2 3 3 0 0 0 0 0 2 0 0 0 0 borderline -succeed 256: 1 1 3 3 3 0 0 0 0 0 2 0 0 2 -2 borderline -succeed 256: 1 2 2 3 3 0 0 0 0 2 2 0 2 2 2 borderline -succeed 256: 1 2 2 3 3 0 0 2 0 2 0 0 0 2 0 borderline -succeed 256: 2 2 2 2 3 0 0 0 0 0 0 2 2 2 2 borderline -succeed 256: 2 2 2 3 3 2 2 0 2 0 0 2 0 0 2 borderline -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= - -In the "borderline" cases, one may insert extra points at precisely located points to create an overlattice, still integral but possibly "even," that does better than the original as far as covering radius smallness. That is the real end of the project, finding all the borderline guys and describing these overlattices.<|endoftext|> -TITLE: Extensions obtained adding torsion points of an elliptic curve -QUESTION [10 upvotes]: When adding to the rational the $p$-torsion points $E[p]$ of an elliptic curve we obtain an extension containing the $p$-th roots of the unity, and whose Galois group can be embedded in $GL(2, \mathbb{F}_p)$. To what extent are such extensions coming from elliptic curves? -I mean, assume $K/\mathbb{Q}$ to be an extension whose Galois group can be embedded in $GL(2, \mathbb{F}_p)$ and containing the $p$-th roots of the unity (which is required to expect a positive answer), is $K$ obtained adding to $\mathbb{Q}$ the torsion points of some elliptic curve defined over the rationals? -Note that I'm not considering a particular Galois representation, but just Galois groups that can be embedded in some way into $GL(2, \mathbb{F}_p)$. Thanks! - -REPLY [14 votes]: This question is discussed very carefully in Section 3 of the paper Mod $p$ representations -on elliptic curves, by Frank Calegari (available here). -In particular, after noting that the answer is positive when $p \leq 5$ (as was already -observed in the comments above), he proves (in Theorems 3.3 and 3.4) that if -$p \geq 7$ then there exists a continuous representation $\rho: Gal(\overline{\mathbb Q}/ -\mathbb Q) \to GL_2(\mathbb F_p)$ with cyclotomic determinant which does not come from -the $p$-torsion of an elliptic curve. (He also notes that the same result was established -by Dieulefait, in the paper Existence of non-elliptic mod $\ell$ Galois representations for every $\ell > 5$.)<|endoftext|> -TITLE: Finiteness of non-abelian cohomology -QUESTION [8 upvotes]: Let $H$ be a finite group acting on a group $G$. Are there general conditions one can place on $G$ so that the set $H^1(H, G)$ is guaranteed to be finite? For instance, does finiteness hold if $G$ is finitely presented, or if $K(G, 1)$ is a finite CW complex? -Added: According to Reid Barton's answer here, $H^1(H, G)$ is identified with $\pi_0(X^{hH})$, where $X$ is the space $K(G, 1)$ and the $hH$ denotes homotopy invariants. Thus it would be enough to know the following: if a finite group $H$ acts on a finite CW complex $X$, is $\pi_0(X^{hH})$ finite? - -REPLY [3 votes]: Take $H=\langle a|a^2=1\rangle$ and $G=\langle b,c|b^2=1\rangle$ with $H$ acting trivially on $G$. Then $H^1(H;G)=\text{Hom}(H,G)=\{g\in G| g^2=1\}$, and this contains the infinite set $\{c^nbc^{-n}|n\in\mathbb{Z}\}$. Thus, it is not enough to assume that $G$ is finitely presented. -Now suppose that there is a finite-dimensional CW complex of homotopy type $K(G,1)$. Let $E$ denote the universal cover. We then see that for every subgroup $G'\leq G$, the quotient $E/G'$ is a model for $K(G',1)$, so for any coefficient ring $R$ we have $H^k(K(G',1);R)=0$ for $k\gg 0$. This means that $G'$ cannot be a finite cyclic group, so we see that $G$ is torsion-free. Thus, if $H$ acts trivially on $G$ we have $H^1(H;G)=\text{Hom}(H,G)=1$. If the action is nontrivial then Ralph's answer becomes relevant.<|endoftext|> -TITLE: Asymptotics of the $q$-harmonic series as $q\to1$ -QUESTION [7 upvotes]: The following (very simply looking!) problem occurs in regularization -of the harmonic series -which can be formally thought of as the limit as $q\to1$, $|q|<1$, of -$$ -h(q):=(1-q)\sum_{n=1}^\infty\frac{q^n}{1-q^n}. -$$ -I can show (with some effort) that -$$ -h(q)=-\log(1-q)+f(q) \qquad\text{as}\quad q\to1, \ |q|<1, -$$ -where $f(q)$ is a bounded function (hint: consider both $h(q)$ -and $h(q^2)$ as $q\to1$). The question is whether the function -$f(q)$ has a limit as $q\to1$ or not; in other words, whether -$$ -h(q)=-\log(1-q)+c+o(1) \qquad\text{as}\quad q\to1, \ |q|<1. -$$ -Then, of course, I am very much interested in the constant $c$. -A straightforward computer experiment is not helpful. - -REPLY [10 votes]: The function $h(q)$ is equal to $$\frac{(1-q) \left(\log \left(\frac{1}{1-q}\right)-\psi _q(1)\right)}{\log \left(\frac{1}{q}\right)},$$ where $\psi _q(z)$ is the $q$-digamma function. According to Mathematica $с$ is equal to the Euler's constant $\gamma$. - -REPLY [9 votes]: Andrew is right, the following limit seems to be what you are looking for -$$\lim_{q\uparrow 1}\left(\log(1-q)-\log q \sum_{n\geq 0}\frac{q^{n+1}}{1-q^{n+1}}\right)=\gamma$$ -See , for example theorem 1 in "Summations for Basic Hypergeometric Series Involving a $q$-Analogue of the Digamma Function" by C. Krattenthaler and H.M. Srivastava. (Though there should be a more canonical source for this somewhere.)<|endoftext|> -TITLE: On the difference between two concepts of even cardinalities: Is there a model of ZF set theory in which every infinite set can be split into pairs, but not every infinite set can be cut in half? -QUESTION [38 upvotes]: An interesting question has arisen over at this -math.stackexchange -question -about two concepts of even in the context of infinite -cardinalities, which are equivalent under the axiom of -choice, but which it seems might separate when choice -fails. -On the one hand, a set $A$ can be even in the sense that it -can be split into pairs, meaning that there is a -partition of $A$ into sets of size two, or in other words, -if there is an equivalence relation on $A$, such that every -equivalence class has exactly two elements. -On the other hand, a set $A$ can be even in the sense that -it can be cut in half, meaning that $A$ is the union of -two disjoint sets that are equinumerous. -Note that if $A$ can be cut in half, then it can be split -into pairs, since if $A=A_0\sqcup A_1$ and $f:A_0\cong A_1$ -is a bijection, then $A$ is the union of the family of -pairs $\{x,f(x)\}$ for $x\in A_0$. And this argument does -not use the axiom of choice. -Conversely, if $A$ can be split into pairs, and if we -have the axiom of choice for sets of pairs, then we may -select one element from each pair, and $A$ is the union of -this choice set and its complement in $A$, which are -equinumerous. -Thus, when the axiom of choice for sets of pairs holds, -then the two concepts of even are equivalent. Note also -that every infinite well-orderable set is even in both -senses, and so in ZFC, every infinite set is even in both -senses. My question is, how bad can it get when choice -fails? - -Is there a model of ZF in which every infinite set can be -split into pairs, but not every infinite set can be cut in -half? -Is there a model of ZF having at least one infinite set -that can be split into pairs, but not cut in half? -What is the relationship between the equivalence of the two concepts of even and the axiom of choice for sets of pairs? - -REPLY [5 votes]: (I have removed my CW answer since it was irrelevant and somewhat wrong, I just did not know that back then.) -First we'll answer the easiest question: We shall construct a model with a 2-amorphous set, which is an amorphous set which can be partitioned into pairs but cannot be cut into two infinite sets. -We start with a model of ZFA where the set of atoms is countable (for instance), write it $A=\coprod_n P_n$ where the $P_n$ are pairs. -Now take $\mathscr G$ to be a group of permutations such that $\pi P_n = P_k$ (that is, the permutation respect this partition), along with the ideal of finite subsets for support. Let $\mathfrak A$ be the permutation model defined by those permutations and the chosen support. -Claim I: If $B\subseteq A$ is infinite and in the permutation model then only for finitely many $P_n$ we have $|P_n\cap B|=1$. -Proof: Assume by contradiction, let $E$ be a finite support of $B$ and $P_k=\{a,b\}$ a pair which meet $B$ at a single point (suppose $a\in B$), as well $P_k\cap E=\varnothing$. Define $\pi(a)=b, \pi(b)=a$ and $\pi(x)=x$ otherwise. It is clear that $\pi$ fixes $E$, but $\pi B\neq B$. $\square$ -Claim II: If $B\subseteq A$ is infinite and in the permutation model, then it is cofinite. -Proof: Suppose otherwise, let $E$ be a support of $B$ and $F$ be a support of $A\setminus B$. We can assume without loss of generality that $E=F$ (take union of both otherwise). By the previous claim we have $\{a_k,b_k\}=P_k\subseteq B$ and $\{a_n,b_n\}=P_n\subseteq A\setminus B$ such that $E\cap (P_n\cup P_k)=\varnothing$. Take $\pi$ to be a permutation for which $\pi(x_k)=x_n, \pi(x_n)=x_k$ (for $x=a,b$) and the identity otherwise. -Again it is clear that $\pi$ fixes $E$ alas $\pi E\neq E$, which is a contradiction. $\square$ -Claim III: The partition $\mathbb P = \{P_n\}$ is in the permutation model (however clearly not a countable set there). -Proof: It is clear that every permutation in our chosen group has $\pi(\mathbb P)=\mathbb P$. $\square$ - -Use any transfer theorem (Jech-Sochor, Pincus, etc.) to have this in a model of ZF, thus answering the question that it is in fact possible to have a set which can be split to pairs but not cut in half. -One can take instead a variation on the second Cohen model. In this model we add countably many real numbers indexed as $\{x_{n,\varepsilon,i}\mid n,i\in\omega,\varepsilon\in 2\}$. We then take $X_{n,\varepsilon}=\{x_{n,\varepsilon,i}\mid i\in\omega\}$ and $P_n=\{X_{n,0}, X_{n,1}\}$. -Cohen's took permutations of $\omega\times 2\times\omega$ which preserve the $P_n$'s. We can instead take permutations which only preserve the partition and have the same result as above. This model, I believe should answer positively the first question (I cannot see why, yet). -In this model we have that the collection of the $X_{n,\varepsilon}$ can be split into pairs, however the index set of these pairs need not be split into pairs itself. Indeed such splitting would induce a partition into sets of $4$ elements, which would also be splittable and so inducing a partition of $8$ elements, and so on to have every $2^n$ size of partition. I doubt that this is the case in the variant I have described above. -It also remains to check these properties on the set of all our generic reals - whether it is countable, or uncountable (but not well-orderable), and can that collection be split too?<|endoftext|> -TITLE: Motives from the fundamental group made nilpotent -QUESTION [10 upvotes]: I am reading the fascinating paper of Deligne on "le groupe fondamental de la droite projective moins trois points", and other stuffs related to anabelian geometry. This suggested the following question, which seems natural to me and that I haven't seen -answered in the literature (but this is perhaps a consequence of my still very low understanding of those matters). -Let $X$ be a proper and smooth, geometrically connected, variety over ${\mathbb Q}$. -Let us assume that $X$ has a rational point $x$ over ${\mathbb Q}$. Let $\bar X = X \times_{\mathbb Q} {\bar {\mathbb Q}}$. Then there is a natural action of $G_{\mathbb Q}=Gal(\bar {\mathbb Q}/ {\mathbb Q})$ on the profinite fundamental group $\pi_1(\bar X,x)$. As in Deligne, let us make $\pi_1(\bar X,x)$ nilpotent of order $N$ (by quotienting it by $[\dots[[\pi_1,\pi_1],\pi_1],\dots]$ where the number of $[$ is $N$), then take its pro-$l$-Sylow (for a given prime $l$), and quotient it by its torsion (which is a normal subgroup): we obtain a nilpotent torsion free pro-$l$ group $P_{l,N}$ which inherits a continuous action of $G_{\mathbb Q}$. Such a group has a finite dimensional Lie algebra over $Z_l$, and let us call $V_{l,N}$ this Lie algebra tensorized by $Q_l$ (just for convenience, to have something over a field). Hence we have a finite dimensional Lie algebra $V_{l,N}$ on which $G_{{\mathbb Q}}$ acts continuously by Lie algebra automorphisms. In particular, forgetting the Lie algebra structure, $V_{l,N}$ is a continuous Galois representation of $G_{\mathbb Q}$. -(From here, I am really on uncharted territory for me, and I coulsd possibly have everything wrong). One of the first aim of Deligne's paper is to interpret $V_{l,N}$ as the $l$-adic realization of a mixed motive. In Deligne's original paper, -a motive is just defined as a system of realizations, which means that what Deligne has to do, and does, is just to construct $l'$-adic analogs of $V_{l,N}$ (for $l'$ an other prime - the construction is just by replacing $l$ by $l'$ from the beginning) and a Betty analog, -and a De Rham analog, and compatibilities between them. If I am not mistaken, -the underlying motive should be mixed, not pure, even if we have assumed $X$ proper and smooth (excepted if $N=1$ where $V_{l,N}$ is just the $l$-adic homology of $X$ and hence -shoould be be pure). In particular, we do not expect $V_{l,N}$ to be semi-simle as a Galois -representation in general. First question: is the above correct? Second question: if it is, - -Do we know any case of $X$, $x$ as above where $V_{l,N}$ is non semi-simple? - -REPLY [9 votes]: As you say, in general the representation should not be semi-simple even if $X$ is smooth projective. One can construct explicit examples as follows: -Let $X$ be any smooth projective curve of genus $g=2$ over $\mathbb{Q}$ with a rational point such that for some prime $p$, $X$ has a regular model over $\mathbb{Z}_p$ with special fibre consisting of a union of two elliptic curves (one can do analogous things for larger $g$). Since the Jacobian of $X$ has good reduction at $p$ it follows that $H^1(X, \mathbb{Q}_l)$ is unramified at $p$. -By a theorem of Takayuki Oda (Galois action on the nilpotent completion of the fundamental group of an algebraic curve. Advances in number theory (Kingston, ON, 1991), 213–232) it follows that since $X$ does not have good reduction at $p$, the Galois representation on $V_{l,N}$ is ramified at $p$ for some $N$. Now the pronilponent completion has a filtration whose associated quotients are quotients of $H_1(X,\mathbb{Q}_l)^{\otimes m}$ for various $m$ and therefore unramified. It follows that $V_{l,N}$ cannot be semisimple.<|endoftext|> -TITLE: An ultrafilter is a set of subsets containing exactly one element of each finite partition: reference request -QUESTION [21 upvotes]: There are probably dozens of ways of defining "ultrafilter". The definition I've seen most often involves first defining "filter", then declaring an ultrafilter to be a maximal filter. -But there's another, shorter way to state the definition: - -Let $X$ be a set. An ultrafilter on $X$ is a set $\mathcal{U}$ of subsets such that for all partitions - $$ -X = X_1 \amalg \cdots \amalg X_n -$$ - of $X$ into a finite number $n \geq 0$ of subsets, there is exactly one $i$ for which $X_i \in \mathcal{U}$. - -I'd be amazed if this wasn't in the literature somewhere, but I haven't been able to track it down. Can anyone help? -Actually, there's an even more economical definition: instead of allowing $n$ to be any natural number, you take it to be 3. Thus, the condition is that whenever $X = A \amalg B \amalg C$, exactly one of $A$, $B$ and $C$ is in $\mathcal{U}$. (The same thing works with 4, or 5, etc., though not with 2.) I'm mostly interested in the version with arbitrary $n$, which seems more natural, but if you've seen the $n = 3$ version in the literature then I'd like to hear about that, too. -Edit To be clear, when I use the word "partition" I don't mean to imply that the sets $X_i$ are nonempty. I just mean a family of pairwise disjoint sets $X_i$ whose union is $X$. They can be empty. - -REPLY [12 votes]: You can find that characterization (even with n = 3), as well as the generalization to $\kappa$-complete ultrafilters, in: Fred Galvin and Alfred Horn, Operations preserving all equivalence relations, Proc. Amer. Math. Soc. 24 (1970), 521-523.<|endoftext|> -TITLE: When did the term "Lie group" first appear? -QUESTION [13 upvotes]: Does anyone know who was the first to coin the term "Lie group"? -The following thesis from 1928 suggests that the term was already in use by that time: "Systems of Two Differential Equations from the Lie-Group Standpoint" -(http://genealogy.math.ndsu.nodak.edu/id.php?id=6129) -I've also found the term in the book "Theory of functionals and of integral and integro-differential equations" by Vito Volterra from 1930. -Does anyone have any idea when the term was first used? Someone suggested that Weyl or Brauer might have been the first to use the term, but I haven't found anything. - -REPLY [16 votes]: Wilhelm Killing's program Zur Theorie der Lie'schen Transformations-Gruppen (Braunsberg, 1886) predates the accepted answer by 5 years. It is reprinted in his Correspondence with Friedrich Engel (1997).<|endoftext|> -TITLE: What is the ring of invariants of GL acting on quaternary cubic forms? -QUESTION [6 upvotes]: Suppose I am looking at $GL(4,K)$ acting on a cubic form in say four variables $x,y,z,w$ over $K$ via the usual induced action on a polynomial. Does anyone know what is/where I can find how to compute the ring of invariants? The case of personal interest is when $K$ is a finite field but the the answer over $\mathbb{C}$ would of course be more than useful. - -REPLY [4 votes]: Don't be misled into thinking that the answer over $\Bbb C$ tells you very much about the answer over your finite field $K$. The space of cubic forms in 4 variables is 20 dimensional. The group $GL(4,K)$ is a finite group and so the ring of invariants you seek has Krull dimension 20. In particular, it has at least 20 generators. Probably, however, it has a few thousand generators if not many more. -There are computer algebra packages (e.g. Magma) that include routines that can compute generators for your ring of invariants in principle, but in practice you will find that they run out of memory long before getting very far on a problem of this size. They will be able to compute vector space bases for the invariants in low degrees. On the other hand, it is doubtful that having a list of thousands of generators (or even just their degrees) for the ring of invariants will help you very much. On the other hand if you want to know something about the properties of the ring of invariants as a ring, then it's possible that may be known.<|endoftext|> -TITLE: Are surface bundles over a surface with non-zero signature necessarily complex (or algebraic)? -QUESTION [12 upvotes]: By "surface bundle over a surface" I mean a compact, oriented 4-manifold $X$ which is the total space of an oriented fiber bundle $X\to B $ over an oriented 2-manifold $B$. Assume that the signature of the 4-manifold $X$ is non-trivial. -Conjecture 1: $X$ and $B$ can be given complex structures such that the map $X \to B$ is holomorphic. -Conjecture 2: $X$ and $B$ can be given the structure of complex algebraic varieties such that the map $X \to B$ is algebraic. -Question: Are the above conjectures true? If not, can you provide a counter example? -Background: The bundle $X\to B$ induces a map $f:B \to M_g$ where $M_g$ is the moduli space of curves and $g$ is the genus of the fiber. The signature of $X$ is given by -$$\sigma(X) = 4\int _B f^*(\lambda)$$ -where $\lambda$ is the first Chern class of the Hodge bundle. Since the signature of $X$ is non-zero, the map $f$ is non-trivial in homology and one might hope that an argument along the following lines is true: Find a representative $\tilde{f}$ in the same homotopy class as $f$ which has minimal energy. Using the fact that $M_g$ has a hyperbolic metric ($g>2$ since $\sigma(X)\neq 0$), prove this minimal energy map is holomorphic or even algebraic. $\tilde{f}$ provides $X$ with the desired structure. -This question is a variant on this recent question: A four-dimensional counterexample? the counter example given there has zero signature. - -REPLY [8 votes]: I seem to have found an answer to my own question. -The first observation is that a holomorphic fibration is automatically algebraic. This is because $M_g$ is quasi-projective and so the map $f:B\to M_g$, which trivially extends to a map to $\overline{M}_g$ is algebraic by GAGA (perhaps one should first pass to some finite cover to take care of stack issues). -However, my conjectures are false --- here is a counterexample. Let $X \to B$ be a surface bundle having non-trivial signature $\sigma$, base genus $h$, and fiber genus $g$. Let $X_N$ be the surface bundle obtained by taking the fiber connect sum of $X$ with the trivial bundle of fiber genus $g$ and base genus $N-h$. So for each $N\ge h$, we get a surface bundle $X_N \to B_N$ of signature $\sigma$ and base genus $N$. If $X_N \to B_N$ were all holomorphic as in my conjecture, then we would obtain a family of complete smooth curves $B_N$ in $M_g$ of fixed degree and unbounded genus, a contradiction. (The degree here is taken with respect to $\lambda$ which is ample on $M_g$ and fixed by the signature formula). -What happens in my minimal energy argument is that as the map $B_N \to M_g$ becomes harmonic, it bubbles off a collapsing component.<|endoftext|> -TITLE: Small residue classes with small reciprocal -QUESTION [50 upvotes]: Let $p$ be a large prime. For any $m \in \{1,\ldots,p-1\}$, let $\overline{m} \in \{1,\ldots,p-1\}$ be the reciprocal in ${\bf Z}/p{\bf Z}$ (i.e. the unique element of $\{1,\ldots,p-1\}$ such that $m \overline{m} = 1 \hbox{ mod } p$). -I am interested in finding $m$ for which for which $m$ and $\overline{m}$ are both small compared with $p$, excluding the trivial case $m=1$ of course. For instance, using Weil's bound on Kloosterman sums and some Fourier analysis, it is not difficult to show that there exists nontrivial $m$ with $\max(m, \overline{m}) \ll p^{3/4}$. But this does not look sharp; probabilistic heuristics suggest that one should be able to get $\max(m, \overline{m})$ as small as $O(p^{1/2})$ or so (ignoring log factors), which would clearly be best possible. Is some improvement on the $O(p^{3/4})$ bound known? For my specific application I would like to reach $O(p^{2/3})$. (I would also be willing to do some averaging in $p$ if this improves the bounds. I'm actually more interested in asymptotics for the number of $m$ with $\max(m,\overline{m})$ bounded by a given threshold, but the existence problem already looks nontrivial.) -I tried playing around with Karatsuba's bounds for incomplete Kloosterman sums, but it was not clear to me how to use them to get both $m$ and $\overline{m}$ into intervals smaller than $p^{3/4}$. - -REPLY [13 votes]: The following argument proves that for almost all $p\le x$, there are $m,n\le p^{1/2+\epsilon}$ with $mn\equiv1 \pmod p$. -Indeed, let -$$ -P=\{x/2x^{1/2+\epsilon}\}. -$$ -Let $q$ be a prime in $(x^{1/2-\epsilon/2},x^{1/2-\epsilon/3}]$. If $1+kp\equiv 0\pmod q$ for some $k\le x^{\epsilon/2}/2$, then $1+kp=qd$ for some $d\le (1+kx)/x^{1/2-\epsilon/2}\le x^{1/2+\epsilon}$. In particular, $p$ is not in $P$. Therefore, if $p$ is counted by $P$, then for each $q\in (x^{1/2-\epsilon/2},x^{1/2-\epsilon/3}]$, it must avoid the congruence classes $\{-\overline{k}\pmod q: 1\le k\le x^{\epsilon/2}/2\}$. Since $p$ is prime is prime, it must also avoid the classes $0\pmod q$ for all primes $q\le \sqrt{x}$. We then apply the arithmetic form of the Large Sieve (see, e.g., page 159 in Davenport) to conclude that -$$ -\#P \ll \frac{\pi(x)}{x^{\epsilon/2}} , -$$ -which proves the claim. -The above result is essentially sharp: the number of $p\in(x/2,x]$ for which there are $m,n\le p^{1/2}(\log p)^c$ with $mn\equiv 1\pmod p$ is $o(\pi(x))$ for small enough $c$. Indeed, we would then have that there is some $k\le (\log x)^{2c}$ such that $1+kp$ can be written as $1+kp=mn$ with $m\le n\le x^{1/2}(\log x)^c$. In particular, $1+kp$ would have a divisor $m\in[0.1k\sqrt{x}/(\log x)^c, \sqrt{1+kx}]$. The number of such primes $p\in(x/2,x]$ is -$$ -\asymp f(k) \frac{x}{\log x} \frac{1+\left(\log\frac{(\log x)^{2c}}{k}\right)^\delta}{(\log x)^\delta(\log\log x)^{3/2}}, -$$ -where $\delta=1-(1+\log\log2)/\log 2$ is the constant appearing in Erdos's multiplication table problem and $f(k)$ is some tame multiplicative function that is usually $\asymp1$ (see http://arxiv.org/abs/math/0401223 and http://arxiv.org/abs/0905.0163. This result is not stated there but it should follow from the methods. The upper bound uses the sieve. For the lower bound, instead of the Bombieri-Friedlander-Iwaniec result on primes in APs to large moduli, we would have to use Zhang's result because we need information about the distribution of primes in progressions $a\pmod q$ with $1+ak\equiv0\pmod q$, so $a$ is not fixed.) Summing over $k\le(\log x)^{2c}$, we find that the number of $p\in(x/2,x]$ for which there are $m,n\le p^{1/2}(\log p)^c$ with $mn\equiv 1\pmod p$ is -$$ -\ll \frac{x}{\log x} \frac{(\log x)^{2c}}{(\log x)^\delta(\log\log x)^{3/2}} . -$$ -Taking $c=\delta/2$ yields the claimed result. -The above line of thought should be able to produce, at least heuristically, the optimal value of $c$ for which the following two propositions hold: -$$ -\#\{p\le x: \exists m,n\le p^{1/2}(\log p)^{c+\epsilon}\ \text{with}\ mn\equiv 1\pmod p\} \sim \pi(x) -$$ -and -$$ -\#\{p\le x: \exists m,n\le p^{1/2}(\log p)^{c-\epsilon}\ \text{with}\ mn\equiv 1\pmod p\} =o(\pi(x)). -$$ -The point is that if $k\neq k'$, then the multiplicative structures of $1+kp$ and of $1+k'p$ should be independent from each other. Therefore the events that $1+kp=mn$ for some $m,n$ in some range and that $1+k'p=m'n'$ for some $m',n'$ in some range should be independent. So a Borel-Cantelli argument would then yield the optimal value of $c$.<|endoftext|> -TITLE: Estimates on the Green function of an elliptic second order differential operator. -QUESTION [8 upvotes]: Let $D$ be a linear differential elliptic operator of second order -with infinitely smooth coefficients acting on real valued functions -on a compact manifold $M$. Let us assume that $D$ has no free term, i.e. $D(1)=0$. -Let us fix a smooth positive measure -(density) $\mu$ on $M$. Does there exist a (integrable) Green -function $G\colon M\times M\to \mathbb{R} $ with the following properties: -(1) $\int_M G(x,y) \cdot D\phi(y) d\mu(y) =\int_M\phi(y) d\mu(y) --\phi(x)$ for any function $\phi$ and $x\in M$ (this is the definition of Green function); -(2) $G$ is infinitely smooth outside of the diagonal; -(3) $G$ is bounded from below. -The last property can be asked in a stronger form: -(3') Does $G$ satisfy the asymptotic estimate near the diagonal: -$$c|x-y|^{2-n}\leq G(x,y)\leq C|x-y|^{2-n}$$ -where $c,C>0$ and $n=\dim M>2$. If $n=2$ there should be a -logarithmic estimate. -I am pretty sure that this is true and should be well known. I would -need a reference. The special case when $D$ is the Laplacian for a -Riemannian metric on $M$ is contained explicitly in some textbooks I -am familiar with. - -REPLY [2 votes]: Green's functions are constructed in Aubin's book for operators such as you mentioned, but with some sign condition on the lowest order term. I have not had a close look but my suspicion is that (3') is fine but for (3) you need a maximum principle.<|endoftext|> -TITLE: When does Zariski closure commute with base change? -QUESTION [5 upvotes]: This should be an elementary question for anyone who knows SGA by heart (alas, not for me). It smells a lot like a descent problem. All schemes are supposed to be noetherian, and all morphisms to be locally of finite presentation. -Let $X$ be a scheme of finite type over a (base) scheme $S$, and let $R \subseteq X(S)$ be a subset of the set of $S$--rational points of $X$. Denote by $\overline R$ the smallest closed subscheme of $X$ whose $S$--rational points contain $R$. -Let $f:S'\to S$ be a (base--change) morphism of schemes, write $X' := X\times_SS'$ and denote by $R'$ the image of $R$ in $X(S') = X'(S')$. Again, let $\overline{R'}$ be the smallest closed subscheme of $X'$ whose $S'$--rational points contain $R'$. Then, $\overline{R'}$ is contained in $\overline R\times_SS'$, and the question is: - -Suppose $f:S'\to S$ is flat. Does the equality $\overline{R'} = \overline R \times_SS'$ hold? - -Clearly some hypothesis on $f$ is needed, and I just guess it's flatness. - -REPLY [6 votes]: It seems that the condition you need is that the generic points of $S'$ go to generic points of $S$; this is much weaker than flatness. -Assuming this condition, we can reduce to the case that $S$ and $S'$ are both $Spec$ of some field and we can also assume that $\overline{R} = X$. If $X$ is a variety over a field $k$ any Zariski dense subset of $X(k)$ is also dense as a subset of $X(K)$ where $K$ is any extension field of $X$, proving the claim.<|endoftext|> -TITLE: Almost-direct product and 1-formality -QUESTION [8 upvotes]: Let $G$ be a finitely presented group. To $G$ is associated in a functorial way a Malcev Lie algebra which can be constructed in several equivalent ways. Roughly speaking, it is the quotient of the completed free Lie algebra on the same number of generators as $G$, by the (formal) logarithm of the defining relations of $G$ (see edit below). -$G$ is called 1-formal if its Malcev Lie algebra is isomorphic as a filtered Lie algebra to the degree completion of a finitely presented Lie algebra with quadratic relations. (If $X$ is a path-connected topological space such that $G=\pi_1(X)$, this is equivalent to say that the Malcev Lie algebra of $G$ is isomorphic to the holonomy Lie algebra of $X$) -It is quite well known that fundamental groups of complementary of hyperplane arrangements in $\mathbb{C}^n$ (e.g. pure braid groups) are 1-formal. -On the one hand, an important fact in the study of these groups is that they are (under some conditions on the underlying arrangement) iterated "almost-direct product" of free groups, and free groups are themselves 1-formal. An almost direct product is a semi-direct product $H\rtimes K$ for which the action of $K$ on the abelianization of $H$ is trivial. On the other hand, according to this interesting survey http://www.arxiv.com/abs/0903.2307 the direct product of two 1-formal groups is again 1-formal, even if no proof is given (it's probably easy..) -So it is temptating to ask: - -is an almost-direct product of - 1-formal groups again 1-formal ? - -Edit: Some details about the construction and its relation with almost direct product. If $G$ is an abelian group, then one can take the tensor product $G \otimes_{\mathbb{Z}} \mathbb{Q}$ leading to a uniquely divisible abelian group. The Malcev construction extends this to any nilpotent group, leading to a uniquely divisible nilpotent group. Let $G^{(0)}=G$, $G^{(n+1)}=[G^{(n)},G]$ be the lower central serie, then the quotients $G/G^{(n)}$ are nilpotent by construction. -The Malcev completion of $G$ is the inverse limit of the $(G/G^{(n)}) \otimes_{\mathbb{Z}} \mathbb{Q}$. It is a pro-unipotent group, hence it has a (pro-nilpotent) Lie algebra $\mathfrak g$. On the other hand, one can define a Lie algebra by -$$gr\ G=\mathbb{Q}\otimes_{\mathbb{Z}} \bigoplus G^{(n)}/G^{(n+1)}$$ -the bracket being induced by the commutator in $G$. It as a graded Lie algebra, and in fact it is the associated graded of the filtered Lie algebra $\mathfrak g$. -A group is called 1-formal if there is an isomorphism of filtered Lie algebras $gr\ G \cong \mathfrak g$. Now, the point is that almost-direct product behave well with respect to the lower central series. If $G=G_1\rtimes G_2$ is an almost-direct product, then it seems to me that a result of Falk and Randell implies that we have -$$G^{(n)}=G_1^{(n)}\rtimes G_2^{(n)}$$ -and -$$gr\ G = gr\ G_1 \rtimes gr\ G_2$$. - -REPLY [16 votes]: Yes, the direct product of two 1-formal groups is again 1-formal, and so is the free product of two 1-formal groups. A proof is given in arxiv:0902.1250, Proposition 9.2. -And no, the almost direct product of two 1-formal groups need not be 1-formal. A proof is given in the same paper, Example 8.2. The group in question is a semi-direct product of the form $G=F_4\rtimes F_1$, where $F_n$ is the free group of rank $n$, which is of course 1-formal. The action of $F_1$ on $F_4$ is given by a certain pure braid $\beta \in P_4$, acting via the Artin representation on $F_4$; thus, the action is trivial on $H_1(F_4)$. For this extension, the ``tangent cone formula" fails: the tangent cone to the characteristic variety $V_2(G)$ is strictly included in the resonance variety $R_2(G)$. In view of Theorem A from the cited paper, the group $G$ is not 1-formal. -It is worth noting that $G$ is the fundamental group of the complement of a certain link of 5 great circles in $S^3$. Alternatively, $G$ can be realized as the fundamental group of the complement of an arrangement of 5 planes in $\mathbb{R}^4$, meeting transversely at the origin (of course, this real arrangement cannot be isotoped to an arrangement of 5 complex lines in $\mathbb{C}^2$). For more details on the construction and properties of such arrangements, see arxiv:math.GT/9712251. In particular, the pure braid $\beta$ is described there in Propositions 4.4, 4.6, and 4.9.<|endoftext|> -TITLE: Uniformly Convex spaces -QUESTION [8 upvotes]: My first question here would fall into the 'ask Johnson' category if there was one (no pressure Bill). I'm interested in constructing a uniformly convex Banach space with conditional structure without using interpolation. The constructions of Ferenczi and Maurey-Rosenthal both use interpolation. -Using existing methods for constructing spaces with conditional structure I think it is possible to construct a hereditarily indecomposable space whose natural basis statisfies a lower $\ell_2$ estimate on any $n$ disjointly supported blocks vectors supported after the $n^{th}$ position on the basis and an upper $\ell_2$ estimate on all finite block sequences. The space $X$ is sure to be reflexive and probably doesn't contain $\ell_\infty$ finitely represented. -I would like to have some way of showing that $X$ is uniformly convex and this is where I'm stuck. Perhaps one could show that $\ell_1$ is not finitely represented in $X$ but as far as I can see this is not good enough (or is it?). -My question: If a space is reflexive and does not contain $\ell_1$ finitely represented is it necessarily uniformly convex? -I suspect the answer is no but I don't have a counterexample. -Another question: Are there any known conditions on a basis, which (1) do not imply the basis is unconditional and (2) do imply the space is uniformly convex? - -REPLY [2 votes]: I think James also showed that if $X$ does not contain almost isometric copies of $\ell_1^2$ (he called such a space uniformly non-square) then $X$ is superreflexive. This is no longer true for $n>2$, as James later constructed a non-reflexive, uniformly non-octahedral (no almost isometric copies of $\ell_1^3$) space, thus also having non-trivial type. -Maybe you can check whether your space is uniformly non-square. Connecting it with your last question I think that you would have to verify that $\exists \delta>0$ such that for any normalized block vectors $x$ and $y$ (but not necessarily disjointly supported) there exist a choice of signs such that $||x\pm y||<2-\delta.$ I don't think this condition implies unconditionality. -Hopefully this makes sense...<|endoftext|> -TITLE: Good books on Dirichlet's class number formula -QUESTION [11 upvotes]: I refrained from asking the technical questions; maybe everyone didn't like my attitude. At least, help me finding good books. -Can anyone suggest a good book that gives a complete reference to "Dirichlet's class number formula" and Class number theory, and explaining each nook and corner of it? Or any reference material which is free? - -REPLY [3 votes]: Well, I don't know any reference that examines ''each nook and corner'', but as Kevin says in a comment above, Washington's ''Cyclotomic fields'' is one good place to start (assuming you have enough grounding in algebraic number theory). -In addition, Lang's ''Algebraic number theory'' contains some things on class number formulas I think; also we have his two-volume (now as one-volume at Springer) book(s) on Cyclotomic fields. But beware, I would say that these/this require a firmer background, than Washington's book. -(Without encouraging illegal activities, I'm sure there are some bootleg versions on the net of the above books.) -Otherwise, there are dozens of nice books on the subject. Also, there must be tonnes of free lecture notes out there in cyberspace.<|endoftext|> -TITLE: Undecidable theories easier than $Q$ -QUESTION [15 upvotes]: Most proofs of undecidability for various theories (pure logic with binary relation, group theory, etc.) show that the natural numbers and Robinson's $Q$, in one form or another, can be encoded appropriately. Hence the decision problem for these theories is as hard as $K$, the halting set. -Are there are recursively axiomatized theories which are undecidable, but yet easier than $K$ (i.e. $K$ is not Turing reducible to deciding to the theory)? - -REPLY [18 votes]: When I was looking around trying to find some inspiration to answer your question, I found the following result of Feferman from 1957: - -For any set $X$ of natural numbers there is a theory $T(X)$ such that: - -The set $X$ and the set of Gödel numbers of consequences of $T(X)$ have the same degree of unsolvability. -If $X$ is r.e. then $T(X)$ is effectively axiomatizable. - - -Because there are nonzero r.e. Turing degrees strictly weaker than $K$, I think this may answer the question. -The result is in the paper "Degrees of Unsolvability Associated with Classes of Formalized Theories", Solomon Feferman, The Journal of Symbolic Logic, Vol. 22, No. 2 (Jun., 1957), pp. 161-175. http://www.jstor.org/stable/2964178<|endoftext|> -TITLE: Blow-up removes intersections? -QUESTION [8 upvotes]: Assume that $\beta:\tilde{X}\to X$ is the blow-up of a nonsinular $\Bbbk$-variety $X$ along a sheaf of ideals $\mathcal{I}$. Let $Y:=Z(\mathcal{I})$. Given nonsingular, closed subvarieties $Z_1,\ldots,Z_r\subseteq X$ such that $\bigcap_i Z_i \subseteq Y$, is it true that $\bigcap_i \tilde{Z}_i=\emptyset$, where $\tilde{Z}_i$ denotes the strict transform of $Z_i$? If not, does this hold if we require $Y$ to be nonsingular and/or the $Z_i$ to intersect transversally? - -REPLY [2 votes]: I have tried to generalize the Exercise referenced by Karl, even though he told me that it shouldn't be possible this way. I think, however, it works: -Edit: I made a mistake concerning $J_i$ - it cannot be equal to $I_i\oplus\bigoplus_{d\ge 1} I_i^dT^d$ because that is not necessarily an ideal - it might not be closed under multiplication by elements from the ring $S$. The version below looks better. - -Proposition. Let $Z_0,\ldots,Z_r$ be closed subschemes of a Noetherian scheme $X$ such that $Z_i\not\subset Z_j$ for $i\ne j$. Let $I_i:=I(Z_i)$ and denote by $\tilde{Z}_i$ the respective strict transform of $Z_i$ under the blow-up $\beta:\tilde{X}\to X$ of $X$ along $I:=\sum_{i=0}^rI_i$. Then, $\bigcap_{i=0}^r\tilde{Z}_i=\emptyset$. -Proof. The statement can be checked locally, so we may assume that $X=\mathrm{Spec}(A)$ is affine. Let $f_i:Z_i\hookrightarrow X$ be the respective closed immersion, so $Z_i=\mathrm{Spec}(A/I_i)$ and $f_i^\sharp:A\twoheadrightarrow A/I_i$. Then, the inverse image ideal sheaf of $I$ under $f_i$ is $I\cdot A/I_i$ and hence, -$\displaystyle\tilde{Z}_i=\mathrm{Proj}\left(\bigoplus_{d\ge 0} \left(I\cdot A/I_i\right)^d\cdot T^d\right)$ -With $S=\bigoplus_{d\ge 0} I^d\cdot T^d$, the homogeneous ideal defining $\tilde{Z}_i$ inside $\tilde{X}=\mathrm{Proj}(S)$ is equal to -$\displaystyle J_i = \bigoplus_{d\ge 0} (I^d\cap I_i)$ -In particular, $J_0+\cdots+J_r\supseteq S_+$, so any point $P\in\tilde{Z}_0\cap\cdots\cap\tilde{Z}_r$ would correspond to a homogeneous prime ideal containing each of the $J_i$ and hence, the irrelevant ideal. There is no such point. - -Did I miss something? Or is this correct?<|endoftext|> -TITLE: Various flavours of infinitesimals -QUESTION [37 upvotes]: I'm not sure if this is a soft question, and whether it may be too broad or, on the contrary, too localized. Well, in Mathematics the concept of "infinitesimal" has been of extreme importance for centuries. -The present mathematical tecnology allows one, according to the context, to formalize this notion in several ways: - -Differential geometry. The classical epsilon-delta formalism of limits in elementary analysis leads to the concept of first-order (or $n$-th order) approximation in Calculus, hence to many standard notions in differential geometry: the differential of a map between smooth manifolds, jets of a map, tangent vectors, differential forms, and Riemannian metrics. -Algebraic geometry. The lack of a useful notion of convergence of sequences due to the coarseness of Zariski topology prevents us from using epsilon-delta arguments to define "infinitesimals". But then one retains the notion of first-order (or $n$-th order) "approximation" in a more formal way, e.g. by means of universal properties of modules and derivations (Kahler differentials...), and that of "infinitesimal space" e.g. by means of local Artinian $\Bbbk$-algebras, and of "infinitesimal neighbourhood" e.g. by completion of local rings, formal schemes etc. (But after all the algebro geometric perspective is just a high brow way of doing the formal derivative of polynomials, which coincides with the "topological one" once we work over $\mathbb{R}$ or $\mathbb{C}$) -Synthetic differential geometry. More recently some mathematicians are exploring the realm of synthetic differential geometry, of which I know nothing except that it kind of unifies the perspectives of the previous two approaches and uses relevant amounts of category theory (please correct me if I'm not being correct). -Non-standard analysis. The concept of infinitesimal element is of course fundamental in non-standard analysis, where the field $\mathbb{R}$ and the use of epsilon-delta arguments is replaced by the introduction of the field ${}^* \mathbb{R}$ of hyperreal numbers, which hosts several hierarchies of infinitesimals (as well as "infinite elements"). -Noncommutative geometry. According to A.Connes (in his book Noncommutative Geometry there's a paragraph on a "Quantized calculus") given an infinite dimensional separable Hilbert space $\mathcal{H}$ and a certain operator $F$ on it, compact operators on $\mathcal{H}$ with characteristic values such that $\mu_n=O(n^{-\alpha})$, $n\to\infty$, can be interpreted as "infinitesimals of order $\alpha$", while the differential of a "complex variable" (read "operator on $\mathcal{H}$") $f$ is just defined to be the commutator $[F,f]$. - -At the risk of seeing my question closed as too vague ("not a real question"), it would be my curiosity to know: - -Is there a theory that encompasses all the above instances of "infinitesimals" within a unique formal picture? Or, on the contrary, are some of the above notions of infinitesimals inherently specific to their field and embody formalizations of different heuristic notions? [A situation as in the second question occurs with the notion of "infinity": it seems to me there's almost no deep relation between the infinity as in $\lim_{n\to\infty}$ and the infinite cardinals of Cantor] - -REPLY [11 votes]: I don't know of the Connes calculus, but the others (including nonstandard analysis à la Robinson) have been brought under a common framework using models of synthetic differential geometry. However: it is important to point out that the infinitesimals used in algebraic geometry (for jet bundles, etc.) are nilpotent infinitesimals, whereas the infinitesimals used in nonstandard analysis are invertible. So in a sense the answer to the question is, "yes and no", but I'm going to concentrate here on the "yes". -This is explored in Models for Smooth Infinitesimal Analysis by Moerdijk and Reyes, which I recommend. This book can be read as "applied sheaf theory" or "applied Grothendieck topos theory", where the art is to choose a site (small category + covering sieves) judiciously to achieve several aims at once. In many of the models, one takes the underlying category of the site to be something like affine spectra of commutative rings, except one is not dealing with commutative rings exactly, but with richer algebraic structures called $C^\infty$-rings. The formal definition of these is in terms of a Lawvere algebraic theory which allows one to apply not just polynomial operations but more general operations based on $C^\infty$ functions. So the underlying category of the site in these models is the opposite of finitely generated $C^\infty$-rings, which Moerdijk and Reyes call $\mathbb{L}$ (for "locus"). -The representing object which gives the locus of invertible infinitesimals is the spectrum of the $C^\infty$-ring given by $C^\infty$ functions $\mathbb{R} - \{0\} \to \mathbb{R}$, modulo the ideal of functions that vanish on some neighborhood of $0$, aka the $C^\infty$ ring of germs at $0$. This is really not much different from nonstandard infinitesimals: usually infinitesimal elements are thought of in some way as germs of functions at infinity, i.e., of functions $\mathbb{R} \to \mathbb{R}$ modulo those which vanish for sufficiently large $x$ (compare here the infinitesimals of Du Bois-Reymond and Hardy). Here Moerdijk and Reyes use $0$ instead of $\infty$. Either way, there are nonarchimedean elements, i.e., nonzero elements less than any $1/n$ in absolute value. -[In nonstandard analysis, one typically refines this idea by considering germs of functions $\mathbb{R} \to \mathbb{R}$ at an "ideal point at infinity", i.e., at a non-principal ultrafilter $U$ on $\mathbb{R}$, or alternatively germs of functions $\mathbb{N} \to \mathbb{R}$ at a non-principal ultrafilter $U$ on $\mathbb{N}$. The more familiar buzzword here is "ultrapower", but see this MO answer by François Dorais, where the implicit message is that an ultrapower along $U$ is really the same as taking a stalk at $U$. (I call $U$ an "ideal point at infinity" because we can think of a non-principal ultrafilter $U$ on $\mathbb{N}$ as a point in the fiber over $\infty$ with respect to the canonical continuous map $\beta(\mathbb{N}) \to \mathbb{N} \cup \{\infty\}$, from the Stone-Cech compactification of $\mathbb{N}$ to the one-point compactification of $\mathbb{N}$.)] -On the other hand, a typical representing object for nilpotent infinitesimals is the spectrum of the $C^\infty$-ring of functions $\mathbb{R} \to \mathbb{R}$ modulo the ideal of squares of elements which vanish at $0$. The "internal hom" represented by this spectrum gives the tangent bundle functor, and other jet bundles can be similarly represented, by using $C^\infty$-rings with different types of nilpotent elements. -The tricky part of all this is to get the right notion of covering sieves, i.e., of sheaves w.r.t. a Grothendieck topology, to achieve disparate aims. One aim would be to embed the usual category of manifolds fully and faithfully in the category of sheaves, so as to preserve "good colimits", such as a manifold $M$ obtained as a colimit along an open covering of $M$. A different aim would be to arrange the topology so that the locus of invertible infinitesimals, as a presheaf on the site category $\mathbb{L}$, is a sheaf w.r.t. the topology. In summary, both aims can be achieved simultaneously so as to accommodate both nilpotent and invertible infinitesimals.<|endoftext|> -TITLE: What invariants of a matrix or representation can be used to find its GL(n,Z)-conjugacy class? -QUESTION [19 upvotes]: First question: For a semisimple invertible $n \times n$ matrix with entries over a field K, its characteristic polynomial completely describes the similarity class of the matrix. For non-semisimple elements, the characteristic polynomial is no longer a complete description of the similarity class, but there exists the rational canonical form, which is a complete invariant. -I'm interested in whether we can find invariants that help determine the similarity class of elements of $GL(n,\mathbb{Z})$ (where $\mathbb{Z}$ is the ring of integers) under the $GL(n,\mathbb{Z})$-conjugation action. Clearly, the invariants for similarity class over $\mathbb{Q}$, but there are semisimple matrices in the same similarity class over $\mathbb{Q}$ but not conjugate over $\mathbb{Z}$: -1 0 -0 -1 -and -0 1 -1 0 -These are clearly conjugate in $GL(2,\mathbb{Q})$ but not in $GL(2,\mathbb{Z})$. To see why they aren't conjugate in the latter, note that on reducing mod 2, the first matrix becomes the identity matrix and the second matrix becomes a non-identity matrix, so they cannot be conjugate mod 2, and hence cannot be conjugate in $GL(2,\mathbb{Z})$. -Second question: For a finite group G, call representations $\alpha, \beta: G \to GL(n,R)$ "locally conjugate" if $\alpha(g)$ is conjugate to $\beta(g)$ for every $g \in G$ via some element of $GL(n,R)$ depending on g. -We say that $\alpha$, $\beta$ are equivalent as representations if we can choose a single element of $GL(n,R)$ that conjugates $\alpha(g)$ to $\beta(g)$ for every $g \in G$. -My question is: does locally conjugate imply equivalent when $R = \mathbb{Z}$? -NOTE 1: When R is a field of characteristic not dividing the order of G, then locally conjugate implies equivalent, and we can prove this by noting that $\alpha$, $\beta$ have the same character (SORRY, the "same character" is enough to complete the proof only in characteristic zero -- in prime characteristics, even those that don't divide the order of the group, having the same character isn't good enough to conclude the representations are equivalent. However, the "locally equivalent implies equivalent" seems to still hold when the characteristic does not divide the order of G using more indirect arguments). However, for $\mathbb{Z}$, the character no longer determines the representation, so the proof used for fields (taking the character) does not work for $\mathbb{Z}$. -NOTE 2: When the characteristic of R divides the order of G, there exist examples of locally conjugate representations that are not equivalent -- in fact, we can construct such examples for the Klein four-group with the field of four elements. I haven't had success with using these to generate counter-examples over $\mathbb{Z}$, though there may be a way. - -REPLY [11 votes]: Here is the requested example of two representations of the Klein 4 group over $\mathbb{Z}$, locally conjugate but not conjugate. -Let $K:= \mathbb{Z}/2 \times \mathbb{Z}/2$ act on $\mathbb{Z}^4$ by permuting the coordinates. Inside $\mathbb{Z}^4$, consider the following two lattices: -$$L_1 := \{ (a,b,c,d) \in \mathbb{Z}^4 : a \equiv b \equiv c \equiv d \mod 2 \}$$ -$$L_2 := \{ (a,b,c,d) \in \mathbb{Z}^4 : a + b + c + d \equiv 0 \mod 2 \}$$ -Verification that $L_1$ and $L_2$ are locally isomorphic: -Let $\sigma$ be the element of $K$ which switches the first two and the last two coordinates. Consider the following bases for $L_1$ and $L_2$: -$$(1,1,1,1),\ (1,-1,1,-1),\ (2,0,0,0), (0,2,0,0)$$ -and -$$(1,1,0,0),\ (1,-1,0,0),\ (1,0,1,0),\ (0,1,0,1)$$ -In both cases, $\sigma$ fixes the first basis element, negates the second and switches the last two. So $L_1$ and $L_2$ are isomorphic modules for $\mathbb{Z}[\sigma]/\langle \sigma^2-1 \rangle$. -Verification that $L_1$ and $L_2$ are not isomorphic: -For each character $\chi$ of $K$, let $L_i^{\chi}$ be the sublattice of $L_i$ on which $K$ acts by $\chi$. Let $M_i = \bigoplus_{\chi} L_i^{\chi}$. Then $L_1/M_1$ has order $2$, and $L_2/M_2$ has order $8$.<|endoftext|> -TITLE: Signs and functoriality of tensor products -QUESTION [9 upvotes]: Let $C,C',D,D'$ be chain complexes of $R$-modules (let's say with upper indexing, so perhaps I should call them cochain complexes, though they're not duals of anything). Let $f\in Hom^\ast(C,C')$ and $g\in Hom^*(D,D')$. Then the standard convention is that $$(f\otimes g)(x\otimes y)=(-1)^{|g||x|}f(x)\otimes g(y),$$ where $|g|$ is the degree of $g$ and $|x|$ is the degree of $x$. As observed on page 171 of Dold, this is consistent with having a degree 0 chain map $$Hom^\ast(C,C')\otimes Hom^\ast(D,D')\to Hom^\ast(C\otimes C',D\otimes D').$$ -What bothers me, though, is that this formula forces $$(h\otimes k)\circ(f\otimes g)=(-1)^{|k||f|}hf\otimes kg,$$ which seems to violate the definition of a bifunctor as given, for example, on page 17 of Kashiwara and Schapira's "Categories and Sheaves", which would seem to require (adapting the notation) $$(1_{C'}\otimes g)(f\otimes 1_D)=(f\otimes 1_{D'})(1_C\otimes g).$$ (Here I suppose we assume that the relevant categories are the category of chain complexes of $R$ modules with $Mor(X,Y)=Hom(X,Y)$ (certainly such things can be composed functorially and the identity behaves properly) and the products of this category with itself). If I'm reading it correctly, this requirement in Kashiwara-Schapira seems to be the same as what Mac Lane is asking for in Proposition II.3.1 of "Categories for the Working Mathematician". -So are we to believe $\otimes$ is not a functor or is there a way to reformulate all of this to be consistent (or am I just getting something wrong)? -Thanks in advance! - -REPLY [14 votes]: There are two options. If you just want an ordinary category of cochain complexes, then you have to take the morphisms to be cochain maps of degree zero. In that context we have $(-1)^{|k||f|}=1$ so there is no problem. -Alternatively, you can have an enriched category of cochain complexes. In more detail, for any symmetric monoidal category $(\mathcal{V},\otimes)$ there is a theory of $\mathcal{V}$-enriched categories. Such a thing has a class of objects, and for each pair of objects $X$ and $Y$, it has an object $\text{Hom}(X,Y)\in\mathcal{V}$. Given a third object $Z$ there is also a composition morphism $c:\text{Hom}(Y,Z)\otimes\text{Hom}(X,Y)\to\text{Hom}(X,Z)$, subject to some obvious axioms. The symmetric monoidal structure on $\mathcal{V}$ includes natural twist isomorphisms $\tau_{PQ}:P\otimes Q\to Q\otimes P$ for all $P,Q\in\mathcal{V}$. When formulating the definition of a bifunctor in an enriched context, you find that you need to use the morphisms $\tau_{PQ}$ in various places. -In the case of interest, we can regard cochain complexes as a category enriched over graded abelian groups. The sign $(-1)^{|k||f|}$ is provided automatically by the relevant twist maps, so the tensor product becomes a bifunctor in the enriched sense.<|endoftext|> -TITLE: Carleson's Theorem (on the Adeles and other exotic groups) -QUESTION [7 upvotes]: I have redone this question: -On $\mathbb R^n$ the Carleson Operator if defined by - $$Cf(x) = \sup_{R>0} \left \vert \int_{B_R(0)} e^{2\pi i x\cdot \xi} \widehat{f}(\xi) d \xi \right \vert. $$ (In the previous version I had an incorrect version of this written down which lead to some stupid conclusions). - -For $n=1$ Carleson proved that this operator is strong (p,p) which is equivalent $L^p(\mathbb R)$ convergence of $S_Rf(x)=\int_{B_R(0)} e^{2\pi i x\cdot \xi}\widehat{f}(\xi)d\xi\to f(x)$ as $R\to \infty$. The modern proof uses boundedness of the Hilbert-Transform and the ``translation trick'' -$$\chi_{[-r,r]}(\xi)\widehat{f}(\xi) = \widehat{[\frac{i}{2}(m_{-r}H m_r - m_r H m_{-r})f] }(\xi).$$ -For $n>1$ Fefferman proved (using a Becosivitch set) that the transform $T$ defined by $$ \widehat{Tf}(\xi) = \chi_{B_1(0)}(\xi) \widehat{f}(\xi)$$ is unbounded in $L^p$ for $p\neq 2$. (There is no Hilbert Transform in higher dimensions, these become the Reisz Transform and the ``translation tricks'' to express $S_R$ in terms of the Hilbert Transform don't work). -The Bochner-Reisz conjecture is about boundedness of smoothened characteristic functions and an MO question about it can be found here Recent progress on Bochner-Riesz conjecture - -I was originally thinking that $Cf(x)$ could be defined easily for locally compact abelian groups and compact groups and that perhaps a ``translation trick'' existed in some generality. This appears to not be the case for two reasons - -The second bullet shows that it is not true in $\mathbb{R}^n$ for $n>1$. -The third bullet shows you that multiplier operator in $\mathbb{R}^n$ are delicate which makes the definition of a Carleson operator for general abelian groups difficult. - -Is the operator $$Cf(x) = \sup_{n\geq 0} \left \vert \int_{\frac{1}{p^n}\mathbb{Z}_p} e^{2\pi i \lbrace x \xi\rbrace } \widehat{f}(\xi) d \mu(\xi) \right \vert. $$ -bounded as an operator $L^r(\mathbb Q_p) \to L^r(\mathbb Q_p)$. (I think one can define a similar operator on the Adeles no?). How does smoothness of multipliers effect the problem in more general groups? -I'll make it a community wiki so people can fix it if it has errors. - -REPLY [4 votes]: There is a p-adic analogue of Carleson's theorem. This was worked out by Hunt and Taibleson and you can find an exposition in Taibleson's book "Fourier analysis on local fields". (NB Taibleson did a lot of work extending Euclidean harmonic analysis results such as Carleson's theorem and boundedness of singular integrals to local fields.) I don't know if an adelic Carleson's theorem has been worked out. -The notion of smoothness in the p-adics is different than in real spaces since the p-adics are totally disconnected. Instead of say differentiability (which can be defined at least formally), one looks at locally constant functions. For more general groups, you need to have an appropriate notion(s) of smoothness to ask how changing the smoothness affects your problem.<|endoftext|> -TITLE: Field with one element example? -QUESTION [18 upvotes]: $$\frac{1}{\mu(B)}\int_B \vert x \vert d\mu(x) = \frac{1}{p+1}$$ -This formula holds for the unit ball in $\mathbb{Q_p}$. This formula also holds for -$\mathbb{R}$ when $p=1$. Should one expect $$\mathrm{Frac}(W_{1^{\infty}} (\mathbb{F_1}))=\mathbb{R}?$$ -What (mathematical) criteria do people use to rule-out field with one element phenomena? What makes point counting formulas better (or worse)? - -REPLY [4 votes]: It is true that if $f(T)\in \mathbb Z[T]$ then, - $$ \frac{1}{\mu_p(B)}\int_{\mathbb Z_p} f(\vert x \vert_p) d\mu_p(x) \to \frac{1}{\vert B \vert}\int_{B}f(\vert x \vert)dx \mbox{ as } p\to 1.$$ -It is not in general true that - $$ \frac{1}{\mu_p(B)}\int_{\mathbb Z_p} \vert f( x )\vert_p d\mu_p(x) \to \frac{1}{\vert B \vert}\int_{B}\vert f( x )\vert dx \mbox{ as } p\to 1.$$ -When $f(x) = x^2-1$ we have - $$\int_{\mathbb Z_p} \vert x^2-1 \vert_p d\mu(x) = \frac{1+p(p-2)}{p}+ \frac{1}{(p+1)p} \to 1/2 \mbox{ as } p \to 1 $$ -and - $$\frac{1}{2}\int_{-1}^1 \vert x^2-1\vert dx = 2/3.$$ -http://imgur.com/a/m6KYA -A couple remarks: - -You should have also written $\mathbb Q_1 = \mathbb{R}$ instead of that terrible notation $\mathrm{Frac}(W_{1^{\infty}}(\mathbb{F}_1))$. -The question as posed is a little stupid since if someone had a procedure for "ruling our phenomena" not only would they would probably have category in mind, but they would be able to compute with it. I think the correct answer is $\mathbb F_1$ numerology is justified when it can be categorified''. This is kind of a weak version of the statement "a conjecture is true when you can prove it". I guess a vague question deserves a vague answer. - -The first part of the question seems to be about Arakelov geometry and replacing the place at infinity with the place 1. It seems that the answer is no. Another bad question would be to ask for more examples where it does make sense. -The last part of the original question can be made more precise: --Are there examples of formulas for $|X(\mathbb F_q)|$ such that $X$ is definable in either monoidal algebraic geometries or Borger's $\Lambda$-ring categorifications of schemes over $\mathbb F_1$ such that the point counting formula's in the category do not agree with $|X(\mathbb F_q)|$ as $q\to 1$? (if you know of another categorification that justifies $GL_n(\mathbb{F}_1)=S_n$ this question applies there too (Maybe Durov's Category?).)<|endoftext|> -TITLE: Axiom of choice: ultrafilter vs. Vitali set -QUESTION [26 upvotes]: It is well known that from a free (non-principal) ultrafilter on $\omega$ one can define a non-measurable set of reals. The older example of a non-measurable set is the Vitali set, -a set of representatives for the equivalence classes of the relation on the reals "the same modulo a rational number". Is it known whether you can have one without the other? -I.e., is ZF consistent with the existence of a set of representatives for the Vitali equivalence relation without having a free ultrafilter on $\omega$? -What about the other direction? I thought I had convinced myself that using an ultrafilter, you can choose representatives for the Vitali equivalence relation, but right now that does not seem clear to me anymore. - -REPLY [12 votes]: Here is a tentative (negative) answer to the question of whether the existence of an $E_{0}$ (Vitali) selector implies the existence of a nonprincipal ultrafilter on $\omega$. I need to check some of the details (I'll say where below) but it seems ok to me. I'm posting this now in hope that someone has any ideas for an easier solution. -Let $P$ be the partial order consisting of countable partial $E_{0}$ selectors, ordered by inclusion. Let (a) be the statement (meant to be applied in the context of Choice) that whenever $D$ is a dense subset of $P$ in $L(\mathbb{R})$ and $F \subseteq P$ is a filter of cardinality less than the continuum ($\mathfrak{c}$), $F$ can be extended to a filter containing a member of $D$. If (a) holds, the continuum is regular, and $|\mathcal{P}(\mathbb{R}) \cap L(\mathbb{R})| = \mathfrak{c}$ (which holds if a measurable cardinal exists, for instance), then one can build $L(\mathbb{R})$-generic filters for $P$. -For all I know, (a) is a consequence of ZFC plus suitable large cardinals. It is clearly a consequence of the Continuum Hypothesis, but that doesn't seem to be of much use. Assuming that the theory of $L(\mathbb{R})$ is fixed by set forcing (which happens for instance if there exist proper class many Woodin cardinals; this hypothesis can be fine-tuned for our purposes), you would get that forcing with $P$ over $L(\mathbb{R})$ does not add a nonprincipal ultrafilter, if you knew that each of the following statements could be forced along with (a) + `$\mathfrak{c}$ is regular" : (1) for every nonprincipal ultrafilter $U$ on $\omega$ there is a Ramsey ultrafilter in $L(\mathbb{R})[U]$; (2) there are no Ramsey ultrafilters. Each of (1) and (2) is forceable along with $\mathfrak{c}= \aleph_{2}$. I'm not sure who showed this; my guess is Blass in the first case, and Kunen in the second. -I don't know anything about producing models of (a) by forcing over models of ZFC, but (a) holds in the $\mathbb{P}\mathrm{max}$ extension of any model of $\mathrm{AD}^{+}$ (such as $L(\mathbb{R})$, if there exist infinitely many Woodin cardinals below a measurable cardinal). This does not seem to give much information on the original question, either, though it shows (via a $\Delta$-system argument) that if $G \subseteq \mathbb{P}\mathrm{max}$ is an $L(\mathbb{R})$-generic filter, then in $L(\mathbb{R})[G]$ there are $L(\mathbb{R})$-generic filters for $P$ whose extensions do not contain any ultrafilter on $\omega$ generated by a tower of length $\omega_{2}$ from the point of view of $L(\mathbb{R})[G]$. -It seems to me that there are natural variations of $\mathbb{P}\mathrm{max}$ for producing models in which the cardinal characteristic $\mathfrak{u}$ is equal to $\aleph_{1}$ (I believe that Woodin has done this, in fact). In these models, (a) seems to hold, and the cardinal characteristic $\mathfrak{g}$ should be $\aleph_{2}$, which gives statement (1), by results of Laflamme. -More importantly, it gives Near Coherence of Filters (NCF), the statement that any two nonprincipal ultrafilters on $\omega$ are finite-to-one reducible to a common ultrafilter. I don't know about getting (2)+(a) with a $\mathbb{P}\mathrm{max}$ variation. However, if there is a $P$-name for an nonprincipal ultrafilter on $\omega$, then one can map this name to two (continuum many, in fact) independent parts of $P$, and thus show that NCF cannot hold in the $P$-extension of $L(\mathbb{R})$ unless there are no nonprincipal ultrafilters there. -This argument seems to have nothing to do with $E_{0}$. If it is correct, then it should show, for instance, that if $L(\mathbb{R}^{\#}) \models \mathrm{AD}^{+}$, then for any Borel action on the reals by a countable group, the partial order which adds a choice function for the orbits, with countable conditions, does not add a nonprincipal ultrafilter on $\omega$. -I expect that this argument is more complicated than necessary in more than one way. One would expect the proof to go though a determinacy-style $\Delta$-system lemma, but I don't see how to do that.<|endoftext|> -TITLE: On what kind of objects do the Galois groups act? -QUESTION [40 upvotes]: I am neither number theorist nor algebraic geometer. I am wondering -whether Galois groups of number fields (say the absolute Galois -group $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$) act on objects which -are not related a priori to the number theory. -I am aware of two such situations of rather different nature: -(1) Grothendieck's dessins d'enfants: -$Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ acts on certain graphs (with extra properties and extra data) on 2-dimensional surfaces. -(2) $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ acts on the profinite -completion of the topological $K$-theory (of sufficiently nice -spaces, e.g. finite $CW$-complexes). -As far as I understand (am I wrong?) the most important and best -studied examples of actions of Galois groups are actions on $l$-adic -cohomology of varieties over number fields. But this is not what I -am looking for: number fields appear in the formulation of the -problem from the vary beginning. - -REPLY [7 votes]: This is not exactly an incarnation of the question you asked, in the sense that is not so much an action of a Galois group but rather an action whose existence is governed by a Galois group of number-theoretic origin, but it seems likely to be of interest. -Let $K$ be a number field, and let $K^{(1)}$ be the maximal unramified abelian extension of $K$. The Galois group of $K^{(1)}/K$ is a subquotient of Gal$(\overline{\mathbb{Q}}/\mathbb{Q})$) which is isomorphic to the class group of $K$. Note that by Minhyong Kim's answer here, we can characterize this subquotient purely Galois-theoretically. Several authors have discovered surprising links between the arithmetic of number fields and actions of groups on spheres. In particular, when $K$ is the real cyclotomic field $K_m=\mathbb{Q}(\zeta_m+\zeta_m^{-1})$, the class group appears to govern the free actions of binary dihedral groups on spheres $S^n$ with $n\equiv 3\pmod{4}$. Let me loosely quote/paraphrase from Lang's "Units and Class Groups in Number Theory and Algebraic Geometry" (bolding mine): - -C. T. C. Wall has already shown to depend in part on the -2-primary component of the ideal class group in real cyclotomic fields $K_m^+$ for suitable $m$...Using the algebraic background of a paper -of Wall, applied to the surgery exact sequence, Thomas gives examples -for the binary dihedral group $D_{4p}$ of order $4p$ operating freely on $S^{4k-1}$ with $k\geq 2$, when the order of $[(K_p^+)^{( 1)}:K_p^+]$ is odd. - -... - -Furthermore, according to Thomas, there exist free actions by $D_{4p}$ which -can be topologically distinguished only by an invariant in the 2-primary part -of the ideal class group of $K_p^+$. - -Perhaps needless to say, the study of these degrees $[(K_p^+)^{( 1)}:K_p^+]$, even their 2-part, is of tremendous interest in algebraic number theory (Vandiver's conjecture, etc.), so the link to actions on spheres is surprising.<|endoftext|> -TITLE: Ping-pong relief map of a given function z=f(x,y) -QUESTION [15 upvotes]: I have an idea to design a type of -Galton's Board -to "draw" a relief map of a given two-dimensional function $z=f(x,y)$. -A typical Galton's Board drops, say, ping-pong balls through a series -of evenly spaced pins into vertical bins to demonstrate that the -balls distribute according to the binomial distribution, approximating -the normal distribution: - -           - - -(See this -this link -for an animation.) -First I would like to generalize this design to approximate -an arbitrary function $y=f(x)$, which leads to my first question: -Q1. -Which class of functions can be represented as a convex combination -of normal distributions? -I know these functions are called -mixture distributions, -but I have not found a description of the total class representable. -I am hoping that (say) any smooth function can be approximated. -Q2. Given a function $f(x)$ to approximate, how could one work backward -to a pin distribution that would realize the approximation? -The result would be a type of user-designed Pachinko machine. -Q3. Can the above be generalized to two-dimensional functions $f(x,y)$? -Presumably the answer is Yes. If so, one could imagine -a potentially mesmerizing -Museum of Math -display in which some famous visage emerges slowly as a ping-pong relief map. - -           - - -Q4. This final thought raises the question of which mathematician's face would be simultaneously -most appropriate and most recognizable. :-) Sir Francis Galton is certainly appropriate... - -REPLY [2 votes]: This is a partial answer to Q2, and suggests to me that there is a physical arrangement which would give a yes answer to Q3. -If you use ordinary pins, you can probably get a dyadic approximation with some arrangement. Let me suggest using weighted pins as a partial solution, and then perhaps someone can implement a close enough approximation to a weighted pin with a series of dyadic pins. -So normalize things so that the function f has integral one over the interval [0,1], and is to be approximated by 2^k bins. Suppose p in [0,1] is the fraction of balls needed to -represent the function on [0, 1/2], equivalently p is the integral of f from [0,1/2]. Then place a weighted pin very high such that it dumps p of the balls toward the pin -over the interval [0, 1/2]. (You may want to put a divider right under this pin so that -the ball doesn't jump to the [1/2,1] side.) Now recurse (k-1) more levels. Working backwards from this to get a horizontal arrangement should be clear, and of course one can use the physics of the situation to change the endpoints from dyadic rationals to something more appropriate to the desired function f. -It may be possible to emulate the bias by ever so slight horizontal adjustments of -the pins, but you need to place the later pins just so that their bias accomodates the various trajectories of the incoming ball. But of course we have infinite precision pins and balls, so what's to worry? -Gerhard "Likes The Unreality of Mathematics" Paseman, 2011.07.06<|endoftext|> -TITLE: "double" semidirect product -QUESTION [8 upvotes]: Let $A$, $B$ and $C$ be discrete countable groups. Let $\alpha$ be an action of $A$ on $B$ and let $\beta$ be an action of $B$ on $C$. - -Question Does there always exist a group $G$ which has $A$, $B$ and $C$ as subgroups and such that the group generated by $A$ and $B$ is $A\ltimes B$ and the group generated by $B$ and $C$ is $B\ltimes C$? - -One obvious situation when such a group $G$ exists is when the action $\beta$ extends to an action of $A\ltimes B$. Then we can take $G=(A\ltimes B)\ltimes C$. But I'm interested in situations when $\beta$ doesn't extend. -For example take $A$ to be the infinite cyclic group with generator $t$, $B$ to be the free group on infinitely many generators indexed by $\mathbb Z$ (denote the generators by $g_n, n\in \mathbb Z$), and $C$ to be the rational numbers. Let $g_n$ act on $C$ by multiplication by $n$ if $n\neq 0$ and by identity for $n=0$, and let $t$ act on $B$ by sending $g_n$ to $g_{n+1}$. - -Question In this specific situation does there exist $G$ as above? - -REPLY [12 votes]: The answer to the specific question is "yes". Let $U$ be the semidirect product of $C$ and $B$. Then as $G$, you take the HNN extension of $U$ with the free letter $t$ and associated subgroups equal to $B$, with the automorphism provided by the shift of generators. The result then follows from the usual properties of HNN extensions. The same argument applies in general when $A$ is cyclic. -In general you can take the semidirect products and take their amalgamated product with amalgamated subgroup $B$.<|endoftext|> -TITLE: Example of a continous functor between locally presentable categories which has no left adjoint -QUESTION [12 upvotes]: It is known a version of Adjoint Functor Theorem for locally presentable categories, which says that a functor between such categories has a left adjoint iff it is continous (i.e. preserves all limits) and accessible (preserves $\lambda$-filtered colimits for some cardinal $\lambda$) (see for this Theorem 1.66 of J. Adamek, J. Rosicky, "Locally Presentable and Accessible Categories", London Math. Soc. Lecture Notes, Cambridge, 1994). My question is then to find an example of a continous functor between locally presentable categories which is not accessible. - -REPLY [15 votes]: The following example was coincidentally mentioned by André Joyal on the categories mailing list today; he attributed it to Mac Lane. For every infinite cardinal number k, let $G_k$ be a simple group of cardinality k. Define the functor ML: Group → Set to be the product of all the representable functors $\mathrm{Hom}(G_k,-)$. Since no group can admit a nontrivial homomorphism from proper-class-many of the $G_k$, this functor does indeed land (or can be redefined to land) in Set. Since it is a product of representables, it is continuous (and of course Group and Set are locally presentable), but it is not itself representable (hence has no left adjoint).<|endoftext|> -TITLE: trace(xy)=trace(yx) in full generality -QUESTION [15 upvotes]: It is well known that, for square matrix $x$ and $y$, we have $\operatorname{tr}(xy)=\operatorname{tr}(yx)$. Here of course the trace of a matrix is just the sum of the elements of the diagonal. -The notion of trace has a lot of generalization. As I know, the most general definition is the following: let $(\mathcal C, \otimes, 1, ^\vee)$ be a rigid symmetric monoidal category, $X$ an object of $\mathcal C$ and $f$ an endomorphism of $X$. Then $\operatorname{tr}(f) \in \operatorname{End}(1)$ is defined by the following composition -$$ 1 \longrightarrow X^\vee \otimes X \stackrel{\operatorname{id}_{X^\vee} \otimes f}{\longrightarrow} X^\vee \otimes X \longrightarrow X \otimes X^\vee \longrightarrow 1 $$ -So my questions is: it is true, in this generality, that $\operatorname{tr}(f\circ g)=\operatorname{tr}(g \circ f)$, for $f$ and $g$ in $\operatorname{End}(X)$? -Ricky - -REPLY [17 votes]: Yes. The string diagram chase can be found on page 8 of Ponto and Shulman - Traces in symmetric monoidal categories.<|endoftext|> -TITLE: Rosenlicht theorem about uniruledeness and zeroes of holomorphic vector field on complex projective manifold -QUESTION [5 upvotes]: I heard that there is a theorem due to Rosenlicht which says the following: - -Theorem. Let $X$ be a complex projective manifold and $V$ a non-trivial holomorphic vector field on $X$. Then $X$ is uniruled,ie,can be covered by rational curves,if $V$ has a zero. - -I have thought for a few days and failed to give myself a proof. Can somebody give me the reference or say something about the idea of proof? -Thanks in advance. - -REPLY [3 votes]: You can look at Lieberman's paper Holomorphic Vector Fields on Projective Manifolds. -His proof is more or less as follows. A result of Grothendieck asserts that $\mathrm{Aut}^0(X)$, the connected component -of the identity of the automorphism group of $X$, is an algebraic group which acts -algebraically on $X$. -Look at the (analytic) subgroup generated by your vector field and let $G$ be its Zariski closure in $\mathrm{Aut}^0(X)$. Notice that $G$ is abelian. -If $p \in X$ is a zero of your vector field then -$p$ is fixed by the action of $G$ on $X$. Thus for $k \in \mathbb N$, $G$ acts on $$\frac{\mathcal O_{X,p}}{\mathfrak{m}_p^k}, $$ where $\mathfrak m_p$ is the maximal ideal of $\mathcal O_{X,p}$. Moreover, if $k \gg 0$ then the action is faithfull. -Thus $G$ is isomorphic to a linear algebraic group and yet another result of Rosenlicht says that a Zariski-closed abelian -subgroup of a linear algebraic group is of the form $(\mathbb Cˆ*, \cdot)^r \times (\mathbb C,+)^s$. The action of the factors of this decomposition generate the sought rational curves. - -Added later: For an alternative proof see Theorem 6.4 of this paper. There it is proved that the existence of a non-zero section of $\bigwedge^q TX$ vanishing at a point suffices to ensure that $X$ is uniruled.<|endoftext|> -TITLE: Coherence for pull-backs and push-forwards -QUESTION [5 upvotes]: Let $p:X \to S$ and $q:Y\to S$ be two objects in the category of ringed spaces over the ringed space -$S$, and let $f:X \to Y$ be a morphism over $S$. -Given a sheaf $\mathcal{F}$ of $\mathcal{O}_Y$-modules, there are at least -two different ways to produce a morphism -$$ -q_*\mathcal{F} \to p_*f^*\mathcal{F} -$$ -of sheaves of $\mathcal{O}_S$-modules by just using canonical operations. We could for instance -apply the adjunction unit for $(f^*, f_*)$ to $\mathcal{F}$, push the map forward to $S$ -and then use the natural isomorphism $q_*f_* \simeq p_*$. That is -$$ -\mathcal{F} \to f_*f^*\mathcal{F} -$$ -$$ -q_*\mathcal{F} \to q_*f_*f^*\mathcal{F} -$$ -$$ -q_*\mathcal{F} \to p_*f^*\mathcal{F}. -$$ -On the other hand, we could start by applying the adjunction co-unit for $(q^*, q_*)$, -pull it back to $X$, use the natural isomorpism $f^*q^* \simeq p^*$ and finally use the -adjuncton $(p^*, p_*)$ to produce a morphism of $\mathcal{O}_S$-modules. That is -$$ -q^*q_*\mathcal{F} \to \mathcal{F} -$$ -$$ -f^*q^*q_*\mathcal{F} \to f^*\mathcal{F} -$$ -$$ -p^*q_*\mathcal{F} \to f^*\mathcal{F} -$$ -$$ -q_*\mathcal{F} \to p_*f^*\mathcal{F}. -$$ -In this case we get the same map (something which at least I find tedious to check; -but I might be thinking of it in the wrong way). -My question is: -Is there a general coherence result which frees us from checking such equalities case by case, -just as in the case of for instance symmetric monoidal categories. I'm thinking of something like: -Start with a commutative diagram of ringed spaces and a map of sheaves of modules over one of -the spaces. Is any map produced from this map, by just applying push-forwards and pull-backs and using -adjunctions, uniquely determined by the strings of symbols in the domain and co-domain of the new map? -(One could probably state a similar question where we throw the tensor product into the mix of canonical operations.) - -REPLY [3 votes]: I would like to revive this old question by answering it with a pointer to my recent preprint Obvious natural morphisms of sheaves are unique, which addresses precisely this problem. In particular, I give an example in the first section that discuses the kind of natural morphism you ask about. I don't yet know how to add tensor products to the mix in general, though there are examples there as well that do concern it when it enters the problem in a non-invasive kind of way. -The summary answer, by the way is: yes, such morphisms are always unique, as long as the start and end functors are both of the form $f_*$ or $\mathrm{id}$ (both $f^*$ or $\mathrm{id}$ also works), or can be reduced to such by means of adjunctions and isomorphisms like $(fg)_* \cong f_* g_*$. If they necessarily have more then you need some conditions ensuring that certain base-change maps are isomorphisms or else there are counterexamples. -I should say that despite the intricacy of the morphisms I consider in that paper, it seems hard in practice to find a map constructed from more than a handful of units and counits and isomorphisms such as $(fg)_* \cong f_* g_*$, so although it is a labor-saving device, it seems so far not to save unbounded labor. I would like to be proven wrong on that assertion, however; I know that when you add the $!$ functors the labor really ramps up, but alas, I haven't been able to add those to the mix either.<|endoftext|> -TITLE: Examples of non-simply connected manifolds with trivial H^1 -QUESTION [12 upvotes]: It is known that, if a topological space is simply connected,its first homology group vanishes. The converse is not true, since for every presentation of a (say, finite) perfect group G we can construct a CW-complex, via generators and relations, having G as a fundamental group. Are there such examples in the class of topological or differentiable manifolds? In other words, does there exist a non-simply connected manifolds with trivial first homology group? - -REPLY [8 votes]: Fake Projective planes -These are smooth complex projective surfaces with the same betti numbers as $\mathbb{CP}^2$, but with infinite fundamental group $\pi_1(X)$ (in fact it is isomorphic to a torsion-free cocompact arithmetic subgroup of $PU(2,1)$).<|endoftext|> -TITLE: What is the relation of the Kuznetsov-Bruggeman trace formula and the Selberg trace formula? -QUESTION [7 upvotes]: I have read that there is an elementary way to show that the above mentioned trace fromulas are equivalent in the sense, that each of them can be derived directly from the other. There should exist a short elegant method by Zagier. Where? -In short, I know how to deduce the Selberg trace formula from Arthur's trace formula at least in principle, how should one proceed to deduce the Kuznetsov formula from the arthur trace formula. What is the utility of the Kuznetsov formula? For which applications is this trace formula more suitable than the Selberg trace formula? - -REPLY [5 votes]: The important point is that the Selberg trace formula includes the contribution of the one-dimensional representations, i.e. the nongeneric spectrum, whereas the Kuznetsov formula does not. Spectrally, the nongeneric spectrum contributes so much that it has a tendancy to obscure the contribution of cusp forms. That is the essential difference between the two trace formulae. -This is especially important with regard to the idea of beyond endoscopy (see Sarnak's letter http://publications.ias.edu/sarnak/paper/487). But see also the work of Frenkel, Langlands, and Ngo. -Rudnick's thesis (http://www.math.tau.ac.il/~rudnick/papers/myphdthesis.pdf) in principle discusses how to pass from the Kuznetsov formula to the trace formula. Essentially the passage is just an application of Poisson summation.<|endoftext|> -TITLE: Rational or elliptic curves on Calabi-Yau threefolds -QUESTION [20 upvotes]: Let $X$ be a Calabi-Yau threefold. From a complex analytic point of view, it is widely believed that it should not be Kobayashi hyperbolic, that is it should always admit some non-constant entire map from the complex plane $f\colon\mathbb C\to X$. -One could be even more ambitious and ask whether a Calabi-Yau threefold always contains a rational or an elliptic curve (or, more generally a non-constant image of a complex torus). -Mostly string theorists have produced lots of examples of such manifolds, mainly by adjunction or crepant resolution of singularities. So my question is: -Is it true that in all known examples of Calabi-Yau threefold one can always find a rational or an elliptic curve (or, more generally a non-constant image of a complex torus)? -Thanks in advance! - -REPLY [9 votes]: Let me give a partial answer. -Most of the known examples of Calabi-Yau threefolds contain rational curves. However, there exist examples of Calabi-Yau threefolds without rational curves. -You can find some of them in the paper by Oguiso and Sakurai Calabi-Yau threefolds of quotient type, Asian Journal of Mathematics 5 (2001). -These threefolds, that the authors call "of Type A", are constructed as the quotient af an Abelian threefold $A$ by a suitable fixed-point free finite group of automorphisms. -Moreover, a Calabi-Yau threefold $X$ is of type A if and only if $c_2(X)=0$, and in this case the Picard number $\rho(X)$ is either $2$ or $3$. -In fact, the authors ask as an open question whether every Calabi-Yau threefold of Picard number $\rho \neq 2,3$ contains rational curves. -I do not know whether Calabi-Yau threefolds of type A contain elliptic curves, but one can probably check this directly, since the construction is very explicit.<|endoftext|> -TITLE: Contest problems with connections to deeper mathematics -QUESTION [80 upvotes]: I already posted this on math.stackexchange, but I'm also posting it here because I think that it might get more and better answers here! Hope this is okay. -We all know that problems from, for example, the IMO and the Putnam competition can sometimes have lovely connections to "deeper parts of mathematics". I would want to see such problems here which you like, and, that you would all add the connection it has. -Some examples: - -Stanislav Smirnov mentions the following from the 27th IMO: -"To each vertex of a regular pentagon an integer is assigned in such a way that the sum of all five numbers is positive. If three consecutive vertices are assigned the numbers $x,y,z$ respectively, and $y <0$, then the following operation is allowed: the numbers $x,y,z$ are replaced by $x+y$, $-y$, $z+y$, respectively. Such an operation is performed repeatedly as long as at least one of the five numbers is negative. Determine whether this procedure necessarily comes to an end after a finite number of steps". -He mentions that a version of this problem is used to prove the Kottwitz-Rapoport conjecture in algebra(!). Further, a version of this has appeared in at least a dozen research papers. -(Taken from Gerry Myerson in this thread https://math.stackexchange.com/questions/33109/contest-problems-with-connections-to-deeper-mathematics) -On the 1971 Putnam, there was a question, show that if $n^c$ is an integer for $n=2,3,4$,… then $c$ is an integer. -If you try to improve on this by proving that if $2^c$, $3^c$, and $5^c$ are integers then $c$ is an integer, you find that the proof depends on a very deep result called The Six Exponentials Theorem. -And if you try to improve further by showing that if $2^c$ and $3^c$ are integers then $c$ is an integer, well, that's generally believed to be true, but it hadn't been proved in 1971, and I think it's still unproved. - -The most interesting part would be to see solutions to these problems using both elementary methods, and also with the more abstract "deeper methods". - -REPLY [2 votes]: IMO 2003-6: - -Show that for each prime $p$, there exists a prime $q$ such that $n^p − p$ is not divisible by $q$ for any positive integer $n$. - -This obviously looks like something you want to apply the Chebotarev Density Theorem to. The condition that is to be satisfied translates to $f(X) = X^p-p$ not having a linear factor over $\mathbb{F}_q$. Now by Chebotarev, there are infinitely many primes $q$ for which $f$ is actually irreducible modulo $q$, since $\operatorname{Gal}(f)\cong\operatorname{Aff}(\mathbb{F}_p)$ contains permutations without fixed points, hence certainly $f$ does not have a linear factor. -[In fact, the condition is not only implied by the irreducibility of $f$, it is equivalent to it. This is a little exercise in field theory: suppose we have $g \mid f$ with $\deg g < \deg f$. Then we have a finite field extension $k/\mathbb{F}_q$ of degree $d -TITLE: Statistics of Number fields -QUESTION [14 upvotes]: A goal which I have been pursuing is to understand how number fields are distributed with respects to their invariants. To be more precise I was captivated by the following question: -Let $N(X,n,G)$ be the number of number fields of dimension $n$ where $G$ is the Galois group of its Galois closure, and their discriminants is bounded by $|X|$ up to isomorphism. This question might be natural to ask: For which group $G\leq S_n$ one might get a positive proportion of number fields when $X\to \infty$. -From Class Field Theory or one can show it more elementary, using Delone-Faddeev correspondence, for $n=3$, $C_3$ has a density 0. And also when $G$ is an abelian group then the answer would be same as $C_3$ by using class field theory. Prof. Manjul Bhargava proved for $n=4$ , $D_4$ has a positive density, and also he showed for $n=5$ no other groups, except $S_5$, could have positive density. -I think, once Manjul told me, one might expect for $n=p$, $p$ is a prime number, the only group which can contribute with positive density is $S_p$. But I could not even find a heuristic that why this should be true. -Is there a heuristic or even a theorem which can support the above expectation? Or more generally what do we know about $N(X,n,G)$? - -REPLY [5 votes]: First of all, the most general conjecture about $N(X,n,G)$ is indeed due to Malle, and there are indeed counterexamples to the original formulation due to Kluners. Seyfi Turkelli has a very nice paper which explains "why" those counterexamples arise by means of a function field analogy, and offers a revised version of Malle's conjecture which seems to me pretty solid. -Why does only $S_p$ give positive density? Because under Malle's conjecture, $N(G,n,X)$ will be asymptotic to $X^{a(G)} \times$(some power of $\log$), where $a(G)$ is at most 1, with equality only when $G$ contains a transposition. So the class of groups expected to have positive density is precisely those containing a transposition. When $p$ is prime, the only transitive subgroup of $S_p$ containing a transposition is $S_p$ itself, so you're done. On the other hand, both $D_4$ and $S_4$ are transitive subgroups of $S_4$ containing a transposition -- indeed, they are the only such, so they're the only two Galois groups that are supposed to arise for a positive density of quartic fields ordered by discriminant.<|endoftext|> -TITLE: The Hölder continuity condition of the Schauder estimates -QUESTION [6 upvotes]: The classical Schauder estimates (see the link) -http://en.wikipedia.org/wiki/Schauder_estimates -Requires $f\in C^\alpha$ in order to get a solution $u\in C^{2+\alpha}$ of the equation -$$\Delta u=f$$ -In fact, we can construct a continuous function f, which is not Hölder of any order on a positive meausre set. -Is it possible to find solution $u\in C^2$? -Or is there any counterexample to show such solution $u$ does not exist？ - -REPLY [11 votes]: For $u$ to be from $C^2$ it is enough that the modulus of continuity of $f$ satisfies the Dini condition. For example, modulus of continuity $\omega(h)=1/(|\log h|+1)^2$ is not marjorised by the Holder modulus of continuity $\omega(h)=C h^\alpha$. -The same goes for coeffitients of elliptic equations. In the work “A priori bounds and some properties for solutions of elliptic and parabolic equations”, Uspekhi Mat. Nauk, 18:4(112) (1963), 215–216, Kruzhkov notes that the problem of continuity of higher derivatives of solutions of elliptic equations is closely connected with the question of when for a function of several independent variables condition of the existence of pure continuous derivatives of order net $l\ge2$ implies the existence of mixed derivatives of the same order. He gives nesessary and suffitient condition for it. Namely, denote $\omega(h)$ the modulus of continuity of the order $l$ pure derivatives of $u$ and $\omega^*(h)=\int_0^h\frac{\omega(t)}t dt$ the second modulus of continuity. Then in order that all mixed derivatives were continuous (in some domain $Q\subset\mathbb R^n$) it is necessary and sufficient that the module continuity $\omega^* (h) $ satisfies the Dini condition. Moreover, if $\omega^* $ satisfies the Dini condition, then all -mixed derivatives of order $l$ in any strictly interior subdomain $Q'\subset Q$ have -modulus of continuity, which does not exceed $Cw^* $. If $\omega^*$ does not satisfy the Dini condition, then there exists a function $u$ which at some point of $Q$ have no mixed derivatives of order $l$.<|endoftext|> -TITLE: Applications of geometric evolution equations. -QUESTION [9 upvotes]: Hi everybody, -I'm looking for applications of geometric evolution equations such as the Ricci flow and the extrinsic flows by Gauss and mean curvature. Applications other than topological applications such as geometrization are what I'm looking for and, even better if the application is to a non-mathematical area, such as Mullins' derivation of mean curvature flow as a model for the motion of `grain boundaries'. -Thanks, -ML - -REPLY [5 votes]: The Willmore flow and the more complicated Helfrich flow are used to model cell membranes.<|endoftext|> -TITLE: Did Joseph Doob prove that random sequences don't exist? -QUESTION [9 upvotes]: In the book "The Mathematical Experience" it says: - -"An infinite [binary] sequence $x_1, x_2, \ldots$ is called random in the sense of von Mises if every infinite sequence $x_{n_1}, x_{n_2}, \ldots$ extracted from it and determined by a policy or rule R is $\infty$-distributed. Now comes the shocker. It has been established by Joseph Doob that there are no sequences that are random in the sense of von Mises." - -A sequence on $\{H,T\}$ is $\infty$-distributed if for each positive integer $k$ and sequence $\vec y \in \{H,T\}^k$ the set $\{n\in {\mathbb N} \colon \langle x_{n},\dots,x_{n+k-1}\rangle=\vec y\}$ has density $2^{-k}$. -But the definition of von Mises seems so natural to me that if a sequence does not satisfy it then the sequence is not random. - -REPLY [7 votes]: There is an excellent article by Sérgio B. Volchan in the American Mathematical Monthly, titled What Is a Random Sequence, which discusses how the von Mises-Wald-Church model of randomness is unsatisfactory. He goes on to explain the proposed candidate for a definition of a random sequence due to Martin-Löf, that of typicality, or "randomness with respect to effective statistical tests". Here randomness is defined with respect to a given measure $\mu$ on infinite binary strings; it turns out to coincide with a natural notion of incompressibility of the sequence. -Anyway, in short: there are other natural candidates for what it should mean for a sequence to be random, that turn out to work pretty well (and are beautiful), and Volchan's paper is a good place to learn about them.<|endoftext|> -TITLE: Why is there a weight 2 modular form congruent to any modular form -QUESTION [13 upvotes]: I got my copy of Computational Aspects of Modular Forms and Galois Representations in the mail yesterday. The goal of the book is "How one can compute in polynomial time the value of Ramanujan's tau at a prime", well, or any other modular form of level 1. It's all very thrilling! -The following fact is essential: for any modular form $f$ of level 1, and any prime $l$, the mod $l$ reduction of the semisimplification of the galois representation attached to $f$ by Deligne is a 2-dimensional subrepresentation of the galois representation of the $l$-torsion of the jacobian of the modular curve of level $l$. -Why? -From what I understand, this is somewhat equivalent (after Shimura and Deligne) to there being a modular form of weight 2 and level $l$ that is congruent to $f$ mod $l$, or something similar. Is this the right statement? Why is it true then? -Searching far and wide for an introduction to this topic yields very little. - -REPLY [6 votes]: If $N\geq 1$ is an integer not divisible by $p$, one can see that any system of Hecke eigenvalues $(a_\ell)$ arising from $S_k(\Gamma_1(N))$ is congruent mod $p$ to a system $(b_\ell)$ arising from $S_2(\Gamma_1(Np^n))$, for some $n$, using an interplay between a theorem of Serre (describing a purely mod $p$ Jacquet-Langlands correspondence), and the more classical, characteristic zero J-L between ${\rm GL}_2$ and the multiplicative group $G$ of the $\mathbf{Q}$-quaternion algebra ramified at $p$ and infinity. -I know that what I describe here is perhaps not the right way of proving the result you are asking, but it seems to me worth to mention. -In his '87 letter to Tate Serre proves: -${\rm Theorem:}$ Systems of mod $p$ Hecke eigenvalues arising from $M_k(\Gamma_1(N))$ are the same as those arising from locally constant function $f:G(A)\rightarrow\overline{F}_p$ that are left invariant under $G(\mathbf{Q})$ and right invariant under a certain open subgroup $K_N$. -Here $G(A)$ is the adelic group associated to $G$. Notice that the functions considered on the quaternion side are independent of the archimedean variable. Moreover, the double coset $G(\mathbf{Q})\backslash G(A)/K_N$ is finite and any mod $p$ system of eigenvalues arising from it can be lifted to characteristic zero. -Therefore applying the theorem and then lifting, we see that for any (char. zero) eigensystem $A=(a_\ell)$ arising from $M_k(\Gamma_1(N))$ there is a (char. zero) eigensystem $B=(b_\ell)$ arising from the space of locally constant functions $f:G(A)\rightarrow\mathbf{C}$ such that $A\equiv B$ mod $P$, where $P$ is a fixed prime of $\overline{\mathbf{Z}}$ lying over $p$. -Assuming that the automorphic form $\Pi_B$ on $G$ associated to $B$ is infinite dimensional, by the J-L correspondence we have that there is a cuspidal automorphic form $\Pi'_B$ on ${\rm GL}_2$ associated to the same eigensystem $B$. The type of $\Pi'_B$ at any finite place other than $p$ is the same as that of $\Pi_B$, while at infinity $\Pi'_B$ is the discrete series of lowest weight $2$. This basically says that there is a cusp form in $S_2(Np^n)$ whose associated system of eigenvalues is $B=(b_\ell)$. -We are only left with deciding when $\Pi_B$ is infinite dimensional, or can be chosen as such. This happens only for systems of eigenvalues of the form $B=(\chi(\ell)(1+\ell))_{\ell\nmid pN}$, where $\chi:\mathbf{Z}/p\rightarrow\mathbf{C}^*$ is any character (in order to show this one has to consider the particular shape of $K_N$, which I did not even define..). The reduction mod $P$ of such eigensystems are all of the form $(\ell^k+\ell^{k+1})_{\ell\nmid pN}$. -Concluding: Let $A=(a_\ell)$ be a sytstem of char. zero eigenvalues arising from $M_k(\Gamma_1(N))$, with $p\nmid N$. Assume that the mod $P$ reduction of $A$ is not of the form $(\ell^k+\ell^{k+1})_{\ell\nmid pN}$. Then, there exists a cusp form in $S_2(\Gamma_1(Np^n))$ such that its associated system of eigenvalues $B$ is congruent to $A$ mod $P$.<|endoftext|> -TITLE: Generalising Vitali Sets to uncountable dense subgroup selectors... -QUESTION [5 upvotes]: Does there exist an uncountable dense subgroup, $\Gamma$, of the additive group $\mathbb{R}$, such that every selector of the partition of $\mathbb{R}$ canonically associated with the equivalence relation $x \in \mathbb{R}$ & $y \in \mathbb{R}$ & $x − y \in \Gamma$ is nonmeasurable? -This question is related to a previous question of mine which has not been answered yet. - -REPLY [8 votes]: I edited my former answer in light of a recent e-mail exchange with Sławomir Solecki. -There is a substantial literature on Vitali-like selectors. A major paper in the area is the following, available here. -Cichoń, J.; Kharazishvili, A.; Węglorz, B. On sets of Vitali's type. Proc. Amer. Math. Soc. 118 (1993), no. 4, 1243–1250. -There are many interesting results in the above paper, including the fact, observed by Joel Hamkins in his answer, that $MA+\lnot CH$ implies the existence of uncountable subgroups $G$ of $\Bbb{R}$ with no measurable $G$-selectors; where a $G$-selector is a choice function that selects one element from each $G$-coset. - -Moreover, Theorem 6 of the above paper shows that under $MA+\lnot CH$ there is even a subgroup $G$ of $\Bbb{R}$ of power $2^{\aleph_0}$ that has no measurable $G$-selector. - -This is to be contrasted with Theorem 2 of the paper, which asserts every analytic subgroup $G$ of reals has a measurable $G$-selector. - -According to Solecki (private communication) it is apparently unknown if $ZFC$ can prove that there is an uncountable subgroup of the reals all of whose selectors are Lebesgue non-measurable. - -However, if one allows extensions of measures, then there is a fundamental distinction to be made (note: the theorem below was first stated and proved with $CH$; but $CH$ was subsequently removed in the paper cited below). - -Theorem (Solecki). The following are equivalent for every subgroup $G$ of reals. -(a) $G$ is countable and dense. -(b) Every $G$-selector is non-measurable with respect to any translation invariant extension of the Lebesgue measure. - -Solecki's result above is a special case of Cor. 2.2. of the following paper of his: -Translation invariant ideals, Israel J. Math. 135 (2003), 93--110<|endoftext|> -TITLE: Can a metric conformal to a Kahler metric be Kahler? -QUESTION [11 upvotes]: Let $X$ be a non-compact complex manifold of dimension at least 2 equipped with a Kahler metric $\omega$. Take a smooth positive function $f : X \to \mathbb R$, and define a new hermitian metric on $X$ by $\tilde \omega = f \omega$. If $f$ is non-constant, then can this new metric ever be Kahler? -If $\dim_{\mathbb C} X = 1$ the new metric is automatically Kahler because of dimension. If $\dim_{\mathbb C} X \geq 2$ and if $X$ is compact the new metric is never Kahler. Indeed, we have that $d \tilde \omega = d f \wedge \omega$ is zero if and only if $df$ is zero by the hard Lefschetz theorem, so $f$ must be constant if $\tilde \omega$ is Kahler. -If $X$ is not compact, then to the best of my knowledge we do not have the hard Lefschetz theorem, but does the conclusion on metrics conformal to a Kahler metric still hold? - -REPLY [10 votes]: You don't have to use hard Lefschetz to conclude $df=0$ from $\omega\wedge df=0$. -This is a linear algebra fact, valid pointwise : if $\alpha \in T_x^*X$ satisfies $\omega_x \wedge \alpha=0$, then $\alpha=0$ (of course, assuming $\dim_R X \geq 4$. -The short argument is that, $\omega_x^{n-1}\wedge : T^*_x X\to \bigwedge^{2n-1} T^*_x X$ is an isomorphism ("pointwise not so hard Lefschetz", so to speak). -This said, as in Francesco's answer, you can have non proportional conformal riemannian metrics that are Kähler with respect to different complex structures, so that the corresponding 2-forms are no longer (pointwise) proportional. - -REPLY [5 votes]: The paper by Apostolov, Calderbank, and Gauduchon that Francesco mentions find different Kaehler structures whose associated Riemannian metrics are conformal to each other. But they correspond to different complex structures $J_+$ and $J_-$. -I believe what Gunnar is asking is whether or not one can have $f \omega$ be closed and thus Kaehler with respect to the same complex structure $J$ associated to $\omega$. The answer is no, and this has nothing at all to do with compactness or the hard Lefschetz theorem. On any almost Hermitian manifold $(M, J, \omega, g)$, it is a fact that the wedge product with the Kaehler form $\omega$ on the space of $1$-forms is injective, regardless of the compactness of $M$, the integrability of $J$, or the closedness of $\omega$. This follows, for example, from the identity -$$ \ast( \omega \wedge (\ast ( \omega \wedge \alpha) ) = - (m-1) \alpha $$ -where $\alpha$ is any $1$-form on $M$, where $\ast$ is the Hodge star operator, and the real dimension of $M$ is $2m$. (One sees that the only requirement is that $m>1$.) - -REPLY [2 votes]: There are examples in real dimension $4$ of manifolds having two conformally equivalent -Kahler metrics, inducing the same conformal structure but with opposite orientation. -See the paper Ambikahler geometry, ambitoric surfaces and Einstein 4-orbifolds by Apostolov, Calderbank and Gauduchon.<|endoftext|> -TITLE: Relation between Gerstenhaber bracket and Connes differential -QUESTION [10 upvotes]: Let $C$ be an arbitrary algebra (more generally, a linear 1-category). The following structures are well-known: -A degree-0 product on the Hochschild cohomology $HH^*(C)$ -$$ - HH^*(C) \otimes HH^*(C) \to HH^*(C) -$$ -$$ - a \otimes b \mapsto ab -$$ -A degree-0 action of Hochschild cohomology on the Hochschild homology $HH_*(C)$ -$$ - HH^*(C) \otimes HH_*(C) \to HH_*(C) -$$ -$$ - a \otimes \gamma \mapsto a\cdot \gamma -$$ -A degree-1 unary operation on Hochschild homology (Connes differential) -$$ - HH_*(C) \to HH_*(C) -$$ -$$ - \gamma \mapsto B(\gamma) -$$ -A degree-1 binary operation on Hochschild cohomology (Gerstenhaber bracket) -$$ - HH^*(C) \otimes HH^*(C) \to HH^*(C) -$$ -$$ - a \otimes b \mapsto a * b -$$ -The above operations satisfy some well-known relations. (Note that I am not attempting to get the signs right.) - -graded commutativity $ab = \pm ba$ -more graded commutativity $a * b = \pm b * a$ -Poisson identity $a * (bc) = (a * b)c + b(a * c)$ -Jacobi identity $a * (b * c) + b * (c * a) + c * (a * b) = 0$ -$B$ is a differential $B(B(\gamma)) = 0$ -various associativities $(ab)c = a(bc)$; $(a * b) * c = a * (b * c)$; $(ab)\cdot\gamma = a\cdot(b\cdot\gamma)$ - -The following relation, expressing the action of a Gerstenhaber bracket on Hochschild homology in terms of the Connes differential, seems to be less well-known. At least I haven't been able to find it in the literature. -$$ - (a*b)\cdot\gamma = ab\cdot B(\gamma) - a\cdot B(b\cdot \gamma) - b\cdot B(a\cdot\gamma) + B(ba\cdot\gamma) -$$ -(Again, I haven't tried to get the signs right.) -Question: Is there a reference for the above relation? -Note: The above relation follows from the fact that the first homology of a certain operad space is 4-dimensional, so there must be some relation between the five degree-1 maps $HH^*(C)\otimes HH^*(C)\otimes HH_*(C)\otimes \to HH_*(C)$ which figure in the relation. -Another note: In cases where $HH^*(C) \cong HH_*(C)$ and there is a BV algebra structure, I think the relation follows from the usual definition of the Gerstenhaber bracket in terms of the BV structure. See the "Antibracket" section of this Wikipedia article. - -REPLY [5 votes]: Hi, -Your formula is due (without the signs!) due to Ginzburg Calabi-Yau algebras (9.3.2) -as explained in Lemma 15 of my paper, Batalin-Vilkovisky algebra structures on Hochschild Cohomology, Bull. Soc. Math. France 137 (2009), no 2, 277-295 -(sorry for quoting myself!) -Here is Lemma 15 -Lemma 15 [17, formula (9.3.2)] Let A be a differential graded algebra. -For any η, ξ ∈ HH ∗ (A, A) and c ∈ HH∗ (A, A), -{ξ, η}.c = (−1)|ξ| B [(ξ ∪ η).c] − ξ.B(η.c) -+ (−1)(|η|+1)(|ξ|+1) η.B(ξ.c) + (−1)|η| (ξ ∪ η).B(c). -In a condensed form, this formula is -(34) $i_{\{a,b\}}=(-1)^{\vert a\vert+1}[[B,i_{a}],i_b]=[[i_{a},B],i_b].$ -See formula (34) of my second paper -Van Den Bergh isomorphisms in String Topology, J. Noncommut. Geom. 5 (2011), no. 1, 69-105. -(sorry for quoting myself again!) -In this paper, I thought I gave a new definition of BV-algebras. -But this definition appears more or less in the -section "Compact formulation in terms of nested commutators." -of the Wikipedia article, you quote! -However, I was unable to find this definition in the bibliography quoted in the -Wikipedia article. -Concerning signs, -in my first paper, I made a mistake, corrected in my second paper. -So (34) is correct and Lemma 15 has some signs problems. -ps: David Ben-zvi is absolutely right. This formula is a consequence of Tamarkin-tsygan -calculus!<|endoftext|> -TITLE: Finite-dimensional subgroups of circle diffeomorphism group -QUESTION [8 upvotes]: Is there a sequence of connected finite-dimensional subgroups Gi of the circle diffeomorphism group G with the following properities: -(a) Gi is contained in Gj for i < j -(b) The union of Gi is dense in G -More rigorously "finite dimensional subgroup of circle diffeomorphism group" means a Lie group H with smooth faithful action on the circle. -In order to make sense of property (b) I have to specify a topology on G. I suspect that all reasonable topologies will yield the same answer, but for the sake of definiteness let's use the -"sup-norm" topology. That is, given two diffeomorphism g1 and g2, I define the distance d(g1, g2) as -supremum over x in S1 of d(g1(x), g2(x)) -Here the latter d is the usual distance on the circle. -This is a metric and it induces a topology. -I suspect that the answer to my question is "no". Moreover, I suspect that there is no H as above with dimension > 3. But I might be wrong... - -REPLY [15 votes]: The answer is indeed no, as described e.g. in the lecture notes by Ghys -http://www.math.ethz.ch/~bgabi/ghys%20groups%20acting%20on%20the%20circle.pdf -Section 4.1 has a list of all connected groups acting faithfully and transitively on the circle or the line. They are -1) $\mathbb{R}$ acting on itself, -2) the circle acting on itself, -3) the affine group of the line acting on the line, and -4) the k-fold cover of $PSL(2,\mathbb{R})$ acting on the circle. -Any faithful action of a connected Lie Group on the circle is made out of these: if $F$ is the set of fixed points, then on each connected component of the complement of $F$ the action must be conjugate to one of the above.<|endoftext|> -TITLE: How to motivate and interpret the geometric solutions of Hamilton-Jacobi equation? -QUESTION [10 upvotes]: Studying the Hamilton-Jacobi equation, I meet a generalization of the notion of its solutions, which is found already in the work of Sophus Lie. - -For an H-J eqn, I mean a first order pde $H\circ dS=0$ in an unknown scalar function $S$ defined on a smooth manifold $M$, where $H\in C^\infty (T^\ast M,\mathbb{R})$. - -If $S$ is a solution then the image $\Lambda$ of its differential $dS$ is included in $H^{-1}(0)$ and has the following properties: - -$\Lambda$ is a lagrangian submanifold of $(T^\ast M,d\theta_M)$, -$\Lambda$ is transversal to the fibers of $\tau_M^{\ast}:T^\ast M\to M$, -the restriction of $\tau_M^{\ast}$ to $\Lambda$ is injective. - -Conversely, if a submanifold $\Lambda$ of $T^\ast M$, included in $H^{-1}(0)$, satisfies the properties 1, 2, and 3, then it is equal to the image of the differential of a solution, unique up to a constant. -But if a submanifold $\Lambda$ of $T^\ast M$, included in $H^{-1}(0)$, satisfies only the conditions 1 and 2, then, around each of its points, it is again equal to the image of the differential of a solution, but this can fail to holds globally. -The idea of Sophus Lie was to give up both conditions 2 and 3. - -Adopting this point of view, we define a generalized (or geometric) solution of $H\cic dS=0$ to be any lagrangian submanifold $\Lambda$ of $(T^\ast M,d\theta_M)$ which is included in $H^{-1}(0)$. - -I don't think that this generalization is only due to the sake of abstractness. -Infact, considering generalized solutions, it is possible, arguing with tecniques from symplectic geometry, to prove the local existence and uniqueness theorem, at the same time, for generalized and usual solutions. -But I am hoping to find "more" practical applications which illustrate the meaningfulness of geometric solutions. I would like to learn if ther is some physical or geometrical problem involving an H.-J. eqn, whose comprehension is sensibly augmented by the consideration of generalized solutions. So my question is: - -What are the possible arguments and applications that motivate and help to interpret the notion of geometric solutions for an Hamilton-Jacobi equation? - -REPLY [4 votes]: The famous KAM tori arose out of HJ considerations. -They are Lagrangian torii. They were found by attempting to solve -the HJ equation generally, and then finding one can only solve it -when certain appropriately irrational frequency conditions hold. - They occur in perturbations of integrable systems, or near `typical' linearly stable periodic orbits in a fixed Hamiltonian systems. -You can read about them in an Appendix to Arnol'd's Classical Mechanics, and -also get some idea from Chris Golé's book 'Symplectic Twist Maps', -or from Siegel and Moser's 'Stable and Random Motion'.<|endoftext|> -TITLE: Surface equivalent of catenary curve -QUESTION [19 upvotes]: A catenary curve -is the shape taken by an idealized hanging chain or rope under the influence -of gravity. It has the equation $y= a \cosh (x/a)$. -My question is: - -What is the shape taken by an idealized, thin two-dimensional sheet, - pinned on a plane parallel to the ground, under the influence of gravity? - -The answer surely depends on how it is pinned to the plane, the boundary -conditions. -Natural options are: - -A disk sheet fixed to a circle. -A square sheet fixed to a square. -A square sheet pinned at its four corners. - -The middle option above would look something like this when inverted: - -           - - -           - -(Image by Tim Tyler at hexdome.com.) - - -I don't think any of these shapes is a -catenoid, -which is the surface of revolution formed by a catenary curve. -Is there a simple analytic description of any of these surfaces, -analogous to the $\cosh$ equation for the catenary curve? -I have been unsuccessful in finding anything but simulations of solutions -of the differential equations. -This question arose in imagining a higher-dimensional version -of the property that an inverted catenary supports smooth rides of a square-wheeled -bicycle -(explored in this MO question). -Thanks for pointers! - -REPLY [16 votes]: A model equation for an inextensible, flexible, heavy surface in a gravitational field was deduced by Poisson Lagrange and later the problem was also studied by Poisson (see the references in the linked papers below). The equilibrium condition for a hanging heavy surface of constant mass density reads -$$\sqrt{1+|\nabla u|^2}\ \nabla\cdot{}\frac{\nabla u}{\sqrt{1+|\nabla u|^2}}=\frac{1}{u+\lambda},\qquad x\in\Omega\subset\mathbb R^2,\qquad\qquad(1)$$ -where $u=u(x)$ is the vertical displacement and $\lambda\in\mathbb R$ is an arbitrary constant (a Lagrange multiplier). (1) is the Euler equation of the variational integral -$$I(u)=\int_{\Omega}u\sqrt{1+|\nabla u|^2}dx,$$ -which can be interpreted as the vertical coordinate of the center of gravity of the surface -$$\mbox{graph}(u)=\{(x,u(x)):\ x\in\Omega\}\subset\mathbb R^2\times\mathbb R.$$ -Equation (1) is to be supplemented with the requirement that the surface has a prescribed area $A$ -$$\qquad\qquad\qquad\qquad\qquad\int_{\Omega}\sqrt{1+|\nabla u|^2}dx=A,\qquad\qquad\qquad\qquad\qquad\qquad\quad\quad(2) $$ -and the Dirichlet boundary condition describing the curve from which the surface is being suspended -$$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\left.u\right|_{\partial \Omega}=g.\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad(3) $$ -One can check formally that a solution to (1)-(3) provides a graph of a heavy surface of prescribed area and boundary with the lowest center of gravity, so this is a precise 2D analogue of the classical catenary problem. -It is known that problem (1)-(3) has no classical solutions for the values of area $A$ outside of some bounded interval $[A_{\min},A_{\max}]$. Moreover, the corresponding variational problem has no global solutions for all $A\in\mathbb R$. A short survey of some old and relatively new results concerning well-posedness of (1)-(3) and its multidimensional analogues can be found in the paper by Dierkes and Huisken, "The N-dimensional analogue of the catenary: Prescribed area", in J. Jost (ed) Calculus of Variations and Geometric Analysis, Int. Press (1996), pp. 1-13. -Addendum. Here is a more recent survey by Dierkes: "Singular Minimal Surfaces" (in Geometric Analysis and Nonlinear Partial Differential Equations, Springer (2003), pp. 177-194).<|endoftext|> -TITLE: Ramification in p-division fields associated to elliptic curves with good ordinary reduction -QUESTION [11 upvotes]: Dear MO, -Let $p$ be a prime and let $E/\mathbb{Q}_p$ be an elliptic curve. Suppose that $E/\mathbb{Q}_p$ has good ordinary reduction at $p$. In his 1972 paper ``Propriétés galoisiennes des points d'ordre fini des courbes elliptiques'' (more specifically, see Corollaire, in p. 274), Serre shows along the way that the inertia subgroup of $\operatorname{Gal}(\mathbb{Q}_p(E[p])/\mathbb{Q}_p)$ is, with respect to a suitable basis of $E[p]$, isomorphic to either a matrix group of the form {$[\ast\ 0; 0\ 1]$} or {$[\ast\ \ast; 0\ 1]$} as a subgroup of $\operatorname{GL}(2,\mathbb{F}_p)$. After this result, Serre remarks that he doesn't know of any simple criterion that would determine whether one is in the first case or the second case. -Question: Nowadays, do we know of a criterion to tell whether one is in the first case or the second case? -A more concrete question: Here is the particular example that I am working with: Let $E/\mathbb{Q}$ be ``1225h1'' in Cremona's tables, given by -$$E : y^2 + xy + y = x^3 + x^2 - 8x + 6. $$ -This curve has a rational $37$-isogeny and therefore $\operatorname{Gal}(\mathbb{Q}(E[37])/\mathbb{Q})$ is a Borel subgroup of $\operatorname{GL}(2,\mathbb{F}_{37})$. The curve $E$ has good ordinary reduction at $p=37$ and I am trying to find out whether the ramification index of $37$ in the extension $\mathbb{Q}(E[37])/\mathbb{Q}$ is just $\varphi(37)$ or rather $\varphi(37)\cdot 37$, where $\varphi$ is the Euler phi function. -The $37$th division polynomial of $E/\mathbb{Q}$ has degree $684$ and it factors (over $\mathbb{Q}[x]$) as a product of $4$ polynomials of degrees $6$, $6$, $6$ and $666$, respectively. The extension of degree $666$ is, well, diabolically large and I can't find the ramification at $37$ computationally... or at least I don't know how to! -Thanks for your help! - -REPLY [12 votes]: Assume $p \ne 2$. The condition for the representation to be tamely ramified (i.e $* = 0$ in the upper right entry of the matrix) is that $j(E) \equiv j_0 \mod p^2$ where $j(E)$ is the $j$-invariant of $E$ and $j_0$ is the $j$-invariant of the canonical lift of the reduction of $E$. This is proved in Gross "A tameness criterion for galois representations..." Duke J. 61 (1990) on page 514. For $p=2$ you need the congruence modulo $8$. Serre gives an algorithm for computing $j_0$ in Lubin-Serre-Tate.<|endoftext|> -TITLE: Reference sought for Conway's observation on stable matchings -QUESTION [5 upvotes]: Looking for a reference on the observation that the set of stable matchings form a distributive lattice. This is attributed to Conway by Knuth in "Marriages Stables" but I would like an explicit reference if possible, likely a text book. An undergrad student needs it for his bibliography in a term paper. - -REPLY [5 votes]: strong text -Conway discovered it when visiting Montreal contemporaneously to Knuth's -lectures on the marriage problem. These lectures are in print, published by Centre de -Recherche Math of Universite de Montreal.<|endoftext|> -TITLE: Reference letters for teaching positions -QUESTION [11 upvotes]: I've been told that when applying for a teaching position, your reference letters can be written by anyone who is familiar with your teaching capabilities in detail. I feel that this primarily just means students, but I wonder if there's some unspoken rule that reference letters should come from people in positions of authority, e.g. professors for whom I've served as TA, administrative staff in the math department, etc. In reality, it's the students who know my teaching capabilities in detail, and perhaps to a lesser degree my friends, whereas professors and administrators have no direct knowledge my teaching abilities whatsoever, and might only have heard things here or there from students, or have read my student evaluations and seen the scores. - -So should reference letters predominantly come from authority figures, or is it okay to have them all come from students? - -REPLY [2 votes]: This is a slightly different response as most people are assuming you are intending to teach at the university level. I respond only for informational purposes for those teaching not at the university level. -As an undergraduate, I took an "advanced" math course with a Ph.D graduate student as the teacher. After finishing his degree, he ended up seeking prep school and community college positions out of interest to teach small groups of students. Not only did these schools ask for "authoritative" recommendations, but a number also asked for multiple student recommendations. -Admittedly, I was asked to write one -- I do not think I really had the perspective then, nor now, to write something. But it seems these less than university positions desire that additional information.<|endoftext|> -TITLE: What is the complexity of this problem? -QUESTION [23 upvotes]: Recently on Dick Lipton and Ken Regan's blog there was a post about problems of intermediate complexity, that is, NP problems that are harder than P but easier than NP-complete. The main message of the post was that most examples of natural questions that have been conjectured to be intermediate have subsequently been shown not to be -- the two most notable exceptions being the discrete logarithm problem (and factoring), and the graph isomorphism problem. There have been related questions here on Mathoverflow and also on TCS stackexchange. -I asked about the status of one particular problem in a comment on the blog post but got no answer, so I'll ask it again here. The problem is the following. You're given a non-singular $n\times n$ matrix over $\mathbb{F}_2$ and asked to determine whether it can be reduced to the identity using at most m Gaussian row operations. It seems very unlikely that this problem is in P, since the problem of finding an explicit matrix that needs a superlinear number of operations is open (and, I think, thought to be hard). On the other hand, if I try to take a problem like 3-SAT and reduce it to this one, I don't have any idea where to start. But that is very weak evidence for the assertion that the problem is not NP-complete, so I'm asking again the question I asked in the blog comment: is anything known, or intelligently conjectured, about the status of this problem? -The question can be reformulated as asking for the length of the shortest path between two vertices in a certain Cayley graph. (NB, that doesn't make it polynomial-time because the number of vertices in the graph is exponential.) I'd be just as happy to be told that the same problem in a different Cayley graph was probably of intermediate complexity: I chose the Gaussian elimination example because I happen to like that graph. - -REPLY [5 votes]: If one relaxes the question to asking the row-reduction distance of arbitrary matrices over $\mathbb{F}_2$ then it can be shown that the problem is $NP$-Complete. That is, consider - -RRD (Row Reduction Distance): -Input: $m\times n$ matrices $M$, $N$ - over $\mathbb{F}_2$, and an integer - $k$ -Output: Whether $M$ can be row-reduces - to $N$ in $\le k$ steps - -The claim is that this problem RRD is $NP$-complete. It is within $NP$, so the hardness is all that remains. To do this, consider the following related problem. - -MHW (Min Hamming Weight): -Input: $P\in\mathbb{F}_2^{m\times -> n}$, $b\in\mathbb{F}_2^n$, integer $k$ -Output: Is $b$ expressible as a linear - combination of $\le k$ rows of $P$. - -I'll first show that MHW reduces to RRD, then show that RRD is $NP$-hard. This together will show RRD is NP-hard. -Let $(P,b,k)$ be an instance of MHW. Create $M$ and $N$ as $(m+1)\times n$ matrices, where $M$ is just $P$ with the zero row appended on, and $N$ is just $P$ with the row $b$ appended on. I'll now show that $b$ is expressible as a linear combination of at most $k$ rows of $P$ iff $M$ is row-reducible to $N$ it at most $k$ steps. -The forward direction of this claim is straightforward. Now for the backward direction. Row-reduction operations over $\mathbb{F}_2$ are just "add this row to that row", or "swap this and that row". It follows that in $\le k$ row-reductions have at most $k$ "source" rows of where the additions come from. Thus, the last row of $N$ is equal to the last row of of $M$ (which is zero) plus at most $k$ other rows of $M$. This is exactly what was wanted, as the last row of $M$ is just $b$. -This completes the proof that MHW reduces to RRD. -Now let's show that MHW is NP-hard. We'll do so with: - -SET-COVER -Input: Sets $S_1,\ldots,S_m\subseteq[n]$, integer - $k$, -Ouptut: Decide if $[n]$ is the union of $\le k$ of the sets $S_i$ - -It is know that Set-Cover is NP-complete. Here we need a slightly stronger version of this fact, where all of the sets are of constant size, and this is still NP-complete. (To see this, one first shows that 3SAT is still NP-complete when each variable appears in at most 3 clauses. Then one runs through the "standard" reduction from 3SAT to Set-Cover, and notices that all of the sets are of constant size). -Now consider a set-cover instance. Note that if we through in all subsets of each $S_i$, the answer to the cover question doesn't change. Note that we can do this as each subset is of constant size, so there aren't too many subsets to add. Thus we get - -Hered-Set-Cover -Input: Let $\mathcal{S}\subseteq -> 2^{[n]}$ be a family of sets, each of - size $O(1)$, that is subset closed. - Let $k$ be an integer. -Ouptut: Decide if $[n]$ is the union - of $\le k$ of the sets from - $\mathcal{S}$. - -and Hered-Set-Cover is NP-complete, as argued above. We'll now reduce Hered-Set-Cover to MHW. Take a Hered-Set-Cover instance, with the family of sets $\mathcal{S}$ and integer $k$. Suppose there are $m$ sets. Then write out $P$ to be the $m\times n$ matrx where the rows are the indicator vectors for the sets in $\mathcal{S}$. The target vector $b$ is the all ones vector, and $k$ is as in the original problem. So if $b$ is a $\le k$ linear combination of the rows, then this immediately gives the set-cover of $\le k$ sets. If there is a set-cover of $\le k$ sets, then we can always pass to subsets so that each element in the ground set is covered exactly once, and thus when we sum up the relevant vectors in $\mathcal{F}_2$ we never run into the issue that 2=0. -So really we are doing a "set-cover with odd covering at each vertex" in the MHW instance. The point is that allowing the subsets in the family makes the exact number of coverings irrelevant, and so we can assume things are covered exactly once. -It seems like there might be a more direct reduction from the problem Exact-Cover (where in set-cover we require that each element be covered exactly once). Indeed, I sort of just untangled the reductions needed to use Exact-Cover. But Exact-Cover doesn't seem exactly right, because if $b$ is a $\le k$ linear combination this doesn't immediately translate to an exact cover. -This approach doesn't seem to address the issue when $M$ and $N$ are full-rank in the RRD problem, as the reduction of MHW to RRD needs non-full-rank, and the MHW problem is solvable in polynomial time when $P$ is full rank.<|endoftext|> -TITLE: Finite-dimensional subgroups of diffeomorphism groups -QUESTION [8 upvotes]: This question is a generalization of my previous question about the circle to arbitrary manifolds. -Is there a smooth manifold M with the following property. -There exists a sequence of connected finite-dimensional subgroups Gi of M's diffeomorphism group G such that -(a) Gi is contained in Gj for i < j -(b) The union of Gi is dense in G -To remove doubt, "finite dimensional subgroup of M's diffeomorphism group" means a Lie group H with smooth faithful action on M. -The answer to my previous question established that S1 is not such a manifold. I suspect that the answer to the general question is still "no". However, the proof would have to be more sophisticated since in the case of S1 we had essentially a closed list of possible "Hs". -There is another closely related question. Fix a smooth manifold M. Consider connected Lie groups H with faithful and transitive smooth action on M. Is there an upper bound of H's dimension? For S1 the answer was "yes, 3". - -REPLY [10 votes]: I guess that the answer to your first question is no, based on the following: If the union of the $G_i$ were dense in $G=\mathrm{Diff}(M)$, then, presumably, for $i$ sufficiently large, the action of $G_i$ would be primitive (i.e., it would not preserve any nontrivial foliation) and locally transitive. The list of the effective, primitive, transitive Lie group actions is known, and, by examining this list, one sees that the dimension of such a group acting on an $n$-manifold is at most $n^2{+}2n$. -The answer to your second question depends on the manifold. First, I suppose you have to restrict to the case in which $M$ actually has a transitive smooth action of a finite dimensional Lie group. (For example, any compact orientable surface $M^2$ of genus $2$ or more is not a homogeneous space.) -It is not hard to come up with cases for which there is no upper bound. For example, let $M = \mathbb{R}^2$ and consider the group $G_d$ that consists of transformations of the form $\phi(x,y) = \bigl(x{+}a,\ y{+}p(x)\bigr)$ where $a$ is any constant and $p$ is any polynomial of degree $d$ or less. Then $G_d$ acts transitively on $M$ for all $d\ge0$ while $\dim G_d = d+2$. Thus, for $M=\mathbb{R}^2$ the dimension of such $H$ can be arbitrarily high. A similar argument with trig polynomials will provide such an example on the torus, which is compact. (N.B.: The action of $G_d$ is not primitive since it preserves the foliation by the lines $x=c$; thus, this example does not contradict my first paragraph.) -On the other hand, for $M=S^2$, there is an upper bound for the dimension of a connected Lie group that acts faithfully and transitively on $M$. That upper bound is 8 ($=2^2+2\cdot2$) and is achieved by $SL(3,\mathbb{R})$ acting on $S^2$ regarded as the space of oriented lines in $\mathbb{R}^3$. In fact, you can say more: Any connected, transitive finite dimensional Lie subgroup of the diffeomorphism group of $S^2$ is conjugate to one of $SO(3)$, $PSL(2,C)$, or $SL(3,\mathbb{R})$. The latter two are maximal and contain the first one as maximal compact. (The easiest proof that I know of these statements uses the classification of primitive actions, at least in dimension $2$.)<|endoftext|> -TITLE: Groups which satisfy Mal'cev's theorem (locally residually finite) -QUESTION [12 upvotes]: Recall that a group $G$ is residualy finite if for every non-zero element $g\in G$ there exists a homomorphism $\sigma:G\rightarrow H$ such that $H$ is finite and $\sigma(g)\neq 0$. Mal'cev's theorem says that if $k$ is a field then any finitely-generated subgroup of $GL_n(k)$ is residually finite. For a proof of this theorem, see Steve D (Smith?)s answer to this question MO:9628. Note that the theorem does not say that $GL_n(k)$ is residually finite. - -Does anyone know any other classes of groups with the property that any finitely generated subgroup is residually finite. - -I would prefer examples with the following two properties: first, the group is not itself residually finite, and second, it is not simply a subgroup of some $GL_n(k)$. So, this excludes, for instance, free groups. -Side question: is there a good name for this property? - -REPLY [4 votes]: There are uncountably many Tarski monsters, which are finitely generated simple infinity groups whose proper subgroups are cyclic. No infinite simple group is residually finite, so Tarski monsters aren't, but their proper subgroups obviously are residually finite. - -REPLY [3 votes]: Perhaps a locally finite simple group would help? How about the subset of all even permutations of the natural numbers with finite support? Unless I am misremembering something, this should be a simple locally finite group that is not a subgroup of GL_n(k). -Gerhard "Email Me About System Design" Paseman, 2011.07.09<|endoftext|> -TITLE: Program transformation as alternative for Hoare logic or temporal logic -QUESTION [6 upvotes]: When trying to prove something about a program, the known techniques are Hoare logic and temporal logics. -An alternative is to transform a program in a mathematical (logical) expression. So, rather that mathematics is used to prove some properties of the program, the program itself is a piece of mathematics. -Loops become transitive reflexive closures. Example, if one has a program that calculates a Fibonacci number. If the program keeps the last two numbers of the Fibonacci sequence in variables, then this be converted by taking the transitive reflexive closure of the relation P, that is true (and only true) for the following situation: -$$ P((x,\space y, \space z), \space (x+1, \space z, \space y+z)) $$ -In the original program, the right value is chosen within the loop. In the transitive reflexive closure, the right value must be selected outside the closure (loop). The transformed program is more like a non-deterministic program. -The transformation of a program in a logical expression, can be done automatically. -Although, this is not rocket science, I can not find any reference for this approach. I am busy with writing an article, where this is a part of (it is not the main subject). But I want to refer to the right articles and look if there is interesting material. -Does someone has interesting references? -Many thanks, -Lucas -Edit: Given the comment of Andreas, some clarification. The goal is to make formal reasoning about the program possible. So, transforming the program in a declarative language is insufficient, because the declarative language may not have means to make conclusions about a program, although the language itself might precisely defined. I was thinking in transforming the program in a FOL + PA expression. After such transformation, formal (that is why I tagged with lo.logic) reasoning can be done about the program. As far to my knowledge, I haven't seen this approach (the methods are always more in the direction of Hoare and temporal logics), although it is not very complicated. In my question I didn't want to restrict to FOL + PA. - -REPLY [4 votes]: Hoare logic and temporal logic might be "the only known techniques for proving programs correct" to you, but there are certainly others! -For example, and this list is not exhaustive: - -equational reasoning about fixpoints, this works in languages like Haskell -properties of programs can be proved via denotational semantics, which in itself is a vast area including domain theory and game semantics, to name just two. -for certain kinds of programs, for example for parametrically polymorphic ones, there are techniques that go under the name "relational parametricity" -you can use various logical interpretations to get correctness of programs: - -a program extracted as a realizer via the realizability interpretation of logic automatically satisfies a certain specification -with tools such as Coq you can use type theory to write programs as proofs, or construct programs and prove them correct all at once -there are other ways of extracting programs from logical statements, one family of which are variants of Gödel's Dialectica interpretation that extract programs from classical logic. - - -Now, regarding your specific question. I think you should look at realizability, type theory, and extraction of programs from proofs. All of these are "logical" methods for developing correct programs, or proving them correct. Some randomly chosen starting points: - -start with something fun and surprising, perhaps Paulo Oliva's tutorial on Programs from classical proofs via Gödel's dialectica interpretation -an accessible paper on realizability interpretation which uses logical methods in computable analysis might be Ulrich Berger's Realisability for Induction and Coinduction with Applications to Constructive Analysis -if you want to use computers to actually show correctness of programs, you could learn Coq and then proceed to Ynot (Hoare logic on steorids) or go straight to Adam Chipala's Certified Programming with Dependent Types. -cool people use Agda instead of Coq. -if you are first-order logic sort of person, you might find Minlog more palatable than Coq and Agda, as it does not throw type theory in your face. - -See you in two years.<|endoftext|> -TITLE: Changing field of study post-PhD -QUESTION [51 upvotes]: I am doing my PhD in algebraic graph theory, for not much more reason than that was what was available. However, I love deep structure and theory in mathematics, and I do not particularly want to be a graph theorist for the rest of my life. -I have heard of mathematicians changing from more theoretical subjects to broad ones like combinatorics, but not the other way round, and am concerned that this is because the time it takes to learn the requisite theory for deeper subjects is hard to come by after grad school. -Has anyone done this, or known of someone who has? Is it at all likely or possible I will be able to get a post-doc position in a subject only marginally connected to my thesis? I have been using small amounts of algebraic number theory, and would ideally like to go on and specialise in something similar. -Any advice much appreciated! -EDIT (June '12) - I came across this old question of mine, and now feel rather silly for having asked it at all. For the record, and for anyone who might be having similar doubts to me: a year or so on I have realised that, as was mentioned below, I should not pigeonhole myself, and that the most interesting problems are often those that lead to other seemingly disparate subfields. Perhaps more to the point, I actually can't now think of a subject I'd rather do than algebraic graph theory! If you find yourself in the position I was in, I advise you to just look for those problems you find most interesting in your "chosen" field - they are bound to lead to other areas. And anyway, surely all that matters is that you are interested and inspired?! In my own research I have used number theory, galois theory, group theory, and even a bit of probability. I might even opine that graph theory is one of the subjects most likely to appear in intra-disciplinary work, which seems to be forming an ever higher proportion of mathematical research. - -REPLY [42 votes]: Speaking as someone whose thesis was also in algebraic graph theory but who has later gone on to do research in other areas, I would say that it is definitely possible to switch fields. The main skills you need are management skills: the ability to manage your own time so that you can spend some time learning a new field while also producing something that others will value, and the ability to manage other people's view of your work. In this regard, I believe that Deane Yang's comment is right on the money. You probably can't afford to drop your initial specialty abruptly and produce no results while you retrain yourself. But if you manage things carefully then anything is possible, regardless of whether you have tenure or are switching to a field that requires a lot of background study. -There are some things you can do to help you solve these management problems. If you can find overlap between your current field and your new field, that will obviously help. Mathematics is so interconnected that this is usually not too hard; in the specific case of algebraic graph theory versus algebraic number theory, the first topic to come to mind is the Ihara zeta function of a graph, but I'm sure there are others. -In my case, I decided it would help to switch from academics to industry/government. I personally found it easier in industry/government to spend x% of my time producing results that pleased my employers and spending the remaining (100-x)% of my time training myself in a new area. Your mileage may vary, of course; in my case, I found that teaching drained me of too much of my energy, but this is not true of everybody. -One last comment I have is that if you take this route, then you will need the ability to maintain a clear sense of your own identity and goals and not be unduly swayed by other people's categorizations. For example, when I switched out of academics, many others regarded me as "leaving mathematics." In fact I was only leaving academics and not leaving mathematics, and it was important for me to ignore other people's view of the matter. As another example, people will want to pigeonhole you as a "something-ist" (and it seems you have been influenced by this point of view, since you use the phrase "being a graph theorist for the rest of my life"); you should resist this pigeonholing, and instead think of yourself just as someone with certain abilities and interests. Thinking of yourself as either a graph theorist or a number theorist is unnecessarily limiting. (Of course you may need to bill yourself as one or the other for the purposes of managing other people's view of you while you're making a transition, but you should not necessarily believe your own propaganda.)<|endoftext|> -TITLE: Lie groups admitting flat (bi)invariant metrics. -QUESTION [8 upvotes]: I would like to see an example of a non-abelian compact lie group admitting a bi/left/right-invariant flat metric. - -Is there any non-abelian compact lie group admitting a flat metric that is bi or one-sided invariant? - -The motivation is the following: as a manifold a torus does admit a flat metric, but it is also a abelian group. The first impression would be that the triviality of the lie algebra could imply the existence of a flat metric, or even be a necessary condition. Of course it is reasonable to ask for some relationship between the metric and the algebraic structure; thus some invariance would be expected (even though I would be happy to see a less restrictive assumption and a more powerful restriction dictated by the algebraic structure in the compact case). -Thus one could ask for an example of a non-abelian lie group admitting a flat metric. -A easy counter example (flat $\Rightarrow$ abelian) is the group of affine transformations of the line. The connected component of the identity is diffeomorphic to the real plane, so as a manifold (forgetting about the group structure) it does admit a flat metric. But if one considers a one-sided invariant metric one gets the Lobachevsky plane. -Since I could not think of a counter example of a non-abelian compact lie group admitting a flat metric (not necessary related to the group structure), I will also be happy to see a counter example of it. I will keep the invariance of the metric on the question though. -P.S.: Since for bi-invariant metrics $R(X,Y)Z=\frac{1}{4}[[X,Y],Z]$ (see e.g.: do Carmo, Riemannian Geometry), it is true in this case that abelian $\Rightarrow$ flat. -EDITED -Due to Igor's comment I will also include another question: - -Is there any non-abelian lie group admitting a flat metric that is bi or one-sided invariant? - -REPLY [2 votes]: I'm a few years late with my answer, but if you're interested in the pseudo-Riemannian case as well, there is a structure theorem due to Baues for Lie algebras with invariant scalar products, see Theorem 5.15 in -http://arxiv.org/abs/0809.0824<|endoftext|> -TITLE: Why is this theorem (about $L(P(\omega_1))^V$ and $L(P(\omega_1))^{V[G]}$) nice? -QUESTION [9 upvotes]: I was recently told that the following (due to M. Viale) is a nice theorem: - -Suppose there are arbitrarily large supercompacts, and $\mathrm{MM}$ holds in $V$. Let $G$ be generic for a proper forcing and $V[G] \vDash \mathrm{MM}$. Then $L(P(\omega_1))^V$ is elementarily equivalent to $L(P(\omega_1))^{V[G]}$. - -My question (borne of ignorance, not skepticism) is: - -Why is this theorem nice, and how does it fit into the bigger picture? - -Some slightly-more-specific questions that refine my main question are: Do the hypotheses of this theorem often come up in natural settings? What's the upshot of the conclusion? Is it that proper forcing which preserves $\mathrm{MM}$, leaves the theory of a small but not-that-small chunk of the universe unchanged? - -REPLY [11 votes]: Let me give a brief answer as the author of the mentioned theorem. I shall first say that (as you might imagine) I was very pleased to see that my theorem raised your interest. Now, as mentioned by Joel in his latest post, your statement of the theorem is not correct. However -there is a typo also in Joel's post, namely the conclusion of the theorem holds for the Chang model $L([Ord]^{<\omega_2})$ and not for the Chang model $L([Ord]^{<\omega_1})$ as posted by Joel. -For this Chang model the result is due to Woodin and was already known, it appears for example in -Chapter 3 of Larson's book on the stationary tower forcing. To complement the very good answer Joel has already given you, I invite you to read the many survey papers related to the philosophical position subsumed by this theorem, here is a sample list: -Woodin's two short papers for the "notices of the AMS" where he exposes the basic ideas behind the generic absoluteness results for $L(\mathbb{R})$: - -Woodin, W. Hugh (2001a). "The Continuum Hypothesis, Part I" (PDF). Notices of the AMS 48 (6): 567–576. http://www.ams.org/notices/200106/fea-woodin.pdf. -Woodin, W. Hugh (2001b). "The Continuum Hypothesis, Part II" (PDF). Notices of the AMS 48 (7): 681–690. http://www.ams.org/notices/200107/fea-woodin.pdf. - -Several of Peter Koellner's papers available at his webpage: -http://www.people.fas.harvard.edu/~koellner/ -Most of them contains a long introductory part which motivates and explains very carefully and plainly the ideas at the heart of the $\Omega$-logic approach to absoluteness result. -It has to be noted that the kind of solution to the continuum problem prospected by my theorem follows Woodin's view as exposed in the papers on AMS notices. Currently Woodin has pursued a different approach towards the solution of the continuum problem that leads him to prospect a view of the universe (Ultimate $L$) which is radically different from the one given by MM. -Finally in case you are interested there are in my webpage several slides of talks I gave on this and related subjects, as well as a proof of the theorem you have mentioned in your question: -http://www2.dm.unito.it/paginepersonali/viale/index.html -(unfortunately my university is currently changing the websites locations, so for some time you may have troubles to consult it....)<|endoftext|> -TITLE: The rain hull and the rain ridge -QUESTION [8 upvotes]: Rain falls steadily on an island, a 2-manifold $M$, which you may -assume, as you prefer, -is: (a) smooth, or (b) a PL-manifold, or perhaps even -(c) a -triangulated irregular network (TIN). -After a time, $M$ is saturated, in the sense that every raindrop -drains into the ocean rather than filling yet-unfilled crevices or basins. -At this point, we have what I will dub the rain hull of $M$, $H_R(M)$, -a uni-directional version of the the reflex-free hull, -explored (by Bill Thurston) in -this MO question. -Q1. -How difficult is to compute the rain hull $H_R(M)$? -My sense is that it might be quite difficult, because it seems -there can be nonlocal influences, as crudely depicted -in this side-view schematic: -Update (13Aug11): I have corrected the figure to more accurately reflect -physical reality. Thanks to Oswin Aicholzer for setting me straight. - -                       - - -Perhaps the computation is NP-hard if $M$ is presented as a PL-manifold? TINs have special properties that might -render the computation polynomial. -Update. -Joel Hamkins has convincingly argued (see below) that the computation is polynomial-time. -Let us assume we have $\overline{M} = H_R(M)$ computed or given. -A raindrop falling on -$p \in \overline{M}$ might follow a unique trickle path -(that is the technical term: e.g., see -"Implicit Flow Routing on Triangulated Terrains" -by deBerg et al.) -to the ocean, or the drop may randomly 'fracture' to follow distinct paths -to the ocean. -Define the rain ridge (my terminology) $R(\overline{M})$ to be the complement of the points of $\overline{M}$ -that have a unique trickle path. -So points on the rain ridge are akin to points on a cut locus, in that -they have two or more distinct paths to $\partial \overline{M}$. -They are, in a sense, continental-divide points, a topic -explored in -this inadequately -answered MO question (inadequately answered by me). -Q2. -What can be said about the structure of the rain ridge $R(\overline{M})$? -Unlike the cut locus, it is not always a tree. All the points in a filled basin are in the rain ridge, for when a raindrop lands in a filled basin, it is -natural to assume it "spreads out" and spills in equal portions over every boundary point of the basin. -But surely there are substantive properties to investigate. Surely the rain ridge $R(\overline{M})$ cannot be -an arbitrary subset of $\overline{M}$? -I finally come to my main question, which I fear has a negative answer: -Q3. -Can an extended metric be assigned to $\overline{M}$ so that its -geodesics are its trickle paths? -An -extended metric -is one that permits $d(x,y) = \infty$ -(e.g., for points not on the same trickle path). -What I am hoping for here is a way to view the rain ridge as a cut locus -of $\partial \overline{M}$, and then apply -a century of knowledge on the cut locus to the rain ridge. -Partly baked ideas, subquestion observations, and random literature pointers all welcomed! -My sense is that the considerable applied-math literature on -watersheds has not approached these questions in their full mathematical generality, -leaving room for delightful theorems. - -REPLY [4 votes]: Since you've explicitly welcomed half-baked ideas... -Regarding question 1, can't we just simulate the rain in polynomial time? -I am thinking of the same-connected-component problem in graph theory. Given two nodes $p$ and $q$ in a graph, we determine whether they are in the same component as follows: color $p$ blue, and then make passes through the graph, coloring blue any node that is directly connected to an already-blue node. Repeat until no new blue nodes arise, and then the component of $p$ consists precisely of the blue nodes. This takes polynomial time in the number of vertices, since the number of passes is bounded by the number of vertices. -In your case, if we imagine a discrete version of the picture, with tiny finite elements and the edge resolution size $n$, that is, an $n\times n\times n$ cube, then can't we simply simulate the rain until the rain hull is saturated? We can follow a single drop along its trickle path, which always has a downward (or at worst horizontal) component by gravity. The complication, as your diagram suggests, are the non-local effects of rain filling up a basin which overspills a border far away, but it seems that this overspill can still be computed in polynomial time. That is, when a drop is added to a local pool, then we allow it to flow horizontally as much as it likes, and we allow it also to flow to nodes at the same altitude, provided there is a blue path from its current location to the new location. Since there are only $n^3$ locations altogether, we can determine whether it is collected or shed in polynomial time. -And if the trickle path of a single rain drop can be computed in polynomial time of $n$, then we simply repeat the process of adding raindrops, from each of the $n^2$ sources overhead, until no new drops are added permanently to the rain hull. Since we can collect at most $n^3$ drops of rain altogether, we get a polynomial bound on the number of simulated raindrops we need to consider. And therefore in the end we can compute the rain hull in polynomial time of $n$. -(Perhaps you will object that your intended input will be much smaller than $n$.)<|endoftext|> -TITLE: What sheaf topoi classify: attribution request -QUESTION [7 upvotes]: Is there an accepted name or attribution by which to refer to the following well-known theorem? - -If C is a small site, then the topos of sheaves on C is the classifying topos for flat cover-preserving functors out of C. - -In the case when C has a trivial topology, the corresponding assertion for its presheaf topos is usually called Diaconescu's theorem. But I don't think I have ever seen a name given to the more general version. - -REPLY [3 votes]: If the site has finite limits, then this can be found in SGA4 Proposition 4.9.4.<|endoftext|> -TITLE: Algorithm for solving systems of linear Diophantine inequalities -QUESTION [8 upvotes]: So, I posted on StackOverflow looking for a reasonably fast algorithm to solve systems of linear Diophantine inequalities and was pointed to this article by Cheng-Zhi Gao and Yu-Lin Dong. The problem is, they give the algorithm on pages 350-351, but part of step (3) appears to be missing. -My question for MathOverflow is, therefore, whether anyone knows either of another such algorithm or has an idea as to what the missing part of Gao and Dong's algorithm is. - -REPLY [6 votes]: GAP provides a function NullspaceIntMat which solves systems -of linear diophantine equations. The documentation says: -25.1-2 SolutionIntMat - -* SolutionIntMat( mat, vec ) ───────────────────────────────────── operation - -If mat is a matrix with integral entries and vec a vector with integral -entries, this function returns a vector x with integer entries that is a -solution of the equation x * mat = vec. It returns fail if no such vector -exists. - -──────────────────────────────── Example ───────────────────────────────── - gap> mat:=[[1,2,7],[4,5,6],[7,8,9],[10,11,19],[5,7,12]];; - gap> SolutionMat(mat,[95,115,182]); - [ 47/4, -17/2, 67/4, 0, 0 ] - gap> SolutionIntMat(mat,[95,115,182]); - [ 2285, -5854, 4888, -1299, 0 ] -──────────────────────────────────────────────────────────────────────────── - -The source code can be found in the file lib/matint.gi included in the GAP distribution. -There is also a function SolutionNullspaceIntMat which additionally -computes a basis of the integral nullspace of the given matrix.<|endoftext|> -TITLE: Titles composed entirely of math symbols -QUESTION [36 upvotes]: I apologize for burdening MO with such a vapid, nonresearch question, but -I have been curious ever since -Suvrit's popular October 2010 -Most memorable titles MO question -if there were any "$E=mc^2$-titles," as I think of them—how Einstein in retrospect might have entitled his 1905 paper -(instead of -"Zur Elektrodynamik bewegter Körper"!)—paper/book titles composed entirely of math symbols. -There are two close misses in the responses to that MO question: -Connes et al.'s -"Fun with $\mathbb{F}_{1}$", -and Taubes's -"${\rm GR}={\rm SW}$: Counting curves and connections." -The only title entirely composed of math symbols with which I'm familiar is the delightful book A=B, by Marko Petkovsek, Herbert Wilf, and Doron Zeilberger. -Can you identify others? -Please interpret this question in a weekend-recreational spirit! :-) - -REPLY [6 votes]: MIP*=RE -Zhengfeng Ji, Anand Natarajan, Thomas Vidick, John Wright, Henry Yuen. -arXiv Abstract. 13 Jan 2020. - -...a negative answer to Tsirelson's problem... our results provide a refutation of Connes' embedding conjecture...<|endoftext|> -TITLE: Doubts on Reproducing Kernel Hilbert Spaces and orthogonal decomposition -QUESTION [5 upvotes]: I'm a CS student and I'm trying to learn RKHS theory to understand the passages made in this paper . -Among the bibliography I'm using there are "On the mathematical fundamentals of learning" and "Learning with Kernels". -I think I've got a good grasp of the relevant theory, but there's still something that bugs me. -As far as I understood a RKHS is a Hilbert space $H\subseteq \mathbb{C}^X$, where $X$ is a generic set of objects, with inner product $\langle f,g\rangle=\int_X\int_X \alpha(x')\beta(x)k(x,x')dxdx'$ with $\alpha,\beta \in \mathbb{C}^X$, $f=\int_X\alpha(x')k(x,x')dx'$ and $g=\int_X\beta(x')k(x,x')dx'$ such that $H=\overline{\mathrm{span}\{k_x|x\in X\}}$ and $\langle f,k_x\rangle=f(x)$ where $k_x=k(.,x)\;\forall x\in X$ and $k$ is the (Mercer) kernel for $H$. Let's call this "continuos-whole" definition. -However, given $D=\{x_1,x_2,...x_n\}\subset X$ a subspace $H_D\subset H$ could be defined by restricting the definition above to $H_D=\overline{\mathrm{span}\{k_{x_i}|x_i\in D\}}=\{f\in \mathbb{C}^X|f=\overset{i=s}{\underset{i=1}{\sum}} \alpha_ik(x,x_i)\quad \alpha_i\in \mathbb{C} \;, r\in \mathbb{N}\}$ and thus define $\langle f,g\rangle_{H_D}=\overset{i=s}{\sum_{i=1}}\overset{j=s}{\sum_{j=1}}\alpha_i\beta_jk(x_i,x_j)$. Let this be the "discrete-finite" definition -Assuming this is correct, the references tend to define $H$ as both being spanned by $\{k_x|x \in X\}$ and having inner product $\langle .,.\rangle_{H_D}$, and that would be possible if and only if $\overline{\mathrm{span}\{k_x|x\in X\}}=\overline{\mathrm{span}\{k_{x_i}|x_i\in D\}}$ which seems bogus to me. -Then, regarding the article, there's a passage of which I'm unsure. Let $H$ be a RKHS, i.e. a Hilbert space s.t. every point-evaluation functional is bounded, which would entail a "whole-continuous" definition. Then $H$ can be orthogonal decomposed in $H=H_D\oplus H_D^\bot$ where $H_D^\bot \bot H_D \rightarrow \langle f,g\rangle=0 \forall f\in H_D \;\wedge\; g\in H_D^\bot$. The paper says that $H_D^\bot=\{g|g(x_i)=0\forall x_i\in D\}$. Starting from the definition of orthogonality, I carried out the following proof: -$\forall f \in H_D \;\wedge\; g \in H_D^\top$ $\langle f,g \rangle=\int_X\int_X \alpha(x')\beta(x)k(x,x')dxdx'=\int_X \beta(x)f(x)dx$ $=\sum_{i \in 1..s}\alpha_i\int_X \beta(x)k(x,x_i)dx=\sum_{i \in 1..s} \alpha_i g(x_i)=0 \forall \alpha_i \in \mathbb{C}\rightarrow g(x_i)=0 \forall i\in 1..s$, for which I applied the reproducing property and the definiton of $g$ in this order. is this correct? -EDIT:I know the definition via the Riesz theorem, however I'm referring to the definiton I found in "Learning with kernels" (pg. 36): -"Definition 2.9 (Reproducing Kernel Hilbert Space) Let $X$ be a nonempty set and by $H$ a Hilbert space of functions $f :X\rightarrow\mathbb{R}$ . Then $H$ is called -a reproducing kernel Hilbert space endowed with the dot product $\langle .,. \rangle$ (and the norm $||f|| : \sqrt \langle f ,f\rangle $ ) if there exists afunction $k :X\times X \rightarrow \mathbb{R}$ with the following properties: - -$k$ has the reproducing property -$\langle f, k(x,. )\rangle= f(x)$ for all $f\in H$ ; -$k$ spans $H$ , i.e. $H=\overline{\mathrm{span}\{k(x,.)| x \in X\}}$ where $\overline{X}$ denotes the completion of the set $X$" - -, and, to my understanding, it is based on the theorems characterising RKHS on Mercer kernels (see pg. 35 of "On the mathematical foundations of Learning"). -The approach they take, as far as I've understood, is inside-out: they start from a Lp space $H=\overline{\mathrm{span}\{k(x,.)| x \in X\}}$ with $k$ being a (Mercer) kernel, for which define a suitable inner product to get a Hilbert space that is also a RKHS. They defined a Mercer kernel to be a function $X^2\rightarrow \mathbb{R}$ that gives rise to a definite positive Gram matrix $K_{ij}=k(x_i,x_j)$ for every $D=\{x_1,x_2..x_n\}\subset X$. - -REPLY [3 votes]: To answer your second question: -$H_D = \operatorname{span}(k_x : x \in D) $ is a finite dimensional (and hence closed) subspace of $H$. A function $f \in H$ is orthogonal to each $k_x, x \in D$ precisely when $\langle f , k_x \rangle = f(x) = 0$. In other words, $H_D^\perp = ( f \in H : f(x_i) = 0, x_i \in D)$ and of course $H = H_D \oplus H_D^\perp$.<|endoftext|> -TITLE: Is there a Serre Tor formula for nonproper intersections? -QUESTION [14 upvotes]: Background: Let $X$ be a smooth complex projective algebraic variety, and let $V$ and $W$ be closed subvarieties. For simplicity, let's assume that $\dim V+\dim W=\dim X$. -Now Serre's famous Tor formula says that if $V\cap W$ has dimension zero, we have: -$$V\cdot W=\sum_{Z\subset V\cap W}\sum_{i=0}^\infty(-1)^i\operatorname{length}_{\mathcal O_{X,z}}\operatorname{Tor}_{\mathcal O_{X,z}}^i(\mathcal O_{X,z}/I_V,\mathcal O_{X,z}/I_W)$$ -where the sum is over the irreducible components of $V\cap W$ (in this case, a finite number of points). -However (according to Wikipedia here), if $Z$ is an irreducible component of $V\cap W$ with positive dimension (this is called a nonproper intersection), then the alternating sum of $\operatorname{Tor}$'s, which I'll call $\mu(Z;V,W)$, is zero. Unfortunately, this means it cannot be used in exactly the same way as before to calculate $V\cdot W$. For example, if $V=W$, then the answer would always be zero, though certainly there exist half-dimensional varieties $V$ with $V\cdot V\ne 0$ (in fact it can even be negative). -Is it possible to remedy this situation? How does one count the intersection multiplicity if the intersection is not proper? -Comment: I know how to compute the self-intersection $V\cdot V$ as the top chern class of the normal bundle evaluated on $[V]$. I'm looking here for an answer that's the most general, i.e. applies to all $V$ and $W$ of complementary dimension, regardless of the dimension of their intersection. - -REPLY [11 votes]: Let $ch_i(F)$ denote the $i$-th coefficient of the Chern character of a sheaf $F$. Then -$$ -V\cdot W = \sum_{i=0}^n (-1)^i ch_n(Tor_i(O_V,O_W)), -$$ -where $n = \dim X$, $O_V = O_X/I_V$, $O_W = O_X/I_W$, and one uses a natural identification of $H^{2n}(X,{\mathbb Z})$ with ${\mathbb Z}$.<|endoftext|> -TITLE: Does random matrix theory make any prediction for the eigenvalue distributions of compact Riemann surfaces? -QUESTION [7 upvotes]: Under RH, Montgomery has proven equidistribution results for the zeros of the Riemann Zeta function, which suggest a close connection of the distribution to certain results in Random matrix theory. Analogues have been proven for zeta function associated finite fields unconditionally. -The Riemann zeros with imaginary part less than $T$ grow like $T \log T$. The zeros of the Selberg zeta function of a compact surface, which are connected to the eigenvalues of the Laplace Beltrami operator, grow roughly like $T^2$, but here the Riemann hypothesis is true except for possible zeros with imaginary part being zero. -Nevertheless, I dare to ask, if there is something similar available for these eigenvalue? - -REPLY [16 votes]: You can look at the following survey by Peter Sarnak: -http://www.math.princeton.edu/sarnak/Arithmetic%20Quantum%20Chaos.pdf -Basically the prediction is that the eigenvalue distribution is Poisson for arithmetic surfaces and GOE for non-arithmetic surfaces. There are some partial results supporting Poisson in the arithmetic case, in particular by Luo and Sarnak. -This survey has a lot of really cool stuff, but is quite dated by now. For one, it does not include the solution to the quantum unique ergodicity conjecture (which among other things got Lindenstrauss the Fields medal).<|endoftext|> -TITLE: Splitting a polynomial with one root -QUESTION [8 upvotes]: Suppose we have an irreducible polynomial $f\in K[x]$. Is there some way to sometimes tell whether $f$ splits completely after adjoining just one root of $f$ to $K$? -I am mostly interested in the case where $K$ is a function field $\mathbb{F}_{q}(t_{1},\ldots,t_{m})$ over some finite field, so it might not be feasible to explicitly compute roots. - -REPLY [5 votes]: Here is the best I can come up with. Consider an algebraic closure $\bar K$ of $K$ and a root $\alpha \in \bar K$. -The number of roots of $f$ in $K(\alpha)$ doesn't depend on $\alpha$: call it $ r(f)$ -Moreover call $s(f)$ the number of the different subfields $K(\alpha)\subset \bar K$ obtained by adjoining roots of $f$ to $K$. Then you have the pleasant equality $$deg(f)=r(f).s(f)$$ -This shows in particular that the number of roots that you get by just adjoining one root divides the degree $deg(f)$ of your polynomial. -For example if $K=\mathbb Q$ and $f(x)=X^8-2$ you have $r(f)=2$ and $s(f)=4$, since the fields you get by adjoining roots of $f$ to $\mathbb Q$ are [with $\omega =\frac{1}{\sqrt 2}(1+i)$]: -$\mathbb Q(\sqrt[4]2)$ -$\mathbb Q(\pm \omega \sqrt[4]2)$ -$\mathbb Q(\pm \bar{\omega} \sqrt[4]2)$ -$\mathbb Q(\pm i \sqrt[4]2)$ -These results are due to Perlis , and although not difficult have found their way in exactly zero books, as far as I am aware.<|endoftext|> -TITLE: Is every algebraic extension of a field of absolute transcendence degree one a separable extension of a purely inseparable extension? -QUESTION [13 upvotes]: Any decent course on field theory will state that in characteristic $p$ an extension of fields $k\subset K$ canonically decomposes as the tower $k\subset K_{sep}\subset K$ with -$K$ purely inseparable over its subfield $K_{sep}$ of elements separable over $k$. -The alert student will then ask if the order can be reversed, that is whether $K$ is separable over its subfield $K_{perfect}$ of elements purely inseparable over $k$, and be told that the the answer is no. -However in all the counterexamples I am aware of, the base field is of absolute transcendence degree two: $trdeg_{\mathbb F_p}(k)=2 $ ( usually $k=\mathbb F_p(x,y)$, the purely transcendental field in two indeterminates over $\mathbb F_p$). -Of course no counterexample can be found with $k$ of absolute transcendence degree zero, because $k$ would then be algebraic over $\mathbb F_p$ and thus perfect. -I have the suspicion that no counterexample can be found with $k$ of absolute transcendence degree one either and that is what I'd like to ask here: -Question Let $k$ be a field with $trdeg_{\mathbb F_p}(k)=1$ and $k\subset K$ an algebraic extension. Is the field $K$ separable over its subfield $K_{perfect}$ in the tower $k\subset K_{perfect} \subset K$ ? -Edit ulrich has solved the question ( with even more general hypotheses!) in the affirmative: congratulations and thank you, ulrich! -For reference purposes and because much information is in the comments to his answer, I'd like to sum up and display the main points of his subtle proof. -General result Let $F\subset k$ be an extension of fields with $F$ perfect and $trdeg_{F}(k)=1$.Then for any algebraic extension $k\subset K$ the field $K$ is separable over its subfield $K_{perfect}$ of purely inseparable elements over $k$. -[My question was the case $F=\mathbb F_p$] -Core result Let $F\subset E$ be an extension of fields with $F$ perfect and $trdeg_{F}(E)=1$. Then an algebraic extension $E\subset K$ is separable as soon as the only elements of $K$ purely inseparable over $E$ are already in $E$. -[This clearly follows from the general result and conversely ulrich reduces the general case to this core result by taking $E=K_{perfect}$ . This core result is false for transcendence degree $trdeg_{F}E \geq2$: cf. my answer, lifted from Bourbaki, here ] -Dimensional lemma If a field $L$ has transcendence degree $1$ over a perfect field $F$, then the purely inseparable extension $L\subset L^{1/p}$ has dimension $1$ or $p$ according as $L$ is perfect or not. -[Ulrich uses it to prove the core result. He considers the tower $E\subset L=E^{sep} \subset K$ and proves that if $L\neq K$, then we would have $L\subsetneq L^{1/p} \subset K$ thanks to the lemma . From there the contradiction to the hypothesis of the core lemma $E \subsetneq E^{1/p} \subset L^{1/p} \subset K$ obtains. (Notice that non-perfection of $L$ implies that of $E$, hence the asserted strict inclusion $E \subsetneq E^{1/p}$.) This dimensional lemma was proved in a particular case in this link furnished by Martin, and generalized by ulrich. This key technical lemma is exactly the reason why the transcendence degree one hypothesis is needed: the lemma is false for higher transcendence degree.] - -REPLY [12 votes]: Yes. One may also replace $\mathbb{F}_p$ by any perfect field $F$. -The reason is that any non-perfect algebraic extension $L$ of $k$ has a unique inseparable extension of degree $p$, i.e. $L^{1/p}$, the field obtained from $L$ by adjoining all $p$'th roots. This follows from the fact that $L^{1/p}$ is of degree at most $p$ over $L$. -Assuming this one can prove the claim as follows: Suppose it is not true. Then we can find subfields $L,M$ of $K$ so that we have inclusions $K_{perfect} \subset L \subset M \subset K$ so that $M$ is inseparable of degree $p$ over $L$. By the above remark we must have $M = L^{1/p}$ so we must have $K_{perfect}^{1/p} \subset L^{1/p}\subset K$. If $K_{perfect}^{1/p} = K_{perfect}$, then it is perfect so $L,M$ as above cannot exist. If not, then it contradicts the definition of $K_{perfect}$.<|endoftext|> -TITLE: (Sh,Sh-map) represents the category of sheaves on a stack. -QUESTION [5 upvotes]: I'm trying to understand the following theorem, but I don't think I'm reading it correctly. -Let $(\mathcal{C},J)$ be a site (with a subcanonical topology). Write $\mathcal{C}/X$ for the groupoid of objects over $X\in \mathcal{C}$. Let $\mbox{Sh}:\mathcal{C}^{op} \rightarrow \mbox{Gpds}$ be the functor taking $X$ to the category of sheaves on $\mathcal{C}/X$ and isomorphisms of sheaves, and let $\mbox{Sh-map}:\mathcal{C}^{op} \rightarrow \mbox{Gpds}$ be the functor taking $X$ to the category whose objects are sheaf morphisms $\mathscr{F} \rightarrow \mathscr{G}$ and whose morphisms are commuting squares of sheaves determined by isomorphisms $\mathscr{F}_1 \stackrel{\sim}{\rightarrow} \mathscr{F}_2$ and $\mathscr{G}_1 \stackrel{\sim}{\rightarrow} \mathscr{G}_2$. These are in fact both stacks on $\mathcal{C}$, and moreover they determine a category-object $(\mbox{Sh},\mbox{Sh-map})$ in the category of stacks. - -Theorem: The category of sheaves on a stack $\mathscr{M}$ is equivalent to the category of morphisms of stacks $\mathscr{M} \rightarrow (\mbox{Sh,Sh-map})$. That is, the objects are the 1-morphisms and the morphisms are the 2-morphisms. - -I'd like to interpret this to mean that the objects of $Shv(\mathscr{M})$ are associated to 1-morphisms $\mathscr{M} \rightarrow \mbox{Sh}$, and that the morphisms of $Shv(\mathscr{M})$ are associated to 2-morphisms in $Hom_{Stacks}(\mathscr{M},\mbox{Sh})$, which in turn should be the same as 1-morphisms $\mathscr{M} \rightarrow \mbox{Sh-map}$. But there a number of problems with this. -First, given a sheaf $\mathcal{F} \in Shv(\mathscr{M})$ I'm having trouble constructing a natural transformation $\mathscr{M} \rightarrow \mbox{Sh}$. Perhaps I shouldn't, but to check this I'm using a test object $X\in \mathcal{C}$. By Yoneda, an object of $\mathscr{M}(X)$ is the same as a 1-morphism of stacks $f:X\rightarrow \mathscr{M}$, and so I obtain an object of $Sh(X)$ (i.e. a sheaf on $\mathcal{C}/X$) via $(\alpha:Y\rightarrow X) \mapsto \mathcal{F}(f\alpha:Y \rightarrow X \rightarrow \mathscr{M})$. That's natural enough. Again by Yoneda, a morphism in $\mathscr{M}(X)$ is a 2-morphism between maps $f,g:X\rightarrow \mathscr{M}$ of stacks, i.e. a section $s:X\rightarrow X\times_\mathscr{M} X$ of the projection from the 2-category fiber product. Out of this, I'm supposed to construct a natural transformation from the sheaf $(\alpha:Y\rightarrow X) \mapsto \mathcal{F}(f\alpha:Y \rightarrow X \rightarrow \mathscr{M})$ to the sheaf $(\alpha:Y\rightarrow X) \mapsto \mathcal{F}(g\alpha:Y \rightarrow X \rightarrow \mathscr{M})$. But the only structure in place to give me such a thing is a morphism in $Stacks/\mathscr{M}$ between $f\alpha$ and $g\alpha$, and I don't see how to construct this. -Second, a 2-morphism between 1-morphisms $f,g\in Hom_{Stacks}(\mathscr{M},\mbox{Sh})$ is a section $s:\mathscr{M} \rightarrow \mathscr{M} \times_{\mbox{Sh}} \mathscr{M}$. Thus for any $(\alpha:X\rightarrow \mathscr{M})\in \mathscr{M}(X)$, we get an object $(\alpha,\beta:X \rightarrow \mathscr{M},\varphi:f\alpha \stackrel{\sim}{\rightarrow} g\alpha)\in (\mathscr{M}\times_{\mbox{Sh}}\mathscr{M})(X)$. On the other hand, a 1-morphism $\mathscr{M} \rightarrow \mbox{Sh-map}$ is for each $\alpha:X \rightarrow \mathscr{M}$ an arbitrary morphism on sheaves on $\mathcal{C}/X$. These can't be the same. -By the way, I've tried to do (what I think is) the right thing and work out the sheaf in $Shv(\mbox{Sh})$ associated to the 1-morphism $\mbox{Id}:\mbox{Sh} \rightarrow \mbox{Sh}$, following Yoneda and all. From the above, it's easy to see what this sheaf should do to morphisms $X\rightarrow \mbox{Sh}$ from a representable stack. But it appears that I need to make choices if I want to say what it does to arbitrary morphisms of stacks $\mathscr{N} \rightarrow \mbox{Sh}$. Perhaps instead I should take a limit or colimit over its application to the full subcategory of representable stacks over $\mathscr{N}$? - -REPLY [3 votes]: The notes you are reading seem to disagree with more commonly accepted language (cf. SGA1 Exp 13, Vistoli's notes, or the Stacks project). Some of this seems to be an attempt at expository ease, e.g., the parenthetical remark in example 8.2 ("We will mention the following technical difficulties but will ignore them for now:") where "for now" really means forever. Oddly enough, one of the mentioned technical difficulties is more or less what prevents $\text{Sh}$ and $\text{Sh-map}$ from having natural stack structures in the sense of the notes - pullback is not strictly functorial. This un-naturality is why the common definition of stack is different - the notion of stack in the notes corresponds to the usual notion of stack in groupoids equipped with a splitting (or cleavage). -The use of the category object $(\text{Sh}, \text{Sh-map})$ is a kludge to replace the usual stack $Sh/\mathcal{C}$ (in categories rather than groupoids) whose objects are sheaves over comma categories, and whose morphisms over any $f: U \to V$ in $\mathcal{C}$ are $f$-maps of sheaves - see Examples 3.20 and 4.11 in Vistoli. The author of the notes employs $\text{Sh-map}$ in order to add non-invertible sheaf maps, because the 2-morphisms in $Hom_{Stacks}(\mathcal{M}, \text{Sh})$ are all invertible. In other words, you have to throw away the 2-morphisms that are given to you by $\text{Sh}$, and use the larger collection of possibly non-invertible two-morphisms afforded by $\text{Sh-map}$. -Once you have done that, I think your main problems are resolved. You've already worked out the object part of getting from a sheaf on $Stacks/\mathcal{M}$ to a natural transformation from $\mathcal{M}$ to $\text{Sh}$. If you have a morphism $\beta: X \to Z$ in $\mathcal{C}$, and $f: Z \to \mathcal{M}$, then $\beta$ induces a morphism of stacks over $\mathcal{M}$. If I'm not mistaken, the sheaf $\mathcal{F}$ takes this to the map in $\text{Sh}$ given by base change: -$$\left( (\alpha: Y \mapsto Z) \mapsto \mathcal{F}(f \circ \alpha) \right) \mapsto \left( \beta^* \alpha: Y \times_Z X \to X) \mapsto \mathcal{F}(f \circ \beta \circ \beta^*\alpha) \right)$$ -Similarly, you can get from a sheaf map on $Stacks/\mathcal{M}$ to a natural transformation from $\mathcal{M}$ to $\text{Sh-map}$. There seems to be a lot of additional checking necessary for proving the equivalence, which I don't feel like doing for you (sorry).<|endoftext|> -TITLE: Class number parity in pure cubic number fields -QUESTION [20 upvotes]: Consider the family of pure cubic number fields -$K = {\mathbb Q}(\sqrt[3]{m})$ for $m = a^3 \pm 3$. -Proposition. If $4 \mid a$ and $m$ is cubefree, then the -class number of $K$ is even. -Proof. Let $\omega = \sqrt[3]{m}$; the element $\alpha = a - \omega$ - has norm $\pm 3$. Since $3$ is completely ramified, the element - $\varepsilon = \alpha^3/3$ is a unit. -If $m = a^3 + 3$, then $\varepsilon = 1 - 3a^2\omega + 3a\omega^2$. - If $4 \mid m$, then $\varepsilon \equiv 1 \bmod 4$, hence - $K(\sqrt{\varepsilon}\,)/K$ is an unramified quadratic extension. -Experiments seem to suggest that if $m = a^3+3$ and $a \equiv 2 \bmod 4$, -then $h$ is also even, but there is no explanation, class field theoretic -or otherwise. In fact, the class number is even for all cubefree values -of $m$ for $a = 2, 4, \ldots, 2 \cdot 88$, but is odd for $a = 2 \cdot 89$. -This cannot be an accident; the parity of the class number in the -case $m = a^3 - 3$ for $a \equiv 2 \bmod 4$ shows a more typical -(i.e. more random) behaviour in that the class number is odd quite -often. -Question: How can this behaviour in the case $m = 8a^3+3$ be explained? -My first guess would be that, for fields in this family, there is -a family of ideals ${\mathfrak a}$ such that ${\mathfrak a}^2$ is -principal, but I can't seem to find anything in this direction. - Edit. Dror's comment made me look at the family of elliptic curves $y^2 = x^3 - m$. -These have rank $\ge 1$, and by the parity conjecture rank $\ge 2$. An inequality due -to Billing now shows that $K$ has even class number. For details, see this -pdf file. -Actually, Paul Monsky stumbled across something similar for pure quartic fields; -see here. - -REPLY [5 votes]: Billing (Beiträge zur arithmetischen Theorie der ebenen -kubischen Kurven vom Geschlecht Eins, R. Soc. Scient. Uppsala (4) 11, -Nr. 1. Diss. 165 S. Uppsala 1938; see Ian Connell's Handbook for elliptic curves for -a modern presentation of the result) proved the following result: -Let $f(x) = x^3 + ax^2 + bx + c \in {\mathbb Z}[x]$ be irreducible, -and consider the elliptic curve $E: y^2 = f(x)$. Let $K$ be the -cubic number field generated by a root $\alpha$ of $f$, and let -$E_K$ be its unit group. Write $({\mathcal O}_K: {\mathbb Z}[\alpha]) =: m_f^2$. -Then -$$ r \ \le \ r_2(K) + r_E(K) + 2n_+ + n_-, $$ -where $r$ is the Mordell-Weil-rank of $E({\mathbb Q})$, -$r_2(K)$ is the $2$-rank of the ideal class group of $K$, -$r_E(K)$ is the ${\mathbb Z}$-rank of the unit group $E_K$ of $K$, -$n_+$ is the number of primes $p \mid m_f$ that split in $K$, -and $n_-$ is the number of primes $p \mid m_f$ that decompose -as $p {\mathcal O}_K = {\mathfrak p}{\mathfrak p}'$ or as -$p {\mathcal O}_K = {\mathfrak p}^2 {\mathfrak p}'$. -If ${\mathbb Q}(\sqrt[3]{m})$ is a pure cubic field with -$m \not\equiv \pm 1 \bmod 9$ cubefree, then the index is trivial, -and $n_+ = n_- = 0$ provides us with the bound -$$ r \ \le \ r_2(K) + 1. $$ -On the other hand, the parity conjecture (see the article by -Liverance pointed out by Dror) implies that the Mordell-Weil -rank of $E$ is even for squarefree values of $m = 8b^3 + 3$, -and the family of nontorsion points -$$ P_b\Big( \frac{2b^3+1}{b^2}, \frac{3b^3+1}{b^3} \Big) $$ -shows that $r \ge 1$. Thus the parity conjecture implies $r \ge 2$, -and Billing's bound finally gives $r_2(K) \ge 1$.<|endoftext|> -TITLE: Simple object in derived category or stable model category? -QUESTION [8 upvotes]: Exist any common definition of simple objects in derived categories, or even better, in stable model categories? -I was only able to find definition for abelian categories. -Thanks. - -REPLY [10 votes]: Sometimes an object $E$ of a $k$-linear category $C$ is called simple if $Hom_C(E,E) = k$. This notion is frequently used in derived categories.<|endoftext|> -TITLE: Jonsson Boolean algebras? -QUESTION [12 upvotes]: Let us say that a mathematical structure of cardinality $\omega_1$ is Jonsson whenever every one of its proper substructures is countable. -There are examples of Jonsson groups due to Shelah or Obratzsov. I am almost sure that there is no Jonsson Boolean algebra but I cannot (dis)prove it by hand. Am I right? -PS. feel free to give any further examples of Jonsson structures or structures which are never Jonsson. - -REPLY [4 votes]: Let $A$ be a unital nonzero Boolean algebra of cardinal $\ge 4$. Then $A$ has a unital Boolean subalgebra of index 2. (This excludes being Jónsson at all cardinals.) -Indeed, $A$ has at least two (unital ring) distinct homomorphisms onto $\mathbf{Z}/2\mathbf{Z}$. This gives rise to a surjective (unital ring) homomorphism onto $(\mathbf{Z}/2\mathbf{Z})^2$. The inverse image of the diagonal yields the desired subalgebra. -Topologically, this corresponds to taking two points in the Stone space and gluing them. - -(Edited Oct 19, 2018) -Actually, no scalar (=associative unital commutative) ring is Jónsson. Here by Jónsson I mean: uncountable and every proper unital subring has smaller cardinal. -First, a domain $A$ is never Jónsson (among unital rings). Indeed, we can consider a maximal algebraically free subset $X$ of $A$; since $A$ is uncountable, we have $|X|=|A|$, and hence we can construct, proper unital subrings of the same cardinal as $A$. -For $p$ prime or zero, say that a scalar ring is $p$-reduced if it embeds into a product of domains of characteristic $p$. Then an uncountable $p$-reduced scalar ring $A$ is never Jónsson. Indeed, write $A\subset\prod A_i$, with $A_i$ domain of characteristic $p$. If some projection has cardinal $|A|$, uncountable, we can pull back some proper subdomain of cardinal $|A|$. Otherwise, each projection has cardinal $<|A|$. If $A$ is a domain, we are done; otherwise, there exists $i,j$ such that the projection on $A_i\times A_j$ is not a domain. Since $A_i$ and $A_j$ have the same characteristic, it follows that this projection is not cyclic, and hence, pulling back the cyclic subring, we obtain a proper unital subring index $<|A|$ in $A$. -Next, a reduced scalar ring $A$ is never Jónsson. Indeed, let $J$ be the set of prime numbers $p$ that are not invertible in $A$. For every $p$, let $P_p$ be the intersection of all prime ideals of $A$ containing $p$ (so $P_p=A$ for $p\notin J$). Then $A/P_p$ is $p$-reduced; if $|A/P_p|=|A|$ then $A/P_p$ is Jónsson and this contradicts the preceding paragraph. So $|A/P_p|<|A|$; hence the inverse image of its cyclic subring has index $<|A|$; since $A$ is Jónsson, this means that $A/P_p\simeq\mathbf{Z}/p\mathbf{Z}$ for every $p\in J$. -By the preceding paragraph, $A$ has at most one prime ideal $P_0$ such that $A/P_0$ has characteristic zero. By the case of domains, $|A/P_0|<|A|$. Hence, again by the argument of pulling back cyclic subrings, we see that $A/P_0$ is an infinite cyclic ring. Hence, if $P_0$ exists, it is contained in $P_p$ for every $p$, and in particular equals the intersection of all prime ideals, i.e., the nilradical, so $P_0=\{0\}$ since $A$ is reduced. Since $A$ is uncountable, we get a contradiction: $P_0$ does not exist. Hence the nilradical $\{0\}$ is equal to $\bigcap_{p\in J}P_j$. That is, the diagonal map $A\to\prod_{p\in J}\mathbf{Z}/p\mathbf{Z}$ is injective. - Consider the composite map $A\to\prod_{p\in J}\mathbf{Z}/p\mathbf{Z}\to (\prod_{p\in J}\mathbf{Z}/p\mathbf{Z})/(\bigoplus_p\mathbf{Z}/p\mathbf{Z})=B$; it has a countable kernel and hence an image of cardinal $|A|$. Then $B$ is a reduced scalar $\mathbf{Q}$-algebra, hence is 0-reduced. Since the image of $A$ in $B$ has cardinal $|A|$, we are done. (Alternative argument for these last few lines: the set of prime ideals of every scalar ring is compact, and it easily follows that if there exists prime ideals such that the quotient have unbounded characteristic, then there exists a prime ideal with quotient of zero characteristic.) -Now let us deal with $A$ arbitrary uncountable scalar ring; let $R$ be its nilradical. If $A/R$ has cardinal $|A|$, then $A$ is not Jónsson, by the reduced case, and if $A/R$ has cardinal $<|A|$ and is non-cyclic, then it has a proper unital subring of index $<|A|$. So we can suppose that $A/R$ is cyclic. -If $A/pA$ is has cardinal $|A|$ for some prime $p$, we can pass to $A/pA$. Otherwise $A/pA$ is has cardinal $<|A|$ for all $p$, and hence $A/nA$ has cardinal $<|A|$ for every $n\ge 1$; in particular, $A$ has characteristic zero, so no nonzero element of $\mathbf{Z}1_A$ is nilpotent: thus $A=R\oplus \mathbf{Z}1_A$. Since this is also true when $pA=0$, we henceforth suppose that $A=R\oplus \mathbf{Z}1_A$ ($A$ being of characteristic zero or prime). -Then observe that for every (non-unital) subalgebra $S$ of $R$, $S\oplus\mathbf{Z}1_A$ is a unital subalgebra of $R$. -If $R\neq 0$, there exists $x\in R$ with $x^2=0\neq x$. The kernel $K$ and image $I$ of the multiplication-by-$x$ map on $R$ are both ideals of $R$, and this multiplication induces a bijection $R/K\to I$. So either $I$ or $K$ has cardinal $|A|$. Hence both $K\oplus\mathbf{Z}1_A$ and $I\oplus\mathbf{Z}1_A$ are proper unital subalgebras of $A$, and at least one of them has cardinal $|A|$. So $R$ is not Jónsson. - -Consequence: if $A$ is an associative unital ring, and is residually of cardinal $<|A|$ (that is, embeddable as unital subring of a product of unital rings of cardinal $<|A|$), then $A$ is not Jónsson. -(Note that Boolean algebras are residually finite, so this is a generalization.) -Indeed, the non-existence of any proper unital subring of countable index implies that every countable quotient of $A$ is cyclic; residual countability then implies that $A$ is commutative, and the previous case discards this. -Of course the argument says more, since it says that every associative unital ring $A$, which is residually of cardinal $<|A|$ (resp. residually countable, resp. residually finite), has a proper unital subalgebra of index $<|A|$ (resp. countable index, resp. finite index), except possibly in the case where $A$ embeds as a unital ring into the quotient of $\widehat{\mathbf{Z}}$ by some closed ideal.<|endoftext|> -TITLE: Searching for an inhomogeneous diophantine approximation algorithm -QUESTION [5 upvotes]: Given two nonzero real numbers $x$ and $y$ such that $y/x$ is irrational, a real number $z$ to be approximated, and a tolerance $\epsilon$, what is an algorithm that will provide coprime integers $a$ and $b$ such that $|ax + by - z| < \epsilon$? -Note that if the restriction that $a$ and $b$ be coprime is lifted, the problem becomes very simple. One possible algorithm is: - -Find $a_1$ and $b_1$ such that $0 < a_1 x + b_1 y < \epsilon$ using the extended Euclidean algorithm. -Let $\displaystyle a = a_1 \left[ \frac{z}{a_1 x + b_1 y} \right]$ and $\displaystyle b = b_1 \left[ \frac{z}{a_1 x + b_1 y} \right],\,$ where $[\cdot]$ is the nearest integer function. - -However, the integers $a$ and $b$ provided by this algorithm are usually not coprime. I'm looking for an algorithm that produces the same kind of approximation but guarantees that $a$ and $b$ are coprime. - -REPLY [4 votes]: The problem isn't really about the existence of algorithms: The required coprime integers $a$ and $b$ can be found by a systematic search if they exist at all, assuming any reasonable interpretation of the word Given in the first sentence of the question. -It will simplify matters to divide the inequality in the first sentence by $x$. -With this in mind, I'll give a proof of the following assertion: -Let $\epsilon>0$.` Suppose $a$ is irrational and $b$ is any real number. Then there are coprime integers $x$ and $y$ such that $|ax-y-b|<\epsilon$. -Proof: The proof has undergone a major rewrite, thanks to Gerry Myerson's helpful comments. -The argument extends a similar result (not mentioning coprimality) proved in Khinchin's book on continued fractions, which is a good reference for the basic facts I'll use here. -Let $p/q$ be a convergent (to be specified later) of the continued fraction expansion of $a$. Then it is well known (see Khinchin) that $p$ and $q$ are coprime, and moreover that $|a-p/q|<1/q^2$. -The latter inequality implies that for some real number $\delta$ with $|\delta|<1$, -$$a=\frac{p}{q}+\frac{\delta}{q^2}.$$ -We will now produce a peculiar-looking estimate for $b$, the reason for which will become apparent shortly. Note that without loss of generality we can and will take $b$ to be positive. -Let $t$ be the largest prime not larger than $bq$. Then by Bertrand's Postulate $t\le bq<2t$. From this we deduce the following chain of inequalities: - $$t/q\le b<2t/q\le t/q+b.$$ - It follows that for some $\gamma$ with $0\le \gamma -TITLE: How to show that an ind-scheme is not a scheme? -QUESTION [6 upvotes]: A standard example of an ind-scheme over a field $\mathrm{k}$ which is not a -$\mathrm{k}$-scheme is $\mathrm{k}((\varepsilon))$. -My question is how to prove that rigorously? To put it more precisely, -let $$\mathrm{k}((\varepsilon)) = \{ a \in \prod_{-\infty}^{\infty}\mathrm{k}: -a_i =0, i \ll 0 \}$$ -An ind-scheme is an injective limit of schemes. So here, -$$\mathrm{k}((\varepsilon)) = -\lim_{i \rightarrow -\infty}\varepsilon^i\mathrm{k}[[\varepsilon]]$$ -But why isn't it an algebraic subset of $\prod_{-\infty}^{\infty}\mathrm{k}$? -EDIT: I seem to have mixed up some notions, and have asked two different questions at once (or maybe even three) so I'll try to make myself clear. -My motivation for the question was to be able to justify the following -"$k((\epsilon))$ is not an algebraic subset of $\prod_{-\infty}^{\infty}k$ so -we define it as $k$-points of an ind-scheme". So the original question is: -why $k((\epsilon)) \subseteq \prod_{-\infty}^{\infty}k$ isn't algebraic -and it is answered by Jason Starr (though I'm not sure if I understand the answer). -We can also define $k((\epsilon))$ as and ind-scheme by -$$k((\epsilon)) = colim_n Spec(k[x_{-n},x_{-n+1}, \ldots])$$ -Now, one can ask, why isn't the ind-scheme we've constructed a scheme after all -(btw. wouldn't it contradict Jason's argument?), and -this question is answered by Scott Carnahan below. Finally, there is a question: if the co-limit exists in the category of schemes which Scott Carnahan addresses below as well ... - -REPLY [3 votes]: Based on the comments, it looks like you have two questions mixed up. There is the question of whether the colimit exists in the category of schemes, and there is the question of whether the ind-scheme described by the colimit in the category of set-valued functors on schemes (or the category of Zariski sheaves of sets, or fpqc sheaves, etc.) is represented by a scheme. The two questions are quite different. Since you mentioned ind-schemes, I'll answer the (easier) question about ind-schemes. -Following Martin's comment, I'm assuming by $k((\epsilon))$, you are referring to $\operatorname{colim}_n \operatorname{Spec}(k[x_{-n},x_{-n+1},\ldots])$ where the colimit is taken in set-valued functors on schemes. It is an ind-scheme in the sense that it is a colimit over a directed system of closed immersions of schemes. This particular ind-scheme is ind-affine, so it can be written as the formal spectrum of the topological ring $A = \varprojlim_n k[x_{-n},x_{-n+1},\ldots]$ (see Beilinson and Drinfeld's Quantization of Hitchin’s integrable system and Hecke eigensheaves section 7.11.2). Since $A$ has a non-discrete localization at zero, and all local rings of schemes at points are discrete, the locally topologically ringed space $\operatorname{Spf} A$ is not isomorphic to a scheme. -I think the directed system $\{ \operatorname{Spec}(k[x_{-n},x_{-n+1},\ldots]) \}_{n \geq 0}$ has a colimit in schemes, given by $\operatorname{Spec} A$, where $A$ is given the discrete topology. $\operatorname{Spec} A$ is certainly the colimit in affine schemes, by the anti-equivalence between affine schemes and commutative rings. Surely there must be a theorem somewhere ...?<|endoftext|> -TITLE: Help motivating log-structures -QUESTION [31 upvotes]: I'm currently reading a thesis that uses log-structures. I should mention that this is my first encounter with them, and the thesis (as well as my expertise) is scheme-theoretic (in fact stack-theoretic) and so the original geometric motivations are lost on me. -Here is my meek understanding. For any scheme, we can give a log-structure. This is a sheaf , $M$, fibered in monoids, on the etale site over a scheme $S$; together with a morphism of sheaves fibered in moinoids $\alpha:M\rightarrow O_S$ such that when it is restricted to $\alpha^{-1}(O_S^{\times})$ it is an isomorphism. -This $\alpha$ is called the exponential map, and for any $t\in O_S(U)$ (for some $U$), a preimage of it via $\alpha$ is called $log(t)$($\in M(U)$). -I am curious about a few things, and puzzled about others. First, in terms of the notation, surely it's no coincidence that these are called exponential maps and log-structures. What is the geometric motivation for it? -Second, these come up in the thesis I'm reading in the context of tame covers. I am puzzled about what, precisely, log-structures contribute. It seems to me, in extremely vague terms (commensurate with my understanding), that the point of log-structures in this context is that if you add this extra information to tame covers it somehow helps you construct proper moduli spaces of covers. -On top of everything I'm also confused about the role of `minimal log-structures' in all of this. -In conclusion, if you can say anything at all about the motivations of log-structures in the geometric setting, or more importantly in the context of tame covers, I would extremely appreciate it. The plethora of notationally different texts on the subject is making it hard to understand the gist of what's going on. -Also, if you have examples that I should have in mind when thinking about it, that would be ideal. - -REPLY [3 votes]: I believe that people refer to $\alpha$ as the exponential map exactly because of the "big example (I)" in Pottharst's notes. Also, a small point, for any scheme we can give many log structures. For instance, in addition to the main example of a log str associated to a closed immersion, the simplest examples are $M = \mathcal O_X^*$ and $M = \mathcal O_X.$ I found K. Kato's paper "Logarithmic structures of Fontaine-Illusie" to be a good introduction. -Logarithmic structures of Fontaine-Illusie. Algebraic analysis, geometry, and number theory (Baltimore, MD, 1988), 191–224, Johns Hopkins Univ. Press, Baltimore, MD, 1989.<|endoftext|> -TITLE: Why separating tangent vectors? -QUESTION [10 upvotes]: Let $X$ be a projective scheme over an algebraically closed field. -It is a well-known theorem that a line bundle $L$ defines a closed embedding into projective space if and only if it separates points and tangent vectors, which means: -(1) For any distinct closed points $P, Q \in X$ there is an $s \in \Gamma(X,L)$ with the property that $s\in m_PL_P$ but $s \not\in m_QL_Q$. -(2) For every closed point $P \in X$ the set $\{s \in \Gamma(X,L)|s_P \in m_PL_P\}$ spans the vector space $m_PL_P/m_P^2L_P$. -Now in Hartshorne (Prop. IV 3.4.) this criterion is applied in the following situation: One wants to show that every curve $C$ can be embedded into $\mathbb{P}^3$. First one embedds the curve $C$ into an arbitrary $\mathbb{P}^n$. Then one chooses some $O \in \mathbb{P}^n$ andprojects the curve down from $O$ into $\mathbb{P}^{n-1}$. -Then one wants to show that this projection map is a closed immersion if and only if -(a) $O$ does not lie on any secant of $C$ -(b) $O$ does not lie on any tangent of $C$. -At this point Hartshorne applies the above criterion. But at this point I do not understand why the fact that $O$ does not lie on any tangent implies $(2)$ of the above criterion. I understood why $(a)$ implies $(1)$ but why does $(b)$ yield that the line bundle separates tangent vectors - -REPLY [6 votes]: [I assume $C$ is smooth] -The point is this: $m_PL_P/m_P^2L_P$ is $1$-dimensional and hence generating it is equivalent to finding a single element that is not zero. The statement that all sections that are in $m_PL_P$ are also in $m_P^2L_P$ is equivalent with the statement that every member of the corresponding linear system that contains $P$, also contains the tangent line to $C$ at $P$. I think you should be able to finish from here. -[A modification of this idea actually works in higher dimensions, so the first sentence is not really necessary, but makes it easier in the curve case.]<|endoftext|> -TITLE: Conic hulls and cones -QUESTION [7 upvotes]: Suppose I have a number of vectors in $\mathbb{R}^n.$ The first question is: what is the most efficient algorithm to compute their "conic hull" (the minimal convex cone which contains them)? The next question is: suppose I have a number of vectors $v_1, \dotsc, v_n,$ as before, and a convex cone $C.$ I want to find the conic hull of $\{v_1, \dotsc, v_n\} \cup C.$ In case it matters, in my application $C$ is the semidefinite cone. By "compute the conic hull", I mean: I want to find the subset of the $v_i$ on the boundary of the hull. -EDIT Thanks for all the comments. It is certainly true that the conic hull is equivalent to the intersection with a plane, and as @Will pointed out, the only problem is finding the plane. In the PSD case, we know that identity is PSD, so this gives us a choice of planes. -As for the algorithm, I had come up with @Matus' algorithm, but was not sure (and still am not) that this is the most efficient, since it looks like there is a lot of recomputation. The fact that the PSD cone is not a polyhedral cone is very true. Notice that you can still ask for the extremal points from the original set, and in fact, the same algorithm works, except that instead of solving a linear program at each step, we need to solve a semidefinite program, which hurts a bit, but is certainly tractable for small dimension. -If you ask for the full convex hull, I am not at all sure of how the answer should even look like, since one will need to describe the "exposed" pieces of the cone. Surely mankind has wondered about this is in the context of, eg, the convex hull of a collection of disks in the plane, or some such. - -REPLY [6 votes]: [Edited: previous version was flaky, sorry; also edited per Matus] -There is a variant of Matus's approach that takes $O(nT_A)$ work, where $A\le n$ is the size of the answer, that is, the number of extreme points, and $T_A$ is the work to solve an LP (or here an SDP) as Matus describes, but for $A+1$ points instead of $n$. -The algorithm is: (after converting from conic to convex hull) maintain an output set $S$, that starts empty, and test each point $v_i$ against $S$ one by one. Solve the LP (or SDP) as Matus describes. If $v_i$ is proven to be in the convex hull of $S$, discard it. Otherwise, the dual certificate gives a direction (at least in the LP case, and something similar should apply in the SDP case) perpendicular to that separating hyperplane, such that the input point that is extreme in that direction is not already in $S$. (While $v_i$ is not in the convex hull of $S$, it may not be extreme itself.) Find that input point and add it to $S$. -Testing each of the $n$ points costs $T_A$, and the $O(n)$ work for finding an extreme point in a given direction yields a new member of $S$, so such tasks need $O(nA)\le O(nT_A)$ work. -This trick and related ones appeared here ("More output-sensitive..."); the notes for the paper give pointers to some related work.<|endoftext|> -TITLE: Volume of fundamental domain and Haar measure -QUESTION [14 upvotes]: In my research, I do need to know the Haar measure. I have spent some time on this subject, understanding theoretical part of the Haar measure, i.e existence and uniqueness, Haar measure on quotient. But I should confess I never felt confident with the Haar measure, essentially because theoretical part did not give me how to construct the Haar measure. Even when I was trying to understand Haar measure on $SL(n,\mathbb{R})$ via Iwasawa decomposition, I could not realize exactly how it works. -I think, or I should say I fell, I am not the only one how has problem with this delicate subject. So I thought I would want to share this with you and see how people think about the Haar measure and how you compute, for instance, the volume $${\rm Vol}(SL(n,\mathbb{Z})\backslash SL(n,\mathbb{R}))$$ - -REPLY [8 votes]: Supplementing the other answer (too long for a comment...) First, for examples of innocent-context computations, there is an on-line computation of volumes of $SL(n,\mathbb Z)\backslash SL(n,\mathbb R)$ and of $Sp(n,\mathbb Z)\backslash Sp(n,\mathbb R)$ - here , written in essentially Siegel's style. The same style of computation can be done adelically, over arbitrary number fields, but still does effectively beg the question of normalization. Nevertheless, such computations show that the normalization can be determined inductively, and, in any case, that the global computation can be done nicely once we have a locally-everywhere normalization of measures. -Yet-another approach is (after Langlands) to look at suitable residues of Eisenstein series (e.g., as in the Boulder conference, AMS Proc Symp IX). In effect, such a computation back-handedly normalizes the Haar measure...<|endoftext|> -TITLE: Large cardinals and constructible universe -QUESTION [8 upvotes]: We know that if $V=L$ holds, then $|\cal{P}(\omega)|=|\cal{P}(\omega)\cap \textrm{L}|=\aleph_1$ whereas, in the presence of a measurable cardinal (in fact, even Ramsey) $|\cal{P}(\omega)\cap \textrm{L}|=\aleph_0$. I remark that the cardinalities are of course computed in (the corresponding) $V$. -The first is just the fact that the constructible universe satisfies CH, while the second has to do with the fact that in the presence of a measurable, $\omega_1^{L}<\omega_1$ i.e. the existence of large cardinals makes the relative $\omega_1^{L}$ "drop" below its "maximum possible" value (which is attained, if you want, in the "extreme case" when $V=L$). -My question is, what can we say, in general, about the beaviour of $\omega_1^{L}$ given axioms of increasing strength above (or equal to, in strength) $V\neq L$? In particular, what happens if we just assume $V\neq L$? - -REPLY [14 votes]: Each of the following implies that (the true) $\omega_1$ is inaccessible in $L$, and hence that there are only countably many constructible reals: - -The proper forcing axiom -There is a Ramsey cardinal -$0^\#$ exists -All projective sets are Lebesgue measurable -All $\Sigma^1_3$-sets are Lebesgue measurable - -(EDIT: These are just some of the well-known examples that came to my mind. This list is neither exhaustive nor canonical.) -The mere existence of a nonconstructible set, or even a nonconstructible real, does not imply that $\omega_1^L$ is countable. There are many forcing notions in $L$ which do not collapse $\omega_1$: adding one or many Cohen reals, destroying Souslin trees, etc. Each such forcing (over L) results in a model where $\omega_1=\omega_1^L$. -In fact, "Martin's axiom plus continuum is arbitrarily large" is consistent with $\omega_1^L=\omega_1$. (But also with $\omega_1^L<\omega_1$.) -ADDED: Preserving $\aleph_1$ of the ground model (which may or may not be the constructible universe $L$) is a key component in many independence proofs concerned with the theory of the reals. The -"countable chain condition", which is enjoyed by all the forcings I mentioned above, is a property of forcing notions that guarantees preservation of $\aleph_1$; there are several other (weaker) properties which also suffice, most prominently (Baumgartner's) "Axiom A" and (Shelah's) "properness".<|endoftext|> -TITLE: Motivation behind Kac's notation for affine root systems -QUESTION [11 upvotes]: I'm reading Kac's Infinite Dimensional Lie Algebras. In Chapter 4, he classifies the affine root systems. Bourbaki classified the affine Coxeter groups, but multiple root systems can give the same Coxeter group. Kac denotes the elements of his classification by $X^{(r)}$, where $X$ is a symbol denoting a finite root system (e.g. $G_2$) and $r$ is $1$, $2$ or $3$. -When $r=1$, everything makes sense to me. $X^{(r)}$ is a root system which gives rise to the affine Coxeter group which Bourbaki would consider to be of type $X$. -When $r$ is $2$ or $3$, I believe I understand the object Kac is defining. However, I do not understand how he chooses its name. For example, look at the root system he calls $A_{2 \ell-1}^{(2)}$. The root system is of rank $\ell+1$, not $(2 \ell-1)+1$ as you would expect. The corresponding Coxeter group is the affine Coxeter group of type $B_{\ell}$. If you look at Macdonald's very readable paper Affine root systems and Dedkind's $\eta$-function, Macdonald denotes this root system by $B_{\ell}^{\vee}$, which makes much more sense to me. -Similar issues apply to every entry in table 'Aff 2'. -I have two questions: - -Why does Kac choose the notation he does? - -Also, - -Is Kac's notation so established that I have to use it? Is there a mainstream alternative? I find Macdonald's much more reasonable, but I'm pretty sure that hasn't caught on. - -REPLY [13 votes]: I think the notation might be explained by the explicit construction of the twisted affine Lie algebras as fixed points of automorphisms of the untwisted ones: the $r$ indicates the order of the chosen automorphism of the extended Dynkin diagram corresponding to $X$ and twised affine Lie algebra is a subalgebra of the affine Lie algebra corresponding to $X$. See Chapter 8 of Kac's book. -This notation is fairly well established, given that Kac's book is the standard reference, but I'm not an expert.<|endoftext|> -TITLE: Maximize the intersection of a n-dimensional sphere and an ellipsoid. -QUESTION [5 upvotes]: I have the conjecture that the volume of the intersection between an $n$-dim sphere (of radius $r$) and an ellipsoid (with one semi-axis larger than $r$) is maximized when the two are concentric, but still did not find a way to prove it. Any suggestion? - -REPLY [8 votes]: By a rotation, you may assume that the axes of the ellipsoid are parallel to the coordinate axes. Choose the first coordinate of the center of the ellipsoid, and translate the ellipsoid so that this coordinate becomes zero, keeping the other center coordinates fixed (this corresponds to symmetrizing about this axis). The volume of the intersection increases when you do this, since the intersection of an interval of fixed length with an interval centered around the origin is maximal when the interval is centered about the origin too (this is the 1-dimensional case of your question). Repeat with the other coordinates, until the ellipsoid is centered at the origin, and the volume of intersection is maximal.<|endoftext|> -TITLE: Gale-Shapley stable marriage theorem: can we entrust matchmaking to monkeys? -QUESTION [13 upvotes]: Disclaimer: This is a question I have not done any real research about. I asked it myself some 5 years ago, and back then I had no idea where to start. Now I have some texts on stable matchings lying around, but from a quick look they don't seem to answer this. -We have $n$ ladies $L_1$, $L_2$, ..., $L_n$ and $n$ gentlemen $G_1$, $G_2$, ..., $G_n$. Each lady ranks all gentlemen in order of preferability (no ties are allowed), and each gentleman does the same to the ladies. A stable marriage means a permutation $\sigma \in S_n$ such that there are no $j\in\left\lbrace 1,2,...,n\right\rbrace$ and $k\in\left\lbrace 1,2,...,n\right\rbrace$ for which $L_j$ prefers $G_k$ to $G_{\sigma\left(j\right)}$ whereas $G_k$ prefers $L_j$ to $L_{\sigma^{-1}\left(k\right)}$. -Okay, I should have said that it is a matching where we cannot find a lady and a gentlemen which prefer each other to their respective matching partners. But is it combinatorics if there are no symmetric groups in it?... -Anyway, this is known to have a simple (but very hard to find) algorithmic proof. What I am wondering is whether the following stupid algorithm can also be forced to terminate: -We choose some arbitrary matching between the ladies and the gentlemen. Then, at each step, we randomly pick a pair that prefers each other to their respective partners, and marry them to each other, simultaneously marrying their respective partners to each other (no matter what they think about it). Repeat until no such steps are possible anymore. -(1) Can this "algorithm" loop endlessly if we choose our pairs in a stupid enough way? -(2) Can we make this algorithm terminate by giving a reasonable choice tactic for the pairs? - -REPLY [12 votes]: For (1), yes, it can loop endlessly. Here's a demonstration with 8 people. -Suppose you have 3 ladies and 3 gentlemen standing in a circle, along with one more lady and gentleman (the pariahs) inside the circle. Here are the preferences you need to know about: everyone around the circle prefers to person to his or her right over the person to his or her left, and prefers either of them to the oppositely-sexed pariah in the center. (The compatibility of people diagonally opposite each other and the opinions of those in the center won't matter for the purposes of this solution.) -Initially, choose a pair of diagonally opposite people and pair each of them up with the appropriate pariahs in the center. Pair up the remaining four people around the circle in adjacent pairs. This is unstable: those matched with pariahs would prefer to be matched with the neighbor to the left, and the feelings are reciprocated (since everyone around the circle prefers right over left). -Performing both of these swaps, however, only serves to rotate the whole picture 120° while preserving the symmetry of the preferences we care about. Do it two more times and you've got the original matching again. - -REPLY [11 votes]: Regarding (2), the answer is still "no". The following counter-example is from: - -Tamura, Akihisa Transformation from - arbitrary matchings to stable - matchings J. Combin. Theory Ser. A - 62 (1993), no. 2, 310–323 - -Consider $n$ men and $n$ women. With indices periodic modulo $n$, the first four choices of each person are: -For $m_i$: First choice is $w_i$, then $w_{i-2}$, then $w_{i+1}$, then $w_{i-1}$. -For $w_i$: First choice is $m_{i+1}$, then $m_{i-1}$, then $m_i$, then $m_{i+2}$. -Start with the matching that pairs $m_i$ to $w_i$ for $1 \leq i \leq n-2$, pairs $m_{n-1}$ and $w_n$ and $m_n$ and $w_{n-1}$. I leave it as an exercise that there is only one unstable pair, swapping them leaves a situation where there is only one unstable pair, and swapping those brings you back to the original situation with indices shifted.<|endoftext|> -TITLE: Limits of reduced schemes question from Eisenbud and Harris -QUESTION [6 upvotes]: My question pertains to exercise II-16 in Eisenbud and Harris' "The geometry of Schemes". For an algebraically closed field $K$ the question is as follows: - -Consider zero-dimensional subschemes $\Gamma \subset \mathbb{A}_K^4$ of degree 21 such that $$V(m^3)\subset\Gamma \subset V(m^4)$$ - where $m$ is the maximal ideal of the origin in $\mathbb{A}_K^4$. Show that there is an 84-dimensional family of such subschemes, and conclude that in general one is not a limit of a reduced scheme. - -What does it mean for a family of subschemes to have dimension 84? I can only think it means up to isomorphism there are 84 such subschemes, but this doesn't seem to work. - -REPLY [8 votes]: Such a subscheme is given by an ideal $I$ such that $m^4 \subset I \subset m^3$. In fact, any vector subspace in $m^3$ containing $m^4$ is an ideal (this is a simple exercise). Since $\dim O/m^3 = 15$ and $\dim O/m^4 = 35$, and we are interested in subspaces $I \subset O$ such that $\dim O/I = 21$, that is in subspaces $I/m^4$ of dimension $35 - 21 = 14$ of the space $m^3/m^4$ of dimension $35 - 15 = 20$. So, the family in question is just the Grassmannian $Gr(14,20)$. Its dimension is $14(20-14) = 14\cdot 6 = 84$.<|endoftext|> -TITLE: Topology on extensions of topological groups -QUESTION [7 upvotes]: Let $G$ and $H$ be two topological groups and let $\mathcal{E}:0 \to G \to E \to H \to 0$ be an extension of abstract groups. -Is there a way to introduce a topology on $E$ such that $\mathcal{E}$ becomes an extension of topological groups? If there is a way, is it unique? -Similarly, let $G$ and $H$ be Lie groups and let $\mathcal{E}:0 \to G \to E \to H \to 0$ be an extension of topological groups. -Is there a way to introduce a smooth structure on $E$ such that $\mathcal{E}$ becomes an extension of Lie groups? If there is a way, is it unique? -Thank you all in advance. - -REPLY [2 votes]: You can describe abstract group extensions of H with G by 2-cocycles of group cohomology. -If you have an extension E, you get an induced H-operation on G, by conjugating in E (take any set-theoretic section of $E\to G$). The extensions of G with H with this H-operation on G are classified up to isomorphism, by the second group cohomology. You get a 2-cocycle corresponding to E by taking any set-theoretic section $s : G\to E$ of $E\to G$ that maps 1 to 1 and write down the 2-cocycle $c : H \times H \to G$ by $c(h,h'):=s(h)s(h')s(hh')^{-1}$. -Then equip the set $G\times H$ with the multiplication $(g,h)(g',h') := (g+h.g'+c(h,h'),hh')$. More explicitly, this is $(g,h)(g',h') = (g+s(h)g's(h)^{-1}+s(h)s(h')s(hh')^{-1},hh')$. -This is again a group extension and it is isomorphic to E. -One can show that all extensions are of the type I just constructed for a given cocycle, up to isomorphism. -The extensions in the same isomorphism class differ only by a coboundary. -A good reference would be Weibel's homological algebra book. -To have an extension with a topological group structure implies that the corresponding cocycle is continuous and in general, this can not be expected. Observe that continuity doesn't follow from the axioms for 2-cocycles and depends on the topological structures of G and H, which you don't want to change. -So I think, it's wrong in general as well as for central extensions, which would be the case of a trivial H-action on G, where you still don't get any continuity for free. At the same time, I don't know of any trivial counter-example. You might take any non-continuous map from the real numbers times real numbers to the real numbers and form the corresponding "twisted" semi-direct product as sketched above. Then you can not get a topological group structure on the extension such that the extension is in the category of topological groups. -As for the smooth case, the same idea applies, where one would need the cocycle to be smooth as well.<|endoftext|> -TITLE: Applications of full integral weight modular forms in elementary number theory -QUESTION [10 upvotes]: Except for Eisenstein series having the divisor functions as their Fourier coefficients, is there any other full integral weight modular form (of some level, preferably full) having arithmetic functions as their Fourier coefficients. -More to the point, my question is, apart from the relations you obtain between $\sigma_3, \sigma_5$ and $\sigma_7$ are there any applications of full integral weight modular forms (preferably cusp forms) to elementary number theory. - -REPLY [2 votes]: Nobody seems to have mentioned "the master" of this subject, and his use of (classical) Eisenstein series to prove things like $p(5n+4) \equiv 0 \ ({\rm mod} \ 5)$ (here $p(m)$ is of course the usual partition function). -Here's his proof (prepared by Hardy), published in Math.Z (1921). B.Berndt published another a bit shorter proof , which employs famous Ramanujan's differential equations. BTW, make sure you are familiar with the Ramanujan "J-series" before you jump to (say) formula (2.2) in Berndt's paper :-)<|endoftext|> -TITLE: What is the fan of the toric blow-up of $\mathbb{P}^3$ along the union of two intersecting lines? -QUESTION [5 upvotes]: Is there a good way to find the fan and polytope of the blow-up of $\mathbb{P}^3$ along the union of two invariant intersecting lines? -Everything I find in the literature is for blow-ups along smooth invariant centers. -Thanks! - -REPLY [3 votes]: Since you already know how to blow up along either ${\mathbb P}^1$ individually, we can concentrate on what's happening nearby the intersection. Which means we can work affinely. -Then the polyhedron for ${\mathbb A}^3$ is the octant $({\mathbb R}_{\geq 0})^3$, and your two lines correspond to two of the three edges. -To blow them up, take a carpenter's plane to those two edges, and shave them off exactly the same amount. The result will have an edge connecting two vertices, one of degree 3, one of degree 4 (the isolated singularity on the blowup). -In coordinates, the polyhedron is {$(x,y,z) : x,y,z \geq 0, x+y \geq 1, x+z \geq 1$}. -If you want the blowup of ${\mathbb P}^3$ not ${\mathbb A}^3$, also impose $x+y+z \leq N$ for some large $N$ (three will do).<|endoftext|> -TITLE: Distributions and measures -QUESTION [6 upvotes]: Hello, -After reading the previous post, I still have some doubts. Let's consider everything on $R$ to avoid complications. - -Can we say that any distribution $\mu\in\mathcal{D}'(R)$ of zero order is a signed radon measure? -Since $\mu\in\mathcal{D}'(R)$ which is non-negative on non-negative test functions $C_c^\infty(R)$ is a positive radon measure, it is natural to ask what is corresponding part for the Schwartz distribution $\mu\in\mathcal{S}'(R)$ which is non-negative on non-negative test functions $\mathcal{S}(R)$? Intuitively, it is the radon measure whose mass grows slowly at infinity. Is there a name for this measure? -For a radon measure $\mu$ on $R$, can we apply the Lebesgue's decomposition locally with a compact set fixed (say $K=[-a,a]$)? Then $\mu_K=\mu_{ac}+\mu_{sc}+\mu_{pp}$. The absolutely continuous part $\mu_{ac}$ corresponds to an absolutely continuous function; the pure point part $\mu_{pp}$ corresponds to sum of delta functions. How about the singular continuous part $\mu_{sc}$? -EDIT: This question can also be put in the following way: Let $f$ be a singular function, what can we say $\int f\psi d x$ with $\psi\in\mathcal{D}(R)$? - -Thank you for your help! :-) -Best -Anand - -REPLY [6 votes]: About the second question: -The Schwartz space $\mathcal{S}(R)$ is a Frechet space, i.e. it's topology is given by a countable sequence of seminorms. Any Schwartz distribution (continuous linear functional on $\mathcal{S}(R)$) has to bounded by one of these seminorms. If the Schwartz distribution is of "order zero" then the only relevant seminorms are $\|f\|_n = \sup (1+|x|)^n |f(x)|$. If $T\in \mathcal{S}'(R)$ is bounded by $\|\ldots\|_n$, then this means $T$ is of the form $\langle T, f\rangle = \int (1+|x|)^{n} f(x) \mu(dx)$ for some finite measure on $R$. -So I would think measures which are Schwartz distributions are measures "of polynomial growth."<|endoftext|> -TITLE: A Question on Koszul duality and $B(\infty)$ structures on $HH^*$ -QUESTION [5 upvotes]: The following theorem is known from a paper "Duality in Gerstenhaber Algebras" by Felix, Menichi, Thomas. Given a simply connected space X of finite type. -There is an equivalence of Gerstenhaber algebras -$HH^*(C_*(\Omega X,\mathbb{Q}), C_*(\Omega X,\mathbb{Q}) \cong HH^*(C^*(X,\mathbb{Q}),C^*(X,\mathbb{Q})$ -On the left hand side we have Pontryagin product on the based loop space and on the right hand side rational cochains. $HH^*$ denotes Hochschild cohomology. -I have never seen anyone speak to the following enhanced statement, which makes me wonder if there is a counterexample or if I am simply missing some literature. -$HCH^*(C_*(\Omega X), C_*(\Omega X) \cong HCH^*(C^*(X),C^*(X))$ -The question is: Is this statement true, false or unknown? -Here we are looking at Hochschild cochains in the homotopy category of $B(\infty)$ algebras. For the background police, a $B(\infty)$ algebra is a type of dg-Gerstenhaber structure, that naturally gives rise to a Gerstenhaber structure by passing to homology. For more info, see the paper of Keller mentioned below. -It is possible to prove this theorem when $C^*(X)$ is equivalent to a graded simply connected Koszul algebra( i.e. X is both formal and coformal). I believe this is due to Keller in a paper called the "Derived Invariance of Higher Structures of the Hochschild complex". - -REPLY [2 votes]: Looking closer at Keller's paper, the result seems to be in there. Namely, in his main theorem in section 3.3, he proves that fully faithful dg-functors $per(A) \to D(B)$ induced by an $A\otimes B^{op}$ module X induce $B(\infty)$ morphisms $\phi_X: HCH^*(B,B) \to HCH^*(A,A)$. Additionally, in the same theorem, he proves that if the map $per(B^{op}) \to D(A^{op})$ induced by X is also fully faithful, then $\phi_X$ is invertible. -These criterion all apply to M a simply connected space, $X= \mathbb{Q}$ the trivial local system, $A=C_*(\Omega(M))$, and $B= C^*(M)$.<|endoftext|> -TITLE: Matrix expression for elements of $SO(3)$ -QUESTION [10 upvotes]: Hi all. Is there any explicit matrix expression for a general element of the special orthogonal group $SO(3)$? I have been searching texts and net both, but could not find it. Kindly provide any references. - -REPLY [14 votes]: There is a good way to derive the sort of thing you're looking for: use the double cover $SU(2) \to SO(3)$. $SU(2)$ is diffeomorphic to the 3-sphere $S^3 \subseteq \mathbb{C}^2$ -$$ SU(2) = \left\{ \begin{pmatrix} a & -\overline{b} \\ b & \overline{a} \end{pmatrix} : |a|^2 + |b|^2 = 1 \right\} $$ -Now $SU(2)$ acts on its Lie algebra $\mathfrak{su}_2$ (which is 3-dimensional) by conjugation. This action preserves the inner product -$$ \langle X, Y \rangle = - \frac12 \mathrm{tr}(XY) = \frac12 \mathrm{tr}(X^*Y)$$ -(which is a scalar multiple of the Killing form of $\mathfrak{su}_2$, FYI) and hence this gives a homomorphism -$SU(2) \to SO(\mathfrak{su}_2) \simeq SO(3)$. (A priori this gives a map to $O(3)$, but $SU(2)$ is connected so the image lands in $SO(3)$. -Now consider the orthonormal basis for $\mathfrak{su}_2$ given by -$$ -e_1 = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}, e_2 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}, e_3= \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}, -$$ -let -$$ -x = \begin{pmatrix} a & -\overline{b} \\ b & \overline{a}\end{pmatrix},$$ -and write down the adjoint action of $x$ on $e_1,e_2,e_3$. For instance you get -$$ -\begin{align} -xe_1x^{-1} & = \begin{pmatrix} i(|a|^2 - |b|^2) & 2ia\overline{b} \\ 2i\overline{a} b & -i(|a|^2 - |b|^2) \end{pmatrix} \\ -& = (|a|^2 - |b|^2)e_1 + i(a\overline{b} - \overline{a}b)e_2 + (a \overline{b} + \overline{a}b)e_3. -\end{align}$$ -This gives you the first column of the matrix representation conjugation by $x$. I'll leave the others to you. But this way you can see where the formulas come from. -This gives exactly David Speyer's answer (possibly modulo some re-ordering of the basis). His four real numbers $a,b,c,d$ would correspond to my complex numbers $a,b$ via -$$a_{mine} = a + i b, \quad b_{mine} = c + id.$$<|endoftext|> -TITLE: On the inverse Galois problem -QUESTION [18 upvotes]: Q: What is the "simplest" finite group $G$ for which we don't know how to realise it as a Galois group over $\mathbf{Q}$ ? -So here the word simplest might be interpreted in a broad sense. If you want something precise -you may take the group of smallest order but I prefer to leave the question as it is. -Also since naturally one classifies finite groups into families one may also ask the following -Q: What is the "simplest" example of a family of finite groups for which the inverse -Galois problem is unknown? - -REPLY [5 votes]: I am not an expert and I might be misrembering some talks I've attended. Anyway, $SL_2(\mathbb{F}_q)$ can be done for prime $q$ by using torsion on non-CM elliptic curves. But I don't think it's been done for general prime powers $q$. Also, what about $SL_3$?<|endoftext|> -TITLE: a Ramsey-type question -QUESTION [8 upvotes]: This question is related to this one but feels more Ramsey-type, so perhaps it is easier. Let $S$ be a finite set, $|S|=k$. Suppose we color all subsets of $S$ in $1000$ colors. What is the maximal (in terms of $k$) guaranteed length $l=l(k)$ of a monochromatic sequence of pairwise different subsets $A_1,A_2,..., A_l$ such that $|A_i\setminus A_{i+1}|+|A_{i+1}\setminus A_i|\le 2$ for every $i$? Clearly if $A$ is a subset of $S$ such that all 2-element subsets of $A$ are monochromatic, then $l(n)\ge |A|-1$ (there is a sequence of 2-element subsets of $A$ which satisfies the above property). So $l(k)$ is at least as big as the corresponding number from the Ramsey theory. Is it much bigger? The number 1000 is of course "any fixed number". - Update 1 Fedor and Tony showed below that $l(k)\ge k/1000$. Thus only the first question remains: What is $l(k)$? Is it exponential in $k$, for example? - Update 2 Although the question I asked makes sense (see Update 1), I realized that it is not the question I meant to ask. Here is the correct question. Same assumptions: $|S|=k$, 1000 colors. We consider monochromatic sequences of pairwise different subsets ${\mathcal A}=A_1,A_2,...,A_l$, where $|A_i\setminus A_{i+1}|+|A_{i+1}\setminus A_i|\le 2$. For each of these sequences we compute $\chi({\mathcal A})=|A_1\setminus A_l|+|A_l\setminus A_1|$. Now the question: what is the maximal guaranteed $\chi({\mathcal A})$ in terms of $k$, call it $\chi(k)$? By Ramsey, this number grows with $k$. Indeed if we color just $s$-element subsets, we will be able (if $k\gg s$) to find a subset of size $2s$ where all subsets of size $s$ are colored with the same color; then we can find a monochromatic sequence of subsets of size $s$ with the above property and $\chi=2s$ because the first and the last subsets in that sequence are disjoint. The question is what is the growth rate of $\chi(k)$. The question is motivated by Justin Moore's answer - here. - -REPLY [3 votes]: Here is a proof of a very weak upper bound for $l(k)$. Consider the colouring of $2^S$ where each set is coloured by its size (mod 1000). A good monochomatic sequence must consist of sets of the same size. Thus we obtain $l(k) \leq \binom{k}{k/2}$. -However, we can be a bit smarter. Instead of colouring all $i$-subsets of $S$ with the same colour, we can use 333 colours and still guarantee that a good monochromatic sequence must use sets of the same size. Thus, we are lead to the problem of 333-colouring $\binom{S}{i}$ to minimize the length of a good monochromatic sequence inside $\binom{S}{i}$.<|endoftext|> -TITLE: Proving that a function's image contains (1/n,...,1/n) -QUESTION [26 upvotes]: This question is a follow-up to a previous question answered by Neil Strickland: -Map from simplex to itself that preserves sub-simplices -Let $B$ denote the closed unit ball in $\mathbb{R}^2$ and let $\Delta_{n-1}$ denote the $(n-1)$-simplex. I have a continuous function $f(x_1,\dots,x_n):B^n \rightarrow \Delta_{n-1}$ defined for all subsets $\lbrace x_1,\dots,x_n\rbrace \subset B$ of size $n$ that satisfy $x_i \neq x_j$ for all pairs $i,j$ (in other words, the function is only defined if all of the $n$ arguments are distinct). This function has the property that, if $\sigma$ denotes a permutation, then $f(\sigma(x_1,\dots,x_n)) = \sigma(f(x_1,\dots,x_n))$. In other words, permuting the arguments of the function merely permutes the output. My question is: are there non-trivial sufficient conditions on $f$ under which the point $(1/n , \dots, 1/n)$ lies in the image of this map? (or, even better, is this always the case?) -Here's one property of the map $f$ that I can add regarding the requirement that arguments be distinct: if $\lbrace \mathbf{x}_k \rbrace$ is a sequence of $n$-tuples (with distinct entries) in $B$ that converges to an $n$-tuple $\bar{\mathbf{x}}$ with (possibly) non-distinct entries, then the limit of $f(\mathbf{x}_k)$ exists if and only if, for each pair of entries $x_i^k$ and $x_j^k$ in the $n$-tuple, the unit direction vector from $x_i^k$ to $x_j^k$ (i.e. $\frac{x_i^k - x_j^k}{||x_i^k - x_j^k||}$) has a limit. - -REPLY [12 votes]: Let me rephrase the question and give a complete answer: -What are conditions for the existence of a $\Sigma_n$-equivariant map -$$f: F_n(B)\to \Delta_{n-1}\backslash(\tfrac{1}{n},\dots,\tfrac{1}{n})\; ?$$ -Here $B\subset \mathbb{R}^2$ denotes the closed unit ball, $F_n(B)$ is the ordered configuration space of $n$ points on $B$ with $\Sigma_n$ acting by permutation and $\Sigma_n$ acts on $\Delta_{n-1}:=\{(x_1,\dots,x_n)\in\mathbb{R}^n: \sum_ix_i=1, 0\leq x_i\leq 1\}$ by permuting the coordinates. -As Neil Strickland pointed out, using an equivariant homeomorphism, we can replace -$\Delta_{n-1}\backslash(\tfrac{1}{n},\dots,\tfrac{1}{n})$ by $\partial\Delta_{n-1}\cong_{\Sigma_n} S^{n-2}.$ This sphere $S^{n-2}$ can be viewed as $S(W_n)$, where $W_n:=\{(x_1,\dots,x_n)\in\mathbb{R}^n: \sum_ix_i=0\}$ and $\Sigma_n$ is again acting by permuting the coordinates. Using equivariant homotopy equivalences, we can replace $F_n(B)$ by $F(\mathbb{R}^2)$ and further use an equivariant model of dimension $n-1$, which Neil Strickland calls $X_0$ and I will call $\mathcal{F}_n(\mathbb{R}^2)$. So an equivalent question is: -What are conditions for the existence of a $\Sigma_n$-equivariant map -$$f: \mathcal{F}_n(\mathbb{R}^2)\to S(W_n)\; ?$$ -Fortunately, there is a complete answer to that questions; it is given by Blagojević and Ziegler: arxiv, springer: -Such a map exists if and only if $n\geq 2$ is a not a prime power -The authors use equivariant obstruction theory and an explicit model $\mathcal{F_n(\mathbb{R}^2)}$, which is called $\mathcal{F}(d,n)$ in their notation. Just plug in $d=2$ and the result is the main theorem of section 4 in the paper on the arxiv. -From the nonexistence of the map we conclude that the barycenter $(\frac{1}{n},\dots,\frac{1}{n})$ is hit: If $n$ is a prime power, then your continous equivariant function $B^n\to\Delta_{n-1}$ will always hit the barycenter and if $n$ is not a prime power, there exist equivariant functions that miss the barycenter.<|endoftext|> -TITLE: Does the random Lorenz gas have a non-trivial diffusion coefficient? -QUESTION [6 upvotes]: For the periodic Lorenz gas Sinai showed that rescaling the trajectory of the tracer particle yields Brownian motion in the limit. Does there exist a similar result for the random Lorenz gas? If not, do people believe that there is such a limit? -By the random Lorenz gas I mean: take circular scatterers distributed uniformly at random in the plane conditioned on the scatterers not overlapping. The scatterers are fixed. The tracer particle is a point that moves with constant speed and has perfectly elastic collisions with the scatterers. An initial condition is chosen at random (say, by picking an initial point away from a scatterer and then picking the initial angle uniformly on $[0,2\pi)$.) -The numerical experiments in Dettmann and Cohen, 2000 suggest that there is diffusive behaviour for the random Lorenz gas. This article by Bunimovich states that it is believed that velocity autocorrelation decays polynomially, but does not mention whether it decays fast enough for there to be a finite diffusion coefficient. - -REPLY [2 votes]: I have been looking at random (and periodic) Lorentz gas models recently in preparing a review which has just appeared at arxiv:1402.7010. As far as I know, there are rigorous results only for the low density (Boltzmann-Grad) limit, and for models where the scatterers are known not to overlap (for example placing a scatterer or empty space randomly at each site of a lattice). In the fixed density case there does not even appear to be a proof that "conditional on the scatterers not overlapping" converges. Choosing a point outside a scatterer at random is problematic since the measure is infinite; better to fix the point at the origin and then choose the scatterers conditional on not overlapping each other or the origin. The low density limit suggests that the velocity autocorrelation decays as $t^{-d/2-1}$, which is fast enough for the diffusion coefficient (its integral) to exist. -Correction: After the above paper appeared online (Commun. Theor. Phys. 62 521-540 (2014).), and following discussions with D. Szasz, I discovered references that do indeed show that the non-overlapping condition converges. See A variational principle for the equilibrium of hard sphere systems, Gallavotti, G and Miracle-Sole, S, Annales IHP A 8 287-299 (1968); appendix B of Observables at infinity and states with short range correlations in statistical mechanics, Lanford III, OE and Ruelle, David, Commun. Math. Phys. 13 194-215 (1969). But the diffusion question remains open.<|endoftext|> -TITLE: A bijection between sets of Young tableaux of two kinds -QUESTION [27 upvotes]: Assuming that the problem of exhibiting a bijection is not considerd a frivolous pursuit, allow me to ask a question troubling me for some time now. -Let $\lambda \vdash n$ denote the fact that $\lambda$ is a partition of $n$. Denote the number of parts by $l(\lambda)$. If $T$ is a standard Young tableau (SYT), we will denote the underlying partition shape by $sh(T)$. -Given a positive even integer $2n$, let -$$ Pe_{2n}=\{ \lambda: \lambda\vdash 2n,\text{ } l(\lambda) \leq3 \text{ and all parts of } \lambda \text{ are even} \}$$ -and -$$ Qe_{2n}=\{ \lambda: \lambda\vdash 2n, \lambda = (k,k,1^{2n-2k}), \text{ }k\geq 1 \}$$ -Using these sets we will define two more sets whose elements are SYTs. -$$ TPe_{2n}=\{T: T \text{ an SYT, } sh(T)\in Pe_{2n} \}$$ -and -$$ TQe_{2n}=\{T: T \text{ an SYT, } sh(T)\in Qe_{2n} \}$$ -$\textbf{Question}$: -Is there a bijective proof exhibiting the fact that the cardinalities of $TPe_{2n}$ and $TQe_{2n}$ are equal? -The second question is very similar. Given an odd positive integer $2n+1$, let -$$ Po_{2n+1}=\{ \lambda: \lambda\vdash 2n+1,\text{ } l(\lambda)=3 \text{ and all parts of } \lambda \text{ are odd} \}$$ -and -$$ Qo_{2n+1}=\{ \lambda: \lambda\vdash 2n+1, \lambda = (k,k,1^{2n+1-2k}), \text{ }k\geq 1 \}$$ -Using these sets we will define two more sets whose elements are SYTs. -$$ TPo_{2n+1}=\{T: T \text{ an SYT, } sh(T)\in Po_{2n+1} \}$$ -and -$$ TQo_{2n+1}=\{T: T \text{ an SYT, } sh(T)\in Qo_{2n+1} \}$$ -$\textbf{Question}$: -Is there a bijective proof exhibiting the fact that the cardinalities of $TPo_{2n+1}$ and $TQo_{2n+1}$ are equal? -I tried quite a few approaches ( Motzkin path interpretations, matching diagrams etc) but did not succeed. I hope somebody here can guide me. -The relevant OEIS entry would be link text -Thanks, -Vasu - -REPLY [10 votes]: Let me explain a way to prove such results. There are (mostly) two types of people out there, those that think of the $n$th Catalan number as $\frac{1}{n+1}\binom{2n}{n}$ and those that think of it as $\binom{2n}{n}-\binom{2n}{n-1}$. The first type of people will likely think of Dyck paths as eqivalence classes or $\mathbb Z/(n+1)\mathbb Z$ orbits of paths, whereas the second type will probably think of Dyck paths as "paths inside an $n\times n$ rectangle that don't cross the origin" (See proof 2 vs proof 3 here). While the first type would probably be happy to count standard young tableaux using the usual form of the hook-length formula, the second type would probably prefer the (equivalent): -Jacobi-Trudi Formula -The number of Standard Young Tableaux on a partition of shape $\lambda=(\lambda_1,\dots,\lambda_k)$ of $n$ is -$$f^{\lambda}=\sum_{\sigma \in S_k}\varepsilon(\sigma)\binom{n}{\lambda_1+k-\sigma(k)\, ,\dots,\,\lambda_k+1-\sigma(1)}.$$ -Where $\varepsilon$ denotes the sign of a permutation. Similar to the Catalan numbers in this presentation one sacrifices obvious positivity but has obvious integrality. For the sake of illustration, in the case $k=3$, this says -$$f^{(a,b,c)}=\binom{n}{a,b,c}+\binom{n}{a+1,b+1,c-2}+\binom{n}{a+2,b-1,c-1}-\binom{n}{a+1,b-1,c}-\binom{n}{a,b+1,c-1}-\binom{n}{a+2,b,c-2}.$$ -The computation: -Let's take the case of three even parts, although this method will work for the odd case as well as the general case of partitions with a fixed (arbitrary) number of rows. If one sums the quantity $f^{(a,b,c)}$ over all triples of integers $a\geq b\geq c\geq 0$ with $a+b+c=n$ then the multinomial coefficients form the formula above telescope and we are left with -$$|TPe_{2n}|=\sum_{k\geq 0}\binom{2n}{k,k,2n-2k}-\sum_{k\geq 0}\binom{2n}{k,k+1,2n-2k-1}$$ -Now we can turn our attention to $TQe_{2n}$. We can count the number of SYT of shape $(k,k,1^{2n-2k})$ from the hook length formula, but instead of presenting it in "type one" as Brian Hopkins did in the other answer, we will use the "type two" presentation $$f^{(k,k,1^{2n-2k})}=\binom{2n}{k}\binom{2n-k-1}{k-1}-\binom{2n}{k+1}\binom{2n-k}{k}$$ -$$=\binom{2n}{k,k,2n-2k}-\binom{2n}{k}\binom{2n-k-1}{k}-\binom{2n}{k-1}\binom{2n-k}{k}.$$ -So in order to get the desired equality we sum over all $k\geq 1$ -$$|TQe_{2n}|=\sum_{k\geq 1}f^{(k,k,1^{2n-2k})}=\sum_{k\geq 1}\left(\binom{2n}{k,k,2n-2k}-\binom{2n}{k}\binom{2n-k-1}{k}-\binom{2n}{k-1}\binom{2n-k}{k}\right)$$ -$$=\sum_{k\geq 1}\binom{2n}{k,k,2n-2k}-\sum_{k\geq 0}\left(\binom{2n}{k}\binom{2n-k-1}{k}+\binom{2n}{k}\binom{2n-k-1}{k+1}\right)+\binom{2n}{0}\binom{2n-1}{0}$$ -$$=\sum_{k\geq 1}\binom{2n}{k,k,2n-2k}-\binom{2n}{k,k+1,2n-2k-1}+\binom{2n}{0,0,2n}$$ -$$=\sum_{k\geq 0}\binom{2n}{k,k,2n-2k}-\sum_{k\geq 0}\binom{2n}{k,k+1,2n-2k-1}=|TPe_{2n}|.$$ -Can it be made bijective? -Yes. All the manipulations with binomial coefficients have obvious bijective interpretations in terms of lattice paths, so one would have to do the same for the Jacobi-Trudi formula mentioned above. This is classical for the case of Catalan numbers, but it can be generalized for any partition using a similar reflection trick. I have been lazy and didn't work out the final bijection for you, but at least this is a start. I am currently writing a note about such bijections between SYT's with bounded parts, so if no one has explained it by then I'll make sure to update this answer.<|endoftext|> -TITLE: To what extent do we know the representations of GL(2,Zp) -QUESTION [8 upvotes]: Consider a local non archimedean field $k$ and its ring of integer $o$. To what extent, do we know the complex irreducible representations of $GL(2,o)$? Is there a specific list giving them all in terms of induction from certain "simple" subgroups? -I have had read, that they have been classified according to their characters, but I am not quite happy with the various presentations so far. -What I'd like to see is pretty simple to explain: - -Classify all irreducibles of the group (Borel subrgoup) of upper triangular matrices $B$ (easy). -Given an irreducible representation $\pi$ of $B$, decompose $Ind_{B}^{GL(2,o)} \pi$ in its irreducibles (hard). - -Is there such a treatment available? If not, what is the reason why not to proceed along these lines? - -REPLY [6 votes]: Paul Broussous has already addressed the question in the title, and since it seems that you are in fact more interested in the last two questions, I will focus on these. -I don't think there is a treatment following steps 1) and 2) available in the literature, but it should be possible to work this out since the representations of $\mathrm{GL}_2(\mathcal{O})$ are known explicitly. On the other hand, it is not clear how useful this would be, because already for $\mathrm{GL}_2$ over a finite field $\mathbb{F}_q$, this is not the preferred approach to the representations (apart from the principal series, of course). This is partly because it quickly becomes unmanageable, and does not allow for an inductive approach in the same way as parabolic induction. -Moreover, over a finite field even step 1) is problematic in general, because the irreps of $B(\mathbb{F}_q)$ are related (via Clifford theory) to the irreps of the upper uni-triangular subgroup $U(\mathbb{F}_q)$, and the classification of the latter is known to be a wild problem (for large enough p, one can describe the irreps of $U(\mathbb{F}_q)$ using a Kirillov orbit method formalism, but it is not clear that this would be helpful for the problem at hand). -Also, in step 2) there is going to be a lot of repetition, that is, two induced representations may have irred constituents in common, and there may be no easy way to tell when this happens and for which irred constituents. -Finally, as you mention in one of the comments, the approach via steps 1) and 2) can only work if there is a good description of the $B$-$B$ double cosets, and this is not available in general. Note however that this is not the only place where wildness enters, as the above remarks on the situation for groups over a finite field shows. One advantage of the approach via Clifford theory and orbits (following Hill), is that one can avoid wild classification problems by restricting the construction to a subcategory of representations, such as the regular representations.<|endoftext|> -TITLE: Mutually generics -QUESTION [7 upvotes]: Given posets $P,Q\in M$, I would like to know under what circumstances there are mutually generic filters $G\subseteq P$ and $H\subseteq Q$ (generic over $M$). Also, what are the characterizations of mutual genericity? And finally, what can we say about the relation between $M[G]$ and $M[H]$ in that case? -I have a slight difficulty finding references to the notion and properties of mutual genericity (whatever they are). - -REPLY [8 votes]: Another fact to add to Joel's list, one which I found surprising when I first came across it, is that adding a single Cohen real adds a perfect set of reals such that any finitely many are mutually generic over each other. To see this simply consider the trace of a Cohen real on the members of an almost disjoint family and use the theorem Joel quotes about products.<|endoftext|> -TITLE: Dimension 1 prime ideals in the intersection of two maximal ideals -QUESTION [17 upvotes]: This question/problem really comes from a fact in algebraic geometry, where it says that given an irreducible variety $V$ ($\dim V \geq 2$) then for any given pair of points $x,y\in V$ there is an irreducible curve $C$ connecting them. -This can proved by invoking Bertini's theorem [See Lazarsfeld R. - Positivity in Algebraic Geometry 1, example 3.3.5]. -Translating this back to coordinate rings we get something like: Given a finitely generated $k$-algebra $R$ which is a domain, $k=\bar{k}$ (alg.closed) then for any pair of maximal ideals $m_1,m_2\subset R$ there is a prime ideal $p\subset m_1\cap m_2 $ with $\dim R/p =1$. -So my question is therefore: Is there a purely ring theoretical argument for this fact? In what generality does it hold? -The reason why i emphasize on purely is because, and i am no expert on this, but apparently the Bertini type argument can be (re-)formulated algebraically, however this, I have been informed, will not be pretty... - -REPLY [4 votes]: This reminds me of what someone once said to me: - -The geometers can always take a hyperplane section. We can't! - -The purpose of this post is to analyze the question to see how close it is to Bertini's theorem (it is not obvious to me). The statement is immediately equivalent to: one can always find a nonzero prime ideal $P \subseteq m_1\cap m_2$ (since if we can, then induction on dimension proves the original). -Now, how can such $P$ exist? Let $U_1 = R-m_1$ and $U_2=R-m_2$. Let $U=\{xy \| x\in U_1, y\in U_2\}$. $U$ is multiplicative and our prime $P$ obviously just has to avoid $U$. So $P$ exists unless the localization $U^{-1}R$ has dimension $0$. But it is a domain, so we have to make sure $U^{-1}R$ is not a field. That statement is equivalent to the existence of some element $f\in R$ such that $f$ does not become an unit in $U^{-1}R$. In other words: - -there are no $a,b \in U$ such that $af=b$. - -Since $b$ is itself a product of elements in $U_1,U_2$, our condition is obvious if $fR$ is a prime ideal and $f\in m_1\cap m_2$. But it is technically weaker, although not clear to me by how much. Note that we do not require $f$ to be linear, which is common for Bertini's type statement. -This analysis would seem to rule out certain clever arguments. But may be one can find some! -For "algebraic" version of Bertini's theorem, I will look at the reference given here. Also, how to rescue Bertini over finite fields using hypersurface instead of hyperplane (which suits your purpose), looks here.<|endoftext|> -TITLE: Reference for a formula expressing the characteristic polynomial of a sum of endomorphisms -QUESTION [9 upvotes]: Let $R$ be a ring, $A$ a (not necessarily commutative) $R$-algebra and $M$ a (left) A-module which is free of finite rank as an $R$-module. If $a\in A$ then multiplication with $a$ on $M$ is an $R$-linear endomorphism of $M$ and as such it has a characteristic polynomial $\chi_a$. I've learned from notes by Bart de Smit which I've accidentally found via Google (http://www.math.leidenuniv.nl/~desmit/notes/charpols.pdf) that the function $\chi_\bullet$ is determined by its values on a generating set of $A$ as an $R$-module. He proves this by deriving suitable formulas for the characteristic polynomial of a sum of two endomorphisms. -I'm looking for a reference for this fact that can be cited more easily than the informal notes above. - -REPLY [7 votes]: One reference is: S. A. Amitsur, On the Characteristic Polynomial of a Sum of Matrices, -Linear and Mult. Algebra 8 (1980), 177-182. (pp. 469-474 in Selected Papers of S. A. Amitsur, -Part 2, AMS 2001.)<|endoftext|> -TITLE: Terminology question: "Transverse" v. "Transversal" -QUESTION [10 upvotes]: Something that's always bothered me is that the word "transversal" is very commonly used as an adjective, but my understanding is that "transverse" is the correct adjective, and that "transversal" is a noun which means "an object which is transverse [to a given object]." So for example you would say "transverse intersection" and "pick a transversal for the line." -However, I could be wrong. I'd like people to answer with their opinion on which is the correct word for the adjective. I'm making this a community wiki since it's too soft to gain reputation over. - -REPLY [19 votes]: "Transversal" is a good old geometry word, a noun, as you say. It goes way back to long before anybody was thinking of transversality in the modern sense. -It grates on me to hear it used as an adjective, and this owes something to the fact that in my impressionable youth I saw one of the chapter-heading quotations in Hirsch's graduate text on differential topology: From Whitehead, "'Transversal' is a noun. The adjective is 'transverse'". No doubt this also had an impact on others who (like me) tend to be fussy about language. -On the other hand, language does drift along, and there's no stopping it, and generally no harm is done. By the time you perceive a serious need to tell the world that some usage is wrong, a case can always be made that it is no longer wrong. -In the case at hand it's understandable that "transversal" has come to be used an adjective; after all, "-al" looks like an adjective ending. (But there are words in English where people have been fooled by that, changing the language. "Bridal" is an example.) -By the way, if we were going to be sticklers on this point, mightn't we want to go back and change "transversality" to "transversity"?<|endoftext|> -TITLE: Is there a "trianguline period ring", or is one expected? -QUESTION [13 upvotes]: Consider a finite-dimensional $\mathbf{Q}_p$-vector space $V$ and a continuous representation $\rho : G_{\mathbf{Q}_p} \to \mathrm{GL}(V)$. Fontaine introduced various $\mathbf{Q}_p$-algebras with $G_{\mathbf{Q}_p}$-actions, notated $B_{\bullet}$ where $\bullet \in \left\{\mathrm{crys}, \mathrm{st}, \mathrm{dR}\dots \right\}$, which "classify" interesting representations $V$; we say $V$ is $\bullet$ if equality holds in the relation $\mathrm{dim}(B_{\bullet} \otimes V)^{G_{\mathbf{Q}_p}} \leq \dim{V}$. -Recently, the notion of a "trianguline" Galois representation has become increasingly important. Avoiding the precise definition, I will say rather perversely that the trianguline representations are roughly the closure of the crystalline locus in the set of all $\rho$'s as above (made precise, this is a theorem of Chenevier which is of course predicated on the actual definition of trianguline). So my question: is there a ring of periods $B_{\mathrm{tri}}$ which classifies trianguline representations in the above manner, and/or is such a ring expected to exist? Is/should it be a suitable "completion" of $B_{\mathrm{crys}}$? - -REPLY [14 votes]: The category of trianguline representation is stable under all the usual representation-theoretic operations (subs, quotients, $\oplus$, $\otimes$), so by some general tannakian formalism, there does exist a ring $B_{tri}$. The rough idea is to look at $Q_p^{alg} \otimes B_{st} \langle \langle \log(t) \rangle \rangle$ where "$\langle \langle \log(t) \rangle \rangle$" means "power series with some non zero radius of convergence" and $t$ is the usual $t$ in this business. It's interesting to note that I first heard about this ring from Fontaine (around 2003-04 maybe - I was still at Harvard) when trianguline representations had not yet been defined. Fontaine told me at the time that repns admissible for this ring should be interesting! A few comments are in order: - -$B_{st}$ does not have the structure of a Banach space, so you need to figure out what "radius of convergence" means -$\exp(\log(t))=t$, so there are relations in the definition of your ring -you need to decide if you want a ring for "trianguline" or "split-trianguline" repns - -I thought about this again a few weeks ago and, if I remember correctly, came to the conclusion that if you take $B = Q_p(\mu_p) \otimes \hat{Q}_p^{nr} \otimes B_e \otimes Q_p \langle \langle \log(t) \rangle \rangle \otimes Q_p[\log(\tilde{p})]$ (whew!), where $B_e=B_{cris}^{\phi=1}$, then $B$-adm reps of $G_{Q_p}$ are trianguline, and conversely split trianguline reps of $G_{Q_p}$ with integer slopes are $B$-adm. This hopefully gives an idea of the kind of ring which one should be looking for.<|endoftext|> -TITLE: In what generality is the Verdier biduality map an isomorphism? -QUESTION [5 upvotes]: Let $X$ be a finite-dimensional, locally compact topological space, and consider the dualizing complex $K_X \in \mathbf{D}^b(X,k)$ (bounded derived category of $k$-sheaves, where $k$ is a noetherian ring). We can define the dualizing functor -$$C \mapsto D(C) = \mathbf{R}\mathcal{H}om(C, K_X),$$ -(derived internal hom), -which leads to a biduality map $C \to D^2(C)$. In SGA 4.5 "Th. de finitude," Deligne shows that, when $k = \mathbb{Z}/n$, the analogous biduality morphism is an isomorphism on the constructible bounded derived category (so, an anti-involution of said category) when one is working with a scheme of finite type over a field or DVR (with $n$ prime to the characteristic). I have heard that the same is true for topological spaces under certain conditions, although I'm not sure what the statement (or proof) should be: first, presumably we are going to want with a nice (Gorenstein?) ring like $\mathbb{Z}/n$, and second, probably there needs to be some analog of the constructible derived category. What is this statement? -I had a look at Kashiwara-Schapira's "Sheaves on Manifolds," but I can't parse the biduality statement given in chapter 3. It's not clear to me how to adapt Deligne's argument to the present case, anyway. - -REPLY [3 votes]: For an analytic space, you can find this on page 118 of Verdier's article "Classe d'homologie -d'un cycle" in Asterisque 36-37. And yes this is on an appropriate constructible derived category with $\mathbb{Z}$-coefficients. I seem to recall that Borel, in his book on intersection cohomology, also discusses this for pseudomanifolds, in case you need something -more general.<|endoftext|> -TITLE: Eigendecomposition after multiplying by diagonal matrix -QUESTION [10 upvotes]: Hello, -If we possess the eigendecomposition of a positive definite matrix: $X = U \Sigma U^T$, is there an efficient way to compute the eigendecomposition of $D X D$ where $D$ is a diagonal matrix? - -REPLY [9 votes]: Write $\Sigma$ as $T^2$, for positive definite $T$. Set $Y = U T$. -So the eigenvalues of $X$ are the squares of the singular values of $Y$, and what you want to compute are the singular values of $DY$. -There is no formula which gives the singular values of $DY$ in terms of those of $Y$ and $D$. However, there is a famous set of inequalities relating the three sets of singular values, called the Horn inequalities. See Bhatia's article Linear Algebra to Quantum Cohomology, particularly Section 11, for a gentle introduction.<|endoftext|> -TITLE: BGG resolution and representations of parabolic subalgebras -QUESTION [8 upvotes]: Everything here is over $\mathbb{C}$. -Let $\mathfrak{g}$ be a finite-dimensional simple Lie algebra and let $\mathfrak{p}$ be a parabolic subalgebra (relative to some fixed Borel subalgebra that is unimportant for this question). -Then $\mathfrak{p}$ has a decomposition -$$ \mathfrak{p} = \mathfrak{l} \oplus \mathfrak{u_+},$$ -where $\mathfrak{l}$ is a reductive subalgebra (the Levi factor of $\mathfrak{p}$) and $\mathfrak{u}_+$ is a nilpotent ideal (the nilradical of $\mathfrak{p}$). -Finally, we can decompose $\mathfrak{g}$ as -$$\mathfrak{g} = \mathfrak{u}_- \oplus \mathfrak{l} \oplus \mathfrak{u}_+$$ -(as $\mathfrak{l}$-modules), where $\mathfrak{u}_-$ and $\mathfrak{u}_+$ are dual to each other via the Killing form of $\mathfrak{g}$. -Assume further that the following (equivalent) conditions hold: - -$\mathfrak{g}/ \mathfrak{p}$ is irreducible as a $\mathfrak{p}$-module; -$\mathfrak{u}_-$ is irreducible as an $\mathfrak{l}$-module; -$\mathfrak{u}_-$ is an abelian Lie algebra; -2 and 3 with $\mathfrak{u}_-$ replaced by $\mathfrak{u}_+$. - -Buzzwords here are "Hermitian symmetric space" and "generalized flag variety." There is a classification of these in terms of root systems but I don't want to use that. -I need to understand the decomposition of ${\bigwedge}^2 \mathfrak{u}_- $ into irreducible modules for $\mathfrak{l}$. -Using the classification of these parabolics, you can just see explicitly what the highest weight of $\mathfrak{u}_-$ is, and then it's not too hard to compute what ${\bigwedge }^2 \mathfrak{u}_-$ is, but I would like a more elegant way to see what's going on here. -I have been informed that there is some version of the BGG resolution that will be helpful for this - this evidently gives the highest weights of ${\bigwedge }^2 \mathfrak{u}_-$ in terms of the dotted action of some elements of the Weyl group on the highest weight of $\mathfrak{u}_-$, but at this point I'm stuck. -I don't know enough (ok, anything really) about the BGG resolution to know where to look for this stuff. Either an explanation or a reference would be much appreciated. -Edit: I would also be happy with a pointer to a nice reference for the BGG resolution in general. - -REPLY [4 votes]: The question involves a fairly long history, going back at least to Bott's Annals paper and two long follow-up papers by Kostant, "Lie algebra cohomology and the generalized Borel-Weil theorem", Ann. of Math. (2) 74 (1961), 329–387, plus the 1963 paper mentioned by rObOt. Those papers predate the work on Verma modules and BGG resolution in the 1970s, which however had Bott's theorem as a byproduct. The BGG arguments made a comparison with Lie algebra cohomology, which in turn involved an easy computation of the 1-dimensional representations of a Cartan subalgebra (Levi subalgebra of a Borel subalgebra) on exterior powers of the nilradical; see 6.4 in my book on the BGG category. This gets less elementary in the parabolic case considered here. -The special situation with Hermitian symmetric spaces is surveyed at the end of Chapter 9 in my book, along with references to what I think are the main developments in that direction. Again Kostant's papers are a key starting point, but the later approaches using the parabolic BGG resolution were explored by people like Lepowsky (a student of Kostant), Boe, Collingwood, Irving, Shelton. Eventually this work enters the more complicated realm of infinite dimensional representations and Kazhdan-Lusztig-Vogan theory. I'm not sure what the best answer is to the narrower question raised here, but most work on Hermitian symmetric spaces has required case-by-case study using the standard classification.<|endoftext|> -TITLE: Euler Characteristic of a Variety -QUESTION [9 upvotes]: Let $Y$ be a "nice" scheme. I am thinking projective varieties over an algebraically closed field, for now, but I am open to more general results. -In terms of singular homology (coefficients in $\mathbb{Z}$), one can define the Euler characteristic $\chi(Y)$. -My question is: - -Can I express $\chi(Y)$ in terms of the Euler characteristic of certain coherent sheaves on $Y$, in terms of sheaf cohomology? - -Most preferably, I would like $$\chi(Y)=\chi(Y,\mathcal{F})$$ for some particular sheaf $\mathcal{F}$. -I am sorry if this is really trivial or widely known, my searching and asking (in the real world) has led me nowhere so far. - -REPLY [7 votes]: First, the answer/reference here might be exactly what you are looking for. -On the other hand, perhaps you just want to relate natural algebro-geometric structure to the classical Euler characteristic. Here is one way to do that: -A pure Hodge structure of weight $k$ is a finite dimensional complex vector space $V$ such that $V=\bigoplus_{k=p+q} H^{p,q}$ where $H^{q,p}=\overline{H^{p,q}}$. This gives rise to a descending filtration -$F^{p}=\bigoplus_{s\ge p}H^{s,k-s}$. Define $\mathrm{Gr}^{p}_{F}(V)=F^{p}/ F^{p+1}=H^{p,k-p}$. -A mixed Hodge structure is a finite dimensional complex vector space $V$ with a real ascending weight filtration $\cdots \subset W_{k-1}\subset W_k \subset \cdots \subset V$ and a descending Hodge filtration $F$ such that $F$ induces a pure Hodge structure of weight $k$ on each $\mathrm{Gr}^{W}_{k}(V)=W_{k}/W_{k-1}$. Then define $H^{p,q}= \mathrm{Gr}^{p}_{F}\mathrm{Gr}^{W}_{p+q}(V)$ and $h^{p,q}(V) =\dim H^{p,q}$. -Let $Z$ be any quasi-projective algebraic variety. The cohomology groups with compact support $H^k_c(Z)$ are endowed with mixed Hodge structures by seminal work of Pierre Deligne. -The Hodge numbers of $Z$ are $h^{k,p,q}_{c}(Z)= h^{p,q}(H_{c}^k(Z))$, and the $E$-polynomial is defined as - $$ - E(Z; u,v)=\sum _{p,q,k} (-1)^{k}h^{k,p,q}_{c}(Z) u^{p}v^{q}. - $$ -From this, one gets the classical Euler characteristic $\chi(Z)=E(Z;1,1)$. -Note: if the counting function of $Z$ over finite fields is a polynomial in the order of the finite field, then $E(Z)$ is exactly the counting polynomial. From this point-of-view, in this case, the Euler characteristic is the number of $\mathbb{F}_1$-points.<|endoftext|> -TITLE: Positive cone of a subgroup of $\mathbb{Z}^n$ -QUESTION [7 upvotes]: This question sounds like it should be very well known, but for some reason I failed to find a decent answer anywhere. Let $G\subset\mathbb{Z}^n$ be a subgroup, and $G_+=G\cap\mathbb{Z}_{\ge0}^n$ be a cone of elements in $G$ whose coordinates are all nonnegative. The semigroup $G_+$ is finitely generated; it can be proved in a couple of ways. My question is, is there an effective upper bound on the number of generators in terms of $n$, or, even better, the rank of $G$? - -REPLY [11 votes]: No, there is no upper bound on the number of generators in terms of $n$. -Let $k$ be arbitrary positive integer. -Consider the subgroup -$$G=\{(x,y)\in\mathbb Z^2\mid x+y\equiv 0\pmod k\}.$$ -Any set of generators of $G_+$ -contains all the elements $(x,y)$ such that $x,y\ge 0$ and $x+y=k$. -Therefore the number of generators of $G_+$ is $k+1$.<|endoftext|> -TITLE: Current status of Waring-Goldbach problem -QUESTION [10 upvotes]: Is the following statement proved? -For any positive integer $k$ there exists positive integer $n$ such that all sufficiently large integers may be represented as $p_1^k+p_2^k+\dots+p_n^k$ for primes $p_1,\dots,p_n$. -This Wiki article claims that some progress is made only for $k$ up to 7, on the other hand, it refers to Hua Lo Keng's monograph, in which this statement is proved, if we may take zero terms instead some prime powers. This zero does not seem to be so essential on the first glance... And on the third hand, this paper of Chubarikov contains announcement of the complete proof, though I did not succeed in finding any reaction on it, even no MathSciNet-review. - -REPLY [9 votes]: This is a corrected version of my original response, incorporating a nice argument by Fedor Petrov. -Hua in his book (cf. review of MR0124306) proved that there are integers $s,K,N>0$ such that every $n>N$ with $n\equiv s\pmod{K}$ is a sum of $s$ $k$-th powers of primes. For any $t>0$ let $M(t)$ denote the set of residues modulo $K$ which can be represented by a sum of $t$ $k$-th powers of primes. Clearly $M(t+1)$ contains $M(t)+p^k$ for any prime $p$, hence $|M(t+1)|> |M(t)|$ unless $M(t+1)$ equals $M(t)+p^k$ for any prime $p$. In this case $M(t)$ is invariant under the shift of $p^k-q^k$ for any two distinct primes $p$ and $q$. The shifts are coprime (e.g. $p^k-q^k$ is coprime with $q$), hence $M(t)=M(t)+1$, and $M(t)$ contains all residues modulo $K$. This argument shows that $|M(K)|=K$, i.e. modulo $K$ every residue class is a sum of $K$ $k$-th powers of primes. If $p$ denotes the largest of the $K^2$ primes used in the latter representation, and $M$ equals $N+Kp^k$, then we have the following. For every $m>M$ there is a sum of $K$ $k$-th powers of primes, denote it by $m'$, such that $m-m'\equiv s\pmod{K}$ and $m-m'>N$. By Hua's theorem, $m-m'$ is a sum of $s$ $k$-th powers of primes, hence in fact every $m>M$ is a sum of $s+K$ $k$-th powers of primes. -To summarize: the statement in Fedor Petrov's original question follows from Hua's theorem.<|endoftext|> -TITLE: Euler product over primes congruent to 3 mod 4 -QUESTION [9 upvotes]: I'm a string theorist and I have come across the following expression in a computation I'm doing (involving a sum over inequivalent Lens spaces): -$$\widehat{\zeta}(s)=\prod_{\mathrm{primes}\ p\equiv 3\ (\mathrm{mod}\ 4)}(1-p^{-s})^{-1}.$$ -I expected this to be a standard sort of construction, and it's clearly closely related to certain standard zeta-functions and Dirichlet $L$-functions (for instance it seems that $\widehat{\zeta}(s)^2/\widehat{\zeta}(2s)=L(\chi_0,s)/L(\chi_1,s)$, where $\chi_0$ and $\chi_1$ are the trivial and nontrivial Dirichlet characters modulo $4$), but with my poor knowledge of this very wide field, I wasn't quite able to find this particular combination. I would like to extend this analytically in $s$ and am particularly interested in any poles along the real axis and their residues. -Any help or references would be much appreciated, and apologies if there was an easily found answer that I missed. - -REPLY [7 votes]: Your functional equation shows that $\widehat{\zeta}(s)^2$ has a meromorphic continuation to $\Re(s)>1/2$ with a simple pole at $s=1$. This also shows that $\widehat{\zeta}(s)$ does not have a meromorphic continuation to a neighborhood of $s=1$, instead it lives on a double cover of that neighborhood branched at $s=1$. -If you want to extend further to the left, you need even higher powers of $\widehat{\zeta}(s)$: precisely the $2^k$-th power for $\Re(s)>2^{-k}$. The poles on the positive real axis will be at the points $s=2^{-k}$, for the relevant powers of $\widehat{\zeta}(s)$. This also tells me that there is no reasonable continuation to the left of the imaginary axis.<|endoftext|> -TITLE: For $\mathfrak g$ A Lie algebra of type $ E_7 $, $\mathfrak h $ a Cartan subalgebra and $\Delta$ the resulting root system, does $ Aut(\mathfrak g,\mathfrak h)\rightarrow Aut(\Delta) $ split over the Weyl group? -QUESTION [13 upvotes]: Given a complex simple Lie algebra $ \mathfrak g $ of type $E_7$, Cartan subalgebra $ \mathfrak h $ and simple roots $\alpha_1,…\alpha_n $, suppose $\pi $ is an involution of the extended Dynkin diagram. I would like to know whether $\pi $ must be induced from an involution of $ \mathfrak g $. Writing $\Delta $ for the root system and W for the Weyl group, since $ Aut(\Delta)/W $ is isomorphic to the group of automorphisms of the Dynkin diagram and this is trivial for $ E_7 $, the answer is "yes" if the map $ Aut(\mathfrak g,\mathfrak h)\rightarrow Aut(\Delta) $ splits over the Weyl group. -That is, it would suffice if there is a subgroup of the automorphism group of $Aut(\mathfrak g, \mathfrak h)$ which is isomorphic to the Weyl group under this mapping. -Is this true? Or do you otherwise know whether every involution of the extended Dynkin diagram for $ E_7 $ must arise from an involution of $ \mathfrak g $? -Thanks very much! - -REPLY [3 votes]: Thanks very much all for these most helpful answers, much appreciated! Regarding the 2nd (and easier) part of the question, as you suspected this is true. In the preprint http://front.math.ucdavis.edu/1111.4028 Katharine Turner and I were needing this for some work on harmonic maps, and the form of the statement we prove there is below. It seems like something that would be known to folks working in this area, but we couldn't find a reference. -Every involution of the extended Dynkin diagram for a simple complex Lie algebra $\mathfrak {g} ^\mathbb {C} $ is induced by a Cartan involution of a real form of $\mathfrak {g} ^\mathbb {C} $. -More precisely, let $\mathfrak {g}^\mathbb {C} $ be a simple complex Lie algebra with Cartan subalgebra $\mathfrak {t} ^\mathbb {C} $ and choose simple roots $\alpha_1,\ldots,\alpha_N $ for the root system $\Delta (\mathfrak {g} ^\mathbb {C},\mathfrak {t} ^\mathbb {C}) $. Given an involution $\pi $ of the extended Dynkin diagram for $\Delta $, -there exists a real form $\mathfrak {g} $ of $\mathfrak {g} ^\mathbb {C} $ and a Cartan involution $\theta $ of $\mathfrak {g} $ preserving $\mathfrak {t} =\mathfrak {g}\cap\mathfrak {t} ^\mathbb {C} $ such that $\theta $ induces $\pi $ and $\mathfrak {t} $ is a real form of $\mathfrak {t} ^\mathbb {C} $. The Coxeter automorphism $\sigma $ determined by $\alpha_1,\ldots,\alpha_N $ preserves the real form $\mathfrak {g} $.<|endoftext|> -TITLE: How to recognize a finite dimensional algebra is Koszul or quadratic? -QUESTION [8 upvotes]: I have a family of finite dimensional algebras that are directed quasihereditary. I think they might be Koszul algebras and I am wondering what approaches there are to check Koszulness or even quadraticity. I know the quivers of these algebras and can compute Ext^n between simple modules for all n, but I do not have a quiver presentation. I know that there are paths of length 2 and of higher lengths between all vertices of the quiver with nonvanishing Ext^2 so I cannot prove or eliminate quadraticity for trivial reasons. Any thoughts? -I should add that I do not have explicit minimal projective resolutions of the simple modules. - -REPLY [4 votes]: One incredibly useful fact is that it suffices to find linear resolutions of standard modules, not simples. These are usually much easier to find "by hand." Strictly speaking this is stronger than Koszul (the term is "standard Koszul") but in practice it seems rare for a quasi-hereditary algebra to be Koszul and not standard Koszul.<|endoftext|> -TITLE: On meromorphic continuation of zeta function(s) and special values at negative integers -QUESTION [14 upvotes]: Euler developped (at least) two different approaches in order to calculate the values $\zeta(-m)$ of the zeta function $$\zeta(s) = \sum_{n\geq 1} \frac{1}{n^s}$$ at non-positive integers. -In one approach Euler obtained the following formula for the zeta function $$\zeta(s) = (1-2^{1-s})^{-1}\sum_{n=0}^\infty \frac1{2^{n+1}}\sum_{i=0}^n \binom{n}{i} (-1)^i \frac1{(i+1)^s} $$ -by applying the so called Euler transformation to the series $\sum_{n=0}^\infty (-1)^n \frac1{(n+1)^s}$ (see for example the Wiki article Euler summation). -One great advantage of this formula is that you can immediately calculate the values at the negative integers because the infinite series reduces to a finite sum in this case. -Moreover this formula gives at the same time a meromorphic continuation of the zeta functions to the complex plane! -In another approach Euler introduced a formal parameter $t$ and showed that $$(1-2^{m+1})\zeta(-m) = (t \frac{d}{dt})^m(\frac{t}{t+1})|_{t=1} = (\frac d {dx})^m(\frac{e^x}{1+e^x})| _{x=0} $$ This is for example described in the very nice book of Hida "Elementary theory of L-functions and Eisenstein series". -Further, this approach could be generalized to "L-functions of totally real number fields" by Shintani, Cassou-Nouges and others (see again Hida's book, e.g.). -My questions are now the following: - -1) Can the first approach described above generalized to other - Dedekind zeta functions? I completely lack an understanding of the Euler - transformation. Is there a way to "understand" Euler's formula - for the zeta function in a broader sense? -2) The Riemannian approach to the meromorphic continuation of the zeta function is based on looking at the Mellin transformation of Jacobi's $\theta$-function. It is well known, by the work of Hecke, that this approach generalizes to arbitrary Hecke L-functions. On the other hand, with this approach the information about the special values at non-positive integers of the corresponding Hecke L-function are more difficult to reveal (at least as far as I know). (EDIT: I'm aware of the fact, that calculating special values is extremely hard and one of the big challenges in number theory, I just try to understand certain structures underlying this more than beautiful area of mathematics...) - -This makes we wondering whether there is a relation between the approaches of Euler and Riemann to the meromorphic continuation of the zeta function. - -Is there a "principle" that relates Euler's analytic continuation to Riemann's ? - -(One major difference is of course that Euler's expression doesn't involve the archimedean factor at all! I also lack a real understanding on why the archimedean Euler factor appears in Riemann's approach.) - -3) Is there a known relation between the two different approaches of Euler for -calculating the values $\zeta(-m)$? Should one expect one? - -Thank you very much for your time. - -REPLY [2 votes]: In partial answer to 2), one can in fact get information about special values of $L$-functions via theta series. See, for example, Villegas and Zagier's paper "Square roots of central values of Hecke L-series" in the book Advances in Number Theory. (Disclaimer: I'm no expert,and it's certainly a lot more complex than Euler's approach to special values of $\zeta(s)$.)<|endoftext|> -TITLE: Hirzebruch Surfaces -QUESTION [6 upvotes]: Good Morning, -I'm trying to prove that two different definitions of the Hirzebruch Surfaces coincide, and am having problems. Let $a \geq 0$. My first definition for the $a^{th}$ surface is -$X_a= \mathbb{P}(\mathcal{O}(a) \oplus \mathcal{O}) \longrightarrow \mathbb{P}^1_{\mathbb{C}}$ -My second definition is as follows. Let $C_a$ be a degree $a$ rational normal curve, ie the image under $\mathcal{O}(a)$ of $\mathbb{P}^1_{\mathbb{C}}$ into $\mathbb{P}^a_{\mathbb{C}}$, and let $D_a$ be the projective cone over $C_a$. That is, $D_a$ is defined by the same equations which define $C_a$, except now $D_a\subseteq \mathbb{P}^{a+1}$. Then $D_a$ is a surface which is smooth except for the possibly singular point $v=[0,...,0,1]$. Define $Y_a$ to be the blow-up of $D_a$ at $v$. -Why are $X_a$ and $Y_a$ isomorphic? (I want to stick to the algebraic or complex category, no smoothness allowed!) -Robert - -REPLY [7 votes]: The cone $D_a$ has a singular point of type $\frac{1}{a}(1,1)$ at its vertex. Blowing up the vertex, the exceptional divisor is a curve $C \subset Y_a$ isomorphic to $C_a$ and whose self-intersection is $\deg \mathcal{O}_{C_a}(-1)=-a$. -Since $Y_a$ is clearly a geometrically ruled surface over a rational curve (the ruling is given by the strict transform af the system of lines of $D_a$) and $C$ is a section of self-intersection $-a$, it follows $Y_a \cong X_a.$ -Conversely, starting from the surface $X_a$ one can consider the unique section $C$ of negative self-intersection, namely $C^2=-a$; then this section can be blown down by Artin contractibility criterion. -The blow-down of $C$ is precisely the map $\varphi$ associated to the complete linear system $|C+aF|$ in $X_a.$ -Indeed $h^0(X_a, C+aF)=a+2,$ -hence $$\varphi \colon X_a \longrightarrow D_a \subset \mathbb{P}^{a+1}.$$ -It is immediate to check that $\varphi$ is birational onto its image $D_a$, that it contracts $C$, that the ruling of $X_a$ is sent into a family of lines passing through the point $\varphi(C)$ and that a general hyperplane section of $D_a$ is a rational normal curve $C_a$ of degree $a$. -Therefore $D_a$ is a cone over $C_a$ and $X_a$ is isomorphic to the blow-up of $D_a$ at its vertex $\varphi(C)$.<|endoftext|> -TITLE: Numerically most robust way to compute sum of products (standard deviation) in floating-point? -QUESTION [6 upvotes]: I stumbled across a paper by Welford (1962), where he proclaims a method that should compute the standard deviation numerically more robust than the naive algorithms (http://www.jstor.org/stable/1266577). -Here, "numerically robust" means that round-off errors are reduced. -He gives a recurrence for the sum of squares $S_n = \sum_{i=1}^n (x_i - \mu_n)^2 = S_{n-1} + \frac{n-1}{n} (x_n - \mu_{n-1})^2 $ , $\mu$ being, of course, the mean. -That way, the standard deviation can be computed iteratively in a single pass. -As far as I understand, he claims that his iterative recurrence formula is numerically more robust, because all terms in it are of the same order (provided the input data are all of the same order). -This is what I don't understand. It seems to me that, as $n$ gets incremented, $S_n$ becomes larger and larger, so more and more significant digits from the second term are lost, aren't they? -Googling a bit further, I have found a paper by -Youngs & Cramer, 1971, who looked at a number of methods, including Welford's, of computing the sum of products / standard deviation more robustly (http://www.jstor.org/stable/1267176). -Conducting a number of experiments, they found that Welford's method does not provide any benefits. So that seems to confirm my doubts about Welford's method. -Now, Youngs & Cramer propose another method, which computes $S_n = S_{n-1} + \frac{n-1}{n}(nx_n - s_n)^2 $, where $s_n = s_{n-1} + x_n$. -Empirically, they found their method to be superior. -Again, I don't understand why this should be the case: isn't there some catastrophic cancellation going on in $(nx_n - s_n)$ ? Don't the terms $S_n$ and the fraction differ in their magnitude more and more, so that more and more digits of the fraction term get rounded off? -I would by most grateful if somebody could shed some light on these questions. -In addition, I'd like to know which is the best method (in terms of roundoff errors) to compute these sums. Surprisingly, I haven't found anything about this in Numerical Recipes (or I overlooked it). -Thank you very much in advance. -Gabriel. - -REPLY [2 votes]: I found a discussion of this exact problem in Higham, Accuracy and stability of numerical algorithms, Section 1.9. -The author suggests an alternative algorithm and claims that it is numerically stable (in the mixed backward-forward sense); proofs for the precise accuracy bounds of the formulas are left as exercises.<|endoftext|> -TITLE: Successive nth powers mod p? -QUESTION [11 upvotes]: While working on a project, I have run into a situation where I have integers x and n so that $x^n \equiv (x+1)^n \equiv (x+2)^n$ mod $p$ for a prime $p$. It seems to me that this an extremely restrictive condition, and I was wondering if there are any results about when (or if?) it can happen, but I couldn't figure out what to search. Any thoughts? What if I add the additional restriction that they are also congruent to $(x+3)^n$, etc. -Thanks! - -REPLY [2 votes]: Here's another approach... one can show for every $n$ there can be solutions for at most finitely many $p$; and for any given $n$ it's not hard to find these $p$ explicitly. -For fixed $n$ the question is when $x^n - (x-1)^n$ and $(x+1)^n - x^n$ can have a common factor (for some integer $x$). Applying the Euclidean algorithm to the two polynomials will yield an integer $N(n)$. For solutions to exist, $p$ must be a factor of $N(n)$, so only finitely many $p$ will do. -(To be perfectly rigorous about this I have to show that the two polynomials have no common factor in $\mathbb{Z}[x]$. But if they did then they would have a common root, say in $\mathbb{C}$. Considering absolute values, we see that the roots of $x^n = (x-1)^n$ all have real part $\frac{1}{2}$ while the roots of $x^n = (x+1)^n$ all have real part $-\frac{1}{2}$. So indeed the polynomials have no common factor, so the Euclidean algorithm will give a constant $N(n)$. -To see that $p$ must divide $N(n)$: the Euclidean algorithm guarantees that $N(n)$ is a linear combination of the two polynomials in $\mathbb{Z}[x]$. So for any value of $x$, $N(n)$ is a linear combination of $x^n - (x-1)^n$ and $(x+1)^n - x^n$. Hence if $x$ is a solution then $N(n)$ is a multiple of $p$.) -A quick calculation by hand gives -N(3) = 2 -N(4) = 30 -N(5) = 44. -Since $p$ cannot be $2$ or $3$, we see that... -For $n=3$ there are no solutions... -For $n=4$ there are solutions only when $p=5$... -For $n=5$ there are solutions only when $p=11$.<|endoftext|> -TITLE: Greatest common divisor of a^{2^n}-1 and b^{2^n}-1 -QUESTION [11 upvotes]: Let a and b be coprime integers. Do we know, expect, or unexpect that there are infinitely many primes p which divide -$gcd(a^{2^n} - 1, b^{2^n}-1)$ -for some n? Certainly any Fermat prime will divide both if I let n get large enough, but one doesn't know whether there are infinitely many of those. - -REPLY [2 votes]: A comment on one of Joe's questions: Let $B$ be any real number. It is known unconditionally that there are infinitely many $m$ for which $\phi(m)$ is a square and for which the smallest prime factor of $m$ exceeds $B$. One can even take $m$ as a product of two primes here; see, e.g., article 4 from -http://www.integers-ejcnt.org/vol11a.html -or an arXiv preprint of Tristan Freiberg. -If we choose $B$ larger than $|a|$ and $|b|$, then $m \mid \gcd(a^{\phi(m)}-1, b^{\phi(m)}-1)$, and so there is a prime $> B$ in the support of $\gcd(a^{n^2}-1, b^{n^2}-1)$.<|endoftext|> -TITLE: Can a finitely generated free group be isomorphic to a non-trivial quotient of itself? -QUESTION [7 upvotes]: I'm never sure about free groups whether a question is easy or not. It feels to me like this is impossible, but I couldn't come up with any argument. -If $F_n$ is a free group on $n$ generators, could there exist a non-trivial $N\triangleleft F_n$ such that $F_n/N \cong F_n$? - -REPLY [14 votes]: No, the free groups are Hopfian because they are residually finite.<|endoftext|> -TITLE: Is it possible to pull back a natural transformation? -QUESTION [5 upvotes]: Suppose that a 2-category $\mathcal{C}$ has strict pullbacks and one has maps $f:F\to C$, $g_0,g_1:G\to C$ and a natural transformation $\gamma:g_1\implies g_0$. Is there a good notion of a pullback transformation $f^*\gamma$. If so, I would expect its codomain to be $$f^*g_1\times_G f^*g_0\to f^*g_1\to F$$ and similarly for the domain. -If there is, is the need for the extra pullback here (over $G$) connected to the second layer of pullbacks in a 2-categorical descent diagram (relative to quotients in a 1-cat, which only require a kernel pair=1 layer of pullbacks)? - -REPLY [6 votes]: There is a good notion of a pullback transformation, but its domain and codomain aren't what you guessed; rather one asks for a map from $f^*(g_0) \to f^*(g_1)$ and a 2-cell filling the resulting triangle over G. The existence of such a pullback transformation is also not automatic from the existence of pullbacks, but requires $f$ to be a fibration (and the existence of all such pullbacks is equivalent to $f$ being a fibration). This characterization of fibrations is studied in Peter Johnstone's paper "Fibrations and partial products in a 2-category".<|endoftext|> -TITLE: Convex bodies with constant maximal section function in odd dimensions -QUESTION [16 upvotes]: In 1970 or so, Klee asked if a convex body in $\mathbb R^n$ ($n\ge 3$) whose maximal sections by hyperplanes in all directions have the same volume must be a ball. The counterexample in $\mathbb R^4$ is trivial and can be described as follows: -Let $f:[-1,1]\to\mathbb R$ be continuous, strictly concave and satisfy $f(-1)=f(1)=0$. For every such function, let $Q_f$ be the body of revolution given by $y^2+z^2+t^2\le f(x)^2$. Then $Q_f$ and $Q_g$ have the same maximal sections in every direction if (and, actually, only if) $f$ and $g$ are equimeasurable, i.e., $|\{f>t\}|=|\{g>t\}|$ for all $t>0$. (Of course, there are plenty of concave functions equimeasurable with $\sqrt{1-x^2}$). -With some extra work, one can construct something like this in $\mathbb R^n$ when $n$ is even though I do not know any similarly nice geometric description of such bodies for $n\ge 6$. -What I (and my co-authors) are currently stuck with is the case of odd $n$ (say, the usual space $n=3$). In view of such simple example in $\mathbb R^4$, I suspect that we are just having a mental block. Can anybody help us out? - -REPLY [4 votes]: To those who are still interested: we've finally made it but it's so ugly that a nice alternative approach will be always welcome :) We are still stuck with Bonnesen's question about the possibility to recover a convex body from the volumes of its maximal sections and projections in odd dimensions, so some help will be appreciated. The even-dimensional case can be found here. I feel a bit like a student asking for help with his homework, of course, but why not? We all get stuck now and then :). This should really be a comment but it's too long to fit the number of characters restriction.<|endoftext|> -TITLE: Turing machines that read the entire program tape -QUESTION [5 upvotes]: Consider a two tape universal Turing machine with a one-way-infinite, read-only program tape with a head that can only move right, as well as a work tape. The work tape is initialized to all zeros and the program tape is initialized randomly, with each cell being filled from a uniform distribution over the possible symbols. What are the possibilities for the probability that the head on the program tape will move infinitely far to the right in the limit? -Obviously, this will depend on the specifics of the Turing machine, but it must always be in the range $[0,1-\Omega)$, where $\Omega$ is Chaitin's constant for the TM. Since this TM is universal, $\Omega$ must be in the range $(0,1)$, so the probability must always be in $[0,1)$. Is this entire range, or at least a set dense in this range and including zero, possible? - -REPLY [5 votes]: Andreas considered the interpretation of your question where we fix the program and then vary the input. Let me now consider the dual version of the question, where we fix the infinite random input and vary the program. Surprisingly, there is something interesting to say. -The concept of asymptotic density provides a natural way to measure the size or density of a collection of Turing machine programs. Given a set $P$ of Turing machine programs, one considers the proportion of all $n$-state programs that are in $P$, as $n$ goes to infinity. This limit, when it exists, is called the asymptotic density or probability of the set $P$, and a set with asymptotic density $1$ will contain more than 99% of all $n$-state programs, when $n$ is large enough, as close to 100% as desired. -What I claim is that for your computational model, almost every program leads to a finite computation. -Theorem. For any fixed infinite input (on the read-only tape), the set of Turing machine programs that complete their computation in finitely many steps has asymptotic density $1$. -In other words, for fixed input, almost every program stops in finite time. -The proof follows from the main result of my article: J. D. Hamkins and A. Miasnikov, The halting problem is decidable on a set of asymptotic probability one, Notre Dame J. Formal Logic 47, 2006. http://arxiv.org/abs/math/0504351. The argument depends on the convention in the one-way infinite tape context that computation stops should the head attempt to move off the end of the tape. The idea has also come up on a few other MO quesstions: -What are the limits of non-halting? and Solving NP problems in (usually) polynomial time? in which it is explained that the theme of the result is the black-hole phenomenon in undecidability problems, the phenomenon by which the difficulty of an undecidable or infeasible problem is confined to a very small region, outside of which it is easy. -The main result of our paper is to show that the classical halting problem admits a black hole. In other words, there is a computable procedure to correctly decide almost every instance of the classical halting problem, with asymptotic probability one. The proof method is to observe that on fixed infinite input, a random Turing machine operates something like a random walk, up to the point where it begins to repeat states. And because of Polya's recurrence theorem, it follows that with probability as close as you like to one, the work tape head will return to the starting position and fall off the tape before repeating a state. -My point now is that the same observation applies to your problem. For any particular fixed infinite input, the work tape head will fall off for almost all programs. Thus, almost every program sees only finitely much of the input before stopping.<|endoftext|> -TITLE: Numbers in the "Fundamentalis Tabula Arithmetica" -QUESTION [15 upvotes]: Does anyone know anything about the "Fundamentalis Tabula Arithmetica" or what could be special about the numbers 106 and 117 that would make Europeans in the 1600s want to know their multiples? Googling reveals nothing about a "Fundamentalis Tabula Arithmetica". Is the word "Fundamentalis" really Latin? - -From: Alex Bellos -Date: Mon, Jul 11, 2011 at 11:01 AM -Subject: [math-fun] tables -To: math-fun -Hi - I've just received an email from a reader who saw in a Cologne museum a document called Fundamentalis Tabula Arithmetica from 1638. It is a table of multiples for all the numbers up to 100 and also the tables for these five numbers: -106, 117, 256, 318 and 365 -It's pretty obvious why 256 and 365 are included. -Might anyone have any thoughts as to why 106, 117 and 318 (3x106) are also included? -Alex - -REPLY [6 votes]: Regarding 318, Vincent Forest Hopper writes on p.75 of "Medieval Number Symbolism" (Dover): -"...various of the Church Fathers began to write figurative interpretations of the biblical texts. They found precedent for giving importance to numbers in the precise directions given for the dimensions of the tabernacle, and in the testimony of the Book of Wisdom that 'God has arranged all things in number and measure.' Early interpretation of scriptural numbers is concerned only with the most prominent of them, such as the 12 springs and 70 palm trees of Elim, and the 318 servants of Abraham..." Something of the attitude of gnosis; that is, of scriptural mysteries hidden from the layman, is to be seen in an interpretation of Barnabas: 'Learn then, my children, concerning all things richly, that Abraham, the first who enjoined circumcision, looking forward in spirit to Jesus, practised that rite having received the mysteries of the three letters. For [the Scripture] saith, 'And Abraham circumcised 10, and 8, and 300 men of his household.' What, then, was the knowledge given to him in this? Learn the 18 first and then the 300. The 10 and 8 are thus denoted. Ten by I and eight by H. You have [the initials of the name of] Jesus. And because the cross was to express the grace [of our redemption] by the letter T, he says also 300. No one else has been admitted by me to a more excellent piece of knowledge than this, but I know that ye are worthy.'" -Added: -The moral here is that the medieval mind set is so very different from what we can imagine that it's hard to guess why they thought individual numbers were significant. For more on 318, see en.wikipedia.org/wiki/Dispute_about_Jesus'_execution_method#Interpretation_as_cross<|endoftext|> -TITLE: Rings with finitely generated nilradical -QUESTION [5 upvotes]: Let $\mathfrak{a}$ be a monomial ideal in a polynomial algebra over some commutative ring $R$. If $R$ is reduced, then the radical $\sqrt{\mathfrak{a}}$ of $\mathfrak{a}$ is again a monomial ideal, and if $\mathfrak{a}$ is moreover finitely generated then so is $\sqrt{\mathfrak{a}}$. If $R$ is not reduced, then the nilradical of $R$ is contained in $\sqrt{\mathfrak{a}}$ and may make the latter somewhat less easy to handle. However, if $\mathfrak{a}$ and the nilradical of $R$ are finitely generated then so is $\sqrt{\mathfrak{a}}$. -This leads to the following question: - -Is there a nice description of rings $R$ such that the nilradical of $R$ is finitely generated? - -Or in more geometric terms: - -Is there a nice description of schemes $X$ such that the associated reduced scheme $X_{{\rm red}}$ is locally of finite presentation over $X$? - -REPLY [6 votes]: Good In a noetherian ring the nilradical, like any ideal, is finitely generated. -Bad Given a ring $A$ and an $A$-module $M$ you can construct an $A$-algebra $R=A\ast M $ whose underlying $A$-module is $A\oplus M$ and in which the multiplication is given by $(a,m).(a',m')=(aa',am'+a'm)$. -The first observation is that $M=0\ast M$ becomes an ideal of the ring $R$ satisfying $M^2=0$ and so definitely nilpotent. -In particular if $A$ is reduced the nilradical of $R$ is exactly $M$. -The second observation is that a subset $G\subset M$ generates $M$ as an ideal of the ring $R$ exactly when $G$ generates $M$ as an $A$-module. -And now you can construct rings with as badly non-finitely generated nilradicals as you like, for example by taking for $M$ a free $A$-module of dimension a huge cardinal.<|endoftext|> -TITLE: Unbased spectral sequences -QUESTION [11 upvotes]: Suppose we have a tower of fibrations of spectra $\{X_k\}_{k\in\mathbb{N}}$ with inverse limit $X_\infty$, and let $F_k$ be the fibre of the map $X_k\to X_{k-1}$. There is then a spectral sequence $E^1_{jk}=\pi_j F_k \Longrightarrow \pi_{j+k} X_\infty$. If we instead have a tower of fibrations of based spaces, then we still have something like a spectral sequence except that some of the entries may be nonabelian groups or just pointed sets, and the sense in which $E^{r+1}$ is the homology of $E^r$ must be modified to take account of this. The details are in the book 'Homotopy limits, completions and localizations' by Bousfield and Kan. Now suppose we have a tower of fibrations of unbased spaces. I think I have heard it said that there is still some kind of spectral sequence building up to $\pi_\ast(X_\infty,a_\infty)$, where the basepoint $a_\infty$ is not given in advance but is chosen iteratively by lifting basepoints $a_n\in X_n$ as we work through the spectral sequence. This makes life difficult because $F_{n+1}$ and $\pi_\ast(X_n)$ are not defined until we have chosen $a_n$. Has any theory of this type been worked out in detail? - -REPLY [7 votes]: As requested, I am reposting this comment as an answer. -Bousfield covers this material in "Homotopy spectral sequences and obstructions," Israel J. Math 66. The discussion is specific to cosimplicial objects (e.g. the discussion of obstruction cocycles in Section 5) and the general method of obtaining "partially" defined spectral sequences without basepoints would have to be extracted. However, I believe that all the necessary content is already there.<|endoftext|> -TITLE: Is canonical class a topological invariant? -QUESTION [13 upvotes]: For a $n$-dim smooth projective complex algebraic variety $X$, we can form the complex line bundle $\Omega^n$ of holomorphic $n$-form on $X$. Let $K_X$ be the divisor class of $\Omega^n$, then $K_X$ is called the canonical class of $X$. -Question: Is homology class of $K_X$ in $H_{2n-2}(X)$ a topological invariant? If it's true, please tell me the idea of proof or some references. If not, please give me the counterexamples. - -REPLY [13 votes]: This answer is about the case of complex surfaces $X$ and their diffeomorphisms (all my diffeos are assumed to be orientation-preserving!). -(1) Examples of self-diffeomorphisms that reverse the sign of the canonical class. -Take $X=\mathbb{C}P^1\times \mathbb{C}P^1$. Let $\tau$ be reflection in the equator of $S^2=\mathbb{C}P^1$. Then $\tau \times \tau$ preserves orientation and acts as $-I$ on $H^2(X)$. It therefore sends $K_X$ to $-K_X$. -One can also realise the automorphism $-I$ of $H^2(X)$ by a diffeomorphism when $X$ is the blow-up of the projective plane at $k$ points, $k = 2,3,\dots,9$. This follows from a result of C.T.C. Wall from -Diffeomorphisms of 4-manifolds, J. London Math. Soc. 39 (1964) 131–140, MR0163323 -Wall says that if $N$ is a simply connected, closed oriented 4-manifold with $b_2(N)<9$, and $X$ is the connected sum of $N$ with $S^2 \times S^2$, then all automorphisms of the intersection form of $X$ are realised by diffeos. To apply this, recall that the 1-point blow-up of $\mathbb{C}P^1\times \mathbb{C}P^1$ is the 2-point blow up of the projective plane. (Wall's strategy, by the way, is to factor the automorphism into reflections along hyperplanes, and to realise those.) -(2) Results from Seiberg-Witten theory. -These results tie complex geometry amazingly closely to differential topology. They say that the unsigned pair $\pm K_X$ is invariant under diffeomorphisms (Witten http://arxiv.org/abs/hep-th/9411102 and others); so too is the Kodaira dimension; so too are the plurigenera (Friedman-Morgan http://arxiv.org/abs/alg-geom/9502026). -In Kodaira dimension $<2$, one can take this further and prove that oriented-diffeomorphic surfaces are actually deformation-equivalent (to be safe, let me specify the simply connected case). But that's not the explanation in general: there are pairs of simply connected general-type surfaces that are diffeomorphic (by diffeos preserving the canonical class), which are not deformation-equivalent (Catanese-Wajnryb http://arxiv.org/abs/math/0405299). -(3) How it happens. -The Seiberg-Witten invariant (for an oriented 4-manifold with $b^+(X)>1$) is a map -$$SW: Spin^c(X)\to\mathbb{Z}$$ -defined on the $H^2(X)$-torsor of $Spin^c$-structures. The overall sign is equivalent to a "homology orientation". It's natural under diffeomorphisms. It's also invariant under "conjugation" $\mathfrak{s}\mapsto \bar{\mathfrak{s}}$ of $Spin^c$-structures. -For algebraic surfaces, there's a canonical spin-c structure $\mathfrak{s}$, so $Spin^c(X)$ is identified with $H^2(X)$. Witten (http://arxiv.org/abs/hep-th/9411102) observed that the elliptic equations that define $SW$ simplify drastically in the algebraic case; in evaluating $SW$ on a cohomology class represented by a complex line bundle $L\to X$, you're led to consider a moduli space of pairs consisting of a holomorphic structure on the line bundle and a holomorphic section of it, with an obstruction bundle on the moduli space. Conjugation-invariance becomes Serre duality. -For general type surfaces, $\pm SW(\mathfrak{s}) = \pm SW(\bar{\mathfrak{s}}) = \pm 1$; all other spin-c structures have vanishing invariant. Since $c_1(\mathfrak{s})=-c_1(\bar{\mathfrak{s}})=-K$, one deduces diffeomorphism-invariance of $\pm K$. For lower Kodaira dimension, a more complicated analysis is needed.<|endoftext|> -TITLE: Uniform approximation of $x^n$ by a degree $d$ polynomial: estimating the error -QUESTION [7 upvotes]: The answer to this question should be well known, but it's a hard question to search for online. -Suppose we want to approximate the function $x^n$ by a polynomial of degree $d$ in the $L_\infty$ norm on $[-1,1]$. What is a good estimate of the error of the best approximator, in terms of $n$ and $d$? -I know this question was solved exactly by Chebyshev for $d = n-1$ (the error is $2^{-d}$ I think). The range of interest for me is $\sqrt{n} \leq d \leq n$ and I don't mind log factors in the estimate. Thus I would be happy to have an estimate for the error of the Chebyshev expansion truncated to degree $d$. -(A bonus would be an answer to the same question for $(1-x^2)^d$.) -Thanks! - -REPLY [8 votes]: For large $n$ and fixed $\epsilon > 0$ there is a polynomial of degree $d = O_\epsilon(\sqrt{n})$ that uniformly approximates $x^n$ to within $\epsilon$ on all of $[-1,+1]$. The polynomial can be taken to be the truncated Čebyšev expansion of $x^n$, as the original proposer (OP) suggested. As $\epsilon \rightarrow 0$, the $O_\epsilon$ constant grows only as $(\log(\epsilon^{-1}))^{1/2}$; for example, $d = 2.576 \sqrt{n}$ suffices to get $\epsilon = .01$ if I computed correctly. -The OP wrote that truncating the Čebyšev expansion will give the correct $L^\infty$ distance to within a log factor. I don't see a priori why this should be, but fortunately the coefficients of the expansion of $x^n$ in Čebyšev polynomials turn out to be elementary and familiar enough to work with explicitly. -It will be convenient to define $T_k(x)$ for all $k \in \bf Z$ as the polynomial such that $T_k(\cos u) = \cos ku$. Then $T_{-k} = T_k$ is a polynomial of degree $|k|$ satisfying $|T_k(x)| \leq 1$ for all $x\in [-1,+1]$. Now the Čebyšev expansion of $x^n$ is simply -$$ -x^n = \frac1{2^n} \sum_{m=0}^n {n \choose m} T_{2m-n}(x), -$$ -which can be checked by writing $x = \cos u = \frac12(e^{iu}+e^{-iu})$ and $T_k(x) = \frac12(e^{iku}+e^{-iku})$. So the coefficients form a binomial distribution, and truncating at degree $d$ eliminates only the tail of the distribution past $d^2/n$ standard deviations. Since each $|T_{2m-n}(x)| \leq 1$, this tail also bounds the truncation error for all $x \in [-1,+1]$, and we conclude that this error can be brought below any positive $\epsilon$ by making $d$ a large enough multiple of $\sqrt{n}$, as claimed. -This might not be the optimal $L^\infty$ approximation (except for $d=n-1$, when its optimality is the result of Čebyšev that you quoted), but it's not too far, because it is the best $L^2$ approximation with respect to the Čebyšev measure $\pi^{-1} dx/ \sqrt{1-x^2}$, and the $L^\infty$ distance is at least as large as the $L^2$ distance. The $L^2$ distance can be computed from the sums of the squares of the coefficients in the tail. -Much the same technique should work for $(1-x^2)^n$; indeed I see that while I was writing this Andrew posted an answer for $(1-x^2)^n$ that looks very similar to what I did for $x^n$. - -REPLY [5 votes]: For $P_n(x)=(1-x^2)^n$, large $n$ and $a>0$ it's possible to produce a polinomial of degree $a\sqrt{2n}$ with difference in $L_\infty$ less than $C(1-\mathrm{erf}\;a)$, where $C$ is an absolute constant. Taking any positive sequence $a(n)\to+\infty$ as $n\to\infty$ leads to $L_\infty$ norm converging to zero. For large $a$ we have $1-\mathrm{erf}\;a \sim e^{-a^2}/(a\sqrt{\pi})\;$. So to obtain the uniform $\varepsilon$ estimate on $[0,1]$ the degree $\sim C(\log \varepsilon^{-1})^{1/2}\sqrt n\ $ is enough. -Namely, consider $P_n$ on the segment $[-1,1]$. Let $x=\sin y$. Now it is enough to approximate the function -$$ -\sin^{2n}y =\sum_{k=0}^n c_n^k\cos 2ky -$$ -on $[0,2\pi]$ by suitable trigonometric polynomials. Here $c_n^0=\frac1{4^{n}}{2n\choose n}$, $c_n^k =(-1)^{n-k}\frac1{2^{2n-1}}{2n\choose n+k}$, $k=1,\ldots,n$. For degree $2m<2n$ we'll take the polynomial -$$ -Q_{2m}(x)=\sum_{k=0}^m c_n^k\cos 2ky. -$$ -Then the $L_\infty$ norm -$$ -\|P_{2n}-Q_{2m}\| {} \le \sum_{k=m+1}^{n} |c_n^k|=\frac1{2^{2n-1}}\sum_{k=m+1}^{n}{2n\choose n+k}. -$$ -The last sum can be easily estimated since it is exactly the sum of the tails in the Bernoulli distribution with probability $p=1/2$ and $2n$ independent trials. For $m=[ a\sqrt{2n}]$ and large $n$ by the central limit theorem it is equal approximately to $2(1-\mathrm{erf}\;a)$. From here the above estimates follow.<|endoftext|> -TITLE: Average over Random Permutations -QUESTION [10 upvotes]: Consider $S_{n}$ the symmetric group and for each $\sigma\in S_{n}$ let $U_{\sigma}$ be its $n\times n$ permutation matrix. Let $A$ be an Hermitian $n\times n$ matrix. I'm interested in computing the average -$$ -\mathbb{E}(A):=\sum_{\sigma \in S_{n}}{w(\sigma) U_{\sigma} A U_{\sigma}^{*}} -$$ -where the $w(\sigma)$ are some positive weight adding up to one. -For instance some natural weights are the ones coming from the Ewens's probability distribution of parameter $\theta>0$ on $S_{n}$ defined as -$$ -w(\sigma)=\frac{\theta^{K(\sigma)}}{\theta(\theta+1)\ldots(\theta+n-1)} -$$ -where $K(\sigma)$ is the number of disjoint cycles of $\sigma$. The case of $\theta=1$ is simply the uniform distribution on $S_{n}$. For the case $\theta=1$, it is known that -$$ -\mathbb{E}(A)=\alpha \frac{ee^{T}}{n} + \Bigg(\frac{\mathrm{Tr}(A)-\alpha}{n-1}\Bigg)\Bigg(I_{n}-\frac{ee^{T}}{n}\Bigg) -$$ -where $e$ is the vector $e^{T}=(1,1,\ldots,1)$ and $\alpha=\frac{e^{T}Ae}{n}$. -My question are: - -Is there anything known about the averages for the more general case of $\theta>0$. -Are there known asymptotics results as $n\to\infty$? - -REPLY [4 votes]: You can calculate this directly. All you need to know is the probability that $(\sigma(i),\sigma(j))=(k,l)$ for each pair $(k,l)$. Write $e_i$ for the $i$th basis vector, $v=\sum_i e_i$. Let $E = \mathbb{E}(A)$. Then: -$$ e_i^T E e_i = \frac{\theta-1}{\theta+n-1} e_i^TAe_i + \frac{1}{\theta+n-1}\mathrm{Tr} A$$ -and -\begin{equation*} -\begin{split} - &(\theta+n-1)(\theta+n-2)e_i^T E e_j\\\\ - &= v^TAv - \mathrm{Tr} A + (\theta-1) \left[e_i^TAv+v^TAe_j+e_j^TAe_i-e_i^TAe_i+e_j^TAe_j\right] + (\theta-1)^2 e_i^TAe_j -\end{split} -\end{equation*} -More details: $E_{ij}$ is the average of $A_{\sigma(i)\sigma(j)}$ over $\sigma$. When $i=j$ you get $E_{ii} = p A_{ii} + \frac{1-p}{n-1}\sum_{k\neq i} A_{kk}$ where $p$ is the probability that $\sigma(i)=i$. To calculate $p$ note that when $\sigma(n)=n$, the restriction of $\sigma$ to $[1,n-1]$ has precisely one less cycle. It follows that $\sum_{\sigma(n)=n}w_n(\sigma) = \frac{\theta}{\theta+n-1} \sum_{\tau\in S_{n-1}} w_{n-1}(\tau)$ where $w_n$ is the weighing above on $S_n$. -Similarly, for $i\neq j$ we have a sum over $A_{kl}$ with different weights. Easy cases include $k=i, l=j$ (this is $\frac{\theta^2}{(\theta+n-1)(\theta+n-2)}$ for the same reason as the diagonal), $k=j, l=i$ (this is $\frac{\theta}{(\theta+n-1)(\theta+n-2)}$ since now there is only one more cycle than the restriction to $[1,n-2]$), $k=i, l\neq i,j$ or $k\neq i,j, l=j$ (this is $\frac{\theta}{\theta+n-1}\left(1-\frac{\theta}{\theta+n-2}\right)$ since we fix one co-ordinate but not the other). -A bit more difficult is the case $k=j, l\neq i,j$ and $i\neq i,j, l=i$. Again for $i=n, j=n-1$ this means that $n, n-1$ are consecutive on a cycle of length at least $3$. The probability that $n$ is on a cycle of length $3$ is the complement of the probability that it is on a cycle of length $1$ or $2$ (that is $\frac{\theta}{\theta+n-1} + (n-1)\frac{\theta}{(\theta+n-1)(\theta+n-2)}$. Given that, the successor to $n$ on the cycle is uniformly distributed, so the probability that $k=j, l\neq i,j$ is $\frac{1}{n-1} \left(1 - \frac{\theta}{\theta+n-1} + (n-1)\frac{\theta}{(\theta+n-1)(\theta+n-2)}\right)$. -Finally, the probability that $k,l$ are distinct from $i,j$ is the complement of the above cases, and if so then the pair $k,l$ is uniformly distributed on the (n-2)(n-3) possibilities. It turns out each of these possibilities has probability $\frac{1}{(\theta+n-1)(\theta+n-2)}$.<|endoftext|> -TITLE: Time-dependent Markov process: infinitesimal generator -QUESTION [6 upvotes]: If one talks about homogeneous Markov diffusion -$$ -\mathrm dX_t = \mu(X_t)\mathrm dt+\sigma(X_t)\mathrm dw_t -$$ -with $\mu,\sigma$ sufficiently differentiable and of appropriate dimensions, there is nice equation for a function $m_f(x,t) = \mathsf E_x f(X_t)$ for $f\in C^2(\mathbb R)$: -$$ -\begin{cases} -\frac{\partial m_f}{\partial t} &= \mu\frac{\partial m_f}{\partial x}+\frac12\sigma^2\frac{\partial^2 m_f}{\partial x^2}, -\\ -m(x,0)&=f(x). -\end{cases} -$$ -On the other hand, while answering on the question Stochastic model I advised to use Fokker-Plank equation for the density of the process rather then the very same equation for $m_f$. -The problem I had is the following. Since $\mu$ and $\sigma$ are time-dependent there, one can construct a process $Z_t = (X_t,Y_t)$ with $\mathrm dY_t = \mathrm dt$ and obtain everything for this process just using theory of homogeneous/time-independent Markov processes (how it is usually written in the books). Unfortunately, if you derive an infinitesimal generator for this process then you obtain -$$ -\mathcal A_Zg(x,y) = \frac{\partial g}{\partial y}+\mu\frac{\partial g}{\partial x}+\frac12\sigma^2\frac{\partial^2 g}{\partial x^2}. -$$ -For sure, now one shoud define $m_f(\tau|x,t) = \mathsf E_{x,t}f(X_{t+\tau})$ where $\tau\in \mathbb R_{\geq 0}$. If I am not wrong then it follows -$$ -\frac{\partial m_f}{\partial \tau} = \frac{\partial m_f}{\partial t}+\mu\frac{\partial m_f}{\partial x}+\frac12\sigma^2\frac{\partial^2 m_f}{\partial x^2}\quad(*) -$$ -which is kind of strange equation. -My question has three parts: - -is an equation $(*)$ correct? if yes, are there developed methods for its solutions? -if this equation is not correct, what is the right equation? -I usually have problems when deal with non-homogeneous Markov processes since trick $Z_t = (t,X_t)$ does not help me. Could you refer me to literature where authors consider non-homogeneous Markov processes in details (rather then saying that this trick will help to use provided theory of homogeneous processes)? - -Link to MSE question: https://math.stackexchange.com/questions/51591/infinitesimal-generator-of-time-dependent-markov-diffusion - -REPLY [5 votes]: You can directly deal with inhomogeneous Markov processes through the Kolmogorov backward and forward equations. I suppose you are asking for the forward equation (i.e. derivative with respect to the time in the future). Let me discuss things on a formal level, which means here I ignore regularity conditions to ensure the existence of generators, derivatives, etc etc. -Let me first give the backward equation, which is probably easier (a specific case of Feynman-Kac formula). Let $u(s,x,t)=\mathbb{E}^{s,x} f(X_t)$ for a nice function $f$ (e.g. infinitely differentiable with compact support), where the expectation is under the measure $\mathbb{P}^{s,x}$ such that $\mathbb{P}^{s,x}\{X_s=x\}=1$. Let $A_s$ be the generator (I am ignoring the detail of how $A_s$ is defined, you can probably figure it out yourself or see references I give below). Then the backward equation states that -$\frac{\partial u}{\partial s}+A_s u =0$. -The forward equation is usually formulated for the density $p(s,x,t,y)$ of the transition kernel $P_{s,t}(x,B)=\int_B p(s,x,t,y)dy$: -$\frac{\partial p}{\partial t}=A_t^* p(s,x,t,y)$ -where $A_t^*$ is the adjoint of the operator $A_t$ (again, there's subtlety in how you definite $A_t$ for forward and backward equations separately, I ignore that here). -For a given diffusion given as an SDE $dX_t=\mu(t,X_t)dt+\sigma(t,X_t)dW_t$ for a Wiener process $W$, it is pretty straightforward that $A_t f(x)=\mu(t,x)f'(x)+\frac{1}{2}\sigma^2(t,x)f''(x)$. -All these are discussed to certain extent in "The Theory of Stochastic Processes, Vol II" by Gikhman and Skorokhod, and "Multidimensional Diffusion Processes" by Stroock and Varadhan (expressed in an integral form instead of derivatives).<|endoftext|> -TITLE: What are the primitives of Lusztig's twisted bialgebra $\mathbf{f}$? -QUESTION [8 upvotes]: I'm trying to understand the coproduct on Lusztig's $\mathbf{f}$ and, apart from the Chevalley generators, I don't know of any more primitive elements. In the $q=1$ case, the Weyl group acts as a Hopf algebra automorphism and maps the Chevalley generators to elements corresponding to the other simple roots, so those are also primitive; but when $q\neq1$, the braid group action is not a coalgebra automorphism, so image of the Chevalley generators under the braid group don't seem to be primitive. Are there more primitive elements, or do the Chevalley generators give a complete basis? - -REPLY [3 votes]: Suppose that $x\in \mathbf{f}$ and $r(x)=x\otimes 1+1\otimes x$. Recall $\mathbf{f}$ is graded by $\mathbb{N}I$, WLOG $x$ is homogenous, $x\in \mathbf{f}_\nu$. We know we get primitive elements if $\nu=i$ for some $i\in I$ so let us assume that $\nu$ is not of this form. -Then $\mathbf{f}_\nu$ is generated by products of the form $yz$ with $y\in \mathbf{f}_\lambda$, $z\in \mathbf{f}_\mu$ with both $\lambda$ and $\mu$ nonzero. Now we consider the symmetric bilinear form on $\mathbf{f}$ and compute -$$(x,yz)=(r(x),y\otimes z)=0$$ -since $x$ is assumed primitive. -This shows that $x$ is orthogonal to all of $\mathbf{f}_\nu$. But the bilinear form is nondegenerate, so $x=0$. Thus the Chevalley generators give a basis of the primitive elements of $\mathbf{f}$.<|endoftext|> -TITLE: Stalks of structure sheaf of fibre product？ -QUESTION [8 upvotes]: What can I say about it？ -Can I say the stalks equal the tensor products of the corresponding factors stalks？ -Thanks！ - -REPLY [10 votes]: Let $X,Y$ be $S$-schemes. Then a point of $X \times_S Y$ corresponds to a pair of points $x \in X, y \in Y$ lying over the same $s \in S$ together with a prime ideal $\mathfrak{p} \subseteq \mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{Y,y}$ which restricts to the maximal ideals in $\mathcal{O}_{X,x}$ resp. $\mathcal{O}_{Y,y}$. The stalk of the structure sheaf in this point is the localization of the tensor product: -$\mathcal{O}_{X \times_S Y,(x,y,\mathfrak{p})} = (\mathcal{O}_{X,x} \otimes_{\mathcal{O}_{S,s}} \mathcal{O}_{Y,y})_{\mathfrak{p}}$. -There are at least two ways to prove these statements: a) Use the universal property of $\text{Spec}(K)$ for a field $K$ to get the points and then use the universal property of $\text{Spec}(R)$ for a local ring $R$ to get their stalks. So this assumes, of course, that you already know that the fiber product exists, but you can recover the description of the elements and the stalks just by using the universal property! But actually, b) you can construct the fiber product as above, also more general in the category of locally ringed spaces. I've written this up here. -Now your actual question seems to be: - -As Hartshorne chapter III.9.2 claim，an Ox-module （need not be quasi coherent) F's - flatness is stable under base change. But the stalks is not the tensor products, how - can I prove the claim? - -The statement is the following: If $f : X \to Y, Y' \to Y$ are morphisms, and $\mathcal{F}$ is a module over $X$ which is flat over $f$, then the pullback of $\mathcal{F}$ to $X \times_Y Y'$ is flat over $X \times_Y Y' \to Y'$. I am pretty sure that Hartshorne understands $\mathcal{F}$ to be quasi-coherent here. Otherwise the sketch of proof also does not make sense. But it is also true in general: -Pick a point in $X \times_Y Y'$, thus a triple $(x,y',\mathfrak{p})$ as described above. Let $y$ be the underlying point in $Y$. Now $\mathcal{F}_{x}$ is flat over $\mathcal{O}_{Y,y}$. By commutative algebra (base change of flat modules), it follows that $\mathcal{F}_x \otimes_{\mathcal{O}_{Y,y}} \mathcal{O}_{Y',y'}$ is flat over $\mathcal{O}_{Y',y'}$. Again by commutative algebra (localizations are flat) $(\mathcal{F}_x \otimes_{\mathcal{O}_{Y,y}} \mathcal{O}_{Y',y'})_{\mathfrak{p}}$ is flat over $\mathcal{O}_{Y',y'}$. But this is exactly the stalk of the pullback of $\mathcal{F}$ in the given point $(x,y',\mathfrak{p})$.<|endoftext|> -TITLE: If X is the coarse moduli space of the algebraic stack M, is there a nice description of Hom(_,X)? -QUESTION [5 upvotes]: Let $\mathcal{M}$ be an algebraic stack, and let $X$ be its coarse moduli space (assume it exists as a scheme). -We know that $h_X(Spec(k))=\mathcal{M}(Spec(k))$ if $k$ is algebraically closed. Is there anything intelligent we can say about $h_X(U)$ for a general scheme $U$? -For example, can you come up with an algorithm for knowing what $h_X(U)$ is that would be considerably easier than constructing $X$? - -REPLY [5 votes]: What André says is absolutely correct. The functor represented by the moduli space does not have any reasonable description, except in very particular cases. Given a map $U \to X$, it can be hard work to decide whether it comes from an object of $\mathcal M(U)$.<|endoftext|> -TITLE: What is the relationship amongst all the different kinds of spectra? -QUESTION [11 upvotes]: The word "spectrum" gets tossed around a lot in mathematics, and there seem to be a number of different concepts to which it applies. There is of course a physical connotation to the word which is commonly associated with scattering processes, rainbows, etc. : http://en.wikipedia.org/wiki/Spectrum -However, in mathematics the term seems to be quite loaded. To name a just a few concepts which could be called spectral, here is a big list: - -Eigenvalues/eigenvectors for normal matrices -Jordan normal form for matrices -Spectrum of a bounded linear operator -Spectrum of a C* algebra -Spectrum of a commutative ring / Scheme -Characters of an abelian group -Irreducible unitary representations of a group -Spectrum of a graph -Spectrum of a Riemannian manifold -Spectral sequences from cohomology theory - -Of course some of these concepts are more general than others. For example, normal matrix < bounded linear operator < c* algebra < commutative ring. However, it doesn't seem like any one of these definitions is sufficiently encompassing to give the whole story. As a counter example, the irreducible representations of a non-commutative group are an instance of the Jordan normal form (for finite groups anyway), but are not really captured by the corresponding notion of the spectra of a commutative ring. Similarly, the spectrum of a graph and a Riemannian manifold don't seem to have much to do with the spectrum of schemes, but yet they are related to the spectra of linear operators. And then there are spectral sequences which are just a bit weird... -I won't profess to completely understand the general idea here, but there do seem to be some patterns. A common theme seems to be `decomposability', for example when finding the eigenvalues/vectors of a matrix one attempts to split apart the domain of the matrix into independent components. This is similar to splitting a space into points; and suggests that there is perhaps a correspondence between eigenvectors/eigenvalues and prime ideals/local rings. The non-commutative picture is of course more complicated, but perhaps the concept might be equally described in terms of irreducible modules and some type of generalized localization at an irreducible module (which is a hazy concept I must admit). In a perfect world, it would be nice if this same concept could even extend to things like the Gauss map from differential geometry, or the Legendre transform from statistical mechanics/convex optimization (of course that might not be feasible). -Also, I marked the question community wiki in case anyone else has some other good examples of `spectral' concepts in mathematics. - -REPLY [5 votes]: I'll extend my comment to an answer. Let $L$ be a complete lattice. Then a prime element of $L$ is an element p such that $a\wedge b\leq p$ implies $a\leq p$ or $b\leq p$. These elements are in bijection with maps from $L$ to the 2-element lattice preserving all sups and finite infs. For example, the prime elements of the lattice of ideals in a commutative ring are the prime ideals. If $A$ is a separable C*-algebra, then the prime elements of the lattice of closed 2-sided ideals are the primitive ideals (kernels of irreducible representations). If $A$ is commutative, these are the maximal ideals. -The prime elements of a lattice $L$ form a space $spec(L)$ called the spectrum of $L$. The topology has as open sets the sets D(a) with $a\in L$ where $D(a)$ consists of all prime elements $p$ with $a\nleq p$. For example, if $L$ is the lattice of ideals, this is the Zariski spectrum. If $A$ is a separable C*-algebra, then the spectrum of the closed 2-sided ideal lattice is the primitive ideal spectrum. So many of your examples are spectra of lattices. -Recall a space $X$ is sober if each irreducible closed subset has a unique generic point. Sober spaces are precisely the spectra of complete lattices. The proof $spec(L)$ is sober amounts to showing that the irreducible closed subsets are the complements of the sets D(p) with p prime and p is generic. Conversely, if X is sober, take the lattice of closed subspaces ordered by reverse inclusion. The prime elements are the irreducible closed subsets, which can be identified with points of X by taking generic points.<|endoftext|> -TITLE: Grothendieck's Galois theory without finiteness hypotheses -QUESTION [24 upvotes]: This is motivated by the discussion here. The usual definition of the etale fundamental group (as in SGA 1) gives automatic profiniteness: Grothendieck's formulation of Galois theory states that any category with a "fiber functor" to the category of finite sets satisfying appropriate hypotheses is isomorphic to the category of finite, continuous $G$-sets for a well-defined profinite group $G$ (taken as the limit of the automorphism groups of Galois objects, or as the automorphism group of said fiber functor). In SGA1, the strategy is to take finite etale covers of a fixed (connected) scheme with the fiber functor the set of liftings of a geometric point. -One problem with this approach is that $H^1(X_{et}, G)$ for a finite group $G$ (which classifies $G$-torsors in the etale topology) is not isomorphic to $\hom(\pi_1(X, \overline{x}), G)$ unless $G$ is finite. Indeed, this would imply that the cohomology of the constant sheaf $\mathbb{Z}$ would always be trivial, but this is not true (e.g. for a nodal cubic). -However, Scott Carnahan asserts on the aforementioned thread that the "right" etale fundamental group of a nodal cubic should be $\mathbb{Z}$, not something profinite. -How exactly does this work? People have suggested that one can define it as a similar inverse limit of automorphism groups, but is there a similar equivalence of categories and an analogous formalism for weaker "Galois categories"? (Perhaps one wants not just all etale morphisms but, say, torsors: the disjoint union of two open immersions might not be the right candidate.) I'm pretty sure that the finiteness is necessary in the usual proofs of Galois theory, but maybe there's something more. - -REPLY [14 votes]: In "The pro-étale topology for schemes" (http://arxiv.org/abs/1309.1198), Bhatt and Scholze introduce the pro-étale fundamental group which seems to give a good answer to your question (see, e.g., Theorem 1.10). The pro-étale fundamental group also compares well against the usual étale fundamental group and the "SGA3 étale fundamental group".<|endoftext|> -TITLE: symmetric integer matrices -QUESTION [16 upvotes]: Suppose I have a symmetric positive definite matrix $M$ with integer entries. I want to decide whether $M = A A^t,$ with $A$ likewise integral. I assume that decision problem is NP-complete, as is the question of finding the $A$ even if an oracle tells you such an $A$ exists. Can someone provide a reference (I would very much like to be wrong about the hardness of the problem...) -EDIT A remark: this question is equivalent to finding a collection of integral vectors (the columns of $A$) with prescribed distances (by the parallelogram law, the inner products give us the distances). If we require $A$ to be a $0-1$ matrix, I am pretty sure that this can encode knapsack, so is NP-complete. It seems that as per Will Jagy and Gerhard Paseman, this question (via Hasse-Minkowski) might only be as hard as factoring (which is generally conjectured to be less than NP-complete), but I haven't yet completely understood what is entailed in the Hasse-Minkowski approach... -Further EDIT -In fact, the Hasse local-to-global principle works fine for small dimensions, since the class number of identity equals one in that case, and one can enumerate solutions by the -Smith-Minkowski-Siegel mass formula. This apparently works only in dimension at most eight. This gives the oracle (the wiki article cited seems to imply that the right hand side can be computed in polynomial time, though I am none-too-certain of this), so this gives the required oracle in small dimensions, though not obviously an algorithm for finding solutions. In dimensions greater than eight we seem to be sunk. - -REPLY [12 votes]: A minor observation: $\left( \begin{smallmatrix} N & 0 \\ 0 & N \end{smallmatrix} \right)$ is of the form $A A^T$ if and only $N$ is of the form $a^2+b^2$. Specifically, if $N=a^2+b^2$ then $\left( \begin{smallmatrix} N & 0 \\ 0 & N \end{smallmatrix} \right) = \left( \begin{smallmatrix} a & b \\ -b & a \end{smallmatrix} \right) \left( \begin{smallmatrix} a & -b \\ b & a \end{smallmatrix} \right)$. The converse is left as an exercise. -So this problem is at least as hard as determining whether or not an integer is a sum of two squares. -We discussed the complexity of determining whether an integer is a sum of two squares here. Nobody really knew the answer, but it seemed that it might be as hard as factoring.<|endoftext|> -TITLE: Edge Covering Shortest path -QUESTION [5 upvotes]: I have a graph of Points G(V,E) and I want to find the shortest path covering all the edge, I want the minimum number of edge repetitions . -Which is the best way to reduce this problem to well know problems like TSP , Hamiltonian circuit , Hamltonian completion ? -Thank you. - -REPLY [5 votes]: That is Chinese Postman Path. Search for Chinese Postman Problem... -E.g., this section from some book looks comprehensive: http://ie454.cankaya.edu.tr/uploads/files/Chp-03%20044-064.pdf<|endoftext|> -TITLE: Haar measures in Solovay's model -QUESTION [18 upvotes]: Haar measure is a measure on locally compact abelian groups which is invariants to translations. For example, the Lebesgue measure on the reals is such measure. -It can be shown without the use of the axiom of choice that the Haar measure exists and it is unique up to a scalar, that is if we want the measure of the unit interval (for example) to be $1$ then it is really unique. -While the measure is defined on Borel subsets, but we can complete the measure in a unique way by adding all the subsets of measure zero sets (and in the case of the real numbers we once again have the Lebesgue algebra) -As with the Lebesgue measure, when the axiom of choice is present there are cases in which non-measurable sets can be constructed. In the Solovay model, however, we have that all subsets of reals are measurables. -Are there any similar results about Haar measures of general LCA groups? Is there a model in which all Haar measures (perhaps under some limitations on the groups) are "full measures" (in the sense that every subset is measurable)? - -REPLY [5 votes]: Solovay's model will have all subsets measurable as long as the group is (locally compact) metrizable. For "big" groups (non-metrizable) the Haar measure is naturally defined on the Baire sets (the least sigma-algebra so that all continuous real-valued functions are measurable), and extension even to the Borel sets may not be unique.<|endoftext|> -TITLE: Rational map having a birational restriction. -QUESTION [7 upvotes]: Let $\sigma\subset |\mathcal{O} _{\mathbb{P} _{\mathbb{C}}^N}(2)|$ be -an $N-$dimensional linear systems of quadrics on $\mathbb{P} _{\mathbb{C}}^N$ ($N\geq 4$), -$F:\mathbb{P} _{\mathbb{C}} ^N\dashrightarrow \mathbb{P} _ {\mathbb{C}}^ N$ the rational map associated to $\sigma$, -and $Q\in \sigma$ a fixed smooth quadric -such that $F|_{Q}:Q\dashrightarrow \mathbb{P} _{\mathbb{C}}^{N-1}$ is birational. -Is $F$ birational? - -REPLY [2 votes]: Nice question. I think that the answer is yes and more general. I have never seen this but seems natural. -${\bf Lemma}$ Let $f\colon \mathbb{P}^n_{\mathbb{C}}\to \mathbb{P}^n_{\mathbb{C}}$ be a rational map of degree $d$, given by $(x_0:\dots:x_n)\to (f_0:\dots:f_n)$, where the $f_i$ are homogeneous of degree $d$. -Suppose that the hypersurface $H\subset \mathbb{P}^n_{\mathbb{C}}$ given $f_0=0$ is irreducible and that the map from $H$ to $\mathbb{P}^{n-1}_{\mathbb{C}}$ given by the restriction (i.e. $(x_0:\dots:x_n)\to (f_1:\dots:f_n)$ on $H$) is birational. -Then, the map $f$ is birational. -${\bf Proof}$ Let $\sigma$ be the linear system associated, which corresponds to hypersurfaces of $\mathbb{P}^n$ of equation $\sum_{i=0}^n a_i f_i=0$. The restriction being birational, the intersection of $n-1$ general elements of $\sigma$ with $H$ give one mobile point, and a non-mobile part that we call $R$. Since every member of $\sigma$ contains $R$ and because elements have all the same degree, the intersection of $n$ general elements of $\sigma$ gives $R$, plus exactly one mobile point (the last part follows from Bézout). This shows that $f$ is birational. -${\bf Remark:}$ We could also view $R$ as a set of points, curves,... with some multiplicities and use intersection form on the blow-up.<|endoftext|> -TITLE: Do etale maps satisfy the following? -QUESTION [6 upvotes]: Let $X\rightarrow Y$ be a finite etale map. Let $R$ be a strict henselian ring with residue field $k$. Say that we have a map $Spec(k)\rightarrow X$ and so also $Spec(k)\rightarrow Y$. Assume that the map $Spec(k)\rightarrow Y$ factors thusly: $Spec(k)\rightarrow Spec(R)\rightarrow Y$. Then the question is: would there be (a unique?) map $Spec(R)\rightarrow X$ that would make this commute? -This would be a little cleaner if I knew how to do commutative diagrams in mathoverflow, but hopefully you get the picture. Intuitively, this means that if you have a point on $Y$ and a point above it on $X$, and if you have a "path" (I use this word very loosely here) near that point on $Y$ then it extends to a "path" near the respective point on $X$. When trying to prove this, the first thing I thought about is that every etale map is formally etale, but in the definition of formally etale they only talk about lifting $1^{st}$ order deformations. -Do you know how to prove (or god forbid disprove) this? - -REPLY [10 votes]: I think this is true. Namely, we can lift each $\mathrm{Spec}(R/\mathfrak{m}^n)\to Y$ to $X$ (because formal etaleness allows the lift to exist for any nilpotent thickening) successively so as to be compatible with all the previous ones (in fact, it has to be, by uniqueness in the lifting property). This successive lifting gives a map $\mathrm{Spec} R \to X$ lifting $\mathrm{Spec} R \to Y$ (to see this, note that without loss of generality, $X, Y$ are affine, and then taking direct limits in the category of affine schemes is the same as taking inverse limits in the category of rings). Thus we get a map from the completion $\\mathrm{Spec} \hat{R}$, hence from $\mathrm{Spec} R$. -I believe this is true even if we just assume $X \to Y$ to be smooth, since then we can still make successive liftings. Note that finiteness of the map $X \to Y$ is not necessary here. This argument seems to require modification because it is not obvious that the map will factor through the henselianization. -For another argument, note that we can reduce to the case where $Y$ is $\mathrm{Spec} R$ (by making a base-change via $\mathrm{Spec} R \to Y$). Then we have to show that there is a section $\mathrm{Spec} R \to X$: this is clear because $X$ is a product of strictly henselian rings finite and etale over $R$, and if one of them also has residue field $k$, it must be isomorphic to $\mathrm{Spec} R$. Here one uses the fact that a finite etale morphism of local rings with the same residue field must be an isomorphism, which follows from Nakayama's lemma and since flatness implies injectivity (for local rings). -(P.S. Just in case this was the question, recall the nilpotent lifting property for an etale morphism $X \to Y$: given any scheme $S$ and subscheme $S_0$ cut out by a nilpotent ideal (one can reduce to the square-zero case by induction), any diagram with $S_0 \to X$ and $S \to Y$ leads to a unique lift $S \to X$.) -(P. P. S. One doesn't need finiteness even in the second argument: if $X \to Y$ is an etale morphism with $Y$ the spectrum of a henselian ring $R$, then the local ring of a point of $x$ lying above the closed point of $Y$ is finite over $R$, by Zariski's Main Theorem: one characterization of henselian rings is that a quasi-finite local algebra over a henselian local ring is finite.)<|endoftext|> -TITLE: Spinor space to Euc. vector space: does there exist a universal bilinear map? -QUESTION [7 upvotes]: Let $S$ be a spin representation of the Euclidean -spin group $Spin(d)$ and let ${\mathbb R}^d$ -be Euclidean $d$-space with $Spin(d)$ action on it -in the canonical way, via the 2:1 cover to $SO(d)$. -(I am being careful here: A spin rep.'', notthe spin rep.'') -Is there, for all $d$, an onto quadratic $Spin(d)$-equivariant map $S \to {\mathbb R}^d$? -If so, is there a `universal' ($d$-independent) construction of this map? -MOTIVATION: For $d=2, 3$ I know these maps. -They are famous in celestial mechanics and yield the -standard regularizations of the Kepler problem, -or, what is the same, of binary collisions in the classical N-body problem. -They turn Kepler for negative energies into a harmonic oscillator. -Case $d=2$. I take $Spin(2)$ to also be $S^1$, but wrapped `twice' around -$S^1 = SO(2)$. $S = {\mathbb C}$. The quadratic map is $w \to w^2$. -This is the Levi-Civita regularization. -Case $d= 3$. This is the standard Hopf map ${\mathbb C}^2 \to {\mathbb R}^3$, -or if you prefer, from the quaternions ${\mathbb H }$ to ${\mathbb R}^3$, sending $q$ to $q k \bar q$. -The astronomers call this Kuustanheimo-Steifel regularization. -WHERE I'VE LOOKED SO FAR: I tried to make sense out -of Deligne's discussion on spinors in the AMS two-volume set -from some Princeton year on string theory from a decade or so ago. -I understand that over the complexes, there is either exactly one or exactly two -spin representations, depending on the parity of $d$. -So even there , we don't get a `universal' d-dependent map. -Over the reals things decompose in a rather complicated -dimension dependent way (mod 8 probably) and there is no clear choice. -I also looked in Reese Harvey's book which I find too baroque and -signature depend to penetrate. -Case $d=4$. Here I am not sure. But I know $Spin(4) = SU(2) \times SU(2)$ -which I can think of as two copies of the unit quaternions, each acting on -``its own'' ${\mathbb H}$. -I guess in this case I better take $S = {\mathbb H} \times {\mathbb H}$ -Then I get the desired quadratic map ${\mathbb H} \times {\mathbb H} \to {\mathbb H} = {\mathbb R}^4$ -as $(q_1, q_2) \mapsto q_1 \bar q_2$. - -REPLY [9 votes]: The answer is no. -For $Spin(5)\simeq USp(4)$, the spin representation of $Spin(5)$ is the defining representation $\mathbf{4}$ of $USp(4)$. The tensor product $\mathbf{4}\otimes \mathbf{4}$ decomposes as $$\mathbf{1}\oplus\mathbf{5}\oplus\mathbf{10}.$$ -The vector representation of $SO(5)$ is this $\mathbf{5}$, but it's in the antisymmetric part of this tensor product decomposition: -$$\wedge^2\mathbf{4} = \mathbf{1}\oplus\mathbf{5}$$ -The $\mathbf{1}$ is there, because of the definition of $USp(4)$. -So, there's no quadratic $Spin(5)$-equivariant map $\mathbf{4}\to\mathbf{5}$.<|endoftext|> -TITLE: System of weights for nilpotent Lie algebras -QUESTION [5 upvotes]: I am studying nilpotent Lie algebra theory. The subject is really new to me and I am studying by myself. I'd love your help with this. -Let $\mathfrak{n}$ be a finite-dimensional nilpotent Lie algebra (over an algebraically closed field of characteristic zero) and let $\operatorname{Der}(\mathfrak{n})$ be the algebra of derivations of $\mathfrak{n}$. The system of weights of $\mathfrak{n}$ is defined as being that of the natural representation of a "maximal torus" $T$ in $\operatorname{Der}(\mathfrak{n})$ and the $\operatorname{rank}$ is the dimension of $T$. By remarkable result due to Gabriel Favre (see [F]), it is known that for a fixed integer $n$ there are finitely systems of weights. Let $T$ be a system of weights, we denote by $\mathrm{N}(T)$ the class of those Lie algebras having the system of weights $T$. -My questions are: - -For a fixed integer $n$, are these system of weights classified? -For a fixed integer $n$, can rank-one system of weights explicitly written? -Are classified rank-one system of weights $T$ such that $\sharp\mathrm{N}(T)=1$ -Is there a good book or resource for learning about this topic and in general, about nilpotent Lie algebras (over $\mathbb{C}$ or $\mathbb{R}$)? - -Any help is much appreciated! -[F] Favre, G.: Système de poids sur une algèbre de Lie nilpotente. Manuscripta Math. 9 (1973), 53-90. - -REPLY [4 votes]: 1.) No, the weight systems are not classified in general. However, the weight systems for complex nilpotent Lie algebras of dimension $n\le 7$ have been computed by Roger Carles, in "Weight systems for complex nilpotent Lie algebras and application to the varieties of Lie algebras", in $1996$. Another reference, in addition to Pasha's references, is this paper of Magnin of $1998$. -2.), 3.) I think no. -4.) There are several books (e.g., by Goze, Onischik, Vinberg, etc.). Also, several articles discuss nilpotent Lie algebras in general, e.g. the article by E. M. Luks, What is a typical nilpotent Lie algebra ? For faithful representations of nilpotent Lie algebras, see, for example, here.<|endoftext|> -TITLE: Conjugacy problem for small braid groups -QUESTION [5 upvotes]: The conjugacy problem for braid groups $B_3$ and $B_4$ can be solved in polynomial time, it is noted in the paper by Birman, Ko and Lee(2001). -That was a result in 2001. Are there any new results on other small braid groups? Is the conjugacy problem in $B_5$ solvable in polynomial time? -Also, I'm still curious on exactly how fast the conjugacy problem in $B_3$ can be solved(polynomial time is too broad). The only paper that described such algorithm is behind a paywall. - -REPLY [3 votes]: This 2008 paper (freely available) explains the fastest known solution of the conjugacy problem in $B_3$. I think the polynomial time complexity of that solution can be easily extracted from that paper.<|endoftext|> -TITLE: Why is there such a close resemblance between the unitary representation theory of the Virasoro algebra and that of the Temperley-Lieb algebra? -QUESTION [41 upvotes]: For those who aren't familiar with the Virasoro or Temperley-Lieb algebras, I include some definitions: -• The (universal envelopping algebra of the) Virasoro algebra is the $\star$-algebra $Vir_c$ generated by elements $L_n$, ($n \in \mathbb{Z}$), subject to the relations -$$ -[L_m,L_n]=(m-n)L_{m+n}+\frac{c}{12}(m^3-m)\delta_{m+n,0}, -$$ -and with $\star$-structure $L_n^* = L_{-n}$. -• The Temperley-Lieb algebra is the $\star$-algebra $TL_{\delta}$ with generators $U_i$ ($i \in \mathbb{Z}$) and relations : - -$U_i^2 = \delta U_i$ and $*$-structure $U_i^* = U_i$. -$U_iU_{i+1}U_i=U_i$ and $U_iU_{i-1}U_i=U_i$ -$U_i U_j=U_j U_i$ for $|i-j|\ge 2$ - -Both $Vir_c$ and $TL_{\delta}$ depend on a parameter. -These are the numbers $c$ and $\delta \in \mathbb{R}$. - -Let's call a representation $\rho$ of a $\star$-algebra on a Hilbert space unitary if $\rho(x^*)=\rho(x)^*$. -We are interested in the unitary representations of $Vir_c$ and $TL_{\delta}$. In the case of the Virasoro algebra, we further restrict to positive energy representations, i.e., unitary representations in which the spectrum of $L_0$ is positive. -Depending on the value of the parameters $c$ or $\delta$, three things can happen: -1. Discrete series (only) of quotient of Verma modules are unitary and positive energy. -2. Continuum of Verma modules are unitary, positive energy representations. -3. The Verma modules are not unitary. -Now here's the striking thing: -$\begin{array}{c|c|c|c|c|c|c} - & \text{Discrete series} & \text{Continuum} & \text{Others} \newline - \hline -Vir_c & c \in \{ 1-\frac{6}{m(m+1)} \vert m = 2,3,4 \ldots \} &c \in [1,\infty) & \text{non-unitary} \newline - \hline -TL_\delta & \delta\in \{ 2\cos\big(\frac\pi m\big)\quad \vert \quad m = 2,3,4 \ldots \} &\delta \in [2,\infty) & \text{non-unitary} -\end{array}$ -The parameters $c$ and $\delta$ belong to a countable set (discrete series) exhibiting an accumulation point, followed by a continuum. - - -Is it pure coincidence that those two algebras exhibit such similar behaviour? -Is there some natural map from $Vir_c$ to $TL_{\delta}$, or vice-versa? -Is there any way of linking the values $c\in 1-\frac{6}{m(m+1)}$ and $\delta\in 2\cos(\frac\pi m)$? -Are there other algebras exhibiting a similar phenomenon? - -REPLY [16 votes]: Overview of an explanation : -Jones-Wassermann subfactors for the loop algebra : -Let $\mathfrak{g} = \mathfrak{sl}_{2}$ be the Lie algebra, $L\mathfrak{g}$ its loop algebra and $\mathcal{L}\mathfrak{g} = L\mathfrak{g} \oplus \mathbb{C}\mathcal{L}$ the central extension : - $$[X^{a}_{n},X^{b}_{m}] = [X^{a},X^{b}]_{m+n} + m\delta_{ab}\delta_{m+n}\mathcal{L}$$ with $(X^{a})$ the basis of $\mathfrak{g}$. - The unitary highest weight representations of $\mathcal{L}\mathfrak{g}$ are $(H_{i}^{\ell},\pi_{i}^{\ell})$ with : - -$\mathcal{L} \Omega = \ell \Omega$ with $\ell \in \mathbb{N}$ the level, and $\Omega$ the vacuum vector. -$i \in \frac{1}{2}\mathbb{N}$ and $i \le \frac{\ell}{2}$, the spin (related to the irreducible representation $V_{i}$ of $\mathfrak{g}$) - -Let $I \subset \mathbb{S}^{1}$ an interval, and $\mathcal{L}_{I}\mathfrak{g}$ the local Lie algebra generated by $(X^{a}_{f})$ with : - -$f(\theta) = \sum \alpha_{n}e^{in\theta}$ and $f \in C^{\infty}_{I}(\mathbb{S}^{1})$ -$X^{a}_{f} = \sum \alpha_{n}X^{a}_{n}$ - -Let $\mathcal{M}_{i}^{\ell}(I)$ be the von Neumann algebra generated by $\pi_{i}^{\ell}(\mathcal{L}_{I}\mathfrak{g})$. -We obtain the Jones-Wassermann subfactor : -$$\mathcal{M}_{i}^{\ell}(I) \subset \mathcal{M}_{i}^{\ell}(I^{c})'$$ of index $\frac{sin^{2}(p\pi/m)}{sin^{2}(\pi/m)}$ with $m=\ell + 2$ and $p=2i+1$. -Its principal graph is given by the fusion rules : -$$H_{i}^{\ell} \boxtimes H_{j}^{\ell} = \bigoplus_{k \in \langle i,j \rangle_{\ell}}H_{k}^{\ell}$$ with $\langle a,b \rangle_{n} = \{c=\vert a-b \vert, \vert a-b \vert+1,... \vert c \le a+b , a+b+c \le n \}$ -Let $\mathcal{R}_{\ell}$ be the fusion ring generated. -Temperley-Lieb case (with $\ell \ge 1$) : -If $i=1/2$ then index=$\frac{sin^{2}(2\pi/(\ell+2))}{sin^{2}(\pi/(\ell+2))} = \delta^{2}$ with $\delta = 2cos(\frac{\pi}{\ell+2})$ and the principal graph is $A_{\ell+1}$. -In this case, the subfactors are known to be completely classified by their principal graph. -The subfactor planar algebra it generates is the Temperley-Lieb planar algebra $TL_{\delta}$. -Jones-Wassermann subfactors for the Virasoro algebra : -Let $\mathfrak{W}$ be the Lie algebra generated by $d_{n} = ie^{in\theta}\frac{d}{d\theta}$ and $\mathfrak{Vir} = \mathfrak{W} \oplus C \mathbb{C}$ its central extension: - $$ -[L_m,L_n]=(m-n)L_{m+n}+\frac{C}{12}(m^3-m)\delta_{m+n,0}, -$$ -Its discrete series representations are $(H_{pq}^{m})$ with : - -$C\Omega = c_{m} \Omega$ with $c_{m}= 1-\frac{6}{m(m+1)}$ for $m=2,3,...$ -$L_{0} \Omega = h^{pq}_{m} \Omega$ with $h^{pq}_{m} = \frac{[(m+1)p-mq]^{2}-1}{4m(m+1)}$ with $1 \le p \le m-1$ and $1 \le q \le p $ - -As for the loop algebra, there are $\mathfrak{Vir}_{I}$ and $\mathcal{N}_{pq}^{m}(I)$ generated by $\pi_{pq}^{m}(\mathfrak{Vir}_{I})$. -We obtain the Jones-Wassermann subfactor : -$$\mathcal{N}_{pq}^{m}(I) \subset \mathcal{N}_{pq}^{m}(I^{c})'$$ of index $\frac{sin^{2}(p\pi/m)}{sin^{2}(\pi/m)}.\frac{sin^{2}(q\pi/(m+1))}{sin^{2}(\pi/(m+1))}$. -Its principal graph is given by the fusion rules : -$$H_{pq}^{m} \boxtimes H_{p'q'}^{m} = \bigoplus_{(i'',j'') \in \langle i,i' \rangle_{\ell} \times \langle j,j' \rangle_{\ell + 1} }H_{p''q''}^{m}$$ with $p=2i+1, q=2j+1, p'=2i'+1, ..., m=\ell+2$ -Let $\mathcal{T}_{m}$ be the fusion ring they generate, it's an easy quotient of $\mathcal{R}_{\ell} \otimes_{\mathbb{Z}} \mathcal{R}_{\ell+1}$, with $\mathcal{R}_{\ell}$ the fusion ring obtained above for the loop algebra. -Temperley-Lieb case (with $m \ge 3$) : -If $(p,q) = (2,1)$, index$=\frac{sin^{2}(2\pi/m)}{sin^{2}(\pi/m)} = \delta^{2}$ with $\delta = 2cos(\frac{\pi}{m})$ and the principal graph is $A_{m-1}$. -As above, the subfactor planar algebra is Temperley-Lieb $TL_{\delta}$. - -$\rightarrow$ We obtain the natural maps $c \leftrightarrow \delta$ - and $\mathfrak{Vir}_{c} \leftrightarrow TL_{\delta}$ that you - expected. - -Generalizations for similar phenomenon : -Here is a list of possibilities : - -take $i$ other than $1/2$ or $(p,q)$ other than $(2,1)$ -take $\mathfrak{g}$ other than $\mathfrak{sl}_{2}$ -take the continuous series -take a $N$-super-symmetric extension of $\mathfrak{Vir}$ : $N=1$ for the Neveu-Schwarz and Ramond algebras. - -References : -- V.F.R. Jones, Fusion en algèbres de von Neumann et groupes de lacets (d'après A. Wassermann), Séminaire Bourbaki, Vol. 1994/95. Astérisque No. 237 (1996), Exp. No. 800, 5, 251--273. -- T. Loke, Operator algebras and conformal field theory for the discrete series representations of $\textrm{Diff}(\mathbb{S}^{1})$, thesis, Cambridge 1994. -- S. Palcoux, Neveu-Schwarz and operators algebras I : Vertex operators superalgebras, arXiv:1010.0078 (2010) -- S. Palcoux, Neveu-Schwarz and operators algebras II : Unitary series and characters, arXiv:1010.0077 (2010) -- S. Palcoux, Neveu-Schwarz and operators algebras III : Subfactors and Connes fusion, arXiv:1010.0076 (2010) -- V. Toledano Laredo, Fusion of Positive Energy Representations of LSpin(2n), thesis, Cambridge 1997, arXiv:math/0409044 (2004) -- R. W. Verrill, Positive energy representations of $L^{\sigma}SU(2r)$ and orbifold fusion. thesis, Cambridge 2001. -- A. J. Wassermann, Operator algebras and conformal field theory. Proceedings of the International Congress of Mathematicians, Vol. 1, 2 (Zurich, 1994), 966--979, Birkhuser, Basel, 1995. -- A. J. Wassermann, Operator algebras and conformal field theory. III. Fusion of positive energy representations of ${\rm LSU}(N)$ using bounded operators. Invent. Math. 133 (1998), no. 3, 467--538. -- A. J. Wassermann, Kac-Moody and Virasoro algebras, 1998, arXiv:1004.1287 (2010) -- A. J. Wassermann, Subfactors and Connes fusion for twisted loop groups, arXiv:1003.2292 (2010)<|endoftext|> -TITLE: Fractional Brownian motion and Laplacian -QUESTION [6 upvotes]: Having read this link on math stackexchange, I would like to submit to your wisdom the following questions. -Is it possible, mutatis mutandis, to repeat the same reasoning for a fractional Brownian motion? -More specifically: a real valued Gaussian process $B^H:=\{B^H(t)\}_{t\geq 0}$ - in a probability space - $(\Omega,\mathscr{F},\mathbb{P})$ is a fractional Brownian - motion (fBm) with Hurst parameter $H \in(0,1)$ if for all $s,t\in \mathbb{R}_+$ - -$B^H(0)=0$, -$\mathbb{E}B^H(t)=0$, -$\operatorname{Cov}[B^H(t),B^H(s)]=\frac{1}{2} - \left(t^{2 H}+s^{2 H}-|t-s|^{2 H}\right)$. - -In addition, the It\^o formula for fBm is written as: -$$ -f(B^H(t))= \displaystyle\int_0^t f'\left(B^H(s)\right)\, d B^H(s) + H \displaystyle\int_0^t f''\left(B^H(s)\right) s^{2H-1}\, ds. -$$ -Taking the expectation in both sides of the above equality, we obtain: -$$ -\mathbb{E}\left[f(B^H(t))\right]= H \displaystyle\int_0^t \mathbb{E}\left[f''\left(B^H(s)\right)\right] s^{2H-1}\, ds. -$$ - I might continue as; - changing the expectation by the conditional expectation $\mathbb{E}_x$ with respect to the event $\{X_0=x\}$ where $X(t)=B^H(t)+ x $, it follows: -$$ -\mathbb{E}_x\left[f(X^H(t))\right]= H \displaystyle\int_0^t \mathbb{E}_x\left[f''\left(X^H(s)\right)\right] s^{2H-1}\, ds. -$$ -And if we put -$$ -m(x,t; H)= \mathbb{E}_x\left[f\left(X^H(t)\right)\right]. -$$ -We get: -$$ -\displaystyle\frac{\partial}{\partial t} m(x,t; H)= H \, t^{2H-1}\displaystyle\frac{\partial^2}{\partial x^2 }m(x,t; H) -$$ -Knowledge that fBm is not a semimartingale nor a Markov process except for cases $H=\frac{1}{2}$. - I have some doubts about the last deduction.1. - -REPLY [4 votes]: Your construction has been carried out for much more general situations. See -Baudoin, F., Coutin, L. Operators associated with a stochastic differential equation driven -by fractional Brownian motions, Stoch. Proc. Appl. 117, 5, 550–574, 2007. ArXiv: math/0509511.<|endoftext|> -TITLE: Can we make Buchberger's algorithm faster for a given ideal if we are allowed to vary the monomial order? -QUESTION [7 upvotes]: Suppose we have a finite set of generators for an ideal $I \subset R := \Bbbk[x_1,\dotsc, x_n]$, where $\Bbbk$ is a field. If we choose a monomial ordering, then Buchberger's algorithm allows us to produce a Groebner basis for $I$. However, the size of the resulting Groebner basis can be enormous, and moreover can vary greatly depending on the monomial order chosen. -Sometimes, we have a reason for desiring a monomial order independent of $I$. (E.g., for elimination, we need an order with certain characteristics of lex; if we want to see of two sets of generators give the same ideal, we obviously want to use the same monomial order for both of them.) However, there are times when we may want to find a Groebner basis for $I$ with respect to some monomial, and we don't really care which. This could be useful, for instance, if we want to find a monomial $\Bbbk$-basis for $R/I$, and thereby (assuming $I$ is homogeneous) calculate the Hilbert polynomial of $I$. - -Are there studies of algorithms and/or heuristics that design a monomial order based on the given generators of $I$ in an effort to produce a smaller Groebner basis for this particular ideal? - -Ideally, it might be possible to choose a monomial order that has a good chance of outperforming grevlex on this particular generating set. At the very least, there should be some sort of heuristics for which grevlex order to choose (i.e., how the variables should be ordered). - -REPLY [5 votes]: You might investigate Singular, a software package for algebraic polynomial computations. -I know little about it, but it does implement a so-called -Hilbert-driven Buchberger algorithm, which (somehow!) finds "an appropriately chosen fast" -ordering of the monomials, specifically to circumvent the problem that "the performance of Buchberger's algorithm is sensitive to the choice of monomial order." -Their documentation provides one example -with a $100 {\times}$ speedup. - -This article by Manuel Kauers in Scholarpedia may help. Here are some quotes: - -Change of Ordering -Some applications require Gröbner bases with respect to a particular ordering of the power products for which Buchberger's algorithm is not as efficient as for other orderings. In such situations it may be advantageous to first compute a Gröbner basis with respect to some ordering where Buchberger's algorithm runs faster and in a second step transform this Gröbner basis to a Gröbner basis for the desired ordering. -Gröbner Walk -Two different techniques for performing such a change of ordering are known. One is known as Gröbner walk. It is based on an interpretation of orderings as regions in a space. If two orderings correspond to regions which overlap, then a Gröbner basis for one of the orderings can be turned into a Gröbner basis for the other by calling Buchberger's algorithm on a small auxiliary problem for which it usually terminates quickly. When the regions for two orderings do not overlap, it is always possible to connect them by a path consisting of orderings where the regions of any two consecutive ones have an overlap. The transformation can then be done step by step, switching in each step to the next ordering on the path. [...] -Linear Algebra -The second technique uses linear algebra. If $G$ is a Gröbner basis for some ordering, - then we have [...] - Using this technique, [...], one can determine the elements of a Gröbner basis with respect to an ordering different from the ordering of $G$. - -See the article for more details and references.<|endoftext|> -TITLE: How are mathematical objects defined from an ultrafinitist perspective? -QUESTION [6 upvotes]: I remember attending a lecture given by an ultrafinitist who denied that curves are a set of points, he would only say that any particular point may or not be on the curve. Similarly for algebraic or analytic objects, the only way I know how to define them is as a set with operations on the elements of that set. Since ultrafinitists cannot use definitions with infinite sets, what sort of definitions do they use? - -REPLY [7 votes]: Point-set definitions are the most common modern way of defining a geometrical object like a line, but they're not the only way. -In Euclid, lines and circles are primitives. -The axiomatization of Euclidean geometry in Tarski 1999 has only points, betweenness, and congruence as primitives, and sets are not even referred to in the axioms except for the axiom of continuity, which basically says lines are Dedekind-complete. I don't imagine that ultrafinitists would even want this kind of continuity, so they'd probably leave this axiom out. (Even if you interpret the axiom as a sheaf of axioms over first-order formulas defining the relevant sets, that would be an infinite sheaf, which I think would be unacceptable.) Tarski's axioms, interpreted using classical logic, imply that there are infinitely many points, but classical logic isn't the appropriate logic for ultrafinitism. So I don't really see any reason to believe that there's any problem with doing an ultrafinitist geometry this way. -As an example, take the unit circle in the Cartesian plane, defined by a first-order formula (not as a set). I can definitely prove to an ultrafinitist's satisfaction that it contains at least four points (i.e., there are at least four points satisfying that formula). Given a line through the origin O (defined by picking some point P outside the circle that the line goes through), the lack of the axiom of continuity means that it's probably not possible to give an ultrafinitist proof that it intersects the circle (i.e., that there is a point between O and P that satisfies the formula defining the circle). However, one can certainly prove that there exist a point on the line and a point on the circle such that the distance between them is no more than 0.0001. There will be points such that it's neither true nor false that the point is on both the line and the circle. -Even if you're working in a system that has infinities, it's not necessary to describe geometry in terms of point sets. The original presentation of the surreal line didn't use sets, and the surreals are too big to be a ZFC set. In smooth infinitesimal analysis, a curve is not representable as a set of points. -Tarski and Givant, 1999, http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.27.9012<|endoftext|> -TITLE: n-th integral cohomology of a non-compact manifold of dimension n -QUESTION [11 upvotes]: Let $M$ be a smooth connected oriented without boundary non-compact manifold -of dimension n. -Let $k$ be a principal ideal, e. g. the integers $Z$ -Let $H_n(M)$ and $H^n(M)$ be the homology and cohomology in degree n of $M$ -with coefficients in $k$. -It is well-known that $H_n(M)=0$. -Is $H^n(M)$ also trivial ? -By the universal coefficient theorem for cohomology, -$H^n(M)=Ext(H_{n-1}(M),k)$. -Therefore if $k$ is a field, $H^n(M)=0$. I would like to know this answer over -a principal ideal: $Z$. -It is also known (Bredon's book) that $H_{n-1}(M)$ is without torsion. -But I believe that $H_{n-1}(M)$ in the non-compact case, is a not a finitely -generated $k$-module. -Therefore I don't know if $H_{n-1}(M)$ is free, i. e. projective. -Since there exists abelians groups without torsion, non-free, e. g. the rationals $Q$ -I suppose that this must be well-known. -But I could not find a reference. - -REPLY [2 votes]: Maybe too late, and close to the Fernando's argument. Let $c$ be an infinite path going through each center of the $n$-simplex of a triangulation. Let $U$ be a neighbourhood of $c$, we can assume $U$ has the homotopy type of $[0,+\infty[$ and $\partial U$ is homeomorphic to $\mathbb R^{n-1}$. Then $V= \overline{X\setminus U}$ retracts on the $n-1$-skeleton, so $H^{n-1}(V)=0$ and we conclude by Mayer Vietoris -$$H^{n-1}(U\cap V) \to H^n(X) \to H^n(U)\oplus H^n(V)$$<|endoftext|> -TITLE: Vladimir Voevodskys 2002 ICM Lecture. -QUESTION [6 upvotes]: Is Vladimir Voevodskys ICM lecture available in videotaped format somewhere? -Strangely it is not at the IMU homepage (but Lafforgues is) http://www.mathunion.org/Videos/ICM2002/ -Was it not taped (Why not?)? -If not is at least a transcript available somewhere? -Is there a least video of the opening ceremony, and the Laudatio(s)? -Are there any taped lectures of Voevodsky lecturing on Motivic Cohomology? - -REPLY [11 votes]: Some of Voevodsky videos are here: -http://video.ias.edu/taxonomy/term/42 -http://www.mathnet.ru/PresentFiles/425/425.flv -http://www.mathunion.org/Videos/ICM98/ICMs/vladimir_voevodsky.html -http://claymath.msri.org/voevodsky2002.mov -The last two talks, Algebraic Cycles and Motives and An Intuitive Introduction to Motivic Homotopy Theory, are perhaps the closest to what you look for. As for the last talk, the notes from it are available here: -http://www.cwru.edu/artsci/phil/Voevodsky.pdf<|endoftext|> -TITLE: From Zeta Functions to Curves -QUESTION [41 upvotes]: Let $C$ be a nonsingular projective curve of genus $g \geq 0$ over a finite field $\mathbb{F}_q$ with $q$ elements. From this curve, we define the zeta function -$$Z_{C/{\mathbb{F}}_q}(u) = \exp\left(\sum^{\infty}_{n = 1}{\frac{\# C(\mathbb{F}_{q^n})}{n} u^n}\right),$$ -valid for all $|u| < q^{-1}$. This zeta function extends meromophicially to $\mathbb{C}$ via the equation -$$Z_{C / \mathbb{F}_q}(u) = \frac{P_{C / \mathbb{F}_q}(u)}{(1 - u) (1 - qu)}$$ -for some polynomial with coefficients in $\mathbb{Z}$ that factorises as -$$P_{C/\mathbb{F}_q}(u) = \prod^{2g}_{j = 1}{(1 - \gamma_j u)}$$ -with $|\gamma_j| = \sqrt{q}$ and $\gamma_{j + g} = \overline{\gamma_j}$ for all $1 \leq j \leq g$. This last point tells us that $Z_{C / \mathbb{F}_q}(u)$ has a functional equation and satisfies a version of the Riemann hypothesis. -What happens if we run this construction in reverse? What if we start with a set of numbers $\gamma_j$, $1 \leq j \leq 2g$, such that $|\gamma_j| = \sqrt{q}$, $\gamma_{j + g} = \overline{\gamma_j}$ for all $1 \leq j \leq g$, and such that the polynomial -$$P(u) = \prod^{2g}_{j = 1}{(1 - \gamma_j u)}$$ -has coefficients in $\mathbb{Z}$? Is there a way of telling whether the function -$$\frac{P(u)}{(1 - u) (1 - qu)}$$ -is the zeta function of some curve $C$? Furthermore, what is this curve exactly? -A simple case of this is if we look at the function -$$\frac{1 - au + qu^2}{(1 - u) (1 - qu)}$$ -for some $a \in \mathbb{Z}$ with $|a| \leq 2 \sqrt{q}$. How do we determine whether this function is the zeta function $Z_{C / \mathbb{F}_q}(u)$ of an elliptic curve $C$ over $\mathbb{F}_q$? If it is indeed equal to $Z_{C / \mathbb{F}_q}(u)$, what is the Weierstrass equation for $C$ (assuming $\mathrm{char}(q) \geq 5$)? - -REPLY [26 votes]: In your last paragraph, you suggest that you are particularly interested in the case of elliptic curves. This is much easier than the general case, which is addressed well by Torsten's answer. -If $q$ is prime, and $|a| \leq 2 \sqrt{q}$, then there is always an elliptic curve with $\zeta$-function $(1-au+qu^2)/(1-u)(1-qu)$. This is the one dimensional case of Honda's theorem. I'll sketch the proof. As you will see, it uses some very sophisticated methods, and it will be very hard to make it effective. -Sidenote: What happens when $q = p^k$ for $k>1$? Then there are $2 p^{k-1}-1$ ways to choose $a$ to be $0 \mod p$. When $a=0$, the elliptic curve must be supersingular. But there are only $\approx p/12$ supersingular elliptic curves over $\mathbb{F}_{p^k}$. So, once $2 p^{k-1}$ is much greater than $p/12$, there will be $a$'s which don't occur. To be honest, I am not clear what happens if you feed one of these $a$'s into Honda's theorem. -Proof Sketch: Let $R$ be the ring $\mathbb{Z}[\phi]/(\phi^2 - a \phi + q)$. $R$ is an order in an imaginary quadratic field (the inequality $a^2-4q<0$ is used to show that this is an imaginary extension.) Let $K$ be the the fraction field of $R$ and let $H$ be its class field. Let $\mathfrak{p}$ be the ideal $(\phi)$ in $\mathcal{O}_K$. Since $\mathfrak{p}$ is principal, it splits in $H$; let $\mathfrak{q}$ lie over $\mathfrak{p}$. So $\mathcal{O}_H/\mathfrak{q} \cong \mathbb{F}_p$. -Let $E$ be an elliptic curve with complex multiplication by $R$; then $E$ can be defined over $H$. ($E$ is only well defined up to a quadratic twist, at this point.) Take a model of $E$ over $\mathcal{O}_H$. Let $E_0$ be the fiber over $\mathfrak{q}$. Then $E_0$ is an elliptic curve over $\mathbb{F}_p$. One can show that the Frobenius acts on $E_0$ by $\pm \phi$ or $\pm \overline{\phi}$, where the bar is the automorphism of $R$ over $\mathbb{Z}$ coming from complex conjugation. By changing the quadratic twist, one can make sure the sign is $+$. Then the trace of the Frobenius is $\mathrm{Tr}(\phi)$, which is $a$, as desired. -Sorry for making this so advanced, I don't know an easier way. I think you can find most of the tools I am using in Silverman's Advanced topics in the arithmetic of elliptic curves.<|endoftext|> -TITLE: Exterior powers in tensor categories -QUESTION [9 upvotes]: Let $\mathcal{C}$ be a cocomplete $R$-linear tensor category. Many notions of commutative algebra can be internalized to $\mathcal{C}$. For example a commutative algebra is an object $A$ in $\mathcal{C}$ together with morphisms $e : 1 \to A$ (unit) and $m: A \otimes A \to A$ (multiplication) satisfying the usual laws. The $n$-th symmetric power $\text{Sym}^n(X)$ of an object $X$ is the quotient of $X^{\otimes n}$ by identifying $x_1 \otimes ... \otimes x_n = x_{\sigma(1)} \otimes ... \otimes x_{\sigma(n)}$, so formally it is defined as a coequalizer of the $n!$ symmetries $X^{\otimes n} \to X^{\otimes n}$. Then $\text{Sym}(X) = \bigoplus_{n\geq 0} \text{Sym}^n(X)$ is a commutative algebra and in fact $\text{Sym}$ is left adjoint to the forgetful functor $\mathsf{CAlg}(\mathcal{C}) \to \mathcal{C}$. -But now what about the exterior power $\Lambda^n(X)$? It is clear how to define $X^{\otimes n}$ modulo $x_1 \otimes ... \otimes x_n = \text{sgn}(\sigma) \cdot x_{\sigma(1)} \otimes ... \otimes x_{\sigma(n)}$ in this context, which one might call the anti-symmetric power $\mathrm{ASym}^n(X)$. The correct definition of the exterior power also has to mod out $$... \otimes a \otimes ... \otimes a \otimes ... = 0,$$ -for example because we want to have that $\Lambda^p(1^{\oplus n}) \cong 1^{\oplus \binom{n}{p}}$. But I have no idea how to internalize this to $\mathcal{C}$, even for $n=2$. The reason is that there is no morphism $X \to X \otimes X$ which acts like $a \mapsto a \otimes a$. Another idea would be to define $\Lambda(X)$ as a graded-commutative algebra object with the usual universal property, classifying morphisms $f$ on $X$ which satisfy something like $f(x)^2=0$, but again it is unclear how to formulate this in $\mathcal{C}$. -If this is not possible at all, which additional structure on $\mathcal{C}$ do we need in order to define exterior powers within them? Is this some categorified $\lambda$-ring structure? This structure should be there in the case of usual module categories (over rings or even ringed spaces). Of course there is no problem when $2 \in R^*$, because then the exterior power equals the anti-symmetric power. The question was also discussed in a blog post. -Here is a more specific (and a bit stronger) formulation: Is there some $R[\Sigma_n]$-module $T$, such that for every $R$-module $M$, we have that $T \otimes_{R[\Sigma_n]} M^{\otimes n} \cong \Lambda^n M := M^{\otimes n}/(... \otimes x ... \otimes x ...)$? Because then we could define $\Lambda^n X := T \otimes_{R[\Sigma_n]} X^{\otimes n}$ for $X \in \mathcal{C}$. -Concerning the "hidden extra structure" in the case of modules: Let the base ring be $\mathbb{Z}$, or more generally a commutative ring $R$ in which $r^2 - r \in 2R$ for all $r \in R$; this includes boolean rings such as $\mathbb{F}_2$ and also $\mathbb{Z}/n$. If $M$ is an $R$-module, then there is a well-defined(!) homomorphism $M^{\otimes~ n-1} \to \text{ASym}^n(M), x_1 \otimes ... \otimes x_n \mapsto x_1 \wedge x_1 \wedge ... \wedge x_n$, and its cokernel is $\Lambda^n(M)$. -Concerning non-linear tensor categories, I've asked here a similar question. - -REPLY [6 votes]: Deligne (Categories Tannakiennes, 1990, p165) defines it to be the image of the antisymmetrisation $a=\sum(-1)^{\epsilon(\sigma)}\sigma\colon X^{\otimes n}\rightarrow X^{\otimes n}$.<|endoftext|> -TITLE: What would the best treatment of Gehring's lemma look like? -QUESTION [10 upvotes]: In a course about elliptic regularity probably one sooner or later stubles into the reverse Holder inequalities, and has to introduce the Gehring lemma, which in one of its many versions improves a bit the regularity of a function, given the knowledge that such function already satisfies local bounds via another more integrable function (for example one bounds the integral of $f^2$ on any ball via a power of $f^{2-\epsilon}$ on the ball of double radius). -Even though I saw a proof of this (by Giaquinta-Modica) and I "believe" the result intuitively, I have the feeling that I didn't catch yet the "juice" of it.. for example I don't have the intuition of the following -1) how far it can be extended, -or -2) if the more extended/general versions are meaningful to teach (e.g. because one can find a clearer proof, or a more natural one, or because they can show connections with other subjects..). -So I would like to ask you if you know how to put this result in a larger context, or about its connections with topics different than elliptic regularity. Also references to nice expositions of it are quite welcome! - -REPLY [11 votes]: I'm not sure about "the larger context", but let me try to dissect the proof a bit so that there will be no mystery left there. -It runs upon 3 main ideas: -1) The "lack of concentration implies better summability" principle. In the nutshell, it is the following. Assume that we have some positive integrable function $f$ with $\int_X f=1$ on some probability measure space and know that for every subset $E$ of measure $\delta$, we have $\int_E f\le c(\delta)$ where $c(\delta)\to 0$ as $\delta\to 0$. Then we can design an increasing function $\Psi$ that grows to infinity and depends on $c$ only such that $\int_X f\Psi(f)<+\infty$. Indeed, assume that $\int_X f=1$. Put $E_k=\{f>2^k\}$. Then $\mu(E_k)\le 2^{-k}$, so, assuming that $\Psi$ is doubling, $\int_X f\Psi(f)$ is comparable to is enough to ensure that $\sum_{k\ge 0} \Psi(2^k)2^k\mu(E_k)\approx \sum_{k\ge 0}\psi(2^k) \int_{E_k}f\le \sum_{k\ge 0}\Psi(2^k)c(2^{-k})$, so we can choose any $\Psi$ for which this series converges. If $c(\delta)=\delta^q$, we can take $\Psi(x)=x^{q-\varepsilon}$ bringing $f$ to some $L^p$ with $p>1$.. -This principle (or something like it) is useful in many settings. In general it often happens that we can substantially improve the integral bounds if we know a priori that the functions aren't concentrated on small sets or there are no strong correlations between their values. I'm not ready to give impressive particular examples, but once you digest the idea, you can easily both recognize and use it in other contests. -2) The "multiscale trick". You want to show that $\int_E f$ is small compared to $\int_X f$ when $\mu(E)$ is small. Instead of going from $E$ to $X$ directly, you show first that there is a set $E_1$ of measure just slightly bigger than $E$ such that $\int_{E_1}f$ is noticeably bigger than $\int_E f$. Then you apply this to $E_1$ and so on until you reach $X$. If the path is long, the small improvement at each stage will result in a huge one in the end (another general principle that I wish our students could digest). So, if, say, we can always find $E_1$ such that $\mu(E_1)\le C\mu(E)$ and $\int_{E_1}f\ge a\int_E f$ with $a>1$ (provided that $\mu(E)$ is small enough, of course), we can get the power bound for the integrals over sets of small measure. -This multiscale reasoning is the core of many results in modern analysis. It can be almost trivial like in our case or extremely sophisticated but the idea is always the same. -3) The "geometric structure theorems". They are often presented as covering lemmas and density and maximal function theorems but what they really say is that arbitrary measurable sets in nice spaces can be viewed as finite unions of simple almost disjoint pieces (balls in $\mathbb R^n$, say). For the Gehring lemma, we just take a set $E$ of small measure and create the union of "low but noticeable density balls" surrounding it (for almost each point $x\in E$ we can find a ball $B$ such that the $\mu(B\cap E)\approx \gamma \mu(B)$ where $\gamma$ is some number between $0$ and $1$ we can choose to be whatever we want (provided that $\mu(E)<\gamma$). -If $B$ is one such ball, we can estimate the square of the average of $\sqrt f$ over $B$ by $\gamma\int_{B\cap E} f+\int_{B\setminus E}f$. Since the first part is small compared to $\int_B f$ when $\gamma$ is tiny, the condition that these averages are comparable implies that second part should be comparable to the entire integral, so if we extend $E\cap B$ to the entire $B$ (multiplying the measure by $\gamma^{-1}$), we increase the integral some fixed number of times. -That would be the end if we could make our balls truly disjoint and covering the entire $E$ (we would just take $E_1$ to be the union of our balls). Since it is not exactly the case, we have to work a bit more. We can use some standard covering lemma to pass to a sequence of "morally disjoint" balls. If, say, we have a doubling measure in a metric space, we can use Vitaly and say that we can find a set of disjoint balls $B$ in our family such that three times larger balls cover $E$. -This still doesn't seem good enough because if we use $B$'s, we are in danger that they capture only a small part of the integral of $f$ over $E$ and if we use $3B$, they will overlap a lot, so we can count the same piece in the extended integral many times but by now we are pretty convinced that the things should work, so we can go over the steps and do the needed tune-up for this particular theorem. -This tune-up can be done in infinitely many ways. I will do it in the way that allows to illustrate one more idea. Let us use the triple balls $B'=3B$. Let $I=\int_E f$, $\mu=\mu(E)$. Let $E_k$ be the set of points outside $E$ covered by exactly $k$ balls $B'$ and let $\mu_k=\mu(E_k)$, $I_k=\int_{E_k} f$. The above consideration show that $\sum_{k\ge 1}I_k\ge cI$ and the doubling amounts to $\sum_{k\ge 1}\mu_k\le C\mu$, whence we can find non-zero $\mu_k$ with $\frac{I_k}{\mu_k}\ge \varkappa \frac I\mu$ with $\varkappa=\frac cC$. But then adding $E_k$ to $E$, we shall increase the ratio $I/\mu^{\kappa}$. Now just notice that if $E$ is a fixed set of positive measure, then this ratio attains its maximum over all supersets of $E$ and we just showed that this maximum cannot be attained on a set of small measure but for sets of large measure, the ratio is trivially bounded. -This "choose the extremal object" idea can also be found almost everywhere. -Of course, the proof of the Gehring's lemma obtained this way is neither the shortest, nor the slickest. My only excuse for bringing it up is that you seem to be dissatisfied with the proof you know, so I took the liberty to build the argument from the a few standard blocks in modern analysis in the most straightforward way. I also avoided a few technicalities related to localization, but those are totally routine and only obscure the main ideas. I also took the liberty to "misinterpret" your request and talk about the "larger context" for the ideas in the proof rather than for the result itself. In a sense, I tried to show that the Gehring lemma is just a combination of 3-4 well-known and widely used principles and it is given a name just because this particular combination is used as a single block often enough to justify creating a formal reference instead of reproducing the proof every single time. Of course, once you realize what the building blocks are, you can freely play with them and get many other statements in the same style, which, IMHO, is what "catching the juice of a theorem" means. I leave it to others to tell where outside PDE this building block comes handy as a whole but I hope I convinced you that the basic steps of which it consists are ubiquitous.<|endoftext|> -TITLE: A dual theory to the theory of currents? -QUESTION [6 upvotes]: The k-currents are defined as dual space to the spaces of all smooth k-forms. -(These monsters are used to work with the minimal k-surfaces.) -Assume I want to look at the generalized k-forms; -they can be defined as certain functionals -on the space of all smooth k-surfaces with boundary. -Did anybody consider such "generalized k-forms"? -In fact I am looking for such generalized 2-forms on $\mathbb R^n$ with values $so(n)$, -this should be some kind of generalized curvature. -(Instead of integral I should consider parallel translation around the boundary. -If $n=2$, it gives me a sign-measure.) -But I am also interested in the linear case. - -REPLY [2 votes]: There is another notion of generalized form called charges. You can find its definition here : -http://www.math.jussieu.fr/~depauw/preprints/dep-moo-pfe.pdf (they are functionals on the space normal currents with a distribution-like topology). -Cochains can be represented as $L^1$ forms with $L^1$ exterior derivative, whereas charges are represented as $\omega + d \eta$, where $\omega$ and $\eta$ are continuous forms. Maybe this kind of generalized form is best suited for your purposes.<|endoftext|> -TITLE: Mayer-Vietoris for sheaf cohomology -QUESTION [6 upvotes]: Assume the following is given: - -$X$ quasi-projective (smooth) variety, -$\overline X$ a projective variety, such that $X$ is the complement of a divisor $D$ in $X$, and -$\mathcal F$ a coherent sheaf on $\hat X$. - -Suppose, we know the cohomology $H^\ast(X,{\mathcal F}|_X)$ of $\mathcal F$ on $X$. Now, for sure, this is not enough to be able to compute $H^{\ast}({\overline X},{\mathcal F})$. -My question is what kind of additional data (localized near D) do we need to know in order to be able to compute $H^\ast({\overline X},{\mathcal F})$? - -REPLY [3 votes]: You need the cohomology of the extension of $\mathcal{F}$ to the completion of $X$ along $D$ (a formal scheme). There is an equivalence of categories that you can look up in -Moret-Bailly, Laurent -Un problème de descente. Bulletin de la Société Mathématique de France, 124 no. 4 (1996), p. 559-585 (numdam) -see also references therein. -From this description, a way of computing cohomology should follow.<|endoftext|> -TITLE: Manifold with all geodesics of Morse index zero but no negatively curved metric? -QUESTION [19 upvotes]: A closed oriented Riemannian manifold with negative sectional curvatures has the property that all its geodesics have Morse index zero. -Is there a known counterexample to the "converse": if (M,g) is a closed oriented Riemannian manifold (Edit: assumed to be nondegenerate) all of whose geodesics have Morse index zero then M admits a (possibly different) metric g' with negative sectional curvatures? -Edit: Motivation for asking this (admittedly naive) question is that Viterbo/Eliashberg have proved that a manifold with a negatively curved metric cannot be embedded as a Lagrangian submanifold of a uniruled symplectic manifold. Actually their proof only seems to use the existence of a nondegenerate metric all of whose geodesics have Morse index zero. I wondered if that was known to be strictly weaker. - -REPLY [12 votes]: As mentioned by Rbega the question should be amended to ask whether it's true that a closed manifold $M$ without conjugate points admits a metric of non-positive (rather than negative) curvature (otherwise a torus is an obvious counterexample). -In that form this is a well-known open problem. The exponential map at any point is a universal covering of $M$ and the geodesics in $\tilde M$ are unique. This does show that $M$ is aspherical but that is a long way from admitting a metric of nonpositive curvature. -There are some partial results suggesting that fundamental groups of manifolds without conjugate points share some properties of fundamental groups of nonpositively curved manifolds. -In particular, there is a result of Croke and Shroeder that if the metric is analytic then any abelian subgroup of $\pi_1(M)$ is embedded quasi-isometrically. By the following observation of Bruce Kleiner the analyticity condition can be removed: Croke and Schroeder show that even without assuming analyticity for any $\gamma\in\pi_1(M)$ its minimal displacement $d_\gamma$ satisfies $d_{\gamma^n}=nd_\gamma$ for any $n\ge1$. This then implies that $d_\gamma=\lim_{n\to\infty} d(\gamma^nx,x)/n$ for any $x\in\tilde M$. This in turn implies that the restriction of $d$ to an abelian subgroup $H \simeq \mathbb Z^n$ extends to a norm on $\mathbb R^n$. This implies that $H$ is quasi-isometrically embedded. -This result implies for example that nonflat nilmanifolds cannot admit metrics without conjugate points and more generally that every solvable subgroup of the fundamental group of a manifold without conjugate points is virtually abelian. -But it's unlikely that any such manifold admits a metric of non-positive curvature. It is more probable that its fundamental group must satisfy some weaker condition such as semi-hyperbolicity but even that is completely unclear. The natural bicombing on $\tilde M$ given by geodesics need not satisfy the fellow traveler property (at least there is no clear reason where it should come from). -So it might be worth trying to look for counterexamples and the first place I would look is among groups that are semi-hyperbolic but not $CAT(0)$. Specifically, any $CAT(0)$ group has the property that centralizers of non-torsion elements virtually split. This need not hold in a semi-hyperbolic group with the simplest example given by any nontrivial circle bundle over closed surfaces of genus $>1$. To be even more specific one can take the unit tangent bundle $T^1(S_g)$ to a hyperbolic surface. Note however that it's known that a closed homogenous manifold without conjugate points is flat so if there is a metric without conjugate points on $T^1(S_g)$ it can not be homogeneous. -Edit: Actually, this last remark is irrelevant as $T^1(S_g)$ can not admit any homogeneous metrics at all.<|endoftext|> -TITLE: Spectrum of an algebra object and Reconstruction of Schemes -QUESTION [8 upvotes]: In "Au-dessous de $\text{Spec}(\mathbb{Z})$", Toen and Vaquié define schemes relative to a complete, cocomplete symmetric monoidal category $C$ using a functorial approach. -In the introduction the authors mention that there should be a description of the underlying topological space $|\text{Spec}(A)|$ of an affine scheme ($A$ algebra object in $C$; always commutative) in terms of ideals, at least if $C$ satisfies some reasonable properties (which one?). But they do not elaborate this. The definition of $|X|$ in the paper can be found in the end of section 2.4 and is rather sketchy and indirect: The category of (functorial) Zariski opens in $X$ is equivalent to the category of open subsets of a topological space $|X|$ by some abstract result. -Question 1. Is there any more concrete description of $|X|$, or at least of $|\text{Spec}(A)|$? -If $C$ was also assumed to be abelian, I would define $|\text{Spec}(A)|$ just as follows: First, an ideal of $A$ is the kernel of a morphism of algebras $A \to B$. Equivalently, it is a subobject $I \subseteq A$ such that $I \otimes A \to A \otimes A \to A$ factors through $I$. It is clear how to define ideal sum and (finite) ideal intersection. If $I,J$ are ideals of $A$, then define the ideal product $I*J$ to be the kernel of $I \to I \otimes A/J$. A prime ideal is a proper ideal $\mathfrak{p}$ of $A$ such that for all ideals $I,J$ of $A$, we have $I * J \subseteq \mathfrak{p} \Rightarrow I \subseteq \mathfrak{p} \vee J \subseteq \mathfrak{p}$. For an ideal $I$, let $V(I)$ be the set of prime ideals $\mathfrak{p}$ satisfying $I \subseteq \mathfrak{p}$. Then as usual we get a topological space $|\text{Spec}(A)|$, the spectrum of $A$. -Question 2. Has this definition already been studied somewhere? -One way to "test" the above definition of the spectrum is to test if $\text{Spec}(\mathcal{O}_X)$ turns out to be $X$; here $X$ is a (nice) scheme and $\mathcal{O}_X$ is our algebra object in $C=\text{Qcoh}(X)$. Now it is not hard to check that we have an injective map $X \to \text{Spec}(\mathcal{O}_X)$ sending $x$ to the vanishing ideal of $\overline{\{x\}}$ and that for noetherian schemes $X$ (actually I only need that $\text{rad}(\mathcal{O}_X)^n=0$ for some $n$), this is an isomorphism. Again I don't know if this is well-known at all. I also wonder what happens if $X$ is more general, say quasi-compact and quasi-separated. -But back to the general setting relative to $C$: -Question 3. If we use the above definition of the spectrum of $A$, how can we define the structure sheaf? - -REPLY [9 votes]: Florian Marty studied this question in his thesis. The relevant chapter is available as arXiv:0712.3676 (otherwise, the thesis is available here). He describes the space $|\mathrm{Spec}(A)|$ as the set of prime ideals endowed with the Zariski topology. He also proves that a basis of this topology is given by the subspaces $|\mathrm{Spec}(A[f^{-1}]|$, from which one deduces easily the description of the structural sheaf of $\mathrm{Spec}(A)$.<|endoftext|> -TITLE: Approximate primitive roots mod p -QUESTION [10 upvotes]: Artin conjectured that if $a$ is an integer which is not a square and not $-1$ then $a$ is a primitive root for infinitely many primes. This conjecture has not been resolved, but partial results are known: Heath-Brown showed that there are at most two prime numbers $a$ for which the conjecture fails. -I'd like to know if a different kind of partial result is known. Let $I(p)$ denote the index of the subgroup of $(\mathbf{Z}/p\mathbf{Z})^{\times}$ generated by 2. Thus $I(p)=1$ if and only if 2 is a primitive root mod $p$. Can one show that there is an infinite sequence of primes in which $I$ remains bounded? - -REPLY [8 votes]: A result of Erdos and Murty asserts that if $\epsilon(p)$ is any decreasing function tending to zero, then $I(p) \leq p^{1/2-\epsilon(p)}$ for almost all primes $p$ (i.e., all but $o(\pi(x))$ primes $p \leq x$). -Kurlberg and Pomerance (see Lemma 20 in the paper mentioned below) show that for a positive proportion of primes $p$, one has the stronger bound $I(p) \leq p^{0.323}$. This follows from a result of Baker and Harman on shifted primes with large prime factors. -The Erdos--Murty paper is #77 at -http://www.mast.queensu.ca/~murty/index2.html -and the Kurlberg--Pomerance paper is -http://www.math.dartmouth.edu/~carlp/PDF/par13.pdf -See also Theorem 23 of this paper (which is conditional on GRH).<|endoftext|> -TITLE: Discrete-compact duality for nonabelian groups -QUESTION [17 upvotes]: A standard property of Pontrjagin duality is that a locally compact Hausdorff abelian group is discrete iff its dual is compact (and vice versa). In what senses, if any, is this still true for nonabelian groups? -I can guess what this means for a compact (Hausdorff) group $G$: the category of unitary representations of $G$ should be discrete in the sense that every one-parameter family of unitary representations consists of isomorphic representations, or something like that. Is this true? Is the converse true? -I am less sure what this means for a discrete group $G$. What does it mean for the category of unitary representations to be compact? I suppose that $\text{Hom}(G, \text{U}(n))$ is a closed subspace of $\text{U}(n)^{G}$, hence compact, hence so is the appropriate quotient space of it... - -REPLY [3 votes]: Just an example, from which I learned a lot. Consider the free group $F$ on two generators. It makes a lot of sense to think of $n \mapsto hom(F,U(n))$ as some sort of dual. It comes with a natural conjugation action of $U(n)$ and natural operations of $\oplus$ and $\otimes$. -However, if one considers the bi-dual, which in this context would be the set of natural transformation $F^{xx}$ from the functor $n \mapsto hom(F,U(n))$ to $n \mapsto U(n)$, compatible with conjugation, $\oplus$ and $\otimes$, then this turns out to be too big and not discrete. First of all, $F^{xx}$ is a polish group and there is a natural homomorphism $F \to F^{xx}$. The whole construction goes under the name Chu duality and works just as in the case of Pontrjagin duality or Tannaka-Krein duality. But $F \to F^{xx}$ is not a homeomorphism. -This follows from the fact that for fixed $n \in \mathbb N$ and a fixed neighborhood $V$ of $1_n \in U(n)$, there exists $w \in F \setminus \lbrace e\rbrace$, such that $\phi(w) \in V$, for all homomorphisms $\phi \colon F \to U(n)$. The same problem appears for every finitely generated group, which is not virtually abelian. -The problem can be cured completely if one takes the infinite-dimensional representations into account. Then, the appropriate bi-dual is equal to $F$ (and the same holds for any locally compact group). An important step in the proof of this assertion is the Gelfand-Raikov theorem.<|endoftext|> -TITLE: Hyperbolic Coxeter polytopes and Del-Pezzo surfaces -QUESTION [23 upvotes]: Added. In the following link there is a proof of the observation made in this question: http://dl.dropbox.com/u/5546138/DelpezzoCoxeter.pdf - -I would like to find a reference for a beautiful construction that associates to Del-Pezzo surfaces hyperbolic Coxeter polytopes of finite volume and ask some related questions. -Recall that a hyperbolic Coxeter polytope is a domain in $\mathbb H^n$ bounded by a collection of geodesic hyperplanes, such that each intersecting couple of hyperplanes intersect under angle $\frac{\pi}{n}$ ($n=2,3,...,+\infty$). Del Pezzo surface is a projective surface obtained from $\mathbb CP^2$ by blowing up (generically) at most $8$ points. -Now, the construction Del Pezzo $\to$ Coxeter polytope goes as follows. -Consider $H_2(X,\mathbb R)$, this is a space endowed with quadratic form of index $(1,n)$ (the intersection form), and there is a finite collection of vectors $v_i$ corresponding complex lines on $X$ with self-intersection $-1$. It is well known, for example that on a cubic surface in $\mathbb CP^3$ there are $27$ lines and this collection of lines has $E_6$ symmetry (if you consider it as a subset in $H_2(X,\mathbb Z)$). Now we just take the nef cone of $X$, or in simple terms the cone in $H_2(X,\mathbb R)$ of vectors that pair non-negatively with all vectors $v_i$. This cone cuts a polytope from the hyperbolic space corresponding to $H_2(X,\mathbb R)$, and it is easy to check that this polytope is Coxeter, with angles $\frac{\pi}{2}$ and $0$ (some points of this polytope are at infinity, but its volume is finite). Indeed, angles are $\frac{\pi}{2}$ and $0$ since $v_i^2=-1$, $v_i\cdot v_j=0 \;\mathrm{or}\; 1$. -Example. If we blow up $\mathbb CP^2$ in two points this construction produces a hyperbolic triangle with one angle $\frac{\pi}{2}$ and two angles $0$. -The connection between algebraic surfaces an hyperbolic geometry is very well-known, and exploited all the time but for some reason I was not able to find the reference to this undoubtedly classical fact (after some amount of googling). So, -Question 1. Is there a (nice) reference for this classical fact? -This question is motivated in particular by the following article http://maths.york.ac.uk/www/sites/default/files/Preprint_No2_10_0.pdf where the polytope corresponding to the cubic surface is used. The authors mention the relation of the polytope to 27 lines on the cubic, but don't say that the relation is in fact almost canonical. -Question 2. The group of symplectomorhpisms (diffeos) of each Del-Pezzo surface $X$ is acting on $H_2(X,\mathbb R)$, let us denote by $\Gamma$ its image in the isometries of corresponding hyperbolic space. What is the relation between $\Gamma$ and the group generated by reflections in the faces of the corresponding Coxeter polytope? -PS It one considers rational surfaces with semi-ample anti-canonical bundles, i.e. surfaces that can have only rational curves with self-intersection $-1$ and $-2$ one gets more examples of Coxeter polytopes; the faces of such polytopes intersect under angles $(\frac{\pi}{2}, \frac{\pi}{3}, \frac{\pi}{4}, 0)$. -Here is a reference on "Algebraic surfaces and hyperbolic geometry" by Burt Totaro (but I don't think that the answer to question one is contained there). - -REPLY [4 votes]: For $Q2$, in my paper joint with T.-J. Li http://arxiv.org/abs/1012.4146 there's a description by reflections of $\Gamma$, and Shevchishin proved the same conclusion in http://arxiv.org/abs/0904.0283 but his language is more Coxeter. We actually dealt with all rational surfaces.<|endoftext|> -TITLE: A name for "not quite saturated" graded modules -QUESTION [8 upvotes]: Let $M$ be a finitely generated graded module over a graded ring $R$. Let $\mathcal{F}$ be the corresponding coherent sheaf on $\operatorname{Proj} R$. There is a natural map of graded $R$-modules -$$\phi \colon M \to \Gamma^*(\mathcal{F}) := \bigoplus_{n} \Gamma(\operatorname{Proj} R, \mathcal{F}(n)).$$ -If I recall Ravi Vakil's notes correctly, $M$ is called saturated if $\phi$ is an isomorphism. - -Is there a term (perhaps semi-saturated, or some such) for modules $M$ such that $\phi$ is injective? - -This concept is appealing for several reasons. For one thing, it is easier to test "semi-saturatedness" than saturatedness; e.g., unless I am mistaken, $\phi$ is automatically injective if $M$ admits any positive-degree homogeneous nonzerodivisor. For another, at least if $R$ is a polynomial ring, $M = R/I$ is "semi-saturated" iff $I$ is a saturated ideal of $R$. (Note that the definition of "saturated ideal" is different from the definition given above for "saturated module", and I do not think the two are equivalent for ideals.) - -REPLY [3 votes]: As Hailong wrote, the injectivity means the vanishing of $H^0_m(M)$. But $\operatorname{depth}(m,M)$ is the least non-vanishing local cohomology, thus the map $M \to H^0_*(\tilde{M})$ is injective iff $M$ has depth $\ge$ 1. -If $M$ has finite projective dimension, e.g. if R is regular, then this happens iff $M$ has projective dimension at most depth $R-1$ by the Auslander-Buchsbaum formula.<|endoftext|> -TITLE: Probability distributions: The maximum of a pair of iid draws, where the minimum is an order statistic of other minimums? -QUESTION [6 upvotes]: General question: What is the distribution for the maximum of 2 independent draws from cdf F(x), when we know that the minimum of those same two draws is the kth order statistic of the minimum of n pairs of independent draws from F(x)? Less technically, what is the distribution of the maximum associated with the kth greatest (of n) minima? -A specific example: -Assume 8 independent draws from cdf F(x), which is defined over 0 to 1. Then, arbitrarily group the draws into 4 pairs. Compare the minimums of each pair. Label the maximum of these minimums as “a”. Label a’s pair (which is by definition > a) as “b”. Now, choose among the other three pairs arbitrarily, and label the two values in that pair as “c” and “d” (where c is the min of the pair and d is the max of the pair). -What are the distributions of b and d? -I know the distribution of a: F(a) = (1-(1-F(x))^2)^4 =Max of 4 draws of the Min of 2 draws of F(x). -I also know the distribution of c: F(c) = mixture of 1st , 2nd, and 3rd order statistics of 4 draws of Min of 2 draws of F(x). I get this by averaging the integrals (wrt x) for the pdfs that result from substituting (k=1, n=4), (k=2,n=4) and (k=3, n=4) into the following equation: -(n!/((k - 1)!(n - k)!))(F(x)^(k - 1))*((1 - F(x))^(n - k))*F'(x) -I don’t know how to define F(b) or F(d) -And help would be greatly appreciated. - -REPLY [3 votes]: This question has been answered by Bogdan Lataianu at this link: -https://stats.stackexchange.com/questions/13259/what-is-the-distribution-of-maximum-of-a-pair-of-iid-draws-where-the-minimum-is<|endoftext|> -TITLE: Lorentzian characterization of genus -QUESTION [6 upvotes]: Suppose we take the "even" indefinite lattice from page 50 in Serre A Course in Arithmetic (1973) -$$ U \; = \; - \left( \begin{array}{cc} - 0 & 1 \\\ - 1 & 0 -\end{array} - \right),$$ -called $H$ in pages 189-191 of Larry J. Gerstein Basic Quadratic Forms. -What I cannot find in any detail is a proof of this arithmetic statement in -SPLAG by Conway and Sloane, page 378 in the first edition(1988), anyway -chapter 15 section 7, that quadratic forms $f,g$ are in the same genus -if and only if $f \oplus H$ and $g \oplus H$ are integrally equivalent. Then -they say this follows from properties of the spinor genus, presumably -including Eichler's theorem that indefinite rank at least 3 means -spinor genus and class coincide. -Also, if f and g do not correspond to "even lattices," I'm not -entirely sure what is being claimed. Oh, I absolutely cannot assume $f,g$ are in any way "unimodular." Very popular, that unimodular. Matter of taste, though. I'm not sure it matters, but my $f,g$ are going to be positive, which is surely the difficult case here. -Everybody with whom I have discussed this regards this as either -obvious or, essentially, an axiom. I would very much like a reference -for this, plus an explanation of what is meant if $f,g$ correspond to -"odd" lattices. For example, it would be wonderful if somewhere this claim and the words Theorem or Proposition or Lemma happened in the same sentence. I think I am making progress on the other bits I -need, essentially ch. 26,27 in SPLAG, but this claim has me snowed, -or perhaps buffaloed, thrown, stumped. As far as books that I own, I do not see the claim being discussed in Jones, Watson, O'Meara, Serre, Cassels, Kitaoka, Ebeling, Gerstein. I stopped by the office of R. Borcherds and discussed related matters for a while, the relevant articles are 1985 The Leech Lattice and 1990 Lattices Like the Leech Lattice, but I don't see the SPLAG claim in an explicit manner. -EDIT... Sexy application: the Leech lattice and all the Niemeier lattices are in the same genus. Pointed out in an MO comment by Noam Elkies, who knows things. - -REPLY [7 votes]: A good reference for this assertion is Cassels's "Rational Quadratic Forms", though you have to dig a bit. Let me see if I can outline the proof. First, I think Conway and Sloane assume $f$ and $g$ are classical integral (i.e. correspond to even lattices). In my copy of SPLAG, at the end of subsection 2.1 of that chapter, they say "so in this book we call $f$ an integral form if and only if its matrix coefficients are integers (i.e. if and only if it is classically integral ...)". -Now suppose $f$ and $g$ are in the same genus. Then so are $f\oplus H$ and $g \oplus H$. Next, we want to show they're in the same spinor genus. This follows from the Corollary of Lemma 3.6 of Chapter 11 of Cassels: "If we show $U_p \subset \theta(\Lambda_p)$ for all $p$, then the genus of $\Lambda$ consists of a single spinor genus". Here $\Lambda = f \oplus U$, where I'm identifying the form and the lattice by a bit of abuse of notation. Since $\theta(\Lambda_p) \supset \theta(H_p)$ (see a few sentences below the corollary), and $\theta(H_p) \supset U_p$ by Lemmas 3.7 and 3.8, we've proved that the genus consists of a single spinor genus. -Finally, since the forms are indefinite of dimension at least $3$, the spinor genus consists of a single class. -To go back is the easier direction (I think): if $f \oplus U$ is equivalent to $g \oplus U$, then they are equivalent over $\mathbb{Z}_p$ for every $p$. Then an analogue of Witt cancellation will do the job (see Chapter 8 of Cassels).<|endoftext|> -TITLE: Affine manifolds -QUESTION [10 upvotes]: An affine manifold is a topological manifold which admits a system of charts such that the coordinate changes are (restrictions of) affine transformations. Let $M$ be a compact affine manifold. Let $G$ be the fundamental group of $M$ and $\tilde M$ be its universal cover. One can show that each $n$-dimensional affine manifold comes with a developing map $D\colon \tilde M \to \mathbb R^n$, and a homomorphism $\varphi \colon G \to {\rm Aff}(\mathbb R^n)$, such that $D$ is an immersion and equivariant with respect to $\varphi$. -An affine manifold is called complete if $D$ is a homeomorphism, in this case: $\varphi$ is injective, $G$ is a Bieberbach group, and $M$ is aspherical, i.e. $\tilde M$ is contractible. The non-complete case seems to be far more complicated. - -Question 1: Is there an easy example, where $D$ is not surjective? -Question 2: Is there an easy example, where $\varphi$ is not injective? -Question 3: Is there an easy example, where $M$ is not aspherical? - -EDIT: As André suggested, let's ask for examples for which $\varphi$ takes values in $SL(n,\mathbb R) \ltimes \mathbb R^n$ or even $SL(n,\mathbb Z) \ltimes \mathbb R^n$, seen as subgroups of ${\rm Aff}(\mathbb R^n)$. - -REPLY [9 votes]: There is a conjecture due to Markus which states that any compact affine manifold has parallel volume (i.e. the linear part of $\varphi$ lies in $\mathrm{SL}(n;\mathbb{R})$) if and only if it is complete. To the best of my knowledge, this conjecture is still open, which goes towards saying that there should be no easy examples to questions 1 and 3 for affine manifolds with parallel volume. -If the fundamental group $G$ of a compact affine manifold with parallel volume is nilpotent, the beautiful Affine manifolds with nilpotent holonomy by Fried, Goldman and Hirsch, Comm. Math. Helv. 56 (1981) proves that Markus' conjecture holds in this case and, thus, there are no examples to questions 1 and 3 with nilpotent fundamental group. The proof is a cunning mixture of representation theory and geometry, so I strongly recommend taking a look at it. The results in this paper also imply that the above examples (to questions 1 and 3) constructed by Andre Henriques cannot be adapted so that the resulting manifolds admit parallel volume (these are, nonetheless, very nice examples to the original question!).<|endoftext|> -TITLE: Singular curves in a 3-fold? -QUESTION [5 upvotes]: Assume that $X$ is a smooth 3-fold and let $C\subseteq X$ a curve with a unique singular point of multiplicity $2$. Does there exist a smooth surface $S$ inside $X$ which contain $C$ ? -Clearly if the multiplicity of $C$ was at least 3 then it would be very easy to find counter-examples. On the other hand, if the multiplicity is 2, it seems that at least infinitesimally it is possible to find such a surface. In case the answer is negative, is it true at least locally (e.g. in an analytic neighbourhood of p)? -Finally any smooth curve is locally complete intersection. What about the curve $C$ above? - -REPLY [4 votes]: Here is a quick proof that any complete local Cohan-Macaulay ring of dimension $1$ and multiplicity $2$ is a hypersurface, so the answer to your last question is always yes. -Call such ring $R$ with maximal ideal $m$. Since $e(R)=2$ and $R$ is CM, there is a regular element $x\in m$ such that $$length(R/xR) = e(R)=2$$ (see Bruns-Herzog, chapter 4). -EDIT: as Graham pointed out below, technically for such $x$ to exist one needs $R/m$ to be infinite. But one can enlarge the field without affecting the conclusion that $R$ is a hypersurface. -Now the left hand side is $length(m/xR) +1$, so $length(m/xR)=1$, so $m/xR$ is one-generated. It follows that $m$ is $2$-generated. Thus $R=A/I$, $A$ is a regular local ring of dimension $2$. But since $R$ is CM, $I$ must be of pure height one, and since $A$ is regular, $I$ is principal. -Note: in fact, the dimension one condition is not necessary, you can take a full length regular sequence and argue the same way (with just a little more work). So any CM singularity with multiplicity $2$ is a hypersurface. -EDIT: Once we know that the local ring at the singular point is a hypersurface, Jason Starr and Qing Liu's nice comments show that there must exist a smooth surface containing $C$, thus completely answer the question.<|endoftext|> -TITLE: Non-oscillatory behaviour in the subadditive ergodic theorem -QUESTION [10 upvotes]: I am currently reading an article in which the author goes to certain lengths which could be avoided if the following result were true: - - -Lemma (proposed): Let $T$ be an ergodic measure-preserving transformation of a probability space $(X,\mathcal{F},\mu)$, and let $(f_n)$ be a sequence of integrable functions from $X$ to $\mathbb{R}$ which satisfy the subadditivity relation $f_{n+m} \leq f_n \circ T^m + f_m$ a.e. for all integers $n,m \geq 1$. Suppose that $f_n(x) \to -\infty$ in the limit as $n \to \infty$ for $\mu$-a.e. $x \in X$. Then $\lim_{n \to \infty} \frac{1}{n}\int f_n d\mu <0$. - - -Via the subadditive ergodic theorem, this effectively states that if $f_n(x) \to -\infty$ almost everywhere then it must do so at an asymptotically linear rate. The supposed lemma would also be equivalent to the statement that if $\frac{1}{n} f_n(x) \to 0$ almost everywhere, then for almost every $x$ the sequence $(f_n(x))$ must return infinitely often to some neighbourhood of $0$ which is not a neighbourhood of $-\infty$. If the sequence $(f_n)$ is additive rather than just subadditive then this last formulation of the result follows from a well-known theorem of G. Atkinson, but the more general subadditive case is less clear. -If the lemma were true then several parts of the paper I am reading would be redundant, which makes me wonder whether it is in fact false. Yet it seems rather plausible. Does anyone know whether this result is true or not? - -REPLY [3 votes]: I believe the Lemma you propose is true, via a relatively straightforward adaptation of the proof given for the additive case in Giles Atkinson, Recurrence of co-cycles and random walks, J. Lond. Math. Soc. (2) 13 (1976), 486-488. (I assume this is the result you were referencing.) This may already be known and written somewhere -- I can't speak to that. In case it's not, here's a proof. -Proof of Lemma. By the subadditive ergodic theorem, there exists a measurable function $f\colon X\to \mathbb{R}\cup\{-\infty\}$ such that $\lim \frac 1n f_n(x) = f(x)$ for $\mu$-a.e. $x\in X$ and $\lim \frac 1n \int f_n\,d\mu = \int f\,d\mu$. So our task is to show that $\int f\,d\mu < 0$. -To this end, given $x\in X$, let $M_x = \{n \mid f_n(x) \geq -1 \}$, and observe that $M_x$ is finite $\mu$-a.e. Thus writing $A_n = \{x\in X \mid \# M_x < n \}$, we see that there exists $N$ such that $\mu(A_N) > \frac 12$. -Furthermore, given $x\in X$, write $L_x = \{ k \mid T^k(x) \in A_N \}$. Since $\mu(A_N) > \frac 12$, we see that $\mu$-a.e. $x$ has -$$ -(*)\qquad\qquad\qquad\# L_x \cap [1,n] \geq \frac n2\qquad\qquad\qquad\qquad\qquad\quad -$$ -for all sufficiently large $n$. We fix such an $x$ and show that $f(x) < 0$. -Let $k_0$ be the smallest element of $L_x$. We define $k_i\in L_x$ recursively with the property that -$$ -f_{k_i}(x) < f_{k_0}(x)-i, -$$ -as follows. Let $J_i$ be the $N$ smallest elements of $L_x \cap (k_i,\infty)$. Because $T^{k_i}(x)\in A_N$, there exists $k_{i+1}\in J_i$ such that $k_{i+1} - k_i \notin M_{T^{k_i}(x)}$. In particular, we have -$$ -f_{k_{i+1} - k_i}(T^{k_i}(x)) < -1. -$$ -Now subadditivity gives -$$ -f_{k_{i+1}}(x) \leq f_{k_i}(x) + f_{k_{i+1} - k_i}(T^{k_i}(x)) < f_{k_0(x)} - i-1. -$$ -The next observation to make is that by $(*)$, we have $k_i \leq 2Ni$ for all sufficiently large $i$. Thus we have -$$ -\frac 1{k_i} f_{k_i}(x) \leq \frac 1{2Ni}(f_{k_0}(x) - i), -$$ -and sending $i\to\infty$, we obtain $f(x) \leq -\frac 1{2N}$. Since this holds for $\mu$-a.e. $x$, we have $\int f\,d\mu \leq -\frac 1{2N} < 0$, which completes the proof. - -Note. It's quite important in this proof that we're dealing with negative values of $f_n$; the proof would fail if we tried to show that $f_n(x)\to+\infty$ for $\mu$-a.e. $x$ implies that $\int f\,d\mu > 0$. I'm not sure if the result is true in this case.<|endoftext|> -TITLE: Ask some matrix eigenvalue inequalities. -QUESTION [5 upvotes]: Let $ \begin{bmatrix} -A& B \\\\ B^* &C -\end{bmatrix}$ be positive semidefinite, $A,C$ are of size $n\times n$. -Are the following plausible inequalities true? I have seen a lot of similar results, but for the following inequalities, I cannot locate them in the literature or find that they have been pointed out to be false. -1 $$\quad \sum\limits_{i=1}^k\lambda_i\begin{bmatrix} -A& B \\\\ B^* &C -\end{bmatrix}\le \sum\limits_{i=1}^k\left(\lambda_i(A)+\lambda_i(C)\right)\quad, $$ -where $1\le k\le n$. -2 Modified -$$\quad \prod\limits_{i=1}^{2k}\lambda_i\begin{bmatrix} -A& B \\\\ B^* &C -\end{bmatrix}\le \prod\limits_{i=1}^k \lambda_i(A)\lambda_i(C) \quad, $$ -where $1\le k\le n$. -3 $$ 2\lambda_i^{1/2}(B^*B)\le \lambda_i(A+C),$$ where $1\le k\le n$. -Here, $\lambda_i(\cdot)$ means the $i$th largest eigenvalue of $\cdot\quad$. Any references or counterexamples are appreciated. - -REPLY [17 votes]: Item 1 is true. This is part of Problem 22 (b) in Section 3.5 of Horn and Johnson [HJ94], which states that for Ky Fan norm ||⋅|| (and in fact for any unitarily invariant norm) and a positive semidefinite block matrix $\begin{pmatrix}A & B \\ B^* & C\end{pmatrix}$, it holds that $\left\|\begin{pmatrix}A & B \\ B^* & C\end{pmatrix}\right\| \le \left\|\begin{pmatrix}A & 0 \\ 0 & 0\end{pmatrix}\right\| + \left\|\begin{pmatrix}0 & 0 \\ 0 & C\end{pmatrix}\right\|$. -([Aud06] contains a proof of a slight generalization of this inequality among other results.) -Item 2 in the original question is false by considering the case where A=C=I/2, B=0, and k=1, where I is the identity matrix. (Did you mean to square the left-hand side?) -Modified item 2 is false; see Willie Wong’s comment on this answer. -Item 3 is false. A simple counterexample is n=2, i=1, -$A=\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}$, -$B=\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}$, -$C=\begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix}$. -Then $2\sqrt{\lambda_1(B^*B)}=2$ but λ1(A+C)=1. -References -[Aud06] Koenraad M. R. Audenaert. A norm compression inequality for block partitioned positive semidefinite matrices. Linear Algebra and its Applications, 413(1):155–176, Feb. 2006. http://dx.doi.org/10.1016/j.laa.2005.08.017 -[HJ94] Roger A. Horn, Charles R. Johnson. Topics in Matrix Analysis. Cambridge University Press, 1994.<|endoftext|> -TITLE: Endpoint Strichartz Estimates for the Schrödinger Equation -QUESTION [6 upvotes]: The non-endpoint Strichartz estimates for the (linear) Schrödinger equation: -$$ -\|e^{i t \Delta/2} u_0 \|_{L^q_t L^r_x(\mathbb{R}\times \mathbb{R}^d)} \lesssim \|u_0\|_{L^2_x(\mathbb{R}^d)} -$$ -$$ -2 \leq q,r \leq \infty,\;\frac{2}{q}+\frac{d}{r} = \frac{d}{2},\; (q,r,d) \neq (2,\infty,2),\; q\neq 2 -$$ - are easily obtained using (mainly) the Hardy-Littlewood-Sobolev inequality, the endpoint case $q = 2$ is however much harder (see Keel-Tao for example.) -Playing around with the Fourier transform one sees that estimates for the restriction operator sometimes give estimates similar to Strichartz's. For example, the Tomas-Stein restriction theorem for the paraboloid gives: -$$ -\|e^{i t \Delta/2} u_0\|_{L^{2(d+2)/d}_t L^{2(d+2)/d}_x} \lesssim \|u_0\|_{L^2_x}, -$$ -which, interpolating with the easy bound -$$ -\|e^{i t \Delta/2} u_0\|_{L^{\infty}_t L^{2}_x} \lesssim \|u_0\|_{L^2_x}, -$$ -gives precisely Strichartz's inequality but restricted to the range -$$ -2 \leq r \leq 2\frac{d+2}{d} \leq q \leq \infty. -$$ -As far as I know, the Tomas-Stein theorem (for the whole paraboloid) gives the restriction estimate $R_S^*(q'\to p')$ for $q' = \bigl(\frac{dp'}{d+2}\bigr)'$ (this $q$ is different from the one above), so I'm guessing that this cannot be strengthened (?). -So my question is: what's the intuition of what goes wrong when trying to prove Strichartz's estimates all the way down to the endpoints using only Fourier restriction theory? - -REPLY [3 votes]: From my less-than-expert (where's Terry when you need him?) point of view, a possible reason seems to be the following (I wouldn't call it something going wrong or even a difficulty): -The statement of restriction estimates only give you estimates where the left hand side is an isotropic Lebesgue space, in the sense that you get an estimate $L^q_tL^r_x$ with $q = r$. This naturally excludes the end-point, which requires $r > q$. -Why is this? The reason is that the restriction theorems only care about the local geometry of the hypersurface, and not its global geometry. (For example, the versions given in Stein's Harmonic Analysis requires either the hypersurface to have non-vanishing Gaussian curvature for a weaker version, or that the hypersurface to be finite type for a slightly stronger version. Both of these conditions are assumptions on the geometry of the hypersurface locally as a graph over a tangent plane.) Now, on each local piece, you do have something more similar to the classical dispersive estimates with $r > q$, which is derived using the method of oscillatory integrals (see, for example, Chapter IX of Stein's book; the dispersive estimate (15) [which has, morally speaking $q = r = \infty$ but with a weight "in $t$", so actually implies something with $q < \infty$] is used to prove Theorem 1, which is then used to derive the restriction theorem). But once you try to piece together the various "local" estimates to get an estimate on the whole function, you have no guarantee of what the "normal direction" is over the entire surface. (The normal direction, in the case of the application to PDEs, is the direction of the Fourier conjugate of the "time" variable.) So in the context of the restriction theorem, it is most natural to write the theorem using the $q = r$ version, since in the more general context of restriction theorems, there is no guarantee that you would have a globally preferred direction $t$. -(Note that Keel-Tao's contribution is not in picking out that time direction: that Strichartz estimates can be obtained from interpolation of a dispersive inequality and energy conservation is well known, and quite a bit of the non-endpoint cases are already available as intermediate consequences of the proof of restriction theorems. The main contribution is a refined interpolation method to pick out the end-point exponents.)<|endoftext|> -TITLE: Why is the Chebyshev function relevant to the Prime Number Theorem -QUESTION [20 upvotes]: Why is the Chebyshev function -$\theta(x) = \sum_{p\le x}\log p$ -useful in the proof of the prime number theorem. Does anyone have a conceptual argument to motivate why looking at $\sum_{p\le x} \log p$ is relevant and say something random like $\sum_{p\le x}\log\log p$ is not useful or for that matter any other random function $f$ and $\sum_{p\le x} f(p)$ is not relevant. - -REPLY [24 votes]: There are several ideas here, some mentioned in the other answers: -One: When Gauss was a boy (by the dates found on his notes he was approximately 16) he noticed that the primes appear with density $ \frac{1}{\log x}$ around $x$. Then, instead of counting primes and looking at the function $\pi (x)$, lets weight by the natural density and look at $\sum_{p\leq x} \log p$. Since we are weighting by what we think is the density, we expect it to be asymptotic to be $x$. -Two: Differentiation of Dirichlet series. If $$ A(s)=\sum_{n=1}^{\infty} a_{n} n^{-s} $$ then $$ (A(s))'=-\sum_{n=1}^{\infty} a_{n} log(n) n^{-s}$$ -The $\log$ term appears naturally in the differentiation of Dirichlet series. Taking the convolution of $\log n$ with the Mobius function (that is multiplying by $\frac{1}{\zeta(s)}$) then gives the $\Lambda(n)$ mentioned above. The $\mu$ function is really the special thing here, not the logarithm. -Expanding on this, there are other weightings besides $\log p$ which arise naturally from taking derivatives. Instead we can look at $\zeta^{''}(s)=\sum_{n=1}^\infty (\log n)^2 n^{-s}$, and then multiply by $\frac{1}{\zeta(s)}$ as before. This leads us to examine the sum $$\sum_{n\leq x} (\mu*\log^2 )(n)$$ (The $*$ is Dirichlet convolution) By looking at the above sum, Selberg was able to prove his famous identity which was at the center of the first elementary proof of the prime number theorem: -$$\sum_{p \leq x} log^2 p +\sum_{pq\leq x}(\log p)(\log q) =2x\log x +O(x).$$ -Three: The primes are intimately connected to the zeros of $\zeta(s)$, and contour integrals of $\frac{1}{\zeta(s)}$. (Notice it was featured everywhere here so far) We can actually prove that -$$\sum_{p^k \leq x} \log p= x - \sum_{\rho :\ \zeta(\rho)=0} \frac{x^{\rho}}{\rho} +\frac{\zeta'(0)}{\zeta(0)} $$ -Notice that the above is an equality, which is remarkable since the left hand side is a step function. (Somehow, at prime powers all of the zeros of zeta conspire and make the function jump.)<|endoftext|> -TITLE: Error term of the Prime Number Theorem and the Riemann Hypothesis -QUESTION [6 upvotes]: I have read that the Riemann Hypothesis is equivalent to -$\pi(x)=\text{Li}(x)+O(\sqrt{x}\log x)$ -Is there an analogous statement saying the Riemann Hypothesis is equivalent to -$\pi(x)=\frac{x}{\log x}+ O(f(x))\quad$ for some $f$ -or -$\pi(x)=\frac{x}{\log x}+ g(x) + O(h(x))\quad$ for some elementary function $g$ and $h$ -I'm guessing that $f$ could not possibly be $\sqrt{x}\log x$ because I plotted -$\frac{\text{Li}(x)-x/\log(x)}{\sqrt x\log x}$ and it looked like it grew without bound as $x$ goes to infinity. - -REPLY [20 votes]: It is not hard to show that -$$\mathrm{Li}(x) = \frac{x}{\log x} \sum_{k=0}^{m - 1}{\frac{k!}{(\log x)^k}} + O\left(\frac{x}{(\log x)^{m + 1}}\right)$$ -for any $m \geq 0$ (just use the definition of $\mathrm{Li}(x)$ and repeated integration by parts). Thus -$$\pi(x) = \frac{x}{\log x} \sum_{k=0}^{m - 1}{\frac{k!}{(\log x)^k}} + O\left(\frac{x}{(\log x)^{m + 1}}\right).$$ -It is not possible to improve on this (this is true unconditionally; you don't even need the Riemann hypothesis). So $\mathrm{Li}(x)$ really is the "better" approximation to $\pi(x)$ compared to $x/\log x$.<|endoftext|> -TITLE: Random manifolds -QUESTION [55 upvotes]: In the world of real algebraic geometry there are natural probabilistic questions one can ask: you can make sense of a random hypersurface of degree d in some projective space and ask about its expected topology where "expected" makes sense because there are sensible measures on the space of hypersurfaces. See Welschinger-Gayet http://arxiv.org/abs/1107.2288 and http://arxiv.org/abs/1005.3228 for recent progress on such questions (e.g. what is the expected Betti number of a random real hypersurface of degree d?). -In geometry more generally you might want to make statements like "a general manifold is aspherical" or "a general manifold has positive simplicial volume". It seems difficult to construct sensible measures for which these questions have answers: to talk about probability you need some way of producing manifolds (and then distinguishing them) in a random way. -However, Cheeger proved that for fixed L there is only a finite set $D_L$ of diffeomorphism classes of manifold admitting a Riemannian metric with sectional curvatures bounded above in norm by L, volume bounded below by 1/L and the diameter bounded above by L (see the first theorem in Peters "Cheeger's finiteness theorem for diffeomorphism classes of Riemannian manifolds" http://www.reference-global.com/doi/abs/10.1515/crll.1984.349.77). This means that you can ask questions like "what is the average total Betti number of a manifold in $D_L$" and "how does this increase with L?" (are there exponential upper bounds?), or one can try to make sense of the limit as $L\rightarrow\infty$ of the proportion of manifolds in $D_L$ with zero simplicial volume. -Are there any known concrete answers to these questions, or other formulations of the questions which lead to answers? - -REPLY [3 votes]: Here are three references that discuss the construction and related properties of "random Riemannian manifolds." The construction is related to random graphs and is "an approach to random Riemann surfaces based on associating a dense set of them - Belyi surfaces - with random cubic graphs." -Random Construction of Riemann Surfaces -Robert Brooks, Eran Makover -http://arxiv.org/abs/math/0106251 -Eran Makover has coauthored two other related papers: - -The length of closed geodesics on random Riemann Surfaces -Eran Makover, Jeffrey McGowan -On the Genus of a Random Riemann Surface -Alexander Gamburd, Eran Makover<|endoftext|> -TITLE: Characterization of locally free modules via exterior powers -QUESTION [7 upvotes]: Let $X$ be a scheme and $\mathcal{F}$ be quasi-coherent module on $X$. It is clear that if $\mathcal{F}$ is locally free of rank $n$, then $\det(\mathcal{F}) := \wedge^n \mathcal{F}$ is invertible, i.e. locally free of rank $1$. But what about the converse? -Question. Assume $\wedge^n \mathcal{F}$ is invertible. Does it follow that $\mathcal{F}$ is locally free (necessarily of finite rank $n$)? -Of course we may assume that $X$ is affine. Then it is enough to prove that $\mathcal{F}$ is flat and of finite presentation, but I don't know how to prove either one. Also it seems to be hard to find counterexamples. - -REPLY [8 votes]: I think that $\mathcal F$ is indeed locally free of rank $n$: -Pick a point $x\in X$. It will be enough to show that there is a neighbourhood -of $x$ on which $\mathcal F$ is free of rank $n$. Now, the exterior power -commutes with pullbacks (aka scalar extensions) so that in particular the fibre -(in the sense of pullback to $\mathrm{Spec}k(x)$) of $\Lambda^m\mathcal F$ at -$x$ equals $\Lambda^m\mathcal{F}_x$. This shows that $\mathcal{F}_x$ is an -$n$-dimensional vector space. After possibly shrinking $X$ we may assume that -there is a an $\mathcal{O}_X$-map $f\colon \mathcal{O}_X^n\to \mathcal F$ which induces -an isomorphism on fibres at $x$. Thus $\Lambda^nf$ is a map between locally free -modules (of rank $1$) that gives an isomorphism on fibres at $x$ and hence is an -isomorphism in a neighbourhood of $x$ so that we may assume that it is a global -isomorphism. The wedge product induces pairings -$\mathcal{F}\times\Lambda^{n-1}\mathcal{F}\to \Lambda^{n}\mathcal{F}$ and -$\mathcal{O}_X^n\times\Lambda^{n-1}\mathcal{O}_X^n\to -\Lambda^{n}\mathcal{O}_X^n$, the latter being a perfect pairing. Composing the -second with $\Lambda^nf$ gives a pairing -$\mathcal{O}_X^n\times\Lambda^{n-1}\mathcal{O}_X^n\to \Lambda^{n}\mathcal{F}$. -As $\Lambda^\ast f$ is multiplicative we get that the composite -$$\mathcal{O}_X^n\xrightarrow{f}\mathcal{F}\to -\mathrm{Hom}(\Lambda^{n-1}F,\Lambda^{n}\mathcal{F})\xrightarrow{\Lambda^{n-1}f^*}\mathrm{Hom}(\Lambda^{n-1}\mathcal{O}_X^n,\Lambda^{n}\mathcal{F})$$ -equals the map induced by the pairing for $\mathcal{O}_X^n$. This is an -isomorphism (as $\mathcal{O}_X^n$ is free of rank $n$ and $\Lambda^nf$ is an -isomorphism) so we get that $f$ is split injective and we may write -$\mathcal{F}$ as `\mathcal{O}_X^n\bigoplus \mathcal G$ for some quasi-coherent sheaf -$\mathcal{G}$. Now, $\Lambda^n(\mathcal{O}_X^n\bigoplus \mathcal{G})$ splits up -as -$$ -\bigoplus_{i+j=n}\Lambda^i\mathcal{O}_X^n\bigotimes \Lambda^j\mathcal{G} -$$ -and $\Lambda^nf$ is the inclusion into the $j=0$ factor. As that inclusion is an -isomorphism, the other factors are zero but $\Lambda^{n-1}\mathcal{O}_X^n\bigotimes -\Lambda^1\mathcal{G}$ has $\mathcal{G}$ as a direct factor and hence $\mathcal{G}=0$.<|endoftext|> -TITLE: Can the Law of the Iterated Logarithm be strengthened? -QUESTION [12 upvotes]: http://en.wikipedia.org/wiki/Law_of_the_iterated_logarithm - -.$\quad$1. Can the independence assumption be weakened, similar to this? - -.$\quad$2. Can the identically distributed assumption be dropped/weakened, in the latter case similar to this? - -.$\quad$3. Can the result be fine-tuned, presumably to something of the form - -$\displaystyle\limsup_{n\to \infty} \frac{\frac{S_n}{\sqrt{2\cdot n\cdot \operatorname{log}(\operatorname{log}(n))}}-1}{f(n)} \; \; $ some_relation_symbol $\;$ some_constant $\qquad$ almost surely $\qquad \; \; $ ? - -REPLY [6 votes]: For your third question, see a paper of Erdös, where he proves an even more precise result (at least, in the special case $S_n=\sum_{i=1}^nY_i$ where $Y_i$ (independent) are $\pm 1$ with probability $\frac 12$), namely that for $\delta>0$, the following holds with probability one: -$$S_n>\left(\frac{2n}{\log\log n}\right)^{1/2}(\log\log n+\frac 34\log\log\log n+\frac 12\log\log\log\log n+\cdots+(\frac 12-\delta)\log^{(k)}n)\qquad\text{for infinitely many $n$}$$ -and -$$S_n>\left(\frac{2n}{\log\log n}\right)^{1/2}(\log\log n+\frac 34\log\log\log n+\frac 12\log\log\log\log n+\cdots+(\frac 12+\delta)\log^{(k)}n)\qquad\text{for only finitely many $n$}$$ -In particular, this implies that: -$$\limsup_{n\to \infty} \frac{\frac{S_n}{\sqrt{2n\log\log n}}-1}{\frac{\log\log\log n}{\log\log n}}=\frac 34$$<|endoftext|> -TITLE: What do we actually know about logarithmic energy ? -QUESTION [13 upvotes]: In potential theory, the $\textit{logarithmic energy}$ of a Radon measure $\mu$ acting on $\mathbb{C}$ is defined by -$$I(\mu)=\iint\log\frac{1}{|x-y|}\mu(dx)\mu(dy).$$ Of course it is not well defined for all measures and may takes values in $[-\infty,+\infty]$. To avoid this annoying fact, one typically restrict to measures which integrate the logarithm around infinity, that is which satisfy the condition (C) -$$ \int \log(1+|x|)\mu(dx)<+\infty,$$ -so that $I(\mu)>-\infty$ thanks to $|x-y|\leq (1+|x|)(1+|y|) $. A fondamental fact is that if $\mu,\nu$ both satisfy (C), have finite logarithmic energy and $\mu(\mathbb{C})=\nu(\mathbb{C})$, then $I(\mu-\nu)\geq0$ and equality holds iff $\mu=\nu$ (we extend naturally the definition of $I$ to signed measures). I understand the condition (C) to be convenient, but maybe not sharp (one can imagine $\mu$ which does not satisfies (C) but with finite logarithmic energy). This leads to my first question : - -What can we says about $I(\mu-\nu)$ when (at least one of) the measures do not satisfy (C) ? - -This question is moreover motivated by the appearance of the logarithmic energies in random matrix theory (in large deviations rate functions) and in free probability (reinterpreted up to a sign as a non-commutative entropy by Voiculescu). In this setting, $I(\mu-\nu)$ is a natural candidate for a relative free entropy, and questions of geometric nature are bothering me : Let $A_{c}$ be the set of signed measures $\mu$ with finite logarithmic energy acting on $\mathbb{C}$ with total mass $\mu(\mathbb{C})=c$ such that if $\mu$ has Jordan decomposition $\mu^+-\mu^-$, then both $\mu^+$ and $\mu^-$ satisfy (C). Then the previous fact yields that $A_0$ is a pre-Hilbert space with scalar product -$$ I(\mu,\nu)=\iint\log\frac{1}{|x-y|}\mu(dx)\nu(dy).$$ Note that $A_0$ is not complete since it is clearly not closed. Moreover, it is not hard to check that $A_0$ acts by translations on $A_1$ which inherits of a structure of affine space, leading to a metric on probability measures with finite logarithmic energy satisfying (C). It is now time for my second question : - -What relations may exist between this metric and metrics compatible with the weak topology ? (e.g Prohorov's ? Levy / Bounded Lipshitz ? Wasserstein's ? ...) Or total variation norm ? - -REPLY [2 votes]: Concerning the first question: We have $I(\mu-\nu)>0$ whenever $I(\mu-\nu)$ is defined, finite or not, and $\mu$,$\nu$ are different signed Radon measures with equal total masses (or rather charges). This, with any finite dimensional Hilbert space in place of the plane, is Example 3.3 in -http://www.ams.org/journals/tran/1997-349-08/S0002-9947-97-01966-1/home.html , -where the assumption $\sigma\neq0$ is missing.<|endoftext|> -TITLE: Image of the trace operator -QUESTION [21 upvotes]: It is well-known that we have the trace theorem for Sobolev spaces. Let $\Omega$ be an open domain with smooth boundary, we know that the map -$$ T: C^1(\bar\Omega) \to C^1(\partial\Omega) \subset L^p(\partial\Omega) $$ -by $Tu(y) = u(y)$ for $y\in\partial\Omega)$ can be extended continuously to a linear map on Sobolev spaces for $p > 1$ -$$ T: W^{1,p}(\Omega) \to L^p(\partial\Omega)$$ -We also know that this map is not surjective, since the Trace Theorem (Sobolev embedding) tells us that when dropping 1 dimension, we have that the image of $T$ actually lives ([Edited May 10 2012] caveat: see my comment on the answer below) in a fractional Sobolev space, -$$ T: W^{1,p}(\Omega) \to W^{1-1/p, p}(\partial\Omega) \Subset L^p(\partial\Omega) $$ -On the other hand, we know that this map $T$ has dense image in $L^p$, just using the density of $C^1$. -Question: Is there a known characterisation of precisely what the image set of $T$ is? A slightly weaker question is: consider‡ $w \in W^{s,q}(\partial\Omega)$ for $1 - 1/p \leq s \leq 1$ and $q \geq p$, does there necessarily exist some function $u\in W^{1,p}(\Omega)$ such that $Tu = w$? -For example, if we assume that $w$ is Lipschitz on $\partial\Omega$, then we can extend (almost trivially) $w$ to a Lipschitz function $C^{0,1}(\bar\Omega)\subset W^{1,p}$ for every $p$. So the case $s = 1, q = \infty$ has a positive answer. Whereas the Sobolev embedding theorem mentioned above tells us that it is impossible to go below $s < 1-1/p$ and $q < p$. -‡ The lower cut-off here is clearly not sharp. The trace theorem combined with Sobolev embedding can be used to trade differentiability with integrability. Out of sheer laziness I will not include the numerology here. One should interpret the conditions on $s,q$ to be that $s \leq 1$, $q \geq p$ plus the requirement that $(s,q)$ is at least as good as what can be guaranteed by Sobolev embedding and the trace theorem. - -REPLY [14 votes]: The image you are looking for equals the Besov space $B_{p,p}^{1-\frac1p} (\partial \Omega )$. See -H. Triebel. Interpolation theory, function spaces, differential operators. Leipzig, 1995 -(in fact, I used the earlier Russian edition, Moscow, 1980).<|endoftext|> -TITLE: Structure of $E(Q_p)$ for elliptic curves with anomalous reduction modulo $p$ -QUESTION [15 upvotes]: For simplicity, take $p\ge7$ a prime and $E/\mathbb{Q}$ an elliptic curve with good anomalous reduction at $p$, i.e., $|E(\mathbb{F}_p)|=p$. There is a standard exact sequence for the group of points over $\mathbb{Q}_p$, -$$ - 0 \to \hat{E}(p\mathbb{Z}_p) \to E(\mathbb{Q}_p) \to E(\mathbb{F}_p) \to 0. -$$ -The assumption that $p$ is anomalous implies that $E(\mathbb{F}_p)=\mathbb{Z}/p\mathbb{Z}$, while the formal group $\hat{E}(p\mathbb{Z}_p)$ is isomorphic to the formal group of the additive group, so we obtain an exact sequence -$$ - 0 \to p\mathbb{Z}_p^+ \to E(\mathbb{Q}_p) \to \mathbb{Z}/p\mathbb{Z} \to 0. - \qquad(*) -$$ -In work I'm doing on elliptic pseudoprimes, the question of whether the sequence $(*)$ splits has become relevant. Questions: - -Are there places in the literature -where the split-versus-nonsplit -dichotomy for anomalous primes comes -up, e.g., in the theory of $p$-adic -modular forms? -Is there a name for -this dichotomy in the literature? -Aside from the obvious observation -that the sequence $(*)$ splits if and only -if $E(\mathbb{Q}_p)$ has a -$p$-torsion point, are there other -natural necessary or sufficient -conditions for splitting? - -REPLY [4 votes]: If you work modulo $p^2$, the sequence $0 \to p\mathbb{Z}/p^2 \to E(\mathbb{Z}/p^2) \to E(\mathbb{Z}/p) \to 0$, splits iff $E$ is a canonical lift (as in Álvaro's answer). This comes up in cryptography (!) in the Smart-Satoh-Araki attack on the ECDLP for anomalous curves. This point is discussed in the Satoh-Araki paper, I believe.<|endoftext|> -TITLE: Haar measure for large locally compact groups -QUESTION [8 upvotes]: In this answer, Gerald Edgar mentions that Haar measure is naturally defined on the $\sigma$-algebra of Baire sets (the smallest $\sigma$-algebra that contains all the compact $G_\delta$ sets) of a locally compact group and that the uniqueness of Haar measure can fail for larger $\sigma$-algebras. I wonder if there is a nice example of this. -Curious, I skimmed through Halmos's classic Measure Theory, and I found that he proves the existence and uniqueness of Haar measure for the slightly larger $\sigma$-algebra generated by all compact sets. (Confusingly, Halmos defines Borel sets to be those in this $\sigma$-algebra; I will stick with the usual definition of Borel sets.) - -Is there a nice example of a locally compact group where the uniqueness of Haar measure fails for the $\sigma$-algebra of Borel sets — the $\sigma$-algebra generated by open sets? - -To dispell some potential confusion (see comments by Keenan Kidwell and Gerald Edgar) Haar measures are not required to be regular (for the purpose of this question). - -REPLY [2 votes]: I may be missing something here but what François was looking for may be the following (it is of course already present in the previous answers, I only put it slightly differently): -Let $B$ be a Borel subset of the group $\mathbb{R} \times \mathbb{R}_d$. Define $B_y = \{x : (x,y) \in B\}$. Denote the Lebesge measure on $\mathbb{R}$ by $\lambda$. -First, let $$\mu(B) = \sum_{y \in \mathbb{R}} \lambda(B_y).$$ Second, let -$$ -\nu(B) = -\begin{cases} -\sum_{y \in \mathbb{R}} \lambda(B_y) \textrm{ if } B \textrm{ only intersects countably many sets of the form } \mathbb{R} \times \{y\},\\ -\infty, \textrm{ otherwise. } -\end{cases} -$$ -It is easy to check that both $\mu$ and $\nu$ are translation invariant $\sigma$-additive measures on the Borel sets, taking finite values on the compact sets and positive values on the nonempty open sets. (And the open sets are inner regular w.r.t. compact sets.) Do you require anything else? -Moreover, $\mu(\{0\} \times \mathbb{R}) = 0$, but $\nu(\{0\} \times \mathbb{R}) = \infty$.<|endoftext|> -TITLE: For models of ZF, if for some $A$ we have $L[A] = L$, what can we deduce about $A$? -QUESTION [7 upvotes]: Suppose $V$ is a model of ZF. Within $V$ we have $L$ which is a model of ZFC, furthermore $L[A]$ is a model of choice for every $A\in V$. -Suppose $A=\emptyset$ then clearly $L[A]=L$, furthermore if $A\in L$ then $A\cap L\in L$, therefore $L[A]=L$. Recall also that if $A' = L[A]\cap A$ then $L[A] = L[A']$. -On the other extreme, suppose $A$ is an amorphous set (an infinite set that every subset of it is either finite or co-finite). Consider $A'=A\cap L[A]$, we have that $A'\in L[A]$ which is a model of choice, so $A'$ cannot be infinite - since infinite subsets of amorphous sets are themselves amorphous. Therefore $A'$ is finite, despite not being able to prove that (at the moment anyway) I have a strong intuition that $A'=A\cap L$ and therefore $L[A]=L$. -(this conjecture stems from noticing that amorphous sets are, as the name suggests, amorphous. There is no actual reason that any element would be "preferred" into $L[A]$ over another, unless it was already in $L$. Since there can only be finitely many constructible elements in an amorphous set this somewhat supports my intuition) -Is this conjecture about amorphous sets true? Can it be extended to weaker infinite Dedekind-finite sets? Suppose $L[A]=L$, as Joel points out in his answer this implies $A\cap L\in L$. - -Suppose $L[A]=L$ for every $A\in V$, is there something to say about $V$ and $L$? (in the sense that $V$ is somewhat minimal over $L$ (that is if it has non-well orderable sets, then this is the only difference of $V$ from $L$)) -Suppose $V$ is somewhat larger than the above description (for example, $V$ is the Feferman-Levy model in which $\omega_1$ is singular and the reals have cardinality $\aleph_1$), is there anything to say about sets for which $L[A]=L$? Can we in some sense generate a model $L\subseteq M\subseteq V$ which behaves as described above (some minimality property)? - -REPLY [7 votes]: Regarding your last question, it is easy to see that if -$A\cap L=\varnothing$, then $L[A]=L$, because at every stage, if we -have agreement $L_\alpha[A]=L_\alpha$ so far, then having -$A$ as a predicate doesn't help us to define any new sets -beyond the empty predicate (since the answer for whether a -set in $L_\alpha$ is in $A$ is always no), and so -$L_{\alpha+1}[A]=L_{\alpha+1}$. -More generally, for your title question, we have the following: -Theorem. $L[A]=L$ if and only if $A\cap L\in L$. -Proof. If $L[A]=L$, then clearly $A\cap L\in L$. -Conversely, if $A\cap L\in L$, then one can show -inductively that $L_\alpha[A]=L_\alpha[A\cap L]$, which is contained in $L$. That is, having $A$ as a predicate over $L_\alpha[A]$ is just as good as having $A\cap L$ as a predicate, if you know $L_\alpha[A]\subset L$. Since $A\cap L\in L$, this means that constructing relative to the predicate $A$ never leaves $L$. QED -Note however that the hypothesis $A\cap L\in L$ does not -imply $L_\alpha[A]=L_\alpha$, because the predicate $A$ may -allow you to define sets more quickly, even when they are -in $L$. -This seems now to answer your conjecture: -Corollary. If $A$ is Dedekind finite (in particular, if $A$ is amorphous), then $L[A]=L$. -Proof. If $A$ is Dedekind finite, then $L\cap A$ must be finite, since otherwise we could construct a countably infinite subset using the $L$-order. Thus, $L\cap A\in L$, and so $L[A]=L$. QED -And your updated question 1 seems to be answered by: -Theorem. The following are equivalent: - -$L[A]=L$ for every set $A$. -$V=L$. - -Proof. If $V=L$, then clearly $L[A]=L$ for every $A$. Conversely, if $V\neq L$, then let $A$ be any $\in$-minimal set not in $L$. Thus, $A\cap L=A\notin L$, and so $L[A]\neq L$. QED<|endoftext|> -TITLE: The NP version of Matiyasevich's theorem -QUESTION [32 upvotes]: By Matiyasevich, for every recursively enumerable set $A$ of natural numbers there exists a polynomial $f(x_1,...,x_n)$ with integer coefficients such that for every $p\ge 0$, $f(x_1,...,x_n)=p$ has integer solutions if and only if $p\in A$. -Now suppose that $A$ is a set of natural numbers with membership problem in $NP$. Is there a polynomial $f$ with integer coefficients such that $f(x_1,...,x_n)=p$ has integer solutions if and only if $p\in A$ and there exists a solution with $||x_i||\le Cp^s$ for some fixed $s, C$, where $||x_i||$ is the length of $x_i$ in binary (i.e. $\sim \log |x_i|$)? Clearly the converse is true: if such a polynomial exists, then the membership problem for $A$ is in NP. - -REPLY [17 votes]: I think this is still an open problem. The idea of a Non-Deterministic Diophantine Machine (NDDM) was introduced by Adleman and Manders. In their paper Diophantine Complexity, they conjecture that the class of problems recognizable in polynomial time by a NDDM are precisely the problems in NP. However, they only prove the following: - -If A is accepted on a NDDM within time $T$, then A is accepted on a NDTM within time $T^2$. -If A is accepted on a NDTM within time $T$, then A is accepted on a NDDM within time $2^{10T^2}$. - -They also show that if R0 is the problem of determining whether all even bits of a natural number are zero, then R0 is recognized in polynomial time by a NDDM if and only if all NP problems are recognized in polynomial time by a NDDM. -PS: Technically speaking, a NDDM is not exactly of the type you ask for in your question. However, one recovers the form you desire using Putnam's trick: the equations $P(x,x_1,\ldots,x_n) = 0$ and $x = x(1 - P(x,x_1,\ldots,x_n)^2)$ have exactly the same solutions.<|endoftext|> -TITLE: How does the right regular of GL(n, R) and GL(n,Qp) decompose? -QUESTION [7 upvotes]: The question is contained in the title. I would guess that this question is already answered in the literature. -Given the reductive group $GL(n)$ over a complete local field, how does the right regular representation on $L^2(G(F))$ or perhaps better on $L^2(G(F)/Z(F))$ or $L^2(G(F)^1)$ decompose, where $Z$ is the center of $G$. Even less intuitive, how does the right regular decomposition on $C_{c}^{\infty}$-version or the Harish-Chandra Schwartz space look? - -REPLY [10 votes]: For the Lie (a.k.a. "archimedean") case, interpreting the $L^2$ question as asking for Plancherel measure, Harish-Chandra in-principle did this for a large class of reductive groups. The early non-compact example was $GL_n(\mathbb C)$ treated by Gelfand and Naimark, where the orbital-integral idea already appeared, in a much simpler guise since there's only one conjugacy class of maximal torus. Evidently they thought that all unitary irreducibles should appear in $L^2(G/Z)$, so that it remained to the 1960s for Knapp-Stein to find the unitary degenerate unitary principal series. Effectively because of the "patching" issue between conjugacy classes, the "real" case, even just $GL(n,\mathbb R)$, is in later papers of Harish-Chandra. Knapp's Princeton book is an easy reference, or Harish-Chandra's collected works (which are out of print?) -In-principle, Harish-Chandra and Arthur further decomposed Schwartz functions, for reductive Lie $G$. Arthur did address compactly-supported and Paley-Wiener space, but in a different style than the question, I think. -For p-adic $G$, as commented in Silberger's Princeton notes, Harish-Chandra had a Plancherel theorem for $L^2$ and corresponding for Schwartz spaces in 1971-73, although Silberger's notes stop short of developing what was known at the time, perhaps because there was no reasonable parametrization of supercuspidals at the time. By now, with Bushnell-Kutzko and Gross-Reeder et alia, in-principle things could be assembled. -The two articles by Moeglin in the Proc AMS 61 Edinburgh conference from 1996 address repns of $GL(n,\mathbb R)$ and $GL(n,\mathbb Q_p)$, with references, but not Plancherel. -But I can't point to a tangible source for unadorned Plancherel for $GL(n)$'s. The Lie case can be obtained by specializing the general cases (the only discrete series on Levis are the holomorphic discrete series on the GL(2)'s). The p-adic case? Someone else? -Edit: Looking at the Edinburgh 1996 volume, I see that Helgason has a piece in which he recaps G-N's and HC's only-one-conjugacy-class (of maximal tori) argument for "discovering" the Plancherel formula, in the Lie case. (To me, this issue of "discovery" is very important.) He also gives the first case where there's the patching issue, $SL_2(\mathbb R)$. -Edit 2: in response to comment/further-question: the parametrization of the "spectrum" is qualitatively very similar to the automorphic $L^2$ spectral decomposition, but also qualitatively simpler. Comparing the Lie and automorphic cases roughly: in the Lie case, the "bulk" of $L^2$ is spherical unitary principal series, with parabolic-induced repns from discrete series on Levis also entering, as well as actual discrete series, if any, on the whole group. In the automorphic case, the "bulk" of $L^2$ is spherical cuspidal repns, by Weyl's Law [sic]; in many situations, provably (and conjecturally) the bulk are unitary principal series. (In general, a naive form of Ramanujan-Petersson conjectures cannot be true, for a variety of reasons... e.g., lifts, as in Howe-PiatetskiShapiro, Corvallis, AMS Proc Sympt 33, 1977/79, ... but Weyl's Law stuff still says that probably most are unitary principal series.) The parabolically induced (Eisenstein series) from afms on Levis is a smaller part of the spectrum. -That is, in broad terms, parabolic induction from special objects on the Levis plays a completely analogous role. (This played a role in some otherwise mysterious choices of terminology in the local repn theory, e.g., "Eisenstein integrals" and "Maass-Selberg relations" in HarishChandra.) -In both cases, there are "atoms" which are to some degree mysterious: locally, discrete series, globally, cuspforms. There are certainly global issues that have much-more-trivial local counterpart, such as Weyl's Law stuff or Ramanujan-Petersson. -Yet again, much more can be said, and there is much literature on details. Not sooo many sources on higher-rank.<|endoftext|> -TITLE: Shortest "painting" of the sphere -QUESTION [5 upvotes]: Let $S$ be the sphere in $\mathbb{R}^3$ and $C:[0,1]\to S$ a continuously differentiable curve on $S$. Let $T:[0,1]\to\mathbb{R}^3$ denote the tangent vector of $C$. Let $P(t)$ be the plane containing $C(t)$ and having normal vector $T(t)$. -Given a size $d$ of the "paint brush" we define the "brush" $b:[0,1]\to \mathcal{P}(S)$ by letting $b(t)$ be the points of $S$ that are at most a distance $d$ (metric on the sphere) from $C(t)$ that are contained in $P(t)$. -We can think of this as saying the "brush" $b(t)$ is an arc on the sphere that is "orthogonal" to the motion $C(t)$ of the "paint brush". -Given $d$ what is the arclength of the shortest curves such that $\cup_{t\in[0,1]} b(t) = S$. This says that the "paint brush" covered the sphere. - -REPLY [17 votes]: The model is that used by Henryk Gerlach and Heiko von der Mosel in their 2010 paper "On sphere-filling ropes" arXiv:1005.4609v1 (math.GT) may be relevant. -Their question is different: What is the longest rope of a given thickness on a sphere? -But their explicit solutions are packings, and it seems they could be converted to -painting paths. -Here is a piece of their Fig. 6: - -REPLY [5 votes]: This question is somewhat related to this recent one. More precisely, the comment by Gjergji Zaimi in the earlier question gives a painting of length $2\sqrt{2}\pi$ for $d=\pi/4$, which, as explained in another comment there, is optimal for a path at constant distance from the sphere. So for $d=\pi/4$ the optimal length should be $2\sqrt{2}\pi$.<|endoftext|> -TITLE: Is the category of toposes cocomplete ? -QUESTION [6 upvotes]: Hello. -[Edits between brackets.] -Does the [1-]category of [elementary] toposes [with logical morphisms] admit any [1-]colimits ? -[By colimit I mean initial object in the category of outgoing cocones. I'm afraid I don't know what pseudo-colimits, bicolimits or strict 2-limits are, but I can say that I'm not interested by any 2-categorical aspect of my question (at least, for now), just by the 1-categorical one.] -Any reference would be greatly appreciated. -Thanks for any answer and please forgive me for my pityful english. - -REPLY [5 votes]: $\mathbf{Log}$, the category of elementary topoi and logical morphisms which preserve everything on the nose is cocomplete; in - -Eduardo J Dubuc, GM Kelly - A presentation of topoi as algebraic relative to categories or graphs - Journal of Algebra (website) - -this category $\mathbf{Log}$ is shown to be the category of algebras of a finitary monad on $\mathbf{Cat}$; it looks like you can even get this over $\mathbf{Grph}$, like the presentation of cartesian closed categories monadic over graphs in Lambek-Scott intro to higher-order categorical logic. This implies then that it is cocomplete, see the nLab page on colimits in categories of algebras, for example. -For the sake of completeness, a nice account (including explicit constructions) of 2-limits in $\mathbf{Log}$ as a locally groupoidal 2-category (1-cells the standard notion of logical morphism, 2-cells between them restricted to be isomorphisms) is given in chapter III of Steve Awodey's PhD thesis, - -S Awodey - Logic in topoi: functorial semantics for higher-order logic - available from his website<|endoftext|> -TITLE: Generalised linking numbers (where they shouldn't be) -QUESTION [5 upvotes]: One can define the linking number of disjointly embedded curves $K,L\subset S^{3}$ in a variety of ways, as is discussed in Chapter 5.D of Rolfsen's "Knots and Links". One way is the Gauss Integral -$$\mathrm{lk}(K,L) = \frac{1}{4\pi}\int_{K\times L}\dfrac{\mathbf{x}-\mathbf{y}}{|\mathbf{x}-\mathbf{y}|^3}\cdot\mathrm{d}\mathbf{x}\times \mathrm{d}\mathbf{y}$$ -(or symbols to that effect). This will be an integer when the curves are closed, and a real number in general. -The integral formula has been generalised to deal with the case of disjointly embedded closed manifolds $K^{k} ,L^{\ell}\subset S^{k+\ell+1}$ (see here and here for example). Presumably these formulas output real numbers when the manifolds $K$ and $L$ have boundaries. -My question concerns the situation of disjointly embedded submanifolds $K^k,L^\ell\subset S^n$, where $k+\ell >n-1$. Is there a useful notion of linking number in this case? For instance, take a surface and a curve in $3$ dimensions (so $k=2$, $\ell=1$ and $n=3$ in the above). Then we could try to define the linking number by somehow "integrating" $\mathrm{lk}(\gamma,L)$ over closed curves in $\gamma\subset K$. -This generalised linking number should be able to measure, say, how many times a curve passes through a length of tube. -Have such things been considered useful before? Or am I just talking nonsense? -Added: As Ryan points out in his comment, I'm not really looking for an isotopy invariant. Also (thanks to Kevin and Tom's answers) I'm slowly coming round to the idea that a single number won't really tell you much about the relative positions of the manifolds, but maybe a matrix valued function (with rows and columns indexed by homology bases for $K$ and $L$ in the appropriate dimensions) might be useful. - -REPLY [5 votes]: Just to mention here the ``linking numbers'' introduced by A.Haefliger [1]: for an embedding $f\colon S^k\sqcup S^l\to S^n$ he defines the linking number as certain element $$\lambda_1(f)\in\pi_k(S^{n-k-1}\vee S^{n-l-1}).$$ -This invariant is stronger than all the homological invariants discussed above and than the $\alpha$-invariant suggested by Paul. However, it is still incomplete isotopy invariant. -[1] A. Haefliger, Enlacements de spheres en codimension superiure a 2, Comm. Math. Helv. 41 (1966-67), 51-72 (in French).<|endoftext|> -TITLE: Free product of Boolean algebras -QUESTION [5 upvotes]: Given a family of Boolean algebras $\mathcal{B}=\{B_i\colon i\in I\}$ with respective Stone spaces $S_i$, the algebra of clopen (both closed and open) subsets of the product space $\textstyle\prod_{i\in I}S_i$ is called the free product of $\mathcal B$. This algebra is typically denoted by $\textstyle\bigotimes_{i\in I}B_i$ (and I will use the standard "tensor" notation for finite free products in the obvious manner). -I am interested in the (possible) Boolean algebras which admit only very particular decompositions in terms of the free product. - -Is there an uncountable Boolean algebra $B$ such that if $B$ is isomorphic to $A\otimes C$ then either $A$ or $C$ is countable? - -REPLY [6 votes]: By Theorem 15.14 of the Boolean algebra handbook, the interval algebra of the real numbers is such an algebra B. -Reference: S. Koppelberg. Handbook of Boolean algebras. Vol. 1. Edited by J. D. Monk and R. Bonnet. North-Holland Publishing Co., Amsterdam, 1989.<|endoftext|> -TITLE: Which graphs are zero-divisor graphs for some ring? -QUESTION [6 upvotes]: Given a (non commutative) ring $R$, we construct a (directed) graph $G_0(R)$ with vertex set $Z(R)\backslash \{0\}$, the zero divisors of $R$ except for $0$. And an edge from $x$ to $y$ whenever $xy=0$. This is called the zero divisor graph of $R$. -My question is, what are the known obstructions to a graph being a zero divisor graph for some ring $R$? - -(Edit) Following the reference by M. Sapir below, I found several articles which give partial answers. By this result of Dolžan and Oblak we learn that there are restrictions on the diameter and girth (they work over semirings). In the case of commutative rings, one knows by a result of Belshoff and Chapman, which zero-divisor graphs are planar, and there are similar results for projective zero-divisor graphs and other genera. Another restriction is that the number of edges must be even. -It seems that a characterisation of the set of zero-divisor graphs is open and complicated enough that is only worth studying over special subclasses of graphs. Because of this I would like to focus just on the second question below, which I have the feeling should have an easy counter-example. - -I'd also like to ask the same question for a weaker notion of graphs associated to a ring. Let $G_{k}(R)$ be defined for every ring $R$ and $k\in R$ as above but with edges going from $x$ to $y$ whenever $xy=k$ (we can start with the vertex set being all of $R$ and then throw away the isolated vertices). -What graphs can be written as $G_k(R)$ for some pair $(R,k)$? - -Are there graphs which are isomorphic to $G_k(R)$ for some $k\neq 0$ but are not isomorphic to any zero divisor graph? - -REPLY [3 votes]: The answer to the second question is "no". Consider first the case of semigroups. Take the bicyclic semigroup $B=\langle a,b \mid ab=1\rangle$. It consists of elements of the form $b^ma^n$, $m,n\ge 0$ (that representation is unique because $ab\to 1$ is a confluent and terminating rewriting system, see also A. H. Clifford and G. B. Preston, "The algebraic theory of semigroups" or http://en.wikipedia.org/wiki/Bicyclic_semigroup). Take $k=1$. The corresponding graph consists of countably many disjoint edges $a^m\to b^m$, $m\ge 0$. Now suppose that this graph is the zero divisor graph of some semigroup $S$ with 0. Then it should have $p,q$ with $pq=0$. Since for every $x\in S$ we have $(xp)q=0$, we should have $xp\to q$ in the zero-divisor graph (the option $xp=0$ does not occur because in our graph there are no edges $x\to p$). Hence $xp=p$ for every $x\ne 0\in S$. Similarly for every $x\ne 0\in S$ we have $qx=q$. Therefore $q=qp=p$, a contradiction. This argument works also for rings. One needs to consider the semigroup ring $FB$ for any field $F$, and take $k=1$. The corresponding graph is not the zero divisor graph of any (associative) ring. -In general if $R$ is a ring with 1 containing two elements $a,b$ such that $ab=1$ but $ba\ne 1$, and $a,b$ are not zero divisors, then the graph corresponding to 1 in $R$ is not a zero divisor graph of any ring $R'$. Indeed, in such a ring $R'$ we would have $ab=0, ba\ne 0$, and $xa=a, bx=b$ for every $x\ne 0$. Hence $a=ba=b$. The fact that $xa\ne 0$ (hence $xa=a$ in $R'$ is true because in $R$ we have $xa\ne 1$ for every $x$ (if $xa=1$, then $x=xab=b$, so $ba=1$, a contradiction). The fact that $xa=a$ is true because in $R$, $zb=1$ implies $z=a$ since $b$ is not a zero divisor. Similar for $bx=b$.<|endoftext|> -TITLE: Is the complete functorial structure for Khovanov--Lee homology known? -QUESTION [6 upvotes]: I'm interested in Lee's modification of Khovanov homology, which I'll denote $\operatorname{Kh}_{\operatorname{Lee}}^\ast$. Below $L$ is a link in $\mathbb R^3$. -The groups $\operatorname{Kh}_{\operatorname{Lee}}^\ast(L)$ for a link $L$ are very simple: there is an isomorphism: -$$\bigoplus_{\text{orientations of }L}\mathbb Q\to\operatorname{Kh}_{\operatorname{Lee}}^\ast(L)$$ -I write the left hand side as I do and not simply as $\mathbb Q^{\oplus 2^{\left|L\right|}}$ because given an orientation of $L$ (and a diagram of $L$), there is in some sense a natural choice of element of $\operatorname{Kh}_{\operatorname{Lee}}^\ast(L)$ (and these elements form a basis of $\operatorname{Kh}_{\operatorname{Lee}}^\ast(L)$). However, this map not natural in the sense of being functorial. -But it's really really close to being functorial! Rasmussen proved (see Khovanov homology and the slice genus Proposition 4.1) that if we identify $\operatorname{Kh}_{\operatorname{Lee}}^\ast(L)$ with $\bigoplus_{\text{orientations of }L}\mathbb Q$, then under a cobordism from $L_1$ to $L_2$, a orientation on $L_1$ is sent to a linear combination of the orientations on $L_2$ which extend to an orientation of the cobordism agreeing with the input orientation on $L_1$. This should be viewed as a sort of approximate functoriality, and the result has been slightly refined by Rasmussen here. -My question is: has $\operatorname{Kh}_{\operatorname{Lee}}^\ast$ been calculated as a functor? In other words, do we know a canonical way of mapping $\bigoplus_{\text{orientations of }L}\mathbb Q$ to $\operatorname{Kh}_{\operatorname{Lee}}^\ast$ so that the maps associated to arbitrary cobordisms are described elementarily in terms of orientations? -This doesn't seem to be a deep problem; it comes down to writing down the generators explicitly in the chain complex defining $\operatorname{Kh}_{\operatorname{Lee}}^\ast(L)$, and checking what happens when we do a Reidemeister move or Morse move. However the calculations are kind of messy, and this is presumably why Rasmussen didn't pursue the point further (at least, not that I am aware of). But the last paper of his referred to above was in 2005, so I'm guessing that someone has probably straightened this out since then. I've been unsuccessful in finding a reference, though. - -REPLY [8 votes]: As the title suggests, the paper Fixing the functoriality of Khovanov homology by Clark, Morrison and Walker fixes the functoriality in Khovanov homology. The same techniques (disoriented surfaces) fix the functoriality in Lee homology. In fact, the paper is written in such a way that it proves functoriality simultaneously for both Khovanov and Lee homology. -(Recall Bar-Natan's "free $\alpha$" version of Khovanov homology. If we set $\alpha = 0$, we get the original Khovanov homology. If we set $\alpha$ to any non-zero complex number, we get something identical to Lee homology, up to Euler characteristic normalizations. Lee's original paper has $\alpha = 1$ (or is it $\alpha = 8$?), but it's actually more convenient to let $\alpha = 2$.) -The upshot is that the Lee theory, after tweaking the definition by using disoriented surfaces, is naturally isomorphic to the simple theory you describe in your question. To a link we assign the vector space generated by orientations of the link, and to an unoriented cobordism we assign the sum, over all orientations of the cobordism, of the induced map between the orientations of the incoming and outgoing boundaries of the cobordism. (If we have set $\alpha$ to something other than 2, then we need to throw in factors of $\sqrt{\alpha/2}$ raised to the Euler characteristic of surfaces.) -If one were only interested in Lee homology, one could simply the proof in the "Fixing functoriality" somewhat, but there would still be many cases of oriented Reidemeister moves and second order oriented Reidemeister moves (movie moves) to check.<|endoftext|> -TITLE: Category with a "metric" for arrow composition -QUESTION [23 upvotes]: Consider a category $\mathcal C$ with a "distance" function $d:\mathcal C^2 \to \mathbb{R}_{\geq 0}$ satisfying the "triangle inequality" -$$d(x \to z)\leq d(x \to y) + d(y \to z)$$ -for every pair of composable arrows $(x\to z)=(x \to y \to z)$. -Let's call $(\mathcal C,d)$ a "metric" category. -The first example is to take any category $\mathcal{C}$, and define -$$d(f)=\begin{cases} 0 & \text{ if $f$ is an isomorphism} \\\ 1 & \text{ otherwise.}\end{cases}$$ -Then the triangle inequality simply translates the statement : "If $f=gh$, then $f$ is an isomorphism if $g$ and $h$ are isomorphisms." -Also, it's clear that every metric space can be made into a metric category in a canonical way. -We can define "open balls" in $\mathcal{C}$: for $c \in \mathcal{C}$, $r\geq 0$, let -$$B(c, r) = \{d \in \mathcal{C} | \text{ there exists }f: c \to d\text{ such that }d(f) < r \}.$$ -In the category of number fields and monomorphisms, we can let $d(K \hookrightarrow L)=\log ([L:K])$. Then the triangle inequality is actually an equality. It's clear that $d$ is a good measure of "how far" $L$ is from consisting of just $K$. The open ball of radius $r$ around $K$ is the set of extensions of $K$ of degree $< e^r$. -Is it possible to endow a big category like $\text{Top}$ or $\text{Grp}$ with a meaningful distance? - -REPLY [12 votes]: 1) In the category of finite sets (or finite groups or finite topological spaces....) let $d(f)$ be the cardinality of the image of $f$. This satisfies the strong triangle condition -$$d(x\rightarrow z)\le\text{min }(d(x\rightarrow y),d(y\rightarrow z))$$ -2) In any category, for each object $x$, let $\xi(x)$ be an (arbitrarily assigned) positive real number and define -$$d(f)=\text{min }\lbrace{\xi(c)|f \hbox{ factors through } c}\rbrace$$ -This also satisfies -$$d(x\rightarrow z)\le \text{min }(d(x\rightarrow y),d(y\rightarrow z))$$ -3) Fix a formal language for describing arrows in your category, let $l(f)$ be the length of the shortest description of $f$, and let $d(f)=l(f)+5$. The triangle inequality follows because $g\circ h$ always has a formal description just slightly longer than the sum of the shortest formal descriptions of $g$ and $h$ (say by putting each of these descriptions between parentheses and inserting a $\circ$ between them, which adds five characters). -In case 3), you have to allow $d$ to take the value infinity, or restrict to categories in which everything has a finite description. -Edited to add: 4) For the category of topological spaces, you can fix a non-negative integer $r$ and let $d(f)= \hbox{rank} (H^r(f,{\mathbb Q}))$ . This requires either allowing $d$ to take the value infinity or restricting to some subcategory where the homology groups are finite dimensional.<|endoftext|> -TITLE: Duality between proper homotopy theory and strong shape theory -QUESTION [5 upvotes]: In the n-lab entry about shape theory one can read that - -Strong Shape Theory [...] has, especially - in the approach pioneered by Edwards - and Hastings, strong links to proper - homotopy theory. The links are a form - of duality related to some of the more - geometric duality theorems of - classical cohomology. - -I would be interested in any reference where I can find a precise formulation of this duality. -EDIT: according to Gjergji Zaimi's answer the duality might be an improvement of Chapman's complement theorem. One can find it as Theorem 6.5.3 on page 230 of the book by Edwards and Hastings ("Cech and Steenrod Homotopy Theories with Applications to Geometric Topology"). Nevertheless, it seems to me that what was meant on the n-lab entry was more a cohomology type duality (like an instance of Verdier duality in the $(\infty,1)$-ctageorical context). Am I completely wrong? - -REPLY [3 votes]: I wrote that part of the nLab entry so can confirm that it is the Edwards and Hastings extension of Chapman's result that was referred to, but my feeling in this is that that result is the geometric form of a lot of the classical cohomological duality results and that there should be more to be said about this ... but I don't know what! Perhaps looking at the Chapman result in the light of modern homotopy theory (say using Lurie's notion of shape) may give an $(\infty,1)$-categorical result. (Note that Batanin did work on strong shape theory and produced an $A_\infty$-structure, which must relate to this. Now I like that set of ideas. Good luck if you try it!) -(You may spur me on to write more on that entry as it has got stalled... Alternatively anyone else is welcome to write more on strong shape, of course.... and to correct any miswording, typos that they find. :-))<|endoftext|> -TITLE: Ramification divisor associated to a cover of a regular scheme -QUESTION [8 upvotes]: Let $S$ be the spectrum of $\mathbf{Z}$ or the spectrum of an algebraically closed field. (Actually, one can take $S$ to be any noetherian integral regular scheme.) -Let $f:X\longrightarrow Y$ be a finite morphism of integral normal projective flat $S$-schemes which is etale above the complement of $B$, where $B\subset Y$ is a closed subscheme of codimension $1$. Suppose that $Y$ is regular. -Example. You could take $f$ to be a finite surjective morphism of normal surfaces such that $Y$ is nonsingular. -Since $Y$ is regular, we have a canonical sheaf $\omega_{Y/S}$. Let $s$ be a nonzero rational section of $\omega_{Y/S}$. -Define the cycle $K_{X/S} := \mathrm{div}(s)$. Note that $K_{X/S}$ is a canonical divisor. -Let $f^\ast s$ be the induced nonzero rational section of the line bundle $f^\ast \omega_{Y/S}$ on $X$ and consider the Weil divisor $\mathrm{div}(f^\ast s)$ on $Y$. -Outside $f^{-1}(B)$, we have that $\mathrm{div}(f^\ast s)$ is the pull-back of $K_{Y/S}$. Therefore, there is a Weil divisor $R_f$, supported on $f^{-1}(B)$, such that $\mathrm{div}(f^\ast s) = f^\ast K_{Y/S} + R_f$. -Question 1. How are the coefficients of $R_f$ defined? -Question 2. Is the Weil divisor $\mathrm{div}(f^\ast s)$ a canonical divisor outside the singular locus of $X$? -Question 3. Is $R_f$ independent of $s$? -Note that I work with cycles and not with classes up to linear equivalence. - -REPLY [3 votes]: Your hypothesis imply that $\omega_{Y/S}$ is an invertible sheaf (because $Y\to S$ is locally complete intersection). -(EDIT) As $f$ is flat at points of codimension $1$ ($Y$ is normal) and we are only interested on codimension 1 cycles, we can restrict $Y$ and suppose that $f$ is flat. -Then the dualizing sheaf $\omega_{X/Y}$ is invertible and you have the adjunction formula -$$\omega_{X/S}=f^*\omega_{Y/S} \otimes\omega_{X/Y}.$$ -The sheaf $\omega_{X/Y}$ is trivial outside of $B$ because $f$ is étale outside of $B$. It can be identified with the sheaf $\mathcal{Hom}_{O_Y}(f_{*}O_{X}, O_{Y})$. -Write $\omega_{X/Y}=O_X(D)$ for some Cartier divisor $D$ on $X$. Its support is contained in $f^{-1}(B)$. For any point $\eta$ of $X$ over a generic point $\xi$ of $B$, the stalk of $\omega_{X/Y}$ at $\eta$ is given by the different ideal of the extension of discrete valuation rings $O_{X,\eta}/O_{Y, \xi}$. The valuation of the different is known to be the ramification index $e_{\eta/\xi}$ minus $1$ when the ramification is tame and bigger or equal to $e_{\eta/\xi}$ otherwise (see Serre: Local fields). So the support of $D$ is equal to $f^{-1}(B)$ and is the ramification locus by definition. -In short, the coefficient of $R_f=D$ at the Zariski closure of $\eta$ is the valuation of the different ideal of $O_{X,\eta}/O_{Y, \xi}$. As for the computation, you can pass to the completions. -A finite extension of complete DVR $R'/R$ is monogenous if the residue extension ($k(\eta)/k(\xi)$ in your case) is separable. If $R'=R[\theta]$, and $P(T)\in R[T]$ is the minimal polynomial of $\theta$, then the different ideal is generated by $P'(\theta)$. See Serre's book for more details.<|endoftext|> -TITLE: Shortest paths on linked tori -QUESTION [7 upvotes]: I will make this question specific at first, and general later. -Suppose we have two linked tori, $T_1$ and $T_2$, -each of radii $(2,1)$, meaning that each torus is the result of sweeping -a circle of radius 1 orthogonally around a circle of radius 2. -$T_2$ is rotated -$90^\circ$ with respect to $T_1$, so that they link snugly together: - -             - - -They touch along two orthogonal circles of radius 1, -$C_1$ and $C_2$. -Let $T_i$ represent the surface of each torus. - -Q1. -What is the structure of a shortest path $\sigma(p_1,p_2)$ on $T_1 \cup T_2$ -between two points, -$p_1 \in T_1$ and $p_2 \in T_2$? - -In general I believe $\sigma(p_1,p_2)$ follows a shortest path -from $p_1$ on $T_1$ to a point $x_1$ on one of the circles, -travels along an arc of that -circle, perhaps switches at their junction to the second circle, -follows an arc to $x_2$, -and finally follows a shortest path on $T_2$ from $x_2$ to $p_2$. -I am wondering whether or not -some version of Snell's law applies -here? Suppose the path from $p_1$ makes an angle $\alpha_1$ -with one circle at $x_1$, -and an angle $\alpha_2$ where it leaves from $x_2$ on -that or the other circle on its way to $p_2$. -Is it the case, with suitable conventions on the definitions -of these angles, that $\alpha_1 = \alpha_2$? - -Answer to Q1: Gjergji Zaimi showed my intuition (concerning following -arcs of the circles $C_1$ and/or $C_2$) is wrong: "The shortest path [is] the one-point union of two geodesics.... -[I]n particular you will have equal angles along the tangency circle.... -This problem is about shortest paths on a single surface in disguise." - - - -Q2. -What is the general principle here, that would apply to several smooth -surfaces contacting along one-dimensional curves? - -Speculations, ideas, references—all welcomed. Thanks! - -Answer to Q2: See Anton Petrunin's remarks on the graded distance function -$|x-y|_n$. - -REPLY [4 votes]: This is mostly related to Q2. -I will make comments on number of switches of geodesic from one space to an other. -Bad case. When you glue two spaces it is useful to consider "graded distance function" -$|x-y|_n$ --- the minimal length of curve form $x$ to $y$, which switch from one space to an other at most $n$ times. -I used it once to prove gluing theorem for Alexandrov spaces. - -It is easy to work with "geodesics" for $|x-y|_n$. -As $n\to\infty$, $|x-y|_n$ converges to the distance in the glued space. -The function $|x-y|_n$ does not satisfy triangle inequality, but it has graded triangle inequality: - -$$|x-z|_{n+m} \le |x-y|_n+|y-z|_m.$$ -Good case. Some times you know that a geodesic will not switch from space to space too many times. -Say you glue $X$ and $Y$ along $A$ and the metric induces on $A$ from $X$ and $Y$ coincide; -in particular doubling of $X$ in $A$. -In this case it is easy to see that at least one geodesic between any two points will not switch more than once. -With little bit more information you may guarantee the same for any geodesic. -Your case. You can see that whole glued space admits a strictly short retracting to each of the circles. That means that any of the geodesic switch at most once on each of two circles. So you have only two switching and that should make you happy.<|endoftext|> -TITLE: Automorphism group of the cartesian product of two graphs. -QUESTION [7 upvotes]: Given two (simple, undirected, finite) graphs $G_1 = (V_1, E_1)$ and $G_2 = (V_2, E_2)$, let their automorphism groups be $Aut(G_1)$ and $Aut(G_2)$. -I'll recall that the cartesian product $G_1 \times G_2$ has vertex set $V_1 \times V_2$ , and two vertices $(a,b) , (x,y) \in V_1 \times V_2$ are adjacent iff $(a,x) \in E_1$ and b=y, or $(b,y) \in E_2$ and a=x. -My question is: does the problem of determining $Aut(G_1 \times G_2)$ in terms of $Aut(G_1)$ and $Aut(G_2)$ has a simple answer? May you suggest some bibliographic reference about this and related problems? Basic texts about graph theory usually barely define the automorphism group, and more algebraically-oriented texts i found did not wuite answered the question. -I tried to find a way to answer the question by myself, but i did not succeed. I'd like to know if the problem is really not-so-trivial, or if i'm simply not smart enough :) -Thanks in advance for any comment. - -REPLY [12 votes]: All you need is in W. Imrich, S. Klavzar: "Product graphs: structure and recognition". -John Wiley & Sons, New York, USA, 2000. See also: Imrich, Wilfried; Klavžar, Sandi; Rall, Douglas F. "Graphs and their Cartesian Products". A. K. Peters (2008). -One issue is that the automorphism group of the Cartesian product of $G$ with itself -is not isomorphic to $\mathrm{Aut}(G) \times \mathrm{Aut}(G)$, but rather to the wreath product of $\mathrm{Aut}(G)$ by $\mathrm{Sym(2)}$. But essentially this is all that can go wrong. The basic issue is to show that if a graph is connected then it has a unique factorization as a Cartesian product of prime graphs. This fact goes back to -Sabidussi, G. (1960). "Graph multiplication". Math. Zeitschrift.(1960). 72: 446–457. -Prime factorization can fail if $G$ is not connected.<|endoftext|> -TITLE: Multiplicity of eigenvalues of the Laplacian on quaternionic projective space -QUESTION [7 upvotes]: Using the classic spherical harmonics theory, one obtains the $k$-th eigenvalue of the $n$-dimensional round sphere $S^n$ to be $k(k+n-1)$, and its multiplicity is $\binom{n+k}{k}-\binom{n+k-1}{k-1}$, see e.g. [Berger, Gauduchon,Mazet, "Le spectre d'une variété riemannienne", Lecture Notes in Mathematics, Vol. 194 Springer-Verlag]. By looking at eigenfunctions of the Laplacian on $S^n$,$S^{2n+1}$ and $S^{4n+3}$ (note they are the unit spheres of $\mathbb R^{n+1}$, $\mathbb C^{n+1}$ and $\mathbb H^{n+1}$) that are respectively invariant under the natural actions of $\mathbb Z_2$, $S^1$ and $S^3$, one can obtain the eigenfunctions hence the $k$-th eigenvalue of the projective spaces $\mathbb R P^n$, $\mathbb C P^n$ and $\mathbb H P^n$ respectively. These are, respectively, $2k(n+2k-1)$, $4k(n+k)$ and $4k(k+2n+1)$. -QUESTION: Compute the multiplicity of the $k$-th eigenvalue $4k(k+2n+1)$ of the Laplacian on $\mathbb H P^n$. -The multiplicity of the $k$-th eigenvalue on the sphere arises as the dimension of the space of homogeneous harmonic polynomials of degree $k$ in $\mathbb R^{n+1}$, which is the difference between the dimensions of the spaces of homogeneous polynomials of degree $k$ and $k-1$ in $\mathbb R^{n+1}$. This follows from a polar decomposition of the first space as direct sum of spaces of homogeneous polynomials. The multiplicity of the $k$-th eigenvalue $4k(n+k)$ of the Laplacian on the complex projective space $\mathbb C P^n$ is given by the difference of the squares of these dimensions, it is $\binom{n+k}{k}^2-\binom{n+k-1}{k-1}^2$. This again follows from a polar decomposition of the space of harmonic homogeneous polynomials on $\mathbb C^{n+1}$. However, I do not know how to extend this idea to decompose the space of harmonic homogeneous polynomials on $\mathbb H^{n+1}$ and hence obtain the multiplicity of the $k$-th eigenvalue of $\mathbb H P^n$. This actually seems to be an exercise of [Berger, Gauduchon, Mazet]. Any natural guesses with similar differences of binomial powers seem to be wrong, since the multiplicity of the first eigenvalue (i.e., $k=1$) of $\mathbb H P^n$ should be $2n^2+3n$. -HINT: Apparently the desired multiplicity should coincide with the dimension of the vector space formed by homogeneous polynomials $f$ in $\mathbb H^{n+1}=\mathbb R^{4n+4}$ with coordinates $(x_\alpha,y_\alpha,z_\alpha,w_\alpha), 1\leq\alpha\leq n+1$, of degree $2k$ such that $Lf=0$, where $L=\sum_j L_jL_j$ and -$$L_1=\sum_\alpha y_\alpha\frac{\partial}{\partial x_\alpha}-x_\alpha\frac{\partial}{\partial y_\alpha}+w_\alpha\frac{\partial}{\partial z_\alpha}-z_\alpha\frac{\partial}{\partial w_\alpha} -$$ -$$L_2=\sum_\alpha z_\alpha\frac{\partial}{\partial x_\alpha}-w_\alpha\frac{\partial}{\partial y_\alpha}-x_\alpha\frac{\partial}{\partial z_\alpha}+y_\alpha\frac{\partial}{\partial w_\alpha} -$$ -$$L_3=\sum_\alpha w_\alpha\frac{\partial}{\partial x_\alpha}+z_\alpha\frac{\partial}{\partial y_\alpha}-y_\alpha\frac{\partial}{\partial z_\alpha}-x_\alpha\frac{\partial}{\partial w_\alpha} -$$ -The space of homogeneous polynomials of degree $2k$ in $4n+4$ variables has dimension $\binom{4n+2k+3}{2k}$, so the desired multiplicity should be this number minus the dimension of the subspace formed by polynomials such that $Lf=0$. Note that the above operators $L_j$, when restricted to the unit sphere $S^{4n+3}$ of $\mathbb H^{n+1}$ give the three vector fields that span the distribution tangent to the Hopf fibers $S^3$. - -REPLY [6 votes]: These dimensions have been calculated explicitly for all the compact rank 1 symmetric spaces. See Cahn and Wolf, "Zeta functions and their asymptotic expansions for compact symmetric spaces of rank one", Commentarii Mathematici Helvetici, vol. 51 (1976), pp. 1-21.<|endoftext|> -TITLE: On the smooth structure of the spaces of $k$-jets -QUESTION [5 upvotes]: I was asking myself, if the following list of conditions is sufficient to determine the usual smooth structure on the spaces of $k$-jets. - -the map $j^k f:M\ni x\to j_x^k f\in J^k(M,N)$ is smooth, for any $f\in C^{\infty}(M,N)$; -the map $(\alpha,\beta):J^k(M,N)\to M\times N$, defined by $(j^k_x f)\mapsto (x,f(x))$, is a smooth summersion, for any smooth manifolds $M$ and $N$; -the composition map $\gamma:J^k(N,O)\times_{N} J^k(M,N)\to J^k(M,O)$, defined by $(j^k_{f(x)} g,j^k_x f)\mapsto j^k_x(g\circ f)$, is smooth, for any smooth manifolds $M$,$N$ and $O$, (here $J^k(M,N)\times_{N} J^k(M,N)$ is the fiber product of $\beta:J^k(M,N)\to N$ and $\alpha:J^k(N,O)\to N$); -the map $(\alpha,\beta)^{-1}(U,V)\to J^k(U,V)$ defined by $j^k_x f\mapsto j^k_x(f|_{U\cap f^{-1}(V)})$ is a smooth isomorphism, for any open subsets $U\subset M$, $V\subset N$; -for any open subsets $U\subset \mathbb{R}^m$ and $V\subset \mathbb{R}^n$, the map $J^k (U,V)\to U\times V\times \bigoplus_{i=1}^k{L^i_{sym}(m,n)}$, given by $j^k_x f\mapsto (x,f(x),Df(x),\ldots,(D^kf)(x))$, is a smooth isomorphism, (here $L^i_{sym}(m,n)$ is the vector space of the $\mathbb{R}^n$-valued symmetric $k$-multinear maps on $\mathbb{R}^m$). - -Probably it is not sufficient, or redundant, but, in such a case, I would know if there is in the literature such a kind of characterization. - -My question is: Once prescribed the usual smooth structure on the $J^k(U,V)$, for arbitrary open sets in euclidean spaces $U$ and $V$ (as in point 5), what kind of conditions are sufficient to uniquely determine the usual smooth structure on $J^k(M,N)$ for all other smooth manifolds $M$ and $N$? - -REPLY [4 votes]: Let $(U,u)$ is a chart for $M$, and $(V,v)$ be a chart for $N$. $u: U\to u(U)\mathbb R^n$ is diffeomorphism. $u(U)$ and $v(V)$ are open subset of $\mathbb R^n$ and $\mathbb R^m$. Then we can identify $$J^k(u(U),v(V))= u(U)\times v(V)\times \Pi_{j=1}^k L^j_{sym}(\mathbb R^n, \mathbb R^m)$$ People give manifold structure on $J^K(M,N)$ by chart $(J^k(U,V), J^k(u^{-1}, v))$. Main aim is to define map $J^k(u^{-1}, v)$. -$$J^k(u^{-1}, v): J^k(U,V)\to J^k(u(U), v(V))\text{ is defined as following:}$$ -Firstly for $u:U\to u(U)$ define $J^k(u,V):J^k(U,V)\to J^k(u(U),V)$ by $ J^k(u,V)j^kf_x= j^k(fog)_{g^{-1}(x)}$. This is a well defined map and $J^k(u,V)^{-1}= J^k(u^{-1},V)$ -Now same way for $v$ define map $J^k(U,v): J^k(U,V)\to J^k(U, v(V)$. Take -$$J^k(u^{-1}, v):= J^k(u^{-1},v)oJ^k(u(U),v)$$ This will be bijective and satisfy coordinate transformation condition: -For details please see: First Chapter 1.1 to 1.8 of Manifolds of differential mapping: P.W. Michor.<|endoftext|> -TITLE: Computing an example of monodromy -QUESTION [6 upvotes]: Consider the double cover $\pi:S^1 \rightarrow S^1, z \mapsto z^2$ and the pushforward of the constant sheaf $\pi_{*}\mathbb{Z}$. This is a locally constant sheaf of rank 2, but not constant (since the space of global sections is rank 1). -Question: if I choose a basis $u,v$ for the stalk at some point $p$, how to compute the monodromy matrix with respect to this matrix? - -REPLY [7 votes]: It looks like this was asked and answered a while ago. But since it floated to the top -again, let me give another answer. -First in classical language, the monodromy exchanges -sheets $ \sqrt{z}\leftrightarrow -\sqrt{z}$. If you imagine taking formal linear combinations -of these, then you should be able to recover the monodromy matrix given -in Francesco's answer. -Here is a more abstract general point of view. -If $Y$ is a connected (sufficiently nice) -space, the category of locally constant sheaves is equivalent to representations -of $\pi_1(Y)$ via monodromy. The pullback of a locally constant sheaf -along a covering space $\pi:X\to Y$ corresponds to -restriction from $\pi_1(Y)$ to $\pi_1(X)$. Pushforward would be the right adjoint which -corresponds to induction in the opposite direction. So in particular $\pi_*\mathbb{Z}$ is the regular representation of -$\pi_1(Y)\to Aut(\mathbb{Z}[G])$ when $\pi$ is Galois with group $G$.<|endoftext|> -TITLE: Entropy of a measure -QUESTION [14 upvotes]: Let $\mu$ be a probability measure on a set of $n$ elements and let $p_i$ be the measure of the $i$-th element. Its Shannon entropy is defined by -$$ -E(\mu)=-\sum_{i=1}^np_i\log(p_i) -$$ -with the usual convention that $0\cdot(-\infty)=0$. -The following are two fundamental properties: - -Property 1: $E(\mu)$ takes its minimum on the Dirac measures. -Property 2: $E(\mu)$ takes its maximum on the uniform probability measure. - -Now, for some application, I am really interested in a possible generalization when $\mu$ is a finitely additive probability measure on the natural numbers. -Question: Is it possible to define a notion of entropy of a finitely additive probability measure on the natural numbers in such a way that it verifies the following properties: - -it takes its minimum on the Dirac measures -it takes its maximum on the finitely additive translation invariant probability measures - -Any reference? Idea? -Thanks in advance, -Valerio - -REPLY [2 votes]: I just wanted to say that topologically, your hand is forced: -Tapio's supremum definition is the way to go. -Give the space $X$ of all maps from $\mathcal{P}(\mathbb{N})$ to $[0,1]$ -the topology of pointwise convergence. -The space $FAM$ of finitely additive probability measures on -$\mathbb{N}$ is then a closed subspace of $X$. -Let $FSM$ be the set of finitely supported measures in $FAM$. -Let $FSA$ be the set of finite subalgebras of of $\mathcal{P}(\mathbb{N})$. -Given $G\in FSA$ and $\mu\in FAM$, it is easy to (definably) choose $\mu_G\in FSM$ that agrees with $\mu$ on $G$: $\mu_G(\{\min(A)\})=\mu(A)$ for every atom $A\in G$. -In $FAM$, each point $\mu$ has a neighborhood base consisting of -sets of the form $U(\mu,G,\varepsilon)$, which I use to denote -the set of $\nu\in FAM$ that agree with $\mu$ on $G$ up to error $\varepsilon$. -Therefore, $FSM$ is dense in $FAM$. -The Shannon entropy Ent is continuous on $FSM$, and -Ent extends uniquely to a continuous map from $FAM$ to $[0,\infty]$ -given by Tapio's $Ent(\mu)=\sup\{Ent(\mu_G): G\in FSA\}$ (using my notation). -To see this, the important step is to check that $Ent(\mu_G)\leq Ent(\mu_H)$ if $G\subseteq H$.<|endoftext|> -TITLE: Simplest example of jumping of cohomology of structure sheaf in smooth families? -QUESTION [20 upvotes]: Using Hodge theory (and the ill-defined Lefschetz principle), one can show that in characteristic 0, given a proper smooth family $X \rightarrow B$, the cohomology groups of the structure sheaf of the fibers are locally constant (as a function on $B$). I'm aware of the existence of a number of counterexamples to the corresponding statement in positive characteristic. Given the success of this question, I want to ask: - -What are the simplest examples of a proper smooth family exhibiting jumping of some cohomology group of the structure sheaf? - -By the above discussion, such an example will necessarily be in positive characteristic. -By "simplest", I mean by one of the following measures. -(best) An example whose proof is as elementary as possible, and ideally short. -An example with a simple conceptual underpinning. (Well, the best answer would do well by both of the first two measures.) -A known example that is simple to state, but may have a complicated proof. (Ideally there should be a reference.) -An expected, folklore, or conjectured example. - -REPLY [5 votes]: By the way, for Enriques this more than just an example - it reflects their classification: the moduli space of Enriques surfaces is connected in any characteristic. All such surfaces arise as desingularizations of quotients $Y/G$, where Y is a (possibly singular) complete intersection of $3$ quadrics in $\mathbb{P}^5$, and $G$ is a finite flat group scheme of length $2$. -In characteristic $p\neq2$, all Enriques surfaces satisfy $h^{01}=h^{10}=0$, and their moduli space is irreducible. Also, over an algebraically closed field of characteristic $p\neq2$, the only possibility for $G$ is $\mathbb{Z}/2\mathbb{Z}$, which is isomorphic to $\mu_2$. -In characteristic $p=2$, we have $h^{01}=0$, $h^{10}=1$ if the surface arises as a quotient by $G=\mu_2$ ("classical"), we have $h^{10}=h^{01}=1$ if $G=\alpha_2$ ("supersingular"), and finally, we have $h^{01}=1, h^{10}=0$ if $G=\mathbb{Z}/2\mathbb{Z}$ ("singular"). The moduli space is connected, but has two irreducible components: one corresponds to $\mu_2$-quotients, the other to $\mathbb{Z}/2\mathbb{Z}$-quotients, and their intersection to $\alpha_2$-quotients. -Thus, the Hodge numbers reflect the position in the moduli space, and a jumping corresponds to a change in type.<|endoftext|> -TITLE: On the average of continuous functions $f:\mathbb{R}^2\rightarrow[0,1]$ -QUESTION [9 upvotes]: Is it true that if the average of a continuous function $f:\mathbb{R}^2\rightarrow[0,1]$ over a unit circle centered around $(x,y)$ is $f(x,y)$ for all $(x,y)\in\mathbb{R}^2$, then $f$ is necessarily constant? - -REPLY [16 votes]: Yes, any such $f$ is constant. In fact, if we relax the condition so that $f$ is only required to be bounded below, but not above, then it is still true that $f$ is constant. This can be proven by martingale theory, as can the statement that harmonic functions bounded below are constant (Liouville's theorem). -Let $X_1,X_2,\ldots$ be a sequence of independent random random variables uniformly distributed on the unit circle, set $S_n=\sum_{m=1}^nX_m$ and let $\mathcal{F}_n$ be the sigma-algebra generated by $X_1,X_2,\ldots,X_n$. Then, $S_n$ is a random walk in the plane, and is recurrent. Your condition is equivalent to $\mathbb{E}[f(S_{n+1})\vert\mathcal{F}_n]=f(S_n)$. That is, $f(S_n)$ is a martingale. It is a standard result that a martingale which is bounded below converges to a limit, with probability one. However, as $S_n$ is recurrent, this only happens if $f$ is constant almost everywhere. By continuity of $f$, it must be constant everywhere. -For the same argument applied to functions $f\colon\mathbb{Z}^2\to\mathbb{R}$, see Byron Schmuland's answer to this math.SE question. - -In general, for a continuous function $f\colon\mathbb{R}^2\to\mathbb{C}$, if $f(x,y)$ is the average of $f$ on the unit circle centered at $(x,y)$ then it does not follow that $f$ is harmonic. So, we cannot prove the result directly by applying Liouville's theorem. As an example (based on the comments by Gerald Edgar and by me), consider $f(x,y)=\exp(ax)$. The average of $f$ on the unit circle centered at $(x,y)$ is -$$ -\frac{1}{2\pi}\int_0^{2\pi}f(x+\cos t,y+\sin t)\,dt=\frac{1}{2\pi}f(x,y)\int_0^{2\pi}e^{a\cos t}\,dt=f(x,y)I_0(a). -$$ -Here, $I_0(a)$ is the modified Bessel function of the first kind. Whenever $I_0(a)=1$ then $f$ satisfies the required property. This is true for $a=0$, in which case $f$ is constant, but there are also nonzero solutions such as $a\approx1.88044+6.94751i$. In that case $f$ satisfies the required property but is not harmonic.<|endoftext|> -TITLE: Surreal Numbers and Set Theory -QUESTION [6 upvotes]: Hello, -I looked through MathOverflow's existing entries but couldn't find a satisfactory answer to the following question: -What is the relationship between No, Conway's class of surreal numbers, and V, the Von Neumann set-theoretical universe? -In particular, does V contain all the surreal numbers? If so, then is there a characterization of the surreal numbers as sets in V? And does No contain large cardinals? -I came across surreal numbers recently, but was surprised by the seeming lack of discussion of their relationship to traditional set theory. If they are a subclass of V, then I suppose that could explain why so few people are studying them. -Thank you, -Alex - -REPLY [5 votes]: I'm probably very late in answering this, but, in addition to what Stefan and Sridar have already said, I wanted to point out that there is another equivalent definition for a surreal number. You can think of a surreal number as a function from some ordinal $\alpha$ into your favourite two-element set (whose elements are usually denoted $-,+$). Then the order among surreal numbers defined this way is the "lexicographical" one, namely, given two surreal numbers, the bigger will be the one with bigger domain, and if the domains are equal, then look at the first place where they differ: the bigger will be the one that has a "$+$" in that place. With this definition, you don't need to worry about equivalence classes, and I personally find it much easier to work with. You can find all of this in Harry Gonshor's book, "An Introduction to the Theory of Surreal Numbers", volume 110 of the London Mathematical Society Lecture Note Series. -Edit: I made a mistake in describing the order, so let me describe the actual order with some more detail: given a surreal number $s:\alpha\rightarrow${$+,-$}`, we abuse notation by saying that $s(\beta)=0$ for all $\beta\geq\alpha$. Now given two surreal numbers $s,t$, look at the first ordinal $\alpha$ such that $s(\alpha)\neq t(\alpha)$ (we consider the possibility that $s(\alpha)$ or $t(\alpha)$ equal $0$). Then we will say that $s < t$ iff $s(\alpha) < t(\alpha)$, with the convention that $- < 0 < +$.<|endoftext|> -TITLE: Consistency strengths related to the perfect set property -QUESTION [10 upvotes]: I want a model of $\mathrm{MA}_{\sigma\mathrm{-centered}}+\neg\mathrm{CH}$ in which every set of reals in $L(\mathbb{R})$ has the perfect set property. In terms of consistency strength, it is known that I need at least an inaccessible: if $\mathrm{PSP}(L(\mathbb{R}))$, then $\omega_1$ is inaccessible in $L$. I haven't been able to find any other lower bounds on the consistency strength of $\mathrm{MA}+\neg\mathrm{CH}+\mathrm{PSP}(L(\mathbb{R}))$. -The best upper bound I can find is the fact, due to Woodin, that a measurable cardinal above infinitely many Woodin cardinals outright implies $\mathrm{Det}(L(\mathbb{R}))$. So, starting with these large cardinals in the ground, I can get what I want by forcing MA (using a "small" forcing). -My question is, do I really need such strong hypotheses? -The ideal answer would be "this is known; the answer can be found in...". -One the other hand, if you can tell me with confidence that it's an open -problem, then at least I'll know that trying to solve it isn't a waste of time. -If it's easier to answer my question for projective sets, please do! -Back in 1964, Solovay proved that Levy-collapsing an inaccessible $\kappa$ to $\omega_1$ forces every set of reals definable by an omega-sequence of ordinals---this includes every set of reals in $L(\mathbb{R})$---to have the perfect set property. The catch is that the Solovay model also satisfies CH. -There's a 1989 JSL paper by Judah and Shelah (http://www.jstor.org/stable/2275017) -that looks at the consistency strength of $\mathrm{MA}_{\sigma\mathrm{-centered}}+\neg\mathrm{CH}$ (and similar forcing axioms) in conjunction with various regularity properties for projective sets: -Lebesgue measurability, the Baire property, and the Ramsey property. -The perfect set property is (from my point of view) conspicuously absent. - -REPLY [5 votes]: $\mathrm{MA}_{\sigma-\mathrm{centered}}+\neg\mathrm{CH}+\mathrm{PSP}(L(\mathbb{R}))$ is equiconsistent with a Mahlo cardinal. -Before Goldstern's comment, I had assumed the perfect set property was important enough to authors to mention if their theorems covered it. With that assumption falsified, I read more carefully, and determined that the Judah-Shelah paper I mentioned in the question had the answer all along. The proof of Lemma 1.1 shows that any forcing extension satisfying the hypotheses of the lemma actually has the same $L(\mathbb{R})$ as some Solovay model. Moreover, the proof of Theorem 3.1 actually builds from a Mahlo cardinal a forcing extension that satisfies $\mathrm{MA}_{\sigma-\mathrm{centered}}+\neg\mathrm{CH}$ and the hypotheses of Lemma 1.1. Therefore, a Mahlo cardinal suffices. A Mahlo cardinal is necessary because $\mathrm{PSP}(L(\mathbb{R}))$ is well-known to imply that $\omega_1^{L[x]}<\omega_1$ for all reals $x$, which in turn implies, by (ii)$\Rightarrow$(i) of 3.1, that if $\mathrm{MA}_{\sigma-\mathrm{centered}}+\neg\mathrm{CH}$ also holds, then a Mahlo cardinal is consistent with ZFC (specifically, $\omega_1$ is Mahlo in $L$ by the proof of (ii)$\Rightarrow$(i)). -(The hypotheses of Lemma 1.1 are that $\kappa$ is inaccessible, $\mathbb{P}$ satisfies the $\kappa$-chain condition, $\mathbb{P}$ forces $\kappa=\omega_1$, and, for every subset $Q$ of $\mathbb{P}$ of size less than $\kappa$, $Q$ extends to $P'\subseteq \mathbb{P}$ such that $|P'|<\kappa$ and the inclusion map completely embeds $P'$ into $\mathbb{P}$.)<|endoftext|> -TITLE: minimal model of a "surface" over $Spec(\mathbb{Z})$ -QUESTION [9 upvotes]: If I have a smooth compact algebraic scheme of dimention $2$ over $Spec(\mathbb{Z})$ whose generic fiber is a surface in minimal model (say of general type). Then: -(a) Is it true that the special fibers are in minimal model as well? (I would guess the answer is no in general but at least in an open subscheme of the base this should be true). -(b) If the asnwer to (a) is negative, does there exist some scheme whose fibers are minimal? (it is clear that in each special fiber I can do this, but I want the resulting scheme to be a smooth scheme over $Spec(\mathbb{Z})$). -I had no luck while searching for references for this sort of questions (since this is not a general threefold) so references on the subject are more than welcome! - -REPLY [9 votes]: The question is local on the base, so you can replace $\mathbb Z$ with a discrete valuation ring $R$. Moreover, the Kodaira dimension and the minimality of surfaces are stable by field extension, so one can suppose the residue field of $R$ is algebraically closed. -Then the positive answer is given in Katsura and Ueno: "On elliptic surfaces in characteristic $p$". Math. Ann. 272 (1985), Lemma 9.4 (see also Lemma 9.6). This holds under the hypothesis that the generic fiber has non-negative Kodaira dimension.<|endoftext|> -TITLE: complete embeddings of boolean algebras and preservation of stationarity -QUESTION [7 upvotes]: Define a complete embedding of Boolean algebra as an homomorphism of Boolean algebras which preserves also the sup and inf operations. Notice that if $\mathbb{B}$ and $\mathbb{D}$ are complete boolean algebras, $i:\mathbb{B}\to\mathbb{D}$ is a complete embedding and -$G$ is $V$-generic for $\mathbb{D}$, then $H=i^{-1}[G]$ is $V$-generic for $\mathbb{B}$. -I'm curious to know if the following can be the case: -Assume $\mathbb{B}$ and $\mathbb{D}$ are complete stationary set preserving boolean algebras. Can there be two distinct complete embeddings $i_0:\mathbb{B}\to\mathbb{D}$, $i_1:\mathbb{B}\to\mathbb{D}$ such that -if $G$ is $V$-generic for $\mathbb{D}$ and $H_j=i_j^{-1}[G]$ are the corresponding $V$-generic filters for $\mathbb{B}$ induced by the respective $i_j$, we can have that there is a name $\tau$ in the forcing language for $\mathbb{B}$ such that: -$\|\tau$ is a stationary subset of $\omega_1\|=1_{\mathbb{B}}$ -$V[G]\models\sigma_{H_0}(\tau)$ is a stationary subset of $\omega_1$ -$V[G]\models\sigma_{H_1}(\tau)$ is non-stationary - -REPLY [6 votes]: Your situation can happen. -Let $\mathbb{B}=\text{Add}(\omega_1,1)$ be the forcing to -add a Cohen subset $S\subset \omega_1$, and let -$\mathbb{D}$ be the forcing that first adds such a set $S$, -and then shoots a club through it $C\subset S$. Note that -$\mathbb{B}$ is countably closed in $V$ and therefore -stationary-set preserving, and the generic Cohen set $S$ -that is added is both stationary and co-stationary. -Further, the forcing $\mathbb{D}$ is stationary-set -preserving over $V$, because by a bootstrap argument we may -find a dense set of conditions $(s,c)$, where $s\subset -\omega_1$ is bounded and $c\subset s\cup\text{sup}(s)$ is -closed, and the set of such conditions in $\mathbb{D}$ is -countably closed. Thus, $V[S][C]$ is -stationary-set-preserving over $V$, even though it is not -stationary-set-preserving over $V[S]$. -Notice that we may completely embed $\mathbb{B}$ into -$\mathbb{D}$ in the natural way, since $\mathbb{D}$ was -described as first adding $S$, and then shooting a club -through it. -But we may also embed $\mathbb{B}$ into $\mathbb{D}$ in a -different way: by first applying the automorphism of -$\mathbb{B}$ that flips all bits. This automorphism in -effect replaces $S$ with its complement, so that under this -embedding, the club gets added to the complement of $S$. -Thus, if $\tau$ is the name of the generic set $S$ added by -$\mathbb{B}$, then $1_{\mathbb{B}}$ forces that $\tau$ is -stationary, and with the first embedding we have that -$\text{val}(\tau,H_0)=S$, which remains stationary and in -fact containing a club in $V[S][C]$, but with the second -embedding we have $\text{val}(\tau,H_1)=\omega_1\setminus S$, which is non-stationary in $V[S][C]$. -The two embeddings correspond as you said to the two fundamentally different ways we can think about the Cohen set being treated by the club-shooting forcing, since either we shoot the club through the set, or through its complement, and this difference radically affects the stationarity of this set. But meanwhile, all the ground model stationary sets are preserved, since the composition forcing has a countably closed dense set.<|endoftext|> -TITLE: When is an integral transform trace class? -QUESTION [29 upvotes]: Given a measure space $(X, \mu)$ and a measurable integral kernel $k : X \times X \rightarrow \mathbb{C}$, the operator -$$ K f(\xi) =\int_{X} f(x) k(x,\xi) d \mu(x),$$ -the operator $K$ is Hilbert Schmidt iff $k \in L^2(X \times X, \mu \otimes\mu)$! -Q1:The main point of this questions, what are necessary and sufficient conditions for it to be trace class? -I know various instances, where -$$ \mathrm{tr} K = \int_X k(x,x) d \mu(x).$$ -Q2:What are counterexamples, where $x \mapsto k(x,x)$ is integrable, but the operator is not trace class? -Q3:What are counterexamples for a $\sigma$ finite measure space, where $k$ is compactly supported and continuous, but the kernel transformation is not trace class and the above formula fails? -Q4: Is there a good survey/reference for these questions. - -REPLY [6 votes]: A remark on (Q3): -There is this famous example of T.Carleman (1916 Acta Math link) where he constructs a (normal ) operator with a continuous kernel such that it belongs to all Schatten p-classes if and only if $p\geq 2.$ -More precisely it's possible to construct $k(x)=\sum_n c_ne^{2\pi i n x}$ continuous and periodic with $\sum_n|c_n|^p=\infty$ for $p<2$. Then $Tf=f\ast k$ acting on $L^2(\mathbb T)$ yields the desired result. -Provided some extra regularity on the kernel, the trace formula works fine (there are a lot of results in the literature) -Regarding (Q4) I personally find C. Brislawn's result very interesting but rather difficult to implement in practice.<|endoftext|> -TITLE: Martingale representation theorem for Levy processes -QUESTION [9 upvotes]: Is there an equivalent of martingale representation theorem for Levy processes in some form? I believe there is no such theorem in generality, but maybe there are some specific cases? - -REPLY [5 votes]: Hi, -Here is a theorem that might answer your question (it is coming from Chesnay, Jeanblanc-Piqué and Yor's book "Mathematical Methods for Financial Markets"). -It is theorem (11.2.8.1 page 621) here it is : -(edit note : be carefull as mentioned by G. Lowther there's a typo in the book regarding the domain of integration in the conditions over $\psi$ (defined hereafter) ) -Let $X$ be an $R^d$ valued Lévy Process and $F^X$ its natural filtration. Let $M$ be an $F^X$-local Martingale. Then there exist an $R^d$-valued predictable process $\phi$ and an predictable function $\psi : R^+ \times \Omega \times R^d\to R$ such that : --$\int_0^t \phi^i(s)^2ds <\infty$ almost surely --$\int_0^t \int_{|x|> 1} |\psi(s,x)|ds\nu(dx) <\infty$ almost surely --$\int_0^t \int_{|x|\le 1} \psi(s,x)^2ds\nu(dx) <\infty$ almost surely -and -$M_t=M_0+ \sum_{i=0}^d \int_0^t \phi^i(s)dW^i_s + \int_0^t \int_{R^d} \psi(s,x)\tilde{N}(ds,dx)$ -Where $\tilde{N}(ds,dx)$ is the compensated measure of the Lévy process $X$ and $\nu$ the associated Lévy measure. -Moreover if $(M_t)$ is square integrable martingale then we have : -$E[(\int_0^t \phi^i(s)dW^i_s)^2]=E[\int_0^t \phi^i(s)^2ds]<\infty$ -and -$E[(\int_0^t \int_{R^d} \psi(s,x)\tilde{N}(ds,dx))^2]=E[ \int_0^t ds \int_{R^d} \psi(s,x)^2\nu(dx)]<\infty$ -and $\phi$ and $\psi$ are essentially unique. -The theorem is not proved in the book but there is a reference to the following parpers : -1/H. Kunita and S. Watanabe. On square integrable martingales. Nagoya J. -Math., 30:209–245, 1967 -2/H. Kunita. Representation of martingales with jumps and applications to -mathematical finance. In H. Kunita, S. Watanabe, and Y. Takahashi, editors, -Stochastic Analysis and Related Topics in Kyoto. In honour of Kiyosi Itô, -Advanced studies in Pure mathematics, pages 209–233. Oxford University -Press, 2004. -Regards<|endoftext|> -TITLE: Torsion in triangle groups -QUESTION [8 upvotes]: A triangle group has a presentation of the form, -$G=\langle a, b; a^{\alpha}, b^{\beta}, c^{\gamma}, abc\rangle, \alpha, \beta, \gamma \geq 2$ -(I believe that these are also called von Dyke groups, or "ordinary" triangle groups, with triangle groups being something slightly different, but names are beside the point). I have been reading the Fine and Rosenberg paper which proves that these groups are conjugacy separable ("Conjugacy separability of Fuchsian groups and related questions"; the proof, and the statement below, can also be found in their book, "Algebraic generalizations of discrete groups"), and in it the authors state, -"The conjugacy classes of elements of finite order...are given by the conjugacy classes {$\langle a\rangle$}, {$\langle b\rangle$}, {$\langle c\rangle$}". -This statement is given without proof or reference. I was therefore wondering if someone could provide either a proof or a reference for this? -I understand where the comment comes from - it is the obvious generalisation of the one-relator groups case (here we are dealing with one-relator products, which generalise one-relator groups). However, I cannot seem to find a proof of the statement in the literature, although I am sure it must be there. Unless, of course, I am simply missing something and the result is obvious... - -REPLY [4 votes]: In fact it's easy, and hopefully enlightening, to give a direct proof. Here's one, using the theory of orbifolds (which happens to be how I think about it). (NB I suspect it's not how Fine and Rosenberger think about it.) -Your triangle group is the fundamental group of an orbifold $O$ with underlying space a 2-sphere and cone points of order $\alpha,\beta,\gamma$. This orbifold has Euler characteristic -$\chi(O)= 2-(1-1/\alpha)-(1-1/\beta)-(1-1/\gamma)=1/\alpha+1\beta+1/\gamma-1$. -For convenience, we will give the proof in the hyperbolic, ie Fuchsian, case where $\chi(O)<0$; a similar proof can be given in the Euclidean ($\chi(O)=0$) case. It seems clear that the statement is false in the spherical (ie $\chi(O)>0$) case, since then $G$ is finite. For more information about orbifolds, see Peter Scott's survey article `The geometries of 3-manifolds'. -The orbifold $O$ is the quotient of the hyperbolic plane $\mathbb{H}^2$ by your triangle group $G$. That is to say, $G$ acts properly discontinuously and cocompactly, but not freely, on $\mathbb{H}^2$. The cone points on $O$ are precisely the images of the points in $\mathbb{H}^2$ with non-trivial stabilizers. Each cone point on $O$ has a preimage in $\mathbb{H}^2$ whose stabilizer is generated by one of the generators $a,b$ or $c$. Call these preimages $x_a,x_b,x_c$ respectively. -Now suppose that $g\in G$ has finite order. By the classification of isometries of $\mathbb{H}^2$, it follows that $g$ fixes a point $y$ in $\mathbb{H}^2$. Thus $y$ has non-trivial stabilizer, and so for some $h\in G$, $y=hx_a$ or $hx_b$ or $hx_c$; wlog, let's say $y=hx_a$. Therefore $h^{-1}gh$ stabilizers $x_a$, and so $g$ is conjugate into $\mathrm{Stab}_G(x_a)=\langle a\rangle$, as required.<|endoftext|> -TITLE: Number of simplicial polytopes with a given f-vector -QUESTION [7 upvotes]: Plenty of very nice literature is available on the characterization of f-vectors of simplicial complexes of diverse sorts (results by Billera, Bjoerner, Kalai, Stanley, among others). I mention, as an example, the Dehn-Sommerville equations, the Upper- and Lower Bound Theorems, for Simplicial Polytopes. -Are there any results on the enumeration of simplicial, convex d-polytopes with a given f-vector? -A slightly simpler question is: How does information about an f-vector (say, specifying the number of vertices, edges and triangles) determine the amount of (simplicial, convex d-) polytopes having this numbers fixed. Are there some cases where the f-vector specifies completely the polytope? -This is related to these posts: -Number of graphs with a given number of nodes, edges and triangles -What is known about the number of permissible simplicial complexes given the number of k-cells? -And the reason I am concerned about this is that in the first of the posts, it has been commented that the problem may be way too difficult, so I was wondering whether imposing the condition that the simplicial complex be a convex polytope may simplify the situation a bit. - -REPLY [4 votes]: These are just some random remarks, with one hopefully useful reference. - -"Are there some cases where the $f$-vector specifies completely the polytope?" - -This is hardly what you are seeking, but for 3-polytopes, $f_2=2f_0-4$ is achieved exactly for the -$f$-vectors of simplicial polytopes. -And of course the stacked and cyclic polytopes achieve the lower and upper bounds respectively. -I don't know if you have seen -Günter M. Ziegler's -"Convex Polytopes: Extremal Constructions and $f$-Vector Shapes" -(IAS/Park City Mathematics Series Volume 14, 2004), -which seems to directly address your questions, albeit as of several years ago. -Here is the PDF. -Here is one tidbit. -He mentions, as a measure of our ignorance, that not even this "suspiciously innocuous -conjecture" of -Imre Bárány is settled: - -For any $d$-polytope, $f_k \ge \min \{f_0, f_{d−1}\}$. - -It is (or was in 2004) only proven for $d \le 6$. -Günter has a particularly careful description of what's known about the $f$-vectors of -4-polytopes, a specialty of his. In particular, the set of these $f$-vectors -"is not the set of all integral points in a polyhedral cone, or even in a -convex set." It has concavities and holes.<|endoftext|> -TITLE: Cartan Matrices of type B and C. -QUESTION [5 upvotes]: I was using the built-in functions for Root Systems in SAGE, and I noticed that the Cartan Matrices for Type $B_n$ and type $C_n$ are interchanged from what I thought they would be, i.e. following the Plates in the back of Bourbaki's Lie Groups and Lie Algebras, vol. 4-6. -Are there different conventions for choosing simple roots from these two root systems? I figured that everyone followed Bourbaki in this regard, but would like to hear about any competing conventions. - -REPLY [10 votes]: This question (which I overlooked for a long time) reflects a natural notational confusion but is easy to answer. The Cartan integers themselves are unambiguous for each root system, but the meaning of the two indices used in writing $c_{i,j}$ is conventional and is reversed in some sources. -For types $B,C$ that reversal leads to transposed matrices, which is what you are seeing. Bourbaki and most other sources now write $c_{i,j} := 2(\alpha_i, \alpha_j)/(\alpha_j, \alpha_j)$ relative to a symmetric bilinear form which historically derives from the Killing form; but sometimes this is instead wrutten $c_{j,i}$. For better or worse, there is no secret supercommittee dictating a single choice. -The labels $A, B, \dots$ for the simple Lie algebras or associated root systems are of course conventional too (going back to Killing and Cartan) but have become standard by now in the literature. The subscript (often $\ell$) indicating the Lie algebra rank is conventionally restricted in an arbitrary way to avoid assigning multiple labels to isomorphic Lie algebras such as $B_2, C_2$. (You could decide to write $C_1$ and discard $A_1$, but hardly anyone does this.) -Above rank 2 the series $C_\ell$ is assigned to the Lie algebras of symplectic groups, where there is a unique long simple root. The series $B_\ell$ belongs to Lie algebras of special orthogonal groups in odd dimensions, where there is a unique short simple root. Numbering the simple roots in a Dynkin diagram is again purely conventional (and not always the same in textbooks), but usually the last simple root $\alpha_\ell$ in these cases is the unique one of its length. -Lie theory has developed in multiple directions, leading to many differences in notation and terminology; but fortunately everyone tends to agree on the underlying ideas.<|endoftext|> -TITLE: Constructing rings with a desired prime spectrum -QUESTION [9 upvotes]: Given a partially ordered set $P$, I'm interested in what is known about when $P$ is the prime spectrum of some (not necessarily commutative, not necessarily unital) ring: i.e., when does there exist a ring $R$ having $\mathrm{Spec}(R) \cong P$ (as an order isomorphism). -Obviously some conditions will be needed on $P$, for example, that every descending chain has a greatest lower bound (because the intersection of a descending chain of prime ideals is prime). From Bergman (personal correspondence, and http://math.berkeley.edu/~gbergman/papers/pm_arrays.pdf), every finite partially ordered set can occur as a subset of the prime ideals of some commutative ring. From other as yet unpublished work, a partially ordered set can occur as precisely the prime spectrum of a (non-commutative, not necessarily unital) ring in case: (a) it has the D.C.C., (b) it is chain-finite, and (c) the set of elements covered by any given element is countable. However, none of those conditions are necessary. -Is anyone aware of any more results on this subject? - -REPLY [4 votes]: In H.A. Priestley, ''Spectral Sets'' (1994), a partially ordered set P is called spectral if it occurs as the specialization order of a spectral topology. The cited paper is a survey of the known results: in particular note Theorem 1.1: a poset is spectral iff it is profinite, iff it is the spectrum of a distributive lattice.<|endoftext|> -TITLE: Is there a name for this type of matrix? (Reference Request) -QUESTION [5 upvotes]: I am working on a problem were I encounter matrices of the form -$X = \begin{bmatrix}\frac{1}{1 - a_ib_j}\end{bmatrix}_{ij}$ -I am aware of Cauchy matrices, which have the form -$X = \begin{bmatrix}\frac{1}{a_i - b_j}\end{bmatrix}_{ij}$ -(sometimes written with a plus rather than a minus). Many of the results I need I can actually obtain by factoring the above matrix as a product of a diagonal matrix with a Cauchy matrix (assuming the $a_i \neq 0$), as in: -$X = \mathbb{diag}(a_i^{-1})\begin{bmatrix}\frac{1}{a_i^{-1} - b_j}\end{bmatrix}.$ -These matrices arise when computing solutions to matrix equations of the form -$X - AXB^T = C$ -which are discrete-time analogs of Sylvester equations: -$AX + XB = C.$ -(Also, related are Lyapunov equations and algebraic Riccati equations). It seems that these must appear in the literature somewhere, but I haven't been able to find them. My question is: - - -Do matrices of the form $X = \begin{bmatrix}\frac{1}{1 - a_ib_j}\end{bmatrix}_{ij}$ have a name in the literature? - -Is anyone aware of good references for general results on these matrices? For example, there are general results on the determinant and inverses of Cauchy matrices. - - - -As I mentioned, I have already found a determinant formula and a formula for the inverse of the matrix using the factorization I mentioned above. But it would be helpful to know of further results if they exist and I would like to properly cite the literature as well. - -REPLY [3 votes]: Given two diagonal matrices $D_1,D_2$, matrices such that $\nabla(X):=D_1X-XD_2$ is low-rank are known in literature as Cauchy-like matrices. This includes your case, as $\operatorname{diag}(a_i^{-1})X-X\operatorname{diag}(b_j)$ is rank 1, assuming $a\neq 0$ as you did. -Cauchy-like matrices with displacement rank (i.e., $\operatorname{rk} \nabla(X)$) equals to 1 are basically Cauchy matrices diagonally scaled from both sides, as you realized in your case, and the formulas for determinant and inverse can be adapted with little effort. -Linear systems with all displacement rank-structured matrices, including Toeplitz, Vandermonde, and Hankel, can be solved using the GKO algorithm (Gohberg-Kailath-Olshevsky, '95). -On related subjects, there are a book by Kailath and Sayed, "fast reliable algorithms for matrices with structure" and an old book by Heinig and Rost, but in many cases you may be better off reading directly the papers. -Shameless self-plug: if you are interested in solving linear systems with this kind of matrices, you may also like two recent papers on the subject, one by Aricò and Rodriguez and one by myself, appeared on Numer. Algo. less than one year ago.<|endoftext|> -TITLE: Explicit embeddings of Cappell-Shaneson knots -QUESTION [13 upvotes]: In 1976 Cappell and Shaneson gave some examples of knots in homotopy 4-spheres and for some time these examples were considered as possible counter-examples to the smooth 4-dimensional Poincare conjecture. -In a series of papers, Akbulut and Gompf have shown most of these Cappell-Shaneson knots actually are knots in the standard $S^4$, the most recent reference being this. -In principle, one should be able to work through their arguments to derive a picture of these 2-knots in the 4-sphere. Has anyone done this, for any of the Cappell-Shaneson knots? -I know various people have created censi of 2-knots, does anyone know if any Cappell-Shaneson knots appear in those censi? (I have a hard time accepting censuses as plural of census, sorry, it sounds so wrong!) -I'd be happy with any fairly explicit geometric picture of a Cappell-Shaneson knot sitting in $S^4$. The two I'm most familiar with is the Whitneyesque motion-diagram, and the "resolution of a knotted 4-valent graph in $S^3$" picture. What I want to avoid is the "attach a handle and fuss about and argue that the manifold you've constructed is diffeomorphic to $S^4$" situation. - -REPLY [5 votes]: I think the explicit embedding of Cappell-Shaneson knot is given in the following paper: -S. Akbulut and R. Kirby, A potential smooth counterexample to in dimension 4 to the Poincare conjecture, the Schoenflies conjecture, and the Andrews-Curtis conjecture, Topology 24 (1985) 375--390. (See Figure 16 of that paper) -The paper of Aitchison and Rubinstein mentioned by Scott Carter figures out that there is an error (on the $\mathbb{Z}/2$-framing of $\gamma$-curve which turns out to be 1) in S. Akbulut and R. Kirby's former paper "An exotic involution on $S^4$, Topology 18 (1979) 1--15. Hence, what S. Akbulut and Kirby really showed (in 1979) is that the specific (or the simplest) Cappell-Shaneson sphere is obtained from the Gluck construction of a smooth 2-knot in standard $S^4$. Figure 16 of 1985 topology paper of S. Akbulut and R. Kirby describes that a smooth 2-knot is obtained from gluing two ribbon disks of a knot $8_9$. -Finally, I would like to say that there is a same stuff given in Figure 6.2, page 17 of Kirby's famous book "The topology of 4-manifolds" Springer Lecture notes in Mathematics 1374.<|endoftext|> -TITLE: Two-cardinal models of the random graph -QUESTION [10 upvotes]: For a first-order theory $T$ and cardinals $\kappa < \lambda$, we say that $M$ is a $(\kappa,\lambda)$-model if it is of size $\lambda$ and has a definable (with parameters) subset of size $\kappa$. -1) Let $T$ be the theory of the countable random graph. Which $(\kappa,\lambda)$-models does it admit? -2) For an arbitrary $T$, what are the sufficient conditions for the existence of $(\kappa,\lambda)$ models for some $\kappa < \lambda$? This is not a question about transfer from some $(\kappa,\lambda)$ to a different $(\kappa',\lambda')$, there are quite a few theorems there. What I am asking for is some kind of a non-structure theorem, (apart from having a Vaughtian pair). - -REPLY [8 votes]: MR1889546 (2003e:03064) -Cherlin, Gregory(1-RTG); Thomas, Simon(1-RTG) -Two cardinal properties of homogeneous graphs. (English summary) -J. Symbolic Logic 67 (2002), no. 1, 217–220. -03C30 (03C65 05C99) -The main result of the paper is the following theorem: If G is the Rado graph or the generic $K_{n}$-free graph, and $\kappa \leq \lambda$ are infinite cardinals, then the following are equivalent: (1) $\lambda \leq 2^{\kappa}$; (2) there is a graph $G^{\prime}$ elementarily equivalent to G of cardinality λ and a vertex $v\in V(G^{\prime})$ for which |Δ(v)|=κ; (3) there is a graph $G^{\prime}$ elementarily equivalent to G of cardinality λ and a vertex $v\in V(G^{\prime})$ for which |Δ′(v)|=κ. (Here Δ(v) is the set of neighbors of v in G∗, and Δ′(v) is its complement.)<|endoftext|> -TITLE: p-adic Gross - Zagier -QUESTION [6 upvotes]: Let E be an elliptic curve defined over Q. Suppose it has complex multiplication by an order of an imaginary quadratic extension K/Q and p is a prime of good ordinary reduction. Also, suppose that the sign of functional equation of L(E,s) is -1 and p splits in K. Accordingly, the anticyclotomic Katz p-adic L-function vanishes. Is there a version of p-adic Gross - Zagier formula for the two variable Katz p-adic L-function in this case? The results of Perrin - Riou seem to exclude this choice of imaginary quadratic extension. - -REPLY [2 votes]: Is there complex Gross - Zagier in this case? Their work also excludes this case. If there is, one can expect the p-adic height of that Heegner point to appear in p-adic version, as in the case of Perrin - Riou. There is a paper of Conrad on complex Gross - Zagier which includes this case. However, I do not know whether complex version follows from that paper. Probably some more work in needed.<|endoftext|> -TITLE: Can the twin prime problem be solved with a single use of a halting oracle? -QUESTION [9 upvotes]: It occurred to me that if it were possible to determine whether a given program halts, that could be used to answer the twin primes conjecture -A) Write a program which takes input n and then counts upward until it's found n pairs of twin primes -B) Write a program which for any input n returns true if A halts and false otherwise -C) Write a program which counts upward running B on every n until B returns false -D) If C halts, there are finitely many twin primes, otherwise infinite. -I was wondering if there was a way to do this without nesting halting problems... ie if you only get one chance to ask whether a program halts, is that sufficient to answer the twin primes conjecture - -REPLY [2 votes]: Let $g(3)=4$, and let $g(n+1)=g(n)!$ for every integer $n \geq 3$. -For an integer $n \geq 3$, let $\Psi_n$ denote the statement: -if a system $$S \subseteq \{x_i!=x_{i+1}: 1 \leq i \leq n-1\} \cup -\{x_i \cdot x_j=x_{j+1}: 1 \leq i \leq j \leq n-1\}$$ has at most -finitely many solutions in positive integers $x_1,\dots,x_n$, -then each such solution $(x_1,\dots,x_n)$ satisfies $x_1,\dots,x_n \leq g(n)$. -We conjecture that the statements $\Psi_3,\dots,\Psi_{16}$ are true. -The statement $\Psi_{16}$ proves the implication: -if there exists a twin prime greater than $g(14)$, -then there are infinitely many twin primes, please see: -A. Tyszka, A common approach to Brocard's problem, Landau's problem, -and the twin prime problem, -http://arxiv.org/abs/1506.08655v21 -That is, assuming the statement $\Psi_{16}$, -a simple single query to $0′$ decides the twin -prime problem.<|endoftext|> -TITLE: The Number of Short Vectors in a Lattice -QUESTION [11 upvotes]: Given a lattice $L = \bigoplus_{i=1}^{m} \mathbb{Z}v_i$ (the $v_i$ are linearly independent vectors in $\mathbb{R}^n$) and a number $c > 0$, can one quickly compute or find a good estimate on the number of lattice vectors $v$ with $|v| \leq c$ without actually enumerating these vectors? The basis $v_1,\ldots, v_m$ of the lattice can be assumed to be LLL reduced. -Also asked at: https://cstheory.stackexchange.com/questions/7488/the-number-of-short-vectors-in-a-lattice - -REPLY [9 votes]: For this problem one typically employs the so-called Gaussian heuristic: - -if $K$ is a measurable subset of the - span of the $n$-dimensional lattice - $L$, then $| K \cap L | \approx -> \mbox{vol}(K)/\det(L)$. - -In particular, the case for $K$ a ball is used in some (enumerative) SVP/CVP solvers. See $\S 5$ of -"Algorithms for the shortest and closest lattice vector problems" -by Hanrot, Pujol and Stehlé.<|endoftext|> -TITLE: non-deterministic turing machines -QUESTION [6 upvotes]: I have one simple question: -There is a set, which can be decided in polynomial time by a (one-band) non-deterministic Turing Machine. -Why should there exist one (one-band) non-deterministic Turing Machine, which decides the same set, but with the additional property: There exists one natural number k, such that all the possible calculations last exactly n^k steps, if the input has the length n? -This is one "without-loss-of-generality-assumption" in the proof of a theorem. (namely the proof of the Theorem of Fagin which says: "ESO captures NP") -I cannot understand it. How can we manipulate the first machine to get the second machine? - -REPLY [6 votes]: This is possible, but it is somewhat tricky to do. Here is an outline of one way to do it... -Start with your original one-tape Turing machine $M_0$ which runs in time $\leq k + n^k$ (say) on input of length $n$. -First create a two-tape Turing machine $M_1$ which simulates $M_0$ on one tape and keeps track of a step-counter on the other tape. The counter is initially set to value $k + n^k$ and is decremented at each simulation step. When the simulation of $M_0$ terminates, $M_1$ keeps doing dummy moves until the counter is exhausted. Thus $M_1$ runs in exactly the same time on every input of length $n$. -Finally, we simulate $M_1$ on a one-tape Turing machine $M_2$ as follows. Think of even cells as belonging to the first tape of $M_1$ and odd cells as belonging to the second tape of $M_1$. To keep track of where the two $M_1$ heads, each symbol will now have a plain and a red variant; there will be only two red variants at any given time and they will mark the two head positions. -It is straightforward to simulate $M_1$ on such a tape, but the simpler ways do not simulate each step of $M_1$ in a constant number of steps since switching from one head to the other requires a variable number of moves. To remedy this, first note that $M_1$ uses less than $\ell + n^\ell$ cells of the tape for some $\ell$ that can be effectively estimated from $k$ and the above transformations. When it starts, $M_2$ reads the input length $n$ and places a freshly minted marker on the $(\ell + n^\ell)$-th cell, beyond any cell required to simulate $M_1$. Whenever $M_2$ simulates a step of $M_1$, it proceeds as follows: - -Starting at the base of the tape, $M_2$ finds the appropriate tape head (red symbol in even/odd position). -$M_2$ then performs the appropriate action to simulate $M_1$, these each take a fixed finite amount of steps which may vary from operation to operation. Once this is completed, $M_2$ dances around a little so that it returns to the original tape position exactly 1001 steps after it arrived there. -Then $M_2$ moves right until it finds the marker at position $\ell + n^\ell$ at which point it turns around and returns to the base of the tape. - -Although this is very inefficient, $M_2$ does correctly simulate $M_1$ and it takes exactly the same amount of time to simulate each step of $M_1$. Furthermore, $M_2$ still runs in polynomial time, though the polynomial is much worse than the original $k + n^k$. -Edit: This answer was simplified from its original version.<|endoftext|> -TITLE: Presentation of the dual of a locally free sheaf -QUESTION [7 upvotes]: Let $\mathcal{F}$ be a locally free sheaf of rank $d$ on a scheme $X$ together with an epimorphism $\mathcal{O}_X^n \to \mathcal{F}$. Now due to abstract reasons (Plücker embedding, Serre's results on coherent sheaves etc.) there is a canonical exact sequence of the form $(\wedge^d \mathcal{F})^{\otimes k_2})^{r_2} \to (\wedge^d \mathcal{F})^{\otimes k_1})^{r_1} \to \mathcal{F}^* \to 0$. Here $\mathcal{F}^*$ is the dual of $\mathcal{F}$. Canonical means that the left morphism is a matrix which is built up out of exterior powers and tensor products of the given global sections of $\mathcal{F}$, in a way that does not depend on any other choices. But how can we get this sequence explicitly? I already know that there is a canonical exact sequence ${(\wedge^d \mathcal{F})^{\*}}^{\binom{n}{d+1}} \to \mathcal{O}_X^n \to \mathcal{F} \to 0$. The naive idea just to dualize it does not work since the arrows reverse their direction. - -REPLY [8 votes]: We have that $\mathcal F^\ast$ is, by the pairing induced by the exterior algebra, canonically isomorphic to $\Lambda^{d-1}\mathcal F\bigotimes(\Lambda^d\mathcal F)^{-1}$. Now, in general if $\mathcal H\to\mathcal G\to \mathcal F\to 0$ is exact then the kernel of the surjective map $\Lambda^\ast \mathcal G\to\Lambda^\ast\mathcal F$ is the ideal generated by the image of $\mathcal H\to\mathcal G$. Hence we get an exact sequence $\Lambda^{i-1}\mathcal G\bigotimes \mathcal H \to \Lambda^i \mathcal H\to\Lambda^i\mathcal F\to0$. Applpy, this your second exact sequence and $i=d-1$ gives a presentation of the desired type for $\Lambda^{d-1}\mathcal F$ and then twist it by $(\Lambda^d\mathcal F)^{-1}$.<|endoftext|> -TITLE: How many integer partitions of a googol (10^100) into at most 60 parts -QUESTION [19 upvotes]: [Ed. Prof. Zeilberger has explained why he was asking this question. In joint work with Sills he had developed one approach to this problem, and he asked this question to see how this method compared to the current state of the art. Thus in order to be most useful, answers should explain a technique for computing the number of partitions of a given number and explain how quickly that technique works on large numbers.] -I am offering $100 (one hundred US dollars) for the EXACT number of integer-partitions of -10^100 (googol) into at most 60 parts. The answer has to come by 23:59:59 Sat. July 30, 2011, -by Email to zeilberg at math dot rutgers dot edu . The first correct answer would get the prize. Please have -Subject: MathIsFun; Computational Challenge for p_60(10^100) ; -Of course, the answer should also be posted on mathoverflow, this way people would know that -it has been answered. -P.S. A quick reminder, the number in question is the coefficient of q^(10^100) in -the Maclaurin expansion of -1/((1-q)(1-q^2)(1-q^3) .....(1-q^60)) - -REPLY [24 votes]: Assuming the generating function is $\frac{1}{\prod\limits_{k=1}^{60}{(1-x^k)}}$ less than two hours of gp/pari computations gave the 5738 digit answer: - p_60(10^100) =8665658129496058121317506067907690810670449746661302178926910025719871763923556519131231633949298593177555906325508652883781373910471029788870613159092577714744699298959630542655868431235383293517785430428105143470764278997663335700807300617251380203960562039097153065595769581604737367932463625728210690290264233462109209449502047552084012897582507856352953372122366503001497382374596961383627310640882732762088247858349594835001209127403970206440358528256158848545988681417786477253718312521371110041268740542243735219559805437741158722226945365260877259574975887931865436323796768414323149281986985984914464330319284659786809566298423522107524417899310454745273151188175974610252942077936146950208305359712142054777129737055172167154830250009128689341504240990938697699836664766773027158935849808737752658068106164680562974177529257816592394170551150368212511196569636986953710445489762266194675667371498652916382246757855404724549249551050717009094858672147217608789006718720506884787394455004007091643855132135139651621331225975297728752925468537064391184304710571495304060528408063669250669285238629086134355322407435134171761568475968928692589193141871394457860040452338822816494929065598690526034793724527295934221923030642592976981967471534200917883341671224479566814875977397855060044124854884439474584405233440389974884143853404781671427917179450402213068278128133886074888631094269651824173935463946919156583175528148765197549031057175521775116161406349098756236876932304423257703509865823023301323725597304651961155779648616409778060057276571719204920795464567652279449220006066581065469740876594092289068107157038464825892428406600236959301341703746084498776058131085678449833066601786737992693269490299786717753440153425351607710881599937870599354494839837353122839086205834048615260000891625148794859744730208722244557491813258521891747595610285413967349313587073698779659708909700215433255607668179291196039371728244140000389178949426021051861859839818897331460171931394086450004696357845089414984367843540352636904465457057773654749456040971381734995124757359666631298760847912535308449515616736968604441632050704778867596525781598712358232672809196421364877939384478237688191269698621239514528426949879332666327025164472115582770495588901755094805715060740734051562197455153591280039537892311729803985899227969211389073221861618671596392840187931047118201439871466291153031838790371938412309450352320669649734449049263844616052051534518547524330157773306828886451185220444310597468986757326571766399226684317567992046862377664619275036166006669354885834049707082083633683219958947974849387331766434272679788446162303401710148513833497703947424492810643267157923118530799927508281149342396047693245885907142659818618019387311297456174138924309945010408363078427509179708084319745805159340262819858402289288465915715967326213600747501221164206825858679011198300448711521631658771498318406160062095551004234795120852506560790358461512275441109372033399737210430368615811639016932864756472965460249682202185277262779005114765549708175865370832373106937632921182032989472096429220430851295127939439281538153122864799674892602198058799896426155965678993551590340071403427524539078672094364606931713341547697701999555293069908909550718426056201012721066251368040019559674361896906373686663877282144371346607514752529569184597907739758667247260060154335272296246524887503438022355530758755283307367629529746226993077599437005882544979449909233309857276353684561356784200765972957595010137779247060457314849695869435850085168102102041759463365048815780355529264690328449616470872332057789770565302238678687921549484988721169086700360112352559884155223431076120569772787852453773467746707378504242409991932751304548288869781067598958742982347717887555708860685713547718217218501551923691952189293114417806673447651378755464496005674811769950049961977329424723922866259851293550283610172957874439230396447174624906850482409537526450008478624067323703032568805487931290125614041112601987671756260465669537075425961038012473359629049415093907883302984727773427984810365503563400553701235744097017813404153001038396275891123674408754895724741849538583834405048819312019888365610393794404748239428920071853866017809100158719528417152679821546618439260707086666330301338440712448755243124269924605813531435085248552427206334328177912736194042513280672498954198663396076953530312246730691801641245056403117403391053359019008681498142528625662093549099886614527390539721662425881986030750710577406102939702594329390605090567543535740023230632807356857650179037912011766789897404394418744798138754987896071320293294777510492110155939657693001085689240339237690313204919679817972605450670368867825700290648437593017433724334771350960557786590484593378054535217606554074698984391044826359846474388374343927299855144027314559591847196268084725788918764471131659745679233559000524730322435952188811312708185043575430917760399760781980251659586238043368238476862951546654226201788124366414677601234034696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- -I am not sure this is correct at all, yet Doron seems happy. -EDIT: Here is how the number was computed. The generating function $\frac{1}{\prod\limits_{k=1}^{60}{(1-x^k)}}$ means the sequence satisfies linear recurrence with constant coefficients. These are known to be efficiently computable assuming arithmetic operations in the range of the result are tractable. A good computational resource for recurrences is the free book "Matters Computational" was: "Algorithms for Programmers" by Jörg Arndt. Basically the method is fast binary exponentiation of a matrix or in $\mathbb{Z}[x]/poly(x)$. The book has code parts of which I used. Got a linear recurrence of order 1830. My gp/pari code is here. A curiosity of the challenge is the result is so small - I wouldn't even try $fibonacci(10^{100})$. -To my knowledge the monetary bounty was donated to Wikipedia by Doron Zeilberger per agreement with the recipient. -EDIT2 A closed form possibly leading to faster approach (avoiding computing the recurrence) appears in the paper -A GENERAL METHOD FOR DETERMINING A CLOSED FORMULA FOR THE NUMBER OF PARTITIONS OF THE INTEGER $n$ INTO $m$ POSITIVE INTEGERS FOR SMALL VALUES OF $m$, W. J. A. COLMAN - -REPLY [4 votes]: Can you do p_60(10^1000)? p_60(10^10000)? – Doron Zeilberger -8.6656581294960581213175060679076908106704497466613.. * 10^5737 Dollar 100 -8.6656581294960581213175060679076908106704497466613.. * 10^58837 Dollar 1000 -8.6656581294960581213175060679076908106704497466613.. * 10^589837 Dollar 10000<|endoftext|> -TITLE: Finite dimensional vector spaces over a complete but not-necessarily-valued field -QUESTION [7 upvotes]: I'm essentially reopening this old question of Ricky Demer which was never fully answered. -Essentially the original question: Suppose we have a topological field $F$ which is complete, Hausdorff, and non-discrete, and we put a Hausdorff topology on $F^n$ so as to make it a topological vector space over $F$; is is this topology necessarily the product topology (and hence complete, and hence closed in anything it embeds in)? -As the link shows, the answer is yes if $\tau$ comes from an absolute value on $F$, and it's easy to see the same argument works if it comes from a field ordering on $F$. -Note that the argument there shows that this question reduces to the following lemma: -Suppose we have a topological field $(F,\tau)$ which is Hausdorff and non-discrete, and we give $F$ a second topology $\tau'$, which is also Hausdorff, such that $(F,\tau')$ is a topological vector space over $(F,\tau)$. Does this force $\tau=\tau'$? What if we assume that $(F,\tau)$ is complete? -(I'm isolating completeness as a separate, possibly-unnecessary condition because the argument there only uses completeness in the reduction to the lemma, not in proving the lemma for valued fields.) -Related to this question in that one way to come up with a counterexample for both simultaneously would be to find a field with two (nondiscrete, Hausdorff) topologies with one strictly finer than the other. - -REPLY [2 votes]: Well, now I feel silly -- on looking through Wieslaw again, I see he does give examples of non-discrete, non-straight fields, just not in that section. For instance, take two absolute values on the rationals; the topology they generate together still make the rationals a topological field, is not discrete, and is obviously finer than either of the ones you started with. Since it isn't even minimal, it can't be straight. -I'm not going to accept my own answer on this since this example presumably isn't complete, and I'd like a complete example. Which might well be in here too, if I keep looking... (Is the completion of this again a field? If so that should work, but I'd need to check if that's true.) -Edit: Nope, the completion of this isn't a field, so I'm still lacking for a complete example. -Edit again: OK, I'm now pretty sure Wieslaw gives one, so the answer is no. Wieslaw gives an example of a complete normed field which is not given by any absolute value (here "normed" means instead of |ab|=|a||b|, we only require |ab|≤|a||b| and |-a|=|a|). Furthermore, he shows given a power-multiplicative norm, you can find an absolute value that generates a coarser topology (so if a power-multiplicative norm wasn't equivalent to an absolute value, it isn't minimal). (Here power-multiplicative only means we require |an|=|an| for positive n, not for negative n.) And after a bit of staring at his example, I'm pretty sure it's power-multiplicative. So unless anyone can show that I've missed something, I'm going to consider this one closed.<|endoftext|> -TITLE: Ph.d thesis assessment -QUESTION [5 upvotes]: Hi! -I would like to put my thesis (in Poisson geometry) online for an assessment, so I can receive feedback to improve the manuscript. Can you help me to find the right place to my request? -Thank you. -Another related question, may one attach some pdf files to questions here? - -REPLY [2 votes]: Hello, -With the help of a nice person, I posted my thesis on Arxiv: -http://arxiv.org/abs/1108.0452 -I will be grateful to any person who could send me comments to improve the manuscript.<|endoftext|> -TITLE: non-rigidity of interior points in polyhedral triangulations? -QUESTION [7 upvotes]: It's well-known that any compact polyhedron $P$ in $\mathbb{R}^n$ (we talk about piecewise-linear setting there, i.e. $P$ is a finite union of compact convex polytopes) can be triangulated into (geometric) simplices, although sometimes it is necessary to add "extra points" in $P$ to serve as vertices of simplices in the triangulation $\mathcal{T}$. E.g. Schönhardt polyhedron requires such extra points. (Here By $\mathcal{T}$ we mean a partition of $P$ into finitely many simplices $T\in\mathcal{T}$ --- more precisely, the interiors $int(T)$ of $T$'s do not intersect, and the closure of $\cup_{T\in\mathcal{T}}int(T)$ equals $P$. -The vertices of $T$'s that are not vertices of $P$ are these "extra points" we talk about.) -It looks correct that one can always construct such a $\mathcal{T}$ so that each extra point in it is "non-rigid", i.e. it can be continuously moved inside an open subset of the face of minimal dimension it is inserted into, so that after such a deformation $\mathcal{T}$ remains a triangulation of $P$. -Is this indeed correct, and can anyone point out a reference? -Added: A weaker form of the question: show that each extra poing in $\mathcal{T} $ is not prescribed, i.e. for any vertex $y$ of $\mathcal{T}$ which is not a vertex of $P$ there exists another triangulation of $P$ which does not have $y$ as a vertex. [This still suffices for our purpose, of showing that a part of certain kind of moment generating functions, for moments of a uniform measure supported on $P$, does not depend upon $\mathcal{T}$ ]. -This is easy to see that $y$ lying in the interior of $P$ is not prescribed---one can directly construct a new triangulation not using $y$, by choosing the points of intersection of the edges on $y$ with a sufficiently small sphere around $y$ and re-triangulating new convex pieces without using $y$. But it is not obvious for $y$ lying in a proper face of $P$. - -REPLY [2 votes]: The answer for arbitrary polyhedra is no. If a 4-dimensional polyhedron has a 3-dimensional Schönhardt polyhedron as one of its faces, there will need to be a new vertex added somewhere within that face, which will not be free to move in an open set. -I believe that the answer is yes in 3d and yes to higher-dimensional polyhedra all of whose faces are already simplices, but I don't have a proof handy.<|endoftext|> -TITLE: Entropy of the Ising model -QUESTION [8 upvotes]: Consider the standard Ising model on $[0,N]^2$ for $N$ large. By that I mean the square-lattice Ising model without external field, inside an $N$-by-$N$ square. What is its entropy for $N$ large? It must behave asymptotically as $c(\beta)N^2$ for some constant $c(\beta)$ depending on the inverse temperature $\beta$. What is $c(\beta)$? Has it been computed? - -REPLY [6 votes]: To expand on Steve Huntsman's comment, the entropy follows from Onsager's result for the free energy per site, $F=$ -$$ --\beta^{-1}\left[\ln 2+ \frac{1}{2}\frac{1}{(2\pi)^2}\int_0^{2\pi}d\theta_1\int_0^{2\pi}d\theta_2 \ln(\cosh2\beta E_1\cosh2\beta E_2 --\sinh2\beta E_1\cos\theta_1-\sinh2\beta E_2\cos\theta_2)\right], -$$ -and the thermodynamic relation, -$$ -S=-\frac{\partial F}{\partial T}, -$$ -for the entropy per site. Here $\beta=1/(k_BT)$ and $E_1$ and $E_2$ are the horizontal and vertical interaction strengths. If you set both interaction strengths equal to 1 and use units where Boltzmann's constant equals 1, then the critical temperature is $2/\ln(\sqrt2 + 1)\approx2.269$. If you plot $S$, you should find that it interpolates between 0 at low temperature and $\ln2$ at high temperature, as expected. At the critical temperature, the graph has infinite slope.<|endoftext|> -TITLE: Probability that a Turing machine is universal? -QUESTION [11 upvotes]: I choose a Turing machine T with n states and an input tape at random. -What can be proven about the probability P_A(n) that it is not decidable whether T will halt for a particular input? What can be proven about P_B(n) that T is universal, that means that there exists an algorithm that takes an obvious encoding of an arbitrary Turing machine A (without input tape) and transforms it into an input for my random Turing machine, so that T halts with this input if and only if A halts? In particular: Is it known that P_A(n) is strictly smaller than P_B(n) for some n? - -REPLY [19 votes]: The answer will depend on which model of Turing machine you -have adopted. -For example, here is one easy thing to say. Suppose that -your Turing machines have alphabet $\{0,1\}$, a set $Q$ -of $n$ states, a single halt state (not counted inside -$Q$), and the ability to move the head left and right, so -that a program is a function $p:Q\times\{0,1\}\to -(Q\cup\{\text{halt}\})\times\{0,1\}\times\{\text{left},\text{right}\}$, -where $p(s,i)=(t,j,\text{left})$ means that when in state -$s$ reading symbol $i$, the program changes to state $t$, -writes symbol $j$ and moves left. In particular, for this -model the total number of programs is $(4(n+1))^{2n}$. -Consider now the collection of programs that have no -transition to the halt state. The total number of such -programs is $(4n)^{2n}$, and the interesting thing here is -that $$\lim_{n\to\infty}{(4n)^{2n}\over (4(n+1))^{2n}}=\lim_{n\to\infty}[{n\over -n+1}]^{2n}=\frac{1}{e^2}$$ which is about 13.5%. -Thus, for this model of computation, at least 13.5% of the -programs never halt on any input and are not universal. -The topic is fun, because we are in effect considering the -behavior of a random program, where each new program line -is chosen randomly from among all the legal program lines. -And such kind of argument is the main theme of my article -(J. D. Hamkins and A. Miasnikov, The halting problem is -decidable on a set of asymptotic probability one, Notre -Dame J. Formal Logic 47, 2006. -http://arxiv.org/abs/math/0504351), which came up also in a -few other mathoverflow questions: Solving NP problems in -(usually) polynomial -time?, -Turing machines the read the entire -tape?. -The main theorem of that article is the following. -Theorem. There is a set $A$ of Turing machine -programs (for machines with one-way infinite tape, single -halt state, any finite alphabet) such that: - -One can easily decide whether a program is in $A$; it is polynomial time decidable. -Almost every program is in $A$; the proportion of all -$n$-state programs that are in $A$ converges to $1$ as $n$ -becomes large. -The halting problem is decidable for members of $A$. - -Thus, there is a decision procedure to decide almost every -instance of the halting problem. The way the proof goes is -to calculate, for any fixed input, the probability that a -Turing machine will exhibit a fatally trivializing behavior -(falling off the left end of the tape before repeating a -state), and observing that in fact this occurs with -probability $1$. Basically, the behavior of a random Turing -machine is sufficiently close to a random walk that one can -achieve the Polya recurrence phenomenon. -A corollary to this proof, answering your question, is that -for this model of computation, the probability that an -$n$-state program is a universal program goes to zero as -$n$ becomes large, since almost every program exhibits the -trivializing behavior, which is incompatible with being -universal. Furthermore, the set of programs with that -behavior is decidable. -The theorem can be extended to other models of computation, -such as the model with two-way infinite tapes and halting -determined by specifying a subset of the states to be -halting states. In this model, as you can guess, machines -are likely to halt very quickly (since each new state has -50% chance of being halting), and so there is a large set of programs for which the halting problem -is decidable for this reason (they halt before they repeat a state). -Let me also mention, since you asked not merely about the -probability of halting, but also about the probability of -decidability of halting, that every computably enumerable -set $B\subset\mathbb{N}$ that is not computable admits -infinitely many $n$ for which $n\notin B$ but this is not -provable in whatever fixed background theory you prefer, -such as PA or ZFC or ZFC + large cardinals. The reason is -simply that if non-membership in $B$ were provable for all -sufficiently large $n$, then we would have a decision -procedure for $B$ by searching either for $n$ to be -enumerated into $B$ or else searching for a proof that $n$ -is not in $B$, and this contradicts our assumption that $B$ -is not decidable. -Thus, every c.e. non-computable set sits in a halo of -undecidability: there are infinitely many numbers $n$ that -are not in $B$, but for which it is also consistent with -your favorite theory that they are in $B$.<|endoftext|> -TITLE: Regular Conditional Probability given a natural filtration of a stochastic process -QUESTION [5 upvotes]: OK, this is kind of re-posting, but I think I can clarify the question more, so it's worth a shot. -Consider a real valued process $(X_t)_{t \leq T}$, cadlag on a probability space $(\Omega, (\mathcal{F}^\circ_t)_{t \leq T}, \mathbb{P}). \mathcal{F}^\circ_t=\sigma(X_s;s\leq t)$ is the uncompleted, natural filtration generated by $X_t$. Unfortunately $X_t$ neither has independent increments, nor is it markov. Since $\Omega$ is a Polish space, $\mathcal{F}^\circ_T$ and also $\mathcal{F}^\circ_t$ are countably generated, so we know, there exists a regular version of the conditional probability of $\mathbb{P}$ for any fixed $t$ for $\mathbb{P}$-a.a. $\omega$, i.e. for fixed $t$, $\mathbb{P}(\cdot|\mathcal{F}_t)(\omega)$ is a prob. measure f.a.a. $\omega$. -Hence we know, that for all $t\in [0,T]\cup \mathbb{Q}$, we find a regular conditional probability f.a.a. $\omega$, depending on $t$. In words: Given almost any path of the process up to time $t$, we can deduce the probablity of events, taking that information into account. -On the remaining $\omega$'s, define some meaningless measure, so we have a measure $\forall \omega$. How can I extend this to all $t$ in a reasonable way? Reasonable means: There is one Null set $N$, so that $\forall t$ $\mathbb{P}(\cdot|\mathcal{F}_t)(\omega)$, $\omega\in N^c$, is a measure Anybody seen anything like this? -I read something like this only for Markov and Feller processes using infinitesimal generators, but this cannot be carried over one to one, because we do not have a transition semigroup. -Maybe I have a deep misunderstanding here. Grateful for any objections, hints and comments. - -REPLY [6 votes]: Let's assume that we are working with the canonical probability space $\Omega = D(\mathbb R)$ of càdlàg functions, and $\mathbb P$ is the law of the process. I would doubt that there is a satisfactory answer at the level of maximal generality you've stated. At the very least, the measure $\mathbb P$ should be Radon. There are extremely general results on the existences of RCPs for Radon measures (cf. Leão, Fragoso and Ruffino, Regular conditional probability, disintegration of probability and Radon spaces). -The RCP is a measure-valued function $P : [0,T] \times \Omega \to \mathcal M(\Omega)$ such that for $\mathbb P$-almost every $\omega$, the measure $P(t,\omega, \cdot)$ is a version of $\mathbb P(\cdot|\mathcal F_t)$. Do you want the function $(t, \omega) \mapsto P(t,\omega,\cdot)$ to simply exist and be measurable? If so, this can be done in the wide generality stated above; see Leão et al. -Recently, I have needed more regularity properties for RCPs, namely, continuity. Consider the space $\mathcal M(\Omega)$ of Radon measures on $\Omega$ equipped with the topology of weak convergence of measures. We say that the RCP is a continuous disintegration (or continuous RCP) when it satisfies the following property: $$\mbox{if $\omega_n \to \omega$, then the measures $P(t,\omega_n,\cdot)$ converge weakly to $P(t,\omega,\cdot)$.}$$ -If the law is Gaussian, then my preprint Continuous Disintegrations of Gaussian Processes gives a necessary and sufficient condition for the law $\mathbb P$ to have a continuous disintegration. I haven't thought about this in the case of càdlàg functions, but I'm pretty sure that this will extend easily. Note that this is just for fixed $t$. -To show that the map $(t, \omega) \mapsto P(t,\omega,\cdot)$ jointly continuous, a little more work is needed. As part of a larger project, Janek Wehr and I have general results in this direction for stationary, Gaussian processes. If this is what you need, I'm happy to discuss this with you further. -Open Question: If the law $\mathbb P$ is not Gaussian but at least is log-Sobolev, then all the same results should hold. This is because log-Sobolev measures satisfy very strong concentration-of-measure properties. I have some ideas how to do this, but I haven't worked out the details because I've been busy with other projects. If anybody is interested in collaborating on extending this work to the log-Sobolev case, please contact me.<|endoftext|> -TITLE: Algebraic axiomatization for AB+BA^T operation on matrices -QUESTION [9 upvotes]: Let us consider a matrix algebra $Mat_{n\times n}(K)$, where $K$ is a field, $char K \neq 2.$ -It is well-known that the axiomatization of commutator operation $[A,B]=AB-BA$ on matrix algebra leads us to the theory of Lie algebras. -Axiomatization of $A\circ B= \frac{1}{2}(AB+BA)$ leads us to Jordan algebras. -Let us consider an operation $A \Box B= \frac{1}{2}(AB+BA^T),$ arising, for example, in control. -How we can describe a class of algebras arising from axiomatization of such an operation? -UPDATE As it was shown by Pasha Zusmanovich below considering only $\Box$ leads us to a trivial variety(of course we can try to proceed to a quasivariety..). -But, if we add a transposition to the signature situation becomes much more interesting. -First of all we have $(A\Box B)^T=A\Box B^T$ and the left unit $I\Box A= A$ -Really, if we consider $T$-invariant subalgebras of some matrix algebra with $\circ$, than we can note that such algebras could be decomposed (as vector spaces) to the direct sum of Jordan algebra and Lie algebra -- symmetric and antisymmetric part,respectively. -Axiomatizing this decomposition we get... -Commutativity for symmetric part: -$$ -(A+A^T)\Box (B+B^T)=(B+B^T)\Box (A+A^T), -$$ -Power-associativity for symmetric part: -$$ -(A+A^T)\Box ((A+A^T)\Box (A+A^T))= ((A+A^T)\Box (A+A^T))\Box (A+A^T) -$$ -Jordan identity for symmetric part: -$$ -((A+A^T)\Box (B+B^T))\Box ((A+A^T)\Box (A+A^T))= (A+A^T)\Box ((B+B^T)\Box ((A+A^T)\Box (A+A^T))) -$$ -Anticommutativity for antisymmetric part: -$$ -A\Box A+A^T\Box A^T=A\Box A^T+A^T\Box A -$$ -Lie identity for antisymmetric part: -$$ -(A-A^T)\Box ((B-B^T)\Box (C-C^T))+(C-C^T)\Box ((A-A^T)\Box (B-B^T))+(B-B^T)\Box ((C-C^T)\Box (A-A^T))=0 -$$ -Commutativity and power-associavity for symmetric part could be seen as averaged commutativity and averaged associativity and(!) commutativity, respectively. -$$ -A\Box B + A\Box B^T + A^T\Box B + A^T\Box B^T= B\Box A + B^T\Box A + B\Box A^T +B^T\Box A^T -$$ -$$ -\sum_{\sigma\in S_3}\sum_{(i,j,k)\in (\varnothing,T)^3}A_{\sigma(1)}^{i}\Box(A_{\sigma(2)}^j\Box A_{\sigma(3)}^k) -=\sum_{\sigma\in S_3}\sum_{(i,j,k)\in (\varnothing,T)^3}(A_{\sigma(1)}^{i}\Box A_{\sigma(2)}^j)\Box A_{\sigma(3)}^k -$$ -Did anyone consider something close to varieties of algebras with identities of that type? - -REPLY [3 votes]: Including the operation $A\mapsto A^T$ can be viewed, in the language of operads, in many different closely related ways: via adjoining a new unary operation $J$ that satisfies $J^2=id$ and $J(ab)=J(b)J(a)$; via considering operads over the semisimple algebra $\mathbb{C}[t]/(t^2-1)$, via splitting everything into symmetric/antisymmetric, and considering the corresponding coloured operads (this looks like what you are doing in the examples of identities you give) etc. For either approach, you will find some papers dealing with similar things, though not necessarily literally the structure you are asking about. One way to try and list the possible identities would be to use operadic Groebner bases, - this way, for example, it is possible to show that for pre-Lie algebras the symmetrised operations does not satisfy any identities (http://arxiv.org/abs/0907.4958), but if there are identities, then one can detect them too, using an appropriate ordering. (It's like Groebner bases in the case of commutative algebras: if you are solving a system of polynomial equations and suspect that on all solutions one of coordinates $z_i$ has finitely many values, Groebner bases can detect that, and produce an equation in one variable that $z_i$ satisfies.)<|endoftext|> -TITLE: Is the singular locus ideal preserved by all derivations? -QUESTION [17 upvotes]: Let $R$ be a commutative ring, with whatever hypotheses let you answer the question -- e.g. Noetherian, local, finitely generated over $\mathbb C$. -Let $I$ be the ideal defining the singular locus in Spec $R$. (With the reduced subscheme structure, or defined using minors of a Jacobian matrix, again whatever helps.) -Is it obvious and/or true that any derivation $d:R \to R$, i.e. additive map satisfying the Leibniz rule $d(ab) = a\ db + b\ da$, has $dI \leq I$? -Morally, $d$ is defining an infinitesimal automorphism of Spec $R$, and the singular locus should be preserved by automorphisms. So I would have hoped that there was a mindless proof using the Jacobian, but I haven't found one. As usual, a reference would be even better than a proof. - -REPLY [12 votes]: The fact that the set-theoretic singular locus is preserved is true only in characteristic zero: for example if you take the curve ($x^p = y^2$) in $\mathbb{A}^2$, then the ideal $(x, y)$ is not preserved by the derivation $d/dx$ in characteristic $p$. On the other hand, the Jacobian ideal in this case, $(y)$, is preserved. -In characteristic zero one can prove that the set-theoretic singular locus is preserved by exponentiating derivations: given a derivation $D$ of $\mathcal{O}_X$ one can consider $e^{t D}$, a derivation of $\mathcal{O}_X[[t]]$, and the composition of this with a character $\mathcal{O}_X \to k$ corresponding to a singular point gives a character $\mathcal{O}_X((t)) \to k$ also with the same dimension of tangent space, hence corresponding to a singular point. Therefore the singular locus is preserved, since the set-theoretic singular locus is the intersection of all such kernels. -This theorem is due to, I believe, Seidenberg: it is the corollary to Theorem 12 in "Differential ideals in rings of finitely generated type" in AJM 1967.<|endoftext|> -TITLE: Expressing Galois actions on fundamental groups explicitly -QUESTION [6 upvotes]: Let $X$ be some variety over $\mathbb{Q}$, and let $\pi_1(X\times_{\mathbb{Q}}\mathbb{C},x)$ denote its (topological) fundamental group. As is well known $Gal(\mathbb{Q})$ acts on this fundamental group. I was browsing old MSRI videos, and in the middle of one of them I saw an intriguing explicit description of this action: -http://www.msri.org/realvideo/ln/msri/1999/vonneumann/schneps/1/main/08.html -(you don't have to know anything from earlier in the talk to understand that page) -As it says there, there was also a talk by Ihara about this. I'm looking, however, for an explanation of this in a more systematic way, in a paper or a book. Do you know of a good reference for this? - -REPLY [4 votes]: Aha, now I think I have a better picture of what you're looking for. I would look at Matsumoto's notes from the Arizona Winter School program on Galois groups and fundamental groups: -http://math.arizona.edu/~swc/notes/files/05MatsumotoNotes.pdf -especially sections 2.2 and 4.1.<|endoftext|> -TITLE: Probabilistic (and other mathematical) methods of physics without the physics? -QUESTION [13 upvotes]: Many of the methods of physics are vastly more general than their use in that discipline. For example, information theory overlaps with a lot of statistical mechanics, and the latter actually developed first. ET Jaynes wrote a famous paper illustrating the connections. However, each is comprehensible without the language and intuition of the other (though I do not deny that a richer understanding comes from knowing both). -What other methods of physics (particularly those with a statistical or computational bent) have interpretations (Please mention useful introductory texts!) that are completely physics free? I understand that various field theories meet this criterion; any good non-physics introductions? - -REPLY [2 votes]: I found Complexity and Criticality by Christensen and Moloney to by quite excellent. It gives a much more computational approach to the percolation phase transition, the ising phase change and issues of self organized criticality (via the sand pile and rice pile model). -As a computer science student, I found this book to be invaluable for my work on phase transitions of NP-Complete problems.<|endoftext|> -TITLE: Are there examples of statements that have been proven whose consistency proofs came before their proofs? -QUESTION [35 upvotes]: I'm wondering if there are examples of statements that have been proven whose consistency proofs came before the proofs of the statements themselves. -More informally, I'm wondering how promising in general is the approach of attempting a consistency proof for a statement when faced with a statement that seems true but difficult to prove. -Background: -If a statement is provable from a set of axioms, then that statement is obviously consistent (assuming the set of axioms is consistent). So provability is stronger than consistency. This might lead one to think that constructing a consistency proof for a statement should be strictly easier than constructing a proof. -Yet consistency proofs (at least oft-cited ones, for example those by Godel and Cohen about the Continuum Hypothesis) seem to require a high level of sophistication (though this might be a byproduct of the fact that consistency proofs like these are for the special class of statements that cannot be proven). -For statements that can be proven then, are there cases where their consistency proofs are easier or came before the proofs themselves? -Update: -Thanks a lot, everyone, for the great answers so far. The number and existence of these examples is interesting to me, as well as the fact that they all rely on the same technique of first proving something using an additional axiom (an approach first suggested by Michael Greinecker). That hadn't occurred to me. I wonder if there are other approaches. - -REPLY [6 votes]: Let $c(X)$ denote the cellularity of the topological space $X$, that is the supremum of the cardinalities of its families of pairwise disjoint non-empty open subsets. -In the early 60's Kurepa asked: - - -Is there a compact Hausdorff space $X$ such that $c(X) -TITLE: Modules over Laurent series rings -QUESTION [9 upvotes]: Let $k[x]$ be the ring of polynomials over a field k in one variable x. A $k[x]$-module is a k-vector space together with a linear endomorphism (the action of x). -The field $k(x)$ of rational functions is the maximal localization of $k[x]$, i.e. field of fractions. (Edit: yes, I was an idiot in what I wrote here first; thanks James.) -The ring $k[[x]]$ of formal power series is the completion of $k[x]$ at the ideal $(x)$. It is natural to consider only $k[[x]]$-modules which are likewise complete, meaning roughly that we can sum "infinite linear combinations" whose coefficients are increasing powers of x. Completeness of a $k[x]$-module automatically makes it a $k[[x]]$-module. -Finally, both $k(x)$ and $k[[x]]$ embed into the ring $k((x))$ of formal Laurent series. I have two questions, which I ask together because they seem related: - -Is there a general ring-theoretic construction, akin to localization and completion, which produces $k((x))$ from $k[x]$? -Is there a natural condition to impose on $k((x))$-modules, akin to completeness for $k[[x]]$-modules, which would enable us to sum infinite linear combinations with coefficients increasing in powers of x? - -REPLY [2 votes]: Sorry, the first version of this answer was broken in a few ways. -For your first question, it seems that there is more than one construction that specializes to what you want. For example, you can take the completion $\hat{X}$ of a variety $X$ along a closed subvariety $Z$, and then take the tensor product $\mathscr{O}_{\hat{X}} \otimes_{\mathscr{O}_X} K_X$, where $K_X$ is the function field. Alternatively, if you have an effective divisor $D$ in a variety $X$, you can take the scheme whose underlying topological space is $D$, and whose sheaf of rings is given by $U \cap D \mapsto \varinjlim_m \varprojlim_n \Gamma(U, \mathscr{O}_X(mD)/\mathscr{O}_X(-nD))$. These two constructions are identical when we are presented with a codimension one subvariety, such as a point in a line. I do not know a succinct name for either construction. -For your second question, you can ask for the modules to have a $k$-linear topology, i.e., there is a basis of neighborhoods of zero formed by $k$-submodules. We can then demand that the action of $k((x))$ is continuous, where $k((x))$ is given the "usual" topology, with $k$ discrete and $ \{ x^n k[[t]] \}_{n \in \mathbb{Z} } $ forms a basis of neighborhoods of zero. A continuous $k((x))$-module is the same as a topological $k$-module equipped with a continuous invertible topologically nilpotent endomorphism.<|endoftext|> -TITLE: Which complete Boolean algebras arise as the algebras of projections of commutative von Neumann algebras? -QUESTION [21 upvotes]: Projections in an arbitrary commutative von Neumann algebra form a complete Boolean algebra. -Moreover, a morphism of commutative von Neumann algebras induces -a continuous morphism of the corresponding complete Boolean algebras. -Thus we have a fully faithful functor F from the category of commutative von Neumann algebras -to the category of complete Boolean algebras and their continuous morphisms. -The category of complete Boolean algebras and their continuous morphisms is a full subcategory -of the opposite category of the category of locales. -Thus the functor F can be seen as implementing the Gelfand-Neumark duality -for commutative von Neumann algebras. -However, to obtain a satisfactory statement of the duality we still need to characterize -in topological terms objects in the essential image of F, -which we call measurable spaces (or locales, think of a localic version of point-set measurable spaces). -What additional topological conditions do we need to impose on a complete Boolean algebra -to ensure that it is the algebra of projections of some von Neumann algebra, -i.e., a measurable space? -It is relatively easy to pin down non-topological conditions. -For example, a complete Boolean algebra comes from a von Neumann algebra -if and only if it admits sufficiently many normal positive measures. -The reason for requiring additional conditions to be topological -is that the resulting definition of a measurable space should be easy -to relate to other parts of general topology. -For example, consider the forgetful functor that sends -a commutative von Neumann algebra to its underlying C*-algebra. -Applying the Gelfand-Neumark duality to both sides we obtain -the forgetful functor from the category of measurable spaces -to the category of compact regular locales -(or compact Hausdorff spaces, if we have the axiom of choice). -A topological definition of a measurable space should allow -for an explicit description of this forgetful functor in terms of open sets. -Other potential applications include functors that send a locale (or a topological space) -to its underlying measurable space, or a smooth manifold to its underlying measurable space. -More speculatively, one could use this definition to replace -ad hoc techniques of classical point-set measure theory with standard tools of general topology. - -REPLY [3 votes]: A pair $(\mathcal{A}, \mu)$ of a $\sigma$-complete boolean algebra $\mathcal{A}$ and a functional $\mu : \mathcal{A} \to [0, \infty]$ is called a measure algebra if $\mu$ is strictly positive and countably additive on disjoint sequences. A measure algebra is semifinite if whenever $\mu(a) = \infty$ there exists a $b < a$ such that $0 < \mu(b) < \infty$. A measure algebra is localizable if it is complete and semifinite. -A measure algebra can be constructed from a measure space by taking the boolean algebra of equivalence classes of measurable sets modulo the null sets. In the reverse direction, by the Loomis-Sikorski Theorem, every $\sigma$-complete boolean algebra is isomorphic to the quotient of the $\sigma$-algebra $\{ A \bigtriangleup B : A \text{ clopen }, B \text{ meager } \}$ by the $\sigma$-ideal of meager sets. An ordinary measure can be defined on the quotient in the natural way, and the concrete measure algebra of the resulting measure space is isomorphic to the original measure algebra. -The preceding construction can also be used to show that the localizable measure algebras are precisely the Boolean algebras of projections of commutative von Neumann algebras. Note the similarities between the definitions of a localizable measure algebra and a normal semifinite weight on a von Neumann algebra. -Due to the semifiniteness of the measure, the problem of characterizing the Boolean algebras that are measure algebras can be reduced to the case of finite measure. Call a $\sigma$-complete boolean algebra $\mathcal{A}$ finitely measurable when there exists a functional $\mu : \mathcal{A} \to [0, \infty)$ making it a measure algebra. Then a complete boolean algebra $\mathcal{A}$ has a functional $\mu : \mathcal{A} \to [0, \infty]$ making it a localizable measure algebra precisely when the set $\{ a \in \mathcal{A} : \mathcal{A}_a \text{ is finitely measurable } \}$ is order-dense in $\mathcal{A}$, where $\mathcal{A}_a$ is the principal ideal generated by $a$. -Unfortunately, even in the finite measure case there is no great solution to this problem. There’s a characterization by Kelley that reduces it to the existence of a strictly positive finitely additive measure and a combinatorial condition (what he calls weakly countably distributive). He also characterizes the existence of a finitely additive measure in terms of intersection/covering numbers. Gaifman wrote a survey paper on this problem, and Jech proved a game-theoretic characterization. -A good reference for most of the facts mentioned about measure algebras is volume 3 of Fremlin’s tome Measure Theory, particularly chapters 32 and 39.<|endoftext|> -TITLE: Projectivity of blowups -QUESTION [6 upvotes]: If I have a quasiprojective variety $X$, and a subscheme $Z$, then the blowup -$$f:Y = Bl(X,Z)\rightarrow X$$ -is projective over $X$, since it is constructed by a relative Proj construction. Can I find a relatively ample bundle on $Y$ that is trivial on -$$f^{-1}(X\backslash Z)?$$ -At first I thought the construction of $O(1)$ in the Proj construction guarantees this property. However, if I have a projective small contraction $Y \rightarrow X$, then it is a blowup map, but then certainly an ample bundle can't be trivial on the birational locus, since it would be trivial on $Y$. -So under what conditions can I find such a relatively ample bundle? -Edit: Karl's answer explains away my confusion: I have to blow up more than the image of the exceptional locus in case of a small contraction. It's still reasonable to ask when it's trivial on the birational locus, in which case Sandor's answer applies. - -REPLY [2 votes]: anon, is your definition of relatively ample local over the base? (There are some different definitions out there, but most people seem to use the one that is local over the base now, compare for instance Ravi Vakil's notes with Hartshorne). -If your definition of relatively ample is local over the base, then there is nothing to show. If you blow-up $Z$, say with ideal sheaf $I_Z$, then $I_Z \cdot O_Y$ is relatively ample. -With respect to a small contraction $\pi : Y \to X$ (at least with $X$ normal), you can NEVER get one of these just by blowing up a scheme supported on the locus where $\pi$ is an isomorphism. In general, the way you get a small resolution, is by blowing up a Weil divisor that is not Cartier. Of course, on a normal variety a Weil divisor is Cartier except on a small close set, and blowing up a Cartier divisor obviously doesn't do anything. -For fun, I'd suggest that you try blowing up the ideal $(x, u)$ on the scheme $\text{Spec} k[x,y,u,v]/(xy-uv)$. $(x,u)$ is a non-Cartier divisor, but it's still a divisor (it's also not $\mathbb{Q}$-Cartier, as Sándor effectively points out above). Blowing it up gives a small resolution.<|endoftext|> -TITLE: Do mapping classes have gonality? -QUESTION [18 upvotes]: (This question was discussed by people at the PCMI workshop on moduli spaces, without any clear resolution, so I thought I'd throw it open to MO.) -The hyperelliptic mapping class group is (by definition) the subgroup of mapping classes commuting with an involution. Not all mapping classes are hyperelliptic. In algebraic geometry, we would say that the map $H_g \to M_g$ (where $H_g$ is the moduli space of hyperelliptic genus $g$ curves) does not induce a surjection on fundamental groups. -What about the trigonal locus $T_g$, parametrizing genus-g curves endowed with a degree-3 map to $\mathbb{P}^1$? Does the map $T_g \to M_g$ induce a surjection, or a finite-index inclusion on fundamental groups? (We do know that $\pi_1(T_g)$, like $\pi_1(H_g)$, surjects onto $Sp_{2g}(\mathbb{Z})$, or at least we know its image is Zariski dense; I'm not sure whether we know its image is finite-index, now that I think of it.) -In topology, we would ask the following question: (equivalent? if not, close to it) Let $\phi$ be a surjection from the free group on $k$ generators to $S_3$, sending each generator to the class of a transposition, and let $\Gamma$ be the (finite-index) subgroup of the $k$-strand braid group consisting of elements which stabilize $\phi$ under left composition. The realization of $\phi$ as a degree-$3$ simply branched cover of a sphere yields a map from $\Gamma$ to some mapping class group $\Gamma_g$, whose image is what we might call the trigonal genus-g mapping class group; the question is whether this is a proper subgroup. -More generally, one could define the gonality of a mapping class $f$ to be the minimal $d$ such that $[f]$ lies in the image of the fundamental group of the space of $d$-gonal curves of genus $g$. Is this an interesting invariant? (i.e. if it is always 2 or 3 it is not so interesting.) - -REPLY [6 votes]: it seems that your question about the possible surjectivity of the map -$$\pi_1(T_g) \to \pi_1(M_g)$$ -has been recently answered positively in http://arxiv.org/abs/1403.7399 (see the very first page of the paper.<|endoftext|> -TITLE: Delooping maps between H-spaces -QUESTION [5 upvotes]: Hi, -this question is related to my question here. Suppose, I have a topological group $G$ and an $A_{\infty}$-space $H$, which is a CW-complex. Furthermore, I have a map $\varphi \colon G \to H$, that induces an isomorphism of groups $[X, G] \to [X,H]$ for finite CW-complexes $X$. - -Is this enough to deloop $\varphi$, - i.e. does there exist a map $B\varphi -> \colon BG \to BH$. Or can I at least - deduce a weak equivalence between $BG$ - and $BH$ from this? - -btw.: What is the "standard" reference for $A_{\infty}$-spaces and $H$-spaces nowadays? Or for Segals $\Gamma$-spaces? - -REPLY [10 votes]: No, it is not: $S^3$ admits uncountably many loop space structures (c.f. Rector "Loop structures on the homotopy type of $S^3$"), but only $12 (= \vert \pi_6(S^3) \vert)$ H-space structures (c.f. James "Multiplication on spheres (II)").<|endoftext|> -TITLE: Shortest irrational path -QUESTION [8 upvotes]: What is the shortest curve $\gamma$ in $\mathbb{R}^2$ - from the origin $o=(0,0)$ to a rational point $p=(a,b)$ - that (a) passes through no other rational point, and - (b) contains no point a rational distance from both $o$ and $p$? - -A rational point is one with rational coordinates. -I am wondering if one can reach $p$ via such a highly -"irrational route." -My guess is that there are curves whose lengths approach -the straight-line distance $|p|$. -Perhaps the curve should be restricted to a specific class: -$C^\infty$, analytic, elliptic, quadratic, circle arc. -(Just avoiding rational points [condition (a)] can be accomplished with a -circle arc of an appropriately chosen radius.) -This is far from my experience, and it may be that all -the variations have trivial, uninteresting answers, in -which case I apologize for the distraction. - -REPLY [8 votes]: A circle passing throiugh the origin and your other point is determined by just one more point. Each rational point therefore determines a unique circle. Hence there are circles with large radius whose arcs ar arbitrarily close to the minimal distance containing no rational point, because there are uncountably many such.<|endoftext|> -TITLE: When should we expect Tracy-Widom? -QUESTION [38 upvotes]: The Tracy-Widom law describes, among other things, the fluctuations of maximal eigenvalues of many random large matrix models. Because of its universal character, it obtained his position on the podium of very famous laws in probability theory. I'd like to discuss what are the ingredients to be present in order expect his apparition. -More precisely, the Tracy-Widom law has for cumulative distribution the Fredholm determinant $$ F(s)=\det(I-A_s) $$ where the operator $A_s$ acts on $L^2(s,+\infty)$ by -$$ -A_sf(x)=\int A(x,y)f(y)dy,\qquad A(x,y)=\frac{Ai(x)Ai'(y)-Ai(y)Ai'(x)}{x-y}, -$$ -$Ai$ being the Airy function. It is moreover possible to rewrite $F$ in a more explicit (?) form, involving a solution of the Painlevé II equation. It is known that this distribution describes the fluctuations of the maximal value of the GUE, and actually of a large class of Wigner Matrices. It curiously also appears in many interacting particle processes, such as ASEP, TASEP, longest increasing subsequence of uniformly random permutations, polynuclear growth models ... (For an introduction, see http://arxiv.org/abs/math-ph/0603038 and references inside. You may jump at (30) if you are in a hurry, and read more about particles models in Section 3). A natural (but ambitious) question is - -You have $N$ interacting random points $(x_1,\ldots,x_N)$ on $\mathbb{R}$, when can you predict that $x_{\max}^{(N)}=\max_{i=1}^N x_i$ will fluctuate (up to a rescaling) according to Tracy-Widom law around its large $N$ limiting value ? - -Assume that the limiting distribution of the $x_i$'s $$ -\mu(dx)=\lim_{N\rightarrow\infty}\frac{1}{N}\sum_{i=1}^N\delta_{x_i}\qquad \mbox{(in the weak topology)} -$$ -admits a density $f$ on a compact support $S(\mu)$, and note $x_\max=\max S(\mu)$ (which can be assumed to be positive by translation). I have the impression that a necessary condition for the appearance of Tracy-Widom is to satisfy the three following points : -1) (strong repulsion) There exists a strong repulsion between the $x_i$'s (typically, the joint density of the $x_i$'s has a term like $\prod_{i\neq j}|x_i-x_j|$, or at least the $x_i$'s form a determinantal point process). -2) (no jump for $x_\max^{(N)} $) $x_\max^{(N)}\rightarrow x_\max$ a.s. when $N\rightarrow\infty$. -3) (soft edge) The density of $\mu$ vanishes like a square root around $x_\max$, i.e. $f(x)\sim (x_\max-x)^{1/2}$ when $x\rightarrow x_\max$. -For TASEP and longest increasing subsequence models, one can see that 1), 2) and 3) hold [since these models are somehow discretizations of random matrix models where everything is explicit (Wishart and GUE respectively)]. For the Wigner matrices, 2) and 3) clearly hold [Wigner's semicircular law], and I guess 1) is ok [because of the local semicircular law]. For ASEP, 1) clearly holds [because of the E of ASEP], 2) and 3) are not so clear to me, but sound reasonable. - -Do you know any interacting particle model where Tracy-Widom holds but where one of the previous points is cruelly violated ? - -Of course the condition 1) is pretty vague, and would deserve to be defined precisely. It is a part of the question ! -NB : I have a pretty weak physical background, so if by any chance a physicist was lost on MO, I'd love to hear his/her criteria for Tracy-Widom... - -REPLY [4 votes]: Disclaimer - I have only recently come to know about this law after taking a class on Random Matrix Theory. And I am no Physicist; I study Computer Science. Thus I do not claim to know or understand the details in your question. However, I will answer your original question, "When should we expect the Tracy-Widom law?" and, "What are the ingredients to be present in order expect his apparition?" So, please pardon me if my answer is not up to the mark you might have been expecting. -The way I understand it best, Tracy-Widom (TW) law is convenient to understand in conjunction with the Marcenko-Pastur (MP) law. The MP law gives a bound for the histogram of the eigenvalues of a random matrix. It is also one of the universal distributions. It looks like follows: - -As you can see, the maximum eigenvalue is bounded. However, if the matrix is not truly random, like for instance in stock price returns, i.e., when there is some correlation among the different stocks, then there are some eigenvalues that go beyond the bound given by the MP law, as shown in my MATLAB simulation below: - -The first plot shows all the eigenvalues, including the maximum eigenvalue which is way over to the right. The second plot removes that maximum eigenvalue and still we can see a few eienvalues beyond the MP law bound. -The meaning of this is understood as follows: whichever stock the one isolated eigenvalue, the maximum eigenvalue, is coming from, it is very, very strongly correlated with all the other stocks. -Anyway, so here comes the Tracy-Widom law. The TW law can be thought of as giving the probability of that maximum eigenvalue lying within the MP law bound. If we look at its graph, it is heavy tailed to the right, meaning that the probability of the maximum eigenvalue being more to the right of the MP law bound becomes lesser and lesser as we keep going to the right, almost nil beyond a point. -So, what is the significance of all this? Well, we know that the MP law and the TW law hold for only random matrices. So, if there is any significant deviation for a matrix, it means that the matrix in question is not in fact a random matrix, like the one of stock returns shown above. This is helpful as the null hypothesis is to assume no correlations among the random vectors, as it was believed for prices of different stocks, until people started doing these spectral analysis discussed here for the covariance (and correlation) matrix of stock returns, and the null hypothesis was rejected based on these evidence. So, deviation from these laws suggests some structure, some pattern in the data. I am aware of this being used, apart from financial engineering, in finding patterns in genetic mutations, among others. -In summary, we should expect Tracy-Widom for truly random matrices, i.e., constructed from measurements of random vectors, without any correlations. So, "the ingredients to be present in order to expect his apparition" is true randomness. If there is some correlation among the random vectors, then there will be some "proportionate" deviation from the law.<|endoftext|> -TITLE: fixed points of products of permutations -QUESTION [7 upvotes]: This question is related to that (if $s$ is co-prime with prime $p$ and a permutation in $S_s$ has order $p$, then it fixes a point). -Let us fix two (finite) numbers $p\gg 1, n\gg 1$. Say, $p=47, n=18999$. Take a -sequence $s_1,s_2,...$ of numbers not divisible by $p$. For each $s$ -compute $p(s)$ as follows. Take two permutations $a, b$ in the symmetric group $S_s$ such that -$\langle a,b\rangle$ is highly (say, $3$-)transitive on $\{1,...,s\}$. Compute the probability $p(s,a,b)$ -that for a word $w(x,y)$ of length $n$, $w(a,b)$ has a fixed point. Then $p(s)$ is the maximum of -$p(s,a,b)$ for all (such) $a,b$. -Question. Is it true that $$\lim_{i\to\infty} p(s_i) = 0?$$ - Note. As Doron Puder explained to me, one cannot replace "highly transitive" by "transitive" because $a,b$ can generate a dihedral group of order $2s_i$ in which case $p(s_i,a,b)$ will be at least $1/5$. F. Ladisch explains in his answer to the previous version of the question below that "2-transitive" is not enough either. - Update Perhaps the "correct" condition instead of "highly transitive" should be "$\langle a,b\rangle$ contains "few" permutations having fixed points, that is the portion of such permutations in $\langle a,b\rangle$ tends to 0 as $i\to \infty$. Examples of Puder and Ladisch contain "many" permutations fixing a point and the arguments were based on that fact. Thus the problem is this: if for many words $w(x,y)$ of length $n$ the permutation $w(a,b)$ has a fixed point, then many elements in $\langle a,b\rangle$ have fixed points, a local-to-global property. - -REPLY [4 votes]: Update: It just occured to me that if you fix the length $n$, then $p(s,a,b)$ is always a fraction with denominator $4\cdot 3^{n-1}$, so only finitely many values are possible. Thus for a sequence $(s_i,a_i,b_i)$ with $\langle a_i, b_i\rangle\leq S_{s_i}$, the sequence $p(s_i, a_i, b_i)$ can converge only if it is constant eventually. It converges to zero only if $\langle a_i, b_i \rangle$ contains no reduced words of length $n$ with fixed points for all but finitely many $i$. So I doubt that only looking at words of length $n$ can tell you much. Also, transitive but nonregular permutation groups always seem to have a large portion of elements with fixed points. -(The following was my original answer to a previous version of the question:) -Take $C_{p-1} \ltimes C_p$ (or somewhat more generally, the affine group of a finite field of order $q=p^k$). This is generated by two elements, has order $(p-1)p$, and all but $p-1$ elements have fixed points. So whithout taking into account the restriction on the length, the probability for fixed points is $\frac{1}{p}$. For $p\gg n$, the probability will be similar, I think. So if you take a sequence of primes tending to infinity for $s_i$, this shows that $\lim_{i\to \infty} p(s_i)$ can be big (probably $1$, in this case). -EDIT: I justed realized that I misunderstood the question, since you are asking for the fraction of reduced words, and not the fraction of group elements itself. However, if we take $a$ and $b$ to be generators of the cyclic groups of order $p-1$ and $p$ respectively, then the reduced words mapping into the regular subgroup $C_p= \langle b \rangle$ are those where the numbers of $a$'s and $a^{-1}$'s occurring is equal $\mod p-1$, that is, equal, if $p\gg n$. The number of these words is independent of $p$, and there are other words, so $\lim_{i\to \infty} p(s_i)\neq 0$. Hope now it's correct.<|endoftext|> -TITLE: RAM simulating another RAM -QUESTION [7 upvotes]: (Cross-posted from cstheory-stackexchange) -The following fact seems to be used implicitly in cs theory, particularly algorithms. Given a RAM machine $M$ running in time $O(f(n))$, another RAM machine $M′$ can simulate $M$ in time $O(f(n))$. This differs from the case for Turing machines, where $M′$ may require $O(f(n) \log(f(n))$ time. -I say this is often used implicitly because many papers will simply say something like "run $M$, but keep track of certain auxiliary information as you do so". This is really simulating $M$, but for RAM machines the distinction is not so important because running times are not (asymptotically) affected. -Is there a reference for this theorem? I am summarizing the situation correctly? -EDIT: -Some questions were raised about what is meant by simulation. More precisely, that there exists a universal RAM $T$ such that, for any RAM machine $M(x)$ running in time $O(f(|x|))$, there is an encoding $e_M$ such that $T(e_m, x) = M(x)$ and $T(e_m, x)$ runs in time $O(f(|x|))$ as well. (The constant term may be different between $T$ and $M$). - -REPLY [2 votes]: I think that the blow-up in time can be much worse than you describe, depending on the specific computational models that are used. For example, consider the function that accepts an input string $w$ of arbitrary length and outputs the double duplicate string $ww$. On a Turing machine with separate input and output tapes having heads that can move independently, then this output can be formed in linear time, essentially copying $w$ to the output tape and then sending the input tape head back to the start, and then copying it again, which takes about $3|w|$ time altogether. But if one instead has the two tape machine model with a single head reading both tapes together, then the former computation can be simulated, but it takes a lot more time. A direct implementation will have the head necessarily making a lot of back-and-forth motions in order to form the second copy of $w$, ferrying small pieces of it back and forth, and the time complexity will rise to something like $O(|w|^2)$. And I expect that one can describe much worse examples. -This kind of issue is precisely what makes the class $P$ of polynomial time computability so conveneint and robust as a notion of feasibility. Because all the various standard computational models can simulate each other with only polynomial time factors, the choice of any particular model of computability does not affect membership in $P$ or its cousins, such as NP, which are closed under these polynomial factors. -But meanwhile, the point is that it doesn't really make sense to talk about, say, an $O(n^2)$ algorithm unless the particular model or class of models of computability has been understood. After all, one model's $O(n)$ can be another's $O(n^2)$ or worse.<|endoftext|> -TITLE: Hodge spectral sequence for algebraic stacks -QUESTION [7 upvotes]: In a beautiful paper Deligne and Illusie have shown the following: Let $f\colon X \to S$ be a smooth proper morphism of schemes in characteristic $p > 0$, let $F\colon X \to X^{(p)}$ be the relative Frobenius and let $b$ be an integer. Assume that $\tau_{\leq b}F_*\Omega_{X/S}$ decomposes in the derived category of coherent sheaves on $X^{(p)}$ (where $\Omega_{X/S}$ denotes the De-Rham complex). Then for $i+j \leq b$ one has: -(1) The sheaves $R^if_*\Omega^j_{X/S}$ are locally free and their formation commutes with arbitrary base change. -(2) The Hodge spectral sequence for $f$ satisfies $E^{ij}_1 = E^{ij}_{\infty}$. -(see Relevements modulo $p^2$ et decomposition du complexe de de Rham, Inv. Math. 89). -As also explained in that paper, the assumption is satisfied for $b = p-1$ and if $X^{(p)}$ can be lifted over a $\tilde S$, where $\tilde S$ is a flat $\mathbb Z/p^2\mathbb Z$-schemes whose reduction modulo $p$ is $S$. The result then can be applied to show the degeneration of the Hodge spectral sequence and the vanishing result of Kodaira-Akizuki-Nakano in characteristic $0$ (see also H. Esnault; E. Viehweg: Lectures on vanishing theorems, Birkhäuser 1992). -Question: Does there exist a reference for a similar result for a morphism of algebraic stacks? -It seems to me that using Chow's Lemma for stacks by Olsson the proof should be completely analogous as the proof of Deligne and Illusie (at least for tame Deligne-Mumford stacks), but a quick browsing did not produce any results except for a paper of Matsuki and Olsson (Kawamata-Viehweg vanishing as Kodaira vanishing for stacks, Math. Res. Letters 12 (2005)), where the above result of Kodaira-Akizuki-Nakano is generalized. There the authors claim that the vanishing result can be proved by the same arguments as in the paper of Deligne and Illusie. I do not doubt this claim but as a reference this is not particularly helpful. -Remark: Kato has also sketched the proof of a generalisation of Deligne's and Illusie's result to arbitrary log schemes. This should of course be a special case of a generalization to algebraic stacks via the work of Olsson. - -REPLY [8 votes]: Did you have a look at Satriano's article de Rham Theory for Tame Stacks and Schemes with Linearly Reductive Singularities?<|endoftext|> -TITLE: Definable Wellordering of the Reals -QUESTION [9 upvotes]: Why are we interested in definable wellordering of the reals? For instance, we have - -Con(ZFC) $\Rightarrow$ Con(ZFC + there is a $\Delta^1_2$-wellordering of $\mathbb{R}$), -Con(ZFC + there is a measurable cardinal) $\Rightarrow$ Con(ZFC + there is a measurable cardinal+there is a $\Delta^1_3$-wellordering of $\mathbb{R}$). - -REPLY [15 votes]: This is a very interesting topic! There are two ways to address the question. They are different, so I treat them separately. -I. -As mentioned by Stefan, knowing that we have a well-ordering of certain complexity gives us a bound on the extent of regularity properties. -For example, the complexity of well-orderings gives us an upper bound on the complexity of pointclasses for which determinacy holds. Abstractly, determinacy of nice pointclasses implies that their members are Lebesgue measurable, etc, so it provides us with a precise framework to talk about "regularity" of sets of reals. -Moreover, in the context of fine structural inner models, the complexity of a well-ordering is really a reflection of the complexity of the underlying "comparison process" used to build the model. We are interested in this complexity as it gives us an upper bound on the kind of reals we can expect to see in these models. -To be more concrete, think of $L$ and its well-ordering: We say that $x\lt y$ (for $x,y$ reals), iff either $x$ appears first (i.e., there is an $\alpha$ such that $x\in L_\alpha$ but $y\notin L_\alpha$), or they appear at the same time, but $x$ is "simpler" (measured by, say, an ordering of formulas and parameters). This ordering is $\Sigma^1_2$ because the models $L_\alpha$ are simple to code by reals, essentially all we say is that $r$ codes a model $(A,E)$ and $E$ is well-founded. -When we go from $L$ to, say $L[\mu]$, it is more complicated to compare the levels where $x$ and $y$ appear. Now we have two set models $L_\alpha[U_1]$ and $L_\beta[U_2]$ and we iterate their respective measures until we reach a model and one of its initial segments. The complexity of describing this is greater than in the case of $L$, as now we need to talk about well-foundedness of the iteration as well as of the relevant models. This complexity increases with the models under consideration, as the iterations become more and more complex (iteration trees). -A good discussion of these ideas can be seen in the paper by Martin and Steel, "Iteration trees." J. Amer. Math. Soc. 7 (1994), no. 1, 1–73. It is also treated at a higher level in Steel's Handbook article. -The reason why the complexity of iterations bounds the complexity of the reals we can obtain is a folklore result in inner model theory. Rather than stating the technical fact, let me illustrate with an example: We can identify the real $0^\sharp$ with a model $(L_\alpha,U)$. This model is countable, and pointwise definable. This translates without much effort in the fact that if $y$ is a real in $L$, then $y\le_T 0^\sharp$, where $\le_T$ is the Turing-reducibility order (essentially, because in fact $y\in L_\alpha$). Hence, if a real is more complex (in the Turing sense) than $0^\sharp$, it cannot be in $L$. It also shows that (if $0^\sharp$ exists) then ${\mathbb R}\cap L$ is countable. -$0^\sharp$ is an example of what we call mice. In a sense, the more complex the mouse, the more reals it contains. If a mouse $m$ is so complex that it contains all reals of certain complexity $\Gamma$, and if the comparison process for a fine-structural model $M$ can be coded in $\Gamma$, then we have a concrete example (namely, $m$) for a real not in $M$ and, in fact, we get that all reals in $M$ are Turing reducible to $m$. -(The ultimate expression of these ideas is the so-called mouse set conjecture, but it would take us far off topic to discuss it here properly.) -II. -There is another reason for being interested in simple well-orderings. This reason appears in practice, and is not guided by fine-structural considerations or by trying to limit the extent of regularity properties. -Typically we are interested in strengthenings of the axioms of set theory by axioms that provide us with "combinatorial tools." Examples of the principles I have in mind include forcing axioms (Martin's Axiom, BPFA, MM, ...), the covering property axiom, real-valued measurability of some cardinal $\kappa\le{\mathfrak c}$, etc. -The combinatorial niceness that these axioms provide usually is a hindrance when it comes to defining well-orderings in simple ways. The reason is that typical coding tools we would use in such a definition are ruled out by the combinatorial principles. I present several examples of this in my paper "Real-valued measurable cardinals and well-orderings of the reals," available at this link, in a section titled "Anticoding results". -It is therefore an interesting technical problem to see whether we can circumvent these obstacles and still obtain (consistent) simple well-orderings. Usually we are not so interested in well-orderings per se, but rather in the possibility of developing coding tools. Typically, we can code arbitrary sets of reals just as well as we can code well-orderings (this, in turn, can be seen as an anti-compactness result). -This line of work, in the context of Martin's axiom, was started by Solovay, and developed by Abraham and Shelah in a nice series of papers: - -"A $\Delta^2_2$ well-order of the reals and incompactness of $L(Q^{MM})$." Ann. Pure Appl. Logic 59 (1993), no. 1, 1–32. -"Martin's axiom and $\Delta^2_1$ well-ordering of the reals." Arch. Math. Logic 35 (1996), no. 5-6, 287–298. -"Coding with ladders a well ordering of the reals." J. Symbolic Logic 67 (2002), no. 2, 579–597. - -(I particularly recommend the introduction to the first paper in the series.) -I have worked on this problem of coding in the context of forcing axioms, and the surprise here is that strong forcing axioms actually provide us with simple definitions of well-orderings (not just consistently). For example: - -Sy Friedman and I showed that if BPFA holds and, say, $\omega_1=\omega_1^L$, then there is a $\Sigma^1_3$ well-ordering. -Velickovic and I showed that if BPFA holds and $C$ is a ladder sequence on $\omega_1$, then there is a $\Delta_1$ well-ordering of the reals in parameter $C$. - -The context here differs from that of the first part of the answer in several ways. For example, we tend not to be interested in projective well-orderings any longer, as decent forcing axioms imply AD${}^{L({\mathbb R})}$ and therefore prevent the existence of such orderings. Also, although we may (and do) ask about third-order definable well-orderings, we are actually more interested in definability over $H(\omega_2)$, as we expect to define not just a well-ordering of the reals but of all of ${\mathcal P}(\omega_1)$.<|endoftext|> -TITLE: Mathematical Advice for Interested Highschool Students -QUESTION [9 upvotes]: This may not be a research level math question, but I believe it is still relevant to Math Overflow. - -What general resources exist for students in highschool who are very interested in Mathematics? What advice would you give to a young student to encourage them, and nurture their interest in mathematics? If a young high school student came to you and said they were very interested in math, and wanted to know what to do to keep learning, what would you tell them? - -Thank you for your help, - -REPLY [6 votes]: There are two outstanding resources in English: -1. R. Courant and Robbins, What's Mathematics?, and -2. Hilbert, D. and Cohn-Vossen, S. -Geometry and the imagination. -Many of my friends (including myself) read these books in their high school years, -and decided to become mathematicians.<|endoftext|> -TITLE: truth vs. provability for ordered fields -QUESTION [8 upvotes]: In Propositions equivalent to the completeness of the real numbers I started by asking "Can anyone point me to a reasonably comprehensive article (or book chapter) explaining which basic theorems of calculus are equivalent to the completeness axiom of the reals and which ones aren't?" and ended by speculating "Perhaps somebody wrote a beautiful Monthly article a few decades ago that explained things so clearly as to make the whole matter seem trivial, with the result that the article was forgotten? :-)" -Since the article I was looking for doesn't seem to exist, I decided to write one myself; the current draft can be found at http://jamespropp.org/reverse.pdf . -One issue I'm a little confused about is the relationship between truth and provability in this context. As I ask on the bottom of page 2 and the top of page 3, is saying "Every ordered ring $R$ satisfying property $P$ satisfies property $P'$" the same as saying "From the ordered field axioms plus the assumption that $P$ holds one can prove that $P'$ holds"? -I believe that they're not the same (because for instance the Riemann Hypothesis might be true but unprovable), but I'd like to hear from people who know more about foundations and model theory than I do. -All kinds of comments on the article are welcome, but comments on the truth-versus-provability issue are especially sought. - -REPLY [3 votes]: In general, the way that people approach these things is to look at provable equivalences over some fixed theory. So, for example, you could prove results of the following form: - -Theory $T$ proves that any object satisfying the ordered field axioms will satisfy property $P$ if and only if it satisfies property $P'$. - -The theory $T$ could be ZFC set theory, or it could be a weaker theory such as second-order arithmetic. The main point of the theory is to give some syntactical tools for manipulating the ordered field axioms and the statements of $P$ and $P'$. For example, if $P$ is the axiom of completeness (every nonempty bonded set has an supremum), the theory $T$ needs to guarantee some sets exist. -To establish positive results of the quoted form, you simply write a proof in $T$ of the desired result. The more difficult thing is to establish negative results, and this is the first time you have to think about semantics. To prove the negation of the quoted statement, it suffices to have: - -A class of interpretations of $T$ such that a statement is provable in $T$ if and only if it is true in every one of these interpretations -And an example of one of these interpretations in which there is an ordered field satisfying $P$ but not $P'$, or vice versa. - -It's clear on a moment's thought that the class of structures we want only really depends on the proof rules we have in $T$, not on our intended interpretation of $T$. In the case that the proof rules are the usual ones, we have a general theorem that the set of all "first-order stuctures" is a sufficient class of interpretations to achieve the first bullet. This works not only for first-order logic, but also higher-order logic and set theory, which have the same sort of proof system. -Finally, let me point out a trivial exercise that underscores the need to look at provability rather than truth. For any effective, consistent theory $T$ that is sufficiently strong, and any statement $\phi$ provable in $T$, there is a statement $\phi'$ that is equivalent to $\phi$ but so that $T$ does not prove $\phi \leftrightarrow \phi'$. Namely, $\phi'$ says "$\phi$ and $T$ is consistent". This sort of method shows that the question in the third paragraph of the question has a negative answer, and this would be true no matter what effective consistent proof system we choose.<|endoftext|> -TITLE: Did Kahler say "a long list of miracles occur"? -QUESTION [7 upvotes]: I've been reading Moroianu's Kahler geometry notes and found a unattributed quote that says that if the Kahler properties hold, then -"a long list of miracles occur" -I am guessing that this quote belongs to Kahler himself, but I can't back this up. Does anyone know? - -REPLY [10 votes]: I will make a CW answer to collect together some information. -Igor Rivin found a published text containing the relevant phrase. -It is in "The unabated vitality of Kählerian geometry," by Jean-Pierre Bourguignon which is included in the collected works of Kähler (Kähler, Mathematische Werke/Mathematical Works, edited by Berndt and Riemenschneider, 2003). -The relevant pasage is (from the text of Bourguignon where 'he' refers to Kähler): - -Quoting his terms, the case $d \omega = 0$ presents itself as "a remarkable exception". This is the condition he supposes throughout the paper whose purpose it is to describe a long list of miracles occuring then. - -This suggest to me that while Bourguignon is first quoting Kähler (the "a remarkable exception") he then stops quoting (and a new sentence started) and describes in his [Bourguignon's] own words the list of result/properties obtained by Kähler as miracolous. -Side note: in this text there are some other verbatim quotes and they are under quotation marks; so except if Bourguignon inadvertently omitted them, he is not quoting. -Furthermore, the paper of Kähler in question "Über eine bemerkenswerte Hermitesche Metrik" does not seem to contain such a phrase (cf. csar). I also searched the above mentioned book for appropriate terms (miracles, the German analog Wunder, and also Mir* in case he should have used Mirakel, which exists but is a bit rare); this did not turn up anything, besides what is mentioned above. -Therefore it seems likely to me that this 'miracles' are due to Bourguignon and not Kähler; and, Moroianu is sort-of quoting Bourguignon. The time-line might seem a bit short, the notes being from 2003 as well as the book, however in view of the fact that Moroianu is a former student of Bourguignon this seems much less surprising, and perhaps even reinforces the idea that Moroianu is quoting Bourguignon. -Final note: in case somebody wants to make really sure, Moroianu is a (it seems now inactive) MO user, so he might, if made aware of the need, give an authentic account.<|endoftext|> -TITLE: What is the purpose of tangential base-points? -QUESTION [10 upvotes]: Let $V$ be an affine complex variety. Let $x \in V$ be a closed point. Then a tangential base-point at $x$ is $x$ together with a regular function $t$ on $V$ that is zero exactly on $x$ (to degree $1$). -My familiarity with this notion is very meek, and I know that there are generalizations for other fields, and for normal crossing divisors (instead of one closed point; in this case they are called tangential morphisms), but I'm not sure I can recite a definition for these. In texts where I've seen this notion introduced, it said that the intuition should be that $t$ gives a direction: look at $t=\epsilon$ for $\epsilon$ going from $0$ to a small real positive number. -This hardly seems like motivation! Why would one want a base-point with a direction? What utility does it have? What problem does it solve? I've seen this in several papers, and I do wish to get over the hump and actually start understanding this. - -REPLY [10 votes]: I'm far from being an expert of the subject, but my naive opinion is that tangential basepoints makes many constructions more canonical and usually behave well with respect to the symmetries of the situation. Grothendieck said in his Esquisse: - -Ceci est lié notamment au fait que les gens s’obstinent encore, en calculant avec des groupes fondamentaux, à fixer un seul point base, plutôt que d’en choisir astucieusement tout un paquet qui soit invariant par les symétries de la situation, lesquelles sont donc perdues en route. - -On the fly translation: - -This is because people working with fundamental groups are still choosing only one basepoint, instead of choosing a whole set of basepoints which is invariant under the symmetries of the situation, which are thus lost on the way. - -Hence, we have to choose several basepoints (ie work with groupoids) and to choose them carefully. -If you take the fundamental example of $P^1(C)$ minus 0,1,$\infty$, there are two automorphisms $t\mapsto 1/t$ and $t\mapsto 1-t$ which induces an action of $S_3$ which is respected by the action of $G_{\mathbb{Q}}$. If you want to keep track of this action, you really want to take $0,1,\infty$ as a set of basepoints. Of course you can't, but you can take the set of tangential basepoints $01,10,1\infty,\infty 1, 0\infty,\infty 0$ which is stable under the action of $S_3$. Now if you decompose loops in appropriate pieces (moves between basepoints, small loops around basepoints) in such a way that the symmetries are again respected, then the action of $G_{\mathbb{Q}}$ can be written in a nice form. - -REPLY [9 votes]: One motivation comes from Grothendieck-Teichmüller theory. The idea, due originally to Grothendieck, is that one can study the absolute Galois group $G_{\mathbb{Q}}$ of $\mathbb{Q}$ by looking at its action on the fundamental groups of varieties defined over $\mathbb{Q}$. More specifically, Grothendieck suggested studying the Teichmüller tower, the collection of moduli spaces $M_{g, n}$ of genus $g$ curves with $n$ marked points together with any natural geometrically meaningful morphisms between them. Since all such morphisms should be defined over $\mathbb{Q}$, the action of $G_{\mathbb{Q}}$ on the fundamental groups $\widehat{\Gamma_{g, n}}$ will be compatible with the induced morphisms between them. -Tangential basepoints arise because most of the geometrically meaningful morphisms that one would like to define between moduli spaces take values not in the target moduli space itself but in the boundary of its compactification. For example, taking a smooth genus 0 curve with $m + 1$ marked points and a smooth genus 0 curve with $n + 1$ marked points and gluing them together along a specified pair of marked point gives a nodal genus 0 curve with $m + n$ marked points: This procedure defines a map $M_{0, m + 1} \times M_{0, n + 1} \to \partial\overline{M_{0, m + n}}$. In order for this to induce a map $\widehat{\Gamma_{0, m + 1}} \times \widehat{\Gamma_{0, n + 1}} \to \widehat{\Gamma_{0, m + n}}$, we are forced to consider tangential basepoints. (Note that the fundamental groups of the compactifications $\overline{M_{g, n}}$ are not interesting, so it does not do us much good to study the action of $G_{\mathbb{Q}}$ on these.)<|endoftext|> -TITLE: ramifications in compositum number fields -QUESTION [8 upvotes]: Suppose {$K_i/\mathbb{Q}$} is a finite set of finite galois extensions of $\mathbb{Q}$ with Galois groups $G_i$. -Suppose we know the ramifications of $K_i$ quite well (e.g., their decomposition groups, inertia groups at some primes), - -What can we say about the ramifications of the compositum field of $K_i$ (e.g., the ramification index, inertia degree of some primes)? Any References? -Particularly, when $K_1\cap K_2=\mathbb{Q}$, we know that $K_1K_2$ has Galois group $G_1\times G_2$. Is the corresponding decomposition group (resp. inertia group) of the form $D_1\times D_2$ (resp. $I_1\times I_2$)? (This is wrong in general, see Álvaro Lozano-Robledo's answer for a counterexample) -How about the case if we remove the requirement that $K_i/\mathbb{Q}$ are Galois? - -REPLY [7 votes]: For (2), the answer is no, not in general. Here is a simple example: take $K_1=\mathbb{Q}(i)$ and $K_2=\mathbb{Q}(\sqrt{-5})$. Then both $K_1/\mathbb{Q}$ and $K_2/\mathbb{Q}$ are (totally) ramified at $p=2$ and $K_1\cap K_2=\mathbb{Q}$, but $F=K_1K_2$ is not totally ramified at $2$. In other words, the extension $\mathbb{Q}(\sqrt{-5},i)/\mathbb{Q}(\sqrt{-5})$ is unramified at $2$ (in fact, $F$ is the Hilbert class field of $K_2$, so it is unramified everywhere). -On the other hand, if you take $K_1=\mathbb{Q}(i)$ and $K_2=\mathbb{Q}(\sqrt{2})$, then $F=K_1K_2$ is totally ramified at $2$ over $\mathbb{Q}$. (Here $F=\mathbb{Q}(\zeta_8)$.) -In both cases, $I_1=D_1=G_1$ and $I_2=D_2=G_2$ (in your notation) but in the first case the inertia in the compositum has order $2$ and in the second case it has order $4$. This shows that one needs to know more than the decomposition and inertia subgroups at a prime in each $K_i$ to understand the ramification index in the compositum.<|endoftext|> -TITLE: Is there a relational countable ultra-homogeneous structure whose countable substructures do not have the amalgamation property? -QUESTION [13 upvotes]: Is there a relational countable ultra-homogeneous structure whose countable substructures do not have the amalgamation property? - -The question can be stated in a fashion not requiring much background: -Let $M$ be a countable ultra-homogeneous relational structure - namely, a countable set equipped with a bunch of relations on its finite cartesian powers, such that any isomorphism between two finite substructures of $M$ extends to an automorphism of $M$. It is classical (and easy to see) that the age of $M$, namely the class $K$ of finite substructures of $M$ (and isomorphic copies thereof) is a Fraïssé class, and in particular has the Amalgamation Property: -Amalgamation Property (AP) - whenever $A,B_i \in K$ for $i=0,1$ and $f_i\colon A \to B_i$ are embeddings, there is $C \in K$ and further embeddings $g_i\colon B_i \to C$ such that $g_0f_0 = g_1f_1$ (i.e., $B_0$ and $B_1$ can be amalgamated in $K$ over $A$). -In all examples I am familiar with, the class of all countable substructures of $M$ also has the same property, but I see no reason why this should be true in general. -Any counterexample (or proof that this does hold in general) will be welcome. -[Of course, there are usually going to be in $M$ countable substructures $A_0,A_1$ with an isomorphism $f\colon A_0 \to A_1$ which does not arise from an automorphism - but this does not exclude the possibility of proper embeddings $g_i\colon M \to M$ such that $g_1^{-1} g_0$ extends $f$.] -ADDENDUM: Notice that if $M$ is saturated then its countable substructures have AP, so a counter-example will have to be non saturated, and in particular non $\aleph_0$-categorical, with an infinite language. -ADDENDUM #2: see also A restatement, in terms of the semi-group product of the left-invariant completion of a Polish group, of http://mathoverflow.net/questions/71389 - -REPLY [9 votes]: EDIT NOTE: Thanks to Emil Jeřábek's comment, (1) has been modified; $X$ in the theorem has been quantified, and the bold sentence in (4) has been added. - -I will first present a counterexample using a structure that has (infinitely many) functions; then I will explain how this functional counterexample can be turned into a relational one. - -We begin with some preliminaries: -(1) Recursively saturated models that have elimination of quantifiers are ultra-homogeneous. This is a basic result in model theory. -(2) If $M_0$ and $M_1$ are models of $PA$ (Peano arithmetic), and $M_0$ is a submodel of $M_1$, then $SSy(M_{0})\subseteq SSy(M)$. -This follows from the definition of $SSy(M)$ (the standard system of $M$). Recall that for a model $M$ of $PA$, $SSy(M)$ is the collection of subsets of $\omega$ that are "coded" by some element of $M$, where "coded" can be defined in various ways, e.g., as: $X \subseteq \omega$ is coded by $c \in M$ if for all $n \in \omega$, $M \models$ “the $n$-th prime divides $c$” iff $n \in X$. -(3) The heart of this counterexample is the following theorem [it is Theorem 2.3.1 (p.40) of the Kossak-Schmerl text on models of Peano arithmetic]. -Theorem. Let $M_0$ be a countable recursively saturated model of $PA$, and suppose $X$ is some fixed subset of $\omega$. Then $M_0$ has elementary end extensions $M_1$ and $M_2$, such that $M_0 \cong M_{1} \cong M_2$, and whenever $M_{3}\models PA$ is an amalgamation of $M_1$ and $M_2$, then $X\in SSy(M_3)$. -(4) Given $M \models PA$, let $M^{+}$ be the EXPANSION of $M$ by the first-order definable functions of $M$. We observe that if $N^{+}$ is a substructure of $M^{+}$, then the reduct $N$ is a model of $PA$ since the universe of $N$ is closed under the functions available in $M^{+}$, and therefore $N$ is an elementary submodel of $M$ because $PA$ has definable Skolem functions. Note, furthermore, that $M^{+}$ eliminates quantifiers, and is also recursively saturated, hence ultrahomogeous. - -(1)-(4) show that for a countable recursively saturated model $M$ of $PA$, the collection of substructures of $M^{+}$ do not satisfy amalgamation. - -More specifically, thanks to the aforementioned theorem in (3), by first choosing some subset $X$ of $\omega$ that is missing from the standard system of $M$, we can be assured of the existence of (end) embeddings $f_{i}:M^{+}\rightarrow M^{+}$ for $i=0,1$ with the property that if there is a structure $N^{+}$, and embeddings $g_{i}:M\rightarrow N^{+}$ for $i=0,1$, with $g_{0}f_{0}=g_{1}f_{1}$, then by (2) and (4) $N^{+}$ is not a substructure of $M^{+}$. - -Now we explain how to obtain a relational counterexample. - -Given a model $A$ in a language with functions, let $\cal{A}$ be the relational structure obtained by replacing each $n$-ary function $f$ in $A$ by the usual $(n+1)$-ary relation known as the graph of $f$. -Let $M$ be a countable recursively saturated model of $PA$. To see that the family of substructures of $\cal{M^{+}}$ do not satisfy amalgamation, we simply observe that if $(X,\cdot \cdot \cdot)$ is a substructure of $\cal{M^{+}}$, and $\overline {X}$ is the closure of $X$ under the functions available in $M^{+}$ , then the inclusion map $i_{X}:X\rightarrow \overline{X}$ is an embedding of the substructure of $\cal{M^{+}}$ determined by $X$ into the substructure of $\cal{M^{+}}$ determined by $\overline{X}$. Therefore, if $AP$ holds in this relational context for some amalgamating substructure with universe $X$, by composing each $g_i$ with $i_{X}$ then $AP$ would also have to hold in the functional context.<|endoftext|> -TITLE: Can you prove that hypergraphs with n-1 edges are partially 2 colorable? -QUESTION [6 upvotes]: I can. But my proof uses a theorem (which I do not reveal yet to avoid influencing you) and it feels like an overkill, so I wonder if there is a simple proof. Now the problem. -Suppose we have a hypergraph on n vertices with n-1 edges. Can we color some (at least one) of its vertices with red and blue such that every hyperedge either contains both colors, or all of its vertices are uncolored? -The motivation is this question. In fact, one can similarly define partial k-polychromatic coloring, where every hypergraph with less than n/(k-1) edges seems to be partially k-colorable, if my proof is correct. Note also that these bounds are tight as shown by disjoint edges of size k-1. And this cannot be improved by requiring some lower bound on the size of each hyperedge, as we observed yesterday with my usual set of friends, Cory, Dani, Keszegh, Nathan and Patkos. -Please do not post links to Beck-Fiala and other well-known, similar theorems! At the moment I am only interested in a short, elementary proof of my question. - -REPLY [4 votes]: The proof of the statement of the question is implicitly contained, along the lines of -the proof by Ilya Bogdanov, in Paul Seymour's MSc thesis, where the relevant part is -published in - -Paul D. Seymour, On the two-colouring of hypergraphs, The Quarterly Journal of Mathematics (Oxford University Press), 1974, Volume 25, Pages 303-312. - -If the hypergraph 2-colouring problem is translated (in the canonical way) into a satisfiability problem (SAT problem), then the statement is precisely that there -exists a so-called "autarky". The theory of autarkies is outlined in my Handbook -of Satisfiability article on autarkies, available at -http://www.cs.swan.ac.uk/~csoliver/papers.html#Handbook2009MUAUT -(see Subsections 11.4.5 and 11.12 for connections to the problem at hand). -An article extending Seymour's result is -http://cs.swan.ac.uk/~csoliver/papers.html#SNSSAT2007<|endoftext|> -TITLE: Average degree of contact graph for balls in a box -QUESTION [13 upvotes]: Imagine you dump congruent, hard, frictionless balls in a box, -letting gravity compress the balls into a stable configuration -(I believe such configurations are called -jammed.) -Assume the box dimensions are much larger than the ball radius -(unlike the image below), -so that boundary effects are minimized. -Define the contact graph of the configuration to have a node for -each ball, and an arc for two balls in contact. - -             - - -             - -Image due to -Hugo Pfoertner - - - -What is the average degree of a node in the contact graph - for balls in $\mathbb{R}^d$, under the scenario above? - -The maximum degree is the -kissing number: -6 in $\mathbb{R}^2$, -12 in $\mathbb{R}^3$, -24 in $\mathbb{R}^4$. -I am interested in two aspects: - -(1) Is the average contact degree known to be significantly smaller - than the kissing number? - -I have seen results on the density of irregular packings (e.g., -about 64% in $\mathbb{R}^3$ vs. 74% in an optimal packing), -but I have not seen this expressed in terms of the structure of the contact -graph. - -(2) For large $d$, is it expected that the average contact degree - increasingly deviates from the kissing number, or the opposite: that - jammed packings approach the densest packings. - -Perhaps one can only hope for an answer here for $d{=}24$, where the -kissing number is known. -But maybe there is a heuristic argument based on how much "room" there is -around a ball in higher dimensions? -Any known structural properties of the contact graph for jammed configurations would be of interest. -Thanks! - -Addendum. The Torquato and Stillinger paper cited by Matthew Kahle -(Rev. Mod. Phys. 82, 2633–2672 (2010)) is a gold mine of information on the topic. -Here is their Fig.14 showing three different "optimal strictly jammed" packings: - -             - - -A configuration is strictly jammed if it is "collectively jammed" and furthermore "disallows all uniform volume-nonincreasing strains of the system boundary." - -REPLY [4 votes]: Torquato and Stillinger have a recent survey article that discusses some questions like this: Jammed hard-particle packings: From Kepler to Bernal and beyond -They are particularly interested in random packings. -A short answer to your question is that for hard spheres, it is generally believed that random jammed packings are close to "isostatic," meaning the average contact number is $2d$, where $d$ is the dimension.<|endoftext|> -TITLE: When are (finite) simplicial complexes (smooth) manifolds? -QUESTION [18 upvotes]: Hi, -is there an algorithm that determines if a given simplicial complex is -a.) a manifold -b.) a smooth manifold -c.) homotopy equivalent to a manifold -d.) a real algebraic variety -? - -REPLY [4 votes]: (b) When does a PL manifold have a smooth structure? -I believe that this is computable. Smoothing theory, closely related to the Hirsch-Smale immersion theory (an h-principle), identifies the set of smooth structures on a PL manifold with the set vector bundles that are reasonable candidates for the tangent bundle in the sense that they are reductions of structure group of the PL tangent bundle. The theory of principal bundles reduces to homotopy theory. The set of homotopy classes of maps from a finite complex to a finite complex with abelian fundamental group and the induced maps between such sets are computable, but it is rarely implemented.<|endoftext|> -TITLE: Manifolds and Polynomials -QUESTION [9 upvotes]: Given a compact smooth manifold $M \subset R^k$ there is a Polynom $f\in R[x_1,..x_n]$ such that the zero set of $f$ is diffeomorphic to $M$. Can the coefficients of $f$ be pertubated slightly to a Polynomial $g \in Q[x_1,..x_n]$ such that the zero set of $g$ is diffeotopic to $M$? Are their conditions on the homology or homotopy on $M$ such that such a pertubation process is possible / not possible? What happens if Q is replaced by an arbitrary number field K? - -REPLY [9 votes]: Yes: proven in Ballico, E., Tognoli, A., Algebraic models defined over $\mathbb{Q}$ of differential manifolds. Geom. Dedicata 42 (1992), no. 2, 155–161. In fact, you can get the zero set to be diffeomorphic to $M$, not just diffeotopic.<|endoftext|> -TITLE: about fixed points of permutations -QUESTION [5 upvotes]: This is another attempt to make a feasible approximation of this question. Two previous (unsuccessful) attempts are here. -Let $n\gg 1$ be a fixed number (say, $n=10^{10}$), $k\gg 1$ a natural number. Let $a,b$ be two permutations from $S_k$. Suppose that for every word $w(x,y)$ of length $\le n$, the permutation $w(a,b)$ has a fixed point. Is it true that every permutation in $\langle a, b\rangle$ has a fixed point? - -REPLY [3 votes]: Let me have a go at this one. We want to show that for any finite set $W$ of reduced words in $a,b$, we can find a subgroup $G = \langle a,b \rangle$ of some symmetric group $S_k$ such that all words in $W$ have fixed points when evaluated in $G$, but $G$ contains fixed-point-free elements. -Suppose inductively that we can do this for some set $W$, and we want to extend the construction to $W \cup \{w \}$ for some new word $w$. We may suppose that $w$ is cyclically reduced and that $w$ acts fixed-point-freely in $G$. -If $w$ has length $n$, then we can define partial permutations $a,b$ in the obvious way on $n$ points such that $w$ fixes the first of these. For any $p>n$, it is then easy to extend these partial permutations to permutations of a set of size $p$ that generate a transitive subgroup of $S_p$. In fact, by an old theorem of Jordan, if a primitive permutation group of degree $p$ contains a $q$-cycle for some prime $q k$ to give a group $H \le S_p$, and let $K < S_{k+p}$ be generated by the disjoint actions of $a$ and $b$ in $G$ and in $H$. So all elements of $W \cup \{w\}$ have fixed points in $K$ and it remains to show that $K$ contains fixed-point free elements. -If all relations in $a,b$ that hold in $G$ held in $H$, then $H$ would be a quotient group of $G$, which is impossible, because $|H|$ is divisible by $p$, but $|G|$ isn't. So the permutations generated by the relations of $G$ generate a nontrivial normal subgroup of $H$ which, by the simplicity of $A_p$, must contain $A_p$. So, there exists a relation $v$ of $G$ such that $vw$ acts fixed-point-freely in $H$. Since $w$ acts fixed-point-freely in $G$, $vw$ is a fixed-point-free elemtn of $K$ as required.<|endoftext|> -TITLE: Replacement for derivations in characteristic p? -QUESTION [12 upvotes]: Let $k$ be a field. -If $f \in k[x]$ is a polynomial, and $d/dx\ f = 0$, then either - -$f$ is constant, or -$char\ k = p$ and $f \in k[x^p]$. - -So "annihilated by all derivations" is perhaps not the right thing to -ask for in characteristic $p$ (though that's what I asked for in -Is the singular locus ideal preserved by all derivations? ). - -What is the right thing to ask for? - -I would like an invariance condition one could state of a subscheme -$Y$ of $X$, that holds for the singular locus, but doesn't hold for (say) any regular closed point on a rational variety. - -REPLY [10 votes]: In principle, there are two possible approaches. One is based on the Hasse derivatives (also called hyperdifferentiations). See http://math.fontein.de/2009/08/12/the-hasse-derivative/ for elementary definitions and properties, and the paper -P. Vojta, Jets via Hasse-Schmidt derivations, ArXiv: math/0407113, -for the use of Hasse derivatives in algebraic geometry. -On the analysis level, there are also the Carlitz derivatives, special difference operators working efficiently just on functions annihilated by usual derivatives, with a rich theory of "differential equations", Fourier series, special functions etc. See -A. N. Kochubei, Analysis in Positive Characteristic, Cambridge University Press, 2009. -However there is no geometry around this approach so far.<|endoftext|> -TITLE: Additivity of projective dimensions, or, help me lower my blood pressure -QUESTION [8 upvotes]: Sorry for the shameless title. I'm rather stuck on a lemma in commutative algebra - namely, I have both a proof and a counterexample! I have tried rather strenuously and frustratingly to find the error here, without success; any help from the community in debugging this would be greatly appreciated. -Suppose $R$ is a Noetherian local ring and $M$ is a finite $R$-module of finite projective dimension ($\mathrm{pd}$ for short); write $I=\mathrm{Ann}_R(M)$ in all that follows. -Claim: Under the above hypotheses, we have $\mathrm{pd}_R(R/I)+\mathrm{pd}_{R/I}(M)=\mathrm{pd}_R(M)$. -Proof of claim: Recall the Auslander-Buchsbaum formula, namely $\mathrm{pd}_R(M)+\mathrm{depth}_R(M)=\mathrm{depth}_R(R)$. Since a sequence $r_1,\dots,r_n \in R$ is $M$-regular if and only if $\overline{r}_1,\dots,\overline{r}_n \in R/I$ is $M$-regular, the $R$-depth and $R/I$-depth of $M$ agree. (This is well-known, see e.g. pp. 130-131 of Matsumura's Commutative Ring Theory). Hence Auslander-Buchsbaum, applied to $R$ and $R/I$, gives the equality -$\mathrm{pd}_R(M)-\mathrm{pd}_{R/I}(M)=\mathrm{depth}_R(R)-\mathrm{depth}_{R/I}(R/I)$. -By the same reasoning as previously, the $R$-depth and the $R/I$-depth of $R/I$ are equal, so the right-hand side of this formula can be rewritten as $\mathrm{depth}_R(R)-\mathrm{depth}_{R}(R/I)$, which is equal to $\mathrm{pd}_{R}(R/I)$ by another (!) application of Auslander-Buchsbaum. $\square$ -Counterexample to claim: Take $R$ local Noetherian, $a\in R$ a nonunit, $M=R/(a) \oplus R/(a^2)$, so $I=(a^2)$. If I have done this right, each of the projective dimensions in my claim is exactly $1$, and I believe $1+1\neq 1$ was known in antiquity. - -REPLY [4 votes]: One problem in the proof is that $\operatorname{pd}_R(R/I)$ might be infinite, so you can't apply Auslander-Buchsbaum the final time. -I can't seem to come up with an example where $M$ has finite projective dimension while $\operatorname{ann} M$ has infinite projective dimension, but they must exist. -Later: The famous Dutta-Hochster-McLaughlin example does the trick. It is a module $M$ of finite projective dimension over $R = k[x,y,z,w]/(xy-zw)$, with annihilator $(x,y,z,w)^3 + (x,z)$.<|endoftext|> -TITLE: Which conjectures only need the Grand Riemann Hypothesis to become genuine theorems? -QUESTION [5 upvotes]: Hello, -I've been interested in number theory for several years, and as time goes by, I read more and more articles in which theorems begin with "Assume the Riemann Hypothesis holds." But up to now, I think I've almost never seen any beginning with "Assume the Grand Riemann Hypothesis holds". So, which are those "theorems" that only need the Grand Riemann Hypothesis to become certain results? - -REPLY [3 votes]: The main result -''Assume that the generalized Riemann hypothesis (GRH) for -zeta functions of number fields holds. There exists a deterministic algorithm that on input positive integers $n$ and $k$, together -with the factorization of $n$ into prime factors, computes the element $T_n$ of the Hecke algebra $T(1, k)$ in running time polynomial in $k$ and $\log n$.'' -of the recent book by Couveignes, Edixhoven, et al. (page 3) -http://www.math.univ-toulouse.fr/~couveig/book.htm -assumes the generalized Riemann hypothesis.<|endoftext|> -TITLE: manifold with given rational homology -QUESTION [7 upvotes]: Is there a smooth compact manifold with rational homology vanishing except -in dimensions $0, 8, 16$ where it is $Q$? What would be a good strategy to find such a manifold? - -REPLY [5 votes]: The following paper might be of interest: Rational Analogs of Projective Planes by Zhixu Su. -It discusses the following question: For which $n$ is there a closed $2n$-manifold $M$ with rational cohomology $H^*(M; \mathbb{Q}) \cong \mathbb{Q}[x]/x^3$ with $x$ of degree $n$? Observe that the multiplicative structure is essentially implied by the additive one by Poincare duality. -There are, of course, the classical examples $\mathbb{CP}^2$, $\mathbb{HP}^2$ and $\mathbb{OP}^2$, but are there more $n$ possible? It is clear that $n$ has to be even for this, but actually $n$ has to be divisible by $4$. Beyond the classical examples, the next example occurs in dimension $32$, where there are already infinitely many (up to homeomorphism). This is, of course, a purely rational result - by the Hopf invariant 1 problem the examples above are the only examples for the integral analogue. -The methods are those sketched in Mark Grant's answer.<|endoftext|> -TITLE: What should be learned in an introductory analytic number theory course? -QUESTION [21 upvotes]: Hello all -- -I have the privilege of teaching an introductory graduate course in analytic number theory at the University of South Carolina this fall. What topics should I definitely cover? -I'm not lacking for good material of course. I intend to cover much of Davenport; there is also Cojocaru and Murty's introduction to sieve methods; there is interesting elementary work by Chebyshev et al. on counting primes; there is also Apostol's excellent book; I could dip into Pollack's new book; and there are many other excellent sources as well. I should also make sure the students master partial summation, big-O, and the kinds of contour integration that come up in typical problems. -I feel prepared to do a good job, and I will also have good people to ask for advice in my new department, but I would cheerfully welcome further advice, opinions, etc. from anyone who would like to offer them. Any thoughts? -Thanks to all. --Frank - -REPLY [2 votes]: This may be a stretch, but what about the Burgess bound for character sums? The Riemann hypothesis over finite fields is a ubiquitous tool in number theory now, and this seems as a good an introduction to it as any. Besides, if you're willing (as I would be!) to simply quote the key estimate for complete character sums, the deduction of Burgess from that is basically a clever application of Hölder's inequality and rearrangements, which of course are equally ubiquitous techniques. -My personal one-sentence summary of (most of) analytic number theory is roughly "Poisson summation, Holder's inequality, combinatorial rearrangement, and the Riemann hypothesis over finite fields form a noncommutative monoid", and I feel it would be ideal if students got a glimpse of all four of these techniques in an introductory course.<|endoftext|> -TITLE: Minimum distance distribution between N random points in a cube and the origin -QUESTION [8 upvotes]: We have $N$ points randomly and uniformly chosen in a cube of side $1$ centered at the origin $O$. This means that the coordinates of the point $P_i$ is a vector of random variables $(X_i,Y_i,Z_i)$ where $X_i$, $Y_i$ and $Z_i$ $\sim U([-0.5,0.5])$, $i=1,\ldots, N.$ Let $D_i$ stand for the euclidean distance between $O$ and the point $P_i$, $D_i = \sqrt(X_i^2+Y_i^2+Z_i^2)$. -I would like to calculate the distribution of $D = \min(D_i)$ or perhaps more simply the mean $E(D)$ of $D$. -I faced many problem by computing the convolution of these nasty square uniform variables. The involved integrals become very aggressive after a while. -Any help would greatly appreciated. Thanks -Thomas - -REPLY [9 votes]: I'm going to give a somewhat heuristic solution that nevertheless gives the right answers. -$E(D)$ is the distance from the origin to the closest of N points. -Let's replace $N$ with a random variable, namely the Poisson with mean $N$. Then your points form a Poisson process of rate $N$ on the unit cube. -Let's furthermore assume that $N$ is large enough that the closest point to the origin is at distance less than $0.5$ from the origin -- that is, $D < 0.5$ -- so we don't have to worry about the geometry of the cube. If there are a lot of points this is reasonable. -So, at least when $N$ is large, your question has approximately the same answer as: consider a Poisson process of rate $N$ in $R^3$. What's the distribution of the distance from the origin to the point closest to the origin? -But this scales with $N$ - a Poisson process of rate $N$ looks like a Poisson process of rate 1, shrunk by a factor of $N^{1/3}$. So it's enough to solve this problem when $N = 1$. We'll get $P(D < x) = F(x)$ when $N = 1$. For generic values of $N$ we'll get $P(D < x) = F(x N^{1/3})$. -So what's $P(D < x)$ when $N = 1$? It's easier to find $P(D > x)$; this is the probability that no point is within the sphere of radius $x$ centered at the origin. But this sphere has volume ${4 \over 3} \pi x^3$ and therefore this probability is $\exp(-4\pi x^3 /3)$. From this we find that the density of $D$ is $f(x) = 4\pi x^2 \exp(-4\pi x^3/3)$ and the expectation of $D$ is -$$ \int_0^\infty x f(x) \: dx = {\pi^{2/3} 2^{1/3} \over 3^{7/6} \Gamma(2/3)} = 0.5539602785 $$ -by using Maple. I conjecture, therefore, that $\lim_{N \to \infty} E(D) N^{1/3}$ is equal to this constant; so $E(D) \approx 0.554 N^{-1/3}$. -To check this I simulated the case N = 1000 -- picking 1000 points in each of 1000 cubes, as follows in R: -cp = function() runif(3,-.5,.5) -dist = function(x) sqrt(sum(x^2)) -x = rep(0, 10^3); -for (i in c(1:10^3)) { - m = 1; - for (j in c(1:10^3)) { - m = min(m,dist(cp())); x[i] = m; - } -} - -The average distance turns out to be 0.05554898, very close to $0.554/10$. For picking 10000 points in each of 1000 cubes I get an average distance of 0.02573758 -, very close to $0.554/10^{4/3}$.<|endoftext|> -TITLE: How far can a particle travel from its origin if it exhibits self-avoiding Brownian motion in two-dimensions? -QUESTION [13 upvotes]: Let's say I have a point-like Brownian particle undergoing two-dimensional diffusion on an infinite plane with the caveat is that the particle can never return to a coordinate that it previously visited (these coordinates become fully reflecting to the particle). As such, the particle may 'trap' itself if its trajectory generates a closed loop. -Assuming the Brownian particle has some diffusion coefficient, $D$, is there a known mean and variance for the maximum distance the particle will travel from its origin before trapping itself? -For the discretized / random-walk version of this question - Is there a known result for the mean maximum distance from origin (and associated variance) for a random walker on an infinite two-dimensional lattice of some degree $m$? - -Edit - To try to better clarify what I mean by the particle being 'trapped', I would define this as the state where the furthest point from the origin in the particle's trajectory is fixed due to the particle being unable to escape from a closed curve. I don't mean to imply that the point-like Brownian circle is "frozen in place", akin to a random walker where all edges from its position connect to previously occupied vertices. I fail to see how this situation is possible without some sort of discretization. -In response to Nate Eldredge's comment, there are certainly some serious issues with defining how the Brownian particle acts at the reflecting boundary its trajectory generates. Loosely I would like to say that the particle reflects at a random angle, and I would appreciate any help with making this better defined. Regarding the second point, that "scaling and the Blumenthal 0-1 law would suggest that the trapping happens immediately", does this also hold true if we assign the Brownian particle some drift parameter $\beta$? -In any case, the discrete or lattice version of this question (mentioned above) is very interesting to me. - -REPLY [8 votes]: As per Ron Maimon's suggestion, I did a little simulation for the lattice version, counting the number of paths -that get trapped at or before $n$ steps. -For example, here is a trapped path of 45 steps: - -Because there is a positive probability at any point of forming a shape like the letter 'G' -(if there is sufficient room), the probability of trapping goes to 1 as $n \rightarrow \infty$. -For $n=100$, the probability is already over $\frac{3}{4}$. -         -Addendum. Incidentally, I have some evidence—not definitive—that the mean path length before -reaching a cul-de-sac is about 71.6 steps.<|endoftext|> -TITLE: Weil-Deligne representations: Two monodromy operators? -QUESTION [17 upvotes]: Let $p$ be a prime number, and $F$ be a finite extension of $Q_p$. To any smooth irreducible representation $\pi$ of $G = Gl_n(F)$ we may associate a sort of ``dual´' representation, called the Zelevinsky dual or Aubert dual, constructed as follows. Let $R$ be the Grothendieck group of smooth $G$-representations of finite length. For any standard parabolic subgroup $P$ with Levi-decomposition $P = MN$ we have the functor of induction $Ind_P^G$ and restriction $Res^G_P$ (both normalized so that they send unitary reps to unitary reps, as in the notes of Casselman on p-adic reps). These functors are both exact, and yield morphisms between the Grothendieck group of $G$ and the Grothendieck group of $M$. Thus, one can define for any $X \in R$ the following object -$$ -i(X) = \sum_P \varepsilon_P Ind_P^G(Res^G_P(X)) \in R, -$$ -where the sum ranges over the standard parabolic subgroups of $G$. Then $i(X)$ is the Aubert dual of $X$. The $\eps_P$ is a sign: $(-1)$ to the rank of $P$. The Zelevinsky dual is almost the same thing as the Aubert dual. Zelevinsky's dual is only defined on the isom classes of smooth irreducible reps. If one applies $i$ to an irreducible representation $\pi$ viewed as an element of the Grothendieck group $R$ of $G$ then there is a sign $\varepsilon$ such that $\varepsilon \cdot i(\pi) \in R$ comes from an irreducible $G$-representation $\iota(\pi)$. This representation $\iota(\pi)$ is the Zelevinsky dual of $\pi$. -The Zelevinsky dual has an explicit description in terms of the Zelevinsky segment classification for smooth irreducible $G$-representations. Historically it came before the Aubert dual. -References: Aubert's paper duality dans le group de Grothendieck de la categorie des representations lisses de longueur finie dún groupe reductif p-adique´' -and Zevinsky's paper:Induced representations of reductive p-adic groups II´'. -See also the IHES paper of Schneider and Stuhler. -The operator $i$ commutes with parabolic induction, and leaves cuspidal representations invariant (look at the formula above). So it is ``interesting´' for $p$-adic representations of $G$ that occur as a proper irreducible subquotient of some parabolic induction. For example, it interchanges the Steinberg representation with the trivial representation. More generally, if $\pi$ is a Speh representation associated to the numbers $(a, b)$, then $\iota(\pi)$ is equal to the Speh representation associated to $ (b, a)$, $n = ab$. -Observe also that, because it interchanges Steinberg with the trivial, the involution $i$ may change the ramification of a representation (in my example, ramified became unramified), but there is a bound to it: for example, semi-stable representations will stay semi-stable. -By the Local langlands correspondence, the Zelevinsky dual corresponds to an involution on the set of Weil-Deligne representations. Let us call this involution $i$. I would like to understand this involution in as many different ways as possible. -To fix normalizations, for me the trivial representation of $G$ corresponds to the Weil Deligne representation with trivial $N$, unramified $\rho$ for which geometric Frobenius acts with eigenvalues $q^{(n-1)/2}, q^{(n-3)/2}, \ldots, q^{(1-n)/2}$ ($q$ is the cardinality of the residue field of $F$). -The first question is also the one of the title. It suffices to think only of semi-stable representations. Thus, let $\pi$ be a semi-stable $G$-representation which occurs in some $Ind_B^G(\chi)$, $B$ being the standard borel of $G$. Because $Ind$ and $Res$ are functors we see that $\varepsilon \cdot i(X)$ also occurs in $Ind_B^G(\chi)$. On the Galois side this means that, if $(\rho, N)$ is a Weil-Deligne representation, then $\iota(\rho, N) =: (\rho', N')$ has $\rho = \rho'$. Thus, in some sense, we have made second new monodromy operator $N'$ on the space $(\rho, N)$. So my question: What is $N'$? Is there a direct construction of $N'$ without using Local langlands and Aubert/Zel duality? How do $N$ and $N'$ relate (I only know this for examples...)? -The second question: -Is there a geometric construction for $N'$ for representations that occur in the cohomology of varieties? - -REPLY [9 votes]: Arthur's theory in Matt's answer applies to local representations that are components of automorphic representations, a strong constraint. For a purely local variant of these ideas you may like to glance at this paper, where a geometric realization of the "Zelevinski correspondence" is given for "elliptic" representations, using a Lefschetz operator on the Lubin-Tate spaces.<|endoftext|> -TITLE: Quantization and noncommutative deformations -QUESTION [14 upvotes]: Hello, -I would like to introduce myself to the theory of quantization and noncommutative deformations of Riemann Poisson structures. In fact, I am familiar with Riemannian and Poisson geometry, but I cannot grasp the principle of the theory above. -As I understand, by reading some introductory texts on the subject, ideas come from physics. This involves replacing the coordinates of the phase space (the cotangent of $\mathbb{R}^3$ endowed with its canonical symplectic structure) by operators (on a Hilbert space) which do not commute with each other and is the description of quantum mechanics in a form similar to classical mechanics. Then there was conflict between the riemannian description (or pseudo-Riemannian) of space in general relativity and the space of quantum mechanics. On small scales, the innermost structure of space-time is broken: the position and velocity of a point do not make sense at the same time, if one is defined precisely the other will not defined (the Heisenberg uncertainty principle). This raises the question how to give a physical meaning to the concept of point? -The Gelfand-Naimark theorem first brought a solution to this "internal conflict" of physics, by establishing a bridge between topology and algebra. So we look at a point $x$ in a space $X$ be fixed, and considering its "shadow" consists of the values $f(x)$ taken by all continuous functions $f$ on $X$. This leads to replace the space $X$ by commutative algebra $C(X)$. -The fact that pseudo-Riemannian geometry is a sufficient description of space-time for most purposes, suggests that noncommutativity might be treated as the limit where Planck's constant $\hbar$ tends to $0$ then I understand it, how to become a constant variable and move towards $0$! hence the idea of deformation quantization, which is to construct noncommutative algebras $\mathcal{A}_\hbar$, by deformation of the Poisson algebra $\mathcal{A}=C^\infty(M)$ (formal series in $\hbar$ with coefficients in $\mathcal{A}$) that converges to $\mathcal{A}$ when $\hbar$ tends to $0$. The work most results in this direction is that of Kontsevich. -1) How Heisenberg's uncertainty principle reviews the classical definition of a point? -2) Why deformation always starts with a Poisson manifold? if it is to deform the phase space of hamiltonian mechanics it suffices to consider symplectic manifolds! -3) Why the deformation of the algebra of functions $C^\infty(M)$ of a Poisson manifold is a way of quantization? in this case what is the Hilbert space, in which the observable $f$ are replaced by a bounded operators $\widehat{f}$? -Alain Conne then directed the program algebraization of differential geometry, in order to then work on "noncommutative spaces". He considered riemannians manifolds with additional structure of Spin. Such structure is canonically attached to the triple $(C^\infty(M),L^2(M),D)$ ; what is $L^2(M)$? and $D$ is the Dirac operator, with some number of properties that can be generalized to spectral triples $(\mathcal{A}, H,D) $: $\mathcal{A}$ a noncommutative algebra, $H$ Hilbert space, and $D$ the Dirac operator. -1) Why in physics we need to replace the Laplacian (which is of order $2$) by a differential operator of order $1$ the Dirac operator (which is the square root of the Laplacian)? -3) How to define, precisely, a spectral triple and how to find the metric using the Dirac operator? -4) Why we deform only Spin manifolds? -On the other hand, in an article \url{http://arxiv.org/abs/math/0504232v2} -Eli Hawkins gave some definitions that I not understand (not being familiar with the language of algebraic geometry). In particular, the definition 1.4 (page 4) and the definition of "metacurvature" (page 5). In particular, how an extension of the algebra of differential forms gives rise to a Poisson bracket!? -If $\mathcal{A}_0$ is an algebra, that means an extension of the form -$$0 \rightarrow\hbar\mathbb{A}\rightarrow\mathbb{A} \rightarrow\mathcal{A}_0$$ -where $\hbar$ is a central multiplier $\mathbb{A}$, and for all $a\in\mathbb{A}$ -$$\hbar^2a=0 \Longrightarrow a\in\hbar\mathbb{A}$$ - -REPLY [2 votes]: Why do deformations always start with a Poisson manifold? This maybe slightly imprecise since I am transporting the story from a knowledge of the deformation theory of algebraic varieties and any corrections are welcome. Here is a small piece of the puzzle that is interesting to me as someone interested in deformation theory. Well there is a general fact in algebra that deformations of a ring or algebra R are governed by the Hochschild cohomology, $HH^*(R)$. That is to say that deformations to first order correspond exactly to classes in $HH^2(R)$ Hochschild cohomology, and deformations to all orders can be understood roughly as placing additional conditions on that class that the deformation extend to all orders, that is to say over $\mathbb{R}[[t]]$. Here the ring in question is $C^\infty(M)$, but it could be the coordinate ring of an algebraic variety as well. Technically speaking we need to think of $C^\infty(M)$ as a topological algebra but let's supress that. Anyways, what's true is that for a $C^\infty(M)$, $HH^2(C^\infty(M)) \cong \Gamma(\wedge^2(T_X))$, that is exactly the bivector fields. The additional condition is exactly that the bivector field be Poisson. Thus Poisson structures are exactly what gives rise to deformations. There is a relation on Poisson structures called gauge equivalence which determines when two deformations are deemed to coincide. -It might be worth noting that this situation is specific to smooth manifolds. In the case of a smooth algebraic variety the situation can be more complicated depending upon what you are trying to deform. There one can as before deform the (sheaf of) functions by algebraic bivector fields, but one can also deform the complex structure of the variety as well. In the algebraic case, we care also about modules over our variety, and there is a third class of deformations of the category of modules, called the gerby deformations. So if you are bored with Poisson structures you can have a look at those!<|endoftext|> -TITLE: Semigroup product of the left-invariant completion of a Polish group (restatement of Question 71389) -QUESTION [6 upvotes]: This is a re-statement, of sorts, of the question Is there a relational countable ultra-homogeneous structure whose countable substructures do not have the amalgamation property?, so far unanswered. -Let $G$ be a Polish group, $d_L$ a compatible left-invariant metric on $G$. -This metric is usually not complete, so let $\hat G$ be the completion of $G$ with respect to $d_L$. -If $(g_i)$ and $(h_i)$ are Cauchy sequences in $(G,d_L)$ then so is $(g_i h_i)$, endowing $\hat G$ with a semigroup structure. -Since any two left-invariant compatible metrics on $G$ are uniformly equivalent, none of this depends on the precise choice of $d_L$. - -Question: Given $a,b \in \hat G$, are there always $c,d \in \hat G$ such that $ca = db$? (No idea why this should be true, but then what is a counter-example?) - -Motivation: $G$ can always be viewed as the automorphism group of some complete separable approximately ultra-homogeneous metric structure $M$, and $G$ is a closed subgroup of $S_\infty$ if and only if $M$ can be taken to be a countable ultra-homogeneous discrete structure (what logicians usually understand by "structure"). -Then $\hat G$ is the semi-group of embeddings of $M$ in itself. -Now the question becomes very close (and in the discrete case, possibly equivalent) to the one cited above: can any two copies of $M$ be amalgamated over a common copy of $M$, with the result embeddable in $M$? - -REPLY [4 votes]: In the end it was the original question which was answered first. -The answer to Is there a relational countable ultra-homogeneous structure whose countable substructures do not have the amalgamation property? by Ali Enayat shows that there exists a countable ultra-homogeneous structure $M$ with embeddings of $f_i\colon M \to M$, $i = 0,1$, which do not amalgamate inside $M$. -Taking $G = \textrm{Aut}(M)$, $G$ is a Polish group (and moreover homeomorphic to a closed subgroup of $S_\infty$), $f_i \in \hat G$, and there are no $g_i \in \hat G$ such that $g_0 f_0 = g_1 f_1$. -This gives the desired counter-example. -(Thank you, Ali!)<|endoftext|> -TITLE: How are these algebraic and geometric notions of homotopy of maps between manifolds related? -QUESTION [17 upvotes]: Let $M$ and $N$ be smooth manifolds, and $f,g: M \to N$ smooth maps. Denote by $(\Omega^\bullet M,\mathrm d_M)$ and $(\Omega^\bullet N, \mathrm d_N)$ the cdgas of de Rham forms in each manifold, and by $f^\ast, g^\ast : \Omega^\bullet N \to \Omega^\bullet M$ the pull-back of differential forms along each map. Note that $\Omega^\bullet$ is a fully-faithful functor from Manifolds to CDGAs. -I have been brought up with two not-obviously-the-same notions of "homotopy" between maps $f,g$: -A geometric homotopy between $f,g$ is a smooth map $H : M \times [0,1] \to N$ such that $H(-,0) = f$ and $H(-,1) = g$. -An algebraic homotopy between $f,g$ is a map $\eta: \Omega^\bullet N \to \Omega^{\bullet - 1} M$ of graded vector spaces such that $f^\ast - g^\ast = \eta \mathrm d_N + \mathrm d_M \eta$. -I believe that the following is true. Any geometric homotopy gives rise to an algebraic homotopy, and two geometric homotopies are homotopic iff the corresponding algebraic homotopies are homologous homotopic. Not every algebraic homotopy comes from a geometric homotopy; rather, it should be required to satisfy some (directly-checkable) condition that says roughly that it's an "antidifferential operator". -Unfortunately, I have been unable to really convince myself of either of the above beliefs. Probably this is textbook material, and so maybe my question is to be pointed to the correct textbook. But really my question is: - -How, explicitly, are the above notions of homotopy between maps related? What extra conditions (if any?) should be put on an algebraic homotopy in order for it to be "geometric"? - -It is somewhat embarrassing not to know the sharp relationship between the above concepts, but this is one of the many parts of mathematics that I have picked up largely from conversations and working on the examples that come from particular research questions, and not from ever formally learning such material. - -REPLY [10 votes]: There is a simple way to understand the implication "geometric implies algebraic homotopy" if you remember that $\Omega^*(M \times I)$ is the (projective) tensor product of $\Omega^*(M)$ and $\Omega^*(I)$ of chain-complexes. Take the chain map $Int:\Omega^*(I) \to C^*(I)$ given by integration, where $C^*(I)$ is the simplicial cochain complex of the 1-simplex $I$ (of total dimension 3). -Then a geometric homotopy, composed with $Int$ gives a chain map -$$ \Omega^*(N) \to \Omega^*(M) \otimes C^*(I)$$ -Unravelling this map into its 3 components, corresponding to the two 0-simplices and the 1-simplex of $I$, you get a triple $(f^*,g^*,\eta)$ which is precisely an algebraic homotopy. -It's very hard to go back since an algebraic homotopy is very weak information, it exists if and only if $f$ and $g$ induce the same map on de Rham cohomology (since these are chain complexes over a field). -One thing to do is to pull back the dga structure on $\Omega^*(I)$ to an $A_\infty$-structure on $C^*(I)$ via $Int$ (and a choice of a homotopy inverse). Instead of just an algebraic homotopy, you could then require an $A_\infty$ map -$$ \Omega^*(N) \to \Omega^*(M) \otimes C^*(I)$$ -which extends your given pair $(f^*,g^*)$ on the boundary (note that by construction, an example comes from a geometric homotopy). If $M$ and $N$ are nilpotent, this should guarantee a homotopy on the "realifications" by rational (or better: real) homotopy theory. -For example, this should detect the Hopf maps between spheres but none of their suspensions (since these are torsion).<|endoftext|> -TITLE: (Non)free differential calculus -QUESTION [14 upvotes]: Let $G$ be a group and $R$ be a commutative ring. Recall that a derivation of the group ring $R[G]$ is a map $\delta : R[G] \rightarrow R[G]$ such that -$$\delta(x+y)=\delta(x)+\delta(y) \quad \text{and} \quad \delta(xy) = \delta(x) \epsilon(y) + x \delta(y)$$ -for $x,y \in R[G]$. Here $\epsilon : R[G] \rightarrow R$ is the augmentation map. -There are a lot of derivations of $R[G]$ if $G$ is a free group, as here we have the Fox free differential calculus. -Question : Do there exist interesting derivations on the group ring (analogous in some way to the Fox free derivatives) for groups that are not free? I'm especially interested in fundamental groups of closed surfaces. - -REPLY [17 votes]: There are always derivations of the form $\delta_b(a) = ab - b\varepsilon(a)$ for $b \in R[G]$, which are called inner derivations. Derivations modulo inner derivations are classified by the first group cohomology with coefficients in the $G$-module $R[G]$, $H^1(G,R[G])$. -In the special case $R=\mathbb Z_2$, Stallings defined a group to have infinitely many ends if $H^1(G,\mathbb Z_2[G])$ has dimension $\geq 2$, in which case it is automatically infinite dimensional. A group has two ends $H^1(G,\mathbb Z_2[G])$ is one-dimensional and one end if $H^1(G,\mathbb Z_2[G])$ vanishes. (See John Stallings, On Torsion-Free Groups with Infinitely Many Ends, Ann. of Math., 1968 vol. 88 (2) pp. 312-334). -Stallings proved that a finitely generated group $G$ has more than one end if and only if the group $G$ admits a nontrivial decomposition as an amalgamated free product or an HNN extension over a finite subgroup. (See Wikipedia.) In that sense, there is a perfect understanding of the class of groups you are looking for. -Fundamental groups of surfaces have one end, so that there exist no interesting derivations, at least for $R=\mathbb Z_2$. Computations for PID's are similar, see Section 13.5 in -Ross Geoghegan, Topological Methods in Group Theory (Graduate Texts in Mathematics), Springer. -This book also contains a whole chapter about the structure of $H^*(G,R[G])$ and its relation to the the topology of the universal cover $EG$ of the classifying space of $G$.<|endoftext|> -TITLE: What is a colimit, really? -QUESTION [7 upvotes]: I can sot of give the definition of a colimit (or limit) as the initial (or terminal) cocone (or cone) under (or over) a certain diagram. Some like to say that colimit (or limit) is a functor and indeed one can define it as a left (or right) adjoint of the diagonal (assuming it exists). But if we use the initial or terminal object definition, it is not so much a functor, since it is well defined, but only well defined up to canonical isomorphism. Some choice will give us a real functor, but that is somewhat contrived. So the question is, if this is not a functor, what kind of categorical gadget is it? Any references will be welcome. - -REPLY [23 votes]: Well, the thing that may or may not be a "real functor" (and which may even fail to exist if the limit(/colimit) does not always exist) is in any case a "profunctor" (that is, a functor into $Set^{C^{op}}$ (or $Set^C$ for colimits) rather than into $C$). The limit of a diagram will actually exist just in case the profunctor's value at that diagram is a representable presheaf (that is, one in the range (up to isomorphism) of the Yoneda embedding). If one makes a choice of such a representation at every diagram, one can factor the entire profunctor through the Yoneda embedding, into a genuine functor. This of course is precisely the choice you want to avoid, but it indicates that one can at least still treat the profunctor as an "anafunctor" in such cases (essentially, a functor whose value at an object/morphism is only determined up to isomorphism, in a coherent way). Further reading on profunctors and anafunctors (for example, at the nLab) may be precisely the sort of thing you are looking for. -In short: profunctors are the way to describe adjoints which may exist only partially, while anafunctors are the way to describe functors whose construction requires a number of arbitrary choices (with anafunctors both avoiding the "evil" in making any single choice and the need for the Axiom of Choice in making so many of them).<|endoftext|> -TITLE: How can I see that $H\mathbb{Z}$ doesn't admit a Bousfield complement? -QUESTION [8 upvotes]: From Ravenel's article "Localization and Periodicity in Homotopy Theory": - -Two spectra $E$ and $F$ are said to be Bousfield equivalent when they give the same localization functor, or equivalently when $E_\ast (X)=0$ iff $F_\ast (X)=0$. The equivalence class of $E$ is denoted by $\langle E \rangle$. There is a partial ordering on the set of Bousfield classes. We say that $\langle E \rangle \geq \langle F \rangle$ if $E_\ast (X)=0$ implies that $F_\ast (X)=0$. Thus $\langle S^0 \rangle$ is the biggest class and $\langle pt \rangle $ is the smallest. Smash products and wedges are well defined on Bousfield classes. A class $\langle F \rangle$ is the complement of $\langle E \rangle$ if $\langle E \rangle \vee \langle F \rangle = \langle S^0 \rangle$ and $\langle E \rangle \wedge \langle F \rangle = \langle pt \rangle$. A class may or may not have a complement. It is easy to find examples of classes (e.g., that of an integer Eilenberg-Mac Lane spectrum) that do not. - -I was trying to figure out why this last statement is true, and at first I wanted to apply cohomotopy to a hypothetical equivalence $H\mathbb{Z} \vee F \simeq S^0$, but then I realized that of course there's no reason that we should have such an equivalence. Is there some other easy approach? - -REPLY [10 votes]: Suppose we have $F$ such that $H\wedge F=0$. We need to show that $H\vee F$ has Bousfield class smaller than that of $S$, or in other words, that there exists $X\neq 0$ with $H\wedge X=0$ and $F\wedge X=0$. I claim that we can take $X=I$ (the Brown-Comenetz dual of the sphere, which is the standard counterexample for everything in this theory). Indeed, we have $\pi_k(I)=0$ for $k>0$, so we can write $I$ as the colimit of its Postnikov pieces $I[-n,0]$. Here $I[0,0]$ is an Eilenberg-MacLane spectrum, as are the fibres of the maps $I[-n,0]\to I[-n-1,0]$, so $F\wedge I[-n,0]=0$ for all $n$ by induction, so $F\wedge I=0$. It is also true that $H\wedge I=0$. This is Corollary B.12 in the memoir 'Morava K-theories and localisation' by Mark Hovey and myself; the proof is a fairly straightforward deduction from the fact that $[BP/p,S]_*=0$, which is contained in Ravenel's Lemma 3.2.<|endoftext|> -TITLE: How elementary can we go? -QUESTION [13 upvotes]: It is a theorem of A. Levy, if $\kappa$ is an inaccessible cardinal, then $V_\kappa\prec_{\Sigma_1} V$ namely $V_\kappa$ is an elementary submodel when considering only $\Sigma_1$ sentences. -One might expect that the "amount" of elementarity will grow quickly as we progress with large cardinal axioms, however for the next step, $V_\kappa\prec_{\Sigma_2}V$ we need to get much higher. In order to assure this level of elementarity a supercompact is enough (is it too strong? judging by the stage this theorem appears in Jech's and Kanamori's textbooks I would say that if it is too strong then it is not strong by that much) -To have $\Sigma_3$ we need to go even further to extendible cardinals (again, this might be too strong. I am not too familiar with this notion yet). - -Is there a known large cardinal notion to give $\Sigma_4$ elementarity of $V_\kappa$? What about larger $n$? -I would expect complete elementarity to fail due to some Kunen inconsistency theorem sort of argument, is this true? -Are there results in the reverse direction? Namely if $\kappa$ is such that $V_\kappa\prec_{\Sigma_k}V$ then $\kappa$ has to be inaccessible/supercompact/extendible/etc - -If we use all sort of set theoretic notions to measure how far $V$ is from an inner model (forcing axioms, large cardinals, how the cardinals behave in the inner model compared to $V$, sharps and covering theorems, etc etc). -Assuming the answer to the first question is not "It is inconsistent.", is there a useful way to use this approach to measure the difference between $V$ and its inner models? - -REPLY [3 votes]: I would like to mention, on the second question, that superhuge cardinals are in fact $\Sigma_5$-reflecting, that is to say inaccessible and $V_\kappa\prec_{\Sigma_5} V$. The reason is first that every superhuge cardinal is extendbile, and so $\Pi_4$ formulas are downward absolute in $V_\kappa$. Suppose we have $\exists x(\phi(x,x_0,\dots,x_n))$, where $\phi(x,x_0,\dots,x_n)$ is $\Pi_4$. Let $z$ be a witness to that, and let $\lambda$ be $\Pi_4$-reflecting such that $z\in V_\lambda$. Then, we can find some $j\colon V\rightarrow M$ such that $M\vDash(\Pi_4\text{ formulas are downward absolute in }V_{j(\kappa)})$ and $j(\kappa)>\lambda$. Then $z\in V_{j(\kappa)}$, and so $M\vDash(V_{j(\kappa)}\vDash\exists x(\phi(x,x_0,\dots,x_n)))$, and so $V_{j(\kappa)}\vDash\exists x(\phi(x,x_0,\dots,x_n))$. As $V_\kappa\prec V_{j(\kappa)}$, we have $V_\kappa\vDash\exists x(\phi(x,x_0,\dots,x_n))$ and so $V_\kappa\prec_{\Sigma_5} V$. -In fact, the set of such cardinals form a normal measure below $\kappa$. Let $D=\{X\subseteq\kappa\mid\kappa\in j(X)\}$. Then by a similar argument $M\vDash(\text{$\kappa$ is $\Sigma_5$-reflecting})$, then $U\in D$, where $U=\{\lambda<\kappa\mid\text{$\lambda$ is $\Sigma_5$-reflecting}\}$.<|endoftext|> -TITLE: Index of Toeplitz Operator via Atiyah-Singer -QUESTION [10 upvotes]: Wikipedia tells me that the (easy) fact that the index of a classical Toeplitz operator expresses the winding number of a function follows from the Atiyah-Singer index theorem. As the Toeplitz operator is not really a pseudodifferential operator, there must be some trick..... -Googleing after a while gave me the impression there is also a (relatively independent somehow) family of Toeplitz-type index formulae (around the work of Boutet de Monvel), which can be reduced to Atiyah-Singer, but this seems to be actually quite a longer journey. Also, that's actually for a very much generalized notion of Toeplitz operator and I got the impression that although it carries the same name, this cannot really be what is meant by the Wiki article. -Also, I got the impression that Atiyah-Singer could be proven by reducing it reversely to Boutet de Monvel's Index Theorem, which then would have to be proven independently. Is that correct and does anybody happen to know a reference? - -REPLY [10 votes]: Actually, Toeplitz operators are pseudodifferential operators of order 0. The Atiyah-Singer index theorem can be formulated in sufficient generality that the Toeplitz index theorem is a special case, but it is likely that such a formulation would be proved by reducing everything to the Toeplitz index theorem. It is a general principle in the area that one should use general theory to reduce an index theorem to a single easily calculated example, and the Toeplitz index theorem is a convenient choice of example because it is probably the most elementary index calculation. -One way to see this is to express the index of an elliptic operator as a pairing between a K-theory class and a K-homology class. Using the Kasparov product and other K-homological tools, the index theorem can effectively be reduced to calculating the index pairing between the Bott class in the K-theory group $K^0(D)$, where $D$ is the open unit disk, and the fundamental class in the K-homology group $K_0(D)$ (i.e. the class of the Dirac operator). The Bott class is the imagine under the boundary map $K^1(S^1) \to K^0(D)$ of the class of the unitary $z \mapsto \overline{z}$ on $S^1$, and it turns out that the boundary of the fundamental class of $D$ is the class of the Toeplitz extension. So by the compatibility of the index pairing with boundary maps, the pairing of the Bott class with the fundamental class is the index of the Toeplitz operator $T_{\overline{z}}$, and this is just 1. -You also ask about (other) generalizations of the Toeplitz index theorem. There are a great many results which follow the following basic pattern: - -Begin with a space $X$ equipped with a nice compactification $\overline{X}$. -Choose a nice family of functions on $X$ (often the kernel of some sort of Laplacian), and form a "Hardy space" $H$ by taking the closure in $L^2(\partial X)$ over their boundary values. -Given a function $f \in C(\partial X)$, form the operator $T_f = PfP$ where $P: L^2(\partial X) \to H$ is orthogonal projection. $T_f$ is Fredholm in nice cases, and its index is often an interesting invariant. - -I'm suppressing a subtlety, which is that it is often necessary to look at matrix valued functions rather than just ordinary functions in order to get something interesting. -For example, one can take holomoprhic functions on strongly pseudoconvex domains in $\mathbb{C}^n$ to obtain a direct generalization of the usual Toeplitz index theorem. Alternatively, one can take the kernel of the Dirac operator on an appropriately chosen Riemannian manifold with boundary. -A good reference for everything I have said is the book "Analytic K-homology" by Higson and Roe - chapters 2 and 11 are particularly relevant.<|endoftext|> -TITLE: Developable 3-manifolds in $\mathbb{R}^4$ -QUESTION [5 upvotes]: Is there a classification of the equivalent of a "developable surface" in $\mathbb{R}^4$? -Analogous to: planes, cylinders, cones, and tangent developables in $\mathbb{R}^3$? -Edit: Here I am imagining "developing" a 3-dimensional manifold embedded in $\mathbb{R}^4$ into -$\mathbb{R}^3$. (Apologies for the earlier misleading version!) -I would appreciate any suggestions for source materials here. My only source -(Edit: now evidently misleading) -is one page (p.342) in Hilbert and Cohn-Vossen (Geometry and the Imagination), in which they say: in $\mathbb{R}^4$ - -there are surfaces that are isometric to the Euclidean plane in the small but are not ruled. - -But now I see from the comments that this must mean a two-dimensional surface embedded -in $\mathbb{R}^4$, which is not exactly what I seek. -A precise definition of developable 3-manifold in $\mathbb{R}^d$ would also be much appreciated. -Thanks for pointers! - -REPLY [6 votes]: This is really just an amplification of Deane's answer. As he points out, it is indeed true that a flat submanifold $M^n\subset \mathbb{R}^{n+1}$ whose second fundamental form is nonvanishing is canonically ruled by $(n{-}1)$-planes. However, just being ruled in this way does not imply that the submanifold is flat. For example, the hyperboloid of one sheet in $\mathbb{R}^3$ is ruled by lines (in two ways, in fact), and it is not flat. -The explicit description of the local flat hypersurfaces in $\mathbb{R}^{n+1}$ is, of course, classical, and it goes like this: Start with a regular curve $p:(a,b)\to S^n\subset\mathbb{R}^{n+1}$, choose a smooth function $g:(a,b)\to\mathbb{R}$, and solve the differential equation $dq = g\ dp$, yielding a smooth curve $q:(a,b)\to\mathbb{R}^{n+1}$. Now consider the set $M$, consisting of the vectors of the form $q(t) + v$ where $v$ satisfies $v\cdot p(t) = v\cdot p'(t) = 0$. This will be a (ruled) hypersurface most places, and, where it is, the induced metric will be flat. Note that this says that such hypersurfaces depend on $n$ functions of 1 variable (the first $n{-}1$ of them come from the curve $p$ and one more comes from the function $g$).<|endoftext|> -TITLE: Wasserstein geometry of measures on manifolds related to the generalized Legendre transform and $d^2/2$-convexity -QUESTION [5 upvotes]: Let $(M,g)$ be a fixed closed Riemannian manifold, normalized to have volume 1. We'll write $d_M(x,y)$ for the (geodesic) distance between two points $x,y\in M$. I'm interested in the following class of functions $\varphi: M\to \mathbb{R}$. - -Call $\varphi$, $d^2/2$-convex when there is $\psi:M\to \mathbb{R}$ such that - $$ -> \varphi(x) = - \inf_{y\in M} \left[ \frac 12 d_M(x,y)^2 + \psi(y) \right] -> $$ -In this case, we say that $\varphi$ is the generalized Legendre transform of $\psi$, which we write $\varphi = \psi^c$ - -In particular, one can show that if a $d^2/2$-convex function is bounded, Lipschitz (and thus differentiable $vol_M$ a.e.) with gradient bounded by the diameter of $M$. Also, $(\varphi^c)^c=\varphi$ if and only if $\varphi$ is $d^2/2$ convex. - -I'm interested in these because of their relation to the theory of optimal transport. I'll briefly describe it here, but in theory one should not need to know anything about optimal transport to answer my questions (although it may help as it seems intimately related). I'm intersted in the 2-Wasserstein metric on $\mathcal{P}(M)$, the space of probability measures on $M$. Lying a bit, I'll say that this is defined to be, for $\mu,\nu\in \mathcal{P}(M)$ probability measures -$$ d^W(\mu,\nu)^2 : = \inf_{F:M\to M, F_*\mu = \nu} \int_M d(x,F(x))^2 dvol_M$$ -Here, $F_*\mu = \nu$ means that for all Borel sets $A$, $\mu(F^{-1}(A)) = \nu(A)$. (In fact this is only really true when $\mu$ and $\nu$ are absolutely continuous wrt the volume measure, and it needs a bit of generalization to be really true. Anyways, what is relevant to the beginning of the question, is that for any measure $\nu \in \mathcal{P}(M)$, there is a unique (up to constants) $d^2/2$-convex function $\varphi$ such that $\nu = exp(\nabla \varphi)_* vol_M$, and it turns out this is the unique minimizer in the above definition of distance with $\mu = vol_M$. That is -$$ -d^W(vol_M,\nu) = \int_M |\nabla \varphi|_g^2 d vol_M -$$ -Furthermore, the geodesic from $vol_M$ to $\nu$ in $\mathcal{P}(M)$ (in the metric space sense) is given by $t\mapsto \exp(t\nabla \varphi)_* vol_M$. -Now in this paper, Sturm (not sure if he was the first to do this) shows that this gives a continuous involution on $\mathcal{P}(M)$, by taking the Legendre transform of the $d^2/2$-convex $\varphi$ (uniquely) associated to $\nu$ and setting $\nu^c := \exp(\nabla \varphi^c)_* vol_M$. - -I'm interested in various geometric properties related to this involution, which can be rephrased in elementary terms as follows (I've included my geometric interpretations in quotations): - -Is it true that for $t\in[0,1]$, $(t\varphi)^c = t (\varphi^c)$? -"geodesics from $vol_M$ are mapped to other geodesics from $vol_M$" -(I think I can prove this using some weird scaling arguments, but I'd like an elementary proof just from the definition, which should probably exist if it is true) -What is the relation (if any) between -$$ -\int_M |\nabla \varphi|_g^2 dvol_M -$$ -and -$$ -\int_M |\nabla \varphi^c|_g^2 dvol_M -$$ -"how close are $d^W(vol_M,\nu)$ and $d^W(vol_M,\nu^c)$?" -For any $d^2/2$-convex function $\varphi$, let $M$ be the supremum of $m$ such that $m\varphi$ is $d^2/2$-convex. One can show that $M\varphi$ is then $d^2/2$-convex (the set of $d^2/2$-convex functions is closed in $H^1$, see the Sturm paper). What does $\exp(M\nabla \varphi)_* vol_M$ look like? Is it in general totally singular, etc? "what do the endpoints of geodesics starting from $vol_M$ look like?" -How should I think of this map geometrically, i.e. on the level of measures? - -I'm most interested in (2), followed by (3), but have included (4) just in case someone has any insight - as far as I can tell this is not very well understood and probably does not have a good answer right now. - -REPLY [2 votes]: Concerning (2), if I don't mix up notations and when $\nu$ is absolutely continuous, $x\mapsto x+\nabla\varphi^c$ is the Brenier map from $\nu$ to $vol_M$. As a consequence, it holds $d^W(vol_M,\nu)=\int_M|\nabla\varphi^c|_g^2 d\nu$, so if $\nu$ is far away from $vol_M$ and $\varphi$ is not too close to be constant, then your two integrals can be very different, and I do not see why they should have a relation.<|endoftext|> -TITLE: Characterizations of Abelian varieties (3-folds) in positive characteristic -QUESTION [14 upvotes]: From this question -Characterizations of complex Abelian varieties (especially 3-folds) among projective nonsingular varieties? -I learned that if $X$ is a smooth complex projective variety of dimension $g$, then $X$ is a torsor over an Abelian variety (its Albanese variety) if and only if $\omega_X \cong \mathcal{O}_X$ and ${\rm h^1}(X; \mathcal{O}_X) = g$. -I would like to know the corresponding statement over an algebraically closed field of positive characteristic. If necessary, we could exclude small primes: -By the Bombieri-Mumford classification of surfaces in positive characteristic, I know that the above statement holds for $g=2$ and characteristics different from 2 and 3. In those small characteristics, one also gets examples of "quasi-hyperelliptic surfaces" (essentially because the Albanese could be a non-reduced group scheme), and they distinguish the two classes via étale cohomology (which as far as I can tell, I cannot calculate for my examples of interest). -For my own purposes, I want the answer for $g=3$, but the answer in general is also welcome. If it helps, I also know that ${\rm h}^i(X; \mathcal{O}_X) = \binom{g}{i}$ for all $i$. - -REPLY [5 votes]: I don't have an answer, but a couple of comments that may be useful: -The group $H^1({\cal O}_X)$ can be identified with the Zariski tangent space of ${\rm Pic}^0(X)$. Since the latter can be non-reduced in characteristic $p$ (in characteristic zero, this is impossible by a theorem of Cartier), the dimension of $H^1({\cal O}_X)$ may be larger than $\dim{\rm Alb}(X)=\dim {\rm Pic}^0(X)=b_1(X)/2$. So maybe you would want to ask whether -$$ \omega_X\cong{\cal O}_X, b_1(X)=2g $$ -implies that $X$ is a torsor over an Abelian variety. -For example, quasi-hyperelliptic surfaces with non-reduced ${\rm Pic}^0(X)$ satisfy $\omega_X\cong{\cal O}_X$, $h^1({\cal O}_X)=2$, but their Albanese variety is $1$-dimensional. -Let $f:X\to{\rm Alb}(X)$ be the Albanese morphism. By a result of Igusa, the pull-back map of global $1$-forms -$$ -f^*:H^0(\Omega^1_{\rm Alb X}) \rightarrow H^0(\Omega_X^1) -$$ -has no kernel, which might be useful. -${\bf Correction:}$ However, even if $f$ is generically finite, Igusa's result does not imply that $f$ is generically etale. In fact, in every positive characteristic $p$, there do exist surfaces of general type, whose Albanese morphisms are purely inseparable of degree $p$ onto an Abelian surface (details upon request!). In these examples, the kernel of $f^*\Omega_{\rm Alb X}^1\to\Omega_X^1$ contains a rank $1$ subsheaf. -Varieties with trivial tangent bundles in characteristic $p$ were studied by Mehta, Nori, and Srinivas (Compositio Math. 64 (1987), no. 2, 191-212), and in case these varieties are moreover assumed to be ordinary (which one should think of as the "generic" or "nice" case), then there exists an Abelian variety $A$ and an etale Galois cover $A\longrightarrow X$.<|endoftext|> -TITLE: Pointwise algebraic models of set theory -QUESTION [15 upvotes]: Let $\mathfrak{M} = \langle M, E \rangle$ be a structure for the language of set theory, and take some $B \subseteq M$ and $m \in M$. Say that $m$ is definable over $B$ iff there is a formula $\phi(x,\overline{y})$ in the language and a sequence $\overline{b}$ from $B$ such that $\mathfrak{M} \models \phi[a, \overline{b}]$ iff $a = m$. 'Definable' means definable over the empty set. Call $\mathfrak{M}$ pointwise definable over $B$ iff each of its elements is. -I learned here on MO that the pointwise definable models of ZFC are precisely the prime models of the theory ZFC + V = HOD. (A model of a theory is prime iff it embeds elementarily into every other model of the theory.) In particular, the minimal transitive model of ZFC is pointwise definable, and every countable model of ZFC has a pointwise definable forcing extension. Another thing I learned, in a paper by Marek and Srebrny called "Gaps in the Constructible Universe", is that if a new set of natural numbers appears at level $L_{\alpha+1}$ of the constructible universe, then $L_{\alpha}$ is pointwise definable. Though I haven't learned much about models that are pointwise definable over $B$, I would be delighted if someone schooled me. -Related to definability is the notion of algebraicity. An element $m$ of $M$ is algebraic over $B$ iff there is a formula $\phi(x,\overline{y})$ and a sequence $\overline{b}$ from $B$ such that $\mathfrak{M} \models \phi[m, \overline{b}] \wedge \psi[\overline{b}]$, where $\psi$ is the formula $\exists_{\leqslant n}x\phi(x,\overline{y})$ for some natural number $n$. One question: - -Do we expect notable differences between the theory of pointwise algebraic models of ZFC (or of weaker foundational set theories) and that of pointwise definable models? - -For instance, is there a pointwise algebraic $L_{\alpha}$ such that no new set of naturals appears at $L_{\alpha+1}$? Is there a model theoretic characterization (along the lines of the prime model of ZFC + V = HOD result) of the class of pointwise algebraic models of ZFC? Another question: - -Are there any open questions, relevant to foundations, about the theories of pointwise definable and pointwise algebraic models of set theory? - -REPLY [15 votes]: Update, May 27, 2013. Cole Leahy and I have now written a joint paper arising from issues originating in this question, and here is an excerpt from the post I made on my blog about it, which is adapted from the introduction of the paper. - - -J. D. Hamkins and C. Leahy, Algebraicity and implicit definability in set theory (also at the arxiv), under review. -We aim in this article to analyze the effect of replacing several natural uses of definability in set theory by the weaker model-theoretic notion of algebraicity and its companion concept of implicit definability. In place of the class HOD of hereditarily ordinal definable sets, for example, we consider the class HOA of hereditarily ordinal-algebraic sets. In place of the pointwise definable models of set theory, we examine its (pointwise) algebraic models. And in place of Gödel's constructible universe L, obtained by iterating the definable power set operation, we introduce the implicitly constructible universe Imp, obtained by iterating the algebraic or implicitly definable power set operation. In each case we investigate how the change from definability to algebraicity affects the nature of the resulting concept. We are especially intrigued by Imp, for it is a new canonical inner model of ZF whose subtler properties are just now coming to light. Open questions about Imp abound. -Before proceeding further, let us review the basic definability definitions. In the model theory of first-order logic, an element $a$ is definable in a structure $M$ if it is the unique object in $M$ satisfying some first-order property $\varphi$ there, that is, if $M\models\varphi[b]$ just in case $b=a$. More generally, an element $a$ is algebraic in $M$ if it has a property $\varphi$ exhibited by only finitely many objects in $M$, so that $\{b\in M \mid M\models\varphi[b]\}$ is a finite set containing $a$. For each class $P\subset M$ we can similarly define what it means for an element to be $P$-definable or $P$-algebraic by allowing the formula $\varphi$ to have parameters from $P$. -In the second-order context, a subset or class $A\subset M^n$ is said to be definable in $M$, if $A=\{\vec a\in M\mid M\models\varphi[\vec a]\}$ for some first-order formula $\varphi$. In particular, $A$ is the unique class in $M^n$ with $\langle M,A\rangle\models\forall \vec x\, [\varphi(\vec x)\iff A(\vec x)]$, in the language where we have added a predicate symbol for $A$. Generalizing this condition, we say that a class $A\subset M^n$ is implicitly definable in $M$ if there is a first-order formula $\psi(A)$ in the expanded language, not necessarily of the form $\forall \vec x\, [\varphi(\vec x)\iff A(\vec x)]$, such that $A$ is unique such that $\langle M,A\rangle\models\psi(A)$. Thus, every (explicitly) definable class is also implicitly definable, but the converse can fail. Even more generally, we say that a class $A\subset M^n$ is algebraic in $M$ if there is a first-order formula $\psi(A)$ in the expanded language such that $\langle M,A\rangle\models\psi(A)$ and there are only finitely many $B\subset M^n$ for which $\langle M,B\rangle\models\psi(B)$. Allowing parameters from a fixed class $P\subset M$ to appear in $\psi$ yields the notions of $P$-definability, implicit $P$-definability, and $P$-algebraicity in $M$. Simplifying the terminology, we say that $A$ is definable, implicitly definable, or algebraic over (rather than in) $M$ if it is $M$-definable, implicitly $M$-definable, or $M$-algebraic in $M$, respectively. A natural generalization of these concepts arises by allowing second-order quantifiers to appear in $\psi$. Thus we may speak of a class $A$ as second-order definable, implicitly second-order definable, or second-order algebraic. Further generalizations are of course possible by allowing $\psi$ to use resources from other strong logics. -The main theorems of the paper are: -Theorem. The class of hereditarily ordinal algebraic sets is the same as the class of hereditarily ordinal definable sets: $$\text{HOA}=\text{HOD}.$$ -Theorem. Every pointwise algebraic model of ZF is a pointwise definable model of ZFC+V=HOD. -In the latter part of the paper, we introduce what we view as the natural algebraic analogue of the constructible universe, namely, the implicitly constructible universe, denoted Imp, and built as follows: -$$\text{Imp}_0 = \emptyset$$ -$$\text{Imp}_{\alpha + 1} = P_{imp}(\text{Imp}_\alpha)$$ -$$\text{Imp}_\lambda = \bigcup_{\alpha < \lambda} \text{Imp}_\alpha, \text{ for limit }\lambda$$ -$$\text{Imp} = \bigcup_\alpha \text{Imp}_\alpha.$$ -Theorem. Imp is an inner model of ZF with $L\subset\text{Imp}\subset\text{HOD}$. -Theorem. It is relatively consistent with ZFC that $\text{Imp}\neq L$. -Theorem. In any set-forcing extension $L[G]$ of $L$, there is a further extension $L[G][H]$ with $\text{gImp}^{L[G][H]}=\text{Imp}^{L[G][H]}=L$. -Open questions about Imp abound. Can $\text{Imp}^{\text{Imp}}$ differ from $\text{Imp}$? Does $\text{Imp}$ satisfy the axiom of choice? Can $\text{Imp}$ have measurable cardinals? Must $0^\sharp$ be in $\text{Imp}$ when it exists? (An affirmative answer arose in conversation with Menachem Magidor and Gunter Fuchs, and we hope that $\text{Imp}$ will subsume further large cardinal features. We anticipate a future article on the implicitly constructible universe.) Which large cardinals are absolute to $\text{Imp}$? Does $\text{Imp}$ have fine structure? Should we hope for any condensation-like principle? Can CH or GCH fail in $\text{Imp}$? Can reals be added at uncountable construction stages of $\text{Imp}$? Can we separate $\text{Imp}$ from HOD? How much can we control $\text{Imp}$ by forcing? Can we put arbitrary sets into the $\text{Imp}$ of a suitable forcing extension? What can be said about the universe $\text{Imp}(\mathbb{R})$ of sets implicitly constructible relative to $\mathbb{R}$ and, more generally, about $\text{Imp}(X)$ for other sets $X$? Here we hope at least to have aroused interest in these questions. - - - -Original answer: -It is a very nice question. -If you restrict to well-founded models, and this includes your $L_\alpha$ examples, then a model is pointwise definable if and only if it is pointwise algebraic. The forward implication is clear. For the backward implication, suppose that $M$ is pointwise algebraic; let us prove that $M$ is pointwise definable by induction on rank. Consider any element $a$, and assume all sets of lower rank are definable in $M$. Since $M$ is algebraic, there is a definable finite collection $a_0,a_1,\ldots, a_n$ that includes $a=a_0$, with this set of minimal finite size. These sets must all have the same rank, since otherwise we could make a smaller definable family including $a$, and so all their elements are definable. Thus, if $a\neq a_i$, there must be some element in one of them that is not in the other. But that element is definable, and so we can again make a smaller family by adding to the definition the requirement that the set must contain (or omit, whatever $a$ does) that new element. This would make a smaller definable set, violating the minimality of the finite set, unless indeed our minimal set had only one element. So $a$ is definable after all. -This argument works only in well-founded models, however, since the induction is not internal. -Update. In a conversation with Leo Harrington at math tea here at -the National University of Singapore, where I am visiting, -we worked out the general ZFC case with the following observation: -Theorem. Every pointwise algebraic model of ZFC is -pointwise definable. -Proof. Suppose that $M\models\text{ZFC}$ and is pointwise -algebraic. Note that this implies that every ordinal of $M$ -is definable, since if we can define a finite set of -ordinals containing some ordinal $\alpha$, then since the -ordinals are definably linearly ordered, $\alpha$ is the -$k^{th}$ member of that set and hence definable. Now, we -argue that every set $A$ of ordinals in $M$ is pointwise -definable. Well, since $A$ is algebraic, it is a member of -a finite definable set. But the lexical order on sets of -ordinals is definable and linear, and so again we may find a -definition of $A$, since it will be the $k^{th}$ element in -that finite set for some $k$. Thus, every set of ordinals -in $M$ is definable in $M$. But by ZFC, every set $a$ is -coded by a set of ordinals, and since that set of ordinals -is definable, it follows that the original set $a$ is also -definable. Thus, every set in $M$ is definable without -parameters. QED -After this, I realized that we can actually omit the use of -choice. -Corollary. Every pointwise algebraic model of ZF is -a pointwise definable model of ZFC+ V=HOD. -Proof. Suppose that $M\models\text{ZF}$ and is pointwise -algebraic. It follows as in the theorem above that every -ordinal of $M$ is definable without parameters. Thus, every -object in the HOD of $M$ is also definable in $M$ without -parameters. If $M$ is not equal to its HOD, then let $A$ be -an $\in$-minimal element of $M-\text{HOD}$. Since $A$ is -algebraic, there is a finite definable set containing $A$. -By minimality, every element of $A$ is in HOD, and so we -have a definable well-ordering on the elements of the -members of the definable set containing $A$. Thus, there is -a definable linear ordering (induced from the lexical order on the -definable HOD order) on the subsets of HOD, and so $A$ is -the $k^{th}$ element of the finite definable set for some -finite $k$, and so $A$ is definable in $M$ without -parameters. In this case, since $A\subset\text{HOD}$, it -would mean that $A$ should be an element of HOD, contrary -to assumption. Thus, $M=\text{HOD}^M$, and so $M$ is a -model of ZFC+V=HOD, and also pointwise definable by -the theorem.QED<|endoftext|> -TITLE: Embedding the product of three circles in the 4-sphere. -QUESTION [10 upvotes]: Let $M$ be a smooth submanifold of the 4-sphere $S^4$. I'm going to demand that $M$ be diffeomorphic to $S^1 \times S^1 \times S^1$. By Jordan-Brouwer separation, $M$ separates the 4-sphere into two compact 4-manifolds $V_1$ and $V_2$, i.e. $V_1 \cup V_2 = S^4$, $V_1 \cap V_2 = M$, $\partial V_1 = \partial V_2 = M$. -The question is, is it possible for the rank of $H_1(V_1, \mathbb Z)$ to be zero? -A little Mayer-Vietoris sequence argument will convince you that $H_i M \simeq H_i V_1 \oplus H_i V_2$ for $i \in \{1,2\}$, the map given by inclusion. -I believe all known embeddings of $(S^1)^3$ in $S^4$ have $rank(H_1(V_i, \mathbb Z)) \geq 1$ for both $i$ -- so one will have rank $1$, the other rank $2$. -Off the top of my head I don't see a reason why that should always be true. -This is a question that came up in a discussion with Jonathan Hillman. - -REPLY [2 votes]: No embedding of a product $M=T_g\times{S^1}$ in $S^4$ can have one complementary component -$X$ with $H_1(X)=0$. -For otherwise, the other component $Y$ would have $H_2(Y)=0$. -But then the inclusions of $M$ and of a wedge of $2g+1$ circles into $Y$ would induce -isomorphisms on the lower central series quotients of the fundamental groups, -by an old theorem of Stallings. -This cannot be so, as $\pi_1(M)$ has a central factor, -whereas the free group $F(2g+1)$ does not.<|endoftext|> -TITLE: What is the state in the WRT TQFT associated to a handlebody? -QUESTION [7 upvotes]: Let $Y^3$ be a handlebody with boundary $\Sigma$. By definition, there is some associated vector $v_{WRT}(Y^3)\in Z(\Sigma)$, the (finite dimensional) Hilbert space associated to $\Sigma$ by the Witten-Reshetikhin-Turaev TQFT. I'd like to understand what this vector is. -In short, $Z(\Sigma)$ is a space of sections of a line bundle over the $\mathrm{SU}(2)$ character variety of $\Sigma$. I am hoping that the section $v_{WRT}(Y^3)$ achieves its maximum value (with respect to the canonical inner product on the line bundle) on the Lagrangian submanifold of the character variety consisting of those representations which extend to $Y^3$. [EDIT: there is a good reason to believe this holds, since then high powers of the section will concentrate on this Lagrangian, giving Volume Conjecture-like convergence to the classical Lagrangian intersection theory as the level of the TQFT goes to infinity] -In more detail, let's discuss an explicit description of $Z(\Sigma)$. There is a natural line bundle $\mathcal L$ over the character variety $X:=\operatorname{Hom}(\pi_1(\Sigma),\mathrm{SU}(2))/\mathrm{SU}(2)$. There is a natural symplectic form on $X$, and choosing a complex structure on $\Sigma$ equips $X$ with a complex structure which together with the symplectic form makes $X$ a Kahler manifold. Then $Z(\Sigma)$ is the Hilbert space of square integrable holomorphic sections of $\mathcal L$ ($\mathcal L$ carries a natural inner product, and the curvature form of the induced connection coincides with the natural symplectic form on $X$). -My question is then: how can one describe $v(Y^3)\in Z(\Sigma)$? Does the corresponding section achieve its maximum value on the Lagrangian subvariety of $X$ comprised of those characters of $\pi_1(\Sigma)$ extending to characters of $\pi_1(Y)$? -Comment: answering this question for an arbitrary $3$-manifold $Y^3$ seems unlikely to yield a clean answer, since it includes as a special case calculating the value of the WRT TQFT applied to $Y$ (and the description of this requires the introduction of a whole bunch of extra stuff, e.g. surgery diagrams for $Y^3$, etc.). This is why I am restricting to the case that $Y^3$ is a handlebody, in hopes that in this special case, there is a clean answer to this question. - -REPLY [4 votes]: I can think of two cases where the Witten-Reshetikhin-Turaev vector $Z_k(Y^3)\in Z_k(\Sigma)$ has been connected to a Lagrangian state as you described: --Laurent Charles and Julien Marché showed that the WRT vector of the figure eight knot complement $Z_k(E_K)$ is a Lagrangian state concentrating on $\mathcal{M}(E_K)=\mathrm{Hom}(\pi_1 E_K, \mathrm{SU}_2)/\mathrm{SU}_2$ and used this to prove Witten's asymptotic expansion conjecture for Dehn-fillings of the figure eight. -(see here part1 and part2 ) --In my thesis, I considered the vectors $Z_k(H,\Gamma)$ where $H$ is a handlebody and $\Gamma$ is a colored trivalent banded graph in $H$ that is a spine of $H.$ Such vectors form a basis of $Z_k(\Sigma),$ and the vector $Z_k(H)$ of the empty handlebody is obtained taking all colors=0. -I showed for generic colors, the vectors $Z_k(H,\Gamma)$ are Lagrangian states concentrating on subsets of the form $\lbrace \mathrm{Tr}\rho(C_i)=x_i \rbrace$ where the curves $C_i$ bound disks in $H$ that are dual to the edges of $\Gamma$. -Both results were obtained from the combinatorial definition of TQFT (Reshetikhin-Turaev, or rather Blanchet-Habegger-Masbaum-Vogel), after choosing an isomorphism $Z_k(\Sigma) \rightarrow H^1(\mathcal{M}(\Sigma),\mathcal{L}^k)$ that turns curve operators in TQFT into Toeplitz operators whose principal symbols are trace functions on the moduli space $\mathcal{M}(\Sigma)$. -Although the problem of showing that the WRT vector $Z_k(H)$ of a handlebody concentrates on the character variety of $H$ seem reasonable, one gets into trouble because of the singularities of $\mathcal{M}(\Sigma)=\mathrm{Hom}(\pi_1\Sigma,\mathrm{SU}_2)/\mathrm{SU}_2,$ and also somewhat because one has to work at the critical levels of the principal symbols of the curve operators. -Laurent and Julien's work avoided this difficulty because $\mathcal{M}(T^2)$ is not really singular, and in my paper because I only considered the geometric quantization of dense open sets of $\mathcal{M}(\Sigma)$ and not the whole moduli space. -It is my feeling that if one was able to give a clean answer to your question, one would not be too far from proving Witten's asymptotic expansion conjecture, at least for $3$-manifolds with a "generic" Heegaard splitting.<|endoftext|> -TITLE: Fundamental Groups of compact Complex manifolds? -QUESTION [13 upvotes]: Hi, -are limitations on the fundamental group for compact complex manifolds known? -Can an arbitrary (finite represantable) group be the fundamental group of a compact -complex manifold? -Thanks - -REPLY [19 votes]: Just to give one more refference, there is now a new proof of this theorem that does not use the deep result of Taubes, the proof is elementary and 8 pages long: -http://arxiv.org/abs/1104.4814 - -REPLY [16 votes]: Every finitely presented group is the fundamental group of a compact complex manifold of dimension $3$. -This is proven in the book by Amoros, Burger, Corlette, Kotschick and Toledo Fundamental groups of compact Kahler manifolds, Corollary 1.66 p. 19. -The rough idea of proof is the following. Let $\Gamma$ be a finitely presented group, and let $Y$ be a smooth closed oriented $4$-manifold with $\pi_1(Y) \cong \Gamma$. Then by a result of Taubes one can find a complex $3$-fold with the same fundamental group by taking the twistor space $Z$ of $X=Y \sharp n \overline{\mathbb{C} \mathbb{P}^2}$ for $n$ sufficiently large.<|endoftext|> -TITLE: Vitali Sets vs Bernstein Sets... -QUESTION [10 upvotes]: AC is enough to guarantee the existence of both Bernstein Sets and Vitali Sets... -However is the existence of Vitali Sets strictly weaker than that of Bernstein Sets? -What about the other way round? - -REPLY [5 votes]: For your second definition of Vitali set, I have a weak partial answer. Namely the existence of a Bernstein set does not imply the existence of a $T$-Vitali set. The answer can be found in logic blog maintained by Andre Nies: -http://dl.dropbox.com/u/370127/Blog/Blog2012.pdf. -Added: A The Logic Blog is on the arXiv. According to a comment below, the following is a better reference than the dropbox link: - -Logic Blog, 2012, Andre Nies (editor), http://arxiv.org/abs/1302.3686 - -Note that a Turing degree does not need to be an addition group. -I don't know whether the existence of a Vitali set implies the existence of a Bernstein set. But it is not difficult to see, under $ZF+DC$, that there is a Vitali set (if it exists) which contains a perfect subset. -For you first definition of Vitali set, I have no idea.<|endoftext|> -TITLE: What are the different theories that the motivic fundamental group attempts to unify? -QUESTION [16 upvotes]: I must preface by confessing complete ignorance in the subject. I've read introductory texts about the theory of motives, but I am certainly no expert. -In http://www.math.ias.edu/files/deligne/GaloisGroups.pdf Deligne talks about (introduces?) the motivic fundamental group. But what is the purpose of this object? -Motives come up in cohomology in order to unify the different Weil cohomologies. But I only know of one way to define the algebraic fundamental group! After all, in cohomology you have a choice of coefficients (in a sheaf), but in the definition of the fundamental group (to my knowledge) there is no equivalent to this. Is the point that just as the algebraic fundamental group classifies etale covers, we can do this for other Grothendieck topologies as well? In what sense would the motivic fundamental group unify these? -The above was really one question: i. What theories does the motivic fundamental group unify, and in what sense does it do so? -I will add two more: -ii. Is the existence of the motivic fundamental group conjectural? -and -iii. What purpose does the motivic fundamental group serve other than unifying? Deligne makes some references to conjectures that arise as prediction related to the motivic fundamental group. What insight does it provide? -I'm well aware that Deligne's text probably has all the answers to these questions, but I find it to be a hard read, so the more I know coming in the more I will take out of it. - -REPLY [13 votes]: As in Birdman's comment, the motivic fundamental group is unifying the notion of monodromy action on the fibers of local systems of "geometric origin." -To explain this, let us start with the case of a field $K$. We have a semisimple $\mathbb{Q}$-linear Tannakian category $\operatorname{Mot}_K$ of (pure) motives over $K$ for which fiber functors are cohomology theories, i.e., it makes sense to have an $L$-valued fiber functor for a field $L$, and this is the same as a Weil cohomology theory for smooth proper $K$-varieties with values in $L$-vector spaces. A motivic Galois group, to my understanding, is attached to a cohomology theory/fiber functor $F$ of $\operatorname{Mot}_K$. -Then the motivic Galois group is the associated group scheme/$L$ whose representations are given by the category $\operatorname{Mot}_{K}\underset{\mathbb{Q}}{\otimes}L$, i.e., it is the group scheme of automorphisms of the fiber functor $F$. So it is "the group which acts on $F$-cohomology of (smooth projective) varieties." Since this category is semi-simple, the motivic Galois group is pro-reductive. E.g., the absolute Galois group (considered as a discrete group scheme) of $K$ acts on $\ell$-adic cohomology, so there is a homomorphism from $\operatorname{Gal}(K)$ to the motivic Galois group of $K$ corresponding to the fiber functor defined by $\ell$-adic cohomology. -For, say, a smooth variety $X$ over $K$, there should be a category of "motivic sheaves" on $X$, or rather, a semi-simple category of pure motivic sheaves contained in an Artinian category of mixed motivic sheaves. You should have e.g. an $\ell$-adic" fiber functor from the mixed category to $\ell$-adic perverse sheaves on $X$ which sends pure guys to (cohomologically shifted) lisse sheaves (alias local systems). E.g., if $K=\mathbb{F}_q$, then this is the category of pure (resp. mixed) perverse sheaves on $X$. If $K=\mathbb{C}$, this should be a full subcategory of pure (resp. mixed) polarizable Hodge modules on $X$. For any smooth proper (resp. just any) map $f:Y\to X$, there should an object in the category of pure (resp. mixed) motivic sheaves on $X$ corresponding to push-forward of the structure sheaf on $Y$. -The motivic fundamental group act on the ``fibers" of pure motivic sheaves on $X$. I.e., for a $K$-point of $X$, you should get a functor to the category of $K$-motives. This is a motivic incarnation of taking the fiber of a local system. Then given our cohomology theory $F$, we obtain a functor from pure motivic sheaves on $X$ to $L$-vector spaces, and the automorphisms of this functor will be the $F$-realization of the motivic Galois group of $X$.<|endoftext|> -TITLE: Measure conjugacy and ergodic decomposition -QUESTION [11 upvotes]: Roughly speaking, this question asks whether there is a measure-conjugacy between two transformations if there are measure-conjugacies between their ergodic components. -Suppose $(X,\mu)$ is a standard probability space and $T$ and $S$ are measure-preserving transformations of $(X,\mu)$. By the ergodic decomposition theorem, there are standard probability spaces $(Y,\nu), (Z,\zeta)$ and Borel maps $\phi:Y \to M_1(X), \psi:Z\to M_1(X)$ (where $M_1(X)$ is the space of Borel probability measures on $X$) such that for a.e. $y \in Y$, $\phi(y)$ is ergodic, $T$-invariant and $\int \phi(y)~d\nu(y)=\mu$. Similarly, for a.e. $z\in Z$, $\psi(z)$ is ergodic and $S$-invariant and $\int \psi(z)~d\zeta(z)=\mu$. -Now suppose there is a measure-space isomorphism $\Omega:(Y,\nu) \to (Z,\zeta)$ such that for a.e. $y \in Y$, $(T,X,\phi(y))$ is measurably conjugate to $(S,X,\psi(\Omega(y)))$. Then is $T$ measurably conjugate to $S$? - -REPLY [3 votes]: This is probably related to the following question (and, most likely, can be obtained from it or from a modification of the argument): let M and N be two finite von Neumann algebras with centers $Z(M)$ and $Z(N)$ and faithful traces $\tau_M$, $\tau_N$. Assume that there is a (trace-preserving) isomorphism $(Z(M),\tau_M)\cong L^\infty(X,\mu)\stackrel{\alpha}{\to} L^\infty(Y,\nu)\cong (Z(N),\tau_N)$ so that the central components $M_x$ and $N_{\alpha(x)}$ are a.e. isomorphic. Does it follow that $M\cong N$? -This was proved in this form by Effros [Trans. Amer. Math. Soc. 121 (1966), 434--454; MR0192360 (33 #585)] with a later (shorter) proof by Elliott [MR0310659 (46 #9757) -An extension of some results of Takesaki in the reduction theory of von Neumann algebras. -Pacific J. Math. 39 (1971), 145–148.] -The proofs rely on existence of Borel structure on von Neumann algebras and Borel selection theorems (so, I would guess, they should go through in your context as well?).<|endoftext|> -TITLE: $\aleph_\omega$ many subsets of $\aleph_\omega$ -QUESTION [8 upvotes]: Consider the following question: -Is there a family $\mathcal{F}$ of subsets of $\aleph_\omega$ that satisfies the following properties? -(1) $|\mathcal{F}|=\aleph_\omega$ -(2) For all $A\in \mathcal{F}$, $|A|<\aleph_\omega$ -(3) For all $B\subset \aleph_\omega$, if $|B|<\aleph_\omega$, then there exists some $B'\in \mathcal{F}$ such that $B\subset B'$. -I am not sure if there is anything special about $\aleph_\omega$, but this was the example that came up. -Any help? - -REPLY [5 votes]: This question has been already answered thoroughly. I just wanted to address the OP's comment "I am not sure if there is anything special about $\aleph_\omega$". -Actually, there is nothing special about $\aleph_\omega$ other than the fact that it's a singular cardinal. Let $\kappa$ be a cardinal and let $S(\kappa)$ be the following statement: - -There is a family $\mathcal{F} \subset [\kappa]^{<\kappa}$ such that $|\mathcal{F}|=\kappa$ and for every $F \in [\kappa]^{<\kappa}$ there is $G \in \mathcal{F}$ such that $F \subset G$. - -Then $S(\kappa)$ holds if and only if $\kappa$ is a regular cardinal. -But things become more complicated if we just consider subsets of $\kappa$ of a fixed cardinality smaller than $\kappa$. For example, let $C(\kappa)$ be the statement: - -There is a family $\mathcal{F} \subset [\kappa]^{\aleph_0}$ such that $|\mathcal{F}|=\kappa$ and for every $F \in [\kappa]^{\aleph_0}$ there is $G \in \mathcal{F}$ such that $F \subset G$. - -Then $C(\aleph_n)$ is true for every $0< n< \omega$, $C(\aleph_\omega)$ is false for essentially the same reason that $S(\aleph_\omega)$ is false, but the truth value of $C(\aleph_{\omega+1})$ depends on your set theory. Namely, if there is an $\aleph_{\omega+1}$-sized family of countable subsets of $\aleph_\omega$ which is cofinal in $([\aleph_\omega]^\omega, \subseteq)$ then $C(\aleph_{\omega+1})$ is true, while if $cof([\aleph_\omega]^\omega, \subseteq) \geq \aleph_{\omega+2}$ (which is consistent with ZFC, modulo large cardinals) then $C(\aleph_{\omega+1})$ is clearly false...<|endoftext|> -TITLE: The parity conjecture -QUESTION [15 upvotes]: The parity conjecture for elliptic curves predicts that the rank of an elliptic curve -defined over the rationals has the same parity as the p-Selmer rank for a prime number p. Could anyone familiar with the recent development sketch what has happened in the last -few years, and what the state of the art is? - -REPLY [6 votes]: Let me add a few remarks to the very nice CW answer already given. - -The parity conjecture (i.e. algebraic rank equals analytic rank modulo 2) is known for all elliptic curves over number fields (not just over $\mathbb{Q}$) under the assumption that Tate-Shafarevich groups of elliptic curves over number fields are finite. The survey by Tim Dokchitser that has already been linked to describes the proof. -There are refined parity conjectures for twists by Artin representations. I will also take this opportunity to explain the content of the $p$-parity conjectures in a little more detail. Let $A/K$ be an Abelian variety and let $\tau$ be an Artin representation of $G_K$. Let $p$ be a prime number. Consider the Pontryagin dual of the $p^{\infty}$-Selmer group of $A/K$ and take the tensor product with $\mathbb{Q}_p$, call this $\chi_p(A/K)$. This is a $\mathbb{Q}_p$-vector space. If we believe that the $p$-primary part of the Tate-Shafarevich group of $A/K$ is finite, then the $\mathbb{Q}_p$-dimension of $\chi_p(A/K)$ is exactly the rank of $A(K)$. If we don't assume this, then we have to allow for the possibility of some copies of $\mathbb{Q}_p/\mathbb{Z}_p$ inside the Tate-Shafarevich group increasing the dimension. In any case, $\chi_p(A/K)$ is a $G_K$-representation, and we can consider the number of copies of $\tau$ inside it: $\langle\tau,\chi_p(A/K)\rangle$. On the analytic side, we have the twisted $L$-function $L(A/K,\tau,s)$ and its root number $w(A/K,\tau)$. The $p$-parity conjecture for twists now predicts that -$$ -(-1)^{\langle\tau,\chi_p(A/K)\rangle} = w(A/K,\tau). -$$ -If we believe in the finiteness of Tate-Shafarevich groups, we could instead work with the $\tau$-isotypical component of the Mordell-Weil group $A(K)$. Anyway, what I wanted to say is that we now know the $p$-parity conjecture for various different twists, here are some examples: DD1, Theorems 1.3, 1.11, 1.12, DD2, Theorem 1.11.<|endoftext|> -TITLE: A geometric/topology notion of Typical Sequences? Power of typical sequences in multiuser channels? -QUESTION [6 upvotes]: The idea of Typical sequences(http://en.wikipedia.org/wiki/Typical_set) is a crucial concept in Shannon's proof of the Noisy channel coding theorem. Unfortunately the notion is not sufficient to settle the capacity of transmission of non-communicating correlated sources over independent channels to a common noisy receiver? Even the problem of two-user interference channel capacity with uncorrelated sources is open. What makes it hard to apply typicality to these cases? -Is it possible to associate a geometry/topology to easily visualize typical sequences(atleast when the alphabets are $1$-dimensional reals - more complicated cases include matrix or non-commutatitve alphabets such as in Multiple Input and Multiple Output systems)? Shannon's proof(http://plan9.bell-labs.com/cm/ms/what/shannonday/shannon1948.pdf) is geometry-less and very abstract. - -REPLY [2 votes]: yes,there is an illustrative proof here : Network information theory by Abbas El gammal and Y.kim<|endoftext|> -TITLE: Distance functions on elliptic curves over number fields -QUESTION [7 upvotes]: My question originates from the book of Silverman "The Aritmetic of Elliptic Curves", 2nd edition (call it [S]). On p. 273 of [S] the author is considering an elliptic curve $E/K$ defined over a number field $K$ and he introduces the notion of a $v$-adic distance from $P$ to $Q$. This is done as follows: -Firstly, let's fix an absolute value (archimedean or not) $v$ of $K$ and a point $Q\in E(K_v)$ (here $K_v$ is the completion of $K$ at $v$). Next let's pick a function $t_Q \in K_v(E)$ defined over $K_v$ which vanishes at $Q$ to the order $e$ but has no other zeroes. Now the $v$-adic distance from $P \in E(K_v)$ to $Q$ is defined to be $d_v(P, Q) := \min (|t_Q(P)|_v^{1/e}, 1)$. We will say that $P$ goes to $Q$, written $P~\xrightarrow{v}~ Q$, if $d_v(P, Q) \rightarrow 0$. Later in the text (among other places in the proof of IX.2.2) the author considers a function $\phi\in K_v(E)$ which is regular at $Q$ and claims that this means that $|\phi(P)|_v$ is bounded away from $0$ and $\infty$ if $P~\xrightarrow{v}~ Q$. -I have a couple of questions about this: - -How does one choose a $t_Q$ that works? In the footnote in [S] it is demonstrated how one could use Riemann-Roch to pick a $t_Q$ that has a zero only at $Q$. It seems to me however that such a procedure will not make sure that $t_Q$ is defined over $K_v$ since $K_v$ is not algebraically closed. -For $\phi$ as above which does not vanish nor has a pole at $Q$, how does one see that $|\phi(P)|_v$ is bounded away from $0$ and $\infty$ as $P~\xrightarrow{v}~ Q$? -Do these $d_v$ have anything to do with defining a topology on $E(K_v)$? I assume not, since I don't see how to make sense of it; but then on the other hand they are called "distance functions"... - -REPLY [7 votes]: Some complement to Joe Silverman's answer. Any algebraic variety over $K_v$ (or any topological field) has a canonical topology induced by that of the base field. This topology can be defined by a distance (far from to be unique). Over $\mathbb{P}^n_{K_v}$, a distance can be given (once a system de coordinates is fixed) by -$$ d((x_0,\dots, x_n), \ (y_0, \dots, x_n))= \dfrac{\max_{i, j} \lbrace |x_iy_j-x_jy_i|_v \rbrace}{(\max_i\lbrace |x_i|_v\rbrace\max_j\lbrace |y_j|_v \rbrace)} -$$ -This is a non-archimedean distance. Concretely, one can see that if there exists an index $r$ such that $|x_i|_v\le |x_r|_v$ and $|y_j|_v\le |y_r|_v$ for all $i, j$, then -$$d(x,y)=\max_i \lbrace |x_i/x_r - y_i/y_r|_v \rbrace. $$ -Otherwise $d(x,y)=1$. -One can describe this distance as following: there is a canonical reduction map $\pi: \mathbb P^n(K_v) \to \mathbb P^n(\mathbb k_v)$ where $k_v$ is the residue field of $K_v$. This map consists, after dividing by a coordinate of maximal absolute value, in reducing the coordinates mod $m_v$. -The fiber $\pi^{-1}(P)$ of a rational point $P\in \mathbb P^n(\mathbb k_v)$ is just an open unit polydisk. Now if $\pi(x)=\pi(y)$, then $d(x,y)$ is the usual distance in the unit polydisk (maximum of the $|x_i-y_i|_v$), and $d(x,y)=1$ otherwise. -For any quasi-projective variety $X$ over $K_v$, the embedding in some projective space induces a distance on $X$ with the above distance on projective spaces. -If $X$ has a smooth quasi-projective model $\mathcal X$ such that canonical map $\mathcal X(O_v)\to X(K_v)$ is surjective (hence bijective), one can define a distance using the reduction map $\pi: X(K_v)\simeq \mathcal X(O_v)\to \mathcal X(k_v)$ similarly to the projective space (the fibers of $\pi$ are analytically isomorphic to a polydisk). If we embedd $\mathcal X$ into a projective space $\mathbb P^n_{O_v}$, then this distance is induced by the above distance on $\mathbb P^n$. -This applies to abelian varieties with their Néron models and the distance is canonical (compatible with homomorphisms of abelian varieties). I don't have access to the books of Lang and of Hindry-Silverman at home, I guess the distance described here has something to do with the good distance function that Joe alludes to.<|endoftext|> -TITLE: A follow up question related to entropy -QUESTION [6 upvotes]: For a self-map $\varphi:X\longrightarrow X$ of a space $X$, many important notions of entropy are defined through a limit of the form $$\lim_{n\rightarrow\infty}\frac{1}{n}\log a_n,$$ where in each case $a_n$ represents some appropriate quantity (see, for example, this answer to one of my previous questions.) Let $h(\varphi)$ denote a typical entropy that is defined by a limit as above and after Ian's example, assume that $h(\varphi)>0$. Does anyone know if limits of the form -\begin{equation} -\lim_{n\rightarrow\infty}\ \ \frac{a_n}{\exp(n\cdot h(\varphi))} -\end{equation} have been studied anywhere? I will appreciate any possible information about such limits. For example, is there a known case where the limit exists? If so, what is the limit called? etc. -EDIT: As pointed out later by Ian, even if we assume $h(\varphi)>0$ this limit may not exist. I was curios to know if there were instances where the limit is known to exist. Or even better, can one characterize self-maps $\varphi$ for which the limit exists? - -REPLY [2 votes]: In the case of the geodesic flow acting on the unit tangent bundle of a compact negatively curved manifold, if $a_n$ is the number of closed geodesics of length at most $n$, and $h$ the topological entropy of the geodesic flow, Margulis proved that $a_n$ is equivalent to $\frac{e^{hn}}{hn}$. -The original article has only 2 pages, but his phd thesis was relatively recently published as a book.<|endoftext|> -TITLE: Steenrod squares in the cohomology of $BO(k)$ -QUESTION [5 upvotes]: Does anyone know of a good reference describing the action of the Steenrod algebra $\mathcal{A}_2$ on the cohomology algebra $$H^\ast(BO(k);\mathbb{F}_2)\cong\mathbb{F}_2[w_1,w_2,\ldots ,w_k]$$ of the classifiying space for $k$-dimensional vector bundles? This is a polynomial algebra on the universal Stiefel-Whitney classes. The action of the squares is determined by Wu's formula $$Sq^i(w_k) = \sum_r \binom{k+r-i-1}{r} w_{i-r}w_{k+r},$$ -where $w_m=0$ for $m>k$. -Most of the references I've found seem to focus on the stable classifying space $BO$. I would like to see a detailed exposition for fixed $k$, in particular, relations on the $Sq^I(w_k)$ following from Wu's formula. -(This paper of Pengelley and Williams seems to contain useful information, but I have a feeling this is something more classical.) -Update: Remark 2.5 in the linked paper seems to be saying that the free unstable $\mathcal{A}_2$-module on $w_k$ injects into $H^*(BO(k);\mathbb{F}_2)$, or in other words that the $Sq^I(w_k)$ are linearly independent for $I$ admissible (the multi-index $I=(i_1,\ldots ,i_p)$ is admissible if $i_\ell\geq 2 i_{\ell+1}$ for $1\leq \ell \leq p-1$). -They refer to a paper of Lannes and Zarati, "Foncteurs dérivés de la déstabilisation", which seems to me to be overkill. I tried to give an elementary proof along the lines of the proof in Thom's paper that the $Sq^I(w_k)$ are linearly independent in $H^*(BO;\mathbb{F}_2)$ for $|I|\leq k$, but so far to no avail. Thom orders monomials in the $w_i$ lexicographically, then shows that the leading monomial in the expansion of $Sq^I(w_k)$ is $w_k\cdot w_I$. Does anyone know a slick proof of this claim (that the $Sq^I(w_k)$ are linearly independent in $H^*(BO(k);\mathbb{F}_2)$ for $I$ admissible)? - -REPLY [2 votes]: In the linked paper, David and I have a self-contained, elementary proof of Theorem 2.3, i.e. the freeness of the Steenrod action, from which follow Remarks 2.4 and 2.5. (It does not use Lannes-Zarati, we mentioned their result only for completeness.)<|endoftext|> -TITLE: Why would the category of Motives be Tannakian? -QUESTION [23 upvotes]: After reading the answer to my previous question: What are the different theories that the motivic fundamental group attempts to unify? -I decided to read up on Tannakian formalism. -Given the category of numerical motives, and assuming Conjecture C of the standard conjectures (the one regarding the grading of numerical motives), one can construct a category that will be Tannakian. This will be done by changing the sign of the ``canonical'' morphism $h^iX\otimes h^jX \cong h^jX \otimes h^iX$ for $ij$ odd . -It seems in texts about motives, that the end goal was always to achieve a Tannakian category. But what motivation is there for this? Why would a category that has to do with motives be the category of representations of an affine group scheme? This seems crazy to me. Is this immitative of some easier, more well-understood, theory in which it make sense to relate cohomology with representations? -Also, is it conjectured what this mysterious affine group scheme is, in the case of numerical motives with the adjustment written above? - -REPLY [5 votes]: My impression is that the language of Tannakian categories is a nice thing for itself; thus applying it to motives may (and will, see below) yield interesting applications. There are several ways of dealing with Tannakian categories, and I am not sure that dealing with explicit affine group schemes appearing this way is the "main" one (since this group scheme is awfully huge and complicated for motives, and you have to "make the category neutral" to ensure its existence). Also, Tannakian categories give a possibility of dealing with the category of motives "abstractly". -Probably the nicest results on the relation of motives to Tannakian formalism are in the case where the base field is either finite or (at least) algebraic over a finite field; you can find them in Milne's http://www.jmilne.org/math/articles/1994aP.pdf. Firstly, the category of numerical motives is known to be Tannakian in this case; see Proposition 1.1. To say more about it one needs certain "standard" conjectures. In particular, the Tate conjecture allows to describe the category of motives over a finite field almost completely in Corollary 1.16, Proposition 1.17 (these two statements are improved in Propositions 3.7 and 3.8), Proposition 2.6, and Proposition 2.22. Moroever, Theorem 3.13 (cf. also Theorem 3.19) gives a complete description of motives over the algebraic closure $\mathbb{F}$ of a finite field. Lastly, Theorem 4.22 gives a very funny functor from the so-called CM-motives over the algebraic closure of $\mathbb{Q}$ into motives over $\mathbb{F}$; this result crucially depends on the language of Tannakian categories (since no "geometric" description of this functor is given).<|endoftext|> -TITLE: Obstructions to formally integrating vector fields in characteristic p? -QUESTION [13 upvotes]: Let $M$ be a smooth scheme over some field $k$ of characteristic $p$, and $\vec X$ a vector field on it. Equivalently, $\vec X$ gives a map $Spec\ k[\epsilon]/\langle \epsilon^2 \rangle \times M \to M$ whose reduction is the identity map $(Spec\ k,m) \mapsto m$. Let $D_n = Spec\ k[\epsilon]/\langle \epsilon^n \rangle$, and $D_\infty = Spec\ k[[\epsilon]]$ be the inverse limit, thought of as the additive formal group. - -What are the obstructions in extending this map $D_2 \times M \to M$ to a formal group action $D_\infty \times M \to M\ ?$ How unique is such an extension? - -Motivation: I'm annoyed by the fact that a function on ${\mathbb A}^1_k$ with first derivative zero, i.e. invariant under the $D_2$-action, need not be constant (it could be a $p$th power). When I asked in Replacement for derivations in characteristic p? -for the right condition to replace it, I was told to take all Hasse derivatives, which I am reinterpreting as having a $D_\infty$- not just $D_2$-action. (I prefer the formal group action to the group action, because it restricts to open sets.) - -REPLY [14 votes]: This is not an answer to the questions but some general comments. One should be aware that the relation between vector fields and Hasse derivations in characteristic $p$ is not at all analogous to the characteristic $0$. It is true that a Hasse derivation in all characteristics is the same thing as an acction of the formal additive group. The difference is whereas in characteristic $0$ the formal additive group is the only $1$-dimensional formal group whereas in positive characteristic there are more, for instance the formal multiplicative group. In positive characteristic the derivation part of a Hasse derivation corresponds to an action of the group scheme that is the kernel of the Frobenius map which is $\alpha_p$. Again there are several group schemes of order $p$ (such as $\mu_p$ which is the kernel of Frobenius on the formal multiplicative group). On the vector field side there thus is a first obstruction on the vector field itself, that it should give rise to an $\alpha_p$-action. Concretely that means that the $p$-th power of the derivation should be $0$. -Note that that means that there may not even be a vector field to start with even though there might be many vector fields on $M$ there may not be any with that property (for instance a smooth proper toric variety with no automorphisms outside of the torus). -In any case if one wants an obstruction theory one should note that given $D_1$, $D_n$ for $n=1,\ldots,p-1$ is just $D_1^n$ so the first undetermined one would be $D_p$. If one has local liftings one should compare two such liftings $D_p$ and $D'_p$ and it follows that their difference $D'_p-D_p$ is a derivation and one gets a torsor of the tangent sheaf as a first obstruction. Unfortonately if $D$ is a derivation $D'_p=D_p+D$ may not fulfil the further condition for being a part of a Hasse derivation namely that its $p$'th power should be zero. One can expand its $p$'th power using the Jacobson formula but it leads to a (seemingly) nasty non-linear problem. Anyway if solvable one can continuer with $D_{p^2}$ which would be the next undetermined term and continue in the same manner but it looks like a nightmare (the fact that I don't think that local liftings may exist makes it even less palatable). -However, it is clear that you do not have unique extensions: Take some $M$ on which an action of $\hat G_a\times \hat G_a$ is given. Then the actions of $\hat G_a$ given by the inclusions $t\mapsto (t,0)$ and $t\mapsto (t,t^p)$ have the same first order action. -Addendum: I think that all in all Hasse-Schmidt derivations are better than Hasse derivations (recall that a Hasse-Schmidt derivation is just a map $D_\infty\times M\to M$, i.e., not necessarily a formal group action). There the liftings of an order $n$ HS-derivation to order $n+1$ is just a pseudo-torsor over the tangent sheaf with local liftings existing so that the obstruction for extension is just an $H^1$. Of course you will have many more of them but as you don't have uniqueness for lifting vector fields to Hasse derivations anyway it seems to matter less. -Addendum (formal multiplicative group): Whereas actions of the formal additive group are quite messy, the formal multiplicative group (as always with tori) is much nicer. As $\mu_p$ is linearly reductive an action of it is nothing but a $\mathbb Z/p$-grading on $M$. This is the best description though in terms of vector fields it corresponds to a derivation $D$ with $D^p=D$. Similarly a $\mu_{p^n}$-action is nothing but a $\mathbb Z/p^n$-grading. Hence, lifting from $\mu_{p^n}$ to $\mu_{p^{n+1}}$ corresponds to a refinement of a $\mathbb Z/p^n$-grading to a $\mathbb Z/p^{n+1}$-grading. Hence an action of the formal multiplicative group is the same thing as a $\mathbb Z_p$-grading. -NB: As a topological group that is; a collection of compatible $\mathbb Z/p^n$-gradings for each $n$. In particular the part of degree $a\in \mathbb Z_p$ is an infinite intersection and could very well be $0$ for all $a$. On the other hand an action of the actual multiplicative group is the same thing as a $\mathbb Z$-grading and the associated $\mathbb Z_p$-grading is just induced by $\mathbb Z\subset \mathbb Z_p$. -However things can be more complicated yet very close to such an action. If we have a derivation $D$ with $D^p=fD$ where $f$ is an invertible function then we don't have a $\mu_p$-action unless $f=1$ and we may not even be able to get a $\mu_p$-action by scaling $D$. However, on the étale cover where we extract a $p-1$'th root of $f$ we do get such a modification. So what we have is a kind of twisted $\mu_p$-action and a sheaf of gradings which is an abelian sheaf of $M$ isomorphic to $\mathbb Z/p$ over the above étale cover (in fact it is the twist of $\mathbb Z/p$ by exactly this cover). This extends without problem to twisted actions of the formal multiplicative group. -Such thingies do occur in practice. My favourite example is the following: Look at the moduli space (i.e., stack if you want to quibble) of ordinary elliptic curves. Formally at each point this space can be identified using Serre-Tate coordinates (which I guess in this $1$-dimensional case is earlier than Serre-Tate) with the formal multiplicative group. We can now consider the (pro)-étale covering (with structure group $\mathbb Z_p^\ast$) trivialising the group of $p$-torsion points of the universal elliptic curve. By Igusa this is connected and on it we have an action of the formal multiplicative group realising formally at each point the Serre-Tate coordinates.<|endoftext|> -TITLE: What to do when your research runs into a computationally challenging problem? -QUESTION [16 upvotes]: Occasionally, but more frequently lately, I would like to perform some hard computations. As an example, yesterday the following question came up: - -What is the projective dimension of the edge ideal of the graph $G$ which is two complete bipartite graphs $K_{a,b}, K_{c,d}$ joined by another edge? - -(I believe I can get the answer by hand, but a confirmation would be nice. Also, I wish to compute other similar examples). -For the above question, my personal computer crashed when $a=b=c=d=6$. I used Macaulay 2 with a special package for such situations: EdgeIdeals. -I have a few vague ideas on how to solve this: email people who are better at computations, try to find access to more powerful computers (a small fee is OK), or carefullly use MO (but may be that only works for people like Kevin Buzzard). Still: - -What can one do in such situations? - -I am looking for more generic answers (that can apply not only for the examples above, but in other situations). For example, a pointer to what powerful computers one can get access to would be helpful. Thank you! - -REPLY [15 votes]: Isn't this particular case easier to prove using the topology of the independence complexes of your $K_{n,m}$ and $K_{s,t}$? -$Ind(K_{n,m})$ is the disjoint union of an (n-1)-simplex and an (m-1)-simplex, and $Ind(K_{s,t})$ is the disjoint union of a (s-1)-simplex and a (t-1)-simplex. So the Stanley-Reisner complex of the disjoint union of those two graphs would be $\Delta=Ind(K_{n,m})\coprod Ind(K_{s,t})$ is the join of those two complexes, which is connected ($\dim\widetilde{H}_0(\Delta)$=0) and has $\dim\widetilde{H}_1(\Delta)=1$. -From Hochster's formula, this gives you a nonzero Betti number at homological stage n+m+s+t-2, so $pd(R/I)\geq n+m+s+t-2$. As this is the entire Stanley Reisner complex, and as the complex is connected, that's as high as the projective dimension could be. -This isn't quite the complex you wanted - but you'd just need to examine what happens to the join of the independence complexes of $K_{n,m}$ and $K_{s,t}$ after deleting the face $\{v,w\}$ corresponding to the edge you added between the graphs and all of the faces containing it. This complex still is connected and has $\dim \widetilde{H}_1(\Delta)=1$, so the projective dimension of both complexes is the same. -I guess a more general answer to this particular question though is - instead of computing the resolutions using edge ideals, you might consider using GAP to compute the homology of subcomplexes of your total complex of the appropriate size. These homology calculations combined with Hochster's formula are often better tools at proving projective dimension or regularity bounds than trying to resolve the ideals themselves.<|endoftext|> -TITLE: Arithmetic-geometric mean of positive matrices -QUESTION [10 upvotes]: Let $A,B$ be positive definite (Hermitian) matrices. Define the Arithmetic-geometric means of positive matrices by $A_0=A, G_0=B$, $A_{n+1}=\frac{A_n+G_n}{2}, G_{n+1}=A_n\natural G_n$, where $A_n\natural G_n$ means the geometric mean defined in http://www.isid.ac.in/~statmath/eprints/2011/isid201102.pdf -Will $\{A_n\}$ and $\{G_n\}$ converge to the same matrix? - -REPLY [13 votes]: Recall that since A and B are (hermitian) positive definite, we can without loss of generality (see below for proof) assume that $A=I$ and $B=D$, where $D$ is some positive diagonal matrix. With this observation, merely recall the convergence theory for the scalar case to conclude that the sequences $\{A_n\}$ and $\{B_n\}$ converge to the arithmetic-geometric mean of $I$ and $D$. - -Note: (Added to improve clarity) -For positive definite $A$ and $B$, let $A=Q\Lambda Q^T$, $S=\Lambda^{-1/2}Q$, and let $U$ diagonalize $S^TQ^TBQS$ to $D$. Then, with $P=QSU$, we have -$$P^TAP = U^TS^TQ^TQ\Lambda Q^TQSU = U^TQ^T\Lambda^{-1/2}\Lambda \Lambda^{-1/2}QU = I,$$ -and by construction, $$P^TBP = U^TS^TQ^TBQSU=D.$$<|endoftext|> -TITLE: Are there interesting semisimple algebras in non-semisimple categories? -QUESTION [14 upvotes]: Are there any interesting examples of semisimple algebras in nonsemisimple categories which don't "come from" a semisimple algebra in a semisimple category? That is, if you want to study semisimple algebra objects can you assume wlog that the underlying category is semisimple? -Here's one way of trying to make this question precise. Suppose that C is a finite tensor category, and A is an algebra object in C. Further suppose that A-mod (the category of left A-module objects in C) is a semisimple category. Does there always exist: a semisimple tensor category C', a monoidal functor F: C'->C, and an algebra object A' in C' such that F(A') = A? (Perhaps I should also require some relationship between A-mod and A'-mod, at the very least A'-mod should be semisimple.) -I would also accept an answer explaining why "A-mod is semisimple" doesn't correctly capture the notion of a semisimple algebra if C is non-semisimple. - -REPLY [2 votes]: If $H$ is a finite dimensional Hopf algebra and $\mathcal C=\mathcal M^H$ is the category of corepresentation then $H$ is an algebra in $\mathcal C$ and $\mathcal{C}_H=\mathcal M_H^H= Vec$, where the last equivalence follows by the fundamental theorem of Hopf modules. Then if $H$ is not semisimple (for example the Taft algebra) still $H$ is a semisimple (in fact simple) in the finite tensor category $\mathcal C$. -Note that $H$ as an object in $\mathcal C$ is a tensor generator, so using arguments of Tannakian reconstruction is possible to prove that you can not have a tensor functor $F:\mathcal C'\to \mathcal C$ and an algebra $A'$ in $\mathcal C'$ such that $F(A')=H$ with $\mathcal C'$ a semisimple tensor category.<|endoftext|> -TITLE: Alternative proofs of the Krylov-Bogolioubov theorem -QUESTION [15 upvotes]: The Krylov-Bogolioubov theorem is a fundamental result in the ergodic theory of dynamical systems which is typically stated as follows: if $T$ is a continuous transformation of a nonempty compact metric space $X$, then there exists a Borel probability measure $\mu$ on $X$ which is invariant under $T$ in the sense that $\mu(A)=\mu(T^{-1}A)$ for all Borel sets $A \subseteq X$, or equivalently $\int f d\mu = \int (f \circ T)d\mu$ for every continuous function $f \colon X \to \mathbb{R}$. This question is concerned with proofs of that theorem. -The most popular proof of the Krylov-Bogolioubov theorem operates as follows. Let $\mathcal{M}$ denote the set of all Borel probability measures on $X$, and equip $\mathcal{M}$ with the coarsest topology such that for every continuous $f \colon X \to \mathbb{R}$, the function from $\mathcal{M}$ to $\mathbb{R}$ defined by $\mu \mapsto \int f d\mu$ is continuous. In this topology $\mathcal{M}$ is compact and metrisable, and a sequence $(\mu_n)$ of elements of $\mathcal{M}$ converges to a limit $\mu$ if and only if $\int fd\mu_n \to \int fd\mu$ for every continuous $f \colon X \to \mathbb{R}$. -Now let $x \in X$ be arbitrary, and define a sequence of elements of $\mathcal{M}$ by $\mu_n:=(1/n)\sum_{i=0}^{n-1}\delta_{T^ix}$, where $\delta_z$ denotes the Dirac probability measure concentrated at $z$. Using the sequential compactness of $\mathcal{M}$ we may extract an accumulation point $\mu$ which is invariant under $T$ by an easy calculation. This proof, together with minor variations thereupon, is fairly ubiquitous in ergodic theory textbooks. In the answers to this question, Vaughn Climenhaga notes the following alternative proof: the map taking the measure $\mu$ to the measure $T_*\mu$ defined by $(T_*\mu)(A):=\mu(T^{-1}A)$ is a continuous transformation of the compact convex set $\mathcal{M}$, and hence has a fixed point by the Schauder-Tychonoff theorem. A couple of years ago I thought of another proof, given below. The first part of this question is: has the following proof ever been published? -This third proof is as follows. Clearly it suffices to show that there exists a finite Borel measure on $X$ which is invariant under $T$, since we may normalise this measure to produce a probability measure. By the Hahn decomposition theorem it follows that it suffices to find a nonzero finite signed measure on $X$ which is invariant under $T$. By the Riesz representation theorem for measures this is equivalent to the statement that there exists a nonzero continuous linear functional $L \colon C(X) \to \mathbb{R}$ such that $L(f \circ T)=f$ for all continuous functions $f \colon X \to \mathbb{R}$. Let $B(X)$ be the closed subspace of $C(X)$ which is equal to the closure of the set of all continuous functions which take the form $g \circ T - g$ for some continuous $g$. Clearly a continuous linear functional $L \colon C(X) \to \mathbb{R}$ satisfies $L(f \circ T)=f$ if and only if it vanishes on $B(X)$, so to construct an invariant measure it suffices to show that the dual of $C(X)/B(X)$ is nontrivial. A consequence of the Hahn-Banach theorem is that the dual of $C(X)/B(X)$ is nontrivial as long as $C(X)/B(X)$ is itself nontrivial, so to prove the theorem it is sufficient to show that the complement of $B(X)$ in $C(X)$ is nonempty. But the constant function $h(x):=1$ is not in $B(X)$ because if $|(g \circ T - g)(x) - 1|<1/2$ for all $x \in X$, then $g(Tx)>g(x)+1/2$ for all $x \in X$ and hence $g(T^nx)>g(x)+n/2$ for all $n \geq 1$ and $x \in X$. This is impossible since $g$ is continuous and $X$ is compact. We conclude that $B(X)$ is a proper Banach subspace of $C(X)$ and the desired functional exists. -The second part of the question deals with the Krylov-Bogolioubov theorem for measures invariant under amenable groups of transformations. Let $\Gamma = \{T_\gamma\}$ be a countable amenable group of continuous transformations of the compact metric space $X$. We shall say that $\mu \in \mathcal{M}$ is invariant under $\Gamma$ if $(T_\gamma)_*\mu = \mu$ for all $T_\gamma$. I believe that by using Følner sequences one may generalise the first proof of the Krylov-Bogolioubov theorem to show that every such amenable group has an invariant Borel probability measure. I seem to recall that the second proof also generalises to this scenario (in Glasner's book, perhaps?). Despite a certain amount of thought I have not been able to see how the third proof might generalise to this situation, even when $\Gamma$ is generated by just two commuting elements. So, the second part of this question is: can anyone see how the third proof generalises to the amenable case? - -REPLY [3 votes]: Please check -Krasnosel'skii, M.A.; Lifshits, Je.A.; Sobolev, A.V. (1990), Positive Linear Systems: -The method of positive operators, Sigma Series in Applied Mathematics, -Krasnoselskii told me that he presented his argument at Bogolubov's seminar in 1949?<|endoftext|> -TITLE: Geometric meaning of a trigonometric identity -QUESTION [15 upvotes]: It follows from the law of cosines that if $a,b,c$ are the lengths of the sides of a triangle with respective opposite angles $\alpha,\beta,\gamma$, then -$$ -a^2+b^2+c^2 = 2ab\cos\gamma + 2ac\cos\beta + 2bc\cos\alpha. -$$ -For a cyclic (i.e. inscribed in a circle) polygon, consider the angle "opposite" a side to be the angle between adjacent diagonals whose endpoints are those of that side (it doesn't matter which vertex of the polygon serves as the vertex of that angle). Then for a cyclic quadrilateral with sides $a,b,c,d$ and opposite angles $\alpha,\beta,\gamma,\delta$, one can show that -\begin{align} -a^2+b^2+c^2+d^2 = {} & 2ab\cos\gamma\cos\delta + 2ac\cos\beta\cos\delta + 2ad\cos\beta\cos\gamma \\ -& {} +2bc\cos\alpha\cos\delta+2bd\cos\alpha\cos\gamma+2cd\cos\alpha\cos\beta \\ -& {}-4\frac{abcd}{(\text{diameter})^2} -\end{align} -And for a cyclic pentagon, with sides $a,b,c,d,e$ and respective opposite angles $\alpha,\beta,\gamma,\delta,\varepsilon$, -\begin{align} -a^2 + \cdots + e^2 = {} & 2ab\cos\gamma\cos\delta\cos\varepsilon+\text{9 more terms} \\ -& {} - 4\frac{abcd}{(\text{diameter})^2}\cos\varepsilon+ \text{4 more terms}. -\end{align} -And for a cyclic $n$-gon with sides $a_i$ and opposite angles $\alpha_i,$ -\begin{align} -\sum_{i=1}^n a_i^2 = {} & \text{a sum of }\binom{n}{2}\text{ terms each with coefficient 2} \\ -& {} - \text{a sum of }\binom{n}{4}\text{ terms each with coefficient 4} \\ -& {} + \text{a sum of }\binom{n}{6}\text{ terms each with coefficient 6} \\ -& {} - \cdots \text{ and so on} -\end{align} -The number of terms depends on $n$ and the power of the diameter on the bottom is in each case what is needed to make the term homogeneous of degree 2 in the side lengths ("dimensional correctness" if you like physicists' language), and the alternation of signs continues. -I showed this by induction. It should work for infinitely many sides, too, by taking limits. Each term would then have a product of infinitely many cosines. -My question is: Is there some reasonable geometric interpretation of the sum of squares of sides of a polygon inscribed in a circle? -Later note:The question was in part inspired by a similar situation in which there is a simple geometric interpretation of the right side of the identity. That looks like this: - -$$ -\begin{align} -& 2\left(\frac{abc}{\text{diameter}}\cos\delta\cos\gamma\cos\zeta\cos\eta\cdots\right) + \text{all other terms with 3 sides} \\ -& {} - 4 \left( \frac{abcde}{(\text{diameter})^3} \cos\zeta\cos\eta\cdots \right) + \text{other terms with 5 sides} \\ -& {} + 6 \left(\text{similarly with 7}\right) - 8(\text{similarly with 9}) + \cdots -\end{align} -$$ - -That's one side of the identity. The other is -$$ -\text{diameter}\cdot (a\cos\alpha + b\cos\beta + c\cos\gamma + \cdots). -$$ -The expression on both sides is 4 times the area of the polygon. That's the simple geometric interpretation. The terms on the "other" side of the identity are signed areas of triangles with a vertex at the center of the circle. Notice that at most one of the cosines can be negative, and that happens precisely if the center of the circle is not in the interior of the polygon. - -REPLY [16 votes]: I don't know of a "reasonable geometric interpretation of the sum of squares of sides of a polygon inscribed in a circle". I do, however, find a proof without explicit induction that to some extent explains the formula, and may be simpler than the proof of original proposer (OP), so it might be of some interest to some readers and perhaps also the OP. It is not simple enough to fit in a comment so I must post it as an answer. -By homogeneity we may assume the circle has diameter 1, which will simplify the formulas. I'll use $j$ rather than $i$ for the index because I'll need $i$ to be $\sqrt{-1}$. -Each $a_j$ is the side of a right triangle with hypotenuse 1 and opposite angle $\alpha_j$. Hence $a_j = \sin \alpha_j$, and the left-hand side is $\sum_{j=1}^n a_j^2$. -Each term in the sum on the right-hand side, call it $S$, is $-\prod_{j=1}^n \cos \alpha_j$ times $(-1)^{k/2} k$ times a product of $k$ factors $a_j / \cos \alpha_j = \tan \alpha_j$ with distinct $j$'s, for some even integer $k>0$. We might as well include $k=0$ because then $(-1)^{k/2} k = 0$. -To access this sum, consider the finite generating function (or generating polynomial) -$$ -P(X) := \prod_{j=1}^n (1 + i \tan \alpha_j \cdot X). -$$ -Expand, sum the coefficients, and take the real part. That almost matches $-S$, except that the $X^k$ coefficients are missing the factor $k$. To get that factor, differentiate and set $X=1$: -$$ --S = \left(\prod_{j=1}^n \cos \alpha_j \right) {\rm Re}\left( P\phantom{|}'(X)|_{X=1}\right) -=\left(\prod_{j=1}^n \cos \alpha_j \right) {\rm Re}\left( P(1) \phantom{/}\sum_{j=1}^n \frac{i \tan \alpha_j}{1 + i \tan \alpha_j \cdot X} \Biggl|_{X=1}\Biggr.\right). -$$ -But $P(1)$ is the product of terms $1+i\tan\alpha_j = e^{i \alpha_j} / \cos \alpha_j$, and (aha) $\sum_j \alpha_j = \pi$ because each $\alpha_j$ is half the angle subtended by the $j$-th side about the center of the circle. Hence $P(1)$ is the real number $-1 / \prod_{j=1}^n \cos \alpha_j$, and $S$ simplifies to -$$ -+ \sum_{j=1}^n \phantom| {\rm Re} \frac{i \tan \alpha_j}{1 + i \tan \alpha_j} -= \sum_{j=1}^n \phantom| \frac{\tan^2 \alpha_j} {1+\tan^2 \alpha_j} -= \sum_{j=1}^n \phantom| \sin^2 \alpha_j, -$$ -QED.<|endoftext|> -TITLE: The Mackey Topology on a Von Neumann Algebra -QUESTION [22 upvotes]: Every von Neumann algebra $\mathcal M$ is the dual of a unique Banach space $\mathcal M_* $. The Mackey topology on $\mathcal M$ is the topology of uniform convergence on weakly compact subsets of $\mathcal M_*$. Is it known whether given a von Neumann subalgebra $\mathcal N \subseteq \mathcal M$, the Mackey topology on $\mathcal M$ restricts to the Mackey topology on $\mathcal N$? -The article below indicates that the answer was unknown at the time of its publication. -Aarnes, J. F., On the Mackey-Topology for a Von Neumann Algebra, Math. Scand. 22(1968), 87-107 -http://www.mscand.dk/article.php?id=1864 - -REPLY [5 votes]: This recent paper says that it is still unknown: https://arxiv.org/abs/1411.1890.<|endoftext|> -TITLE: Counter-examples to Krull's intersection theorem -QUESTION [9 upvotes]: The more general form of Krull intersection theorem says: - -Let $R$ be local and Noetherian and $I \subset R$ a proper ideal. If $M$ is finitely generated over $R$, and $N=\cap_1^{\infty} I^iM$, then $IN=N$. - -What is the simplest counter-examples when one (and only one) condition among: $R$ local, $R$ Noetherian or $M$ finitely generated is dropped? So this is three questions I guess. -Sorry if this is too easy for this site. It has been a while, you know! -LATER: Andrea's answer gave a counter-example to the stronger statement: there is an element $r \in I$ such that $N(1-r)=0$. I believe it is not a counter-example to the form stated above, see David Eisenbud's book on commutative algebra, the Example after Corollary 5.5 and Exercise 5.6. -However, Dustin Cartwright pointed out that one can safely drop the "local" hypothesis. So there are only two questions left. - -REPLY [7 votes]: Here's a variation on user2035's answer. -Let $R=\mathbb Q[x, z, y_1, y_2, \ldots]/ \langle x-zy_1, x-z^2y_2, \ldots \rangle$. Then take $M=R$ and $I=\langle z \rangle$, and then $\cap_{i=1}^n I^n = \langle x \rangle$, but $z\langle x \rangle \neq \langle x \rangle$. Obviously, in this case, $R$ is not Noetherian. -Now keep $M$ as above, but replace $R$ with the subring $\mathbb Q[z]$. The intersection theorem fails in the same way as above. In this case, $R$ is Noetherian, but $M$ is not a finitely generated $R$-module.<|endoftext|> -TITLE: Computing the q-series of the j-invariant -QUESTION [7 upvotes]: It is a fundamental fact, often quoted these days in its connection with Monstrous Moonshine, that the q-expansion (i.e., the Laurent expansion in a neighborhood of $\tau = i\infty$) of the j-invariant is given by -$$j(\tau) = \frac{1}{q} + 744 + 196884q + 21493760q^2+\dots, \quad q = e^{2 \pi i \tau}.$$ -I do not recall, however, ever seeing a modern treatment, nor even a hint, of how one might go about obtaining this expansion. Does anybody know a nice way to compute these coefficients? (I mean a way which does not invoke Moonshine, not that I'd expect that to make the computation more pleasant.) Is there a standard way to do it? -I did find one approach published by H.S. Zuckerman in the late 1930s*, which makes use of a "fifth order multiplicator equation" for $j(\tau)$ -- distilled from Fricke and Klein's Vorlesungen uber die Theorie der elliptischen Modulfunktionen -- and an identity of Ramanujan for the generating function of partition numbers of the form $p(25n + 24)$. Is this typical? -*Zuckerman, Herbert S., The computation of the smaller coefficients of $J(\tau)$. Bull. Amer. Math. Soc. 45, (1939). 917–919.} - -REPLY [13 votes]: In addition to the power-series methods using Eisenstein series and $\Delta$, and the modular equation methods using, e.g., $h_5$ given in the other answers and comments, there are transcendental methods that allow you to compute individual coefficients numerically without working with the whole series. -H. Petersson gave an explicit formula for each $q$-expansion coefficient of $j$ in: Über die Entwicklungskoeffizienten der automorphen Formen, Acta Math. 58 (1932), 169–215. Rademacher gave an independent derivation of the formula using his enhancement of the Hardy-Ramanujan circle method. If we write $j(\tau) = \sum_{n \geq -1} c_n q^n$ with $c_{-1} = 1$, then for $n \geq 1$: -$$c_n = \frac{2\pi}{\sqrt{n}} \sum_{k = 1}^\infty \frac{A_k(n)}{k} I_1 \left( \frac{4\pi \sqrt{n}}{k} \right), \qquad \text{where} \quad A_k(n) = \sum_{ab \equiv -1 \pmod k} e^{\frac{2\pi i}{k}(na+b)},$$ -and $I_1$ is a modified Bessel function of the first kind. While this is an infinite sum, in practice you only need to add the first few terms because you already know the coefficients are integers. -More generally, if you want coefficients of any (weakly holomorphic) modular form of non-positive weight, you can apply the circle method, and you need only input the principal parts of the expansions at cusps. The convergence seems to be faster for forms of "more negative" weight, e.g., the weight -12 form $\Delta^{-1}$ converges spectacularly quickly. If you are just looking for expansions of Hauptmoduln such as the Monstrous Moonshine functions $T_g$, there is a formula for any genus zero group (that looks eerily similar to the formula for $j$) at the end of Venkov's 1983 ICM lecture.<|endoftext|> -TITLE: Is there a "Basic Number Theory" for elliptic curves? -QUESTION [61 upvotes]: Tate's thesis showed how to profitably analyze $\zeta$ functions of number fields in terms of adelic points on the multiplicative group. In particular, combining Fourier analysis and topology, Tate gave new and cleaner proofs of the finiteness of the class group, Dirichlet's theorem on the rank of the unit group, and the functional equation of the $\zeta$-function. Weil's textbook Basic Number Theory re-presented algebraic number theory from the adelic perspective, showing how adelic methods could provide simple and unified proofs of all the results proved in a first course in algebraic number theory (and perhaps in a second one as well.) -I have heard rumors that one can similarly rewrite the theory of elliptic curves in adelic terms, and that doing so gives intuition for the BSD conjecture. Franz Lemmermeyer's paper Conics, a poor man's elliptic curves provides a brief sketch. Is there a survey paper or textbook which lays this picture out in full, as Weil did for the multiplicative group, pointing out the connections between the adelic and the classical language at each step, and ideally discussing the connections with BSD? -Note: This question has a peculiar history. See this meta thread if you are interested, but feel free to ignore the past and just answer the question if you are not. - -REPLY [6 votes]: Hello, It's nice that there has been some more interest in Pell conics. I decided a couple days ago after reading a few of the interesting discussions here to post on arXiv what I've learned, arXiv:1108.1610, about Franz Lemmermeyer's result that Sha for conics is isomorphic to a subgroup of the narrow class group of a quadratic field.<|endoftext|> -TITLE: Is there a ``path'' between any two fiber functors over the same field in Tannakian formalism? -QUESTION [8 upvotes]: I will take the approach of this question: Tannaka formalism and the étale fundamental group -and think of the etale fundamental group as Tannakian formalism for $\mathbb{F}_1$. Then our "Tannakian category" is the category of finite etale covers, and each fiber functor is a functor from this category to $Sets$ (thought of as finite dimensional spaces over $\mathbb{F}_1$). -For the etale fundamental group, it is true that for any two fiber functors (given by two different geometric points) there is a ``path'' between them. Meaning: there is a natural isomorphism between these two functors. -My question is whether this ``independence of the basepoint'' result applies to Tannakian formalism as well: -Is it true that for any two fiber functors $H_1, H_2: \mathcal{C}\rightarrow Vec_K$, there is a natural isomorphism $H_1 \cong H_2$? - -REPLY [8 votes]: The obstruction to the existence of such an isomorphism is a (bi)torsor, that has been studied in various real-life situations. An example extracted from -On the relation between Nori Motives and Kontsevich Periods -Annette Huber, Stefan Müller-Stach -http://arxiv.org/abs/1105.0865 -"As already explained by Kontsevich, singular cohomology and algebraic de Rham -cohomology are both fiber functors on the same Tannaka category of motives. By -general Tannaka formalism, there is a pro-algebraic torsor of isomorphisms between them. The period pairing is nothing but a complex point of this torsor." -Basically, by tannaka duality, you can build a counter-example out of any couple of non-isomorphic objects of a gerbe.<|endoftext|> -TITLE: Number of closed walks on an $n$-cube -QUESTION [13 upvotes]: Is there a known formula for the number of closed walks of length (exactly) $r$ on the $n$-cube? If not, what are the best known upper and lower bounds? -[Edit] Note: the walk can repeat vertices. - -REPLY [4 votes]: Although this is an old question, I wanted to record what I think is a very cute elementary technique for obtaining the summation formula appearing in Qiaochu Yuan's answer. Maybe it is ultimately similar to Ira Gessel's answer: it also uses generating functions, but it avoids use of exponential generating functions. -I saw this technique in this mathstackexchange answer, but have never seen it elsewhere. -Here's the argument. -First of all, we note that it's easy to see, as mentioned in the answer of Derrick Stolee, that the number of closed walks of length $r$ in the $n$-hypercube is $2^n$ times the number of words of length $r$ in the alphabet $[n] := \{1,2,...,n\}$ in which every letter appears an even number of times. So we want to count words of this form. -For a word $w$ in the alphabet $[n]$, let me use $\bf{z}^w$ to denote $\mathbf{z}^w := \prod_{i=1}^{n} z_i^{\textrm{$\#$ $i$'s in $w$}}$, where the $z_i$ are formal parameters. For a set $A \subseteq [n]^{*}$ of such words, I use $F_A(\mathbf{z}) := \sum_{w \in A} \mathbf{z}^{w}$. -For $i=1,\ldots,n$ and ${F}(\mathbf{z})\in\mathbb{Z}[z_1,\ldots,z_n]$ define $$s_i(F(\mathbf{z})) := \frac{1}{2}( F(\mathbf{z}) + F(z_1,z_2,\ldots,z_{i-1},-z_{i},z_{i+1},\ldots,z_n)),$$ -a kind of symmetrization operator. We have the following very easy proposition: -Prop. For $A\subseteq [n]^{*}$, $s_i(F_A(\mathbf{z})) = F_{A'}(\mathbf{z})$ where $A' := \{w\in A\colon \textrm{$w$ has an even $\#$ of $i$'s}\}$. -Thus if $A := [n]^r$ is the set of words of length $r$, and $A'\subseteq A$ is the subset of words where each letter appears an even number of times, we get -$$ F_{A'}(\mathbf{z}) = s_n(s_{n-1}(\cdots s_1(F_{A}(\mathbf{z})) \cdots ) ) = s_n(s_{n-1}(\cdots s_1((z_1+\cdots+z_n)^r) \cdots ) ) $$ -$$= \frac{1}{2^n}\sum_{(a_1,\ldots,a_n)\in\{0,1\}^n}((-1)^{a_1}z_1 + \cdots + (-1)^{a_n}z_n)^r.$$ -Setting $z_i := 1$ for all $i$, we see that -$$\#A'=\frac{1}{2^n}\sum_{j=0}^{n}\binom{n}{j}(n-2j)^r,$$ -and hence that the number of closed walks we wanted to count is -$$\sum_{j=0}^{n}\binom{n}{j}(n-2j)^r,$$ -as we saw in Qiaochu's answer. -Incidentally, this gives a combinatorial way to compute the eigenvalues of the adjacency matrix of the $n$-hypercube (see Stanley's "Enumerative Combinatorics" Vol. 1, 2nd Edition, Chapter 4 Exercise 68).<|endoftext|> -TITLE: What is the precise relationship between Langlands and Tannakian formalism? -QUESTION [23 upvotes]: As anyone who's been reading the forums closely can see, I've been averaging a question a day about Tannakian formalism for the past few days. It's quite an interesting concept! -In any case, I wish now to relate it to yet another topic of which I have only a tenuous grasp: the Langlands program. As I understood more and more about Tannakian formalism, it seemed more and more like it has something to do with Langlands. A google search confirms this. There are several sources that group these two things together. Here is a sample: -http://www.claymath.org/cw/arthur/pdf/automorphic-langlands-group.pdf -http://www.institut.math.jussieu.fr/~harris/Takagi.pdf -But my understanding of Langlands is weak. I'm certainly familiar with Class Field Theory, and to some limited extent with Taniyama-Shimura. I always found Langlands difficult to penetrate. But now that I know that there is a relationship between Langlands and Tannakian formalism, I am hopeful that this will give me a bird's eye view of Langlands. -So the question is: Does Tannakian formalism simplify the statement of Langlands, or at least motivate it? Does it have to do with the motivic Galois group (defined to be the group predicted from Tannakian formalism on the category of numerical motives)? How precisely is Tannakian formalism used in Langlands? -In light of these ideas, I ask a secondary question: is there a relationship between the standard conjectures and Langlands? (does one imply the other?) - -REPLY [15 votes]: Though there are several automorphic papers discussing the Tannakian outlook (notably Ramakrishnan's article in Motives (Seattle 1991, AMS) and Arthur's A note on the Langlands group, (referred to above) there is as yet no formulation of Langlands correspondence between Galois representations and automorphic representations as an equivalence of Tannakian categories. There are (at least) two outstanding fundamental questions on the Tannakian aspects of the Langlands correspondence. -1) What is the definition of the category of automorphic representations for any number field? -here one means automorphic representations for GL, any $n \ge 0$. -2) How to endow the category in 1) with a tensor structure so as to render it Tannakian? -here the postulated Tannakian group is the "Langlands Group" which is much larger -than the motivic Galois group (not all automorphic representations correspond to Galois -representations..only algebraic ones do - see work of Clozel and more recent work of Buzzard-Gee). -An interesting point: Arthur's paper postulates the Langlands group as an extension of the usual -Galois group by a (pro-) locally compact group whereas the motivic Galois group is an extension -of the usual Galois group by a pro-algebraic group. An illustration of the difference is provided -by the case of abelian motives; the Langlands group is the abelianisation of the Weil group -whereas the motivic group is the Taniyama group (see references below). -But the Tannakian outlook, despite its present inaccessibility, has already made a profound impact. See Langlands paper "Ein Marchen etc" (where the Tannakian aspect was first written out with many consequences for the Taniyama group (Milne's notes)) as well as - Serre's book Abelian l-adic representations for many references. -Nothing seems to be known regarding the second question. However, see (page 6 of) these comments of Langlands: -"Although there is little point in premature speculation about the form that the final theory connecting automorphic forms and motives will take, some anticipation of the possibilities has turned out to be useful. Motivic $L$-functions, in terms of which Hasse-Weil zeta functions are expressed, are introduced in a Tannakian context. -.... -An adequate Tannakian formulation of functoriality and of the relation between automorphic representations and motives ([Cl1, Ram]) will presumably include the Tate conjecture ( [Ta] ) as an assertion of surjectivity. The Tate conjecture itself is intimately related to the Hodge conjecture whose formulation is algebro-geometrical and topological rather than arthmetical. ..." -The references here are to Clozel and Ramakrishnan's papers and then Tate's paper for the Tate conjecture. -This is just a rough answer from a novice..for a precise and detailed answer, let us wait for the experts!<|endoftext|> -TITLE: In an Erdős–Rényi random graph, what is the threshold for the property "every edge is contained in at least one triangle"? -QUESTION [16 upvotes]: Let $G(n,p)$ denote the Erdős–Rényi random graph, where $n$ is the number of nodes and $p$ is the probability for each edge. I'm interested in precisely what range of $p$ the random graph has at least one edge not contained in any triangle. -One easily checks that if $$p \ge \left( \frac{2 \log n + \omega}{n} \right)^{1/2}, $$ where $\omega \to \infty$ arbitrarily slowly, then every pair of vertices is a.a.s. connected by a path of length $2$, so it follows that every edge is contained in a triangle. -This can be sharpened though. Let $X$ denote the expected number of edges not in any triangle. -Then $$E[X] = {n \choose 2} p (1-p^2)^{n-2},$$ and if I did my calculation correctly, then if $$p \ge \left( \frac{(3/2) \log n + (1/2) \log \log {n} + \omega}{n} \right)^{1/2}$$ -then there are a.a.s. no edges not contained in any triangles, since $E[X] \to 0$ as $n \to \infty$. -My guess is that this inequality is more-or-less sharp. What I'd like to show then is that if $p$ is much smaller, then there are a.a.s. edges not contained in any triangle. -Suppose for example that $$p \ge \left( \frac{(3/2) \log n - C \log\log{n} }{n} \right)^{1/2}.$$ -Is it true that for large enough constant $C>0$ we have a.a.s. that at least one edge not contained in any triangle? The expected number of such edges is tending to infinity as a power of $\log{n}$, but obviously that's not enough. -I have tried using Janson's inequality, for example, but I am stuck because the events I am trying to count are not pairwise independent even though they are "almost independent." - -REPLY [10 votes]: I just stumbled across this question and see that it is five years old, but since I know the reference I thought I might as well share it. This threshold is determined in the paper "Local Connectivity of a Random Graph" by Erdos, Palmer, and Robinson (JGT Vol 7 1983 pp. 411-417) -- see Theorem 2 and discussion preceding it.<|endoftext|> -TITLE: Are proper classes objects? -QUESTION [23 upvotes]: Many of us presume that mathematics studies objects. In agreement with this, set theorists often say that they study the well founded hereditarily extensional objects generated ex nihilo by the "process" of repeatedly forming the powerset of what has already been generated and, when appropriate, forming the union of what preceded. -But the practice of set theorists belies this, since they tend—for instance, in the theories of inner models and large cardinal embeddings—to study classes that, on pain of contradicting the standard axioms, are never "generated" at any stage of this process. In particular, faced with the independence results, many set theorists suggest that each statement about sets—regardless of whether it be independent of the standard axioms, or indeed of whether it be formalizable in the first order language of set theory—is either true or false about the class $V$ of all objects formed by the above mentioned process. For them, set theory is the attempt to uncover the truth about $V$. -This tendency is at odds with what I said set theorists study, because proper classes, though well founded and hereditarily extensional, are not objects. I do not mean just that proper classes are not sets. - -Rather, I suggest that no tenable distinction has been, nor can be, made between well founded hereditarily extensional objects that are sets, and those that aren't. - -Of course, this philosophical claim cannot be proved. - -Instead, I offer a persuasion that I hope will provoke you to enlighten me with your thoughts. - -Suppose the distinction were made. Then in particular, $V$ is an object but not a set. Prima facie, it makes sense to speak of the powerclass of $V$—that is, the collection of all hereditarily well founded objects that can be formed as "combinations" of objects in $V$. This specification should raise no more suspicions than the standard description of the powerset operation; the burden is on him who wishes to say otherwise. -With the powerclass of $V$ in hand, we may consider the collection of all hereditarily well founded objects included in it, and so on, imitating the process that formed $V$ itself. Let $W$ be the "hyperclass" of all well founded hereditarily extensional objects formed by this new process. Since we can distinguish between well founded hereditarily extensional objects that are sets and those that aren't, we should be able to mirror the distinction here, putting on the one hand the proper hyperclasses and on the other the sets and classes. -Continuing in this fashion, distinguish between sets, classes, hyperclasses, $n$-hyperclasses, $\alpha$-hyperclasses, $\Omega$-hyperclasses, and so on for as long as you can draw indices from the ordinals, hyperordinals, and other transitive hereditarily extensional objects well ordered by membership, hypermembership, or whatever. It seems that this process will continue without end: we will never reach a stage where it does not make sense to form the collection of all well founded hereditarily extensional objects whose extensions have already been generated. We will never obtain an object consisting in everything that can be formed in this fashion. -For me, this undermines the supposed distinction between well founded hereditarily extensional objects that are sets, and those that aren't. Having assumed the distinction made, we were led to the conclusion of the preceding paragraph. But that is no better than the conclusion that proper classes, including $V$ itself, are not objects. Indeed, it is worse, for in arriving at it we relegated set theory to the study of just the first two strata of a much richer universe. Would it not have been better to admit at the outset that proper classes are not objects? If we did that, would set theory suffer? In particular, how would it affect the idea that each statement about sets is either true or false? - -REPLY [14 votes]: Let me offer another answer in counterpoint to Andreas's answer, by pointing out a number of cases in set theory where it seems that a second-order treatment of classes, as in Goedel-Bernays set theory, seems fruitful in contrast to the definable-classes-only approach. - -First, much of our understanding of large cardinals is based upon a felicitous use of large cardinal embeddings $j:V\to M$, which are all proper classes. And set-theorists routinely quantify over the meta-class collection of such embeddings. For example, a cardinal is measurable if it is the critical point of such an embedding; it is strong if for any $\theta$ there is such an embedding with $V_\theta\subset V$; it is supercompact if such embeddings can be found with $M^\theta\subset M$. Although in each case we have a first-order combinatorial equivalent to the large cardinal concept in terms of the existence of certain kinds of measures or extenders on certain sets, nevertheless it is the embedding characterizations that have a robust and strongly coherent power that unifies our understanding of the large cardinal concepts. This seems to be a case in which treating the embeddings as objects has deepened our knowledge. -The Kunen inconsistency result, the assertion that there is no nontrivial elementary embedding $j:V\to V$, becomes trivial when one treats all classes as definable. One can easily rule out all such definable $j$, if one only cares to consider the case in which $j$ is first-order definable with parameters, and one needs neither the axiom of choice nor any infinite combinatorics to do it. (Just argue like this: the question of whether a given formula $\varphi(x,y,p)$ with parameter parameter $p$ defines such a $j$ is a first-order property of $p$, and so one can define $\kappa$ to be the least possible critical point of such a $j$ arising from any $p$, and this contradicts the fact that $\kappa\lt j(\kappa)$, since $j(\kappa)$ would also be defined this way.) The various formalizations of the Kunen inconsistency is explained in the first part of our recent paper on Generalizations of the Kunen Inconsistency. Note that Kunen formalized his theorem in Kelly-Morse set theory, in order to have a way of expressing the elementarity of $j$, but it turns out to be possible to formalize this in GBC. -Class forcing is vital to much of our understanding of the relative consistency of global assertions, such as the full class version of Easton's theorem or the fact that supercompact cardinals are relatively consistent with GCH and with V=HOD. But a development of class forcing is most direct in a context, such as Goedel-Bernays set theory, where classes can be treated as objects. For example, when forcing the GCH with class forcing, the generic class will not be definable in the forcing extension, so this is a case where in order to achieve the extension we most naturally want to consider non-definable classes.<|endoftext|> -TITLE: Which Lie algebra admit symplectic forms -QUESTION [7 upvotes]: Hello, -I am interested in symplectic Lie groups, I will sketch out their definition. -A symplectic Lie group is a given pair $(G,\omega)$, where $G$ is a Lie group and $\omega$ is a left invariant symplectic form on $G$. -The algebraic situation is as follows. If $\mathcal{G}$ is the Lie algebra of $G$ and $\omega$ is a non-degenerate $2$-cocycle on $\mathcal{G}$, that is $\omega\in\wedge^2\mathcal{G}^*$ such that for any $x,y,z\in\mathcal{G}$ -$$\omega([x,y],z)+\omega([y,z],x)+\omega([z,x],y)=0$$ -The left invariant $2$-form associated to $\omega$ is a symplectic form on $G$. -My questions are: -1) What are the Lie algebras that admit such $omega$? The algebra can not be semi-simple, according to an article by Chu, Symplectic homogeneous spaces. -2) What is the link with the existence of an invertible solution of the classical Yang-Baxter equation $[r,r]=0$ for $r\in\wedge^2\mathcal{G}$ where $[\,,\,]$ is the Schouten bracket? -Thanks for any help. - -REPLY [2 votes]: It's a bit late now, but there is also a more recent and very exhaustive paper by Cortés and Baues which adresses and summarizes many related questions: -http://arxiv.org/abs/1307.1629<|endoftext|> -TITLE: knot complement -QUESTION [5 upvotes]: I was told that a knot complement in $S^3$ is an Eilenberg-Mc Lane space. -And that it is quite easy to see this. However I am not able to find out why. -could you help? Thanks - -REPLY [10 votes]: Papakyriakopoulos, C. D. On Dehn's lemma and the asphericity of knots. Ann. of Math. (2) 66 (1957), 1–26.<|endoftext|> -TITLE: Where are $+$, $-$ and $\infty$ in bordered Heegaard-Floer theory? -QUESTION [15 upvotes]: Here goes my first MO-question. I've just read Lipshitz, Ozsváth and Thurston's recently updated "A tour of bordered Floer theory". To set the stage let me give two quotes from this paper. - -Heegaard Floer homology has several - variants; the technically simplest is - $\widehat{HF}$, which is sufficient - for most of the 3-dimensional - applications discussed above. Bordered - Heegaard Floer homology, the focus of - this paper, is an extension of - $\widehat{HF}$ to 3-manifolds with - boundary. -[...] -the Heegaard Floer package contains - enough information to detect exotic - smooth structures on 4-manifolds. For - closed 4-manifolds, this information - is contained in $HF^+$ and $HF^-$; the - weaker invariant $\widehat{HF}$ is not - useful for distinguishing smooth - structures on closed 4-manifolds. - -Since I am mainly interested in closed 4-manifolds, I have not paid too much attention to the developments in bordered Heegaard-Floer thoery. But right from the beginning I have wondered why only $\widehat{HF}$ appears in the bordered context. So my question is: -Why are there no $^+$, $^-$ or $^\infty$ flavors of bordered Heegaard-Floer theory? Are the reasons of technical nature or is there an explanation that the theory cannot give more than $\widehat{HF}$? -I assume there are issues with the moduli spaces of holomorphic curves that would be relevant to defining bordered versions of the other flavors of Heegaard-Floer theory, but I am neither enough of an expert on holomorphic curves to immediately see the problems nor could I find anything in the literature that pins down the problems. -Any information is very much appreciated. - -REPLY [12 votes]: A biased answer, based on Auroux's work http://arxiv.org/abs/1003.2962. -Auroux makes a connection between bordered Floer theory and an alternative approach, due to Lekili and myself, which is (still) under development, but which should include the $\pm$ and $\infty$ versions. We do have a preliminary paper out: http://arxiv.org/abs/1102.3160. -A general set-up: Say you have a compact symplectic manifold $(X,\omega_X)$; and a codim 2 symplectic submanifold $D$, whose complement $M$ is exact: ${\omega_X}|_M=d\theta$, say. -Key example: $X=Sym^g(F)$, where $F$ is a compact surface of genus $g$, and $\omega_X$ a suitable Kaehler form; $M=Sym^g(F-z)$, where $z\in F$. -Forms of Floer cohomology: There are various forms of Floer cohomology one can consider. -(i) As in $\widehat{HF}$ Heegaard theory, one can consider $HF^\ast_M(L_0,L_1)$, the Floer cohomology in $M$ of a pair of (exact) compact Lagrangian submanifolds of $M$. When $L_0$ and $L_1$ are spin, this can be defined as a $\mathbb{Z}$-module. -(ii) As in $HF^-$ Heegaard theory, one can consider the filtered Floer cohomology $HF^\ast_{X,D}(L_0,L_1)$ of a pair of compact Lagrangians $L_i\subset M$ as before. The coefficients are in $\mathbb{Z}[[U]]$. The differential counts holomorphic bigons in $X$, weighted by $U^n$ where $n$ is intersection number with $D$. -(iii) One can consider non-compact Lagrangians $L_i\subset M$ which go to infinity nicely (following the Liouville flow). These have wrapped Floer cohomology $HW^\ast(L_0,L_1)$, as well as "partially wrapped" variants. Wrapping concerns how one chooses to perturb $L_0$ at infinity. This version takes place in $M$, and (AFAIK) can't naturally be extended to something that takes place in $X$. -Invariants for 3-manifolds with boundary. A basic idea is that a 3-manifold $Y$ bounding $F$ should define a (generalized) Lagrangian submanifold $L_Y$ where $X=Sym^{g(F)}F$, as in the "key example" above. The collection of filtered Floer modules $HF^*_{X,D}(\Lambda, L_Y)$ as $\Lambda$ ranges over Lagrangian submanifolds of $M$ (more precisely, the module, over the compact filtered Fukaya category of $(X,D)$, defined by $L_Y$) should be an invariant of $Y$. -If one is interested only in the simpler groups $HF^*_M(\Lambda,L_Y)$, one can (in principle) determine these by looking at the finite collection of (partially wrapped) groups $HW^*(W_i,L_Y)$, where $W_i$ ranges over the thimbles for a certain Lefschetz fibration $M\to \mathbb{C}$. That is, one thinks of $L_Y$ as defining a module over the algebra $A_{LOT}$ formed by the sum of groups $HW^*(W_i,W_j)$. This follows from a deep theorem of Seidel about generating Fukaya categories by thimbles, adapted by Auroux. -The algebra $A_{LOT}$ is (part of) what Lipshitz-Ozsvath-Thurston assign to a parametrized surface, and the module is what they call $\widehat{CFA}(Y)$. They arrived at it by a quite different route. They don't bother with constructing $L_Y$ itself, only the module it defines. Because they use the groups of type (iii) to form their algebra, their approach only works in $M$, not $X$. For that reason, they only capture the hat-theory. -The great advantage of LOT's approach is its finiteness and computability. Lekili and I do construct $L_Y$. We can guess at finite collections of "test Lagrangians" sufficient to compute the module $HF^*_{X,D}(\cdot, L_Y)$, but have not yet proved that they are sufficient.<|endoftext|> -TITLE: Analysis of Misere Nim? -QUESTION [8 upvotes]: My friend likes to impress people by playing 3-5-7 which has three piles of counters of sizes 3, 5 and 7. You can remove any number of coins from a single pile, the last player to move loses. -ooo -ooooo -ooooooo -This is a winning position for the first player, but With a solid understanding of the game tree she wins nearly every time playing second. She says, it reduces to knowing a few winning positions. -Two piles of the same size is second player win, in the jargon of Combinatorial Game Theory. Here is the pile (5,5). -ooooo -ooooo -If first player moves to (5,n) for n > 1, second player can imitate on the other pile, moving to (n,n). However, if first player moves to (5,1), second player moves to 1 and wins. -The other winning positions she remembers is (3,4,1) and (4,5,1). She can win once she recognizes these positions. Eventually (after losing many times) I told her that (n, n+1,1) is a losing position for any n... - -If our game were played in normal play convention (player moving last wins), but real life Nim is played as a misere game. Probably the analysis is similar to normal-play Nim with some modification. -Recently there was a theory of Misere quotients where each game has a commutative monoid assoicated with it. What does that monoid looks like here? Is it finitely generated? - -REPLY [13 votes]: Let $\oplus$ denote the bitwise xor operation on natural numbers. It is both well-known and easy to verify that a Nim position $(n_1,\dots,n_k)$ is a second player win in misère Nim if and only if some $n_i>1$ and $n_1\oplus\cdots\oplus n_k=0$, or all $n_i\le1$ and $n_1\oplus\cdots\oplus n_k=1$. -If I understand it correctly, this translates to the following structure. Let $A=(\omega,\oplus,0)$ (in other words, $A$ is the direct sum of countably many copies of the two-element abelian group), let $A_0=\{0,1\}$ be its submonoid, and let $B=(\{0,1\},\lor,0)$ be the two-element semilattice. Then the underlying monoid of the misère quotient of Nim is the submonoid $Q=(A\times\{1\})\cup(A_0\times\{0\})$ of the product monoid $A\times B$, the misère quotient itself being $(Q,\{(0,1),(1,0)\})$. If $(n_1,\dots,n_k)$ is a Nim position, its value in $Q$ is $(n_1\oplus\cdots\oplus n_k,u)$, where $u=0$ if $n_i\in\{0,1\}$ for each $i$, and $u=1$ otherwise. $Q$ has (as a commutative monoid) the infinite presentation $\langle \{a_n:n\in\omega\},b\mid a_n+a_n=b+b=0,a_n+b=a_n+a_0\rangle$. Finitely generated submonoids of $Q$ are finite, hence $Q$ itself is not finitely generated. -(Note that the monoid corresponding to normal play Nim is just $A$.)<|endoftext|> -TITLE: Billiard dynamics under gravity -QUESTION [38 upvotes]: Has the dynamics of billiards in a polygon subject to gravity been -studied? -What I have in mind is something like this: - -           - - -Still Snell's Law applies so the angle of incidence equals the -angle of reflection, and the collision is perfectly elastic, -but the path followed by the ball between -contacts is a parabola. -I am wondering if such a system can somehow be converted into one -without gravity, so that our understanding of, e.g., the dynamics of -billiards in a square may be applied. -To be specific: - -What initial conditions lead to a periodic path in a square? - -For example, suppose the square has corners $(0,0)$ and $(1,1)$. -Starting at $p_0=(0,\frac{1}{2})$, with -vertical velocity zero and horizontal velocity that first lands -the ball at $p_1=(\frac{1}{2},0)$ -(say, $v_x=\frac{1}{2}$, gravity $=1$), -produces (I believe) a periodic path bouncing -between the three points $\{ p_0, p_1, p_2 \}$, where $p_2=(1,\frac{1}{2})$: - -                     - - -I'd appreciate literature pointers—Thanks! - -Addendum. -Following fedja's intriguing comment, here a path when gravity is tilted $30^\circ$: - -REPLY [6 votes]: This problem is very similar to what is known in the physics literature (including in real atom-optics experiments) as the "wedge billiard." -Self-edit: Another comment - -Formulas for linear stability of billiards with potentials (such as gravity) are given in -Holger Dullin's paper here: - -Holger R Dullin, "Linear stability in billiards with potential." - Nonlinearity, Volume 11, Number 1, - 11, 151. 1998. -Abstract: A general formula for the linearized Poincaré map of a billiard with a potential is derived. ...<|endoftext|> -TITLE: Inverse Image and Tensor Product -QUESTION [5 upvotes]: Given a morphism $f$ of Schemes $X \to Y$ and two sheaves $\mathcal F$, $\mathcal G$ of modules on $Y$, -is it right that the tensor product of $\mathcal F$ and $\mathcal G$ as modules commutes with the inverse image (not the module pullback but only the inverse image $f^{-1}$) -construction? Here I mean one time tensor product over $\mathcal O_Y$ and the other time over $f^{-1}\mathcal O_Y$. -Regards! - -REPLY [10 votes]: The tensor product $F \otimes G$ of sheaves of modules has the following universal property: -$\hom(F \otimes G,H) = \text{bilin}(F \times G,H)$ -Here, the right hand side is the set of bilinear sheaf homomorphisms $F \times G \to H$. You should keep this in mind instead of the explicit construction involving sheafifications! The idea behind tensor products is not their construction, but rather that they classify bilinear maps. This idea is familiar from commutative algebra and should not be lost in algebraic geometry. -Now we also have a canonical identification -$\text{bilin}(F \times G,H) = \hom(F,\underline{\hom}(G,H))$. -The pullback $f^*$ has also a universal property, it is left adjoint to the pushforward $f_*$. Thus, we get: -$\hom(f^* F \otimes f^* G,-) \cong \hom(f^* F,\underline{\hom}(f^* G,-)) \cong \hom(F,f_* \underline{\hom}(f^* G,-))$ -$\cong \hom(F,\underline{\hom}(G,f_* -)) \cong \hom(F \otimes G,f_* -) \cong \hom(f^* (F \otimes G),-)$ -and the Yoneda lemma gives us $f^* F \otimes f^* G \cong f^* (F \otimes G)$. -PS: Try to prove $F \otimes (G \otimes H) \cong (F \otimes G) \otimes H$ using sheafifications and compare the proof with the one using trilinear maps.<|endoftext|> -TITLE: On Brown representability theorem -QUESTION [6 upvotes]: The classical Brown representability theorem is for set valued functors. Is there a version for abelian group valued functors, and ring valued functors? -In other words say we have an abelian group valued functor F on the category of CW top. spaces, satisfying the necessery condition that F maps colimits to limits. What extra conditions on F do we need to ensure that the classifying object is an H-space. Actually Brown doesn't state this, but at a brief glance his paper seems to prove that F just needs to satisfy excision that is we have exact sequences -$$0 \to F (V \cap W) \to F(V) \oplus F (W) \to F (V \cup W) \to 0,$$ for V,W open sets in X. -Is this right? What about the case of ring valued functors, when are they representable by (E_\infty? whatever that is)-ring space. - -REPLY [4 votes]: For the group case, this is an exercise in Chapter 9 of Switzer's "Algebraic Topology: Homotopy and Homology" (bottom of page 157). -I think the rough idea is as follows. Suppose your functor $F$ on pointed CW complexes is representable as $F(-)=[-,Y\;]$, and takes values in the category of groups. You wish to show that $Y$ is a group-up-to-homotopy. To get a multiplication on $Y$ use the naturality part of Brown's theorem (Theorem 9.13 in Switzer). The functor $F\times F$ is represented by $Y\times Y$, and the group multiplication is a natural transformation $F\times F\to F$, which is therefore represented by a map $m\colon\thinspace Y\times Y\to Y$, unique up to homotopy. -Now check that the group axioms for $F(-)$ yield the desired properties of $m$ (associative, unital, with inverses up to homotopy).<|endoftext|> -TITLE: W*-completion of a C*-algebra? -QUESTION [16 upvotes]: tl;dr: Is there such a thing as a W*-completion of a C*-algebra, and if so, where can I read about it? -I'm wondering about the relationship between (abstract) C*-algebras and W*-algebras. On the one hand, every W*-algebra is a C*-algebra. On the other hand, it seems to me that it should be possible to complete any C*-algebra to a W*-algebra. (Categorially, this would be a reflection.) In the case of commutative algebras, I even think that I know how this works: every commutative C*-algebra is the algebra of continuous functions on some compact Hausdorff space, and we extend this to the W*-algebra of essentially bounded Borel-measurable functions on the space (considered up to equality almost everywhere). [Warning: this is determined in the comments to be wrong.] -So, is this correct? Does it work for noncommutative algebras as well? Is there a good algebro-analytic (without passing through topology) description of this? Is there a good reference, especially online? -Also, in the commutative case, it seems that every state (positive normal linear functional) on a C*-algebra extends uniquely to its W*-completion, so they have the same space of states. Is this correct? Does it extend to the noncommutative case? -Another question is how this relates to concrete algebras (those given as algebras of operators on some Hilbert space). One way to complete a C*-algebra would be to pick a concrete representation and take its weak closure (or double commutant). But I expect that this will depend on the representation chosen. (And my analysis is bad enough that I can't check this for even the commutative case.) -I'd appreciate any help even for the main question, never mind this stuff about states and representations! - -REPLY [13 votes]: The universal enveloping W* algebra of a C* algebra is discussed in detail in chapter III.2 of volume 1 of Takesaki's work on "Theory of operator algebra". It is universal in the sense that any map to an W* algebra factors through it (modulo some mumbling about topologies), and as mentioned above is given by the double dual of the C* algebra. This is called the Sherman-Takeda theorem, and was announced by Sherman in 1950 and proved by Takeda in 1954.<|endoftext|> -TITLE: Cherlin's "Main Conjecture" -QUESTION [11 upvotes]: Cherlin's "Main Conjecture" from his 1979 paper "Groups of Small Morley Rank" is the following: Every simple $\omega$-stable group is an algebraic group over an algebraically closed field. Zilber was simultaneously doing similar work (as Cherlin notes). The finite Morley rank version of the conjecture is sometimes called the Cherlin-Zilber conjecture or the Algebraicity conjecture. -There is an extensive literature for the finite Morley rank case. I am not asking about the finite rank case. I am interested in the status of the conjecture for the infinite rank case. Cherlin notes that this conjecture would imply that any simple $\omega$-stable group is of finite rank - modulo the finite rank version of the conjecture, this is essentially the content of the infinite rank version of the conjecture. -Further, Cherlin notes that one could formulate a linear version of the conjecture, in which the group acts as a subgroup of the linear transformations of some vector space. -What is the status of the infinite rank version of Cherlin's "Main Conjecture"? -Edit: -See the comments made by S. Thomas below. (Thanks for the clarification Simon +1). - -REPLY [4 votes]: I think the 'status' might be described as : Pillay has shown using used Selah's work that the free group is not CM-trivial. -All known counterexamples to Zilber's conjecture are CM-trivial. A non-abelian simple group of finite Morley rank is not CM-trivial. We therefore suspect that the current methods based upon Hrushovski's counterexamples cannot produce even an infinite rank counterexample who is a simple group.<|endoftext|> -TITLE: Looking for a mathematically rigorous introduction to game theory -QUESTION [23 upvotes]: I am looking for the best book that contains a mathematically rigorous introduction to game theory. -I am a group theorist who has taken a recent interest in game theory, but I'm not sure of the best place to learn about game theory from first principles. Any suggestions? Thanks! - -REPLY [3 votes]: An Introductory Course on Mathematical Game Theory by González-Díaz, García-Jurado and Fiestras-Janeiro -Reviews can be found here: 1, 2, 3. -Taken from the second review: - -The book is self-contained and written very - rigorously, but on the other hand, it is also very friendly to the reader, containing a lot of explanations - and interpretations of game theory notions, as well as very many examples describing and - analyzing various economic and other models with an application to game theory results. This - makes it very suitable for graduate and advanced undergraduate courses on game theory.<|endoftext|> -TITLE: Lattices in SOL -QUESTION [6 upvotes]: Consider a semi-direct product $\mathbb{Z}^2\rtimes_A\mathbb{Z}$, where $A\in SL_2(\mathbb{Z})$ and $|Tr(A)|>2$. It is clear that it is isomorphic to a lattice in the 3-dimensional solvable Lie group SOL. To what extent do these examples exhaust lattices in SOL? (i.e., up to a suitable equivalence relation, is every lattice in SOL of this form?) -The question comes from a desire to understand better the Eskin-Fisher-Whyte result on quasi-isometric rigidity of SOL: every finitely generated group quasi-isometric to SOL is virtually a lattice in SOL. - -REPLY [7 votes]: To add to Igor Rivin's answer: it seems that all the lattices in SOL are isomorphic as abstract groups to $\mathbb{Z}^2\rtimes_A\mathbb{Z}$ for hyperbolic $A\in SL_2(\mathbb{Z})$. If I am reading the paper correctly, it is in Theorem 2.1 of the paper linked in Igor's answer. -I think that this fact can also be easily derived from the following theorem (Corollary 3.5 in Raghunathan's book, which is due to either Auslander or Mostow): -If $G$ is a connected solvable Lie group, and $N$ is its maximum connected (normal) closed nilpotent Lie subgroup, then for any lattice $\Gamma$ in $G$, $\Gamma \cap N$ is a (cocompact) lattice in $N$. -Thus you always have the short exact sequence -$$1 \to \Gamma \cap N \to \Gamma \to \Gamma/(\Gamma \cap N) \to 1.$$ -The fact that $\Gamma \cap N$ is cocompact in $N$ implies that $\Gamma/(\Gamma \cap N)$ is a discrete subgroup of $G/N$. -If $G = SOL = \mathbb{R}^2 \rtimes \mathbb{R}$, then $N \approx \mathbb{R}^2$, and so $\Gamma \cap N \approx \mathbb{Z}^2$. Also since $G/N \approx \mathbb{R}$, -$\Gamma/(\Gamma \cap N) \approx \mathbb{Z}$. So the short exact sequence above reads -$$1 \to \mathbb{Z}^2 \to \Gamma \to \mathbb{Z} \to 1.$$ -Such a sequence must split, so $\Gamma$ is a semidirect product. -The linked paper does something much more detailed and impressive, sort of like the classification of crystallographic groups.<|endoftext|> -TITLE: Is the support of an Artinian module finite? -QUESTION [11 upvotes]: Let $R$ be a commutative Noetherian ring, $M$ is an Artinian $R$-module. Is the set $Supp_R(M)$ finite? -Thanks. - -REPLY [6 votes]: It is well-known that for finitely generated Artinian $M$, the support is a finite set of maximal ideals. Since $\mathrm{Supp}(M)=\bigcup_{M'\subseteq M\ \mathrm{f.g.}}\mathrm{Supp}(M')$, we know that $\mathrm{Supp}(M)$ consists of maximal ideals. For any finite set $S\subseteq\mathrm{Supp}(M)$, it again follows easily from the finitely-generated case that $f_S\colon M\to\prod_{s\in S}M_s$ is surjective. In particular, for a strictly increasing chain $S_1\subset S_2\subset\dots$ of finite subsets of $\mathrm{Supp}(M)$, the chain of submodules $\ker(f_{S_i})$ is strictly decreasing, so $\mathrm{Supp}(M)$ must be finite.<|endoftext|> -TITLE: A Closed Form for the Diagonal Matrix Nearest an Arbitrary Square Matrix -QUESTION [8 upvotes]: If I have a square matrix in $\mathbf{A} \in \mathbb{R}^{n \times n}$, I want to find another diagonal matrix $\mathbf{D} \in \mathbb{R}^{n \times n}$ that minimizes the residual $ \min_\mathbf{D} || \mathbf{A-D} ||^2 $, where the norm here is the induced norm $\max_{x\neq 0} \frac{\mathbf{||Ax||_2}}{\mathbf{||x||_2}}$. Is there a currently-known closed form for this optimization? - -REPLY [3 votes]: The case of the $2$-norm may well not have a closed form. However, in the Frobenius norm the problem has a trivial answer: $D_A = \mathrm{diag}(A)$. Since $\| A \|_2 \leq \| A \|_F \leq \sqrt{n} \| A \|_2$, one then has that -$$ -\frac{1}{\sqrt{n}} \| A - D_A \| \leq \min_{D} \| A - D \| \leq \| A - D_A \|, -$$ -which, at the least, gives a bound.<|endoftext|> -TITLE: Is every representable map a submersion? -QUESTION [12 upvotes]: Recall that a morphism $f:C \to D$ in a category $\mathscr{C}$ is representable if for all maps $g:E \to D$ in $\mathscr{C},$ the pullback $C \times_{D} E$ exists. -Let now $\mathscr{C}$ be the category of smooth manifolds. Then any submersion is representable. Is the converse true? I have heard from various people that the converse is true, but the only reference I have found is David Metzler's Topological and Smooth Stacks. -However, the proof he gives there is not complete, for it assumes implicitly that if $M \times_N L$ is a pullback of manifolds, then the induced map $$M \times_N L \to M \times L$$ is a a smooth embedding. I do not see how this is automatic. -I do have a sketch of a proof that this map must be a topological embedding (using diffeological spaces) but is it necessarily an immersion? I would like to argue this using curves, however, this is difficult without knowledge of how to differentiate them in the pullback. -Does anyone have either have a proof or a counterexample for this statement? - -REPLY [4 votes]: Consider $f(x)=x^3$ on the real line, $C=D=\mathbb R$. Then for any smooth $g:E\to \mathbb R$ -the pullback $\mathbb R\times_{f,\mathbb R,g}E$ is a smooth manifold diffeomorphic to the graph of $g$, but is is not a submanifold of $\mathbb R\times E$ in general. -So this is not really a counterexample. - -Edit: - -The following shows, that not every pullback is embedded into the product. -Consider the topological pullback $N = \lbrace(x,y)\in \mathbb R^2: x^2=y^3\rbrace$ -of the two smooth mappings $x^2, x^3: \mathbb R\to \mathbb R$ which is Neill's parabola, and consider the manifold $P=\mathbb R$ with the two mappings $x^3, x^2:\mathbb R\to \mathbb R$ which give a topological homeomorphism $P\to N$: -$$ -\begin{array}{ccccc} -P=\mathbb R & \xrightarrow{(x^3,x^2)} & N & \rightarrow & \mathbb R \newline -& & \downarrow & & \downarrow x^2 \newline -& & \mathbb R & \xrightarrow{x^3} & \mathbb R -\end{array} -$$ -Claim: The triple $(P,x^3,x^2)$ has the universal property of a pullback. -Namely, let $M$ be a smooth manifold and let $f,g:M\to \mathbb R$ be smooth mappings with -$f^2= g^3$. Note that then $g\ge 0$. I claim that $f_1:=f^{1/3}:M\to \mathbb R$ -is a smooth mapping which gives a smooth factorization $f_1:M\to P$. -Indeed, by convenient calculus (see 1) it is sufficient to show, that -$f_1\circ c: \mathbb R\to \mathbb R$ is smooth for each smooth curve $c:\mathbb R\to M$. -But $(f_1\circ c)^2=g\circ c$ is smooth and $(f_1\circ c)^3 = f\circ c$ is smooth, so by the theorem of Joris (http://mathoverflow.net/questions/127724), $f_1$ is smooth. QED<|endoftext|> -TITLE: How much of P versus NP's difficulty stems from having to rule out the existence of Turing machines that "accidentally" solve, say, 3-SAT efficiently? -QUESTION [5 upvotes]: It seems like there is a sense in which a Turing machine that demonstrates P=NP could be said to "accidentally" exist. I'm wondering the extent to which the possibility of such machines is the main reason P not equal to NP is so hard to prove. A priori can such machines exist, or are there heuristic or known reasons they can't? -By a Turing machine that accidentally solves P=NP, I mean a Turing machine that solves an NP-complete problem, say 3-SAT, in polynomial time. But its description might bear no logical or discernible relation to 3-SAT. To make this precise, maybe the right definition of an accidental solution is that it's not provable that the machine is a solution, even though one could observe empirically that it is (because it would be). -This would be analogous to observing empirically that a solution to 3-SAT happens to be encoded in the digits of pi starting at some digit (under some natural enumeration of all 3-SAT inputs and with pi written in base 2), or in the digits of some other irrational number whose digits can be computed in polynomial time. -If the existence of accidental solutions is consistent with ZFC, say, then P not equal to NP couldn't be proven. If this is in fact one of the things that makes P versus NP difficult, does it make sense to consider a variant of the P versus NP problem that restricts P not just to Turing machines that accept certain languages in polynomial time, but also to ones for which this is provable? Does this make the problem any easier? -Lines of thinking like this might be common knowledge to complexity theorists, but I haven't seen it expressed in writing. -This question is a little reminiscent of this MO question about the Goldbach conjecture. - -REPLY [6 votes]: Yes, it makes sense to consider such variants of the problem. Apart from the complexity-theoretic motivation, they arise quite naturally in the study of weak fragments of arithmetic (bounded arithmetic): for example, it is known that Buss’ theory $S_2$ (or equivalently, $I\Delta_0+\Omega_1$) is finitely axiomatizable if and only if $S_2$ proves that the polynomial hierarchy collapses (in an explicit way, i.e., there is a $\Sigma^P_n$-algorithm $M$ such that $S_2$ proves that $M$ solves a $\Sigma^P_{n+1}$-complete problem). -Unfortunately, the answer to the second part of your question is no, it does not seem to make the problem any easier, even if we ask for provability in an extremely weak theory (such as PV, and similar fragments of bounded arithmetic). -If you want to learn more about these issues, you can consult Bounded Arithmetic, Propositional Logic, and Complexity Theory by Krajíček, or Logical foundations of proof complexity by Cook and Nguyen.<|endoftext|> -TITLE: A little help with the unmixedness theorem? -QUESTION [5 upvotes]: I have two smooth subvarieties $Y$ and $Z$ of a smooth variety $X$. Their intersection $Y \cap Z$ has two irreducible components, both of the expected dimension and generically reduced. I want to conclude that $Y \cap Z$ is reduced by the unmixedness theorem. Is this right? - -REPLY [11 votes]: Dear Nick -- First of all, if a ring satisfies Serre's criterion $S1$ and is "generically reduced", i.e., the stalk at every generic point is a field, then the ring is reduced. This is explained, for instance at the top of p. 183, Section 23 of Matsumura's "Commutative Ring Theory". Second, if $Y$, resp. $Z$ is a closed subscheme of a regular, locally Noetherian scheme which is itself regular, then it is everywhere locally cut out by a regular sequence, cf. Theorem 21.2(ii), p. 171, of Matsumura. Finally, if also $Y\cap Z$ has the "expected codimension", then $Y\cap Z$ is also locally cut out by a regular sequence, and thus Cohen-Macaulay, by Theorem 17.4, p. 135 of Matsumura. (In fact it is even LCI by Theorem 21.2 again.) A Cohen-Macaulay scheme satisfies Serre's criterion $Sn$ for every integer $n\geq 0$. Thus your scheme $Y\cap Z$ is reduced.<|endoftext|> -TITLE: Totally rational polytopes -QUESTION [17 upvotes]: Define a convex polytope in $\mathbb{R}^d$ as -totally rational (my terminology) -if its vertex coordinates are rational, its edge lengths -are rational, its two-dimensional face areas are rational, etc., -and finally its (positive) volume is rational. -So: -rational coordinates, and the measure of every $k$-dimensional face, -$1 \le k \le d{-}1$, is rational, and the $d$-dimensional volume -is positive and rational. -(Scaling could then convert all these rationals to integers.) -For example, the hypercube with vertex coordinates $\{0,1\}^d$ -is totally rational. -Similarly an axis-aligned box with integral vertex coordinates -is totally rational. - -Q1. Are there other classes of totally rational polytopes, - classes defined for all $d$? - -In particular, - -Q2. Do there exist totally rational simplices in $\mathbb{R}^d$ for - arbitrarily large $d$? - -Pythagorean triples yield totally rational triangles. -I am not even certain that the -Heronian tetrahedra -described in -this MathWorld article -are totally rational, because it is unclear (to me) if they can be realized -with rational vertex coordinates. -All this is likely known, in which case key search phrases or other -pointers would be welcomed. Thanks! - -Addendum. Gerry Myerson's useful summary of Problem D22 in Unsolved Problems In Number Theory -answers Q2: The problem is open! -Q1 remains (apparently) interesting; see the comments by Steve Huntsman and Gerhard Paseman. - -REPLY [16 votes]: Guy, Unsolved Problems In Number Theory, problem D22: Simplexes with rational content. "Are there simplexes in any number of dimensions, all of whose contents (lengths, areas, volumes, hypewrvolumes) are rational?" -Guy notes the answer is "yes" in 2 dimensions, by Heron triangles. Also "yes" in three dimensions: "John Leech notes that four copies of an acute-angled Heron triangle will fit together to form such a tetrahedron, provided that the volume is made rational, and this is not difficult." The smallest example has three pairs of opposite edges of lengths 148, 195, and 203. -There is much more discussion, more examples, and several references. So far as I can see, there is no discussion of dimensions exceeding 3.<|endoftext|> -TITLE: Higher dimensional version of the Hurwitz formula? -QUESTION [24 upvotes]: In Hartshorne IV.2, notions related to ramification and branching are introduced, but only for curves. The main result is the Hurwitz formula. -Now if you have a finite surjective morphism between nonsingular, quasi-projective varieties, then the notion of ramification (divisor) would still make sense and we can also still talk about the degree of a canonical divisor. It also seemed to me like no result in IV.2 really uses the fact that $X$ and $Y$ are of dimension $1$. So I ask, can I replace $f$ by a finite, dominant, separable morphism $X\to Y$ of nonsingular, quasi-projective varieties of arbitrary dimension? That is, of course, up to and including Proposition 2.3. -If this is so, can we say anything about the degree of a canonical divisor in dimension greater than one? Maybe in special cases? - -REPLY [10 votes]: Here is a lighthearted attempt at generalizing the discussion of Hurwitz' formula in Hartshorne to higher dimensions. -Let $f:Y\to X$ be morphism of schemes over a field $k$. Assume that $X$ and $Y$ are integral and smooth of dimension $n$, and that $f$ is finite, dominant and separable. -Consider the exact sequence -\begin{equation*} -f^* \Omega_X \to \Omega_Y \to \Omega_{Y/X} \to 0. -\end{equation*} -Separability of the extension of fields $k(X)\subseteq k(Y)$ is equivalent to the vanishing $\Omega_{k(Y)/k(X)}=0$. Thus the map $f^*\Omega_X\to \Omega_Y$ is surjective at the generic point of $Y$. It follows that it is also injective there, since the two sheaves involved are locally free of the same rank. We conclude that $f^*\Omega_X\to \Omega_Y$ is injective everywhere, since $Y$ is integral. -Exterior powers of injective maps of modules that are finite and free over a ring are again injective. Thus the natural map $f^*\omega_X\to \omega_Y$ is injective. Tensoring with $\omega_Y^{-1}$, we obtain an invertible ideal sheaf $f^*\omega_Y\otimes \omega_Y^{-1} \subseteq \mathscr O_Y$. The corresponding effective Cartier divisor $R$ is called the ramification divisor of the cover $f:Y\to X$. By construction there is a natural isomorphism $f^*\omega_X(R)\xrightarrow\sim \omega_Y$. -Let $P\in Y$ be a point of codimension 1. We next show that -\begin{equation*} -\operatorname{length}_{\mathscr O_{Y,P}}\mathscr O_{R,P} = \operatorname{length}_{\mathscr O_{Y,P}} (\Omega_{Y/X})_P. -\end{equation*} -(I apologise in advance for the inelegant proof.) -This will imply the Weil divisor associated with $R$ is -\begin{equation*} -\sum_{F\subseteq Y} (\operatorname{length}_{\mathscr O_{Y,F}} \Omega_{Y/X}) \cdot F, -\end{equation*} -where the sum runs over all prime divisors of $Y$. In particular, the complement of $R$ in $Y$ is the largest open subset restricted to which $f$ is unramified. -Denote $A:=\mathscr O_{Y,P}$ and let $t\in A$ be a uniformizer. Choose bases around $P$ for the rank-$n$ locally free sheaves $f^*\Omega_Y$ and $\Omega_Y$. The map $(f^*\Omega_Y)_P\to (\Omega_X)_P$ is then given by a matrix $\alpha$, which we may assume to be in Smith normal form: -\begin{equation*} -\alpha = -\begin{bmatrix} -t^{a_1} & & \\ -& \ddots & \\ -& & t^{a_n} -\end{bmatrix}, -\end{equation*} -where the $a_i\ge 0$ are integers. Here there are no zeroes along the diagonal because $\alpha$ tensored with $\operatorname{Frac}(A)$ must be surjective. It easy to see that -\begin{equation*} -\mathscr O_{R,P} \cong A/(\det \alpha) = A/(t^{\sum a_i}), -\end{equation*} -while -\begin{equation*} -(\Omega_{Y/X})_P\cong \oplus_{i\ge 0}^n A/(t^{a_i}), -\end{equation*} -so these two modules have the same length $\sum a_i$. -Let $r(P)$ be the ramification of $f$ at $P$: this is the valuation of the image of any uniformizer of $\mathscr O_{X,f(P)}$ in $\mathscr O_{Y,P}$. Assume that the characteristic of $k$ does not divide $r(P)$, that the finite field extension $k(f(P))\subseteq k(P)$ is separable, and that the finitely generated one $k\subseteq k(f(P))$ is separably generated. We show that under these circumstances -\begin{equation*} -\operatorname{length}_{\mathscr O_{Y,P}} (\Omega_{Y/X})_P = r(P)-1. -\end{equation*} -Denote -\begin{align*} -r &:= r(P), \\ -A &:= \mathscr O_{X,f(P)}, \\ -B &:= \mathscr O_{Y,P}. -\end{align*} -Thus we have inclusions $k\subseteq B\subseteq A$. Denote the maximal ideal of $A$ by ${\frak m}_A$, its residue field by $k_A$, and similarly for $B$. Thus ${\frak m}_A = (t_A)$ and ${\frak m}_B = (t_B)$ with $t_B = u t_A^r$, where $u\in A^\times$ is a unit. -By hypothesis there is a transcendence basis $\bar f_1,\dotsc, \bar f_{n-1}$ of $k_B$ over $k$ such that the extension $k_B/k(\bar f_1,\dotsc,\bar f_{n-1})$ is separable (hence has no relative differentials). Choose lifts $f_1,\dotsc, f_{n-1}\in B$ of the elements of this transcendence basis. Looking at the exact sequence -\begin{equation*} -{\frak m}_B/{\frak m}_B^2 \xrightarrow{d} \Omega_B \otimes k_B \to \Omega_{k_B} \to 0 -\end{equation*} -and applying Nakayama we see that the map $B^{\oplus n} \to \Omega_B$ that sends -\begin{equation*} -e_i \mapsto -\begin{cases} -df_i & \text{if }i\le n-1\\ -dt_B & \text{if }i = n -\end{cases} -\end{equation*} -is surjective. Let $K$ denote its kernel. Then $\operatorname{Tor}_1^B(\Omega_B,k_B)$ surjects onto $K\otimes_B k_B$. From the fact that $\Omega_B$ is free, it follows that $K=0$. Thus the free $B$-module $\Omega_B$ has $df_1,\dotsc, df_{n-1}, dt_B$ as a basis. -From the fact that $k_A/k_B$ is separable, it follows that $k_A/k(\bar f_1,\dotsc, \bar f_{n-1})$ is as well, so by the preceding argument the free $A$-module $\Omega_A$ has $df_1,\dotsc, df_{n-1}, dt_A$ as a basis. -The relative differentials $\Omega_{A/B}$ are thus the quotient of $\Omega_A$ by the submodule generated by $df_1,\dotsc, df_{n-1}, dt_B$. Denote by $M$ the itermediate quotient of $\Omega_A$ by $df_1,\dotsc, df_{n-1}$. Then $M$ is freely generated by the image of $dt_A$. Write -\begin{equation*} -dt_B = t_A^r du + u r t_A^{r-1} dt_A -\end{equation*} -in $\Omega_A$. In $M$ we have $du = f dt_A$ for some $f\in A$, so $dt_B = u' t_A^{r-1}dt_A$, for some $u'\in A$. From the fact that $r\ne 0$ in $k$, it follows that $u'$ is a unit. Hence the map -\begin{equation*} -A/(t_A^{r-1}) \xrightarrow{\cdot dt_A} \Omega_{A/B} -\end{equation*} -is an isomorphism and $\operatorname{length}_A \Omega_{A/B} = r-1$ as claimed.<|endoftext|> -TITLE: Do the elementary properties of mixed volume characterize it uniquely? -QUESTION [22 upvotes]: Background -Take 2 convex sets in $\mathbb{R}^2$, or 3 convex sets in $\mathbb{R}^3$, or generally, $n$ convex sets in $\mathbb{R}^n$. "Mixed volume" assigns to such a family $A_1, \ldots, A_n$ a real number $V(A_1, \ldots, A_n)$, measured in $\mathrm{metres}^n$. -As I understand it, mixed volume is a kind of cousin of the determinant. I'll give the definition in a moment, but first here are some examples. - -$V(A, \ldots, A) = \mathrm{Vol}(A)$, for any convex set $A$. -More generally, suppose that $A_1, \ldots, A_n$ are all scalings of a single convex set (so that $A = r_i B$ for some convex $B$ and $r_i \geq 0$). Then $V(A_1, \ldots, A_n)$ is the geometric mean of $\mathrm{Vol}(A_1), \ldots, \mathrm{Vol}(A_n)$. -The previous examples don't show how mixed volume typically depends on the interplay between the sets. So, taking $n = 2$, let $A_1$ be an $a \times b$ rectangle and $A_2$ a $c \times d$ rectangle in $\mathbb{R}^2$, with their edges parallel to the coordinate axes. Then -$$ -V(A_1, A_2) = \frac{1}{2}(ad + bc). -$$ -(Compare and contrast the determinant formula $ad - bc$.) -More generally, take axis-parallel parallelepipeds $A_1, \ldots, A_n$ in $\mathbb{R}^n$. Write $m_{i1}, \ldots, m_{in}$ for the edge-lengths of $A_i$. Then -$$ -V(A_1, \ldots, A_n) = \frac{1}{n!} \sum_{\sigma \in S_n} m_{1, \sigma(1)} \cdots m_{n, \sigma(n)}. -$$ -(Again, compare and contrast the determinant formula.) - -The definition of mixed volume depends on a theorem of Minkowski: for any compact convex sets $A_1, \ldots, A_m$ in $\mathbb{R}^n$, the function -$$ -(\lambda_1, \ldots, \lambda_m) \mapsto \mathrm{Vol}(\lambda_1 A_1 + \cdots + \lambda_m A_m) -$$ -(where $\lambda_i \geq 0$ and $+$ means Minkowski sum) is a polynomial, homogeneous of degree $n$. For $m = n$, the mixed volume $V(A_1, \ldots, A_n)$ is defined as the coefficient of $\lambda_1 \lambda_2 \cdots \lambda_n$ in this polynomial, divided by $n!$. -Why pick out this particular coefficient? Because it turns out to tell you everything, in the following sense: for any convex sets $A_1, \ldots, A_m$ in $\mathbb{R}^n$, -$$ -\mathrm{Vol}(\lambda_1 A_1 + \cdots + \lambda_m A_m) = \sum_{i_1, \ldots, i_n = 1}^m V(A_{i_1}, \ldots, A_{i_n}) \lambda_{i_1} \cdots \lambda_{i_n}. -$$ -Properties of mixed volume -Formally, let $\mathscr{K}_n$ be the set of nonempty compact convex subsets of $\mathbb{R}^n$. Then mixed volume is a function -$$ -V: (\mathscr{K}_n)^n \to [0, \infty), -$$ -and has the following properties: - -Volume: $V(A, \ldots, A) = \mathrm{Vol}(A)$. (Here and below, the letters $A$, $A_i$ etc. will be understood to range over $\mathscr{K}_n$, and $\lambda$, $\lambda_i$ etc. will be nonnegative reals.) -Symmetry: $V$ is symmetric in its arguments. -Multilinearity: -$$ -V(\lambda A_1 + \lambda' A'_1, A_2, \ldots, A_n) = \lambda V(A_1, A_2, \ldots, A_n) + \lambda' V(A'_1, A_2, \ldots, A_n). -$$ -(These first three properties closely resemble a standard characterization of determinants.) -Continuity: $V$ is continuous with respect to the Hausdorff metric on $\mathscr{K}_n$. -Invariance: $V(gA_1, \ldots, gA_n) = V(A_1, \ldots, A_n)$ for any isometry $g$ of Euclidean space $\mathbb{R}^n$ onto itself. -Multivaluation: -$$ -V(A_1 \cup A'_1, A_2, \ldots, A_n) = V(A_1, A_2, \ldots) + V(A'_1, A_2, \ldots) - V(A_1 \cap A'_1, A_2, \ldots) -$$ -whenever $A_1, A'_1, A_1 \cup A'_1 \in \mathscr{K}_n$. -Monotonicity: $V(A_1, A_2, \ldots, A_n) \leq V(A'_1, A_2, \ldots, A_n)$ whenever $A_1 \subseteq A'_1$. - -There are other basic properties, but I'll stop there. -Questions -Is $V$ the unique function $(\mathscr{K}_n)^n \to [0, \infty)$ satisfying properties 1--7? -If so, does some subset of these properties suffice? In particular, do properties 1--3 suffice? -If not, is there a similar characterization involving different properties? -(Partway through writing this question, I found a recent paper of Vitali Milman and Rolf Schneider: Characterizing the mixed volume. I don't think it answers my question, though it does give me the impression that the answer might be unknown.) - -REPLY [15 votes]: Sorry to answer my own question, but asking this in public seems to have spurred me into thought. -As auniket suspected, the answer is "yes" in the strongest sense I'd hoped: properties 1-3 do characterize mixed volume. In fact, something slightly stronger is true: $V$ is the unique function $(\mathscr{K}_n)^n \to \mathbb{R}$ satisfying - -$V(A, \ldots, A) = Vol(A)$ -$V$ is symmetric -$V(A_1 + A'_1, A_2, \ldots, A_n) = V(A_1, A_2, \ldots, A_n) + V(A'_1, A_2, \ldots, A_n)$. - -In other words, we don't need multilinearity, just multiadditivity. -The proof is along the lines suggested by auniket. -Fix $n$ and $A_1, \ldots, A_n \in \mathscr{K}_n$. Write $\mathbf{n} = \{1, \ldots, n\}$, and for sets $R$ and $S$, write $\mathrm{Surj}(R, S)$ for the set of surjections $R \to S$. -I claim that for all subsets $S$ of $\mathbf{n}$, -$$ -\sum_{f \in \mathrm{Surj}(\mathbf{n}, S)} V(A_{f(1)}, \ldots, A_{f(n)}) -$$ -is uniquely determined by the properties above. The proof will be by induction on the cardinality of $S$. When $S = \mathbf{n}$, this sum is -$$ -n! V(A_1, \ldots, A_n), -$$ -so this claim will imply the characterization theorem. -To prove the claim, take $S \subseteq \mathbf{n}$. Then -$$ -Vol(\sum_{i \in S} A_i) = \sum_{f: \mathbf{n} \to S} V(A_{f(1)}, \ldots, A_{f(n)}) -$$ -by the three properties. This in turn is equal to -$$ -\sum_{R \subseteq S} \sum_{f \in \mathrm{Surj}(\mathbf{n}, R)} V(A_{f(1)}, \ldots, A_{f(n)}). -$$ -By the inductive assumption, all but one of the summands in the first summation - namely, $R = S$ - is uniquely determined. Hence the $S$-summand is uniquely determined too. This completes the induction, and so completes the proof. -The proof makes it clear that $V(A_1, \ldots, A_n)$ is some rational linear combination of ordinary volumes of Minkowski sums of some of the $A_i$s. It must be possible to unwind this proof and get an explicit expression; and that expression must be the one auniket gave (which also appears in Lemma 5.1.3 of Schneider's book Convex Bodies: The Brunn-Minkowski Theory). -This all seems rather easy, and must be well-known, though I'm a bit surprised that this characterization isn't mentioned in some of the things I've read. Incidentally, I now understand why it doesn't appear in the paper of Milman and Schneider mentioned in my question: they explicitly state that they want to avoid assuming property 1.<|endoftext|> -TITLE: Jones Polynomial of the trace closure of the fundamental braid -QUESTION [6 upvotes]: The fundamental braid $\Delta_n \in B_n$ is simply a twist by $\pi$ applied to the entire row of $n$ strands. In terms of Artin generators, it is given by -$$ -\Delta_n = (\sigma_1 \sigma_2 \cdots \sigma_{n-1})(\sigma_1 \sigma_2 \cdots \sigma_{n-2})\cdots (\sigma_1 \sigma_2) \sigma_1~. -$$ -The square of $\Delta_n$ (i.e., the full $2\pi$ twist) generates the center of $B_n$. -I have a rather simple (and quite possibly trivial) question about these braids. What is the Jones polynomial of the trace closure of $\Delta_n$? Do the trace closures of the $\Delta_n$ result in some well-known link family? -I have tried computing the J.P. in the obvious way using the Kauffman bracket; some simplifications are possible, but so far nothing sufficient to lead to a general formula. - -REPLY [18 votes]: Calculation of the Jones polynomial of this link is a (good) exercise in representation -theory. As you have observed by Schur's lemma, in any irreducible representation it is, up to a scalar, -a square root of the identity. This scalar can be obtained by a determinant argument. -So we reduce to the situation where the eigenvalues are 1 and -1. The trace is then -just the difference between the multiplicities. This can be determined by specialisation -to the case t=1 where the representation is a symmetric group representation and all -such questions are well known. So you have the trace in all irreducible representations going into -the Jones representation and you just add them up with their weights. -Unfortunately I was persuaded not to include this method in my first paper on the polynomial -when I already had the Jones polynomial of torus knots. For Homflypt it is a little more -complicated but carried out in detail in my Hecke algebras annals paper, and again in -the paper with Marc Rosso where we compute arbitrary quantum invariants of torus knots. -It's all a lot simpler for the Jones polynomial itself where there are so few irreducible -representations. -Have fun, Vaughan Jones<|endoftext|> -TITLE: A limit to Shoenfield Absoluteness -QUESTION [20 upvotes]: Shoenfield's Absoluteness Theorem states that if $\phi$ is any $\Sigma^1_2$ sentence of second-order arithmetic, then $\phi$ is absolute between any two models of $ZF$ which share the same ordinals. This means that such $\phi$ are unaffected by forcing, or by Axiom of Choice-related considerations (since for any $V\models ZF$, the corresponding subclass $L$ is a model of $ZFC$). -What I'm looking for is a simple example of either a $\Sigma^1_3$ or $\Pi^1_3$ sentence $\psi$ which is not absolute in this way - ideally, such a $\psi$ which is true in some $V\models ZFC$ and false in a generic extension $V[G]$. This is a purely pedagogical question for me - I'm trying to internalize Shoenfield Absoluteness, and I feel that a nice example of why it can't be made much stronger would help. -Thank you all in advance. - -REPLY [4 votes]: Jensen and Solovay dealt with the question. For example $0^\#$ is a $\Delta^1_3$ real number, which is obviously not absolute because it cannot go deeper than $L[0^\#]$, and clearly it can't be absolute enough to be in $L$. -Jensen and Solovay point out that assuming a measurable cardinal, then there is such example. But what if one doesn't want to assume large cardinals? -Solovay constructed the following example. $M$ is a transitive countable model of $V=L$. Then there is $N=M[a]$ where $a$ is a real number and there is a $\Pi^1_2$ predicate $S(x)$ such that $N\models(\exists!x\subseteq\omega)S(x)$, and $N\models S(a)$, and moreover $N=M[a]$. -Jensen extended this result and showed that we can have that every $y\in\mathcal P(\omega)^L$ is recursive in $a$. -Both these results appear in the paper: - -Jensen, R. B.; Solovay, R. M. "Some applications of almost disjoint sets." Mathematical Logic and Foundations of Set Theory (Proc. Internat. Colloq., Jerusalem, 1968) (1970) pp. 84–104.<|endoftext|> -TITLE: Congruence subgroups as abstract groups -QUESTION [11 upvotes]: This is probably well know, and maybe even trivial, but not to me. Consider for concreteness the subgroup -$$ -\pm\Gamma_0(3)=\left\{\begin{pmatrix}a & b \\ c & d\end{pmatrix}:\;a,b,c,d\in\mathbb{Z},ad-bc=\pm1, c\equiv 0\pmod 3\right\} -$$ -of $GL_2(\mathbb{Z})$. This has of course index 4 in $GL_2(\mathbb{Z})$. The first (possibly completely ridiculous) question is - -Does $\pm\Gamma_0(3)$ contain a subgroup isomorphic to $GL_2(\mathbb{Z})$? - -It's not even obvious to me that the two are not isomorphic as abstract groups. The second question is - -Does $GL_2(\mathbb{Z})$ contain subgroups that are isomorphic to $\pm\Gamma_0(3)$ with finite index other than 4? If the answer is yes, then what is the least common multiple of all such indices? E.g. is there a subgroup of index 3 (or 5, or 7, or...) in $GL_2(\mathbb{Z})$ isomorphic to $\pm\Gamma_0(3)$? Or will all such indices be multiples of 4? - -An answer or technique that is applicable to other congruence subgroups and to other values of 2 would be a great bonus, but for now I would happily settle for an answer to this concrete question. - -REPLY [9 votes]: Your question is a great example of the usefulness of the multiplicativity of Euler characteristic. The fractional Euler characteristic of $\textrm{GL}_2\mathbb{Z}$ is -$$\chi(\textrm{GL}_2\mathbb{Z})=\ \ {-\frac{1}{24}}$$ This means by definition that any torsion-free index-$k$ subgroup of $\textrm{GL}_2\mathbb{Z}$ has Euler characteristic $-\frac{k}{24}$. For a proof (indeed a number of different proofs), see this Math Overflow question. -The Euler characteristic is multiplicative for finite-index subgroups, and it's immediate from the definition that the same is true for fractional Euler characteristics. Therefore since $\pm\Gamma_0(3)$ has index 4 in $\textrm{GL}_2\mathbb{Z}$, we have -$$\chi\big(\!\pm\!\Gamma_0(3)\big)=\ \,-\frac{1}{6}$$ -This demonstrates that $\pm\Gamma_0(3)$ is not isomorphic to $\textrm{GL}_2\mathbb{Z}$, and moreover any finite-index subgroup isomorphic to $\pm\Gamma_0(3)$ must be of index 4. You can certainly continue this analysis: for example, whenever $p$ is prime we have $\chi(\pm\Gamma_0(p))=-\frac{p+1}{24}$, so none of the groups $\pm\Gamma_0(p)$ are isomorphic to each other, or to $\Gamma_0(p)$, since $\chi(\Gamma_0(p))=-\frac{p+1}{12}$ (I leave it to you to work out what happens with $\Gamma_0(n)$ when $n$ is composite). However, you can only take it so far: there are certainly congruence subgroups that are non-isomorphic but can't be distinguished by their Euler characteristic.<|endoftext|> -TITLE: Finding colinear points in F_q^n -QUESTION [8 upvotes]: Forgive me if this is well known, it's not really my field, but it's a problem I've run across and thought about a bit. -Let $\mathbb{F}_q$ be a finite field with $q$ elements, let $n\ge2$, and let $A,B,C$ be subsets of $\mathbb{F}_q^n$ each containing $N$ points. How hard is it to determine if there is a triple $(a,b,c)\in A\times B\times C$ such that $a$, $b$, and $c$ are colinear? More specifically: - -Is this problem NP hard? -Is there an algorithm to solve the problem in time $O(N^\kappa)$ for some small $\kappa$? (I'd be - especially interested if $\kappa$ is strictly smaller than $\frac{3}{2}$.) -Or am I missing something and there's an obvious polynomial-time algorithm to solve this problem? - -Note that the decision problem and the computational problem are polynomial-time equivalent. Thus suppose you can solve the decision problem in time $F(N)$. Write -$$ - A=A_1\cup A_2,\quad B=B_1\cup B_2,\quad C=C_1\cup C_2 -$$ -and solve the decision problem for the 8 sets $A_i\times B_j\times C_k$. That takes -time $8F(N/2)$. If any of the decision problems returns a YES answer, then repeat the process with that particular $A_i,B_j,C_k$. After about $\log_2(N)$ iterations, you'll be down to sets containing only one element, which gives the colinear triple. -The case I'm most interested in is $n=2$. Obviously there are various generalizations, for example one could take $t$ sets and ask if there is a $t$-tuple lying in a linear space of dimension $t-2$. -One final related (easier?) question. If $A,B,C$ are simply taken to be subsets of $\mathbb{F}_q$, how difficult is it to determine if there is a triple $(a,b,c)$ satisfying $a+b+c=0$? There are obvious collision algorithms, but are there better algorithms? - -REPLY [3 votes]: The problem is NP-hard if your sets are represented by functions as you describe in your comment. This is because you can encode some NP-hard problem with $F_A$. Suppose that $B$ and $C$ are both singletons. Let $F_A(k)$ return a point not on the $BC$ line if $k$ is not the solution of some NP-hard problem and a point on the $BC$ line if it is. This way there are three collinear points if and only if the NP-hard problem has a solution. -This also shows that the related (easier?) version is also NP-hard.<|endoftext|> -TITLE: Compactness of the class of connected sets with perimetre smaller than 1? -QUESTION [8 upvotes]: Hello! -It is my first post, so please be indulgent! -Here is the problem: I am in the class S of closed subsets of [0,1]^2 that are connected and have perimeter less or equal to 1. -I endow this space with the Hausdorff metric (or equivalently Fell topology), that says that for two compacts A, B, d(A,B)<=r iff every point of A is at distance less than r of B, and the other way around. -Question: Is this space compact? -Equivalent question: Is the function "perimeter" lower semi continuous on this set? (It is equivalent because the Hausdorff metric is compact for the class of all compact sets, and the class of connected sets is closed). -In other words, given a sequence of connected sets with perimeter at most 1, is it possible to see suddenly some perimeter appear at the limit? (which would be an answer "no" to the question). -The perimeter here is the 1-dimensional measure of the boundary, i.e. the infimum over all coverings of the boundary by balls of the sum of the diameters of the balls. -Remark: If one drops the assumption of connectedness, it is not true anymore (consider for instance a set of points -thus with zero perimeter- that becomes dense in the square). So this assumption is very important! -Thak you in advance! - -REPLY [12 votes]: No. Let $B$ be a closed ball of radius 1/2 and $I$ a diameter of $B$. Construct a Cantor-like set $K\subset I$ of lengths $\mathcal H^1(K)=0.9$ (where $\mathcal H^1$ denotes the 1-dimensional Hausdorff measure). We have $I\setminus K=\bigcup I_i$ where $I_1,I_2,\dots$ are disjoint open subintervals of $I$ and $\sum_{i=1}^\infty\mathcal H_1(I_i)=0.1$. Let $B_i$ be the open ball for which $I_i$ is a diameter, and let $X_k=B\setminus\bigcup_{i=1}^k B_i$. Then each $X_k$ is a closed connected set and the sequence $\{X_k\}$ Hausdorff converges to $X=\bigcap_{k=1}^\infty X_k=B\setminus\bigcup_{i=1}^\infty B_i$. -The boundary of $X_k$ is a union of $k+1$ disjoint circles $\partial B$ and $\partial B_i$, $1\le i\le k$, hence -$$ - \mathcal H^1(\partial X_k) = \mathcal H^1(\partial B)+\sum_{i=1}^k \mathcal H^1(\partial B_i) = \pi+\sum_{i=1}^k\pi\mathcal H^1(I_i) \le \pi+0.1\pi =: C. -$$ -However the boundary of $X$ contains $K$, hence $\mathcal H^1(\partial X)\ge\mathcal H^1(\partial B)+\mathcal H^1(K)=\pi+0.9> C$. -Thus the space of connected sets of perimeter $\le C$ is not closed.<|endoftext|> -TITLE: Mnemonic for how left and right duals interact with Homs -QUESTION [8 upvotes]: Suppose you have a rigid monoidal category. This means you have a left dual and a right dual. By Frobenius reciprocity, you can move objects from one side of a Hom to the other side at the price of dualizing (e.g. something like $\mathrm{Hom}(X \otimes Y, Z) = \mathrm{Hom}(X, Z \otimes Y^*$). However this comes in four flavors: you can tensor on the left or on the right and you can move from the input to output or vice-versa. For each of these flavors you need to remember whether to take left dual or right dual. Half of these are easy to remember: if you know how to move from input to output, then you have to use the other kind of dual to move back. But I can never remember consistent conventions for everything. -So does anyone have some easy-to-remember clear mnemonic for dealing with this? - -REPLY [6 votes]: I find that thinking in string diagram pictures is easiest for me. The identification of homs comes from taking a map $X \otimes Y \to Z$ and bending one of the strings around to the other side, as in the picture below. - -What we get is a map $X \to Z \otimes Y^{\ast}$. How do you know that this is $Y^{\ast}$ and not ${^{\ast}}Y$? Well, I call $Y^{\ast}$ the left right dual (maybe other people call it the right left dual), and it's the one where the arrow on the string goes from right to left, at least the way I draw the diagrams. The other way to remember it is that the ${\ast}$ goes on the inside in the evaluation pairing (and hence on the outside in the coevaluation). -I don't think people will ever agree on conventions for which way string diagrams go, or which one is the left dual and which one is the right dual, but I can at least be internally consistent with these conventions.<|endoftext|> -TITLE: Really rigid varieties -QUESTION [7 upvotes]: Are there (complex, projective) varieties $X_0$ with the following property? - -Every (flat, say) family $\mathfrak X\to T$ over a reasonable and reasonably big class of bases with a fiber over a closed point equal to $X_0$ is a trivial family, in the sense that $\mathfrak X$ is $T\times X_0$. - -If one asks this with $T$ restricted to local Artinian rings, say, one gets the class of (infinitesimally) rigid varieties, but I wonder if there are "globally rigid" examples. -Later: As Francesco and unknowngoogle observe, it is quite unreasonable to ask for such strong rigidity... :) - -REPLY [12 votes]: If $X_0$ is projectively embedded, we can use a Gr\"obner basis for its defining ideal to make a Rees family over ${\mathbb A}^1$ most of whose fibers are $X_0$, but whose central fiber is a monomial scheme. So $X_0$ must already be a monomial scheme. Since you said it's a variety, it must be a projective space. If you then allow me to reembed (by a Veronese), I could break that projective space, too, unless it's a point. So my answer: $X_0$ must be a point.<|endoftext|> -TITLE: Hom(A,C) ⊗ Hom(B,D) injects into Hom(A⊗B,C⊗D): when? why? -QUESTION [15 upvotes]: Sorry for asking a linear algebra question on a research forum, but this seems to be either a case of extreme blindness on my side, or a case of a result lying much deeper than it seems. -The following theorem is "easily seen" according to a text I have been reading (more precisely, it is part of Proposition I.1.2 in that text): -Theorem 1. Let $A$, $B$, $C$, $D$ be four vector spaces over a field $k$. Then, the canonical map -$\mathrm{Hom}\left(A,C\right)\otimes\mathrm{Hom}\left(B,D\right) \to \mathrm{Hom}\left(A\otimes B,C\otimes D\right)$, -$f\otimes g\mapsto \left(a\otimes b\mapsto f\left(a\right)\otimes g\left(b\right)\right)$ -is injective. -I see how this is trivial if $A$ and $B$ are finite-dimensional. I also see that it is indeed easy if $C$ and $D$ are finite-dimensional. But without finite-dimensionality conditions I have nowhere to start. The $\mathrm{Hom}$ functor does not commute with direct sums, while $\otimes$ does not commute with direct products (or does it over a field?), so there seems to be no easy way to reduce it to finite-dimensional cases. How can we proceed then? -Also, is there any application of the above theorem outside of the two cases I mentioned? -To make this more interesting, how much is saved if we let $k$ be a commutative ring with $1$, and require (say) flatness instead of freeness? - -REPLY [5 votes]: Here is a alternative proof, which is not as short as a-fortiori's but it seems to be more conceptual and includes the following lemma which is useful in its own right: (Everything takes place over a field) - -Lemma. The natural map $V \otimes \prod_i W_i \to \prod_i (V \otimes W_i)$ is injective. - -Proof: Choose a basis $B$ of $V$. Then the map corresponds to the natural map $(\prod_i W_i)^{(B)} \to \prod_i (W_i^{(B)})$, given by a kind of transposition $((w_{ib})_{i})_{b} \mapsto ((w_{ib})_{b})_{i}$, obviously injective. - -Corollary. The natural map $\prod_i V_i \otimes \prod_j W_j \to \prod_{i,j} V_i \otimes W_j$ is injective. - -Proof: Apply the Lemma twice. - -Lemma. The natural map $\hom(V',V) \otimes \hom(W',W) \to \hom(V' \otimes W',V \otimes W)$ is injective. - -Proof: Choose basis $B,C$ of $V',W'$. Then the map corresponds to the natural map $V^B \otimes W^C \to (V \otimes W)^{B \times C}$, which is injective by the Corollary.<|endoftext|> -TITLE: When can a finite map be blown up to a flat one? -QUESTION [5 upvotes]: Let $f:X\to Y$ be a generically finite proper morphism of varieties. There is some locus in $Y$ over which the fiber of $f$ is positive dimensional, so we blow it up, along with the preimage of it in $X$ to get a map $\tilde{f}:\tilde{X}\to\tilde{Y}$ which has finite fibers. -Are there any nice conditions that will guarantee that the map $\tilde{f}$ is flat? - -REPLY [5 votes]: Have you seen "Critères de platitude et de projectivité. Techniques de "platification'' d'un module" by Raynaud-Gruson (1971)? In particular 5.2.2. I think it is very close to what you want (this was explained to me by Bhargav Bhatt not too long ago). It doesn't say that any blow-up works, but there is one that's ok. -Basically, there exists a blow-up $Y' \to Y$ (you can assume $Y'$ is normal, ie normalize the blow-up) such that the appropriate component(s) of the fiber product $Y'' \to Y' \times_Y X$ give us a map $Y'' \to Y'$ which is flat. This is proven in the modern setting by Hilbert-Scheme arguments usually (also see for example various papers talking about universal flattening).<|endoftext|> -TITLE: 2-cocycle twists of braided Hopf algebras -QUESTION [13 upvotes]: 2-cocycle twists of Hopf algebras -Let $H$ be a Hopf algebra over a field $k$. Then a (left, unital) 2-cocycle on $H$ is a map -$$ f: H \otimes H \to k$$ -such that -$$ f(x_{(1)},y_{(1)})f(x_{(2)} y_{(2)}, z) = f(y_{(1)}, z_{(1)}) f(x, y_{(2)} z_{(2)}) $$ -(in Sweedler notation) and -$$ f(x,1) = \varepsilon(x) = f(1,x) $$ -for $x,y,z \in H$. Also one usually wants that $f$ is invertible for the convolution product, i.e. that there is some functional -$$ \bar{f} : H \otimes H \to k $$ -such that -$$ f(x_{(1)}, y_{(1)}) \bar{f}(x_{(2)}, y_{(2)}) = \varepsilon (x) \varepsilon (y) = \bar{f} (x_{(1)}, y_{(1)}) f(x_{(2)}, y_{(2)}). $$ -The cocycle can be used to twist the multiplication of $H$ as follows: -$$ x \cdot_f y = f(x_{(1)}, y_{(1)}) x_{(2)} y_{(2)} \bar{f}(x_{(3)}, y_{(3)}), $$ -The cocycle condition ensures that $\cdot_f$ is associative, and the fact that $f$ is unital means that the old identity element is still the identity element for $\cdot_f$. -You can also twist the antipode of the Hopf algebra in such a way as to get a new Hopf algebra structure on $H$, where the comultiplication and counit are the same as before. You have to check that the original comultiplication and counit are still algebra homomorphisms with respect to $\cdot_f$. -Twisting the algebra structure only -You can modify this construction in such a way as to obtain only an algebra instead of a Hopf algebra. This is done by defining -$$ x \cdot_f y = f(x_{(1)}, y_{(1)}) x_{(2)} y_{(2)}. $$ -Again, $f$ being unital means that the identity element is the same. -The braided setting -All of this can be done in exactly the same way whenever $H$ is a Hopf algebra object in a braided monoidal category. All the maps can be interpreted as string diagrams, and the proofs that $\cdot_f$ is associative and that the comultiplication and counit are algebra maps for the twisted multiplication go through in exactly the same way by manipulating the string diagrams. -My question is: -Does anybody know a reference for 2-cocycle twists of a Hopf algebra object in a braided monoidal category? I've looked in Majid's book, and he does talk a great deal about Hopf algebras in braided categories (what he calls braided groups), and also about 2-cocycle deformations, but I don't think he puts the two concepts together. Please correct me if I'm wrong. Thanks! - -REPLY [9 votes]: The two concepts - twisting a Hopf algebra one-sided to an algebra and two-sided to a new Hopf algebra - are actually intimately connected and play an important role in several areas of current research. The former are known (as always, there's restrictions on the equivalence, e.g. in infinite dimension) as Galois objects, the latter as Doi-twist. -A Galois object over a Hopf algebra $H$ is an $H-$comodule algebra $_HA$ with a nondegeneragy condition (can be bijective); mostly (if a cleaving exists, s.a.) this means $A=H^\sigma$ is a one-sided 2-cocycle deformation. It can be completed to a Bigalois object $_HA_L$ (a second compatible comodule structure on the other side) between $H$ and what turns out to be the (two-sided Hopf algebra) Doi-twist $L=H_\sigma$. These can be multiplied by the contensor product if they fit together: -$$_HA_L\;\;and\;\;_LB_M\rightarrow\;_H(A\Box_L B)_M$$ -This forms the Bigalois gruppoid (gruppoid because you dont stick with one Hopf algebra $H$, but "hop accross" the Doi twists $L,M,...$ and can only multiply "fitting ends"), which is the best analogon you have to a cohomology group (which is no group for the reason above!) and usually much more organized to actually calculate with...the basics can be found e.g. in Schauenburg's paper or more elaborate in Susan Montgomory's "Hopf algebras and their actions on rings". There are many worked out examples (e.g. for generalized Taft algebras), stuff like a Küneth-formula etc.pp. -On the other hand, Doi-twists are a very "mild" operation, that do not change the module-category....that's way the question "what Doi twists are there of $H$?" often appears as the question for quasi-isomorphisms $H\rightarrow L$ - this may also help googeling ;-) -Now directly to the question... -Though I've never seen it, I'd definitely agree that you get no problem, as all the concepts (both approaches) just use the braided category structure ("braided" doesn't mean it's nontrivial, just that you got the structure map)...or as you say, are expressable in string diagrams. -Are you interested in a specific use of these deformations? Do you have a specific braided category in mind? -...and a good source of examples? -My favourite examples of braided categories "in-use" are the Yetter-Drinfel'd modules $V$ over a finite group $G$, which means: $G$-graded, $G$-action, such that $g.V_h=V_{ghg^{-1}}$. Of course this taste is influenced by it being my field of work ;-) They include the "super- and color-spaces" ($\mathbb{Z}_2,\mathbb{Z}_3$), but of course mus more... -In these categories you have much-studied braided Hopf algebras resembling (but maybe finite-dimensional!) a universal enveloping of a Lie algebra, the Nichols algebras. These are glued to the groupring itself to form new quantum groups by a Radford/Majid-construction, and often deformed by a Doi twist to get a complete classification (Andruskiewitsch/Schneider)...well, usually AFTER the relations are found it is proven, that it's really a Doi twist... -I think one could use these known Doi twists of the Hopf algebra to write down deforming 2-cocycles just on the Nichols algebra in the braided sense. I've tried this for the "generic cases" (linking- and root-relations), but there once you get rid of the groupring, they're trivial (because in the deformed relations some $1-g$ appears...I could even imagine that's why they're the "easyer" ones??). However I've found several exceptional later-worked-out cases in literature (e.g. selflinkings Daniel Didt's Dissertation p. 46-49) where this is no longer the case. Here you have "real" Nichols-with-Nichols-to-Nichols relations and in my opinion these deformations have to correspond to nontrivial braided 2-cocycles on the Nichols algebra part alone...(?) -If there's still interest in the question (it's been almos a year) I'm sure we could work something out :-)<|endoftext|> -TITLE: Which commutative algebras admit a nonzero Poisson bracket? -QUESTION [7 upvotes]: Let $A$ be a commutative algebra, not necessarily unital, over a field $k$ (of characteristic not equal to $2$, or even equal to $0$, if it helps). A second-order formal deformation of $A$ is a $k[h]/h^3$-bilinear associative product $\star$ on $A[h]/h^3$ such that quotienting by $h$, we obtain the original product on $A$. Writing such a product as -$$a \star b = ab + h m_1(a, b) + h^2 m_2(a, b), a, b \in A$$ -it's not hard to verify that $\{ a, b \} = m_1(a, b) - m_1(b, a)$ is a Poisson bracket on $A$, that is, a Lie bracket satisfying the Leibniz rule $\{ a, bc \} = \{ a, b \} c + b \{ a, c \}$. Given a nonzero Poisson bracket on $A$, it is interesting to ask whether we can find a formal deformation (replace $k[h]/h^3$ with $k[[h]]$) which gives rise to it as above ("deformation quantization"). -But of course we can't ask this question until we have a nonzero Poisson bracket in the first place. So: - -Which commutative algebras admit a nonzero Poisson bracket? - -If there is no reasonable description in general feel free to restrict to the finitely-generated case or smooth functions on manifolds etc. -What I know: any polynomial algebra in $2$ or more variables admits a nonzero Poisson bracket (take the symmetric algebra on a nonabelian Lie algebra). Any nonzero Poisson bracket gives a nonzero element of the alternating part of the second Hochschild cohomology $H^2(A, A)$, so if this group is trivial then no such brackets exist. I doubt this implication can be reversed in general, but I don't know a counterexample. If you do, I have a math.SE question you should answer! - -REPLY [4 votes]: Smooth functions on a manifold always admit lots of Poisson structures. Indeed, one can construct Poisson structures on $\mathbb R^n$ with support in $(-1,1)^n$ and any prescribed value $\pi^{ij}e_i \wedge e_j$ of the Poisson-Bivector at the origin. This is done in two steps: -At first one constructs $n$ commuting vector fields $X_i$ with support in $(-1,1)^n$. -Then on just defines the $\pi:=\pi^{ij} X_i\wedge X_j$.<|endoftext|> -TITLE: Is there a "derived" Free $P$-algebra functor for an operad $P$? -QUESTION [8 upvotes]: Recall that an operad (in vector spaces, say) $P$ consists of a collection of vector spaces $P(n)$ for $n\geq 0$, such that $P(n)$ is equipped with an action by the symmetric group $S_n$, with maps $P(n) \otimes P(k_1) \otimes \dots \otimes P(k_n) \to P(k_1+\dots k_n)$ for any $n,k_1,\dots,k_n$, subject to natural associativity constraints. Recall also that a $P$-algebra for an operad $P$ consists of a vector space $V$ and for each $n$ a map $P(n) \to \hom(V^{\otimes n},V)$ subject to natural associativity constraints. Recall finally that for any operad $P$, the functor from vector spaces to $P$-algebras that is adjoint to the Forgetful functor is: -$$ \operatorname{Free}: V \mapsto \bigoplus_n \bigl(P(n) \otimes V^n \bigr)_{S_n} $$ -where by $W_{S_n}$ I mean the coinvariants of the $S_n$-module $W$, i.e. $W_{S_n} = W \otimes_{\mathbb{K} S_n} 1$. -Let me switch to working with chain complexes rather than vector spaces. Then I can do a more refined operation than taking coinvariants. Namely, I can use the derived functor of coinvariants: you replace $W$ by a projective resolution, tensor, and consider the result up to quasiisomorphism. Which is to say "the complex that computes $\operatorname{Tor}$". - -Question: What is the meaning of the ($\infty$-)functor that takes derived coinvariants rather than coinvariants in the above formula? What relationship does it play to the operad, etc.? - -The type of answer I'm looking might be the following: "The $(\infty,1)$-functor so described is adjoint to the functor that forgets from the $(\infty,1)$-category of strongly homotopy $P$-algebras to the $(\infty,1)$-category of chain complexes." -Another direction that you could take the following question might be: - -Question 2: What type of "Koszul duality" statements are there using the "derived coinvariants" version of $\operatorname{Free}$? - -REPLY [6 votes]: Yes this is an expression of the free algebra functor, left adjoint to the forgetful functor, see Section 3.1.3 of Lurie's Higher Algebra. -Koszul duality in the (oo,1)-setting is treated very nicely in the work of John Francis, see in particular math/1104.0181, where eg the relation with deformation theory is explained (the tangent complex to an augmented O-algebra carrying an action of the Koszul dual operad, etc)<|endoftext|> -TITLE: How many sequences of rational squares are there, all of whose differences are also rational squares? -QUESTION [14 upvotes]: After commenting on a -question - of Joseph O'Rourke's, I thought it interesting that a number theory result (artihmetic progressions of rational squares cannot be arbitrarily long) had applications to geometry (don't look at mostly regular cones of regular hypercubes for totally rational polytopes). (I hope I got the above right and that it is indeed an application; I proceed on that basis.) Of course I am also impressed by the fact that there is no known geometric proof of the fact that finite geometries satisfy both or neither of the configurations of Pappus and of Desargues. -So of course, a natural question would be to consider applications of number theory to geometry; I'm not going to do that here. Instead, I will ask for assistance with Joseph's program by asking a question about rational squares. -The first question that occurred to me was " (1) Is there a sequence of integer squares whose differences are also integral squares?" For sake of interest I require all squares to be nonzero, though later they may be rational and not just integral. -Before posting this question, I saw the answer was yes, and that indeed there were at least countably many such sequences, although I don't know if there are infinitely many tails. So I nominate question first': - -(1') How many infinite sequences of integer squares are there, all of whose first differences are also integer squares? - -There is the potential to be uncountably many such, especially if there are (is?) an uncountable infinity of tails. But wait! There's more! - -(2) Fix an integer $k$ with $k > 1$. How many infinite sequences of integer squares are there, all of whose first through $k$th differences are also integer squares? - -Recall that for a sequence $a_i$, the first difference is the sequence $b_i = a_{i+1} - a_i$, and the $(k+1)$st difference is the first difference of the $k$th difference. I suspect that for $k$ large enough, the answer will be zero. However, those are just warm ups for this question: - -(3) How many sequences of rational squares are there such that for every positive integer $k$ all $k$th differences are also rational squares? - -Motivation: I think it is a cool set of questions. Also I think that if Joseph is going to get a family of rational polytopes of arbitrarily high dimension, he will find such sequences useful (I am thinking volume of a pyramid being base times height times some rational number in combination with a multidimensional Pythagorean-type expression), and that such a family will imply the existence of such sequences, but I do not see the converse as the polytopes have to satisfy additional relations. As usual, reference requests and related problems are welcome. -Gerhard "Ask Me About System Design" Paseman, 2011.08.03 - -REPLY [2 votes]: Here is an attempt at a cleaner exposition for problem (1');  although I take full credit/blame for the exposition, it is based on ideas posted by Barry Cipra, Gjergji Zaimi, and joro. -Let me define S, also known as the square sequence graph.  The vertices will be all positive integers which are squares of integers greater than 2, and edges will be directed: $(a,b)$ will be an edge from $a$ to $b$ iff the quantity $b-a$ is the square of a positive integer.  Paths through S will thus be monotonically increasing. -S has many vertices with out degree at least 2.  Indeed, if $d$ is not the square of a prime the square of either an odd composite number or of a number with at least three not necessarily distinct prime divisors (thanks to Gerry Myerson for an earlier counterexample), it has at least two factorizations of the form $mn$ where $m-n$ is even and positive, and each such factorization leads to an arrow from $d$ to $((m+n)/2)^2$.  Further, considerations mod 4 show that odd squares can only have arrows to other odd squares in S.  -Each infinite sequence in the question (1') corresponds to a path through S.  Barry Cipra suggested how to find uncountably many such paths.  Let $d$ be any vertex in S which is 9 mod 10, and $d > 10$. There is an arrow from $d$ to $c = ((d + 1)/2)^2$.  This is a number which is an odd nontrivial multiple of 5 as well as being a square, and has two or more arrows leading from it.  One of the arrows from $c$ leads to a still larger square $((c+1)/2)^2$ which is 9 mod 10.  Another leads to a different square (corresponding to a factorization where $m = c/5$ and $n = 5$) which leads to another square which is 5 mod 10. -Any path which starts out with a large square mod 5 has a choice of passing through an infinite number of other squares mod 5, where after each such square, the path may go to a square which is 9 mod 10 before going to a square which is 5 mod 10.  Since this subset of paths is determined by which subset of these countably infinite set of choices to make, Barry has shown us a subset which has a bijection (thanks, Gjergji) with a set of infinite binary sequences.  In the set theory that I like doing this, this means there are at least continuum many such sequences. -I invite others to play with S and find out properties of infinite sequences of squares with square first differences. -The related graph R using rational squares holds promise also.  It may be possible to use S to show that for $k=2$, the answer to question (2) is 0, which is my intuition.  It should be clear that all (with at most one exception) of the kth differences of an integral square sequence must be even for them to be all integral squares. -Gerhard "Ask Me About System Design" Paseman, 2011.08.04<|endoftext|> -TITLE: Lebesgue Measurability and Weak CH -QUESTION [19 upvotes]: Let $LM$ denote "all subsets of $\Bbb{R}$ are Lebesgue measurable", and -$WCH$ (weak continuum hypothesis) denote "every uncountable subset of $\Bbb{R}$ can be be put into 1-1 correspondence with $\Bbb{R}$". -[Warning: in other contexts, weak CH means something totally different , i.e., it sometimes means $2^{\aleph_{0}} < 2^{\aleph_{1}}$]. -We know that $LM$ and $WCH$ both hold in Solovay models. By forcing a (Ramsey) ultrafilter over a Solovay model one can also arrange $WCH+\neg LM$ (due to joint work of Di Prisco and Todorčević, who showed that the perfect set property holds in the generic extension). -This prompts my question ($DC$ below is dependent choice). - -Question: Is it known, relative to appropriate large cardinal axioms, whether there is a model of $ZF+LM+DC+\neg WCH$? - -My question arose from an FOM-question of Tim Chow, and my answer to it; see also Chow's response. - -REPLY [9 votes]: This is an expansion of my comments above. In the paper with Zapletal that I reference, we assume a proper class of Woodin cardinals and force over $L(\mathbb{R})$ with a partial order of countable approximations to a certain kind of MAD family (which Jindra named an "improved" MAD family). Although I have yet to write out the details, I believe that the resulting model satisfies LM (it clearly satisfies DC), and that, in this model, $\mathbb{R}$ cannot be injected into the generic MAD family. The arguments I have in mind are straightforward applications of the arguments given in the paper. -The paper of Horowitz and Shelah referenced in my second comment works from the assumption of a strongly inaccessible cardinal and, as I understand it, adds the construction of a generic MAD family to Solovay's argument. As shown in their paper, DC + LM hold in the resulting model. I wrote to Haim and asked if $\mathbb{R}$ can be injected into the generic MAD family in this model, and he said no. He says he'll update their paper to include a proof of this (so they'll probably have a proof out before we do).<|endoftext|> -TITLE: Calculation of Lyapunov Exponent from Time Series -QUESTION [5 upvotes]: I am currently doing research in non-linear dynamical systems, and I require to calculate Lyapunov exponents from time series data frequently. I found a MatLab program lyaprosen.m that does this for me, but I am not very sure of its validity, as I do not get the same results from it, as some results in some papers. Does anyone have any alternative tools to calculate Lyapunov exponents from time series data? - -REPLY [3 votes]: TSTOOL is the state of the art: http://www.physik3.gwdg.de/tstool/index.html<|endoftext|> -TITLE: How to resolve a disagreement about a mathematical proof? -QUESTION [50 upvotes]: I am having a problem which should not exist. I am reading what I believe to be an important paper by a person - let me call him/her $A$ - whom I believe to be a serious and talented mathematician. A lemma in this paper is proven by means of an argument which, if correct, is a highly elegant piece of mental acrobatics in the spirit of Grothendieck, where a complicated situation is reduced to a simple one by embedding the objects of study in much larger (but ultimately better) object. Unfortunately, the beauty of this argument is - for me - marred by a doubt about its correctness. In my eyes, the argument rests upon a confusion of two objects which are not equal and should not be, but have the same name by force of an abuse of notation going awry. A dozen of emails exchanged with $A$ did not clear up the situation, and I start feeling that this is unlikely to improve; what is likely is that after a few more mails the correspondence will degenerate into a flamewar (as any prolonged arguments with my participation seem to do, for some reasons unknown). The fact that $A$ is not a native English speaker adds to the difficulty. -At this point, I can think of several ways to proceed: - -Let go. There is a number of reasons for me not to choose this option; first of all, I really want to know whether the proof of the lemma is correct or not (even though there seems to be a different proof in literature, although not of that beauty), but this has also become, for me, a matter of idealism and an exercise in tenacity (in its cheapest manifestation - it's not like writing emails is hard work...). -Construct a counterexample. This is complicated by the fact that I am attacking the proof, not the theorem (which seems to be correct). Yet I think I have done so, and $A$ failed to really address the counterexample. But given the frequent misunderstandings between us (not least because of the language barrier) I am not sure whether $A$ has realized that I am talking counterexamples at all - and whether there is a way to tell this without switching to what will be probably understood as an aggressive tone. -Request $A$ to break down the argument into simple steps, eschewing abuse of notations. This means, in the particular case I am talking about, requesting $A$ to write two pages in his/her free time and respond to some irritating criticism of these pages with the prospect of seeing them destroyed by a counterexample. I am not sure this counts as courteous. Besides, the paper is about 10 years old - most authors do not even bother answering questions on their work of such age. -Go public (by asking on MO or similarly). This is something I really want to avoid as long as there is no other way. Neither criticizing $A$ as a person/scientist, nor devaluing the paper (which consists of far more than the lemma in question...) is among my goals; besides I cannot rule out as improbable that the error is on my side (and my experience shows that even in cases when I could rule this out, it still often was on my side). -Have a break and return to the question in a month or so. I am expecting to hear this (seems to be a popular answer to lots of questions...) yet I am not sure how this can be of any use. - -These ideas are all I could come up with and none of them sounds like a good plan. What am I missing? Is my problem a common one, and if yes, does it have a time-tested solution? Can it be answered on this general scale? Is it a real problem or an artefact of my perception? -PS. This is being posted anonymously in order to preserve genericity (of the author and, more importantly, of $A$). - -REPLY [24 votes]: There are three separate issues here. -1) How to clarify whether the proof is correct? You should start with making a serious good will effort to understand what is written (which amounts to redoing all the bad notation, splitting things into small steps, etc. to the best of your abilities). If this fails, you should state as clearly as you can what exactly the problem with the argument is and hope that some expert will figure out who is right. Of course, first you should send the full account of your effort to the author reproducing all the parts of the proof you understand and showing clearly where you are stuck and why. Just to say "your notation is bad here so..." won't accomplish anything: at best, he'll make a local correction that will move the real issue somewhere else and you can play this shifting game forever. -If he still fails to address the issue after that, request the help of some third party -sending the same account of your effort together with the paper. Again, it is important that you demonstrate your good will and decent understanding of what is written in the paper before you raise your objection. Without this, you just won't be taken seriously. Make sure that you understand everything that precedes the unclear/incorrect step and that you make it clear to everyone whom you want to ask that you understand it. Nobody pays attention to people coming out of nowhere with zero credentials and doubtful qualifications. If your first words are "I don't understand ... and I think it is wrong", the most likely answer will be "Go learn ... ". However if your first words are "This argument starts with using ... to establish ...", your general credibility goes up immediately (provided that what you are saying makes sense). The more times the person agrees with you on the issue before you raise the question, the more likely he is to take you and your objection seriously. -It is your moral duty to make a real effort trying to understand the proof before making -any public comment on its correctness but it is also your moral duty to report a problem -with a proof when you are convinced that you see one. Note that it is completely normal -in mathematics to make bad mistakes occasionally and it is completely normal to fail -to understand correct arguments now and then. The priority/reputation chase has distorted the general attitudes beyond recognition, of course, but the heart of the matter is still the search for the knowledge, not building/preserving/destroying reputations and relationships. -Even if you are wrong on the account that the proof has a gap, you may be right -on the account that it is unclear (assuming that you have a decent education in the subject, the fact that you fail to understand the argument is a clear indication that the paper is written not in an ideal way). So, the clarification may help innumerable poor souls (like graduate students) who may have the same difficulty but just do not dare to ask questions. You risk to look like a fool, of course, but the only way to avoid looking like a fool occasionally is to always be one. -2) How to avoid the confrontation? At some point there may be no way and all that you'll be able to write to the author something like "It seems very difficult for us to understand each other. Since the issue is principal, the best we can do is to seek an opinion of a third party. I'd appreciate your suggestions of whom we should ask. I'm thinking of (put the list of experts you know)". This may not save your good relationships but, at least, will clear you from any "doing things behind the curtains" charges. After that, send your doubts to both the people on your list and the people on his list, if he provides one. If he doesn't, it is his problem. There are three possibilities: 1) you'll be backed by some expert, which will make the author harder to ignore you; 2) someone will explain to you why the proof is correct, and 3) everyone will ignore your request. In that last case you may have to seek the opinion of general public but not before you double check your argumentation. -3) Is MO an appropriate place for this discussion? It is not what it was intended for but if you finally decide to seek the general public opinion and post your objection on arXiv or somewhere else (in the way I outlined above; let me emphasize once more that unless you are Terry Tao or Tim Gowers you should demonstrate both good understanding of the matter and your good will before anyone will bother to take a look at your objection in honest) I see nothing wrong with making a short post containing the corresponding link in this thread. -In brief, if you really want to figure things out, I would advise that: -a) you make a good effort putting all your thoughts together in written. Create a clear "case" starting from the beginning where you agree with A on everything and talk in the same terms and stopping where you have an objection. -b) present this full writeup to A and wait for his comment. -c) if it doesn't result in anything meaningful, present this writeup to a few experts or, -at least, people whom you feel to be more knowledgeable in the subject than yourself. -d) if you are totally ignored, ask yourself why that may be the case and, if you see nothing wrong with your written argumentation (if you see nothing wrong with what you keep in your mind, there is a good chance that you are just blind), present it to the general public. -I don't think asking a graduate student of A is a good idea. First, many graduate students are totally incompetent in anything beyond their thesis project and in such cases, you can just as well use a parrot for communication. Second, if the student is actually good and you are right, you'll put the student in the position where he will have to tell his adviser that the paper is wrong. This doesn't go well with many people.<|endoftext|> -TITLE: Matrix inversion lemma with pseudoinverses -QUESTION [11 upvotes]: The utility of the Matrix Inversion Lemma has been well-exploited for several questions on MO. Thus, with some positive hope, I'd like to field a question of my own. -Suppose we pick $n$ values $x_1,\ldots,x_n$, independently sampled from $N(0,1)$ (mean 0, unit variance gaussian). Then, we form the (rank 3 at best) positive semidefinite matrix: -$$A = \alpha ee^T + [\cos(x_i-x_j)],$$ -where $e$ denotes the vector of all ones, and $\alpha > 0$ is a fixed scalar. -For $n \ge 3$, simple experiments lead one to conjecture that: -$$e^TA^\dagger e = \alpha^{-1},$$ -where $A^\dagger$ is the Moore-Penrose pseudoinverse of $A$ (obtained in Matlab using the 'pinv' function). -This should be fairly easy to prove with the right tools, such as a Matrix inversion lemma that allows rank deficient matrices or pseudoinverses. So my question is: - -How to prove the above conjecture (without too much labor, if possible)? - -REPLY [12 votes]: In fact more generally for any positive semidefinite matrix $A = \sum_{i=0}^k e_i e_i^T$ with $e_i$'s linearly independent, we have that $e_i^T B e_i = 1$, where $B$ is the Moore-Penrose pseudoinverse of $A$. This applies here since almost surely your matrix $A$ is of this form with $k=3$ and $e_1 = \sqrt \alpha e$. -Proof: Let $E$ be the linear span of the $e_i$'s. If I understood correctly the notion of Moore-Penrose pseudoinverse, $B$ is described in the following way: as a linear map, $B$ is zero on the orthogonal of $E$, and on $E$ it is the inverse of the restriction of $A$ to $E$. Let $\beta_{i,j}$ be defined by $B e_i = \sum_j \beta_{i,j} e_j$, so that $e_i^T B e_i = \sum_j\beta_{i,j} e_i^T e_j$. Expressing that $A B e_i = e_i$, we get in particular that $\sum_j\beta_{i,j} e_i^T e_j = 1$, QED.<|endoftext|> -TITLE: What are some proofs of Godel's Theorem which are *essentially different* from the original proof? -QUESTION [38 upvotes]: I am looking for examples of proofs of Godel's (First) Incompleteness Theorem which are essentially different from (Rosser's improvement of) Godel's original proof. -This is partly inspired by questions two previously asked questions: -(Proofs of Gödel's theorem) -(When are two proofs of the same theorem really different proofs) -To give an example of what I mean: The Godel/Rosser proof (see http://www.jstor.org/pss/2269059 for an exposition) shows that any consistent sufficiently strong axiomatizable theory is incomplete. The proof uses a substantial amount of recursion theory: the representability of primitive recursive functions and the diagonal lemma (roughly the same as Kleene's Recursion Theorem) are essential ingredients. The second incompleteness theorem - that no consistent sufficiently strong axiomatizable theory can prove its own consistency - is essentially a corollary to this proof, and a rather natural one at that. On the other hand, in 2000 Hilary Putnam published (https://doi.org/10.1305/ndjfl/1027953483) an alternate proof of Godel's first incompleteness theorem, due to Saul Kripke around 1984. This proof uses much less recursion theory, instead relying on some elementary model theory of nonstandard models of arithmetic. The theorem proven is slightly weaker, since Kripke's proof requires $\Sigma^0_2$-soundness, which is stronger than mere consistency (although still weaker than Godel's original assumption of $\omega$-consistency). -Kripke's proof is clearly sufficiently different from the Godel/Rosser proof that it deserves to be considered a genuinely separate object. What makes the difference seem really impressive, at least to me, is that Kripke's proof yields a different corollary than that of Godel/Rosser. In a short paragraph, Putnam shows (and I do not know whether this part of his paper is due to Kripke) that Kripke's argument proves that there is no consistent finitely axiomatizable extension of $PA$. This is not a result which I know to follow from the Godel/Rosser proof; moreover, the Second Incompleteness Theorem, which is a corollary to Godel/Rosser's proof, does not seem easily derivable from Kripke's proof. -Motivated by this, say that two proofs of (roughly) the same theorem are essentially different if they yield different natural corollaries. Clearly this is a totally subjective notion, but I think it has enough shared meaning to be worthwhile. -My main question, then, is: - -What other essentially different proofs of something resembling Godel's First Incompleteness Theorem are known? In other words, is there some other proof of something close to "every consistent axiomatizable extension of $PA$ is incomplete" which does not yield Godel's Second Incompleteness Theorem as a natural corollary? - -I am especially interested in proofs which don't yield the nonexistence of consistent finitely axiomatizable extensions of $PA$, either, and in proofs which do yield some natural corollary. I don't particularly care about the precise version of the First Incompleteness Theorem proved: if it applies to systems in the language of second-order arithmetic, if it assumes $\omega$-consistency, or if it only applies to systems stronger than $ATR_0$, say, that's all the same to me. However, I do require that the version of the incompleteness theorem proved apply to all sufficiently strong systems with whatever consistency property is needed; so, for example, I would not consider the work of Paris and Harrington to be a good example of this. -The only other potential example of such an essentially different proof that I know of is the proof(s) by Jech and Woodin (see https://andrescaicedo.files.wordpress.com/2010/11/2ndincompleteness1.pdf), but I don't understand that proof at a level such that I would be comfortable saying that it is in fact an essentially different proof. It seems to me to be rather similar to the original proof. Perhaps someone can enlighten me? -Of course, entirely separate from my main question, my characterization of the difference between the specific two proofs of the incompleteness theorem mentioned above may be incorrect. So I'm also interested in the following question: - -Is it in fact the case that Kripke's proof does not yield Second Incompleteness as an natural corollary, and that Godel/Rosser's proof does not easily yield the nonexistence of a consistent finitely axiomatizable extension of PA as a natural corollary? - -REPLY [2 votes]: There’s also an “invariant definability” argument. I’ll sketch it quickly below, and then give an analysis to explain why I think it’s meaningfully different. Embarrassingly I can't find a source for it at the moment; I recall seeing it as a footnote in Kreisel's Model-theoretic invariants paper, but it doesn't seem to be there. Multiple authors have written on invariant definability (which this answer is not-so-secretly an advertisement of) so I haven't yet been able to conduct an exhaustive search for the reference, but when I find it I'll update this. Incidentally, this argument was referred to at the beginning of another answer of mine. -Below, “definable” means “definable without parameters.” For more pleasant language I'll call this argument the "Tarskian argument." - -Argument -Let $T$ be an “appropriate” theory of arithmetic (say, $T\supseteq R$). We tweak Tarski’s undefinability theorem very slightly as follows. For $X\subseteq\mathbb{N}$ and $M\models T$, say that $X$ is -pseudo-definable in $M$ if $X=D\cap \mathbb{N}$ for some definable $D\subseteq M$. We then have: - -$(*)\quad$ Suppose $M\models T$. Then $Th(M)$ is not pseudo-definable in $M$. - -The standard proof still works: supposing to the contrary that $\theta$ pseudo-defined $Th(M)$ in $M$, let $m$ be the Godel number of the formula $\eta(x)\equiv$ "$\theta$ fails on the number of the sentence gotten by plugging $x$ into the formula with number $x$" - appropriately formalized - and consider the sentence $\eta(\underline{m})$ (applying representability appropriately). -With this in hand we argue as follows. Suppose $S\supseteq T$ is computable and satisfiable. Then by representability we have that $S$ is pseudo-definable in $M$ for every $M\models T$. Taking $M\models S$, we have by $(*)$ that $S\not=Th(M)$, so $S$ is not complete. - -Analysis -Now let me argue in favor of the Tarskian argument being a genuine variation. -First, there’s an easy negative observation: it applies to Willard’s self-verifying theories and so cannot yield the second incompleteness theorem as a direct corollary. More subjectively, the argument is fairly non-constructive, and doesn’t (as far as I can see) quickly yield a specific undecidable sentence. -Of course, this is also a feature of the many standard computability-theoretic arguments. I think there’s still a difference here - this time a positive one - due to the way the Tarskian argument interacts with the notion of invariant definability. There are a couple ways to frame this - see e.g. the beginning of the article Abstract Computability and Invariant Definability by Moschovakis for some discussion - and I'll use the following: - -Definition: - -An arithmetic context is a set $\mathfrak{C}$ of models of Robinson's arithmetic $R$. - -For an arithmetic context $\mathfrak{C}$, a set $A\subseteq \mathbb{N}$ is $\mathfrak{C}$-invariantly definable if there is some formula $\varphi$ with $\varphi^M\cap\mathbb{N}=A$ for all $M\in\mathfrak{C}$. - - - -(Here I’m indulging in the usual abusive conflation of $\underline{k}^M$ and $k$.) - - -For a theory $E$ and an arithmetic context $\mathfrak{C}$, say that $E$ is $\mathfrak{C}$-satisfiable if some member of $\mathfrak{C}$ satisfies $E$. - - -Then the Tarskian argument in fact gives: - -Proposition: Suppose $\mathfrak{C}$ is an arithmetic context. Then no $\mathfrak{C}$-satisfiable theory is $\mathfrak{C}$-invariantly definable. - -(Since the computable sets are invariantly definable in every arithmetic context and every theory extending $R$ yields an arithmetic context, this is a generalization of essential undecidability.) -The point is that in general the $\mathfrak{C}$-invariantly definable sets need not support a good computability theory in any sense: by judiciously terrible choice of $\mathfrak{C}$, we can make the set of $\mathfrak{C}$-invariantly definable sets have basically no structural properties besides forming a Boolean algebra. So, for example, I don’t see how to whip up an analogue of the “inseparable c.e. sets” argument for arbitrary arithmetic contexts. -Of course, pathological arithmetic contexts are uninteresting, so it’s hard to argue that this aspect is actually valuable in any way. But it is - as far as I can tell - a nontrivial feature.<|endoftext|> -TITLE: Manin's algebraic geometry textbook? -QUESTION [17 upvotes]: Context: I recently chatted with a postdoc from russia, and we somehow got on the topic of learning mathematics and textbooks, and he told me about a wonderful textbook by Yuri Manin, on algebraic geometry which was written at the beginning of the seventies and wildely used for a long time in the former SU. He was so full of praise that I decided to check it out. -Now the problem I checked our library catalog and amazon.com and can't find anything that I can identify as this AG-textbook. -Question: -So does this book exists? What's it's title? Was it maybe not translated?. -Q2: -If anyone has anything interesting to say about the book this would also be appreciated. -PS: I also speak german so if you know a german translation please also post. - -REPLY [11 votes]: There now is a book by Manin Introduction to the theory of schemes (Translated from the Russian, edited and with a preface by Dimitry Leites) recently published by Springer based on Manin's lectures on algebraic geometry mentioned in David Roberts' post.<|endoftext|> -TITLE: Factoring some integer in the given interval -QUESTION [8 upvotes]: I'm posting this question here (rather than on CSTheory) since it seems to require much more knowledge about number theory than algorithms. - -Let N be a positive integer. Is there an efficient (i.e. probabilistic polynomial time) algorithm which, on input a sufficiently large N, outputs the full factorization of some integer in the interval $[N - O(\log N), N]$? -Note that the running time of the algorithm is measured in $|N| = O(\log N)$. - -Cross-posted https://math.stackexchange.com/questions/54580. -Spin-off: https://math.stackexchange.com/questions/54719. - -REPLY [6 votes]: I think Felipe Voloch has the right sense about the problem: you should expect to encounter a prime or a number which is a small number times a prime. Since you suggested probabilistic and did not mention that you wanted ONLY the desired number output, here is a start on your desired program. -Pick a desired small bound B, which will be the largest of the primes to be sieved out. Make B compatible with the desired running time of your eventual program, but I like Felipe's suggestion of $O(\text{log}^2N)$. Now for each prime $p$ up to B, compute the remainder of $N$ after dividing by the largest power of $p$ that is smaller than $N$. Use this to populate an array of length of your interval with powers of $p$ which are the factors of the corresponding numbers. This should take (B/log(B)) times $O(\text{log}N)$ time and space. -At this point you have several options. The simplest one is to perform the divisions and list out the cofactors as well as the small factors. Even with B smaller than Felipe's suggestion, you will most likely have printed out the complete factorization of one of the numbers. Use whatever time you have remaining to find it, either by doing quick primality tests on all the candidates or slow primality test on some appropriate subset. An alternative is to continue eliminating small factors, for once $p$ is larger than your $O(\text{log}N)$, you won't need to worry about picking powers, but can switch to doing gcd with 0 or 1 numbers in the interval. -In short, there is a way to make an efficient version which is morally equivalent to trial division, and still have time left over to pick with high chance of success the completely factored number. -EDIT 2011.08.10 I got curious, so I asked a -question and did some computations with B=2, with the expectation that I would find very few intervals of the form $[N - \text{log}_2N ,N]$ which did not contain a prime or a power of 2 times a prime, or a power of 2. The data so far are contrary to my expectations: if I haven't messed up the programming, there are more than 200 such $N$ less than $10^5$. Although I still think the small intervals will contain a B-smooth number for B not too large, this recent data puts a measure of doubt in my mind. -END EDIT 2011.08.10 -Gerhard "Ask Me About System Design" Paseman, 2011.08.04<|endoftext|> -TITLE: Constructing Steiner Triple Systems Algorithmically -QUESTION [11 upvotes]: I want to create STS(n) algorithmically. I know there are STS(n)s for $n \cong 1,3 \mod 6$. But it is difficult to actually construct the triples. For STS(7) it is pretty easy and but for larger n I end up using trial and error. Is there a general algorithm that can be used? - -REPLY [4 votes]: Since this thread just got bumped to the front page, historically the very first proof (by T. P. Kirkman, On a Problem in Combinatorics, Cambridge Dublin Math. J. 2 (1847) 191-204, 1847.) of the existence of an ${\rm STS}(v)$ for all $v \equiv 1, 3 \pmod{6}$ is completely algorithmic, where you start with a singleton as the point set and an empty set as its block set (i.e., the trivial design ${\rm STS}(1)$) and successively construct an ${\rm STS}(3)$, ${\rm STS}(7)$, ${\rm STS}(9)$, and so forth by applying the same algorithms recursively to the smaller ${\rm STS}$s you have at hand. So, you conjure up ${\rm STS}$s from thin air one after another algorithmically for all admissible orders. -A modernized version of this technique is called the doubling construction. This construction can be found in a very accessible textbook "Design Theory" by C. C. Lindner and C. A. Rodger from CRC Press (in Section 1.8 of the second edition). -The doubling construction actually consists of two separate construction techniques to cover all $v \equiv 1, 3 \pmod{6}$. If you want a single algorithm to cover all orders, the same textbook also explains such a technique (originally by R. M. Wilson, Some partitions of all triples into Steiner triple systems, Lecture Notes in Math., Springer, Berlin, 411 (1974) 267-277) in Section 1.6 (in either edition). -Edit: Here's the first half of the doubling construction: -Assume that you have an ${\rm STS}(v)$ with point set $V$ and block set $\mathcal{B}$. First, you copy all points; if $a \in V$, you make a new point $a' \not\in V$ so you have another set $V'$ of the same size. You add one extra point, say $\infty$, and use -$$W = \{\infty\}\cup V\cup V'$$ -as the new point set. Now, for each block $\{a,b,c\} \in \mathcal{B}$, you create new blocks $\{a',b',c\}$, $\{a',b,c'\}$ and $\{a,b',c'\}$. Then you join $\mathcal{B}$ and all these pseudo-copied new blocks as well as new $v$ blocks $\{\infty, a, a'\}$, where $a \in V$. So, the new block set $\mathcal{B}'$ is $$\mathcal{B}' = \mathcal{B}\cup\{\{a',b',c\}\ \vert\ \{a,b,c\} \in \mathcal{B}\} \cup \{\{\infty, a, a'\} \ \vert \ a \in V\}.$$ You can easily check that the ordered pair $(W, \mathcal{B}')$ is an ${\rm STS}(2v+1)$. -The latter half of the construction produces an ${\rm STS}(2v+7)$ from an ${\rm STS}(v)$ in a little more complicated way. Applying these two algorithms recursively gives you an ${\rm STS}(v)$ for all $v \equiv 1, 3 \pmod{6}$, covering all $v$ satisfying the necessary conditions for the existence of an ${\rm STS}(v)$.<|endoftext|> -TITLE: Lie locally nilpotent associative algebras -QUESTION [6 upvotes]: Let $A$ be an associative algebra over a field. Then $A$ can be regarded as a Lie algebra via the Lie bracket defined by $[a,b]=ab-ba$ for every $a,b\in A$. -The algebra $A$ is called Lie locally nilpotent if it is locally nilpotent as a Lie algebra. Also, $A$ is said to be locally Lie nilpotent if every finitely generated associative subalgebra of $A$ is nilpotent as a Lie algebra. -Clearly, if $A$ is locally Lie nilpotent then it is Lie locally nilpotent. Is the converse true? - -REPLY [3 votes]: I have to mention that for associative algebras over fields of characteristic not 3 the question has now a positive answer. This is a consequence of a very recent paper by Dias and Krasilnikov: https://arxiv.org/pdf/1709.05728.pdf<|endoftext|> -TITLE: Is $Ded(\kappa)\kappa$, then $D(\kappa,\kappa^\mu)$ holds, which implies in particular that $Ded(\kappa)\ge \kappa^\mu$. -Questions (1) Can we prove that $Ded(\kappa)< Ded(\kappa)^\omega$ is consistent? -(2) If $\mu$ a cardinal between $\omega$ and $\kappa$, can we prove that $Ded(\kappa)=\kappa^\mu$ is consistent? -Note 1: Following Keisler $Ded(\kappa)$, $Ded(\kappa)^\omega$ are two of the six possible "stability functions", the other four being $\kappa$, $\kappa+2^\omega$, $\kappa^\omega$ and $2^\kappa$. Stability functions give us the number of types of a theory $T$ over models of power $\kappa$. For more on this consult The Stability Function of a Theory by -H. Jerome Keisler, The Journal of Symbolic Logic, Vol. 43, No. 3 (Sep., 1978), pp. 481-486 -Note 2: There is a similar question posted on MathOverflow (Given a cardinal k, what's the biggest dense linear order with a dense subset of size k?) that asks for the consistency of $Ded(\kappa)<2^\kappa$ (answer is positive) - -REPLY [5 votes]: I just found the following paper on arXiv: "On non-forking spectra" by Artem Chernikov, Itay Kaplan and Saharon Shelah ( http://arxiv.org/abs/1205.3101 ). They claim that it is consistent that $Ded(\kappa)< Ded(\kappa)^\omega$, therefore answering this question positively.<|endoftext|> -TITLE: Ultrapowers by normalized ultrafilters -QUESTION [5 upvotes]: Suppose $j\colon V\to M$ is an elementary embedding and $\kappa$ is the critical point of $j$, then $\kappa$ is measurable, and we can define the ultrafilter $U$ over $\kappa$ as: $$A\in U\iff \kappa\in j(A)$$ -This is a normal ultrafilter. Despite not seeing an actual example, I am aware that $\operatorname{Ult}(V,U)$ may not be $M$ itself (where $\operatorname{Ult}(V,U)$ is the transitive collapse of the ultrapower of $V$ by $U$). -I have two questions in this regard: - -When does $\operatorname{Ult}(V,U)=M$? (Except, of course, the trivial case where $M$ was already defined as that ultrapower) -Suppose $\kappa$ is supercompact, and $M$ witnesses some $\lambda$-supercompactness, we can of course repeat the above construction, is there any extra properties we can say about $\operatorname{Ult}(V,U)$ in relation to $M$? (some further closure properties, or so...) - -REPLY [4 votes]: I would answer question 1 by saying that -$M=\text{Ult}(V,U)$ if and only if $j$ is isomorphic to an -ultrapower by some normal measure. This is another way of -saying that normal measures are minimal with respect to the -Rudin-Kiesler order. The point is that an embedding is the -ultrapower by a normal measure if and only if it is the -ultrapower by its induced normal measure, if and only if -$\kappa$ as a seed generates the whole embedding, in the -sense that every element of $M$ has the form -$j(f)(\kappa)$. -For question 2, suppose that $j:V\to M$ witnesses the -$\lambda$-supercompactness of $\kappa$, so that $\kappa$ is -the critical point of $j$ and $M^\lambda\subset M$. Your -ultrafilter $U$ is what is sometimes called the induced -normal measure of $j$, and since the seed hull -$X_\kappa=\{j(f)(\kappa)\mid f:\kappa\to V\}$ is an -elementary substructure of $M$, we may collapse it and form -a commutative diagram, with $V\to M_U\to M$, where -$j_U:V\to M_U=\text{Ult}(V,U)$ is the ultrapower by $U$ and -$k:M_U\to M$ is the inverse collapse of $X_\kappa$ and -$j:V\to M$ is the composition $j=k\circ j_U$. - -insert triangular commutative diagram here - -Since $M^\lambda\subset M$ in $V$ and $M_U\subset V$, it -follows immediately that $M^\lambda\cap M_U\subset M$. -Thus, $M$ remains $\lambda$-closed for sequences in $M_U$. -However, when $\lambda\gt\kappa$, then it will never be the -case that $M\subset M_U$. This is because $M_U$ is an -ultrapower by a filter on $\kappa$ and therefore $j_U$ is -continuous at ordinals of cofinality $\kappa^+$. Therefore, -$j_U(\kappa^+)$ has true cofinality $\kappa^+$ in $V$ and -hence also in $M$, but it is regular in $M_U$, being a -successor cardinal there. Thus, although the quotient map -$k:M_U\to M$ is a $\lambda$-closed embedding, it can never -be an internal embedding from the perspective of $M_U$, -since $M$ is not a subclass of $M_U$. -However, if you are willing to consider the case -$\lambda=\kappa$, then it is possible for $k:M_U\to M$ to -be internal to $M_U$, for we may simply let $j$ be the -ultrapower by $U\times U_1$ for any other measure on -$\kappa$, so that $k$ is simply the ultrapower in $M_U$ by -$j_U(U_1)$. In this case, $M$ is $j_U(\kappa)$-closed in -$M_U$, since it is an internal ultrapower there by a -measure on $j_U(\kappa)$.<|endoftext|> -TITLE: What is the complex structure on the boundary torus of a hyperbolic knot complement? -QUESTION [8 upvotes]: Let $K$ be a hyperbolic knot in $\mathbb S^3$. Restrict the corresponding representation $\pi_1(\mathbb S^3\setminus K)\to\operatorname{PSL}(2,\mathbb C)$ to the fundamental group of the boundary (the peripheral subgroup) to get a map $\pi_1(\partial N_\epsilon K)\to\operatorname{PSL}(2,\mathbb C)$. Now $\pi_1(\partial N_\epsilon K)$ is just $\mathbb Z\oplus\mathbb Z$, with two generators $m$ (meridian) and $\ell$ (longitude). The image of $\pi_1(\partial N_\epsilon K)$ is parabolic, and so (up to overall conjugation) is of the form: -$$ -am+b\ell\mapsto\left(\begin{matrix}1&a+b\lambda\cr 0&1\end{matrix}\right) -$$ -for some complex number $\lambda\in\mathbb C\setminus\mathbb R$. -What is known about this invariant $\lambda(K)$ of hyperbolic knots? Has anyone defined/studied it? - -REPLY [11 votes]: The conformal structure on the cuspidal torus is usually called the "cusp shape." -See Adams, Hildebrand, Weeks Hyperbolic invariants of knots and links and McReynolds, Arithmetic cusp shapes are dense, for starters. -See also work on "geometric inflexibility" by Neumann and Reid. -Also check out the work of Marc Lackenby and Jessica Purcell.<|endoftext|> -TITLE: Schemes do not form a stack in the etale topology? -QUESTION [7 upvotes]: As I understand, one of the reasons for "bootstrapping" to the category of algebraic spaces before constructing the category of Artin stacks is that algebraic spaces form a stack in the etale (at least) topology, while schemes do not, even though one frequently has (at least in other contexts) to use the fact that, say, affine or quasi-affine schemes do form a stack even in the fpqc topology (by descent theory for quasi-coherent sheaves). As a result, I'm curious: what is the simplest example of non-gluable (say, etale) descent data for schemes? -To clarify, I'm looking for an example of an fpqc morphism $Y' \to Y$, a scheme $X' \to Y'$ together with the usual patching after pull-back to $Y' \times_Y Y'$ that does not come from a scheme over $Y$. - -REPLY [11 votes]: I'm pretty sure this example works. I do not include any proof that this is the simplest example, and it may not be, but it's not too complicated. -Let $L_1$ and $L_2$ be two rational curves in $\def\P{\mathbb P}\P^3$ which intersect in two points. A standard example of a proper non-projective variety $X$ is obtained by blowing up $L_1$ and $L_2$, but doing it in one order at one intersection point and in the other order at the other intersection point (I think this example is explained at the end of Hartshorne). -There is an involution $\sigma$ of $\P^3$ which switches the two lines and the two intersection points. Let $U\subseteq \P^3$ be the open locus where $\sigma$ acts freely, and let $Y=U\times_{\P^3}X$. Then $Y/\sigma$ is an algebraic space (over the scheme $U/\sigma$) which is not a scheme. It becomes a scheme after the etale base change $U\to U/\sigma$.<|endoftext|> -TITLE: Are associated bundles representable in schemes? -QUESTION [7 upvotes]: I have seen the following claim without proof in more than one paper, but it is sufficiently general that I suspect it is stated too strongly to be true: - -Let $G$ be an affine group scheme (say, over a field of characteristic zero), let $X$ be a scheme (smooth over the same field), and let $P \to X$ be a $G_X$-torsor. If $Y$ is a scheme with a $G$-action, then the associated bundle $P \times^G Y$ is a scheme. - -I can use fpqc descent to show that this is true when $Y$ is affine (or some other effective fpqc descent class), and I can use Zariski descent when $P$ is Zariski-locally trivial. In full generality, I suspect one can assemble known counterexamples of descent to falsify this claim, but I have been unable to do so. -Question: Is there a counterexample known? Failing that, is there a proof of the claim? -I'm somewhat more interested in the case where $G$ is connected, but here is a candidate that I don't know how to prove: Take $X$ to be a smooth curve with a nontrivial étale double cover $P$ (with $G$ constant of order 2), and set $Y$ to be Hironaka's 3-fold with involution (whose quotient sheaf is not a scheme). - -REPLY [7 votes]: This is false; there is a counterexample (with $G$ a finite group of order $2$) in my notes on descent theory http://homepage.sns.it/vistoli/descent.pdf, subsection 4.4.2.<|endoftext|> -TITLE: Graphs with few induced subgraphs -QUESTION [18 upvotes]: Which graphs have the property that the number of $i$-vertex induced subgraphs is at most $i$ for some $i < n/2$ (where $n$ is the number of vertices)? -To avoid cases I am not interested in, I want the graph to be connected and non-bipartite and its complement to have the same properties. -Of course every graph has this property for $i=2$, and every triangle-free graph (or complement of one) for $i=3$. What about larger values of $i$? -Edit. -More background information on the problem can be found here. - -REPLY [4 votes]: Thanks, Doug. But, of course, I am not satisfied with this answer! I would like the graph to be vertex-transitive, and edge-transitive. I already know that its diameter is at most 4 (in my application). -I will come a bit cleaner about the application. This arises in a question of Joao Araujo about semigroup theory, which reduces to the following: Which groups are $(i,i+1)$-homogeneous but not $i$-homogeneous? (The condition $(i,i+1)$-homogeneous means that given any $i$-set $I$ and $j$-set $J$, there is an element of the group mapping $I$ to a subset of $J$. -Assume that $2i+1\le n$. Such a group is easily seen to be primitive and have at most three orbits on $2$-sets. If it is $2$-homogeneous, then it must be $2$-transitive, and then one has the possibility of using the classification of such groups. -It is easy to see that such a group has at most $i$ orbits on $i$-sets; so if it is not $2$-homogeneous, then all the symmetrised orbital graphs have the property of my original question. -By the way, five examples of groups with this property are known: the cyclic and dihedral groups of degree 5, and the affine groups $\mathrm{AGL}(1,7)$, $\mathrm{ASL}(2,3)$ and $\mathrm{AGL}(2,3)$ (with $n=7,9,9$ for the last three).<|endoftext|> -TITLE: Conjugacy for $p$-adic matrices of finite order -QUESTION [13 upvotes]: Say $p$ is an odd prime, and take two matrices $A,B\in GL_n({\mathbb Z}_p)$ of finite order $m$. Is it true that they are conjugate in $GL_n({\mathbb Z}_p)$ if and only if their reductions mod $p$ are conjugate in $GL_n({\mathbb F}_p)$? -Edit: Thank you all for the answers and a very insightful discussion! As the answer is NO, it is important for me to know whether conjugacy in $GL_n({\mathbb F}_p)$ at least implies conjugacy in $GL_n({\mathbb Q}_p)$. I cannot see this from the answers, so I think I will ask this as a separate question. - -REPLY [2 votes]: Here is a trial proof for the question over $Q_p$. -Write $J[f(x)^k]$ for the general Jordan form of a irreducible $f$, being $k$ identical blocks joined by 1's in general (minimal polynomial of block is $f^k$). -Let $A$ be finite order over $Q_p$, so $A\sim\oplus J[f(x)]$ where the $f$ have $f|\Phi_m$ (cyclotomic polynomials) and finiteness implies the $f$ are irreducible (not powers). -Note $\bar f$ determines $m$ up to $p$-powers, writing $m=up^v$ for $(u,p)=1$. Further note, if $\bar\Phi_u=\prod \bar g$ then $\bar\Phi_{up^v}=\prod\bar g^{\phi(p^v)}$, -and what is more, the corresponding Jordan block to $\bar g^{\phi(p^v)}$ does not split, -in other words this is the minimal polynomial. This follows since the reduction (mod $p$) of the companion matrix of $f$ is itself a companion matrix (ones above the diagonal) over a field $F_p$, and so has its minimal and characteristic polynomials equal to $\bar f=\bar g^{\phi(p^v)}$. -So, every reduction to $\bar f$ from the $A\sim\oplus J[f]$ decomposition has $\bar f(x)=\bar g(x)^{\phi(p^v)}$ for some irreducible $\bar g|\bar\Phi_u$, that lifts to $g|\Phi_u$. What is more, $\bar A\sim\oplus J[\bar g(x)^{\phi(p^v)}]$. -From this, $\bar A\sim\oplus J[\bar g(x)^{\phi(p^v)}]$ determines the general Jordan form of $A$ uniquely as something like $A\sim\oplus J[\Phi_{pu}^{g-part}(x^{p^{v-1}})]$. The general Jordan form classifies the conjugacy type over a field, as is $Q_p$. -Note that, $\Phi_3\Phi_6$ and $\Phi_6^2$ give 4x4 matrices with order 6, failing for $p=2$.<|endoftext|> -TITLE: Coderivations of S(V) correspond to linear maps S(V) -> V. Only over characteristic 0? -QUESTION [7 upvotes]: Definition. Let $k$ be a commutative ring. Let $V$ be a $k$-module. We turn the symmetric algebra $\mathrm{S}\left(V\right)$ of $V$ into a graded Hopf algebra by defining the comultiplication -$\Delta : \mathrm{S}\left(V\right) \to \mathrm{S}\left(V\right) \otimes \mathrm{S}\left(V\right)$ -by -$\Delta\left(v_1v_2...v_n\right) = \sum\limits_{i=0}^n \sum\limits_{\sigma\in\mathrm{Sh}\left(i,n-i\right)} \left(v_{\sigma\left(1\right)}v_{\sigma\left(2\right)}...v_{\sigma\left(i\right)}\right) \otimes \left(v_{\sigma\left(i+1\right)}v_{\sigma\left(i+2\right)}...v_{\sigma\left(n\right)}\right)$, -where $\mathrm{Sh}\left(i,n-i\right)$ denotes the set $\left\lbrace \sigma\in S_n \mid \sigma\left(1\right) < \sigma\left(2\right) < ... < \sigma\left(i\right) \text{ and }\sigma\left(i+1\right) < \sigma\left(i+2\right) < ... < \sigma\left(n\right) \right\rbrace$ of all $\left(i,n-i\right)$-shuffles. The counit of this Hopf algebra is simply the projection from $S\left(V\right)$ onto $k$. -Definition. Let $k$ be a commutative ring, and $C$ be a $k$-coalgebra. A coderivation of $C$ means a $k$-linear map $c:C\to C$ such that $\Delta \circ c = \left(c\otimes\mathrm{id} + \mathrm{id}\otimes c\right)\circ \Delta$. -Remark. Coderivations behave, in some sense, dually to derivations (which is not surprising since the condition $\Delta \circ c = \left(c\otimes\mathrm{id} + \mathrm{id}\otimes c\right)\circ \Delta$ is a kind of dual to the Leibniz identity, when the latter is written in pointfree notation): First of all, if $c:C\to C$ is a coderivation, then $c^{\ast} : C^{\ast}\to C^{\ast}$ is a derivation. The converse holds at least when $C$ is finite-dimensional. As an exercise in reversing arrows, the reader can prove that $\varepsilon\circ c=0$ for every coderivation $c$ of a coalgebra (in analogy to the equality $d\left(1\right)=0$ which holds for every derivation $d$ of an algebra). -Theorem. Let $k$ be a field of characteristic $0$. Let $V$ be a $k$-vector space. Then, the maps -$\mathrm{Hom}\left(S\left(V\right),V\right) \to \mathrm{Coder}\left(S\left(V\right)\right)$, -$X\mapsto \mu \circ \left(\mathrm{id}\otimes X\right) \circ \Delta$ (where $\mu$ is the multiplication morphism $S\left(V\right)\otimes S\left(V\right)\to S\left(V\right)$) -and -$\mathrm{Coder}\left(S\left(V\right)\right) \to \mathrm{Hom}\left(S\left(V\right),V\right)$, -$c\mapsto \pi_1\circ c$ (where $\pi_1$ is the projection from $\mathrm{S}\left(V\right)=\bigoplus\limits_{i\in\mathbb N}\mathrm{S}^i\left(V\right)$ onto the addend $\mathrm{S}^1\left(V\right)=V$) -are mutually inverse isomorphisms. -This is how I understand Chapter 5, Theorem 4.19 in Eckhard Meinrenken, Clifford algebras and Lie groups. (Unfortunately, the statement of the theorem in Meinrenken's text is obscured by the fact that one direction of the isomorphism - the easy one - is not written out explicitly.) The proof given in this text uses the characteristic-$0$ assumption: first, by assuming "WLOG" that a generic element of $\mathrm{S}\left(V\right)$ has the form $e^v$ for some $v\in V$ (this is made formal by going over to formal power series, but stripped of this formality, this is exactly what is known as umbral calculus for over a century), and second, by using the fact that the primitives of $\mathrm{S}\left(V\right)$ all come from $V$. -Question. Does the above theorem hold in arbitrary characteristic? -I am sure this is intimately related to the question whether $\mathrm{S}\left(V\right)$ is the cofree graded cocommutative coalgebra over $V$ (or at least whether it is "cogenerated" in degree $1$, whatever this means precisely!). Unfortunately, the only case when I am sure of this is the characteristic-$0$ case, so this is of no help to me. Loday-Valette does not seem to care for positive characteristic too much, either, and it is difficult for me to find any other source. - -REPLY [3 votes]: Let us assume $k$ has characteristic $p$. The problem is (to me at least) easier to understand by dualising (assume that we are only looking at homgeneous derivations so that we can take the graded dual and have no problems at least if $V$ is finite-dimensional, things will go wrong even here). Then the statement would say that any linear map $V\to V\subset S(V^\ast)^\ast$ has a unique extension to a derivation of $S(V^\ast)^\ast$. However, $S(V^\ast)^\ast$ is the divided power algebra on $V$ which is not generated as an algebra by its degree $1$ elements so the uniqueness doesn't follow as for the symmetric algebra. In fact for $V=kx$ we have that the divided power algebra is generated as a commutative algebra by $x_n:=\gamma_{p^n}(x)$ and relations $x_n^p=0$. Hence one can arbitrarily choose the value of a derivation on the $x_n$ which makes it very clear that the derivation is not determined by its value on $x_1$.<|endoftext|> -TITLE: What Dirichlet doesn't tell... -QUESTION [10 upvotes]: Let $n>1$ be an integer, and let us consider the set $P(n)$ of all prime numbers $p$ such that $p$ is not congruent to $1$ modulo $n$. Dirichlet's Density Theorem tells us that $P(n)$ has a natural density, equal to -$$1-\varphi(n)^{-1}$$ -where $\varphi(n) = |(\mathbb Z /n)^\ast|$ is Euler's totient. -From the Frobenian point of view, saying that $p$ is congruent to $1$ modulo $n$ is to say that the ideal $(p)$ splits completely in the cyclotomic field $\mathbb Q(\zeta_n)$. -From Chebotarev's point of view, saying that $p$ is congruent to $1$ modulo $n$ is to say that the Frobenius element over $p$ in $\operatorname{Gal}(\mathbb Q(\zeta_n)|\mathbb Q) \simeq (\mathbb Z /n)^\ast$ is the identity. -So far so good, now let us consider the set $P$ of all prime numbers $p$ which are not congruent to $1$ modulo $n^2$ for any $n>1$, that is -$$P := \bigcap_{n>1}P(n^2) = \bigcap_{\ell\mathrm{ prime}}P(\ell^2)$$ -Supposing that "the events $P(\ell^2)$ are uncorrelated" for different $\ell$'s, we can phantasise about the density of $P$, hoping it might be (at least up to a rational factor, I don't vouch for it) -$$\operatorname{dens}(P) = \prod_\ell 1-\frac{1}{\ell(\ell-1)} \quad = 0.37395581361920228805...$$ -a number called Artin's constant (it appears in Artins primitive root conjecture, which is similar in nature). The question whether $P$, or similarly constructed sets of primes, have a density and whether it is the expected one goes far beyond the density theorems of Dirichlet, Frobenius and Chebotarev. The corresponding Galois extension would be the maximal cyclotomic extension of $\mathbb Q$, which is ramified everywhere. - -Can you name this problem? Have you seen it before? Where? - -Hooley (1967) has shown that Artins primitive root conjecture follows from GRH. In principle, the problem of determining the density of $P$ should be simpler. - -Under GRH, is it true that the density of $P$ exists and is equal to Artin's constant? - -REPLY [5 votes]: Let me sketch a proof. If you fix a bound $z$, then the events $P(\ell^2)$ for different $\ell \leq z$ are uncorrelated; this is just a consequence of the prime number theorem for progressions and the multiplicativity of Euler's function. This reduces the problem to "understanding the tails"; in other words, we have to show that as $z\to\infty$, the relative upper density of the primes divisible by $\ell^2$ for some $\ell > z$ tends to zero. -Consider the primes $p \leq x$. By the Brun--Titchmarsh inequality, the number of such $p$ for which $p-1$ is divisible by $\ell^2$ for some prime $\ell$ with $z < \ell \leq x^{1/4}$ (say) is $$ \ll \sum_{z< \ell \leq x^{1/4}} \frac{x}{\phi(\ell^2)\log{(x/\ell^2)}} \ll \pi(x) \sum_{z > \ell} \frac{1}{\ell^2} \ll \frac{\pi(z)}{x}. $$ Also, the number of $p$'s with $p-1$ divisible by $\ell^2$ for some $\ell > x^{1/4}$ can be estimated trivially: We just count how many $n \leq x$ are divisible by some $\ell^2$ with $\ell > x^{1/4}$, which is clearly at most $\sum_{\ell > x^{1/4}} \lfloor x/\ell^2\rfloor \ll x^{3/4}$. This is negligible for us. Hence, the relative upper density in the previous paragraph is $\ll 1/z$. So it does indeed tend to zero as $z\to\infty$.<|endoftext|> -TITLE: Unconditional nonexistence for the heat equation with rapidly growing data? -QUESTION [46 upvotes]: Consider the initial value problem -$$ \partial_t u = \partial_{xx} u$$ -$$ u(0,x) = u_0(x)$$ -for the heat equation in one dimension, where $u_0: {\bf R} \to {\bf R}$ is a smooth initial datum and $u: [0,+\infty) \times {\bf R} \to {\bf R}$ is the smooth solution. Under reasonable growth conditions on $u_0$ and $u$ (e.g. $u_0$ and $u$ are at most polynomial growth), there is a unique solution $u$ to this problem given by the classical formula -$$ u(t,x) = \frac{1}{\sqrt{4\pi t}} \int_{\bf R} e^{-|x-y|^2/4t} u_0(y)\ dy.$$ -However, as is well known, once one allows $u_0$ or $u$ to grow sufficiently rapidly at infinity, then smooth solutions to the heat equation are no longer unique, as demonstrated first by Tychonoff in 1935. -My question then concerns the corresponding question for existence : does there exist smooth initial data $u_0$ for which there are no global smooth solutions $u$ to the initial value problem for the heat equation? -One obvious candidate for such "bad" data would be a backwards heat kernel, such as -$$ u_0(x) = e^{|x|^2/4}.$$ -One can verify that -$$ u(t,x) = \frac{1}{(1-t)^{1/2}} e^{|x|^2/4(1-t)}$$ -is a smooth solution to the initial value problem for the heat equation with initial datum $u_0$ up to time $t=1$, at which point it blows up (rather dramatically). However, this does not fully solve the problem due to the aforementioned lack of uniqueness; just because this particular solution $u$ blows up, there could be some other more exotic smooth solution with the same data which somehow manages to retain its smoothness beyond the time $t=1$. This seems highly unlikely to me, but I was not able to demonstrate such an "unconditional non-existence" result - the absence of any growth hypotheses at infinity seems to destroy most methods of controlling solutions, and could potentially create some strange scenario in which one could continually keep singularities from forming by pumping in infinite quantities of energy from spatial infinity in just the right manner. But perhaps there is some literature on this problem? - -REPLY [52 votes]: It is true that for any initial datum $u_0\in C^\infty(\mathbb{R})$ there exists a solution $u\in C^\infty(\mathbb{R}^+\times\mathbb{R})$ to the heat equation with initial condition $u(0,x)=u_0(x)$. As you point out, this will not be unique. -I can give a method of constructing such solutions now. The idea is to show that we can write $u=\sum_{n=1}^\infty f_n$ where $f_n$ are carefully constructed solutions chosen such that the partial sums $\sum_{n=0}^mf_n(0,x)$ eventually agree with $u_0(x)$ any bounded subset of the reals, and for which $f_n$ tends to zero arbitrarily quickly in the compact-open topology. First a bit of notation. I use $\mathbb{R}^+=[0,\infty)$ for the nonnegative reals. For a space $X$ then $C_0(X)$, $C^\infty(X)$, $C^\infty_0(X)$, and $C^\infty_K(X)$, represent the continuous real-valued functions on $X$ which are respectively vanishing at infinity, smooth, smooth and vanishing at infinity, and smooth with compact support. Let $(P_t)_{t\geq 0}$ be the kernels -$$ -\begin{align} -&P_t\colon C_0(\mathbb{R})\to C_0(\mathbb{R}),\\\\ -&P_tu(x)=\frac{1}{\sqrt{4\pi t}}\int_\mathbb{R}e^{-(x-y)^2/4t}u(y)\,dy -\end{align} -$$ -for $t > 0$, and $P_0u=u$. This is the Markov transition function for Brownian motion (more precisely, for standard Brownian motion scaled by $\sqrt{2}$, because of the normalization used here). For $u\in C_K^\infty(\mathbb{R})$, then $f(t,x)=P_tu(x)$ is in $C_0^\infty(\mathbb{R}^+\times\mathbb{R})$, and is a solution to the heat equation with initial condition $f(0,x)=u(x)$, agreeing with the classical solution stated in the question. I'll also consider initial conditions $u\in C^\infty_K((a,\infty))$ (for $a\in\mathbb{R}$) by setting $u(x)\equiv0$ for all $x\le a$. The first step in the construction is to find initial conditions supported in $(a,\infty)$ so that $P_tu(0)$ approximates any given continuous function of time that we like. - -1) For any $a > 0$ and $h\in C_0((0,\infty])$, there exists a sequence $u_1,u_2,\ldots\in C^\infty_K((a,\infty))$ such that $\sqrt{t}P_tu_n(0)$ converges uniformly to $h(t)$ (over $t > 0$) as $n\to\infty$. - -Consider the closure, $V$, in $C_0((0,\infty])$ (under the uniform norm) of the space of functions $t\mapsto\sqrt{4\pi t}P_tu(0)$ for $u\in C^\infty_K((a,\infty))$. Note that the limit as $t\to\infty$ does always exist and is just the integral of $u$. Then $V$ is a closed linear subspace of $C_0((0,\infty])$. Consider a sequence $u_n\in C_0((a,\infty))$ tending to the delta function $\delta_b$ at a point $b > a$, in the sense that $u_n$ all have support in the same compact set, and converge in distribution to $\delta_b$. Then, from the expression defining $P_t$, $\sqrt{4\pi t}P_tu_n(0)$ converges uniformly over $t > 0$ to $\exp(-b^2/4t)$. So, the function $t\mapsto\exp(-b^2/4t)$ is in $V$. As the set of functions of the form $t\mapsto\exp(-b^2/4t)$ for $b > a$ is closed under multiplication and separates points, the locally compact version of the Stone-Weierstrass theorem says that $V=C_0((0,\infty])$. Then, (1) follows. - -2) For any $u\in C^\infty_K((0,\infty))$ and $a,T > 0$, there exists a sequence $u_1,u_2,\ldots\in C^\infty_0((a,\infty))$ such that $f_n(t,x)\equiv P_t(u+u_n)(x)$ converges uniformly to zero (along with all its partial derivatives to all orders) over $t\in[0,T]$ and $x\le0$. - -Choosing $0 < \epsilon < a$ so that the support of $u$ is contained in $(\epsilon,\infty)$, (1) implies that we can choose $u_n\in C^\infty_K((a,\infty)$ so that $\sqrt{t}P_tu_n(\epsilon)$ converges uniformly to $-\sqrt{t}P_tu(\epsilon)$ over $t\ge0$ as $n\to\infty$. Then, $f_n(t,x)\equiv P_t(u+u_n)(x)$ is a bounded solution to the heat equation with boundary conditions $f_n(0,x)=0$ for $x\le\epsilon$ and $f_n(t,\epsilon)=P_t(u+u_n)(\epsilon)$. It is then standard that the solution is given by an integral over the boundary, -$$ -f_n(t,x)=\int_0^t\frac{\epsilon-x}{\sqrt{4\pi (t-s)^3}}e^{-(\epsilon-x)^2/4(t-s)}\sqrt{s}P_s(u_n+u)(\epsilon)\frac{ds}{\sqrt{s}} -$$ -for $x\le0$. Differentiating this wrt $x$ and $t$ an arbitrary number of times, and using dominated convergence as $n\to\infty$, it follows that $f_n(t,x)$ (and all its partial derivatives) converge uniformly to zero over $x\le0$ and $t\in[0,T]$ as $n\to\infty$. - -3) Suppose that $T > 0$, $0 < a < b$ and $u\in C^\infty_K(\mathbb{R})$ has support contained in $(-\infty,-a)\cup(a,\infty)$. Then, there exists a sequence $u_n\in C^\infty_K(\mathbb{R})$ with supports in $(-\infty,-b)\cup(b,\infty)$ such that $P_t(u+u_n)(x)$ (and all its partial derivatives) tends to 0 uniformly over $\vert x\vert\le a$ and $t\in[0,T]$. - -Set $u^+(x)=1_{\{x > 0\}}u(x)$ and $u^-(x)=1_{\{x < 0\}}u(x)$. Applying (2) to $u^+(x+a)$, there exists a sequence $u^+_n\in C^\infty_K(\mathbb{R})$ with supports in $(b,\infty)$ such that $P_t(u^++u^+_n)(x)$ tends to 0 uniformly over $x\le a$ and $t\in[0,T]$. Applying the same argument to $u^-(-x-a)$, there exists $u^-_n\in C^\infty_K(\mathbb{R})$ with support in $(-\infty,-b)$ such that $P_t(u^-+u^-_n)(x)$ tends to zero uniformly over $x\ge -a$ and $t\in[0,T]$. A sequence satisfying the statement of (3) is $u_n=u^+_n+u^-_n$. - -4) If $u\in C^\infty(\mathbb{R})$ then there exists $f\in C^\infty(\mathbb{R}^+\times\mathbb{R})$ solving the heat equation with initial condition $f(0,x)=u(x)$. - -We can inductively choose a sequence $u_n\in C^\infty_K(\mathbb{R})$ such that $\sum_{m=1}^nu_m(x)=u(x)$ for $\vert x\vert\le n+1$ and $n\ge1$. Let $u_1=u$ on $[-2,2]$ and then, for each $n\ge2$, apply the following step. - -As $\tilde u=u-\sum_{m=1}^{n-1}u_m$ is zero on $[-n,n]$, it has support in $(-\infty,-a)\cup(a,\infty)$ for $a=n-1/2$. Choosing $b=n+1$, by (3), there exists $v\in C^\infty_K(\mathbb{R})$ with support in $(-\infty,-b)\cup(b,\infty)$ such that $P_t(\tilde u +v)(x)$ along with all its partial derivatives up to order $n$ are bounded by $2^{-n}$ over $\vert x\vert\le n-1/2$ and $t\le n$. Take $u_n=\tilde u + v$. - -Setting $f_n(t,x)=P_tu_n(x)$, then $f_n$ are smooth functions satisfying the heat equation and the initial conditions $\sum_{m=1}^nf_m(0,x)=u(x)$ for $\vert x\vert\le n+1$. Also, by the choice of $u_n$, for $n > 1$ then $f_n$ together with all its derivatives up to order $n$ is bounded by $2^{-n}$ on $[0,n]\times[1-n,n-1]$. As $\sum_n2^{-n} < \infty$, the limit $f=\sum_nf_n$ exists and is smooth with all partial derivatives commuting with the summation. Then $f$ has the required properties.<|endoftext|> -TITLE: The generalized word problem vs. the uniform generalized word problem -QUESTION [12 upvotes]: Before asking my question, let me define my terms. -Let $G$ be a finitely generated group with fixed generating set $A$. Let $\psi\colon (A\cup A^{-1})^{\ast}\to G$ be the canonical projection where $X^*$ denotes the free monoid on a set $X$. - -Definition 1. A subgroup $H$ of $G$ has a decidable membership problem if there is a Turing machine which on input a word $w\in (A\cup A^{-1})^{\ast}$ outputs "Yes", if $\psi(w)\in H$, and "No", otherwise. - -The generalized word problem is said to be decidable for $G$ if each finitely generated subgroup of $G$ has decidable membership problem. - -Definition 2. $G$ has decidable uniform generalized word problem if there is a Turing machine which on input words $w_1,\ldots, w_k,w\in (A\cup A^{-1})^{\ast}$ outputs "Yes", if $\psi(w)\in \langle \psi(w_1),\ldots,\psi(w_k)\rangle$, and "No", otherwise. - -One easily verifies that decidability of these problems is independent of the chosen finite generating set. -Trivially decidablity of the uniform generalized word problem implies decidability of the generalized word problem. The difference between the two problems is that the uniform generalized word problem asks that there is a computable map that takes a finite subset of $G$ to a Turing machine which decides membership in the subgroup it generates. -My question is: - -Does anybody know an example of a group with decidable generalized word problem but undecidable uniform generalized word problem? - -In case no example is known, here is a proposed attack that I am not capable of implementing. A finitely generated group is a Tarski monster if it is infinite but all its proper subgroups are finite. Olshanskii has constructed such "beasts". - -Observation 1. A Tarski monster $G$ has decidable generalized word problem iff it has decidable word problem. - -For the non-trivial direction, the Turing machine which always says "Yes" solves the membership problem in $G$. If $H$ is a proper subgroup, there is a Turing machine who knows a finite list of words representing all the elements of $H$ and so it can take an input word and use the word problem for $G$ to decide if the input word belongs to $H$. - -Observation 2. A Tarski monster $G$ with decidable word problem has decidable uniform generalized word problem iff one can decide given elements $w_1,\ldots, w_k\in (A\cup A^{-1})^{\ast}$ whether $G=\langle \psi(w_1),\ldots, \psi(w_k)\rangle$. - -Indeed, if the uniform generalized word problem is decidable, then one can check whether each letter of the generating set $A$ of $G$ belongs to $\langle \psi(w_1),\ldots, \psi(w_k)\rangle$. -For the converse, given words $w_1,\ldots, w_k,w\in (A\cup A^{-1})^{\ast}$ one first checks whether $\psi(w_1),\ldots,\psi(w_k)$ generates $G$. If so, the Turing machine outputs "Yes". Otherwise, $\psi(w_1),\ldots,\psi(w_k)$ generates a finite subgroup. Using decidability of the word problem, the Turing machine can compute the finite subgroup $\langle \psi(w_1),\ldots, \psi(w_k)\rangle$ (by taking products until nothing new can be obtained) and then use the word problem again to determine if $\psi(w)$ belongs to this subgroup. - -Is there a Tarski monster with decidable word problem, but for which on cannot decide whether a given finite set of elements generate it? - -Update - -Derek has shown that Tarski monsters will always have decidable uniform generalized word problem as soon as they have decidable word problem because the word problem already gives that one can decide if a finitely generated subgroup is the whole group. I'm a bit embarrassed I didn't see that. My original question still seems open though. - -REPLY [4 votes]: The answer is almost certainly not known (not published), but here is a way to construct a possible example. Take a non-recursive r.e. set of numbers $I$ which is given as the set of positive values of a polynomial $p(x_1,...,x_n)$ with integer coefficients. We can assume by a theorem of Matiyasevich that in addition $p$ takes each of its values only finite number of times (see Matiyasevich's book on the 10th Hilbert problem). Consider the infinitely generated Abelian group $A$ generated by $\{a_1,a_2,..., b_1, b_2...\}$ given by the commutativity relators and relators of the form $a_i=b_j^{i}$ provided $p(gd(i))=j$ where $gd(i)$ is the $n$-tuple of natural numbers with the Goedel number $i$. Note that for every f.g. subgroup $H$ of $A$, the membership problem in $H$ is decidable, but the uniform membership problem is not decidable. The first part of this statement follows from the finiteness assumption about $p$: -Indeed, consider any finitely generated subgroup $H$ generated by $w_1,...,w_k$ and an element $g$. For $g$ to belong to $H$, its letters should belong to a finite set (because of the finiteness condition). So the membership question becomes a question about a finitely generated Abelian group which is LERF. -The second part follows from the fact that $I$ is not recursive. Indeed, the question $a_i\in \langle b_j\rangle$ is undecidable. -Now embed $A$ into a finitely generated group using small cancelation, namely, take the free group $F_2$ and an infinitely many words $u_1,u_2,..., v_1, v_2,...$ in $F_2$ satisfying, say, $C'(1/12)$, then consider the presentation obtained by replacing each $a_i$ in the presentation of $A$ by $u_i$ and each $b_i$ by $v_i$. The resulting group will have undecidable uniform membership problem, but I believe (it needs to be checked!), that the membership problem in every finitely generated subgroup will be decidable. I do not have time to check it myself now, and it may be non-trivial, but intuition tells me that it is the case.<|endoftext|> -TITLE: Algebraic geometric measure theory -QUESTION [7 upvotes]: Suppose I have $V\subset \mathbb{C}^n$ be the zero set of a polynomial $P(z_1, \dotsc, z_n),$ with bounded height of coefficients (where height is, to fix something, $|\log|a||$) and degree $d.$ Suppose I now have a ball $B=B(z_0, r) \subseteq \mathbb{C}^n.$ Is there an upper bound on $2n-2$ dimensional measure of $B\cap V?$ -EDIT A quasi-answer: Wirtinger's formula (see Griffiths and Harris, p. 31) seems to indicate that the Fubini-Study volume of a $k$-dimensional sub variety $V$ of $\mathbb{P}^n$ equals $\deg(V) \mathrm{vol}(\mathbb{P}^k).$ For real algebraic varieties, there seems to be only a Cauchy-Crofton derived inequality, as suggested in the answer. - -REPLY [3 votes]: There is an explicit upper bound based on a 2-d version of the Crofton formula. Namely, the area of $B \cap V$ is the integral of the number of points of intersection $W \cap (B \cap V)$ over the space of all affine 2-planes $W \subseteq \mathbb{R}^{2n}$. Since the real algebraic variety $V$ has degree $\leq d^2$ the number of points of intersection is at most $d^2$. So an upper bound is $d^2$ times the measure of the space of affine $2$-planes meeting $B$. It seems to me that, unless I have misunderstood, the bound on the coefficients is unnecssary.<|endoftext|> -TITLE: A missing paper by Auslander? -QUESTION [10 upvotes]: I was reading Auslander's talk at the 1962 ICM (beginning of Section 2 on this page). At the end, the reference began: - -[1] M. Auslander, Modules over unramified regular local rings, Illinois. J. Math. 5 (1961), pp. 631–647. -[2] M. Auslander, Modules over unramified regular local rings II, Illinois. J. Math. (To appear). - -The trouble is, I can not find [2] on MathSciNet or Google. I also looked at his Selected works without luck. Of course, there are many papers that are forthcoming but never materialized, but this one even has the journal name attached to it! So - -What happened to Auslander paper [2] above? - -Note that [1] is an influential paper (still being quoted very recently). I enjoyed it a lot, and would really like to read [2]. -UPDATE: both Professor Buchsbaum and Professor Reiten have graciously replied to my query about this missing paper. Unfortunately, neither of them know what happened. - -REPLY [2 votes]: His "Selected Works" (AMS: http://www.ams.org/bookstore-getitem/item=CWORKS-10) lists on Chapter II, two articles with the same title "Modules over unramified regular local rings"<|endoftext|> -TITLE: Decomposition of linear partial differential operators -QUESTION [7 upvotes]: I was wondering about the following: -Let $M$ be a smooth, second-countable (possibly noncompact) manifold and let $E$ and $F$ be smooth vector bundles over $M$. - -Can every smooth linear partial differential operator $P$ from $E$ to $F$ of finite order be written as $\sum_{i=0}^n (T_i)_* ∘ P_i$ for certain smooth linear partial differential operators $P_0, \dots, P_n$ from $E$ to $E$ and certain vector bundle homomorphisms $T_0, \dots, T_n$ from $E$ to $F$? -Can every smooth linear partial differential operator $P$ from $E$ to $E$ of finite order be written as a finite sum of compositions of smooth linear partial differential operators from $E$ to $E$ of order at most 1? - -Of course, locally (i.e., on a chart domain over which the vector bundles trivialize) the answer to both questions is yes, but this does not seem to be of much help when trying to answer the questions globally. - -REPLY [3 votes]: If the (say, $d$-dimensional) base manifold $M$ is parallelizable (i.e. $TM\to M$ is trivial), then the answer to both questions is yes even globally, provided we choose a (say, torsion-free) covariant derivative $\nabla^M$ on $TM$ and a covariant derivative $\nabla^E$ on $E\to M$. We may then combine these connections by means of the Leibniz rule to define a covariant derivative on $\otimes^k T^*M\otimes E\to M$ for all $k$. Let us denote this common extension of $\nabla^M$ and $\nabla^E$ simply by $\nabla$. One may then define iterated covariant derivatives $\nabla^k$ of any order $k\geq 0$ on $E\to M$ recursively as follows: $$\nabla^0\phi=\phi\ ,\,\nabla^1\phi=\nabla^E\phi\ ,\,\nabla^{k+1}\phi=\nabla(\nabla^k\phi)\ ,\quad\phi\in\Gamma(M,E)\ ,$$ so that $\nabla^k\phi\in\Gamma(M,\otimes^kT^*M\otimes E)$ for every $k\geq 0$. Let us denote the contraction of $\nabla^k\phi$ with $k$ smooth vector fields $Y_1,\ldots,Y_k\in\mathfrak{X}(M)$ by $\nabla^k_{Y_1,\ldots,Y_k}\phi\in\Gamma(M,E)$. The key fact we need (apparently due to R. Palais, but I am not sure), holding even when $M$ is not parallelizable, is: - -Any linear partial differential operator of order $r$ $P:\Gamma(M,J^rE)\to\Gamma(M,F)$ may be written uniquely as $$\tag{1}\label{e1}P\phi=\sum^r_{k=0}A_k\nabla^k\phi\ ,\quad\phi\in\Gamma(M,E)\ ,$$ where $A_k\in\Gamma(M,S^k TM\otimes E^*\otimes F)$ for each $k=0,1,\ldots,r$, $S^kTM\to M$ is the symmetric, rank-$k$ contravariant tensor sub-bundle of $\otimes^kTM\to M$ and $E^*\to M$ is the dual bundle to $E\to M$. Particularly, one may replace $\nabla^k$ in \eqref{e1} by its symmetrized part. - -The coefficients $A_k$ of the representation \eqref{e1} of $P$ may of course be seen as vector bundle morphisms from $\otimes^kT^*M\otimes E\to M$ to $F\to M$ covering $\mathrm{id}_M$. This is "almost" what you want, what gets in the way is precisely the potential non-triviality of $TM\to M$ - that is where the hypothesis of parallelizability of $M$ enters. -Recall now that the latter amounts to being able to choose $d$ smooth vector fields $X_1,\ldots,X_d\in\mathfrak{X}(M)$ such that $\{X_1(p),\ldots,X_d(p)\}$ is a basis of $T_pM$ for every $p\in M$ - that is, a (global) linear frame in $TM$. Let $\omega^1,\ldots,\omega^d\in\Omega^1(M)$ be the corresponding linear coframe (i.e. $\omega^i(X_j)=\delta^i_j$) - from \eqref{e1} one may write $$\tag{2}\label{e2}P\phi=\sum^r_{k=0}\sum^n_{j_1,\ldots,j_k=1}A_k(\omega^{j_1}\otimes\cdots\otimes\omega^{j_k})\nabla^k_{X_{j_1},\ldots,X_{j_k}}\phi\ ,\quad\phi\in\Gamma(M,E)\ ,$$ where $A_k(\omega^{j_1}\otimes\cdots\otimes\omega^{j_k})\in\Gamma(M,E^*\otimes F)$ is the contraction of $A_k$ with $\omega^{j_1}\otimes\cdots\otimes\omega^{j_k}\in\Gamma(M,\otimes^kT^*M)$. Formula \eqref{e2} is precisely the affirmative answer to your first question. From there to answering affirmatively your second question is just a(n essentially combinatorial) matter of rewriting $\nabla^k_{X_{j_1},\ldots,X_{j_k}}\phi$ in \eqref{e2} in terms of $\nabla^E_{X_{j_1}}\cdots\nabla^E_{X_{j_{m+1}}}\phi$ and $\nabla^M_{X_{l_1}}\cdots\nabla^M_{X_{l_m}}X_i$ for $i,j_1,\ldots,j_{m+1},l_1,\ldots,l_m=1,\ldots,d$, $m=1,\ldots,k-1$ using Leibniz's rule (the first-order operators you want are, of course, $\nabla^E_{X_j}$ with $j=1,\ldots,d$). -For most applications to global analysis, however, \eqref{e1} suffices, with the advantage that it holds in complete generality. There is, of course, the possibility of using a partition of unity subordinated to a finite atlas (an atlas with $d+1$ charts always exists whenever $M$ is second countable by topological dimension theory, see e.g. the Corollary to Theorem 1.2.I, pp. 20-21 of the book of W. Greub, S. Halperin and R. Vanstone, Connections, Curvature, and Cohomology, Volume I (Academic Press, 1972)) and write $P$ in each chart using coordinates, as suggested by you in your question and Dmitri Pavlov in his comment above, but another advantage of \eqref{e1} and \eqref{e2} is that they are both coordinate-free representations of $P$. -There is a middle course to circumvent the hypothesis of parallelizability which relies on the following remark: if $U\subset M$ is the (open) domain of a chart $x:U\to\mathbb{R}^d$, then $TU$ is trivial (just pick the coordinate vector fields $X_j(p)=(T_px)^{-1}\frac{\partial}{\partial x^j}$, $j=1,\ldots,d$). Let now $\{(U_\alpha,x_\alpha)\ |\ \alpha=1,\ldots,d+1\}$ be a finite atlas of $M$ as in the previous paragraph, $\{f_\alpha\ |\ \alpha=1,\ldots,d+1\}$ a partition of unity subordinate to the open cover $\{U_\alpha\ |\ \alpha=1,\ldots,d+1\}$ of $M$ and $X_{1,\alpha},\ldots,X_{d,\alpha}$ be the linear frame on $U_\alpha$ obtained from $x_\alpha$ as above for each $\alpha$, with corresponding linear coframe $\omega^{1,\alpha},\ldots,\omega^{d,\alpha}$. We can then write $$P\phi=\sum^{d+1}_{\alpha=1}\sum^r_{k=0}\sum^n_{j_1,\ldots,j_k=1}A_k(\omega^{j_1,\alpha}\otimes\cdots\otimes\omega^{j_k,\alpha})\nabla^k_{X_{j_1,\alpha},\ldots,X_{j_k,\alpha}}(f_\alpha\phi)\ ,$$ recalling that $$\nabla^k\phi=\sum^{d+1}_{\alpha=1}\nabla^k(f_\alpha\phi)$$ and, by Leibniz's rule, $$\nabla^k_{Y_1,\ldots,Y_k}(f_\alpha\phi)=\sum_{I\subset\{1,\ldots,k\}}\nabla^{M|I|}_{Y_I}f_\alpha\nabla^{n-|I|}_{Y_{I^c}}\phi\ ,\quad Y_I=(Y_j)_{j\in I}\ .$$ This leads us to the following, not-so-appetizing formula $$\tag{3}\label{e3}P\phi=\sum^{d+1}_{\alpha,\beta=1}\sum^r_{k=0}\sum^n_{j_1,\ldots,j_k=1}\sum_{I\subset\{1,\ldots,k\}}(\nabla^{M|I|}_{X_{j_I,\alpha}}f_\alpha)A_k(\omega^{j_1,\alpha}\otimes\cdots\otimes\omega^{j_k,\alpha})(f_\beta\nabla^{n-|I|}_{X_{j_{I^c},\alpha}}\phi)\ ,$$ -where $Y_{j_I,\alpha}=(Y_{j_l,\alpha})_{l\in I}$. Since $\mathrm{supp}\nabla^kf_\alpha\subset\mathrm{supp} f_\alpha$ for all $k,\alpha$, formula \eqref{e3} is globally well defined and yields a positive answer to your first question even in the absence of parallelizability. To answer positively your second question, one proceeds as in the parallelizable case with each term $\nabla^{n-|I|}_{Y_{j_{I^c}}}\phi$ in \eqref{e3}. At this point, yet another advantage of working with \eqref{e1} instead should become clear - namely, economy and simplicity.<|endoftext|> -TITLE: Why are there so many smooth functions? -QUESTION [46 upvotes]: I do understand that my question might seem a little bit ignorant, but I thought about it a lot and still can't wrap my head around it. -Analycity imposes very strong conditions on a map, from elementary ones like "locally zero implies globally zero", to a little bit more deep like the Hurwitz formula (in the complex case). Neither of above are true if we just assume smoothness. -On one hand, it is quite easy to prove that smooth functions are dense in any "reasonable" function space (I guess it depends on what one considers reasonable, though…) - just convolve with smooth approximations of identity. Also, (although I do take it on faith), any map of two manifolds is homotopic to a smooth one and two homotopic smooth maps are actually smooth-homotopic. -Because of above facts, it seems to me that smooth functions are "abundant" and are actually very close to topology, ie. mere continuity. -On the other hand, when I recall basic calculus course, it always seemed like being differentiable even once is a "miracle", and being differentiable infinitely many times is a very, very strong condition, even more so in several variables. -Why are objects so constrained, ie. smooth functions, so useful and also, so malleable? - -REPLY [6 votes]: On the other hand, in the Banach space $C[0,1]$ with the supremum norm, the set of functions which do not have (finite) derivative at any of the points of $[0,1]$ is residual. This was first proved by Stefan Banach and Stefan Mazurkiewicz in 1930s. A proof following J. C. Oxtoby can be found here (if you read Spanish): -http://www.ciens.ula.ve/matematica/publicaciones/libros/por_profesor/wilman_brito/Teorema_de_Baire_Aplicaciones.pdf -For an even more refined result (a combination of several ones), look up Banach-Mazurkiewicz-Jarnik theorem (Theorem 7.2.1) in the following book: -Marek Jarnicki, Peter Pflug: Continuous nowhere differentiable functions. Monsters of analysis. Springer Monographs in Mathematics, 2015.<|endoftext|> -TITLE: Erdos-Szekeres in high dimensions -QUESTION [8 upvotes]: All the point sets in this post are in general position. A set of points in $R^d$ is in general position if every $k+1$ points are affinely independent for $k \le d$. If the set contains at least $d+1$ points it is enough to demand that every $d+1$ points in the set affinely span $R^d$. -In the plane a set of three or more points is in general position if NO TRIPLE OF POINTS IS COLLINEAR. -We say that a set of points is in convex position if they all lie on he boundary of their convex hull. e.g. in the plane they are the vertices of a convex $n$-gon. -We say that a set of points is in cyclic position if their convex hull is a cyclic polytope and they all lie on the boundary of their convex hull. -In the paper that Erdos and Szekeres invented Ramsey theory; they proved that any set of $ES(n)$ points in the plane contains a subset of n in convex position. They gave decent bounds on $ES(n)$. Their lower bound is conjectured to be the truth, $ES(n)=2^{n-2}+1$. -There are many extensions and generalizations (yet the upper bound after 75 years is $\frac{1}{2}$ their original upper bound, no asymptotic improvement in all these years). -There are two higher dimensional versions of this result: There are functions $ES_d(n)$ and $C_d(n)$ such that among any set of $ES_d(n)$ points in $R^d$ there is a subset of n in convex position, and among any $C_d(n)$ points there are n in cyclic position. -$ES_k(n)\leq ES_d(n)$ whenever $k\geq d$ is easy to see, and there is some interesting lower bounds on $ES_d(n)$. -Here is the question: are there decent bounds on C_d(n)? -Where the function C_d(n) is the smallest number such that "among any C_d(n) points in general position there are n in cyclic position". And being in cyclic position means having the combinatorial type of the cyclic polytope. -I'm more interested in the upper bound, anything that does not use Ramsey theorem is of interest. References anyone? - -REPLY [3 votes]: Andrew Suk gives new bounds for general d and a pretty good bound for d=3 -http://arxiv.org/abs/1305.5934<|endoftext|> -TITLE: Ideal classes and integral similarity -QUESTION [12 upvotes]: Matrices $A$ and $B$ are integrally equivalent if there is an invertible integer matrix $L$ and $L^{-1}AL=B$. Suppose $f(t)$ is an integer polynomial with no repeated factors. Latimer and MacDuffee proved that the number of integral similarity classes of matrices with characteristic polynomial equal to $f(t)$ is equal to the number of non-singular ideal classes of $\mathbb{Z}[\theta]$. (And it's not clear to me what a non-singular ideal is, or was in 1933.) -If for each irreducible factor of $f$ the corresponding number field has class number 1, and if the ring of algebraic integers in it is equal to $\mathbb{Z}[\theta]$, then it follows that two matrices with characteristic polynomial $f$ are integrally equivalent. (This provides another way to verify Tracy Hall's computation as reported in example, which is what started me down this rabbit hole.) But the above two assumptions on $K$ are strong, and I am trying to find out what is known when these conditions are weakened. Hence: -Questions - -Can someone point me to a reference (or more) concerning ideal classes in $\mathbb{Z}[\theta]$ when this order is not maximal? Even just a proof of the fact that the number of ideal classes in $\mathbb{Z}[\theta]$ is finite? [I am assured that this is a fact, and it appears to follow from the usual proof for Dedekind domains; I am hoping that any source that treats this explicitly will offer further information.] -Is there any characterization of the non-invertible ideals in $\mathbb{Z}[\theta]$? [I am aware of results in Harvey Cohn's "A Classical Invitation..." about ideals coprime to the conductor.] -Will the theory simplify if I assume that $\theta$ is totally real? - -Remark -I am interested in cospectral graphs - non-isomorphic graphs whose adjacency matrices are similar. Experimental evidence suggests that almost all graphs have irreducible characteristic polynomials. Haemers has conjectured that the proportion of graphs on $n$ vertices that are determined by their characteristic polynomials goes to 0 as $n\to\infty$. My suspicion is that pairs of cospectral graphs are not normally integrally equivalent. I am hoping that if I learn more number theory, I might be able to confirm -this. - -REPLY [9 votes]: A very belated answer to 1), but: I just saw that this is treated very nicely in Curtis and Reiner's Representation Theory of Finite Groups and Associative Algebras. Theorem 20.6 therein not only works with non-maximal orders $\mathbb{Z}[\theta]$ but even does the non-commutative case as well. And the argument is very clean and simple. -2) is a nice question by the way, and one that I have wondered about over the years. As Franz says, you need to be careful about what you mean by the "ideal class group" in this case. For any domain $R$ the quotient of the monoid of the nonzero ideals of $R$ by the submonoid of nonzero principal ideals gives you a monoid, say $H(R)$: let's call it the ideal class monoid. This is the monoid that your question 1) asks for a proof of the finiteness of when $R$ is an order in a number field $K$. (Well, your question asks about the monogenic case, but the proof of course doesn't need that.) It is known that $H(R)$ is a group iff $R$ is a Dedekind domain, so in the arithmetic case iff $R$ is integrally closed in its fraction field. In every other case there are non-invertible ideal classes, so to get a group you should pass to the group of units of the monoid $H(R)$: by definition this group is the Picard group of $R$ (it is also the group of isomorphism classes of rank one projective $R$-modules under tensor product, or if you like of isomorphism classes of line bundles on the scheme $\operatorname{Spec} R$). If e.g. $R$ is Noetherian, integrally closed, there is also a divisor class group $\operatorname{Cl}(R)$ whose nonvanishing is the obstruction to $R$ being a UFD. There is a canonical injection $\operatorname{Pic}(R) \hookrightarrow \operatorname{Cl}(R)$ which need not be an isomorphism: the most famous example is probably $\mathbb{C}[x,y,z]/(xy-z^2)$ which has two-element divisor class group but trivial Picard group. The map is an isomorphism if $R$ is "locally factorial", so e.g. if $R$ is nonsingular, hence always in dimension one. -The one example I know of $H(R)$ for a nonmaximal order is $R = \mathbb{Z}[\sqrt{-3}]$, where the monoid $H(R)$ has order $2$. There is up to isomorphism exactly one order $2$ commutative monoid which is not a group: the nonidentity element $\bullet$ must be "absorbing": $x \bullet = \bullet$ for all $x$. In this case a representative for the absorbing element is the prime ideal $\mathfrak{p} = \langle 2, 1+ \sqrt{-3} \rangle$. This is the unique non-principal prime ideal in $R$ (something which cannot happen in a Dedekind domain!). I would be interested to know if anything whatsoever is known or conjectured about which finite non-cancellative monoids arise as $H(R)$ as $R$ ranges over orders in algebraic number fields. -If $R$ is a nonmaximal order in a number field $K$ with maximal order $\tilde{R}$, there is a nonzero ideal $\mathfrak{f}$, the conductor, which is the set of all $x \in \tilde{R}$ such that $\tilde{R} x \in R$. This is an ideal in both $\tilde{R}$ and $R$, and in fact it is the -largest ideal of $\tilde{R}$ which is also an ideal of $R$. The primes $\mathfrak{p}$ of $R$ containing $\mathfrak{f}$ are precisely the singular points on $\operatorname{Spec} R$. Every ideal prime to $\mathfrak{f}$ is invertible and factors uniquely into primes, and every invertible ideal can be adjusted in its equivalence class to be prime to $\mathfrak{f}$. Thus the noninvertible ideals are the ones with support which cannot be "moved off of $\mathfrak{f}$". This material can be found in Neukirch's Algebraic Number Theory.<|endoftext|> -TITLE: Plagiarism in the community -QUESTION [16 upvotes]: I am a graduate student thinking about some stuff. -1) If I have an idea for a result but it is not yet complete, is it advisable to mention it to others (especially more experienced people)? Is it possible that he/she will take it, develop it quicker than I do and publish it? How should one behave in these situations? -2) Is there a rule of thumb about when (and how) to mention your ideas to others in the community? -3) How often does it happen that a student's idea is hijacked by a more experienced mathematician? - -REPLY [6 votes]: To be honest, I don't like the question very much, but I will answer it by -plagiarizing, I mean quoting, Hailong's succinct answer: - -"There is a nonzero risk. - But the rewards ... far outweigh the risk". - -To be more explicit, talking to an expert might save you a great deal of effort and -pain, as in "Look at this paper", or "Try this", or more importantly "Here is a counterexample..." It is possible, and there are some famous examples, to solve a -hard problem without discussing it, but this is an exception, especially for someone -at your stage. -Anyway, you should discuss the specifics with advisor. Unless that's the person -your worried about. In which case, you should switch advisors, or perhaps professions... - -REPLY [5 votes]: The math community is rather small, bad people are known (and there's only few of them), -stories about them circulate. -Just ask around: "Do you think that it's a good idea that I go talk to professor X?" -If professor X happens to be an (one of the very few) unethical mathematician, people will warn you against talking to him/her. - -REPLY [3 votes]: Ask yourself: do you want to be part of a community, or do you want to do math by yourself in the woods? -There is no wrong answer: doing either is a fine and potentially rewarding lifestyle, but giving up on the community means giving up on both the positives and negatives that it brings, and most answers here make clear that the positives outweigh the negatives.<|endoftext|> -TITLE: Quotients of number rings -QUESTION [14 upvotes]: Hi, -Here's a question that comes up every now and then. Of course, the quotient of a number ring (ring of integers of a number field) by an ideal $I$ is a finite (Artin) ring. If we take $I$ to be the power of a prime, we obtain a finite local (Artinian) ring. Is there a characterization of finite local rings which arise in this way? -In particular, can we obtain the quotient rings $k[x]/x^n$, where $k$ is a finite field? -Much thanks! - -REPLY [4 votes]: This answer consists of a few remarks to put the "necessary condition" in context. Recall: - -Theorem: If $R$ is a Dedekind domain and $I$ is a nonzero ideal of $R$, then $R/I$ is a principal Artinian ring (i.e., an Artinian ring in which every ideal is principal). - -Using the factorization of ideals into primes, the Chinese Remainder Theorem, and the easy fact that a finite product of rings is principal Artinian iff each factor is principal Artinian, one reduces to the case of $I = \mathfrak{p}^a$ a prime power. The ideals of $R/I$ correspond to the ideals of $R$ containing $\mathfrak{p}^a$, i.e., $R, \mathfrak{p},\ldots,\mathfrak{p}^a$. So $R/I$ is certainly Artinian. Moreover, $R/I = R_{\mathfrak{p}}/I_{\mathfrak{p}}$ is also a quotient of a DVR, hence of a PID, and any quotient of a principal ring is principal. -In my experience it is traditional to emphasize not the previous result per se but the following consequence. - -Theorem (C.-H. Sah) For an integral domain $R$, the following are equivalent: - (i) $R$ is a Dedekind domain. - (ii) Every ideal $I$ of $R$ can be generated by "$1+\epsilon$" elements: for all $0 \neq a \in I$, there exists $b \in I$ such that $I = \langle a,b \rangle$. - -The implication (i) $\implies$ (ii) is immediate from the above. The converse implication is also short but somewhat tricky, so I refer to $\S 20.5$ of my commutative algebra notes for the proof. -So it should be "well known" that the answer to the OP's question is not all finite rings. Note also that the example $\mathbb{F}_q[x,y]/\langle x,y \rangle^2$ of a finite non-principal ring appears as an Exercise in $\S 16.3$ of my notes. I am very sorry to say that my 264 pages of notes have no discussion of Zariski tangent spaces whatsoever: caveat emptor, to be sure.<|endoftext|> -TITLE: Equivalent metrics on Fréchet spaces and Lipschitz maps -QUESTION [7 upvotes]: Lipschitz maps are defined over metric space as maps $f:(X,d_X) \to (Y,d_Y)$ such that -$$ d\left( f(x),f(x^\prime) \right)_Y \le k d(x,x^\prime)_X \ \forall x,x^\prime \in X, $$ -where $k$ is a positive constant. We usually say that $f$ is a contraction if $k<1$. -It is well know that a different equivalent metric on $X$ does not preserve contractions, i.e. a map can be a contraction with respect to a metric but not with respect to an equivalent one. -In the Banach space setting, where the spaces $X$ and $Y$ are endowed with a norm defining the topology, there is a somehow "canonical" distance given by -$$ d(x,x^\prime) = \lVert x-x^\prime \rVert .$$ -With this distance, Lipschitz maps can be characterized as maps satisfying, for some $k>0$ -$$ {\left\lVert f(x) - f(x^\prime) \right\rVert}_Y \le k {\left \lVert x-x^\prime \right \rVert}_X \ \forall x, x^\prime \in X. $$ -It is obvious that, if $f$ satisfies the above relation, then is a $k$-lipschitz map. - -In the Fréchet space setting, the topology is defined by a countable family of semi-norms $({\lVert\cdot\rVert}_n)$. The classical example of metric inducing the same topology is given by -$$ d(x,x^\prime) = \sum_{n=0}^\infty {2^{-n}} \frac{{\lVert x-x^\prime\rVert}_n}{1+{\lVert x-x^\prime\rVert}_n} . $$ -In analogy with the Banach case, I would like to characterize (at least some) Lipschtiz maps between Fréchet spaces as maps satisfying -$$ {\left\lVert f(x) - f(x^\prime) \right\rVert}_n \le k {\left \lVert x-x^\prime \right \rVert}_n \ \forall x, x^\prime \in X,\ \forall n \in \mathbb{N}. $$ -Again, maps satisfying the last equation are Lipschitz maps with respect to the metric defined above, but the Lipschitz constant is not $k$ anymore, and in particular contraction with respect to the semi-norms (i.e. maps satisfying the last equation with $k<1$) are not contraction with respect to the metric. -Are there equivalent distances on $X$ and $Y$ such that every contraction with respect to the semi-norms is a contraction with respect with the new distance? If this is not possibile for every contraction, is it possible for a specific one? - -REPLY [4 votes]: Use $\sum 2^{-n}(\|x-y\|_n \wedge 1)$ for the distance on $Y$ and $\sum 2^{-n}(\|x-y\|_n \wedge 2)$ for the distance on $X$.<|endoftext|> -TITLE: Restricted Lie algebras of low dimension -QUESTION [6 upvotes]: Over the decades there has been a lot of papers devoted to the classification of Lie algebras of low dimension. Do you know any paper dealing with the problem of determining (up to restricted isomorphisms) restricted Lie algebras $(L,[p])$ of low dimension over a field of characteristic $p>0$? - -REPLY [3 votes]: Only for information, I would like mentioning that a paper of C. Schneider and H. Usefi just on this topic recently appeared on arXiv: http://arxiv.org/abs/1404.1047<|endoftext|> -TITLE: On rational points of conics -QUESTION [6 upvotes]: Let $K$ be a field of characteristic $0$. Let $C/K$ be a be a quasi-projective conic defined over $K$. Let $L/K$ be a finite dimensional field extension of odd dimension. Assume that -$C(L)$ is not empty. -Q: Then is true that $C(K)$ is not empty? If so, then how does one prove it? - -REPLY [2 votes]: So if we use Riemann-Roch for smooth projective curves over $K$ the problem becomes easy. So without lost of generality we may assume that $C/K$ is smooth and projective since a conic -admits a point over $K$ iff $C(K)$ is infinite (this is because the existence of a parametrization over $K$). -If $C(K)$ is not empty we are done. So now assume that $C(K)$ is empty. We will try to reach a contradiction. Let $Q$ be a point in $C(L)$ with minimal field of definition -$L$ and let $[L:K]=m\equiv 1\pmod{2}$. Choose a point $P_1$ in $C(\overline{K})$ that lives in a quadratic extension of $K$. Such a point exists since we have a conic and $C(K)$ is empty. Let $P:=P_1+P_1^{\sigma}\in Div_K(C)$. Thus we have -$deg(P)=2$ and $deg(Q)=m\geq 3$. We way thus write $m=2a+1$ for -some positive integer $a$. Now let -$$ -D:=[Q]-a[P]\in Div_K(C) -$$ -We have $deg(D)=1$. Now let us consider the line bundle $L_D$ on $C$ where -$$ -L_D=\{f\in K(C):div(f)\geq -D\} -$$ -By Riemann-Roch, we have that $dim_K(L_D)=2$ and thus there exists a non-constant function -$f\in L_D$. Note that the map -$$ -C(\overline{K})\rightarrow P^1(\overline{K}) -$$ -given by $x\mapsto [f(x),1]$ has degree $deg(div(f)_{\infty})$. So in general, it is not an embedding. Now let us work over $\overline{K}$ so that $[Q]=[Q_1]+[Q_2]\ldots+[Q_m]$ -and $[P]=[P_1]+[P_2]$. Since -$$ -div(f)\geq -D, -$$ -$f\in K(C)$ (so $deg(div(f))=0$ and $div(f)$ is $G_K$-invariant) we must have that over $\overline{K}$ -$$ -div(f)=a[P_1]+a[P_2]+[P_3]-[Q_1]-[Q_2]-\ldots -[Q_m] -$$ -where $deg([P_3])=1$. This forces $P_3$ to be defined over $K$. This contradicts the fact that $C(K)$ was empty.<|endoftext|> -TITLE: Packing moebius bands -QUESTION [9 upvotes]: I know that in the smooth category the following is true. There are at most countable many embedded moebius bands in euclidean 3-space. Is this also true in topological category? - -REPLY [10 votes]: There are at most countably many disjoint embeddings of homeomorphic images of a non-orientable hypersurface in $\mathbb R^k$. This is theorem 2 in "An uncountable family of disjoint spatial continua in Euclidean space" by V.K. Ionin and Yu.G. Nikonorov.<|endoftext|> -TITLE: Perfect set property for projective hierarchy -QUESTION [9 upvotes]: Is there any paper discussing the consistency strength (or possible equivalents, maybe large cardinals) of just assuming the perfect set property for certain levels of the projective hierarchy? - -REPLY [11 votes]: The classical argument Andres mentions can be taken one step further to show that if every uncountable $\Pi_1^1$ set contains a perfect subset then $\aleph_1$ is inaccessible in $L$. -This follows from the same fact for $\Sigma^1_2$-sets since, by the uniformization theorem, -every $\Sigma^1_2$-set is the injective projection of a $\Pi^1_1$-set.<|endoftext|> -TITLE: Partitioning $\mathbb{R}$ into $\aleph_1$ Borel sets -QUESTION [14 upvotes]: I just ran into this deceptively simple looking question. - -Is it always possible to partition $\mathbb{R}$ (or any other standard Borel space) into precisely $\aleph_1$ Borel sets? - -On the one hand, this is trivial if the Continuum Hypothesis holds. Less trivially, this also follows from $\mathrm{cov}(\mathcal{M}) = \aleph_1$, $\mathrm{cov}(\mathcal{N}) = \aleph_1$, $\mathfrak{d} = \aleph_1$, and similar hypotheses. However, I can't think of a general argument that allows one to split $\mathbb{R}$ into precisely $\aleph_1$ pairwise disjoint nonempty Borel pieces. -On the other hand, PFA or MM might give a negative answer but I don't see a good handle from that end either. - -REPLY [10 votes]: Here is another example from recursion theory: -Take a chain $\{x_{\alpha}\}_{\alpha<\omega_1}$ from Turing degrees. -For each $\alpha<\omega_1$, let $A_{\alpha}$ be the collection of the reals neither in $\bigcup_{\beta<\alpha}A_{\beta}$ nor Turing-computing $x_{\alpha}$. -Then $\{A_{\alpha}\}_{\alpha<\omega_1}$ is a Borel partition of $\mathbb{R}$.<|endoftext|> -TITLE: Less discriminating discriminants -QUESTION [7 upvotes]: Suppose that I want to know whether a polynomial $P(z)$ has a root with multiplicity at least three. This is obviously an algebraic condition, but is there some reasonably concise set of conditions defining the variety (in the space of coefficients)? This must have been studied by the ancients. It is clearly necessary that the discriminant vanish, and also that the resultant of the polynomial $P$ and the second derivative $P^{\prime\prime}$ vanish, but, just as obviously, not sufficient... -EDIT Abhinav certainly gives a nice answer to the question, but the question I would REALLY like to know the answer to is: what is the degree of the variety as a function of the partition (as in @Gjergji's answer). Maybe I should read the reference... - -REPLY [4 votes]: For partitions of the special form $\lambda=(k,1,1,\dots,1)$ (i.e., hook shapes) there is a very explicit description of the ideal of definition of the corresponding set of polynomials (i.e., those polynomials with a root of order $\geq k$). Namely, it is the span of the set of (symmetric) Jack polynomials $f_\mu$, with parameter specialized to $-1/k$, and where $\mu$ ranges over all partitions satisfying -$$\mu_i-\mu_{i+k-1} \geq 2$$ for all $i$. This is the main result of the paper http://arxiv.org/pdf/math/0112127 by Feigin, Jimbo, Miwa, and Mukhin. -For the application to degree, one might note that this computes the Hilbert function of this ideal quite explicitly in combinatorial terms, though for your problem you are probably interested in the grading where each elementary symmetric function has degree one, and I don't know, off the top of my head, how to extract that information. Maybe I'll post a separate question so we can see if anywhere else here knows.<|endoftext|> -TITLE: Automorphism groups of virtually cyclic groups -QUESTION [6 upvotes]: Let $V$ be a virtually cyclic group. -Then is $Aut(V)$ also a virtually cyclic group? -This is true when $V$ is a finite group (zero-ended) and when $V = C_\infty, D_\infty$ (both two-ended). - -REPLY [8 votes]: In Finitely generated groups with virtually free automorphism groups, by M.R. Pettet, it is proved in theorem 3.4 that the automorphism group $Aut(G)$ of a finitely generated group is virtually cyclic if and only if both $Z(G)$ and $G/Z(G)$ are virtually cyclic.<|endoftext|> -TITLE: Proofs of Mordell-Weil theorem -QUESTION [12 upvotes]: I would like to ask if there exist pedagogical expositions of the Mordell-Weil theorem (wikipedia). What parts of number theory (algebraic geometry) one should better learn first before starting to read a proof of Mordell-Weil? - -REPLY [7 votes]: Here is a short proof of the weak Mordell-Weil theorem for Abelian varieties over a number field using étale cohomology (easily adopted to finitely generated fields). The construction of the height paring can be found in Hindry-Silverman, or in [Brian Conrad, http://math.stanford.edu/~conrad/papers/Kktrace.pdf ], section 9 (Conrad even proves a more general theorem, the Lang-Néron theorem). -Let $K$ be a number field, $A/K$ be an Abelian variety and $S$ a finite set of places of $K$. Let $X = \mathrm{Spec}\mathcal{O}_{K,S}$ and $\mathscr{A}/X$ the Néron model of $A/K$. By the Néron mapping property, it suffices to show that $\mathscr{A}(X)/n = A(K)/n$ is finite for some $S$ and $n > 1$. -By enlarging $S$ by the set of primes lying over $n$, one has a short exact Kummer sequence $0 \to \mathscr{A}[n] \to \mathscr{A} \to \mathscr{A} \to 0$, inducing in (étale) cohomology $0 \to \mathscr{A}(X)/n \hookrightarrow H^1(X,\mathscr{A}[n])$. So it suffices to show that $H^1(X,\mathscr{A}[n])$ is finite. (This group is related to the Selmer group. The cokernel $H^1(X,\mathscr{A})[n]$ is related to the $n$-torsion of the Tate-Shafarevich group.) -There is a finite étale Galois covering $X'/X$ such that $\mathscr{A}[n] \times_X X' \cong (\mathbf{Z}/n)^{2g} \cong \mu_n^{2g}$. The Hochschild-Serre spectral sequence $$H^p(\mathrm{Gal}(X'/X), H^q(X',\mathscr{A}[n] \times_X X')) \Rightarrow H^{p+q}(X,\mathscr{A}[n])$$ induces $$0 \to H^1(\mathrm{Gal}(X'/X), H^0(X',\mathscr{A}[n] \times_X X')) \to H^1(X,\mathscr{A}[n]) \to H^0(\mathrm{Gal}(X'/X), H^1(X',\mathscr{A}[n] \times_X X')).$$ Since $\mathrm{Gal}(X'/X)$ and $H^0(X',\mathscr{A}[n] \times_X X')$ are finite, the left hand group is finite, so it suffices to show that $H^1(X',\mathscr{A}[n] \times_X X') \cong H^1(X',\mu_n^{2g})$ is finite. But the short exact Kummer sequence $1 \to \mu_n \to \mathbf{G}_m \to \mathbf{G}_m \to 1$ induces $$1 \to \mathbf{G}_m(X')/n \to H^1(X',\mu_n) \to H^1(X',\mathbf{G}_m)[n] \to 0.$$ The left hand group is finite by the finite generation of the $S$-unit group, and the right hand group is finite by the finiteness of the $S$-class number (Hilbert 90: $H^1(X',\mathbf{G}_m) = \mathrm{Pic}(X') = \mathrm{Cl}(X')$).<|endoftext|> -TITLE: Polynomial Vector Fields on the 3-Sphere -QUESTION [5 upvotes]: EDIT(3): I am looking for a basis for the Lie algebra of polynomial vector fields on $S^3$. -EDIT(2): I am fairly certain now that my question is more along the lines of, what does the Lie algebra of polynomial vector fields on $S^3$ look like? -EDIT: my question is really the following. The Lie algebra of the diffeomorphism group of $S^1$ is the Witt algebra. What is the corresponding Lie algebra for $S^3$? -(1) What is the diffeomorphism group of the 3-sphere? My reason for asking is that I want to know if there is an analogue of the Witt (=centerless Virasoro) algebra in three dimensions. I am aware of the $W_n$ series in Cartan's classification, but this is not the generalization I am looking for. -(1.0) Alternatively/equivalently, I would like to know how to describe "regular" (smooth?) sections of the tangent bundle on $S^3$ - it seems like it might be helpful (to me, at least) to think of the fibers as copies of $sl(2)(\cong su(2))$. It's been a while since I looked at principal fiber bundles, but this definitely reminds me of one. -(0) Currently I'm trying to think of all of this stuff in terms of [unit] quaternions. This seems promising to me for a number of reasons, so if you can tell me anything about the above in such terms or point me towards papers/preprints/steles about the above in quaternionic language, that would be fantabulous. -(Obligatory "sorry if this is vague to the point of madness" and "sorry if this has been answered previously; I did my best to check the related questions.") - -REPLY [5 votes]: Complementing André's answer, here's another possible definition of polynomial vector fields on $S^3$. You think of $S^3$ as the unit sphere in $\mathbb{R}^4$ and consider polynomial vector fields on $\mathbb{R}^4$ which are tangent to the sphere; that is, which annihilate the function $\sum x_i^2$. -One thing to point out, which may or may not be relevant to the applications you have in mind but which I mention since you did mention the Virasoro algebra, is that the structure of the diffeomorphism algebras (or algebras of polynomial vector fields) in dimension greater than 1 is very different than in dimension 1. For example, you don't have a nice decomposition such as the one -$$ -\mathfrak{Vir} = \mathfrak{Vir}^- \oplus \mathfrak{Vir}^0 \oplus \mathfrak{Vir}^+ -$$ -for the Virasoro algebra, and this in turn hinders the construction of positive energy representations. -I am aware on some work on this topic in the mathematical physics literature; e.g., -Fock space representations of the algebra of diffeomorphisms of the $N$-torus, F Figueirido and E Ramos -and also papers by TA Larsson.<|endoftext|> -TITLE: Torus knots in Euclidean space -- a symmetry argument -QUESTION [9 upvotes]: Consider a $(p,q)$ torus knot $K$ in 3-dimensional Euclidean space $\mathbb R^3$ where $p,q \geq 2$ and $\operatorname{GCD}(p,q)=1$. -Let $\operatorname{Isom}(\mathbb R^3,K)$ be the isometries of $\mathbb R^3$ that preserve $K$. -It's a fairly standard argument using theorems about uniqueness of Seifert fiberings to prove that it's impossible for $\operatorname{Isom}(\mathbb R^3, K)$ to contain subgroups isomorphic to both $\mathbb Z_p$ and $\mathbb Z_q$. Of course, $\operatorname{Isom}(S^3,K)$ can and does for the standard embeddings of torus knots in $S^3$. In some sense the core issue is that when this does happen, the $\mathbb Z_p$ and $\mathbb Z_q$ subgroups of $\operatorname{Isom}(S^3,K)$ have disjoint fixed point sets. -My question: is there a reasonably elementary proof $\operatorname{Isom}(\mathbb R^3, K)$ does not contain subgroups isomorphic to both $\mathbb Z_p$ and $\mathbb Z_q$ that avoid the use of Seifert-fiber space techniques? I'm particularly interested if any "quantum topology" invariants can make this kind of symmetry argument. I thought a little about this, at least I'm not seeing how one could use the Alexander polynomial. - -REPLY [3 votes]: It seems to me that the isometry group of a $(2,q)$ torus knot can be a dihedral group, thus containing both $\mathbb{Z}_2$ and $\mathbb{Z}_q$. The standard realization (on a torus of revolution, take the curve having the right homotopy class and whose latitude and longitude move at constant speed) should do the trick. It is indeed $q$ symmetric with respect to an axis passing in the hole of the torus, and seems $2$ symmetric (with respect to any axis that is orthogonal to the first one and meets the knot) to me. -Edit: as precised by Ryan, the knot is in fact oriented and we only consider symmetries preserving this orientation. -Further edit: let me give a partial elementary proof under this assumption. Assume that there are a $\mathbb{Z}_p$ and a $\mathbb{Z}_q$ in the symmetry group of the knot. Since they preserve its orientation, they act by translation and therefore their actions on the knot commute. Since the knot is not planar and the symmetries are linear, they must commute globally. Except the ruled out case of $p=q=2$, this implies that the two subgroups of isometries have the same axis. From here the knot can be cut off into $pq$ isometric pieces glued by rotations, and I guess that someone more used to knots than me can produce a contradiction.<|endoftext|> -TITLE: How does one see Hecke Operators as helping to generalize Quadratic Reciprocity? -QUESTION [7 upvotes]: I posted this question on math stackexchange: -https://math.stackexchange.com/questions/56040/how-does-one-see-hecke-operators-as-helping-to-generalize-quadratic-reciprocity -and got 10 upvotes but no answers. I interpret this as evidence that maybe I've matured to mathoverflow. Here is what I wrote: -My question is really about how to think of Hecke operators as helping to generalize quadratic reciprocity. -Quadratic reciprocity can be stated like this: Let $\rho: Gal(\mathbb{Q})\rightarrow GL_1(\mathbb{C})$ be a $1$-dimensional representation that factors through $Gal(\mathbb{Q}(\sqrt{W})/\mathbb{Q})$. Then for any $\sigma \in Gal(\mathbb{Q})$, $\sigma(\sqrt{W})=\rho(\sigma)\sqrt{W}$. Define for each prime number $p$ an operator on the space of functions from $(\mathbb{Z}/4|W|\mathbb{Z})^{\times}$ to $\mathbb{C}^{\times}$ by $T(p)$ takes the function $\alpha$ to the function that takes $x$ to $\alpha(\frac{x}{p})$. Then there is a simultaneous eigenfunction $\alpha$, with eigenvalue $a_p$ for $T(p)$, such that for all $p\not|4|W|$ $\rho(Frob_p)=a_p$. (and to relate it to the undergraduate-textbook-version of quadratic reciprocity, one need only note that $\rho(Frob_p)$ is just the Legendre symbol $\left( \frac{W}{p}\right)$.) -Now I'm trying to understand how people think of generalizations of this. First, still in the one dimensional case, let's say we are not working over a quadratic field. What would the generalization be? What would take the place of $4|W|$? Would the space of functions that the $T(p)$'s work on still thes space of functions from $(\mathbb{Z}/N\mathbb{Z})^{\times}$ to $\mathbb{C}^{\times}$? What is this $N$? -Now let's jump to the $2$-dimensional case. Here we have the actual theory of Hecke operators. However, as I understand it, there is a basis of simultaneous eigenvalues only for the cusp forms. Now I'm finding it hard to match everything up: are we dealing just with irreducible $2$-dimensional representations? Instead of $\rho$ do we take the character? Would we say that for each representation there's a cusp form such that it's a simultaneous eigenfunction and such that $\xi(Frob_p)=a_p$ (the eigenvalues) where $\xi$ is the character of $\rho$? This should probably be for all $p$ that don't divide some $N$. What is this $N$? Does it relate to the cusp forms somehow? Is it their weight? Their level? -In other words: -Questions -$1$. What is the precise statement of the generalization (in the terminology above) of quadratic reciprocity for the $1$-dimensional case? -$2$. What is the precise statement of the generalization (in the terminology above) of quadratic reciprocity for the $2$-dimensional case? -Edit -Actually, now that I have the attention of experts, let me add two more questions: -$3$. Does Langlands predict anything for $1$-dimensional representations with infinite image? -and -$4$. I very much want to understand Hecke operators better. For example, why are the $T(p)$'s that I gave above the $1$-dimensional analogue of the usual Hecke operators? I've heard something about Hecke correspondences, and I wonder what is a good reference I can sink my teeth into about that. - -REPLY [5 votes]: This is what Langlands reciprocity is about, you can read about it in many surveys. Very briefly it says that every $d$-dimensional (continuous) Galois representation is associated to an automorphic form on $\mathrm{GL}_d$ so that their $L$-functions agree. For a Galois representation the $L$-function is given in terms of the characteristic polynomials of the Frobenius elements, while for an automorphic form the $L$-function is given in terms of Hecke eigenvalues (more precisely of Langlands-Satake parameters that can be read off from Hecke eigenvalues or vice versa). So Langlands reciprocity says that for any $d$-dimensional Galois representation there is a great harmony of the images of the Frobenius elements: they yield a function on $\mathrm{GL}_d$ with fantastic properties (namely an automorphic form). Your case is $d=2$. And yes, $N$ is the level.<|endoftext|> -TITLE: Decomposable Banach Spaces -QUESTION [26 upvotes]: An infinite dimensional Banach space $X$ is decomposable provided $X$ is the direct sum of two closed infinite dimensional subspaces; equivalently, if there is a bounded linear idempotent operator on $X$ whose rank and corank are both infinite. The first separable indecomposable Banach space was constructed by Gowers and Maurey. It has the stronger property that every infinite dimensional closed subspace is also indecomposable; such a space is said to be HI or hereditarily indecomposable. There do not exist HI Banach spaces having arbitrarily large cardinality (although Argyros did construct non separable HI spaces), but I do not know the answer to: -Question: If the cardinality of a Banach space is sufficiently large, must it be decomposable? -Much is known if $X$ has some special properties (see Zizler's article in volume II of the Handbook of the Geometry of Banach Spaces). -Something I observed (probably many others did likewise) around 40 years ago is that the dual to any non separable Banach space is decomposable; I mention it because it is not in Zizler's article (in his discussion of idempotents he is interested in getting more structure--a projectional resolution of the identity) and I did not publish it because it is an easy consequence of lemmas J. Lindenstrauss proved to get projectional resolutions of the identity for reflexive spaces. - -REPLY [11 votes]: According to the recent preprint by Koszmider, Shelah and Świętek under the generalised continuum hypothesis there is no such bound. In particular, one cannot prove the existence of such a bound working merely within the ZFC.<|endoftext|> -TITLE: Connected components of large induced subgraphs of hypercubes -QUESTION [5 upvotes]: Let $H$ be the $n$-dimensional hypercube, i.e. $\{0,1\}^n$ with edges between two vertices if and only if they differ in exactly one co-ordinate. We say that an edge is in direction $i$ if its endpoints differ in exactly the $i$'th co-ordinate. Suppose $V$ is a subset of $H$ such that $|V| > 2^{n-1}$. Is it true that at least one connected component of the graph induced by $V$ contains edges in all $n$ direction? - -REPLY [4 votes]: Yes, this is true. Thanks to Sukhada Fadnavis and Seva for pointing out in the comments that the argument I had written here was wrong. Instead I will point you to the paper where this is proved - -"Bulky subgraphs of the hypercube", by Andrei Kotlov, Europ. J. Combinatorics (2000) 21, 503-507 - -As far as I can tell from looking at the literature, it is not known if there are configurations of more than $2^{n-1}$ vertices for which one can not find $n+1$ of them which induce a tree with an edge in every direction. This would be a strengthening of the result in question.<|endoftext|> -TITLE: Fast Vandermonde matrix multiplication over finite field -QUESTION [7 upvotes]: Let $V_{i,j}=x_i^j$ where $x_i\in\mathbb F_q$ for $1\le i\le n,1\le j\le n$ be a Vandermonde matrix over finite field $\mathbb F_q$. -I wish to know the currently known fastest algorithms for computation of -1) $Vx$ where $x\in\mathbb F_q^{n\times1}$; -2) $V^Tx$ where $V^T$ is the transpose of $V$; -3) $V^{-1}x$; -4) $(V^T)^{-1}x$. -Can you also provide some references for the above algorithms? - -REPLY [5 votes]: All problems can be solved in $O(M(n)\log(n))$ base field operations, where $M(n)$ is the time it takes to multiply polynomials in degree $n$ (so using FFT, this is quasi-linear). -This is in Chapter 3 of Pan's Structured Matrices and Polynomials.<|endoftext|> -TITLE: Generalisations of Riemann-Roch for surfaces -QUESTION [5 upvotes]: Let $X$ be a smooth projective algebraic surface (over $\mathbb{C}$ ). For all $L\in \mathrm{Pic}(X)$, we have -$$\chi(L)=\chi(\mathcal{O}_X)+\frac{1}{2}(L^2-L\cdot \omega_X).$$ -This is the famous Riemann-Roch theorem in the flavour I like the most. It usually comes together with the following two formulas: -$$\chi(\mathcal{O}_S)=\frac{1}{12}(K_X^2+\chi_{top}(X)),$$ -the Noether's formula and -$$2p_a(C)-2=C^2+C\cdot K_X,$$ -the genus formula for $C$ an irreducible (possibly singular) curve. -Is there a similar (or maybe the same) version for -a) Smooth quasi-projective surfaces. -b) Projective surface with quotient singularities, or A-D-E singularities. - -REPLY [13 votes]: If $X$ is proper with rational singularities (and quotient and A-D-E (=Du Val) singularities are rational), then you can do most cohomology computations on a resolution. -Let $\pi:Y\to X$ be a resolution of singularities (not necessarily minimal). Then if $X$ has rational singularities, then $R^i\pi_*\mathscr O_Y=0$ for $i>0$. Let -$L$ be a line bundle on $X$. Then by the above vanishing, $h^i(Y,\pi^*L)=h^i(X,L)$, so we have that -$$\chi(Y,\pi^*L)=\chi(X,L).$$ -Note that this actually holds in any dimension. -It follows that one has a sort-of-RR on $X$ (surface): -$$\chi(X,L)=\chi(\mathcal{O}_X)+\frac{1}{2}((\pi^*c_1(L))^2-(\pi^*c_1(L))\cdot K_Y).$$ -(Remark: it is not a bad idea to distinguish when we talk about a line bundle and when we talk about the associated Cartier divisor! Besides having the formulas be well-defined one has to remember that for example -the push forward of a divisor via a birational map is also a divisor, while the push forward of a line bundle is not necessarily a line bundle. And even if it is, the divisor associated to the push forward of a line bundle is not necessarily the same as the push forward of the divisor associated to the line bundle.) -As for your question about the quasi-projective case, Christian already said that it is tricky. In general, when it comes to cohomology, experience shows that it is better to compactify and figure out the difference than trying to develop a handicapped theory for quasi-projective varieties directly. (See Deligne's way of doing Hodge theory on open varieties). -The genus formula on a surface is essentially equivalent to Riemann-Roch, so as soon as the formula makes sense, it will follow.<|endoftext|> -TITLE: embedding of local representation into automorphic representation -QUESTION [9 upvotes]: Assume $v$ is a place of a number field $k$, finite or not. Let $\pi_v$ be an irreducible admissible generic representation of $GL_n(k_v)$. Is it always true that we can find some irreducible generic automorphic representation $\Pi$ of $GL_n(\mathbb{A}_k)$ with $v$-component exactly isomorphic to $\pi_v$? -A form of the famous generalized Ramanujan conjecture says that if $\Pi$ is cuspidal, then every component is tempered. So the above question is kind of converse to Ramanujan conjecture. -It is known that if $v$ is a finite place, and $\pi_v$ is supercuspidal, then $\Pi$ always exists, and in fact we can take $\Pi$ to be a cuspidal representation. -Many thanks for any answer or references related to this question. - -REPLY [7 votes]: I don't know if you're still interested in the question, but Arthur proves something like this in his upcoming book on representations of classical groups (http://www.claymath.org/cw/arthur/pdf/Book.pdf): -Lemma 6.2.2 says that for $G=SO(n)$ or $Sp(2n)$, a local field $F\neq\Bbb C$, and a square-integrable irreducible representation $\pi$ of $G(F)$, there is a global field $K$, a place $v$, and an automorphic representation $\Pi$ in the discrete spectrum that has $\pi$ as the $v$-component and is spherical at all other finite places. -(He needs the lemma to use the trace formula over the constructed global field $K$.) -And I think I've seen something like this also for other groups somewhere else, at least for $GL(n)$ (with weaker conditions on the remaining places).<|endoftext|> -TITLE: Eigenspace of Euclidean distance matrix. -QUESTION [6 upvotes]: What is the necessary and sufficient condition (if there is any) that $n$ orthonormal vectors $v_1,v_2,\cdots,v_n$ are eigenvectors of a Euclidean distance matrix. When $n=2$, the orthonormal vectors are easily charcterized, i.e., $(1/\sqrt{2}, 1/\sqrt{2})$ and $(1/\sqrt{2}, -1/\sqrt{2})$. - -REPLY [6 votes]: This isn't an answer, but it's too long for a comment. As you're maybe aware, a real $n \times n$ matrix $M$ is a Euclidean distance matrix if and only if the following conditions hold: - -$M_{ij} \geq 0$ for all $i, j$ -$M_{ii} = 0$ for all $i$ -$M$ is symmetric -$M$ is conditionally negative definite, that is, -$$ -x^t M x \leq 0 -$$ -whenever $x \in \mathbb{R}^n$ with $\sum_i x_i = 0$. - -This was shown in: I. J. Schoenberg, Metric spaces and positive definite functions, Transactions of the AMS 44 (1938), 522-536.<|endoftext|> -TITLE: 2-norm of the upper triangular "all-ones" matrix -QUESTION [9 upvotes]: Let $M_n$ be the $n\times n$ matrix -$$ -(M_n)_{ij}=\begin{cases}1 & i\leq j,\\\\ 0 &i>j.\end{cases} -$$ -Is there around an explicit expression or at least an asymptotic for $\left\Vert M_n \right\Vert$? The norm is the usual Euclidean induced norm $\left\Vert M \right\Vert=\rho(M^TM)^{1/2}$. -I apologize if this a stupid question... - -REPLY [6 votes]: [This may be largely an alternate version of Noam's answer, but the extra context could be -interesting.] -Let $N$ be the $m\times m$ matrix with $N_{i,i+1}=1$ for $i=1,\ldots,m-1$ -and all other entries zero. Then the matrix -$$ - A = \begin{pmatrix}0&I+N\\\\ (I+N)^T&0 \end{pmatrix} -$$ -is the adjacency matrix of the path on $2m$ vertices. Now -$$ - A^{-1} = \begin{pmatrix} - 0&(I+N)^{-1}\\\\ (I+N)^{-T}&0 - \end{pmatrix} -$$ -and since $N^n=0$, -$$ - (I+N)^{-1} = I-N+\cdots+(-1)^{n-1}N^{n-1} -$$ -Let $D$ be the $2m\times 2m$ diagonal matrix with $D_{i,i}=(-1)^{i-1}$. Then it -easy to check that -$$ - D^{-1}AD = \begin{pmatrix} - 0&M\\\\ M^T&0 - \end{pmatrix} -$$ -where $M$ is the matrix from the question. The 2-norm we want is the -square of the largest eigenvalue of $D^{-1}AD$, which is the square of -the largest eigenvalue of $A$, which is the square of the reciprocal of the $n$-th -eigenvalue of the path on $2n$ vertices (which is its smallest positive eigenvalue). -The eigenvalues of the path on $n$ vertices are $2\cos\left(\frac{j\pi}{n+1}\right)$ for $j=1,\ldots,n$. -More on this appears in my old paper ``Inverses of trees''. (We can view $M$ as the incidence -matrix of a chain, and so some of the above extends to a larger class of posets.)<|endoftext|> -TITLE: Finite, Étale Morphism Of Varieties -QUESTION [13 upvotes]: I have a, probably very simple, question: My intuition tells me that the following statement should be true, but I couldn't find it anywhere and I wanted to make sure I am not missing something. -Let $\pi:Y\to X$ be a finite, étale morphism of nonsingular varieties over some algebraically closed field $\Bbbk$. Is it true that every point $P\in X$ has an affine neighbourhood $U$ such that $\pi^{-1}(U)$ consists of $\deg(\pi)$ irreducible components, each of which is isomorphic to $U$ via $\pi$? -Of course, if it is true, I would also be happy if you could provide a proof, in literature or otherwise. - -REPLY [21 votes]: No, it's not true. Consider the map $x\mapsto x^2$ as a map from $X=\mathbb A^1-\{0\}$ to itself. The "problem" is that Zariski neighborhoods are too big. Any open subset of $X$ has exactly one irreducible component (in general, an open subset cannot have more components than the ambient space), so there is no hope to get the preimage of an open set to have two components. -However, if you refine your topology to allow "etale open neighborhoods" (i.e. you allow pullback by etale maps $U\to X$, not just open immersions $U\to X$), then the answer is yes. Perhaps the easiest way to prove that is to pull back by $\pi$ itself, after which you can "peel off" the diagonal component of $Y\times_X Y$. Now you have a finite etale map $(Y\times_X Y -\Delta)\to Y$ which has degree one less. Repeat until the degree is $1$, at which point you have an etale map $U\to \cdots \to Y\to X$ so that $Y\times_X U$ is $deg(\pi)$ copies of $U$. - -REPLY [7 votes]: This is almost never true. For example if $X,Y$ are irreducible, then for any (affine) Zariski open $U$ of $X$, the inverse image is open in $Y$ and hence irreducible.<|endoftext|> -TITLE: Sums of two cubes -QUESTION [13 upvotes]: In his solution of the equation $x^3 + dy^3 = 1$, Nagell -comes across the equation -$$ u^3 + 6u^2v + 3uv^2 - v^3 = w^3. $$ -He then observes that -$$ (u^3 + 6u^2v + 3uv^2 - v^3) U^3 = V^3 + W^3 $$ -for -$$ U = u^2+uv+v^2, \quad - V = u^3+3u^2v-v^3, \quad - W = 3u^2v+3uv^2, $$ -and then appeals to Fermat's Last Theorem for the exponent $3$. -My question: what is known about representing a binary cubic form -$$ au^3 + bu^2v + cuv^2 + dv^3 $$ -as a sum of two rational cubes in the polynomial ring ${\mathbb Z}[u,v]$? -Where is Nagell's solution coming from? -P.S. Dehomogenizing by setting $v = 1$ gives us an elliptic surface. Still I wonder how Nagell dreamed up his solution. - -REPLY [3 votes]: A paper by Bruce Reznick and Jeremy Rouse, preprint available at http://front.math.ucdavis.edu/1012.5801 completely solves the equation $x^3+y^3=f^3+g^3$, where -$f$ and $g$ are homogeneous rational functions in $x$ and $y$.<|endoftext|> -TITLE: Does there exist a model of chains on oriented manifolds with both a strict intersection pairing and strict functoriality for closed embeddings? -QUESTION [8 upvotes]: Let $M$ be a smooth oriented $n$-dimensional manifold. My favorite model of $\operatorname{Chains}_\bullet(M) \otimes \mathbb R$ is the space of smooth compactly-supported de Rham forms on $M$, shifted in degree by $[n]$. I like it because the intersection pairing of chains is well-defined: it corresponds after the grade-shifting to the wedge product of forms. It is deeply problematic in one way, though: this model of chains is functorial only for submersions of manifolds. In particular, it is not functorial for the diagonal embedding $M \hookrightarrow M^{\times 2}$, which would generate a coproduct dual to (dg commutative) the product of forms. -My first question is whether there exists any model of chains which has both a strict intersection pairing and a strict coproduct? I do not need a model for all manifolds — I only need something defined on, say, nice open subsets of $\mathbb R^n$, and I only need functoriality for maps as nice as embeddings. For example, does some version of "semialgebraic chains" do the trick (I'm happy working only with algebraic manifolds, and algebraic maps thereof)? For my application, I don't expect to have any problems with things like the possible difference between $\operatorname{Chains}_\bullet(M^{\times 2})$ and $\operatorname{Chains}_\bullet(M)^{\otimes 2}$. I'm also perfectly happy (possibly even more happy) to work with a model of $\operatorname{Cochains}^\bullet$ rather than $\operatorname{Chains}_\bullet$ — my formulas can be read from left to right or from right to left. -But I do not expect such a model to exist. Naive attempts to define a push-forward along non-submersions creates chains with $\delta$-function singularities, and indeed you should expect a version of Sweedler's theorem that any cocommutative coalgebra has grouplike elements, which in my case would be "$\delta$-functions" or "points". But if I have $\delta$ functions, then in order not to create extra cohomology I would need to have step-functions among my $1$-chains, and it is hard to come up with a model that does not ultimately lead to the presence of non-(locally)-constant idempotents. But any idempotent in a dg commutative algebra is necessarily closed (Exercise!). -So my main question is: - -Is the possibility of both intersection product and coproduct obstructed in some way? I.e. is there a reasonable proof that such models (where, understand, part of the proof requires making the statement of the problem precise) do not exist? - -Note that "take your model of $\operatorname{Chains}_\bullet$ to be $\operatorname{Homology}_\bullet$" is not an adequate answer. Here's one reason: $\operatorname{Homology}_\bullet$ has "Massey products", which is to say it is an $A_\infty$ algebra and _not_ a strictly associative algebra. (Question: is it strictly commutative?) -My final question is for suggestions for the next best thing. For example, even with singular chains in a manifold, there is an intersection pairing that is defined up to contractible space of choices, and is homotopy-commutative: namely, you perturb your chains slightly and thereby make the intersection transverse. But this particular structure is very large, and nigh impossible for me to write explicit formulas for. So if I am right that a strict model is obstructed, then my next hope would be a model in which the homotopy-commutative structure can be given completely explicitly in a small hands-on combinatorial way. - -REPLY [2 votes]: After work on this and other problems, I have come to the conclusion that the correct answer to my question is "No, there does not exist a model of chains with simultaneously strict intersection and diagonal maps". Even in characteristic 0, the problem is obstructed in dimension 1, as I prove in http://arxiv.org/abs/1308.3423. To conclude the answer to this question from what I prove in that paper, one must decide also whether the Frobenius axiom is required to hold strictly — the paper proves that even a "homotopy Frobenius" algebra structure does not exist (in the properadic sense) on Chains(R), and so certainly there is no strict Frobenius algebra structure. But the linked paper does not imply that you can't strictify both the associativity and coassociativity while keeping the Frobenius axiom weak.<|endoftext|> -TITLE: Do there exist groups with word problems in arbitrary P-degrees? -QUESTION [10 upvotes]: This has been posted on TCS stack exchange for a while here and hasn't gotten any answers, so I'm trying again here. -It has been known for a long time that, given any r.e. Turing degree, there is a finitely presented group whose word problem is in that degree. My question is whether the same thing is true for arbitrary polynomial time Turing degrees. Specifically, given a decidable set, $A$, does there exist a finitely presented group, with word problem, $W$, such that $W\leq_T^P A$ and $A\leq_T^P W$? I would also be willing to relax finitely presented to recursively presented. -I suspect that the answer is yes, and I have heard others say they read this somewhere, but I haven't been able to chase down a reference. -EDIT: As per the comments, here, $\leq_T^P$ means polynomial-time Turing reducibility. See here for more info. - -REPLY [7 votes]: The answer is "yes" (for finitely presented groups). It follows from the main result of Sapir, Mark V.; Birget, Jean-Camille; Rips, Eliyahu Isoperimetric and isodiametric functions of groups. Ann. of Math. (2) 156 (2002), no. 2, 345–466. - Edit. Here is the combination of me explanations in comments below. -You should look at the construction of the group in the paper. First there is a modification of a Turing machine so that the input configurations contribute to the time function the most. Then an S-machine is constructed which by prop. 4.1 is working in polynomial time comparing to the Turing machine and has the same property re input vs arbitrary configurations. Then a group is constructed using the S-machine. Then the Dehn function of the group is estimated (you need the estimate from above). The last step is done by the snowman decomposition. The snowman decomposition decomposes every van Kampen diagram into a linear number of discs and a diagram without hubs (whose area is at most cubic). The perimeters of the discs are linear in terms of the length of the boundary of the diagram. Given a boundary label of a disc, deciding that the disc with this boundary label exists is essentially the same as to decide that certain word belongs to L. Thus the snowman decomposition is a certificate that the word problem is in $L^P$. - -REPLY [7 votes]: I think the question is also answered positively by the main result in a paper of -mine -- Efficient computation in groups and simplicial complexes. -Trans. Amer. Math. Soc. 276 (1983), no. 2, 715–727 -- where it is shown that any Turing -machine may be simulated by a finitely presented group in linear space and cubic time.<|endoftext|> -TITLE: History of the triangle inequality -QUESTION [10 upvotes]: I am currently preparing a talk that revolves around the triangle inequality. -Because this inequality is so well-established, I do not want to (in my talk) belabor too much upon the importance it enjoys. For example, I learned some useful views here. But, these concerns are currently too advanced---for my purposes, I am seeking first some historical background; specifically, - -Approximately when, where, and how did the concept of a triangle inequality get formalized, and its importance recognized? - -EDIT -It seems that the above question is not precise or clear enough. How about the slightly clarified question: - -When was it realized (was it Fréchet's 1906 paper cited in the comments?) that the triangle inequality should be a fundamental axiom for defining distances? - -REPLY [2 votes]: Here is a suggestion, to get the idea across in an informal way---it is what I always tell the students when I introduce the triangle inequality: I tell them that its essential content, and the way it gets used, is that "things close to the same thing are close to each other".<|endoftext|> -TITLE: What are the best known bounds on the number of partitions of $n$ into exactly $k$ distinct parts? -QUESTION [7 upvotes]: For example, if $n = 10$ and $k = 3$, then the legal partitions are -$$10 = 7 + 2 + 1 = 6 + 3 + 1 = 5 + 4 + 1 = 5 + 3 + 2$$ -so the answer is $4$. By choosing $k$ random elements of $\{1,\ldots,2n/k\}$, one can easily construct about $(n/k^2)^k$ such partitions. For $k \approx \sqrt{n}$ this is not far from best possible, since the total number of partitions is (by Hardy and Ramanujan's famous theorem) asymptotically -$$\frac{1}{4 \sqrt{3} n} \exp\left( \pi \sqrt{ \frac{2n}{3} } \right).$$ -Can one do much better than $(n/k^2)^k$ for smaller k? -To be precise, writing $p^*_k(n)$ for the number of such partitions, is it true that, for some constant $C$, -$$p^*_k(n) \leqslant \left( \frac{Cn}{k^2} \right)^k$$ -for every $n,k \in \mathbb{N}$? - -REPLY [2 votes]: On further reflection, there seems to be a very simple (and nice) solution to my question. I'll sketch a proof of the following theorem. -Theorem: -There is a constant $C$ such that -$$\frac{1}{Cnk} \left( \frac{e^2 n}{k^2} \right)^k \leqslant p_k^*(n) \leqslant \frac{C}{nk} \left( \frac{e^2 n}{k^2} \right)^k.$$ -The upper bound follows from the recursion -$$p_k^*(n) \leqslant \frac{1}{k} \sum_{a=1}^n p^*_{k-1} (n-a)$$ -by a simple induction argument. To see the recursion, simply note that since the elements of the partition are distinct, we count each one exactly $k$ times. -For the lower bound, we use the probabilistic method. Motivated by the calculation above, let's choose a random sequence $A = (a_1,\ldots,a_k)$ by selecting each $a_j$ independently according to the distribution -$$\mathbb{P}(a_j = a) \approx \frac{(k-1)(n-a)^{k-2}}{n^{k-1}}.$$ -Discard the (few) sequences with repeated elements, and note that the expected value of $\sum a_j$ is $n$. We claim that the probability that $\sum a_j = n$ is roughly $1/(n \sqrt{k})$, and that each such sequence appears with probability at most -$$\left( \frac{k-1}{en} \right)^k.$$ -It follows that there are at least -$$\frac{1}{Cn \sqrt{k}} \left( \frac{en}{k-1} \right)^k$$ -such sequences. Dividing by $k!$ gives the desired bound on the number of sets.<|endoftext|> -TITLE: On the functoriality of scalar extensions of local rings (edited) -QUESTION [5 upvotes]: Note. I have edited my question to make it more transparent, following some very good comments that I received. I am sorry if it is a bit long. -A local homomorphism of local rings $(A,\mathfrak{m})\stackrel{\varphi}{\longrightarrow}(B,\mathfrak{n})$ is called a scalar extension (terminology due to Hans Schoutens) if: - -$\varphi(\mathfrak{m})B=\mathfrak{n}$, and -$\varphi$ is a flat extension. - -Let's fix a field $K$ (algebraically closed, if you wish) and let $\mathscr{C}_K$ be the category of Noetherian local rings whose residue field is a subfield of $K$, with morphisms being local homomorphisms. -Question A. Is there a functorial way of producing scalar extensions with a prescribed residue field? More precisely, is it possible to define a functor $F_K:\mathscr{C}_K\rightarrow\mathscr{C}_K$ in such a way that for every $A\in\mathscr{C}_K$ the local ring $F_K(A)$ is a scalar extension of $A$ with residue field $K$? -Here are some things that I know about this question: -(1) Grothendieck proved that scalar extensions with prescribed residue field always exist: -Theorem. (EGA III, Proposition 10.3.1, page 20). Let $(A,\mathfrak{m})$ be Noetherian local ring with residue field $k$, and let $K$ be a field extension of $k$. Then there exists a scalar extension $(A,\mathfrak{m})\stackrel{\varphi}{\longrightarrow}(B,\mathfrak{n})$ from $A$ to a Noetherian local ring $B$, with the property that $B/\mathfrak{n}$ is $k$-isomorphic to $K$. -Grothendieck's construction of the desired scalar extension depends on various 'choices' that he makes in his proof, and hence, does not produce a unique answer. For this reason I think it is hopeless to get a functor there. -(2) Various mathematicians have used a different method to construct scalar extensions with prescribed residue field, which seems 'more hopeful' to be functorial. In [b] (pp. 776-777) Kunz calls a special case of this construction the constant field extension. A version of this construction in the equicharacteristic case appears in [a] (pp. 18-19, 6.3). A more detailed description of this method can be found in [c] (pp. 4-7) and in [d] (pp. 36-38). I describe it in the equicharacteristic case: Given a local ring $(A,\mathfrak{m},k)$ and a field extension $K$ of $k$, take a coefficient field $k\hookrightarrow\hat{A}$ and complete $\hat{A}\otimes_kK$ with respect to the ideal $\mathfrak{m}(\hat{A}\otimes_kK)$. This is your $F_K(A)$. It is easy to see that this $F_K(A)$ is an scalar extension of $A$ with residue field $K$. ($F_K(A)$ depends on the choice of a coefficient field of $\hat{A}$, but is unique up to isomorphism). -Question B. Is the $F_K(\:\cdot\:)$ that was just described a functor from $\mathscr{C}_K$ to $\mathscr{C}_K$? To clarify the question, if $A_1\stackrel{\psi}{\longrightarrow} A_2$ is a local homomorphism of Noetherian local rings in $\mathscr{C}_K$, then does $\psi$ extend to a local homomorphism $B_1:=F_K(A_1)\rightarrow B_2:=F_K(A_2)$? -I can see how the method described in [c] provides an affirmative answer in equicharacteristic $0$ to Question B (it comes down to the fact that in equicharacteristic $0$ every maximal subfield of a complete local ring is a coefficient field) but I don't see how the method of [c] would still work in equicharacteristic $p>0$. I haven't checked the mixed characteristic case, yet, because I thought the equicharacteristic case is easier and if it cannot be settled positively, then there is even less hope for the mixed characteristic. -References. -a. M. Hochster and C. Huneke, $F$-regularity, test elements, and smooth base change, Trans. Amer. Math. Soc., 346 (1994). -b. E. Kunz, Characterizations of regular local rings of characteristic $p$, Amer. Jour. of Math., 41 (1969). -c. H. Schoutens, Classifying singularities up to analytic extensions of scalars, Ann. of Pure and Applied Logic, 162, (2011) (also available on the Arxiv, here). -d. H. Schoutens, The use of ultraproducts in commutative algebra, Lecture Notes in Mathematics, 1999, Speringer (2010). - -REPLY [4 votes]: I post this as an answer since it is too long, actually answers Question B and sheds some light on Question A. The example is taken from Eisenbud, Commutative Algebra, Exercise 7.17b. -Let $A_1=\mathbf F_p(t)$, $A_2=\mathbf F_p(u)[[x]]$, $\psi\colon A_1\to A_2$, $t\mapsto u^p+x$. On the residue fields, $\psi$ induces an isomorphism $\mathbf F_p(t)\cong\mathbf F_p(u^p)$. If $K$ is any extension field of $\mathbf F_p(u)$, the $F_K$ in Question B has $F_K(A_1)=K$ and $F_K(A_2)=K[[x]]$ with the obvious homomorphisms $A\to F_K(A)$. However, the diagram -$$\begin{array}{ccc} \mathbf F_p(t) & \xrightarrow{\psi} & \mathbf F_p(u)[[x]] \\ \downarrow && \downarrow \\ -K && K[[x]]\end{array}$$ -cannot be completed since $t$ becomes the $p$-th power $u^p$ in $K$, whereas it is mapped to $u^p+x$ in $K[[x]]$ which is not a $p$-th power. -As for Question A, depending on the exact interpretation, any $F_K$ should satisfy $F_K(k)=K$ for subfields $k\subset K$, so this example shows that $k[[x]]\to F_K(k[[x]])$ cannot just be the canonical homomorphism $k[[x]]\to K[[x]]$ in general. -EDIT: The above assumes, contrary to what you said in the comments, that the embedding of the residue field into $K$ is part of the data in $\mathscr C_K$. Otherwise, you can still apply the same argument if you assume $K$ algebraically closed (so that the image of $t$ is still a $p$-th power). However, I think that in this case even the simpler problem of choosing a natural embedding into $K$ for all fields in $\mathscr C_K$ is already impossible except in trivial cases.<|endoftext|> -TITLE: The motorcyclist's challenge -QUESTION [8 upvotes]: n walkers ${A}_{i}$ start out from X to Y simultaneously with speeds ${a}_{i}$, $i=1,2,...,n$. ${a}_{i}\neq {a}_{j}$ if $i\neq j$. At the same time, a motorcyclist M with speed $m=1$ starts out from Y to carry them (as shown in the illustration attached). The motorcyclist can pick up any person he meets. He can carry at most one person. But he's free to drop off the person he carries at any time. Walkers just walk straight forwards (from X to Y), while motorcyclist is free to drive either forwards or backwards. -The motorcylist's goal is to find a plan under which all the walkers arrive at Y simultaneously, if such a plan exists. We call such a plan a feasible plan. -Define the configuration of the problem to be the n-tuple $({a}_{1},{a}_{2},...,{a}_{n})\in {\mathbb{R}}_{++}^{n}$. -Define ${B}^{n}\subseteq{\mathbb{R}}_{++}^{n}$ to be the set of all feasible configurations (i.e., configurations for which a feasible plan exists). -Trivially, we know that ${B}^{1}={\mathbb{R}}_{++}$ and ${B}^{2}={\mathbb{R}}_{++}^{2}$. Also it's easy to see that ${B}^{n}\subseteq {(0,1)}^{n}$ for $n>2$. However, it is nontrivial to find ${B}^{n}$ for $n>2$. Without loss of generality, assuming ${a}_{1}<{a}_{2}<...<{a}_{n}$, I find that ${B}^{3}=\{({a}_{1},{a}_{2},{a}_{3})| 2{a}_{2}\leq{a}_{1}+1,\ {a}_{3}<1 \}$, which is neat! - -When I tried to figure out ${B}^{4}$, however, I found it to be significantly harder than ${B}^{3}$. So I have two related questions here: -1.How should we tackle ${B}^{n}$? Is there some systematic method we can use to characterize ${B}^{n}$ for successive n's? -2.If not, can we speculate what ${B}^{n}$ will look like, for example by giving some upper and lower bound on it which we are able to characterize? - - -EDIT: $n=4$ is EXHAUSTING, and honestly I have no idea where to start. Maybe it's just impractical to find ${B}^{n}$ for $n>3$. An interesting way to simplify the model is to assume that M can teleport himself anywhere he wants (as long as the motorcycle is not loaded). This assumption may allow a cleaner analysis and nicer answer. - -REPLY [2 votes]: Well, this is not really an answer. It is motivated by fedja's comment. In the original problem we do allow motorcyclist (M) to drive walkers backward. If not, I think I can say something more: -If driving walkers backward is not allowed, the best M can do is trying to successively drive slower walkers forward enough to correct positions before the fastest walker reach Y. Conceivably in a feasible configuration then, if he is visionary enough, M can figure out a plan in which he never goes forward unloaded (in this plan, when M placed someone in correct position, he immediately turns back to pick another walker). If this is true, then half of time M is empty (going backward), and half of time it's loaded (going forward). -Let $T$ be the total time lapse of this plan, $L$ be the distance between XY, ${t}_{i}$ be the time on the motorcycle for walker $i$, then we have: -$\sum{t}_{i}=T/2$ -${t}_{i}+(T-{t}_{i}){a}_{i}=L\ $ for $i=1,2,...,n$ -which is n+1 equations for n+1 unknowns. What is perhaps a little surprising about the plan is that the total time $T$ doesn't depend on whom M decides to pick first, and whom to pick later. Maybe this is not very relevant to my original questions, but I can't help remark on it.<|endoftext|> -TITLE: Can distribution theory be developed Riemann-free? -QUESTION [25 upvotes]: I imagine most people who frequent MO have been indoctrinated into the point of view that the Riemann integral can be safely discarded once one has taken the time to develop the Lebesgue integral. After all the two integrals agree more or less whenever they are both defined, and the Lebesgue theory is well known to be more robust and flexible in a lot of important ways. -However, I have recently encountered an apparent counter-example to the extreme view (which perhaps nobody actually holds) that the Riemann integral is entirely dispensable as a technical tool. The context is the theory of distributions. It is not uncommon that when one wants to generalize an operation from test functions to distributions that there are two natural choices: the operation can either be defined "directly" or by specifying how it pairs with test functions. Here are two basic examples: - -The first example involves the convolution of a distribution $F$ with a test function $\psi$. The direct definition is given by $F \ast \psi(x) = \langle F, \psi_x \rangle$ where $\psi_x(y) = \psi(x-y)$. The definition by pairing stipulates that for any test function $\phi$, $\langle F \ast \psi, \phi \rangle = \langle F, \phi \ast \psi_0 \rangle$. -The second example involves the Fourier transform of a (tempered) distribution $F$. The direct definition is given by $\hat{F}(\xi) = \langle F, e_\xi \rangle$ where $e_\xi(x) = e^{2 \pi i \xi x}$. The definition by pairing just sets $\langle \hat{F}, \psi \rangle = \langle F, \hat{\psi} \rangle$ for any appropriate test function $\psi$. - -In both of these examples, and others like them, all of the authors that I have consulted (including Folland and Taylor) prove that the direct definition agrees with the definition by pairing by carrying out a calculation with Riemann sums. -So I am left wondering if there decent proofs of these results for ordinary Lebesgue-abiding citizens. This question is a little problematic since the Lebesgue integral and the Riemann integral agree on the relevant space of functions, but if there isn't a good affirmative answer then it seems to me that there should be a convincing explanation why measure theoretic tools aren't strong enough to make the argument work. - -REPLY [2 votes]: In the fifties and sixties an elementary treatment of distributions which goes all the way up to the Fourier transform for tempered distributions and convolutions was developed by several mathematicians, in particular, J. Sebastião e Silva. Elementary means using only the tools of a univariate freshman analysis course (in the one dimensional case), and of a second term course for the multivariate case. Neither the Lebesgue nor the Riemann integral is required and there is no use of functional analysis (in particular, duality theory for non-normed locally convex spaces). The only facts about integration used are those concerning the integral of continuous functions on a compact interval and its simple properties. This can be developed (as in Dieudonné's texts) by first integrating step functions with intervals as sets of constancy and then extending to uniform limits thereof (of course, here you need the fact that continuous functions on compact intervals are uniformly continuous but it is hard to see how one can get any useful theory of analysis without this). -The starting point is to construct, for a compact interval $J$, a superspace $C^{-\infty}(J)$ of $C(J)$ and a linear operator $D$ thereon with the following properties: -1) if $x$ is continuously differentiable, then $Dx$ is the classical derivative; -2) for each distribution $x$, there exist an integer $n$ and a continuous function $X$ so that $x=D^n X$; -3) if $x$ is a distribution such that $D^n x= 0$, then $x$ is a polynomial of degree less then $n$. -Of course, it follows from the Schwartz theory that such objects exist but the point here is that one can construct a model using only the tools mentioned above. Furthermore, this system is categorical in the sense that there is only one model. This is important philosophically since it means that one can deduce from it anything that can be deduced from the Schwartz approach. -In order to construct such a model, one only requires the tools mentioned above (see below). It is then fairly standard to extend to the case of multivariate distributions and distributions on open subsets (rather than compact intervals). This has been carried out in detail by Sebastião e Silva, who went on to develop a theory of definite integrals, integrals of distributions depending on a parameter and hence the Fourier transform, and convolutions for -distributions, all at this elementary level. -Let me close with a brief description of his construction of $C^{-\infty}([0,1])$. -Let $I$ denote the operation which associates to each continuous function that primitive which vanishes at zero. Then one considers, for each $n$, the space of $n+1$-tuples of continuous functions $(x_0,....,x_n)$ and takes its quotient with respect to those families for which $I^n x_n + I^{n-1} x_{n-1} +... +x_0=0$. The required space of distributions is then the union of those over $n$. The differentation operator is the right shift. -This is, of course, a very cursory description but a precise, elementary and easily readable treatment with many examples can be found in the texts of Sebastião e Silva and these are readily available online (at jss100.campus.ciencias.ulisboa.pt , go to "Publicações", then to "Textos didacticos" vol. III, Theory of Distributions). -Apologies: I can't get the special portuguese symbols to work but the texts are in english by the way.<|endoftext|> -TITLE: Where can I find the divisor class groups of du Val singularities? -QUESTION [9 upvotes]: The du Val singularities are the simplest type of surface singularities. Each type of du Val singularity has a divisor class group. Specifically, let $X$ be a surface with an isolated singularity at $P$; then the (analytic or étale) local ring at P depends only on the type of the singularity, and has a divisor class group. -The most familiar example is the quadric cone (A1 singularity), found in many algebraic geometry textbooks. A line $L$ passing though the vertex of the cone is not locally principal, but $2L$ is, and we find that the divisor class group has order $2$. (Note: in general an A1 singularity will be étale locally, but not Zariski locally, isomorphic to the vertex of the cone. As far as I can see, there's no reason in general to expect the generator of the divisor class group to come from a divisor on the ambient surface; we may well have to pass to an étale (or analytic) neighbourhood.) -In a beautiful article, Lipman (Pub. Math. IHES 1969) studied these and computed the (finite) divisor class group of each du Val singularity. However, as far as I can see, he does not give explicit generators like we have in the example of the quadric cone. -So: - -Is there in the literature an explicit description (i.e. with explicit generators) - of the divisor class groups of the du Val singularities? - -REPLY [4 votes]: I do not know a reference, but here is my guess. I will use the notion of the wikipedia article. The order is: type, class group, the generator ideals. ($i^2=-1$, and I assume char. 0 for simplicity) -$A_n$, $\mathbb Z/(n+1)$, $(w+ix, y)$. -$D_n$ ($n$ even), $\mathbb Z/(2)\oplus \mathbb Z/(2)$, $(w,y), (w, x+iy^{(n-2)/2})$. -$D_n$ ($n$ odd), $\mathbb Z/(4)$, $(w,y)$. -$E_6$, $\mathbb Z/(3)$, $(x, w+iy^2)$. -$E_7$, $\mathbb Z/(2)$, $(w,x)$. -$E_8$, $0$.<|endoftext|> -TITLE: A Presentation for Rubik's cube group? -QUESTION [27 upvotes]: Let $G$ be Rubik's cube group. It is generated by the rotations by 90 degrees $L,R,D,U,F,B$ (left, right, down, up, front, behind), but what relations beyond $L^4=R^4=...=B^4=1$ do they satisfy? Thus I would like to know a presentation of the group as -$G = \langle L,R,D,U,F,B ~:~ ?\rangle$. -After playing aroumd I'have also found the relations $LR=RL$, $(LU)^{105}=1$, $(LRFB)^{12}=1$, $(LRFBFB)^4=1$, $(LRLRFBFB)^2=1$ (of course together with the symmetric relations). -From "The Mathematics of Rubik's cube" by W. D. Joyner I know that $G$ is generated by two elements and presentations are known, but I have not found one. Besides, I'm only interested in the standard generating set above. Remark that there is a well-known abstract group-theoretic description of $G$, it is the kernel of the homomorphism $(S_{12} \ltimes (\mathbb{Z}/2)^{12}) \times (S_{8} \ltimes (\mathbb{Z}/3)^{8}) \to \mathbb{Z}/2 \times \mathbb{Z}/2 \times \mathbb{Z}/3$ which maps $(a,x,b,y) \mapsto (\text{sign}(a) \text{sign}(b),\sum_i x_i,\sum_j y_j)$. - -REPLY [9 votes]: This discussion: https://web.archive.org/web/19990202074648/http://www.math.niu.edu/~rusin/known-math/95/rubik seems to culminate in a presentation (due to Dan Hoey). I did not read it carefully, I must admit. The presentation is quite complicated. For the 2x2x2 group there is this: -http://cubezzz.dyndns.org/drupal/?q=node/view/177<|endoftext|> -TITLE: Does the Zariski closure of a maximal subgroup remain maximal? -QUESTION [5 upvotes]: Let $k$ be an algebraically closed field and let $G\leq\rm{GL}_n(k)$ be a linear group. Assume that $M< G$ is a maximal subgroup (in the abstract group sense). Denote by $\bar{G}^Z$ the Zariski closure of $G$ in $\rm{GL}_n(k)$. Is it true that if $\bar{M}^Z\neq \bar{G}^Z$ then it is a maximal subgroup in the algebraic groups sense? If yes, would it be a maximal subgroup in the abstract group sense? -(I asked this question on MathSE five days ago and got no response, so I re-posted it here) -Thanks in advance for any help. - -REPLY [10 votes]: The answer is no. Assume that $G=\text{SO}(2,\mathbf{R})\ltimes\mathbf{R}^2\subset\text{GL}_3(\mathbf{C})$ and $M=\text{SO}(2,\mathbf{R})$. Then $M$ is maximal in $G$. However the Zariski closures are $\text{SO}(2,\mathbf{C})\subset\text{SO}(2,\mathbf{C})\ltimes\mathbf{C}^2$, so $\text{SO}(2,\mathbf{C})$ is not maximal in the algebraic sense because it stabilizes a line $L$ in $\mathbf{C}^2$ and is thus contained in $\text{SO}(2,\mathbf{C})\ltimes L$. -There are obvious similar examples with $G$ countable. However it seems more subtle to cook up an example with $G$ finitely generated. -(Edit: I understand the confusing assumption "$G\leq\rm{GL}_n(k)$ be a linear group" as "let $G$ be an arbitrary subgroup (in the abstract sense) of the group $\text{GL}_n(k)$.)<|endoftext|> -TITLE: Why are operads useful? -QUESTION [33 upvotes]: The question is not about where operads are used, I know that. It is about what makes them useful. For example, van Kampen diagrams are useful in combinatorial group theory because these are planar graphs and so one can use planar geometry (say, the Jordan lemma) to investigate the word problem in complicated groups. Similarly, asymptotic cones are useful in geometric group theory because they allow to study large scale properties of a discrete object (a group) by looking at small scale properties of a continuous object. I would like to know a similar answer for operads. - Update Many thanks to everybody for your answers. Unfortunately I can accept only one. So I just accept the first answer. - -REPLY [7 votes]: There surely are many answers to this question... For me, one of the key reasons is that there are lots of situations where existing geometric and algebraic structures exhibit some kind of associativity. (Geometrically, think of gluing pants, like in Jeff's comment, algebraically think of composing operations and/or cooperations.) The notion of an operad allows to formalise this observation, and treat objects like that as associative algebras in a certain monoidal category, - and since we know an awful lot of ways to approach usual associative algebras, this gives intuition of how to approach problems for these, more tricky objects. I think the analogy with your examples is quite clear. - -REPLY [2 votes]: I'm not an expert by the way I could give you an answer based on my personal experience in my study of category theory. - -Operads allow to make lots of constructions and to encode information of many mathematical object into an algebraic structure, this algebraic structure allows to work in a simpler way with the above mentioned objects. - -In pratical I think operads are similar to homotopy/homology groups, which encode homotopical information of topological spaces and enable to distinguish such spaces in a simple way, studying algebraic structures. This is useful because is more simple classifying groups rather then topological spaces. -More in general I think the usefulness of all such structures derive by the fact that usually (concrete) algebraic structures are easier to work with, but I emphasise that these are just my thoughts.<|endoftext|> -TITLE: Ordinary n-uple Points and Resolution of Singularities on a Surface -QUESTION [6 upvotes]: Let $X$ be an algebraic variety over some algebraically closed field $\Bbbk$ and let us assume $\dim(X)=2$, i.e. $X$ is an algebraic surface. -First, I would like to know the definition of an ordinary $n$-uple singular point on $X$, because in the Literature I know, it is only defined with respect to curves. Wolfram Mathworld has a definition, but $X$ does not always admit an embedding into $\mathbb{P}^3$, i.e. it is not necessarily defined by a single equation $f(x,y,z)$, so I am really not sure how this generalizes. -Second, I have been told that such singularities can be resolved by "blowing up once" - I would really like to know why that is, i.e. I am looking for a paper or textbook with this statement in it. If it is trivial to proove, of course, that request may be void. - -REPLY [4 votes]: A germ $(X, p)$ of isolated surface singularity is called an ordinary n-tuple point if -$$\hat{\mathcal{O}}_p=\mathbb{C}[[ x^n, x^{n-1}y, \ldots, xy^{n-1}, y^n]],$$ -see for instance Miyaoka's paper The maximal number of quotient singularities on surfaces with given numerical invariants , Section 1. -Equivalently, this can be seen as: -$\mathbf{1)}$ the singularity type (at its vertex) of the cone $C_n$ over the rational normal curve of degree $n$ in $\mathbb{P}^n$. In particular the embedding dimension is $n+1$, so it cannot be realized as a isolated singularity in $\mathbb{P}^3$ unless $n=2$ (ordinary double point); -$\mathbf{2})$ the cyclic quotient singularity $\frac{1}{n}(1,1)$, i.e. the quotient of $\mathbb{C}^2$ by the action of the group $\mathbf{Z}/n \mathbf{Z}$ given by $$\xi \cdot (x,\ y) \to (\xi x, \ \xi y),$$ where $\xi$ is a primitive $n$-th root of unity. In particular it is a rational singularity. -The fact that such singularities can be resolved by "blowing up once" is a standard computation. Or, if you prefer, just note that the blow up of the cone $C_n$ at its vertex is the Hirzebruch surface $\mathbf{F}_n$, which is smooth. This also shows that the minimal resolution of an ordinary $n$-tuple point is given by a smooth rational curve with self-intersection $-n$.<|endoftext|> -TITLE: Where to start with research regarding maslov index/class -QUESTION [5 upvotes]: Hi, -I am a physicist and currently doing my bachelor thesis about geometric quantization. -In the book by Bates and Weinstein I encountered the Maslov index, which seems to be very important :-). -But unfortunately my education didn't include anything in the direction of algebra beyond the scope of basic linear algebra. My present research about this topic showed that the Maslov class is an element of the integral cohomology of the manifold. But I couldn't find an introduction or something like this to 'integral cohomology' and I was lost in the big realm of cohomology. -(For example: What is the difference of de Rham cohomology, Cech cohomology, Cech - de Rham cohomology and the needed integral cohomogogy). -So could someone provide me with a "path" along the topics I have to study to be able to understand Maslov classes. -Thanks in advance, -Tobias! - -REPLY [3 votes]: There are different incarnation of the Maslov index. The one that I prefer is the one proposed in Arnold's paper suggested by Igor Rivin. The paper by Cappell-Lee-Miller suggested by Greg Friedman is also an excellent source. (These two papers helped me understand this concept but they addressed primarilty to a mathematical audience.) -Maslov introduced his index in his investigation of asymptotics of certain oscillatory appearing in quantization problems. I suspect this is closest to what had in mind. It is sometime known as the Hormander index. Section 3.4 of Duistermaat's book Fourier Integral Operators has a rather efficient description of the Maslov index. As an aside, the operators introduced and investigated by Maslov are special examples of Fourier integral operators.<|endoftext|> -TITLE: What is the role of contact geometry in the hamiltonian mechanics? -QUESTION [20 upvotes]: Let us assume someone is interested in the study of Hamiltonian mechanics. -What are good examples to illustrate him of the usefulness of contact geometry in this context? -On one hand the Hamiltonian mechanics was time ago expressed in the language of symplectic geometry, but, on the other hand, the contact geometry is often presented like the brother of the symplectic geometry. -My question is: - -In the hamiltonian mechanics, not necessarily only for Hamiltonian of mechanical type, what is the role played by the contact geometry? - -Any kind of suggestion is welcome. - -REPLY [8 votes]: Form the contact 1-form $\Theta = p \, dq -H \, dt$ on extended phase space $T^* Q \times {\mathbb R}$, the second factor being time and parameterized by $t$, the function $H = H(q,p,t)$ being the time-dependent energy -or Hamiltonian, and $p \, dq$ denotes the usual canonical one-form on $T^* Q$ pulled back to extended phase space by the projection onto the first factor. -(Assume $H \ne 0$ to get $\Theta$ contact.) Then the Reeb vector field for this 1-form, i.e the kernel of $d \Theta$, is the time-dependent -Hamiltonian vector field, up to scale. -Arnol'd has a nice discussion of this in his Mathematical Methods in Classical Mechanics.<|endoftext|> -TITLE: Conjugacy for p-adic matrices of finite order II -QUESTION [12 upvotes]: Question: Say $p$ is an odd prime, and take two matrices $A,B\in GL_n({\mathbb Z}_p)$ of finite order $m$. Is it true that if their reductions mod $p$ are conjugate in $GL_n({\mathbb F}_p)$ then they are conjugate in $GL_n({\mathbb Q}_p)$? - -This is a follow-up to the question whether $GL_n({\mathbb F}_p)$-conjugacy implies $GL_n({\mathbb Z}_p)$-conjugacy, for which the answer is "No". -For $p=2$ this is not true, although I would be very much interested to know whether conjugacy mod 4 implies conjugacy over ${\mathbb Q}_2$. -(Note that if the answer is "Yes", then by character theory any two representations $\rho, \rho': G\to GL_n({\mathbb Z}_p)$ that are equivalent mod $p$ are equivalent in $GL_n({\mathbb Q}_p)$.) - -REPLY [3 votes]: Edit: Sorry, just realized this is in in the wrong direction. What was asked for was an example of matrices over $\mathbb Z_p$ which are conjugate over $\mathbb F_p$ but not over $\mathbb Q_p$; I swapped $\mathbb Q_p$ and $\mathbb F_p$, but then of course the assertion is clear anyway (I'll look at it again later). - -After looking at Geoff's answer and §34C in "Curtis and Reiner: Methods of Representation Theory, Volume I" which deals with integral representations of $C_{p^2}$, I came up with a counterexample (I would recommend to use GAP or Maple to verify it; doing the computations by hand would be insane). -So, here is the counterexample: Consider the following two matrices in $\mathbb Q_3^{10\times 10}$: -$$ -A=\left(\begin{array}{rrrrrrrrrr} -1&0&0&0&1&0&0&0&0&0\newline -0&0&0&1&1&0&0&0&0&0\newline -0&1&0&0&1&0&0&0&0&0\newline -0&0&1&0&1&0&0&0&0&0\newline -0&0&0&0&0&0&0&0&0&-1\newline -0&0&0&0&1&0&0&0&0&0\newline -0&0&0&0&0&1&0&0&0&0\newline -0&0&0&0&0&0&1&0&0&-1\newline -0&0&0&0&0&0&0&1&0&0\newline -0&0&0&0&0&0&0&0&1&0\newline -\end{array}\right) -$$ -and -$$ -B = \left(\begin{array}{rrrrrrrrrr}% -1&0&0&0&0&0&0&0&0&0\newline -0&0&0&1&0&0&0&0&0&0\newline -0&1&0&0&0&0&0&0&0&0\newline -0&0&1&0&0&0&0&0&0&0\newline -0&0&0&0&0&0&0&0&0&-1\newline -0&0&0&0&1&0&0&0&0&0\newline -0&0&0&0&0&1&0&0&0&0\newline -0&0&0&0&0&0&1&0&0&-1\newline -0&0&0&0&0&0&0&1&0&0\newline -0&0&0&0&0&0&0&0&1&0\newline -\end{array}\right) -$$ -Then $A$ and $B$ are conjugate over $\mathbb Q_3$ but their reductions to $\mathbb F_3$ are not conjugate. Moreover, both have finite order (their order is $9$). -To see that their reductions mod 3 are not conjugate just compute the rank of -$A-\textrm{id}$ and $B-\textrm{id}$ in $\mathbb F_3^{10\times 10}$. The rank is 8 in the first case and $7$ in the second, so they cannot be conjugate. -To see that they are conjugate over $\mathbb Q_3$ just compute minimal and characteristic polynomial. The minimal polynomial is $x^9-1$ in both cases, which implies that the matrices are semisimple. Therefore they are conjugate if and only if their characteristic polynomials coincide. But the characteristic polynomial is $(x-1)(x^9-1)$ in both cases.<|endoftext|> -TITLE: Why are operads so closely connected to mathematical physics? -QUESTION [10 upvotes]: Mark Sapir's question inspired me to ask the question in the title. A lot of mathematicians who have done work related to mathematical physics (e.g Kontsevich, Stasheff, Getzler, Manin, etc.) have done work with operads, but I don't really grok why operads have anything to do with physics. I wonder if anyone has ideas about why there is such a close connection. -Added: Examples of how operads are used in physics are welcome, but just like in Mark Sapir's question, I am much more interested in a general reason that explains why they appear. Especially why multiple operads appear, not some particular operad. Note that operads can be defined in various categories (topological spaces, vector spaces, etc) and I'm most interested in operads in the category of vector spaces. But if there is an answer that covers more than one category, that would be very nice. - -REPLY [9 votes]: I'm not a mathematical physicist, so parts of this may be wrong. -In quantum field theory, one encounters operators that are supported at points in spacetime, or at least are very local. For example (if we ignore uncertainty), one may have a "photon creation at $x$ with momentum $p$ and helicity $\xi$" operator, which changes the universe so that there is an extra photon at $x$ with momentum $p$ and helicity $\xi$. If you have encountered raising and lowering operators when studying the quantum mechanics of a harmonic oscillator, this is a generalization to the situation where you have harmonic oscillators at every point (and in interacting theories, the oscillators may coupled nonlinearly). -If we set up multiple operators supported at different points, there may be no canonical way to compose them. For example, if we have creations at spacelike separated points, any choice of time-ordering can be permuted by a boost. However, we can consider a parameter space of all possible ways to compose such operators, and we can consider how the compositions change when we continuously vary the positions of the creation events in spacetime. Furthermore, if we put a subset of such creations close to each other, we may view their composition as a single operator supported on the neighborhood, that depends on their relative position inside that neighborhood. Operads are precisely the objects that encode such families of composition laws and iterations. -One example where these arise is in the theory of vertex operator algebras, which (so I'm told) describe the chiral parts of conformal field theories. Given an compact complex algebraic curve and a vertex operator algebra, you can choose points with frames, and insert states (i.e., apply certain operators) at those points. There is a procedure by which you can construct from these data a space of chiral correlation functions, and this space will vary smoothly with the points and frames (and also with the complex moduli of the curve). If you put some points with frames close to each other and ignore the global geometry of the curve, the configurations of points with frames form a space in the framed $E_2$ operad, and the spaces of correlation functions are something like an algebra over the operad. This is a genus zero restriction of the compatibilities we see in conformal field theory, and can be strengthened to the statement (proved here under some extra hypotheses) that modules of a vertex operator algebra form a framed $E_2$ (in particular, braided) tensor category. - -REPLY [8 votes]: It's not really possible to give a precise answer to this question, so I apologize for being vague here. One answer is because a lot of multiplications in physics are associated with moving two things close to each other and looking at the result as a single object. The archetype for this is the operator product expansion in quantum field theory. This naturally leads to thinking about a little n-spheres operad with various decorations. Most examples that come to mind right now really reduce to that.<|endoftext|> -TITLE: Mazur-Tate-Teitelbaum p-adic L-function -QUESTION [7 upvotes]: I start to read the paper "On p-adic analogues of the conjectures of Birch and Swinnerton-Dyer" by Mazur, Tate and Teitelbaum (referred as [MTT]) to learn how we can associate p-adic L-function to certain eigenforms. -For an eigenform $f$ (with certain conditions) and a choice of the root $\alpha$, the authors define the $V_f$-valued measure $\mu_{f,\alpha}$ in page 13. Here $V_f=C_p\otimes_\bar{Q}L_f\bar{Q}$ is a finite dimensional $C_p$-vector space -defined in line 5 of page 13. And $L_f\bar{Q}$ (I think) is the $\bar{Q}$-vector subspace of $C$ generated by elements of $L_f$, where $L_f$ is defined right before the proposition in page 6. -Finally, they define the p-adic L-function - $$L_p(f,\alpha,\chi,s):=L_p(f,\alpha,\chi\chi_s)$$ -as in page 19. -Apparently, this function has values in the $C_p$-vector space $V_f$. But in the rest of the paper, it seems to me that they treat this as a $C_p$-valued function. On the other hand, other references, such as Greenberg-Stevens' paper in Inventiones 1993, and the earlier paper ``Arithmetic of Weil Curvess'' by Mazur--Swinnerton-Dyer use the two periods $\Omega_f^{\pm}$ to get out two measures $\mu_f^{\pm}$. Then use them to define a $C_p$-valued function. -Question: do we need to change the functions in [MTT] into $C_p$-valued functions, say, by using the above periods as in Greenberg-Stevens? - -REPLY [7 votes]: In order to get $\mathbb{C}_p$-valued functions, you need to choose a basis for the vector space $V_f$. If $f$ corresponds to an elliptic curve, there is a reasonably canonical way of doing this (using the periods of a Neron differential), as in the paper by Mazur and Swinnerton-Dyer. If $f$ is a more general modular form it is much less clear what the "right" normalisation is for a basis of $V_f$. This is an important issue, though, because one needs to fix such a normalisation to make sense of whether or not two L-values are congruent modulo a prime. The question has been studied in detail by Vinayak Vatsal in his paper "Canonical periods and congruence formulae" (Duke Math Journal 98 no. 2, 1999), which determines a canonical normalisation up to p-adic units.<|endoftext|> -TITLE: Inverse of the Riemann zeta function -QUESTION [8 upvotes]: I'm wondering if there is any information on the inverse of the Riemann zeta function (not it's reciprocal, but its functional inverse). This would obviously be a multi-valued function. - -REPLY [3 votes]: The question is about the value distribution of $\zeta(s)$; it is considered (without speaking of inverse) in some detail in Chapter XI of Titchmarsh's book.<|endoftext|> -TITLE: Asymptotics for primality of sum of three consecutive primes -QUESTION [8 upvotes]: We consider the sequence $R_n=p_n+p_{n+1}+p_{n+2}$, where $\{p_i\}$ is the prime number sequence, with $p_0=2$, $p_1=3$, $p_2=5$, etc.. -The first few values of $R_n$ are: -10, 15, 23, 31, 41, 49, 59, 71, 83, 97, 109, 121, 131, 143, 159, 173, 187, 199, 211, 223, 235, 251, 269, 287, 301, 311, 319, 329, 349, 371, 395, 407, 425, 439, 457, ... -Now, we define $R(n)$ to be the number of prime numbers in the set $\{R_0, R_1 , \dots , R_n\}$. -What I have found (without justification) is that $R(n) \approx \frac{2n}{\ln (n)}$. -My lack of programming skills, however, prevents me from checking further numerical examples. I was wondering if anyone here had any ideas as to how to prove this assertion. -As a parting statement, I bring up a quote from Gauss, which I feel describes many conjectures regarding prime numbers: -"I confess that Fermat's Theorem as an isolated proposition has very little interest for me, because I could easily lay down a multitude of such propositions, which one could neither prove nor dispose of." - -REPLY [6 votes]: I verified Noam's calculation of the factor $\lambda=2.30096\ldots$ and Álvaro's computations, extended the latter and calculated the corresponding ratios: -$$ -\begin{array}{|c|c|c|c|c|} -n & R(n) & 2n/\log n & \lambda n / \log n & R(n)\log n/n\\\\ -\hline -10 & 7 & 9 & 10 & 1.61181\\\\ -100 & 44 & 43 & 50 & 2.02627\\\\ -1000 & 339 & 290 & 333 & 2.34173\\\\ -10000 & 2437 & 2171 & 2498 & 2.24456\\\\ -100000 & 18892 & 17372 & 19986 & 2.17502\\\\ -1000000 & 157183 & 144765 & 166549 & 2.17156\\\\ -10000000 & 1346797 & 1240841 & 1427564 & 2.17078\\\\ -30000000 & 3784831 & 3484987 & 4009410 & 2.17208\\\\ -\end{array} -$$ -(The values in the third and fourth columns are rounded to the nearest integer, the values in the last column are rounded to 5 digits after the decimal point.) -I don't think we can deduce anything from the ratio in this form, however, since it shows convergence in the "random" fluctuations but not with respect to the asymptotic approximations made, e.g. dropping a term $\log\log n$, which at this stage is still comparable to $\log n$; a more detailed analysis will be required to test the independence hypothesis in this case. -[Update:] With reference to Noam's comments below, here are some data for the relative frequencies of the sum of three consecutive primes being divisible by the first four odd primes. These are averaged over samples of $400,000$ primes beginning at powers of ten, which are given in the first column; note that these refer to the numbers $x$ themselves, not the indices $n$ of the primes. -$$ -\begin{array}{|c|c|c|c|c|} -\log_{10}x&3&5&7&11\\\\ -\hline -8 &0.183&0.165&0.130&0.087\\\\ -9 &0.189&0.169&0.131&0.087\\\\ -10&0.195&0.170&0.133&0.087\\\\ -11&0.198&0.172&0.133&0.088\\\\ -12&0.203&0.173&0.133&0.088\\\\ -13&0.208&0.175&0.134&0.087\\\\ -\hline -\text{limit?}&0.250&0.188&0.139&0.090 -\end{array} -$$ -I also looked at the joint distribution of the residues modulo $3$ for the three primes. There's a significant preference for avoiding repeated residues; for instance, at $x=10^9$, the repeating patterns $1,1,1$ and $2,2,2$ have relative frequencies around $0.095$, the alternating patterns $1,2,1$ and $2,1,2$ have relative frequencies around $0.150$, and the remaining mixed patterns have relative frequencies around $0.128$, which is almost completely explained by $1,1$ and $2,2$ having relative frequencies $0.445$ and $1,2$ and $2,1$ having relative frequencies $0.555$. I'm trying to work out a probabilistic model for these effects.<|endoftext|> -TITLE: What is the topology on hom-sets of spectra? -QUESTION [6 upvotes]: In Segal's paper on $\Gamma$-spaces, he gives a functor $Spectra \rightarrow \Gamma-Spaces$ defined by taking a functor $E$ and sending it to the $\Gamma$-space $AE$ with $AE(n) = Mor(S \times \cdots \times S, E)$, where $S$ is the sphere spectrum. Now, since this is supposed to define a $\Gamma$-space, in particular the sets $Mor(S \times \cdots \times S, E)$ should be topological spaces... but they don't seem to come with any obvious topology, at least not obvious to me. -On the other hand, it seems like there should be some sort of spectrum that acts like $Mor(S \times \cdots \times S, E)$; could he mean, possibly, the 0th space of this spectrum? -EDIT: The reference is -Segal, Graeme Categories and cohomology theories. Topology 13 (1974), 293--312 - -REPLY [9 votes]: If $X = (X_n)$ and $Y = (Y_n)$ are spectra, one can define a morphism just to be a collection of maps $x_n \to Y_n$ commuting with the suspensions. Thus the set of morphisms between $X$ and $Y$ is a subset of $\prod_n Map(X_n,Y_n)$ - and we give it just the subspace topology. -An alternative way is the simplicial set approach: we define an $n$-simplex of the mapping space between $X$ and $Y$ to be a map $X\wedge \Delta[n]^+ \to Y$. If we want a topological space back, one can geometrically realize. -If you want mapping spectra, it is perhaps more reasonable to go to symmetric spectra. You can find an exposition of mapping spectra (and also of mapping spaces) in Stefan Schwede's book project on symmetric spectra: http://www.math.uni-bonn.de/people/schwede/SymSpec.pdf 2.24 & 2.25. -If you're interested in the relationship between Gamma-spaces and spectra from a homotopical view, you might also be interested in the Bousfield-Friedlander paper: http://club.pdmi.ras.ru/~topology/books/bousfield-friedlander.pdf<|endoftext|> -TITLE: Mercer's Theorem: uniform $L_\infty$ bound on eigenfunctions? -QUESTION [6 upvotes]: I recently came across a statement of Mercer's theorem in Hermann Koenig's book: Eigenvalue distribution of compact operators. It is interesting that in addition to the usual statement of Mercer's theorem (uniform convergence of kernel in basis of eigenfunctions for continuous kernels on bounded domains) it states that the eigenfunctions are uniformly bounded in the supremum norm (with eigenfunctions normalized in the two norm). His proof does not address this point directly. Does anybody know if this is true and if so how to show a uniform bound on the eigenfunctions $f_n(x)$ over both space $x$ and the index $n$? - -REPLY [7 votes]: This is not true as stated. For example, on compact (connected) Lie groups, or homogeneous spaces for them, such as spheres, the ratio of sup-norm to $L^2$-norm on an eigenspace for Laplacian/Casimir is proportional to the square root of the dimension of the eigenspace. (The standard argument for spherical harmonics, as given, e.g., in Stein-Weiss, generalizes.) Thus, when the multiplicities grow, as they do, polynomially, the ratio grows, to say the least. -In some relatively exotic situations, sharp understanding of such a comparison would prove very serious things, like the Lindelof Hypothesis on zeta and L-functions. :) -Hormander, Seeger-Sogge, and others have made extensive studies of the Laplacian on Riemannian manifolds... On compact Riemannian manifolds, the resolvent is compact, so fits into the question here. -Nevertheless, certainly, one hopes to understand such relationships!<|endoftext|> -TITLE: Rank of quaternionic matrix -QUESTION [7 upvotes]: Is there a method of founding of rank of quaternionic matrix by Dieudonne's determinants ? - -REPLY [4 votes]: I think that the method is exactly the same as in the commutative case. The rank of a quaternionic matrix (or more generally, over any not necessarily commutative field) is equal to the maximal size of a minor with non-zero Dieudonne determinant. The proof of this fact is essentially the same as in the commutative case. The key point is: given a square quaternionic matrix, its rows (equivalently, columns) are linearly independent if and only if the Dieudonne determinant does not vanish. -Unfortunaly I do not have a reference for the answer to your question. But non-commutative linear algebra and Dieudonne determinants are discussed in the book "Geometric algebra" by -E. Artin.<|endoftext|> -TITLE: Translation functors for category $\mathcal O$ -QUESTION [6 upvotes]: Let $\mathfrak g = \mathfrak n^- \oplus \mathfrak h \oplus \mathfrak n$ be a semisimple Lie algebra over $\mathbb C$ and let $\mathcal U$ be its enveloping algebra. Category $\mathcal O$ is the category of finitely generated $\mathcal U(\mathfrak h)$-semisimple modules such that the action of $\mathcal U(\mathfrak n)$ is locally finite. (In particular, finite dimensional modules are in category $\mathcal O$.) One can show that $\mathcal O$ is a direct sum of subcategories $\mathcal O_{\chi_\lambda}$, each of which consists of modules where the center $Z(\mathfrak g)$ of $\mathcal U$ acts by a fixed (EDIT: generalized) character $\chi_\lambda:Z(\mathfrak g) \to \mathbb C$. -If $L$ is a finite dimensional module, then we can define a shift functor $T_\lambda^\mu: \mathcal O_{\chi_\lambda} \to \mathcal O_{\chi_\mu}$ by the formula $T_\lambda^\mu(M) = pr_\mu\circ (L \otimes M)$. (Here we wrote $pr_\mu$ for the projection $\mathcal O \to \mathcal O_{\chi_\mu}$.) These translation functors have many nice properties and in certain cases are nice enough to be equivalences of categories. However, in general their behavior depends quite subtly on the geometry of the root system and on $\mu, \lambda$. -I had a reference request and a naive question: - -1) Are there other definitions of translation functors, perhaps ones that behave more simply? -2) Are there simple sufficient conditions which show that two blocks $\mathcal O_{\chi_\lambda}$ and $\mathcal O_{\chi_\mu}$ are not equivalent? In particular, if $\lambda - \mu \in \Lambda_r$ (the root lattice) is it possible for $\mathcal O_{\chi_\lambda}$ and $\mathcal O_{\chi_\mu}$ to be in-equivalent? - -(Retag as appropriate.) -EDIT: As Victor kindly pointed out, $\mathcal O_{\chi_\lambda}$ consists of modules where the endomorphisms $z-\chi_\lambda(z)$ are nilpotent, not necessarily 0. Also, to attempt to clarify (1), I was just wondering if there are natural functors between different blocks of $\mathcal O$ that don't come from tensoring with a finite dimensional module. - -REPLY [6 votes]: It's probably relevant to note that the translation functors are shadows of "obvious" functors: the categories of $\lambda$- and $\lambda+\nu$-twisted $D$-modules on the flag varieties are equivalent for any $\lambda$ and any integral $\nu$, with the functor being simply tensor product by the line bundle $L_{\nu}$. (Thus the category of twisted D-modules depends on a parameter $exp(\lambda)$ in the dual torus GROUP, not just Lie algebra). The translation functors result from the fact that the global sections from D-modules to representations does not commute with these obvious equivalences (eg global sections is sometimes an equivalence to representations of the associated central character and sometimes not, though it always has both left and right adjoints which we can use to go up and down).<|endoftext|> -TITLE: Solving for Moore Penrose pseudo inverse -QUESTION [5 upvotes]: I have a system to solve, set up as : -$$Ax = b$$ -with a square rank deficient matrix $A$. The paper suggests to use a Moore Penrose pseudo inverse, which in my case can be computed using the traditional inverse : -$$ A^+ = (A+\frac{ee^T}{n})^{-1} - \frac{ee^T}{n} $$ -where $e$ is a vector containing only ones, and $n$ is the dimension of the matrix. This matrix comes from the solution of a Multidimensional Scaling problem using the SMACOF method (the Guttman transform). -However, in my case, my matrix $A$ is very sparse (and rank deficient) : what method can I use to efficiently solve the original system without making my matrix dense (as would be the case by using an SVD, by using the above formula for the pseudo inverse, by computing $A^TA$ or by QR factorization) ? -$A$ is also symmetric, has a positive diagonal, and the other values are either -1 or 0, and such as the sum of each row (resp. column) is 0. -Preferably, since I'll need to solve for multiple right hand sides with this same matrix, I would like to avoid performing the resolution from scratch for each right hand side. I would also like to get exactly the same result as the one obtained with the Moore Penrose pseudoinverse. -Thanks. - -REPLY [3 votes]: Are you looking for a fast practical method, or a fast theoretical method? If the former, there are very fast solvers based on sparse Cholesky or sparse LDL decomposition (both of which can be reused for many $b$). You should check out Tim Davis' beautiful book called something like "Sparse direct solvers". -There are very few theoretical bounds, since the running time depends heavily on the sparsity pattern.<|endoftext|> -TITLE: Number theory and NP-complete -QUESTION [6 upvotes]: What are some of the natural number theory problems that are np-complete? I am looking for examples not in lattices and geometric number theory. Examples in analytic/algebraic number theory are ok. - -REPLY [15 votes]: You can take a look at the papers by Adleman and Manders (not always in this order) from the 70s (at least "Computational complexity of decision procedures for polynomials", "NP-complete decision problems for quadratic polynomials", "Diophantine complexity"), and the references therein. -One example of the problems they show to be NP-complete is the following decision problem. - -Input: three natural numbers $a,b,c$ (in base $2$), -Output: YES/NO depending whether the equation $ax^2 + by = c$ has solutions (for $x,y$) in the natural numbers. - -REPLY [6 votes]: See pages 249-251 of Garey and Johnson, Computers and Intractability, for a dozen NP-complete problems in Number Theory. -EDIT: A couple of examples, by request. -AN2, Simultaneous incongruences. Given a collection $\lbrace(a_1,b_1),\dots,(a_n,b_n)\rbrace$ of ordered pairs of positive integers with $a_i\le b_i$ for $1\le i\le n$, is there an integer $x$ such that for all $i$, $x\not\equiv a_i\pmod{b_i}$? -AN4, Comparative divisibility. Given sequences $a_1,a_2,\dots,a_n$ and $b_1,b_2,\dots,b_m$ of positive integers, is there a positive integer $c$ such that the number of $i$ for which $c$ divides $a_i$ is more than the number -of $j$ for which $c$ divides $b_j$?<|endoftext|> -TITLE: Categoricity in second order logic -QUESTION [10 upvotes]: Hi, -It's shown by an easy cardinality argument that there are complete second-order theories that are not categorical (have more than one model up to isomorphism). Anyone knows of a concrete example of such a theory? -Thanks in advance - -REPLY [14 votes]: A theory of the type you are asking cannot be that concrete, because: - - -By an old result of Victor Marek, it is consistent with the axioms of $ZFC$ that the second order theory of every countable structure (in a countable vocabulary) is categorical. See this FOM-post of mine for a reference. -In the above FOM-post, I conjectured that Marek's result can be extended to all Borel structures. This conjecture was verified by Harvey Friedman in this FOM-post. -In yet another FOM-post, Solovay showed that it is consistent with $ZFC$ that as soon as a second order theory $T$ is both axiomatizable and complete, then $T$ is categorical. See also this other related FOM-post of Solovay.<|endoftext|> -TITLE: If C is a fusion category over a field of nonzero characteristic and dim C = 0, is Z(C) ever fusion? -QUESTION [7 upvotes]: If $C$ is a fusion category and $\dim(C) \neq 0$ (the latter is automatic in characteristic zero, but not in nonzero characteristic), then the Drinfel'd center $Z(C)$ is fusion. More generally, if $C$ is a fusion category and $M$ is a semisimple module category over $C$, then the dual of $C$ over $M$ is fusion if $\dim(C)$ is nonzero. But if $\dim(C)=0$ these results need not hold. For example $\operatorname{Vec}(G)$ in characteristic $p$ when $p | \#G$ acts on $\operatorname{Vec}$ and its dual is $\operatorname{Rep}(G)$ which is not semisimple. Similarly, $Z(\operatorname{Vec}(G))$ is not semisimple when $p | \#G$. -I want to know if $C$ and $Z(C)$ both being semisimple implies that $\dim(C) \neq 0$. -(Feel free to assume everywhere that the base field is algebraically closed if you'd like to.) - -REPLY [6 votes]: We eventually sorted this out, and it appears as (one direction of) Theorem 3.6.7. in Dualizable Tensor Categories (joint with Christopher Douglas and Chris Schommer-Pries). Note that (for C semisimple) separability is equivalent to semisimplicity of Z(C) (see Corollary 3.5.9.), so Theorem 3.6.7 does address exactly this question.<|endoftext|> -TITLE: The topology of open semi-algebraic sets (appl.: totally positive matrices) -QUESTION [8 upvotes]: Let $P_1,\ldots,P_r$ be polynomials over ${\mathbb R}^N$. I am interested in the homotopy type of the semi-algebraic set defined by -$$ P_j(x_1,\ldots,x_N)>0,\qquad j=1,\ldots,r . $$ - -Is there a general theory about that? - -Here are motivating examples: - -If $r=1$ and $P$ is a homogeneous polynomial, hyperbolic in the direction of some vector $e\ne0$, the connected component of $e$ in $P>0$ is convex (Garding) and therefore has a trivial topology. -If ${\mathbb R}^N=M_n({\mathbb R})$, $r=1$ and $P(M)=\det M$, then $P>0$ is $GL_n^+({\mathbb R})$. Thanks to the polar decomposition, it is homeomorphic to $SO_n({\mathbb R})\times SDP_n$ and its homotopy type is that of $SO_n({\mathbb R})$. For instance, the fundamental group is ${\mathbb Z}$ if $n=2$ and ${\mathbb Z}/2{\mathbb Z}$ if $n\ge3$. -If ${\mathbb R}^N=M_n({\mathbb R})$, $r=n$ and $P_j$ is the $j$th principal minor, then the set $X$ defined by all $P_j(M)>0$ is homeomorphic to $L_n\times U_n^+$, where $L_n$ is the set of lower triangular matrices with unit diagonal and $U_n$ consists of upper triangular matrices with strictly positive diagonal ($LU$ factorization). The homotopy type of $X$ is thus trivial. - - -I am interested in the special case of the set $TP_n$ of totally positive $n\times n$ matrices. It is defined by the inequalities involving minors - $$ M\begin{pmatrix} i_1 & \ldots & i_s \\ j_1 & \ldots & j_s \end{pmatrix} > 0 $$ - for every $s=1,\ldots,n$ and every increasing sequences $k\mapsto i_k$ and $k\mapsto j_k$. - -I point out that this latter question can be reduced to that of the homotopy type of appropriate subsets $\ell_n$ and $u_n^+$ of $L_n$ and $U_n^+$, defined by the minor equalities above with either the restriction $i_s\le j_1$ (upper triangular case) or $j_s\le i_1$ (lower triangular case). It is known that $TP_n=\ell_n\cdot u_n^+$. - -REPLY [3 votes]: The homotopy groups of the semigroup of determinant one totally positive matrices are all trivial. A proof of this is in the paper (example 6.2) -A. J. Santana and L.A.B. San Martin: The homotopy type of Lie semigroups in semi-simple Lie groups. Monatshefte fur Mathematik, v. 136, n. 2, p. 151-173, 2002. -More generally in this paper it is proved (Theorem 4.15) that the homotopy type of a Lie semigroup in a semi-simple Lie group is that of a compact subgroup. The theorem is proved under an assumption of existence of a "large" subsemigroup necely generated by exponentials. It applies also e.g. to the semigroup of nxn matrices with positive entries and determinant one. This semigroup has the homotopy type of the compact group SO(n-1).<|endoftext|> -TITLE: How to compute KL-divergence when PMF contains 0s? -QUESTION [19 upvotes]: From the Wikipedia page on Kullback-Leibler divergence, the way to compute this metric is to utilize the following formula: - -The way I understand this is to compute the PMFs of two given sample sets and then use the above formula to compute the KL-divergence. I am not quite sure I am getting this though. Let us say I have the following set of values: -0 0 0 0 0 1 1 0 0 0 0 0 1 2 0 0 0 0 0 20 0 0 0 - -I am trying to compute the KL-divergence for a sliding window size of 7. So, I start of by choosing the bin size for my histogram as 1 and (max, min) values as (20, 0) based on the entire dataset. So that when I have the sliding window as follows: -0 0 0 0 0 1 1 0 0 0 0 0 1 2 0 0 0 0 0 20 0 0 0 -|-----------| - |-----------| - -I compute the PMFs across these two sliding windows as: -0 0 0 0 0 1: 0.81818182, 0.18181818 0 0 ... 0 -0 0 0 0 1 2: 0.66666667, 0.16666667, 0.16666667 0 0 ... 0 - -Both these PMFs satisfy the first condition from the Wiki page: - -The K-L divergence is only defined if P and Q both sum to 1 and if - Q(i) > 0 for any i such that P(i) > 0. - -but failing with the second condition. Can someone please tell me where I am going wrong? - -REPLY [24 votes]: I'll give a short answer and a long answer. -The short answer is that the KL divergence on multinomials is defined when they have only nonzero entries. When there are zero entries, you have two choices. (1) Smooth the -distributions in some way, for instance with a Bayesian prior, or (similarly) taking the convex combination of the observation with some valid (nonzero) distribution. (The standard Bayesian approach is to use a Dirichlet prior, which amounts to treating each entry as a fraction $n_i / m$ where $m=\sum_i n_i$, and $n_i$ should be integer (but with your provided data this may get messy), and replacing these fractions with for instance $(n_i + 1) / (m+|x|)$ where $|x|$ is the number of atoms in the discrete distribution; the "convex combination" smoothing choice is similar, if $x$ is your observation and $\alpha \in (0,1]$, return $\alpha U_{|x|} + (1-\alpha)x$ where $U_{|X|}$ is the uniform distribution on $|x|$ points.) (2) Employ heuristics throwing out all the values that do not make sense, as suggested above. While I acknowledge that (2) is a convention, it doesn't really fit with the nature of these distributions, as I will explain momentarily. -The longer answer is the mathematical reason why KL divergence can't handle these zeros, which requires information about Exponential family distributions (multinomial, gaussian, etc). Every exponential family distribution is defined relative to some base measure, and it must be nonzero everywhere on that base measure: this is true with multinomials, it is true with gaussians (covariance matrix must be full rank), etc. This arises because these distributions are the solution to an optimization problem which breaks down in the presence of those zeros. Anyway, so what needs to happen is that the relative base measure is the "tightest possible": in the case of multinomials, it is the uniform distribution on the nonzero entries, and in the Gaussian case, it is Lebesgue measure restricted to the affine subspace corresponding to the eigenspace of the provided covariance, shifted by the provided mean. The KL divergence (written as an integral) only makes sense if both distributions are relative to the same "tightest fit" measure. -To summarize, the invalidity of the formula in the presence of zeros isn't just some unfortunate hack, it is a deep issue intimately tied to how these distributions behave. The smoothing/Bayesian solution is thus better motivated: it nudges the distributions into validity. But many people simply choose to throw out those values (i.e., by erasing $0\ln(0)$ or $a\ln(a/0)$ and writing $0$ in its place).<|endoftext|> -TITLE: finding the parity of a permutation in little space -QUESTION [37 upvotes]: Suppose we have a permutation $\pi$ on $1,2,\ldots,n$ and want to determine the parity (odd or even) of $\pi$. -The standard method is find the cycles of $\pi$ and recall that the parity of $\pi$ equals the parity of the number of cycles of even length. However this seems to require $\Theta(n)$ bits of additional memory to carry out in linear time. (As you trace each cycle, you need to mark its elements so that you can avoid tracing the same cycle again.) -Alternatively, using only $O(\log n)$ bits (a few integer indexes), the inversions can be counted. This takes $\Theta(n^2)$ time, using the naive algorithm. -So my question: can the parity be determined in $O(n)$ time and $O(\log n)$ bits? -I have in mind that the input is a read-only array $p$ where $p[i]=\pi(i)$. An interesting variation is to allow the array to be modified. But you can't assume each array element has more than enough bits to hold the integer $n-1$ (not $n$, because I want only the values $1,\ldots,n$ to be representable; you don't get an additional value to use as a marker) or else you need to count the extra bits as working space and linear time becomes easy. - -REPLY [2 votes]: I can suggest an algorithm with $\tilde{O}(\sqrt{n})$ space and $\tilde{O}(n \sqrt{n})$ time complexities. One can divide the array into $\sqrt{n}$ chunks of similar size and compute number of inversions in each. After that we compute for each chunk how much inversions elements left to the chunk make with the chunk. -The algorithm can be modified to use $\tilde{O}(s)$ space and $\tilde{O}(\frac{n^2}{s})$. -UPD: Smart people suggest paper http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.29.2256 -Parity of permutation can be expressed in terms of parity of numbers of its cycles (n - c mod 2). The paper provides at least two algorithms for computing them. The basic idea is to determine the unique cycle leader for each cycle. For example, the minimal element on a cycle can be such a leader. One can easily obtain an algorithm, which uses $O(n^2)$ time and $O(\log n)$ memory, for computing number of cycles. -One of the algorithms picks 5-wise independent hash-function and chooses the minimal element by this function as a leader. Traverse each cycle and if you find an element with smaller hash-value than initial, stop traversing. Otherwise, traverse cycle up to closing, and then increment counter of cycles. -The algorithm works in $O(n \log n)$ time and $O(\log n)$ memory. -The authors also provide a complicated deterministic algorithm with $O(n \log n)$ time and $O(\log^2 n)$ memory.<|endoftext|> -TITLE: Question on a projective bundle associated to a vector bundle -QUESTION [5 upvotes]: Following is quoted from -Nakayama, On Weierstrass models-, -" Let $S$ be a complex surface, and $L$ a line bundle on it. consider $P=\mathbb{P}(\mathcal{O}_S\oplus L^2 \oplus L^3)$. Let $a$ and $b$ be arbitrary sections of $L^{-4}$ and $L^{-6}$ and let $(x,y,z)$ be the canonical sections of $\mathcal{O}_P(1)\otimes L^{-2}$, $\mathcal{O}_P(1)\otimes L^{-3}$ and $\mathcal{O}_P(1)$ respectively which correspond to the natural injections of $L^2$, $L^3$ and $\mathcal{O}_S$ into $\mathcal{O}_S\oplus L^2 \oplus L^3$. Then the Weierstrass model is given by equation $y^2z=x^3+axz^2+bz^3$ in $P$ and is an elliptic fibration over $S$ ..." -My question: When I do calculations for my self, I see that the embeddings given above should correspond to canonical sections of $\mathcal{O}_P(1)\otimes L^{2}$, $\mathcal{O}_P(1)\otimes L^{3}$ and $\mathcal{O}_P(1)$, not the one he is saying and therefore we should consider $(a,b)$ as sections in dual of what he has said and at the end we should get an equation in $\mathcal{O}(3)\otimes L^6$ not in $\mathcal{O}(3)\otimes L^{-6}$ ? How does he get $x$ (and similarly $y$) from embedding $ L^2 \rightarrow \mathcal{O}_S \oplus L^2 \oplus L^3$?? -My calculation: -We have the exact sequnce -$$0\rightarrow \mathcal{O}_P(-1) \rightarrow \mathcal{O}\oplus L^2 \oplus L^3 \rightarrow Q \rightarrow 0$$ -over $P$, from which we get -$$0 \rightarrow Hom(\mathcal{O}(-1),\mathcal{O}(-1)) \rightarrow Hom(\mathcal{O}(-1),\mathcal{O}\oplus L^{2} \oplus L^{3}) \rightarrow T_{W/S} \rightarrow 0$$ -Here the last object is the relative tangent bundle and the first map is given by three sections $(z,x,y) \in Hom(\mathcal{O}(-1),\mathcal{O}\oplus L^{2} \oplus L^{3})\cong \Gamma(\mathcal{O}(1)) \oplus \Gamma(\mathcal{O}(1)\otimes L^{2}) \oplus \Gamma(\mathcal{O}(1)\otimes L^{3})$ as above. These are analogue of $(x,y,z)$ coordinate on projective space. - -REPLY [4 votes]: There are two competing definitions for $\mathbb P(\mathcal E)$, one classifies subbundles of $\mathcal E$ of rank 1, the other classifies quotients. With the latter (as used in EGA or Hartshorne), you have a canonical quotient $p^*\mathcal E\to\mathcal O(1)$ instead of a canonical subbundle $\mathcal O(-1)\to p^*\mathcal E$. As long as $\mathcal E$ is locally free of finite rank, there is no big difference, you can just dualize $\mathcal E$ to pass from one to the other.<|endoftext|> -TITLE: Ext groups and Serre duality -QUESTION [11 upvotes]: Hi, -I have a question related to Serre Duality: -if I have a smooth projective variety $X$ with dualizing sheaf $\omega$ and two coherent sheaves $F$ and $G$ on $X$, then how can I get a canonical map -$Ext^{i}(F,G)\rightarrow Ext^{n-i}(G, F \otimes \omega)^{\star}$ ? -I know that in the case that $F$ and $G$ are locally free, one gets it and then it is an isomorphism. -But I don't see how you get it for coherent ones. -Thank you - -REPLY [12 votes]: This works directly when one of $F, G$ is locally free. I am not sure whether it is true when both are merely assumed to be coherent (e.g. I don't see how to get the map). (In general, even the generalization of Serre duality -- Grothendieck duality -- tells you how to hom out of $\mathbf{R}\Gamma \mathcal{F}$ (or more generally derived push-forward) of a sheaf $\mathcal{F}$ into some complex of abelian groups, and this doesn't seem to tell you about $\mathrm{Ext}$ functors up top, in $X$, though perhaps I'm missing something). See below for the extension without local freeness hypotheses. -Namely, there is a map $H^n(X, \omega) \to k$ (the "integration" map*). To get the map -$$\mathrm{Ext}^i(F, G) \to \mathrm{Ext}^{n-i}(G, F \otimes \omega)^*$$ (which is natural), we need a pairing -$$\mathrm{Ext}^i(F, G) \times \mathrm{Ext}^{n-i}(G, F \otimes \omega) \to k.$$ -To do this, we can use the Yoneda product to pair these to $\mathrm{Ext}^n(F, F \otimes \omega)$. If $F$ is locally free, then this naturally maps to $\mathrm{Ext}^n(O_X, F \otimes F^{\vee} \otimes \omega)$, which in turns maps to $H^n(X, \omega)$ (by coevaluation) and thus to $k$. If $G$ is locally free, we can similarly write both sides as $\mathrm{Ext}^i(F \otimes G^{\vee}, \omega)$ and $\mathrm{Ext}^{n-i}(O_X, G^{\vee} \otimes F \otimes \omega)^*$, and we get the pairing and isomorphism just as in Hartshorne. -Now if we fix one of $F, G$, we get a $\delta$-functor in the other. So if the natural transformation is an isomorphism when both are locally free, it is an isomorphism when one is locally free and the other merely coherent (since on a projective scheme, every coherent sheaf has a locally free presentation, and we can use the "finite presentation trick"). -*Here the comparison is as follows: on a compact complex manifold $X$ of dimension $n$, if $\omega$ denotes the sheaf of holomorphic $(n,0)$-forms, we have (Dolbeaut isomorphism) $$H^n(X, \omega) = \frac{(n,n)\mathrm{-forms}}{\overline{\partial}\mathrm{-exact\ top forms}}$$ and so we can define the map as integration, legitimately (because a $\overline{\partial}$-exact top form is exact in the usual sense, this is well-defined). -Edit: As above, the obstacle to defining the map was that there was no natural trace morphism $$\mathrm{Ext}^n(F, F \otimes \omega) \to H^n(X, \omega);$$ given one, the same arguments would answer your question for the case of $F, G$ both only assumed coherent. As Donu Arapura observes below, there is a natural way to define the trace. Reason: $F$ can be replaced by a bounded complex of locally frees in the derived category (it is a "perfect" complex) since we are working over a smooth variety (in particular, this means that any locally free resolution can be truncated at a finite stage to still yield a locally free one, by Serre's theorem on the finiteness of global dimension). For a bounded complex of locally frees $K^\bullet$, we can define a map -$$\mathrm{Ext}^n(K^\bullet, K^\bullet \otimes \omega) \to H^n(X, \omega)$$ -by taking the "partial trace." One can think of the former as consisting of maps $K^\bullet \to K^\bullet \otimes \omega[n]$, or $\mathbf{R}\underline{Hom}(K^\bullet, K^\bullet) \to \omega[n]$. (By the conditions on $K^\bullet$, the derived internal hom is the same as the usual sheaf hom.) -Since there is a natural map from $\mathcal{O}_X$ to the derived internal hom (given by the identity), we can define the trace. To show that map you are interested in becomes an isomorphism, we use the same "finite presentation" trick.<|endoftext|> -TITLE: Asymptotics/growth for coefficients of algebraic power series -QUESTION [5 upvotes]: Assume that the formal power series $a(z)=\sum\limits_{n\geq 1} a_n z^n$ satisfies an algebraic equation with polynomial coefficients (that is, there exists a nonzero polynomial $F(z,y)$ such that $F(z,a(z))=0$), and a finite number of terms of $a(z)$ determine uniquely the whole series as a solution to that equation. -Assume also that the sequence $\{a_n\}$ is increasing (and if necessary, one can even assume that all $a_i$ are integers), and grows polynomially, that is for some $C,d$ we have $a_n>1$. -Is it true that there exists "a limit value of the exponent $d$"? More precisely, is it true that the infimum of exponents $d_1$ such that $a_n< C_1 n^{d_1}$ for some $C_1$ (depending on $d_1$) and all $n>>1$ coincides with the supremum of exponents $d_2$ such that $a_n > C_2 n^{d_2}$ for some $C_2$ (depending on $d_2$) and all $n>>1$? - -REPLY [5 votes]: This should follow from Theorem VII.8 in Flajolet and Sedgewick's Analytic Combinatorics (freely available at the link). As I understand the argument, you can obtain a straightforward general form for asymptotics of coefficients of algebraic generating functions by expanding in Puiseux series around the dominant poles. -This implies the dominant term is a finite sum of terms of the form $n^d \zeta^n$ where $|\zeta| = 1$ for some $d \in \mathbb{Q}$, and the increasing growth condition should rule out the terms with $\zeta \neq 1$ messing up the asymptotics.<|endoftext|> -TITLE: If the discriminant of a binary quadratic form has high valuation, is the form "almost a square". -QUESTION [6 upvotes]: For a binary quadratic form $ax^2+bxy+cy^2$ over a field (characteristic not 2), the discriminant $b^2-4ac$ is 0 if and only if that form is the square of a linear form. I am curious about an analogue of this fact over a ring with a valuation. - Rough version of question. Let -$$f=ax^2+bxy+cy^2$$ -be a binary quadratic form over a ring $Q$ with a valuation. If the valuation of the discriminant $b^2-4ac$ is large (so heuristically $b^2-4ac$ is ``almost 0''), does this imply that $f$ is "almost a square"? - More precise version of question. Here's the precise case I am interested in. Let $R$ be a valuation ring, where 2 is a unit. Let $Q$ be the standard graded ring $Q=R[s,t]$ over $R$, i.e. $\text{deg}(s)=\text{deg}(t)=1$ and $Q_0=R$. The ring $Q$ inherits a valuation from $R$. Let $f$ as above, where $a$,$b$, and $c$ are all homogeneous elements of $Q$, and where $\text{deg}(a)+\text{deg}(c)=2\cdot \text{deg}(b).$ - Claim: If $b^2-4ac$ has valuation at least $\nu$, then there exist $d,e,f$, and $a',b',c'$ in $Q$, such that -$$ -ax^2+bxy+cy^2=d(ex+fy)^2 + (a'x^2+b'xy+c'y^2), -$$ -where $a',b',$ and $c'$ all have valuation at least $\nu/2$. - Update: Let me add some clarification and an example. First of all, note that $Q$ is not a valuation ring, since $Q=R[s,t]$. So an element with valuation $0$ is not necessarily a unit. For instance, $s^2$ is an element of $Q$ with valuation $0$ that is not a unit. -Also, here's an example. Let $R=\mathbb Q[[u]]$ and let $f=u^3x^2+u^3y^2$. Then $b^2-4ac=-4u^6$, which has valuation $6$. We can (trivially) write $f=(0)^2+f$ as the sum of a square and something with valuation $6/2=3$. So I think that the $\nu/2$ bound in the claim is optimal. - -REPLY [2 votes]: Speaking to the rough version, not the precise version, I think you will need some conditions on Q for the statement to be true. For instance, suppose Q is $R[T,U,V]/(TU-V^2)$, and let f be the form -$T x^2 + 2 V xy + Uy^2$. -The discriminant of f is 0, but I don't think this can be expressed as a constant multiple of a square of a linear form, which gives a negative answer to the statement in the case where nu is infinite.<|endoftext|> -TITLE: Are bundle gerbes bundles of algebras? -QUESTION [13 upvotes]: The category of line bundles (possibly with connection) -on a smooth manifold M can be defined in two different ways: -The first definition uses transition functions that satisfy a cocycle condition -(possibly with additional data of a 1-form that defines a connection), -and the second definition uses invertible vector bundles (possibly with connection), -where vector bundles are defined -algebraically as dualizable modules over the algebra of smooth functions -or geometrically as vector spaces in the category of smooth submersions over M -and connections are defined algebraically as certain linear maps satisfying the Leibniz identity, -geometrically in terms of subbundles, or topologically as certain functorial field theories. -The above two categories (with or without connection) are equivalent to each other: -There is a canonical fully faithful functor from the first category to the second one. -This functor is essentially surjective and hence an equivalence of categories, -even though the construction of an inverse functor requires a choice. -The first definition was categorified by Michael Murray in 1994, -the result being the bicategory of bundle gerbes (with or without connection). -Can one categorify the algebro-geometric definition of a line bundle -in such a way that there is an equivalence from Murray's bicategory of bundle gerbes -to the bicategory of categorified line bundles, -thus obtaining a “chart-free” definition of bundle gerbes? -Naïvely, one might expect that vector bundles should categorify to bundles of algebras. -Just as any fiber of a line bundle is noncanonically isomorphic to the vector space of scalars, -any fiber of a categorified line bundle should be -noncanonically Morita-equivalent to the algebra of scalars. -Furthermore, if a line bundle over M is equipped with a connection, -then any path in M gives an isomorphism between the corresponding fibers. -These isomorphisms can packaged in a 1-dimensional topological field theory, -and in fact a theorem by Florin Dumitrescu, Stephan Stolz, and Peter Teichner -shows that any 1-TFT comes from a vector bundle with connection, -thus giving an alternative definition of a vector bundle with connection. -Similarly, if a bundle gerbe over M is equipped with a connection, -then any path in M should give a Morita equivalence between the corresponding fibers. -Moreover, a bigon in M should give an isomorphism between the corresponding Morita equivalences. -Thus, one should be able to package the above parallel transports in a local 2-TFT. -Invertible morphisms between bundle gerbes should correspond to bundles of invertible bimodules -(i.e., Morita equivalences) and non-invertible morphisms (introduced by Konrad Waldorf in his paper -More Morphisms between Bundle Gerbes) -should correspond to bundles of non-invertible bimodules. -Several papers in the literature are closely related to the above question. -For example, Corollary 4.9 in the paper by Urs Schreiber and Konrad Waldorf -Connections on non-abelian gerbes and their holonomy -proves that the bicategory of bundle gerbes with connection over M -is equivalent to the bicategory of transport functors with values in some bicategory. -This statement would almost resolve the above question if not for the fact that -the target bicategory has only one object, which prevents one from considering -constructions like the algebra of global sections of a bundle gerbe -(the analogous construction for line bundles (the vector space of global sections) -plays an important rôle in geometric quantization and other areas of mathematics). -In fact, the algebra of global sections of a bundle gerbe is interesting enough -to warrant its own question: -Can one construct the algebra of global sections of a bundle gerbe -that categorifies the vector space of global sections of a line bundle? -Finally, I am also interested in the answers to the above questions -for the case of arbitrary vector bundles (respectively non-abelian bundle gerbes) -and not just line bundles. - -REPLY [5 votes]: There is a bicategory of Dixmier-Douady bundles of algebras which is equivalent to the bicategory of bundle gerbes. In particular, sections into these bundles form algebras. -The price you pay is that the bundles are infinite-dimensional; for that reson I am not sure if that picture persists in a setting "with connections". -I do not know a good source for the bicategory of Dixmier-Douady bundles or for the equivalence. Everything depends certainly on the type of morphisms you consider between the bundles; they clearly have to be of some Morita flavor. You may look into Meinrenken's "Twisted K-homology and group-valued moment maps", Section 2.1.1 and 2.1.4. In Section 2.4 Meinrenken indicates indirectly that his bicategory of Dixmier-Douady bundles is equivalent to the one of bundle gerbes, by transfering the notion of a multiplicative bundle gerbe (which depends on the definitions of 1-morphisms and 2-morphisms) into his language. -Side remark: a bundle gerbe is not the direct generalization of transition functions of a bundle. There is one step in between, namely a bundle 0-gerbe: instead of open sets, it allows for a general surjective submersion as the support for its transition functions. If you take bundle 0-gerbes instead of transition functions, the functor you mentioned at the beginning of your question has as canonical inverse functor. See my paper with Thomas Nikolaus "Four equivalent versions of non-abelian gerbes". -Added (after thinking a bit more about the question): If you want to categorify the vector space of sections into a vector bundle, you first have to fix a categorification of a vector space. An algebra is one possible version of a "2-vector space", probably due to Lurie. Another version, due to Kapranov-Voevodsky, is to define a 2-vector space as a module category over the monoidal category of vector spaces (add some adjectives if you like). -Let us define a section of a bundle gerbe $\mathcal{G}$ over $M$ to be a 1-morphism $s: \mathcal{I} \to \mathcal{G}$, where $\mathcal{I}$ is the trivial bundle gerbe. Then, sections form a category, namely the Hom-category $Hom(\mathcal{I},\mathcal{G})$ of the bicategory of bundle gerbes (the one with the "more morphisms" defined in my paper which was mentioned in the question). -The category $Hom(\mathcal{I},\mathcal{G})$ of sections of $\mathcal{G}$ has naturally the structure of a module category over the monoidal category of vector bundles over $M$. Indeed, a vector bundle is the same as a 1-morphism between trivial gerbes, i.e. an object in $Hom(\mathcal{I},\mathcal{I})$. Under this identification, the module structure is given by composition: -$$ -Hom(\mathcal{I},\mathcal{G}) \times Hom(\mathcal{I},\mathcal{I}) \to Hom(\mathcal{I},\mathcal{G}). -$$ -The functor which regards a vector space as a trivial vector bundle induces the claimed module structure over vector spaces. - -Summarizing, sections of bundle gerbes do not directly form algebras, but they form Kapranov-Voevodsky 2-vector spaces.<|endoftext|> -TITLE: Can we axiomatize Omnific Integers without the Surreal Number system? -QUESTION [21 upvotes]: Omnific integers are the counterpart in the Surreal numbers of the integers. The surreal numbers are usually defined using set theory, and then the omnific integers are defined as a particular subset (or rather subclass) of them. My question is, does it have to be this way? Is it possible to give a first-order axiomatization of the Omnific integers and their arithmetic, without having to define the surreal numbers themselves? I know they form a proper class, so there is a risk that they may be "too big" to describe. But Tarksi gave a first-order axiomatization for the ordinal numbers, which also form a proper class, so at least we have some hope. -The reason I'm interested is because of this question I asked a while back, about finding a nonstandard model of (Robinson) arithmetic whose field of fractions forms a real closed field. The Omnific integers form such a nonstandard model, so I want to find out whether we can axiomatize them. -Any help would be greatly appreciated. -Thank You in Advance. -EDIT: To be clear, I don't want an axiomatization of the Omnific Integers that's based on something else, like the real numbers, the surreal numbers, or set theory. I want a theory along the lines of Peano Arithmetic. -EDIT 2: As Emil said, it seems that a recursive axiomatization of the Omnific integers is impossible. So might we define them in some other way, without reference to the surreal numbers (or the real numbers)? - -REPLY [16 votes]: Here’s a couple of observations from my comment above. -First, the theory of omnific integers is a complete extension of Robinson’s arithmetic, hence it is not recursively axiomatizable. This makes it rather unlikely that we can describe its full axiomatization in any reasonable way. -Surreal numbers No form a real-closed field, and omnific integers Oz are its subring, hence they make an ordered ring. In fact, it is known that Oz is an integer part of No (i.e., for any surreal number $r$ there exists a unique omnific integer $n$ such that $n\le r< n+1$), which—by a well-known result of Shepherdson—means that Oz is a model of IOpen (the theory of discretely ordered rings + induction for open formulas in the language of ordered rings). Moreover, the fraction field of Oz (namely, No) is real-closed; this can be expressed by a first-order axiom schema (let’s call it A), with one axiom for each degree. (This set of axioms can be simplified: in the presence of A, IOpen is equivalent to the theory of discretely ordered rings + division with remainder.) -On the other hand, Oz does not satisfy induction for larger classes such as $E_1$ (bounded existential formulas), nor does it satisfy algebraic axioms such as normality or gcd. The reason is that such axioms contradict A (or even its corollary that $\sqrt2$ is in the fraction field of Oz).<|endoftext|> -TITLE: Self-intersection of exceptional divisor -QUESTION [7 upvotes]: Suppose that $X$ is a smooth threefold, and $C \subset X$ a smooth curve. Let $Y$ be the blowup of $X$ along $C$, with exceptional divisor $E$. What is the intersection number $E^3$ on $Y$? (in terms of the genus and normal bundle of $C$, etc) -I assume that I could extract the answer from Theorem 6.7 of Fulton's book on intersection theory, were I better familiar with the contents of chapters one through five -- I'd be happy to hear either a direct method or a pointer about how to get it from Fulton! - -REPLY [4 votes]: Let $Y\subset\mathbb{P}^n$ be a smooth variety, and let $\epsilon:X = Bl_Y\mathbb{P}^n\rightarrow\mathbb{P}^n$ be the blow-up of $\mathbb{P}^n$ along $Y$. -Let $\widetilde{H}$ be the pull-back of the hyperplane section $H$ of $\mathbb{P}^n$, and $E$ be the exceptional divisor. If $H_Y =H\cdot Y$ we have -$$\widetilde{H}^{h-i}E^i = p^*H_Y^{n-i}\cdot i^*E^{i-1} = H_Y^{n-i}\cdot p_*i^*E^{i-1}.$$ -Recall that $E = \mathbb{P}(N_{Y/\mathbb{P}^n})$, and $i^*E = -e$, where $e = c_1(\mathcal{O}_E(1))$. Let use denote by $s_j$ the Segre classes of $N_{Y/\mathbb{P}^n}$, and let $c = codim_{\mathbb{P}^n}(Y)$. We have the following intersection numbers: - -$\widetilde{H}^n = 1$; -$\widetilde{H}^{n-i}\cdot E^i = 0$ for $i < c$; -$\widetilde{H}^{n-i}\cdot E^i = (-1)^{i-1}s_{i-c}H_Y^{n-i}$ for $i\geq c$. - -Let $C\subset\mathbb{P}^3$ be a smooth curve of degree $d$ and genus $g$. By the exact sequence -$$0\mapsto T_{C}\mapsto T_{\mathbb{P}^3|C}\rightarrow N_{C/\mathbb{P}^3}\mapsto 0$$ -we get $s_1(N_{C/\mathbb{P}^3}) = -c_{1}(N_{C/\mathbb{P}^3}) = -4d-2g+2$. Then $\widetilde{H}^3 = 1$, $\widetilde{H}^2\cdot E = 0$, $\widetilde{H}\cdot E^2 = -s_0H_Y = -d$, and $E^3 = s_1 = 2-2g-4d$. For instance we can compute the cube of the anti-canonical divisor: -$$(-K_{Bl_C\mathbb{P}^3})^3 = (4\widetilde{H}-E)^3 = 64-12d+4d+2g-2 = 62-8d+2g.$$<|endoftext|> -TITLE: When is an HNN extension a free group? -QUESTION [6 upvotes]: Let $A$ be a free group and $G = A*_t$. When is $G$ also a free group? -Suppose $t y t^{-1} = z$ and there is a splitting $A = B*C$ so that $y \in B$ and $z \in C$ and $z$ is a member of some basis of $C$ then clearly $G$ is free. Is this the only case that $G$ is free? - -REPLY [10 votes]: Yes. This is a theorem of Shenitzer. For a modern treatment see, for instance, this recent paper of Louder. I give a proof of a similar fact in section 2 of this preprint.<|endoftext|> -TITLE: Spin structure on mapping torus -QUESTION [8 upvotes]: I would like to know if, given a spin manifold $X$ and an orientation-preserving diffeomorphism $f : X \longrightarrow X,$ we can naturally endow the mapping torus $M_f = X \times [0, 1] / (x, 0) \sim (f(x), 1)$ with a spin structure. -In the case that interests me particularly, $X$ is simply the two-dimensional torus and $f$ is a classifying map for an automorphism of ${\mathbb Z}^2.$ -Thank you for any answer ! - -REPLY [3 votes]: You can do this iff the spin structures $\mathfrak{s}$ and $f^*(\mathfrak{s})$ are isomorphic. -When $X$ is the 2-torus the set of Spin structures is naturally in bijection with $\mathbb{Z}/2 \oplus \mathbb{Z}/2$, but $SL_2(\mathbb{Z})$ does not act in the usual way. In fact it doesn't act linearly at all, but affinely: -$$ -\begin{bmatrix} -A &C \\ -B& D -\end{bmatrix} : \begin{bmatrix} -u \\ -v -\end{bmatrix} \mapsto \begin{bmatrix} -A &C \\ -B& D -\end{bmatrix} \cdot \begin{bmatrix} -u \\ -v -\end{bmatrix} + \begin{bmatrix} -AC \\ -BD -\end{bmatrix}. -$$ -Using this formula you can check if your $f$ preserves a given Spin structure.<|endoftext|> -TITLE: Picard group of $\mathcal{M}_{0,n}$ -QUESTION [7 upvotes]: Let $\mathcal{M}_{0,n}$ be the complement of the boundary of the Mumford-Knudsen compactification of the moduli space of genus zero, n-pointed curves. -Is $Pic(\mathcal{M}_{0,n})$ trivial? - -REPLY [12 votes]: Yes. By fixing the three points $\{0,1,\infty\}$ one sees that $M_{0,n}$ is isomorphic to an open subscheme of $\mathbb{A}^{n-3}$ which has trivial Picard group. Since it is smooth, the Picard group of any open subscheme is also trivial.<|endoftext|> -TITLE: Paradoxical Decompositions -QUESTION [17 upvotes]: Question: Does Con($ZF$) imply Con($ZF$ + $DC$ + "there is no paradoxical Banach-Tarski decomposition of the unit ball")? - -Here Con($X$) is the consistency of $X$; $DC$ is dependent choice. -Motivation for the Question: Since the "paradoxical" sets in the Banach-Tarski theorem have to be non-measurable, Solovay's model of "every subset of reals is measurable" shows: -Theorem. Con($ZF$+ Inacc) implies Con($ZF$ + $DC$ + "there is no paradoxical Banach-Tarski decomposition of the unit ball"). -In the above Inacc is the statement "there is an inaccessible cardinal". -Note that Shelah's model [Israel Journal of Math, 1984], constructed only -from Con($ZF$), in which $DC$ holds and all sets of reals have the Baire -property, is of no help in answering the above question since Dougherty and -Foreman [ J. Amer. Math. Soc., 1994] have shown that there are paradoxical -decompositions of the unit ball using pieces which have the property of -Baire. I do not know how much choice is needed in their construction. -This question was posed a while ago on an FOM-posting of mine, but remained unanswered. - -REPLY [9 votes]: A positive answer is proved in S. Wagon's book "The Banach-Tarski Paradox", Theorem 13.2. Specifically, the statement proved there is: -Con(ZF) $\leftrightarrow$ Con(ZF + DC + GM), -where GM is the existence of an isometry-invariant measure on all subsets of $\mathbb R^n$ taking the value $1$ on the unit cube.<|endoftext|> -TITLE: Current status of a conjecture of Bloch -QUESTION [9 upvotes]: In the seminal paper $K_2$ and algebraic cycles, Bloch make the following conjecture : - -Suppose $A$ is a local Noetherian integral domain with quotient field $F$ - -$K_2(A)$ → $K_2(F)$ is injective - -Assume in addition $A$ is normal, $K_2(A)$ = $∩_pK_2(A_p)$ where $p$ runs through all height 1 prime ideals in $A$. - - - -What is the current status of this conjecture? -I only know that the first statement is true for discrete valuation ring by a theorem of Dennis and Stein. Can we prove it for a local algebra over a field? -Moreover, the second claim in this conjecture is a Hartogs like statement, so we want it still to hold without the local assumption, could this be true? For example，can we prove it for Dedekind domain or more specifically coordinate ring of a smooth affine curve over a (finite) field? - -REPLY [11 votes]: The second statement is false (even if we modify it by replacing $K_2(A)$ by its image in $K_2(F)$). A counterexample is $A=k[x,y,z]_(x,y,z)/(z^2-xy)$. See J. Reine Angew. Math. 381 (1987), 37–50.<|endoftext|> -TITLE: Profinite topologies on a group generated by different families of subgroups -QUESTION [6 upvotes]: Let G be a finitely generated group. Suppose we have two families F1 and F2 of finite index subgroups of G, and each family has trivial intersection and is filtered from below (i.e. for any two elements in the family their intersection contains some third element). -These families generate two profinite topologies on G. (Taking the subgroups in the families as basis of open neighborhoods around identity). -Suppose the completions wrt to these families produce isomorphic profinite groups. -Can we say that these families generate the same topology on G? -(Equivalently, given any N∈F1 is there N2∈F2 such that N1≤N2 and vice versa.) -What if one family is a subfamily of the other? - -REPLY [10 votes]: No. $\widehat{\mathbb{Z}} \times \mathbb{Z}_p$ is isomorphic to $\mathbb{Z}_p \times \widehat{\mathbb{Z}}$, and the two topologies on $\mathbb{Z} \times \mathbb{Z}$ are different, though homeomorphic. -Am I interpreting your question too narrowly? -Also, in the case of a subfamily, you have a map $G'' \to G'$ from one completion to the other. If the subfamily is cofinal in the other, then the topologies are the same, and otherwise there will be a nontrivial kernel. Even then, I don't see why $G''$ and $G'$ couldn't still be isomorphic. -EDIT: By Proposition 2.5.2 of Ribes-Zalesskii's book "Profinite Groups," a finitely generated profinite group $G$ is hopfian in the sense that any continuous epimorphism $G \to G$ is an isomorphism. -So, in the situation of a subfamily, it turns out the answer is yes.<|endoftext|> -TITLE: Lemma on infinitely generated projective modules -QUESTION [6 upvotes]: Is it true that every finitely generated submodule of a non-finitely generated projective over a (not necessarily commutative!) ring is contained in a proper summand? -N.B.: I asked this already on math.stackexchange.com without much luck. - -REPLY [7 votes]: The lemma is at least true, if the projective module has an uncountable projective base (sometimes also called a dual base). -Proof: Let $P$ be a projective $R$-module with uncountable projective base $(x_i, f_i)$, $(i\in I)$ and $M = \sum_{k=1}^nRy_k \subseteq P$. Define inductively -$$I_0 = \lbrace i \in I \mid \exists 1 \le k \le n: f_i(y_k) \neq 0 \rbrace$$ -$$I_{n+1} = I_n \cup \lbrace i \in I \mid \exists j \in I_n: f_i(x_j) \neq 0 \rbrace$$ -$$J = \cup_{n\ge 0}I_n\hspace{140pt}$$ -Set $Q = \sum_{j \in J}Rx_j \le P$. Since $y_k = \sum_{i \in I}f_i(y_k)x_i$ it follows from $I_0 \subseteq J$ that $M \le Q$. -Next I want to show -$$x_j = \sum_{i \in J}f_i(x_j)x_i \quad\text{ for each } i \in J \hspace{80pt}(\ast)$$ -Let $j \in I_n$. Write $x_j = \sum_{i \in I}f_i(x_j)x_i$. If $f_i(x_j) \neq 0$ it follows $j \in I_{n+1} \subseteq J$. Thus $(\ast)$ is shown. Define -$$\kappa: P \to Q, x \mapsto \sum_{i \in J}f_i(x)x_i.$$ -$\kappa$ is $R$-linear and from $(\ast)$ one concludes $\kappa|Q = \text{id}_Q$. Thus $Q$ is a direct summand of $P$ and since $Q$ is countably generated, $Q$ is a proper subset of $P$. -BTW: In the great example from F. Ladisch, $P$ has a countable projective base (see Lam's book).<|endoftext|> -TITLE: Is the endomorphism algebra of a dualizable bimodule necessarily finite dimensional? -QUESTION [18 upvotes]: Let $k$ be field. Let $A$, $B$ be $k$-algebras, and let ${}_AM_B$ be a dualizable bimodule. -Pre-Question (too naive): Is the algebra of $A$-$B$-bilinear endomorphisms of $M$ necessarily finite dimensional? -Answer: No. Take $A$ some infinite dimensional commutative algebra, and $M={}_AA_A$. Then $End({}_AA_A)=A$ is not finite dimensional. - -Question: Assume that $A$ and $B$ have finite dimensional centers. - Is it then true that the algebra of $A$-$B$-bilinear endomorphisms of $M$ has to be finite dimensional? - -Special case for which I know the answer to be positive: -If $k=\mathbb C$ or $\mathbb R$, and if we're in a C*-algebra context, then I know how to prove that $End({}_AM_B)$ is finite dimensional. But my proof relies on certain inequalities, and it does not generalize. - -Definitions: -A bimodule ${}_AM_B$ is called left dualizable if there is an other bimodule ${}_BN_A$ (the left dual) and maps $r:{}_AA_A\to {}_AM\otimes_BN_A$ and $s:{}_BN\otimes_AM_B\to {}_BB_B$ such that $(1\otimes s)\circ(r\otimes 1) = 1_M$ and $(s\otimes 1)\circ(1\otimes r) = 1_N$. -There's a similar definition of right dualizability. I'm guessing that right dualizability is not equivalent to left dualizability, but I don't know a concrete example that illustrates the difference between these two notions. -In my question above, I've just used the term "dualizable", which you -should interpret as - -"both left and right dualizable". -There's also the notion of fully dualizable, which means that the left dual has its own left dual, which should in turn have its own left dual etc., and similarly for right duals. Once again, I'm a bit vague as to whether all these infinitely many conditions are really needed, or whether they are implied by finitely many of them. -PS: I hope that I didn't mix my left and my right. - -REPLY [4 votes]: This is more or less orthogonal to Kevin's answer; he puts a stronger restriction on $A$ and $B$, while I put a stronger restriction (perhaps too strong!) on ${_A}M_B$. I likewise apologize for the lack of pictures. -If we assume that - -${_A}M_B$ has a 2-sided dual ${_B}N_A$, and -the duality pairings exhibit ${_B}N \otimes_A M_B$ as a retract of ${_B}B_B$, - -then $\operatorname{End}({_A}M_B)$ is dualizable as a $Z(A)$-module, and hence, if $Z(A)$ is finite dimensional, it will follow that $\operatorname{End}({_A}M_B)$ is finite dimensional. -To prove this, we exhibit a duality pairing between ${_{Z(A)}}\operatorname{End}({_A}M_B)$ and $\operatorname{End}({_B}N_A)_{Z(A)}$. -The evaluation map $\epsilon: {_{Z(A)}}\operatorname{End}({_A}M_B) \otimes_k \operatorname{End}({_B}N_A)_{Z(A)} \to {_{Z(A)}}Z(A)_{Z(A)}$ is given by first mapping to $\operatorname{End}({_A}M \otimes_B N_A)$ and then pre- and post-composing by the appropriate unit and counit, respectively, from the duality pairings between $M$ and $N$ to get an endomorphism of ${_A}A_A$, i.e., an element of $Z(A)$. -The coevaluation map $\eta: k \to \operatorname{End}({_B}N_A) \otimes_{Z(A)} \operatorname{End}({_A}M_B)$ is given by identifying the codomain with $\operatorname{End}({_B}N \otimes_A M_B)$, which follows from condition 2, whence $\eta$ is just the unit of this algebra. -That $\epsilon$ and $\eta$ give a duality pairing follows from condition 2. (The nontrivial morphism in one of the triangle diagrams sends a map $f: {_A}M_B \to {_A}M_B$ to the identity on ${_A}M_B$ tensored with the "trace" of $f$ via the duality pairings. Condition 2 tells us that the identity string can absorb the trace bubble, and similarly for the other triangle diagram.) -Unfortunately, condition 2 (or its "mirror," which would show dualizability as a $Z(B)$-module) seems very strong, but it is crucial for the above proof.<|endoftext|> -TITLE: A unique ultrafilter extending a union of filters? -QUESTION [12 upvotes]: Original Question: -Let $\mathcal{P}(\omega)/fin$ denote the Boolean algebra formed from $\mathcal{P}(\omega)$ by modding out by the ideal $fin$ of finite subsets of $\omega$. As a first pass at the intended question, consider the following: -Question 0: Are there two filters $F$ and $G$ in $\mathcal{P}(\omega)/fin$ such that there is a unique ultrafilter extending $F \cup G$? -The answer, of course, is yes: consider the case where $F$ is already an ultrafilter, and $G$ is some filter such that $G \subseteq F$. We might therefore ask (what seems to be) a harder question. Given an ideal $I$ in $\mathcal{P}(\omega)/fin$, notice that -$$\{a \in \mathcal{P}(\omega)/fin \: : \: a \geq I\}$$ -(where $a \geq I$ iff $(\forall b \in I)[a \geq b]$) is a filter; call such filters regular filters (I made up this terminology, and I would be glad to know if there is already a word for such objects). Now we can ask: -Question 1: Are there two regular filters $F$ and $G$ such that there is a unique ultrafilter extending $F \cup G$? -Via Stone duality, this question can be rephrased (I believe) in topological terms: -Question 1$'$: Are there two regular closed subsets $C,D \subset \omega^{\*}$ such that $C \cap D$ is a singleton? -Here, a regular closed set is simply a set which is equal to the closure of its interior, and $\omega^{\*} = \beta \omega \setminus \omega$, the space of all non-principal ultrafilters on $\omega$ (i.e. the Stone space of $\mathcal{P}(\omega)/fin$). If $C$ and $D$ are witnesses to a positive answer to question 1$'$, then $int(C)$ and $int(D)$ must be disjoint, in which case the ideals corresponding to these open sets form a gap; this is the basis for my original interest in this question. - -Update: -Based on the suggestions given by Andreas Blass, it turns out we have the following consistency result. -Theorem: Under CH, there exist regular filters $F$ and $G$ such that $F \cup G$ extends to a unique ultrafilter. -Proof. (sketch) -Let $\{c_{\alpha} \: : \: \alpha < \omega_{1}\}$ be an enumeration of all elements of $\mathcal{P}(\omega)/fin$. Choose elements $a_{0}, b_{0}$ such that $a_{0} \land b_{0} = 0$, $a_{0} \lor b_{0} < 1$, and furthermore such that either $c_{0} \leq a_{0} \lor b_{0}$, or else $\lnot c_{0} \leq a_{0} \lor b_{0}$. -Now suppose that for $\gamma < \omega_{1}$, we have constructed increasing sequences $\{a_{\alpha} \: : \: \alpha < \gamma\}$ and $\{b_{\alpha} \: : \: \alpha < \gamma\}$ such that, for all $\alpha < \gamma$, -(a) $a_{\alpha} \land b_{\alpha} = 0$; -(b) $a_{\alpha} \lor b_{\alpha} < 1$; -(c) either $c_{\alpha} \leq a_{\alpha} \lor b_{\alpha}$, or $\lnot c_{\alpha} \leq a_{\alpha} \lor b_{\alpha}$. -First suppose that $\gamma = \eta + 1$ is a successor ordinal. Let $d \in \{c_{\gamma}, \lnot c_{\gamma}\}$ be such that -$$a_{\eta} \lor b_{\eta} \lor d < 1,$$ -let $\{d_{a}, d_{b}\}$ be a (nontrivial, if possible) partition of $d \land \lnot(a_{\eta} \lor b_{\eta})$, and set $a_{\gamma} = a_{\eta} \lor d_{a}$ and $b_{\gamma} = b_{\eta} \lor d_{a}$. Then it is easy to see that $\{a_{\alpha} \: : \: \alpha < \gamma + 1\}$ and $\{b_{\alpha} \: : \: \alpha < \gamma + 1\}$ are increasing sequences satisfying (a) through (c). -Suppose now that $\gamma$ is a limit ordinal. Observe that the sequence $\{\lnot(a_{\alpha} \lor b_{\alpha}) \: : \: \alpha < \gamma\}$ is countable and strictly decreasing. It follows that there exists a nonzero lower bound of this sequence; equivalently, there exists an $e < 1$ such that $e \geq a_{\alpha}$ and $e \geq b_{\alpha}$ for all $\alpha < \gamma$. Moreover, since $\{a_{\alpha} \: : \: \alpha < \gamma\}$ and $\{b_{\alpha} \: : \: \alpha < \gamma\}$ are both countable, they cannot form a gap; hence there exist $a, b \in \mathcal{P}(\omega)/fin$ such that, for all $\alpha < \gamma$, $a \geq a_{\alpha}$ and $b \geq b_{\alpha}$. Replacing $a$ and $b$ by $a \land e$ and $b \land e$, if necessary, we may assume that $a, b \leq e$. Now we can repeat the argument given for the case where $\gamma$ is a successor, replacing $a_{\eta}$ by $a$ and $b_{\eta}$ by $b$. -Thus we obtain $\{a_{\alpha} \: : \: \alpha < \omega_{1}\}$ and $\{b_{\alpha} \: : \: \alpha < \omega_{1}\}$. I claim that these form a gap. If not, then there is some $\beta < \omega_{1}$ such that $c_{\beta} \geq \{a_{\alpha} \: : \: \alpha < \gamma\}$ and $\lnot c_{\beta} \geq \{b_{\alpha} \: : \: \alpha < \gamma\}$; on the other hand, we know that either $c_{\beta} \leq a_{\beta} \lor b_{\beta}$, or $\lnot c_{\beta} \leq a_{\beta} \lor b_{\beta}$, each of which readily yields a contradiction. -I claim also that for every two-element partition $\{p, q\}$ in $\mathcal{P}(\omega)/fin$, one element, say $p$, is such that $\{a_{\alpha} \land p \: : \: \alpha < \omega_{1}\}$ and $\{b_{\alpha} \land p \: : \: \alpha < \omega_{1}\}$ do not form a gap. Indeed, each such partition must be of the form $\{c_{\beta}, \lnot c_{\beta}\}$ for some $\beta < \omega_{1}$. Without loss of generality, suppose we have $c_{\beta} \leq a_{\beta} \lor b_{\beta}$; then it is not difficult to see that -$$a_{\beta} \geq \{a_{\alpha} \land c_{\beta} \: : \: \alpha < \omega_{1}\}$$ -and likewise -$$b_{\beta} \geq \{b_{\alpha} \land c_{\beta} \: : \: \alpha < \omega_{1}\},$$ -from which it follows that these sequences do not form a gap. $\blacksquare$ -So I suppose the name of the game here is consistency results, such as - -Is a negative answer to Question 1 consistent with ZFC? Can a positive answer be proved under any assumptions weaker than CH? - -As this seems to be a rather slippery problem, I would welcome any suggested reading on this topic. And thanks again for the helpful replies already given; they are much appreciated. - -REPLY [6 votes]: The following is due to Alan Dow: -In any model obtained by adding $\aleph_2$ many Cohen reals to a model -of $\mathsf{CH}$ the statement is false. -We force with $\mathbb{P}=\operatorname{Fn}(\omega_2,2)$ and we let -$\dot{\mathcal{I}}$, $\dot{\mathcal{J}}$ and $\dot u$ be -$\mathbb{P}$-names such -that $\dot{\mathcal{I}}$ and $\dot{\mathcal{J}}$ are forced to be ideals -and $\dot u$ is forced to be the unique ultrafilter that extends the -two associated regular filters. -Now let $M$ be an elementary substructure of a suitable large $H(\theta)$ -that has cardinality $\aleph_1$ and that is closed under $\omega$-sequences. -Let $\delta=M\cap\omega_2$ and -$\mathbb{P}_M=\operatorname{Fn}(\delta,2)$. -By elementarity the $\mathbb{P}_M$-names -$\dot{\mathcal{I}}\cap M$ and $\dot{\mathcal{J}}\cap M$ are forced to be -ideals and $\dot u\cap M$ is forced to be the unique ultrafilter that extends -the two associated regular filters. -Work in $V[G\cap M]$. -We write $\mathcal{I}_M$, $\mathcal{I}_M$ and $u_M$ for the interpretations -of $\dot{\mathcal{I}}\cap M$, $\dot{\mathcal{J}}\cap M$ and $\dot u\cap M$ -in this model. -Note that every element of $u_M$ meets elements of $\mathcal{I}_M$ -and $\mathcal{J}_M$. -On the other hand if $\mathcal{I}'$ and $\mathcal{J}'$ are countable -subfamilies of $\mathcal{I}_M$ and $\mathcal{J}_M$ respectively then some -infinite subset, $x$, of $\omega$ separates the elements of $\mathcal{I}'$ from -those of $\mathcal{J}'$. -This implies that, without loss of generality, for every countable subfamily -$\mathcal{J}'$ of $\mathcal{J}_M$ there is an element $x\in u_M$ such that -$x\cap y$ is finite for all $y\in\mathcal{J}'$. -Using this one can construct a sequence -$\langle b_\alpha:\alpha<\omega_1\rangle$ -in $\mathcal{J}_M$ -such that every element of $u_M$ contains all but countably many of -the $b_\alpha$. -In $V[G]$ consider the next Cohen real $c$, added by -$\operatorname{Fn}([\delta,\delta+\omega,2)$, say and assume, -without loss of generality, that $c\notin u$. -There must be a set $Y\subseteq c$ that separates -$\lbrace x\cap c:x\in\mathcal{I}\rbrace $ from $\lbrace y\cap c:y\in\mathcal{J}\rbrace $. -In $V[G\cap M]$ we take names, $\dot c$ and $\dot Y$, for $c$ and $Y$, -note that these are $\operatorname{Fn}(C,2)$-names for some countable set $C$. -For every $\alpha<\omega_1$ it is forced that $\dot Y\cap b_\alpha$ is finite; -by pigeon-holing there will be one $p\in\operatorname{Fn}(C,2)$ and one -$n\in\omega$ such that $p$ forces $\dot Y\cap b_\alpha\subseteq n$ for -uncountably many $\alpha$. -Still in $V[G\cap M]$ let -$Y_p=\lbrace k:(\exists q\le p)(q \Vdash k\in\dot Y)\rbrace $. -Then $b_\alpha\cap Y_p\subseteq n$ for uncountably many $\alpha$ so that -$Y_p\notin u_M$. -On the other hand for every $x\in\mathcal{I}_M$ it is forced that -$x\cap\dot c\subseteq \dot Y$ (mod finite); this implies that -$x\setminus Y_p$ is finite for all $x\in\mathcal{I}_M$, so that $Y_p\in u_M$.<|endoftext|> -TITLE: Every real function has a dense set on which its restriction is continuous -QUESTION [58 upvotes]: The title says it all: if $f\colon \mathbb{R} \to \mathbb{R}$ is any real function, there exists a dense subset $D$ of $\mathbb{R}$ such that $f|_D$ is continuous. -Or so I'm told, but this leaves me stumped. Apart from the rather trivial fact that one can find a dense $D$ such that the graph of $f|_D$ has no isolated points (by a variant of Cantor-Bendixson), I don't know how to start. Is this a well-known fact? - -REPLY [48 votes]: It is a theorem due to Blumberg (New Properties of All Real Functions - Trans. AMS (1922)) and a topological space $X$ such that every real valued function admits a dense set on which it is continuous is sometimes called a Blumberg space. -Moreover, in Bredford & Goffman, Metric Spaces in which Blumberg's Theorem Holds, Proc. AMS (1960) you can find the proof that a metric space is Blumberg iff it's a Baire space.<|endoftext|> -TITLE: $\Sigma_n$ version of HOD -QUESTION [5 upvotes]: Fix a natural number, $n \geq 1$. Consider the class, M, of all sets hereditarily ordinal-definable using some $\Sigma_n$ formula. Since there is a universal $\Sigma_n$ formula, M is definable. Is M necessarily a model of ZF? It seems to me that it is closed under Godel operations and almost universal for the same reasons that HOD is, and therefore a model of ZF. But I feel like I'm missing something, since I've never heard anything about this model. -If it is a model of ZF, where can I learn more about it? Has anybody done any research about it? How does it relate to HOD? -Note: I require $n \geq 1$ because the formula witnessing that HOD is almost universal is $\Sigma_1$. - -REPLY [6 votes]: It follows from the Reflection Principle that every ordinal definable set is ordinal definable by a $\Sigma_2$-formula in the language of set theory. Indeed, if $A = \{x : \phi(x,\bar\alpha)\}$, then $$\exists\gamma(\gamma \in \mathrm{Ord} \land \bar\alpha \in \gamma \land \forall x(x \in A \leftrightarrow x \in V_\gamma \land V_\gamma \vDash \phi(x,\bar\alpha))).$$ Therefore, there is some $\gamma \in \mathrm{Ord}$ such that $$x \in A \leftrightarrow \exists U(U = V_\gamma \land x,\bar\alpha \in U \land U \vDash \phi(x,\bar\alpha)),$$ which is $\Sigma_2$ since $U = V_\gamma$ is $\Pi_1$.<|endoftext|> -TITLE: The relationship between group cohomology and topological cohomology theories -QUESTION [25 upvotes]: I was recently trying to learn a little bit about group cohomology, but one point has been confusing me. According to wikipedia (http://en.wikipedia.org/wiki/Group_cohomology and some other sources on the internet), given a (topological) group $G$, we have that the group cohomology $H^n(G)$ is the same as the singular cohomology $H^n(BG)$ (with coefficients in a trivial $G$-module $M$). Moreover, it says that given any group $G$, if we don't care about its topology, we can always give it the discrete topology and look at the cohomology of $K(G,1)$. This seems to suggest that when $G$ has topology that we do care about, we can just look at $BG$ with whatever topology $G$ is supposed to have. The relevant citation in this section is a reference to a book called Cohomology of Finite Groups, but I was wondering if this result would work with groups such as $U(1)$ which are not finite? Moreover, it would seem then that there is some sort of natural way to define group cohomology to detect the topology of the group; for example maybe look at continuous $G$-modules and continuous cochains. However, I heard that when doing this, one has to be careful because in general, the category of continuous $G$-modules might not have enough injectives. Also, I found this article by Stasheff (http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.bams/1183540920) which seems to suggest that for continuous cohomology, the we might not have the equality $H^n(BG)=H^n(G)$ between singular and group cohomology. I was wondering if someone could explain these connections to me (including what "continuous cohomology" is) and/or clarify what is happening? It would also be great if someone could tell me how one might compute something like $H^n(U(1);M)$ where $U(1)$ carries the discrete topology. Thanks. - -REPLY [6 votes]: Continuous cohomology of a topological group G used to mean using the usual chain complex -of multivaraible functions F:G^n \to A with the usual coboundary -except that f is required to be continuous. in the smooth case, continuous or smooth gives the same cohomology. In the Lie case, van Est established very nice relations between this cohomology, -the Lie algebra cohomology and the purely topological cohomology of the underlying space G<|endoftext|> -TITLE: Ends as a "cotrace" operation on profunctors -QUESTION [8 upvotes]: As mentioned here, there is a trace operation on the monoidal category of profunctors given by taking coends: for any profunctor $F : A\times X \nrightarrow B \times X$, there is a profunctor $Tr^X(F) : A \nrightarrow B$ defined by -$$Tr^X(F)(a,b) = \int^x F((a,x),(b,x))$$ -I am interested in the sort of dual operation, where we take an end rather than a coend: -$$CoTr^X(F)(a,b) = \int_x F((a,x),(b,x))$$ -Note that $CoTr^X(F)$ has the same type ($A \nrightarrow B$) as $Tr^X(F)$. -My question is, - -To what extent can the operation $CoTr^X$ be seen as a "cotrace"? - -I've found a bit of information googling on "cotraces", but nothing very comprehensive. Specifically, I'd like to know the following: - -First of all, is there a commonly-accepted definition of "cotrace", and what is its relationship to the trace? -Is there a string-diagrammatic definition? -One way to view the operation $CoTr^X$ is as a limited form of closure for the "external monoidal" structure on profunctors, in the sense that -$${\bf Prof}(G, CoTr^X F) = {\bf Prof}(G \times Hom_X, F)$$ -holds naturally in $G : A \nrightarrow B$. That is, we can view $CoTr^X(F)$ as "$Hom_X \multimap F$". Is this part of the general definition of cotrace (assuming the answer to (1) is positive), or is it a special feature of this particular operation on Prof? - -REPLY [3 votes]: Your exact set of questions appears to have languished unanswered for some time, but I can offer at least a partial answer. -You appear to have rediscovered the notion of Tambara modules. -The comonad $CoTr^X$ is talked about in some depth in Doubles for Monoidal Categories by Pastro and Street. They also dig into the left adjoint of this construction, which is a monad on $Prof$, such that the "strong" profunctors are just its algebras. They talk a fair bit about point #3 as well, though as $Hom$ is the unit for profunctor composition, they can just fuse it away.<|endoftext|> -TITLE: Which curves have stable Faltings height greater or equal to 1 -QUESTION [11 upvotes]: Let $Y$ be a smooth projective connected curve of genus $g>0$ over $\overline{\mathbf{Q}}$. Let $h_{\textrm{Fal}}(Y)$ be the Faltings height of $Y$. -Question 1. Can one classify or describe the curves $Y$ such that $h_{\textrm{Fal}}(Y) \geq 1$? -Question 2. For any $g>0$, does there exist a curve $Y$ of genus $g$ such that $h_{\textrm{Fal}}(Y) <1$? -Essentially, I would like to know which curves one is excluding by looking at curves $Y$ such that $h_{\textrm{Fal}}(Y) \geq 1$. -A result of Bost says that the stable Faltings height of an abelian variety $A$ over $\overline{\mathbf{Q}}$ of dimension $g$ is bounded from below by $-\frac{1}{2}\log(2\pi)g$. -By the Northcott property of the Faltings height, the set of curves of genus $g$ with $h_{\textrm{Fal}}(Y) <1$ is finite. This means that I'm looking at the finite set of curves of genus $g$ with Faltings height not in the interval $$[-\frac{1}{2}\log(2\pi)g,1)\subset[-2/5g, 1).$$ -Added: To answer Junkie's question, I'm aware of only one definition of the Faltings height of a curve over $\overline{\mathbf{Q}}$. There are several equivalent definitions, though. -Let $X$ be a smooth projective curve of genus $g>0$ over $\overline{\mathbf{Q}}$. Let $K$ be a number field such that $X$ has a semi-stable regular model $p:\mathcal{X}\to \mathrm{Spec} O_K$ over the ring of integers $O_K$ of $K$. Then, the Faltings height $h_{\mathrm{Fal}}(X)$ of $X$ is the arithmetic degree $$h_{\mathrm{Fal}}(X):=\frac{\widehat{\mathrm{deg}} Rp_\ast \mathcal{O}_{\mathcal{X}}}{[K:\mathbf{Q}]},$$ where we endow the determinant of cohomology with the Arakelov-Faltings metric. This is well-defined, i.e., independent of the field $K$. By Serre duality, it coincides with $$h_{\mathrm{Fal}}(X)=\frac{\widehat{\mathrm{deg}} p_\ast \mathcal{\omega}_{\mathcal{X}/O_K}}{[K:\mathbf{Q}]}.$$ It also coincides with the Faltings height of the Jacobian. All of this is explained in Section 4.4 of -https://arxiv.org/abs/math/0605244 -For a curve over a number field, there is also the important relative Faltings height. This invariant depends on the number field $K$, though. - -REPLY [15 votes]: Dear Ariyan, the elliptic curve with equation $$y^2=x^3+6$$ has Faltings height -$$-(3/2)\log(\Gamma(1/3)/\Gamma(2/3))+(1/4)\log(3)=-0.748752...;$$ the curve -of genus $2$ with equation $$y^2+y=x^5$$ has Faltings height -$$ -h_{\rm Fal}(C_{\bar{\bf Q}})=2\log(2\pi)- -{1\over 2}\log\big(\Gamma(1/5)^5\Gamma(2/5)^3\Gamma(3/5)\Gamma(4/5)^{-1}\big) -$$ -$$ -\approx --1.452509239645644650317707042; -$$ -For the first example, see Deligne, "Preuve des conjectures de Tate et Shafarevich", Séminaire Bourbaki. For the second one, see Bost, Mestre, Moret-Bailly, "Sur le calcul explicite des 'classes de Chern' des surfaces arithmétiques de genre $2$", Séminaire sur les Pinceaux de Courbes Elliptiques (Paris, 1988). Astérisque No. 183 (1990), 69–105. -Another explicit formula that should allow you to produce elliptic curves of arbitrarily large -Faltings height is the inequality -$$ -|h(j_E)-12h_{\rm Fal}(E)|\leqslant 6\log(1+h(j_E))+47.15 -$$ -See paragraph 5. of the article "Serre's uniformity..." by Bilu and Parent for references. -Something else you can do is make numerical experiments with formula in Conj. 3 of the article of Colmez, "Hauteur de Faltings..." (Compositio), which is true (without $\log(2)$ factor !, see A. Obus, arXiv:1107.0684) if the CM field is abelian over $\bf Q$. In that case, the Artin $L$-functions become Dirichlet $L$-functions and can be computed explicitly in terms of values of the Gamma function using the Hurwitz formula. -This is not a complete answer but I hope that it helps.<|endoftext|> -TITLE: Overlapping Gershgorin disks -QUESTION [16 upvotes]: We all know Gershgorin's Circle Theorem, which I will summarise for convenience. Let $A=(a_{ij})$ be an $n\times n$ complex matrix. Define the disks $D_1,\ldots,D_n$ by $$D_i = \Bigl\{ z : |z-a_{ii}|\le \sum_{j\ne i} |a_{ij}|\Bigr\}.$$ Then each eigenvalue of $A$ lies in one of the disks. Moreover, if a connected component of the union of the disks contains $k$ disks, then exactly $k$ eigenvalues of $A$ lie in that union. -My question is when a stronger statement is true. When is it possible to list the eignvalues $\lambda_1,\ldots,\lambda_n$ in such an order that $\lambda_i\in D_i$ for all $i$? -What is a small counterexample for general matrices? Is there a counterexample for real symmetric matrices? Is there a nice family of matrices for which there is no counterexample? -Note that by Hall's marriage theorem, the stronger statement is equivalent to saying that for each $k$, the union of any $k$ disks includes at least $k$ eigenvalues. - -REPLY [6 votes]: Let $A$ be a Hermitian matrix. Let $c$ be column $j$ of $A$, but with element $j$ set to zero, and let $E = ce_j^T + e_j c^T$, where $e_j$ is a standard basis vector. Note that $A-E$ has $a_{jj}$ as an eigenvalue, and a straightforward computation gives that -$$\|E\|_2 = \|c\|_2 \leq \|c\|_1$$ -Because $A$ and $A-E$ are both Hermitian, we know that there is an eigenvalue of $A$ within $\|E\|$ of each eigenvalue of $A-E$. In particular, there is an eigenvalue $\lambda$ of $A$ such that $|\lambda - a_{jj}| < \|c\|_1$ -- that is, there is an eigenvalue of $A$ in the $j$th Gerschgorin disk.<|endoftext|> -TITLE: What exactly does the weight filtration in Hodge theory have to do with the Weil conjectures? -QUESTION [21 upvotes]: Let $X$ be a variety over $\mathbb{C}$, say separated. According to Deligne's results, there is a "mixed Hodge structure" on the total cohomology $H^\bullet(X(\mathbb{C}), \mathbb{Z})$. One component of this is a "weight filtration" on $H^\bullet(X(\mathbb{C}), \mathbb{Q})$. I haven't read Deligne's "Theorie de Hodge" and don't really understand all this, but I believe that in the case of a smooth projective variety, this reduces to usual Hodge theory and the weight filtration is the filtration by grading, and the extension to singular varieties comes by some sort of simplicial resolution by smooth objects. -Let $Y_0$ be a variety over a finite field $\kappa$. Given a mixed perverse sheaf $K_0$ on $Y_0$, there is a canonical (and functorial) weight filtration on $K_0$, such that the sucessive subquotients are pure complexes of increasing weight (in the sense of Weil II). -What do these to have to do with each other? In section 6 of BBD (asterisque 100), it seems that the authors are using the functoriality of the weight filtration over finite fields to deduce results about the weight filtration over $\mathbb{C}$. Namely, I'd be interested if, given a perverse sheaf $K$ (say, of geometric origin) on a smooth, proper scheme $X$ over $\mathbb{C}$ which can be "spread out" to perverse sheaves of "reduction of $X$ mod a prime*" there is some way in which the weight filtration on the cohomology of $K$ (actually, I'm not sure that this exists, it seems to in the constant case at least) can be -viewed as a completion of the weight filtrations in finite characteristic. -Here is the specific result in BBD: Let $f: X \to Y$ be a separated morphism of schemes of finite type over $\mathbb{C}$. Suppose that the stalks of $R^n f_* \mathbb{Q}$ are $H^n(X_y, \mathbb{Q})$, and that these form a local system. Then the weight filtration on these stalks form a locally constant filtration of the local system $R^n f_* \mathbb{Q}$. -This appears to be proved by reducing mod a prime, where one has a Frobenius and the perverse weight filtration makes sense. -(One reason to think these might be related is that if $X_0$ is a proper smooth scheme over $\mathbb{\kappa}$, then the cohomologies $H^i(X, \mathbb{Q}_l)$ have weight $i$ by the Weil conjectures, and this has some correspondence with how the weight filtration was defined for projective, smooth schemes over $\mathbb{C}$.) -*Which is done by reducing the field $\mathbb{C}$ of definition to some finitely generated ring over $\mathbb{Z}$, and then working from there. - -REPLY [4 votes]: see Deligne's ICM 1970 address (Theorie de Hodge I) as well as his ICM 1974 address.<|endoftext|> -TITLE: When can a contractible 2-complex be embedded in R^3? -QUESTION [15 upvotes]: Let $X$ be a contractible 2-dimensional simplicial complex. Are there nice necessary and sufficient conditions for $X$ to be embeddable in $\mathbb R^3$? Clearly it is necessary that the link of every vertex be a planar graph. Is this sufficient? - -REPLY [14 votes]: If your complex is finite, then figure out the possible ways of -thickening it to a 3-manifold. The possible thickenings are -determined by the various embeddings of the links of the vertices -into $S^2$, then seeing if these induce compatible thickenings -over the edges (determined by the same cyclic ordering over -the link of the edge) and faces of the complex. If it can be thickened -this way, then it must be a ball since it is a contractible 3-manifold.<|endoftext|> -TITLE: Does every irreducible representation of a finite group G embed into its group algebra? -QUESTION [7 upvotes]: Let $G$ be a finite group, and let $F$ be a field. Is there a simple proof that every irreducible representation of -$G$ embeds into the group algebra $F[G]$? I am specially interested in the case when $gcd(|F|,G)\neq 1$? - -REPLY [16 votes]: A group algebra of a finite group over a field is a Frobenius algebra. See http://en.wikipedia.org/wiki/Frobenius_algebra -Essentially, it means there is a nice non-degenerate bilinear form on the algebra (send (a,b) to the coefficient of 1 in ab). In a Frobenius algebra the dual of the right regular module is isomorphic to to the left regular module. Since the injective indecomposables are the duals of the right projective indecomposables, it follows the injective indecomposables are direct summands in the left regular module. Since each simple embeds in its injective envelope (i.e., the dual of its right projective cover), it follows each simple module embeds in the left regular module. - -REPLY [6 votes]: When $F$ is algebraically closed (or, more generally a splitting field), the answer is -positive in the sense that every irreducible right $FG$-module is isomorphic to a minimal -right ideal of the group algebra $FG$ (I stick to my preferred notation). Possibly the simplest argument I know, which really dates back to Richard Brauer is as follows: -(I assume the structure of semi-simple algebras known, which is reasonable for MO). -Let $V$ be an irreducible right $FG$ module, and let $\sigma:FG \to {\rm End}_{F}(V)$ -be the associated representation. Let $\tau$ be the $F$-valued trace afforded by this representation. Form the element $i(\tau) = \sum_{g \in G} \tau(g^{-1})g$, which is -a central element of the group algebra $FG.$ Let $I_{\tau}$ denote the (two-sided) ideal -$i(\tau)FG$ of $FG.$ Notice that for each $h \in G$, we have -$i(\tau).h = \sum_{g \in G} \tau(g^{-1}h)g$. Hence for any $x \in FG$, -we have $i(\tau)x = \sum_{g \in G} \tau(g^{-1}x) g.$ In particular, $i(\tau)x = 0$ -whenever $x \in J(FG)$, since nilpotent endomorphisms have trace 0. More generally, -the annihilator of $V$ (which is a maximal two-sided ideal of $FG$) annihilates $i(\tau)FG$. -Since $\{ g\sigma: g \in G \}$ spans ${\rm End}(V)$, we see that an element $x$ of $FG$ -annihilates $V$ if and only if $\tau(g^{-1}x) = 0$ for each $g \in G$, so the annihilator -of $i(\tau)FG$ is no larger than the annihilator of $V$. Hence $i(\tau)FG$ is isomorphic -to the simple algebra ${\rm End}_{F}(V)$ just as as right $FG$-module, and the latter module is isomorphic to the direct sum of ${\rm dim}_F(V)$ copies of $V$ as right $FG$-module. -Hence a minimal right ideal of $FG$ contained in $i(\tau)FG$ is isomorphic to $V$.<|endoftext|> -TITLE: Stirling Number of first kind : Implementation -QUESTION [9 upvotes]: Hi everybody, -Does there exist an explicit formula for the Stirling Numbers of the First Kind which are given by the formula -$$ -x(x-1)\cdots (x-n+1) = \sum_{k=0}^n s(n,k)x^k. -$$ -Otherwise, what is the computationally fastest formula one knows? - -REPLY [7 votes]: There is an explicit formula : $s(n,m)=\frac{(2n-m)!}{(m-1)!}\sum_{k=0}^{n-m}\frac{1}{(n+k)(n-m-k)!(n-m+k)!}\sum_{j=0}^{k}\frac{(-1)^{j} j^{n-m+k} }{j!(k-j)!}.$ For once, it is not in Wikipedia (en), but in the french version of it (and I posted it there myself, if I may so brag)<|endoftext|> -TITLE: An image of the hierarchy of algebraic structures -QUESTION [9 upvotes]: Hello! Does anybody know an image of a graph featuring the hierarchy of algebraic structures? Something rather complete. -So far I've found similar images describing the hierarchies of classes/categories in various programming languages. For example - -Haskell's basic algebra library -Coq's math classes -So far the largest Axiom's abbreviated and full name graph of its categories (incomplete?) -category hierarchy of Sage - -REPLY [7 votes]: This is sort of a delayed response, but I came upon this question while googling for the visualization of the hierarchy of structures used in the Coq proof of the Feit-Thompson Odd Order theorem. That visualization is located here, image copied below:<|endoftext|> -TITLE: References on semismall maps -QUESTION [6 upvotes]: Where can I find references on semismall maps, in the sense of Goresky and MacPherson? I don't want to restrict to the case where the base is $\mathbb C$ (an arbitrary alg. closed field would be fine), or maps $f:X\to Y$ from a smooth variety $X.$ In particular, I'd like to find the proof (if the statement is correct, which I'm not sure) that $Rf_*$ takes an irreducible (middle) perverse sheaf $F$ supported on $X$ to a perverse sheaf; I can only do this when $X$ is smooth and $F$ is a lisse sheaf, or when all the fibers of $f$ have dimension at most one. -Recall: A proper surjective morphism $f:X\to Y$ is called $semismall$ if $\dim X\times_YX=\dim X.$ -Thank you. - -REPLY [6 votes]: The statement that $Rf_*$ takes a middle perverse sheaf to a perverse sheaf is not true for arbitrary perverse sheaves: -For example, let $Y$ be a smooth surface and $X$ the blow up of $Y$ at a point. Let $E$ be the exceptional divisor and $F$ the constant sheaf on $E$ placed in degree $-1$ (to make it perverse). The natural morphism $f:X \to Y$ is small but $Rf_* F$ is not perverse since it has cohomology sheaves supported on a point in degrees $-1$ and $1$. -EDIT: The other counterexample, suggested in a previous edit, does not appear to work.<|endoftext|> -TITLE: How would a motivic proof of the Riemann hypothesis over finite fields go? -QUESTION [13 upvotes]: It is well known that Grothendieck had a different idea than Deligne about how one should go about proving the Riemann hypothesis for finite fields. However, since Grothendieck's desired proof never came to fruition, I find it hard to look up what his proposed proof was. -It is clear from texts on the subject that he wanted to use motives in some way or other, and that he wanted to prove the standard conjectures first. Given the standard conjectures, is there an easy proof the Riemann hypothesis over finite fields? What is it? Did he want other things to be true also? What is the sketch he had in mind -- can you write a proof of the Riemann hypothesis with some conjectural black boxes like the standard conjectures? - -REPLY [18 votes]: The standard conjectures imply directly that the category of motives over the finite field $\mathbb{F}_q$ is a polarizable (hence semisimple) Tannakian category. Using only that, we have the following result. -PROPOSITION: Let $X$ be a motive of weight $m$ over $ \mathbb{F}_q$, and let $\alpha\mapsto\alpha^t$ be the involution of $\mathrm{End}(X)$ defined by a Weil form $ \varphi$. The following statements hold for the Frobenius endomorphism $\pi =\pi_X$ of $X$: -(a) $\pi\cdot\pi^t=q^m$; hence $\mathbb{Q}[\pi ]$ is stable under the involution $\alpha\mapsto\alpha^t$; -(b) $\mathbb{Q}[\pi ]\subset\mathrm{End}(X)$ is a product of fields; -(c) for every homomorphism $\rho\:\mathbb{Q}[\pi ]\rightarrow \mathbb{C}$, $\rho (\pi^t)=\iota (\rho\pi )$, and $|\rho\pi |=q^{m/2}$. ($\iota$ is complex conjugation) -PROOF: (a) By definition, $\varphi$ is a morphism $X\otimes X\to T^{\otimes (-m)}$ ($T$ is the Tate object). It is invariant under $\pi$, and so $$\varphi (\pi x,\pi y)=\pi (\varphi (x,y))=q^m\varphi (x,y)=\varphi (x,q^my).$$ But $\varphi (\pi x,\pi y)=\varphi (x,\pi^t\pi y)$, and because $ \varphi$ is nondegenerate, this implies that $\pi^t\cdot\pi =q^m$. Therefore $\mathbb{Q}[\pi ]$ is stable under $ \alpha\mapsto\alpha^t$, and we obtain (a). -(b) Let $R$ be a commutative subalgebra of $\mathrm{End}(X)$ stable under $\alpha\mapsto\alpha^t$, and let $r$ be a nonzero element of $R$. Then $ s=rr^t\neq 0$ because $\mathrm{Tr}(rr^t)>0$. As $s^t=s$, $\mathrm{Tr}(s^2)=\mathrm{Tr}(ss^t)>0$, and so $s^2\neq 0$. Similarly $s^4\neq 0$, and so on, which implies that $s$ is not nilpotent, and so neither is $r$. Thus $R$ is a finite-dimensional commutative $\mathbb{Q}$-algebra without nonzero nilpotents, and the only such algebras are products of fields. -(c) In an abuse of notation, we set $\mathbb{R}[\pi ]=\mathbb{R}\otimes_{ \mathbb{Q}}\mathbb{Q}[\pi ]$. As in (b), this is a product of fields stable under $\alpha\mapsto\alpha^t$. This involution permutes the maximal ideals of $\mathbb{R}[\pi ]$ and, correspondingly, the factors of $ \mathbb{R}[\pi ]$. If the permutation were not the identity, then $\alpha\mapsto\alpha^ t$ would not be a positive involution. Therefore each factor of $\mathbb{R}[\pi ]$ is stable under the involution. The only involution of $\mathbb{R}$ is the identity map (= complex conjugation), and the only positive involution of $\mathbb{C}$ is complex conjugation. Therefore we obtain the first statement of (c), and the second then follows from (a). -This (conjectural) proof of the Riemann hypothesis for motives is very close to Weil's original proof for abelian varieties (Weil 1940).<|endoftext|> -TITLE: Why did Gabriel invent the term "quiver"? -QUESTION [28 upvotes]: A quiver in representation theory is what is called in most other areas a directed graph. Does anybody know why Gabriel felt that a new name was needed for this object? I am more interested in why he might have felt graph or digraph was not a good choice of terminology than why he thought quiver is a good name. (I rather like the name myself.) -On a related note, does anybody know why quiver representations, resp. morphisms of quiver representations, are not commonly defined as functors from the free category on the quiver to the category of finite dimensional vector spaces, resp. natural transformations? -Added I made this community wiki in case this will garner more responses. -My motivation for asking this is that one of my students just defended her thesis, which involved quivers, and the Computer Scientist on the committee remarked that these are normally called directed graphs and using that term might make the thesis appeal to a wider community. Afterwards, some of us were wondering what prompted Gabriel to coin a new term for this concept. - -REPLY [15 votes]: Gabriel actually gave a short explanation himself in [Gabriel, Peter. Unzerlegbare Darstellungen. I. (German) Manuscripta Math. 6 (1972), 71--103]: - -Für einen solchen 4-Tupel schlagen wir die Bezeichnung Köcher vor, und nicht etwa Graph, weil letzerem Wort schon zu viele verwandte Begriffe anhaften. - -Attempt at translation: For such a 4-tuple we suggest the name quiver, rather than graph, since the latter word already has too many related concepts connected to it. -(This is community wiki, so anyone can add a proper English translation.)<|endoftext|> -TITLE: Why does the definition of modularity demand weight 2? -QUESTION [19 upvotes]: Allow me to quote a definition from Gelbart in "Modular Forms and Fermat's Last Theorem": -Definition. Let $E/\mathbb{Q}$ be an elliptic curve. We say that $E$ is modular if there is some normalised eigenform -$$ f(z) = \sum_{i=1}^{\infty} \ a_ne^{2\pi inz} \in S_2(\Gamma_0(N),\epsilon), $$ -for some level $N$ and Nebentypus $\epsilon$, such that -$$ a_q = q + 1 - \#(E(\mathbb{F}_q)) $$ -for almost all primes $q$. -This is the basic question of the post: - -Why is the weight of $f$ taken to be 2? Can I instead take 3, or 4, or 5, or even 19/2, without disturbing the peace? - -I am aware of other definitions of modularity, some of which don't mention modular forms at all, but nonetheless I feel that weight 2 lurks beneath all of these. -I think one approach would involve differentials, and the construction of Eichler-Shimura, but I'm not so sure. Further, perhaps there are several reasons which fit together to tell a nice story. -Is it a corollary of this question that it doesn't matter what the weight is? -Finally, can I replace $E$ above with any abelian variety, and ask the same question? - -REPLY [3 votes]: I started to write this as a comment to the original question, but it became too long. It is not disjoint from the previous expert responses, but it emphasizes a particular viewpoint. -One can (and one should) shift normalize any automorphic (in particular any modular) $L$-function so that the functional equation relates $s$ to $1-s$. Then the gamma factors identify the archimedean component of the underlying automorphic form much like the Euler factors identify the non-archimedean components. In particular, if the $L$-function of a holomorphic cusp form is so normalized (i.e. $s$ is related to $1-s$), then the gamma factors determine the weight (and vice versa). I say gamma factors, because the usual single gamma factor can be factored into two gamma factors by the doubling formula for the gamma function (for a modular $L$-function the "true gamma factors" form a pair). -So you can ask your question as follows. If we shift normalize the $L$-function of an elliptic curve so that the functional equation relates $s$ to $1-s$, then why are the gamma factors always the same? In order to ask this question you need to assume already that the $L$-function obeys a rather specific functional equation with gamma factors, and then the question inquires what the gamma factors can be. -It is possible that the fairest answer to this question (i.e. your question) is as follows: -It was an experimental fact that the gamma factors are always the same, hence the precise form of the modularity conjecture was formulated, which then turned out to be right, namely it was proved by great efforts of great mathematicians. -Once again, normalization of all automorphic $L$-functions is key in this discussion, it should not be underestimated. It is not the usual normalization favored by algebraic people. -Added: In a leisurely style one could say the following. The first miracle is that the $L$-function of an elliptic curve is entire and satisfies some functional equation. The second miracle is that the $L$-function is automorphic, as suggested by the functional equation. More specifically, it looks like a $\mathrm{GL}_2$ automorphic $L$-function. Not only it is $\mathrm{GL}_2$, but it comes from a very specific modular form, namely a holomorphic form, let's call this the third miracle. Then, as a final miracle, this holomorphic form is always of weight 2 whose level can also be specified in terms of the elliptic curve.<|endoftext|> -TITLE: error estimates for multi-dimensional Riemann sums -QUESTION [7 upvotes]: Suppose that $f$ is a continuous function of bounded variation from $R^2$ to $R$ that's negative outside of some bounded set, and let $F=\max(f,0)$. Let $S_n$ be the Riemann sum for the integral of $F$ over $R^2$ obtained by summing the values of $F$ at all points in the lattice $(Z/n)^2$ and dividing by $n^2$. What sort of bounds can be given for the difference between $S_n$ and the integral of $F$ over $R^2$? ($O(1/n)$ or $O(1/n^2)$ or what?) -Also how can this basic bound be improved if one knows more about $f$, e.g. that it is smooth or concave? -I'm restricting the question to functions on $R^2$ for definiteness, but I'd like to know the more general situation for $R^n$. - -REPLY [8 votes]: With the hypotheses given, one can't do better than $O(1/n)$ decay. Consider for instance the function $\frac{1}{n} \cos^2(2\pi n x_1)$ smoothly localised to a ball for some large $n$. This has a total variation norm of $O(1)$, but for this specific value of $n$, the Riemann sum will be off by $O(1/n)$. -Of course, this function depends on $n$. For an $n$-independent example, one could then consider the Weierstrass type function $\sum_{n=1}^\infty \frac{1}{j^2 n_j} \cos^2(2\pi n_j x_1)$ smoothly localised to the unit ball, where $n_j$ goes rapidly to infinity. This is still continuous and of bounded variation, but now the Riemann sum will be off by about $O(1/j^2 n_j)$ at scale $1/n_j$. -In dimensions $d$ greater than 1, the situation is much worse; one can't do much better than $O(1)$, basically because of the failure of the Sobolev embedding $W^{1,1} \subset L^\infty$ in higher dimensions. For instance, one can consider a function $f$ that consists of a bump function of height 1 localised to a ball of radius $O( n^{-d/(d-1)} )$ at each lattice point on $\frac{1}{n} {\bf Z}^d \cap B(0,1)$. This has total variation norm $O(1)$ and is bounded by $O(1)$, but the Riemann sum is off by $O(1)$. By superimposing several such examples together as in the Weierstrass type example we can then construct an $n$-independent function of bounded variation and continuous of compact support whose Riemann sum error decays as slowly as one pleases. -Once one does have enough regularity (in, say, a Sobolev class) to control local $L^\infty$ oscillation, then one can estimate the error term in the Riemann sum by partitioning space into cubes, using some sort of local Sobolev inequality on each cube, and summing up. This for instance gives an $O(1/n)$ error term in the one-dimensional bounded variation case. -One can also analyse Riemann sums by Littlewood-Paley theory. Functions whose Fourier transform is supported on frequencies much smaller than $n$ have excellent agreement between the integrals and their Riemann sums (particularly if one uses quadrature to improve the accuracy of the latter), and functions whose Fourier transform are supported on frequencies much larger than $n$ have a negligible integral. So the error term is basically the same thing as the Riemann sum of the high-frequency component of the function $f$. -Concavity should be very helpful, ruling out the oscillatory counterexamples mentioned above and giving some new bounds on first and second derivatives of $f$ that can be plugged into the local Sobolev inequality method, but I don't immediately see what the best bounds would be with this hypothesis.<|endoftext|> -TITLE: What is the normalizer of the circle in the diffeomorphism group of the 2-sphere? -QUESTION [9 upvotes]: What is the normalizer of $SO(2)$ in $\mathrm{Diff}(S^2)$? -Remarks: - -We let $SO(2)$ act on $S^2$ via the rotation about the $z$-axis. -It is immediate that each element of the normalizer must map any parallel -(i.e. a fiber of the projection of $S^2$ onto the $z$-axis) -to a parallel. -Of course, the normalizer -contains $O(2)$, as well as the following elements that ``push along the meridians''. Let $(\phi, r)\in [0,2\pi]\times [0, \pi]$ be coordinates on $S^2$ -with parallels given by $r=\mathrm{const}$. Define a self-map of $S^2$ -by $H(\phi, r)=(\phi, r+h(r))$ where $h$ is some smooth function subject -to the boundary conditions ensuring that $H$ is a diffeomorphism -(e.g. $h$ vanishes near $0$ or $\pi$). Then $H$ commutes with the -$SO(2)$-action, that is given by translation in $\phi$. - -REPLY [2 votes]: Let's call your coordinates $\phi$ and $\theta$, as is more usual. Thus $(x,y,z)=(sin \phi\ cos\theta, sin\phi\ sin\theta, cos\phi)$. -Yes, in the centralizer of $SO(2)$ there is the group that leaves $\theta$ unchanged, $(\phi, \theta)\mapsto (f(\phi ),\theta)$ where $f$ is a diffeomorphism from $[0,\pi]$ to itself that is nice enough at the endpoints. Also there is the group that leaves $\phi$ unchanged, $(\phi, \theta)\mapsto (\phi ,\theta +g(\phi))$ where $g$ is a smooth map from $[0,\pi]$ to $\mathbb R/2\pi \mathbb Z$. The centralizer is the semidirect product of these, and the normalizer is bigger by a factor of two. -EDIT Let's consider the related and slightly easier problem of maps $F$ from the plane to itself that commute with all of $SO(2)$, in other words maps $F:\mathbb C\to \mathbb C$ such that $F(re^{i\theta} -)=F(r)e^{i\theta}$. Such a function is determined by its restriction to $\mathbb R$. That restriction must be odd, $F(-x)=-F(x)$, by considering $\theta=\pi$. Thus if $F$ is smooth then the restriction can be written as $x\mapsto xG(x^2)$ for a smooth $G:[0,+\infty)\to\mathbb C$. Conversely, given any such smooth $G$ we can write $F(z)=zG(|z|^2)$ and get a smooth map $\mathbb C\to \mathbb C$, the unique such map commuting with rotations and having $x\mapsto xG(x^2)$ as its restriction to $\mathbb R$. -If $F$ is a diffeomorphism then $G(0)$ is different from $0$ and also $G(u)$ is different from $0$ when $u>0$. Thus $G$ can be written $G(u)=a(u)e^{ib(u)}$ where $a>0$ and $b$ are smooth real functions of $u\ge 0$. The only further constraint on $a$ or $b$ is that $x\mapsto xa(x^2)$ must be a diffeomorphism from the positive reals to itself.<|endoftext|> -TITLE: What reasonable choices of morphisms are there for the category of Poisson algebras? -QUESTION [12 upvotes]: The first definition of the category of Poisson algebras that comes to mind is that a morphism between Poisson algebras is an algebra homomorphism that is also a Lie algebra homomorphism with respect to the Poisson bracket. This definition does not seem to be easily compatible with how people actually use Poisson algebras (in particular rings of functions on Poisson manifolds): - -A Poisson-Lie group is not a group object in the opposite of the category of Poisson algebras because inversion negates the Poisson bracket. -The standard choice of bracket on the tensor product of two Poisson algebras is not a categorical coproduct (if I have the correct general definition: it's defined by the requirements that it restricts to the given brackets on two Poisson algebras $A, B$ and that every element of $A$ Poisson-commutes with every element of $B$). - -This suggests to me that if we used a different choice of morphisms, we might get actual group objects and an actual categorical coproduct. So are there any nice choices that do this? -I read somewhere on MO that the correct definition of a morphism between Poisson manifolds is a Lagrangian submanifold of their product. How does this generalize to Poisson algebras? Does it fix the two issues above? (I'm a little more pessimistic about the second issue, so if there's a different general principle that leads to the standard choice, I would be interested in hearing about that as well.) -Edit: The discussion in my previous question about Poisson-Lie groups seems relevant, and perhaps it shows that the above point of view is misguided. Any Poisson algebra $A$ admits an "opposite" $A^{op}$ given by negating the Poisson bracket, and then inversion in a Poisson-Lie group is a "contravariant morphism" rather than a morphism. This suggests to me that it might make more sense to look for a bicategory of Poisson algebras similar to the bimodule bicategory. - -REPLY [6 votes]: As you suggest in your question and Todd Trimble mentions in a comment, one interesting choice of morphism between Poisson manifolds is that of a coisotropic correspondence: if $M, M'$ are Poisson manifolds, depending on exactly how you work you either think about coisotropic submanifolds in $\bar M \times M'$, or maps $N \to \bar M \times M'$ with coisotropic image, where $\bar M$ is the same manifold as $M$ but with the opposite Poisson structure (and I give $\bar M \times M'$ the product Poisson structure that you're rightly not fond of). Then it is a straightforward fact that a correspondence $N \subseteq M\times M'$ which is the graph of a smooth map $M \to M'$ is coisotropic in $\bar M \times M'$ iff the map is a Poisson map. -Note that this all generalizes the category in which objects are symplectic manifolds and morphisms are Lagrangian correspondences --- then a correspondence that is the graph of a smooth function is the graph of a symplectomorphic open embedding iff it is Lagrangian. It also has just as many bad properties. Notably, only composition between generic morphisms is defined, as in the non generic case some intersections may not be transverse. So to make it into a category requires the same kind of $A_\infty$ work (or Wehrheim-Woodward method, or...). I know that some of Alan Weinstein's recent papers discuss this category. -This category generalizes easily to the algebraic case that you ask about. Recall that an ideal in a Poisson algebra is coisotropic if it is a Lie subalgebra for the bracket (not necessarily a Lie ideal!), and that a submanifold of a Poisson manifold is coisotropic iff its vanishing ideal is coisotropic. So what I'm suggesting is that if $P,P'$ are Poisson algebras, and writing $\bar P$ for $P$ with the opposite Poisson structure, then one interesting notion of "morphism" $P \to P'$ is a coisotropic ideal in $\bar P \otimes P'$. -Dima Shlyakhtenko has suggested more or less the same category in another answer. There is the following philosophy: Poisson manifolds / algebras are a sort of "infinitesimal" piece of noncommutative algebra, and under this rough relationship coisotropic submanifolds are supposed to correspond to (left, say) modules. Then coisotropic correspondences are roughly the same as bimodules. Recall that from an algebra point of view, bimodules are a fairly natural notion of morphism: they are precisely the left adjoints (say, or right adjoints, or adjunctions) between the corresponding categories of modules. The module theory of an algebra knows a lot about the algebra, including its Hochschild homology and cohomology (and hence its center, its perturbative deformation theory, and so on). -Of course, it is far from the case that the tensor product of algebras has much to do with the (co)product in any category. Rather, remembering only the Morita theory of algebras helps to explain what is their tensor product: it is the tensor product in the 2-category of (nice) categories with left-adjoints as morphisms, in the sense of being universal for "bilinear" maps. One can be quite precise about this: the 2-category of algebras and bimodules is a categorification of the 1-category of abelian groups. Actually, if you remember the underlying algebra, then that's the same as remembering its module theory along with the data of a "rank-1 free module", and so this is a categorification of the 1-category of abelian groups with a distinguished element. (Morita theory is like linear maps that ignore the distinguished element.) -Incidentally, it is now straightforward to invent the notion of "sesquialgebra", which is an algebra object in the 2-category of algebras and bimodules, or equivalently a closed monoidal category structure on the module theory of said algebra. The same notion in Poisson manifolds is an algebra object in the category of Poisson manifolds and coisotropic correspondences, so this includes the Poisson Lie monoids. Alan Weinstein and collaborators a few years ago tried to write down a good notion of "Hopfish algebra" for controlling when this map would be invertible, but my opinion is that their paper doesn't quite get it right. What you should do is the following. Recall that a functor between monoidal categories is strong-monoidal if it comes equipped with a natural isomorphism between the two ways of composing the functor and the corresponding monoidal structures (and maybe extra data for associativity, etc.). A strong monoidal functor between closed monoidal categories also determines a natural transformation between "inner homs", which need not be a natural iso. If it is, call the monoidal functor "hopfish" or "strongly closed". A bialgebra is a sesquialgebra with a marked right adjoint to Vect (equivalently, a marked "rank 1 free algebra", the image of the 1-dimensional vector space under the corresponding left adjoint) which is strong monoidal; a Hopf algebra is a bialgebra in which the strong monoidal functor is hopfish.<|endoftext|> -TITLE: A characterisation of well-ordering ? -QUESTION [8 upvotes]: It is easy to prove that if $E$ is well-ordered, and if $f$ is a strictly increasing map from $E$ to $E$, then, for all $x$ in $E$, $f(x) \ge x$ (just consider the sequence $x$, $f(x)$, $f(f(x))\dots$). But is the converse true, i.e. for any totally ordered set $E$ which is not well-ordered, does it exist a strictly increasing map $f$ from $E$ to $E$ and an element $a$ in $E$ such that $f(a) < a$ ? Even assuming choice, I couldn't find a proof (or a counterexample) ; Cantor-Bendixon (or its generalisation to surreals) seems involved, but it could be a red herring. Any hint? - -REPLY [10 votes]: There are dense subsets $X$ of the real line with the usual order (hence not well-ordered) such that the only strictly increasing map from $X$ to itself is the identity. Here's a sketch of the construction. First note that, for any dense set $X$ of reals, an increasing map from $X$ to itself extends to an increasing map on the reals (not necessarily uniquely, beacuse there may be countably many jumps). Such extensions $f$ are determined by their values at the rationals plus some information about jumps; in particular, there are only continuum ($\mathfrak c$) many possibilities. Well-order the set of all such possibilities $f$ in a sequence of length $\mathfrak c$ (the initial ordinal). Build the desired $X$ in $\mathfrak c$ stages, putting one number into $X$ and one into the complement of $X$ at each step, choosing these numbers so as to defeat one possible $f$ at each step.<|endoftext|> -TITLE: What are the statistics of prime knots in 3d Random walk? -QUESTION [9 upvotes]: This question on physics stackexchange https://physics.stackexchange.com/questions/12973/the-entropic-cost-of-tying-knots-in-polymers has a formulation which is perhaps more appropriate for this forum. -Given a Brownian motion for time t, link the ends to infinity by horizontal lines parallel to the x-axis, going in opposite directions. The walk will not intersect those lines generically, since a 2d random walk is marginally recurrent. You have then closed a loop on the one-point sphere compactification, and it makes sense to ask what knot you made. -There is a scaling problem, so that the knot you get might be very wild. But one can fix this by asking the right question in the limit. Approximate the Brownian motion with small randomly oriented straight line segments. Then, for long walks, the resulting knot will have a prime decomposition, and it is is very plausible to me that the number of prime knots of each type in the prime decomposition will converge to a fixed distribution in the limit of long walks. -Does this distribution exist? -Is there a more efficient method than simulation to get the distribution? - -REPLY [7 votes]: I am not sure how your 2D random walk relates to knots but physicists have investigated -random knotting in 3D. You may be aware of this already. I learnt about this when the following paper was presented at a meeting in Warsaw. -MR1634449 (99e:57010) Deguchi, Tetsuo ; Tsurusaki, Kyoichi . -Numerical application of knot invariants and universality of random knotting. - Knot theory (Warsaw, 1995), - 77--85, Banach Center Publ., 42, Polish Acad. Sci., Warsaw, 1998. -The hypothesis is that if you have a model for random knots of length L then -the probability that you realise a fixed knot K scales as -$C(K) (L/N)^{m(K)} \exp(-L/N)$ -where $C(K)$ depends on the knot and the model, $m(K)$ depends on the knot but not -on the model and $N$ depends on the model and not on the knot. -This paper also conjectures that $m(K)$ is additive under connected sum. -I don't know what has happened in this area since then. If this hypothesis is correct -then $m(K)$ becomes a fascinating knot invariant.<|endoftext|> -TITLE: different N=2 SUSY structures on the chiral de Rham complex of a Calabi-Yau manifold? -QUESTION [11 upvotes]: The context -In a beautiful paper, Malikov-Schechtman-Vaintrob defined a canonical sheaf of vertex algebras equipped with a differential on any manifold $X$ (either in the $C^\infty$, complex analytic or algebraic context). They called it the chiral de Rham complex (it is called this way because the ordinary de Rham complex embed into the chiral de Rham complex, and this embedding is a quasi-isomorphism), and denoted it $\Omega^{ch}_X$. -They also proved in the complex analytic setting that $\Omega^{ch}_X$ carries the structure of a conformal vertex algebra. Moreover, if $X$ is Calabi-Yau (in the weak sens: $X$ admits a global holomorphic volume form) then $\Omega^{ch}_X$ admits the structure of a topological vertex algebra (such are structures are in 1-1 correspondance with $N=2$ superconformal vertex algebra structures, aren't they?). -In another paper (also very nice), Ben-Zvi-Heluani-Szczesny proved that in the $C^\infty$ context, we have that: - -if $X$ is Riemannian then $\Omega^{ch}_X$ admits a $N=1$ superconformal vertex algebra structure. -if the metric is Kähler and Ricci-flat then $\Omega^{ch}_X$ inherits a $N=2$ superconformal structure. - -The question(s) -My question is then - -What is the relation between those - $N=2$ superconfromal structures when - $X$ is Calabi-Yau. - -From what I understand, when $X$ is kähler the complex analytic chiral de Rham complex embbed into the $C^\infty$ chiral de Rham complex, and the $N=1$ superconformal structure of Ben-Zvi-Heluani-Szczesny restricts to the conformal structure of Malikov-Schechtman-Vaintrob. -But it seems that the $N=2$ superconformal structure of Ben-Zvi-Heluani-Szczesny does not restrict to the one of Malikov-Schechtman-Vaintrob in the case when $X$ is Calabi-Yau unless the metric is flat. - -Does anybody understand what is going on there? - -In yet another paper Heluani contructs yet another $N=2$ superconformal structure on any kähler manifold $X$, which commutes with the one constructed by Ben-Zvi-Heluani-Szczesny when $X$ is Calabi-Yau. - -Is this new $N=2$ superconformal structure related to the one constructed by - Malikov-Schechtman-Vaintrob ? If not, then do the three $N=2$ structures commute ? - -REPLY [4 votes]: Ooops, I should start reading this site. In some sense all of these N=2 structures are the same. Unfortunately we now understand the situation well better than we did then. -To simplify the answer you may think of the $C^\infty$ Chiral de Rham (CDR) as the tensor product of a holomorphic CDR with a anti-holomorphic one (both commuting, and I presume this is the embedding you mentioned in the post). Then [MSV] constructed two commuting copies of N=2: one in each sector. The N=2 structure of [BZHS] would be their difference. -The unfortunate misleading comment in [BZHS] about the N=2 structure there agreeing with [MSV]'s only with flat metrics is due to the following. The fields in [BZHS] are written in general coordinate systems that's why they depend on a choice of a global holomorphic volume form in the CY (N=2) case. This dependence enters there as derivatives of $\sqrt{\det g}$. The fields in [MSV] are written in the coordinates where the volume form is taken to be constant and so this term does not appear. One of the advantages of having the fields in general coordinates was to treat the N=4 case which is the main point of [BZHS]. -To compare the two N=2 structures of [MSV] (holomorphic + anti-holomorphic) with the two in the second paper you mentioned (I'll call that [H]) the situation is subtler. In short: I claim that there is a much more natural embedding of holomorphic and anti-holomorphic CDR into the smooth one than the obvious one, and this has to do with different topological twistings. -Having the fields in general coordinates makes it clear that the two different N=2 structures in [H] correspond to the complex and symplectic structures of the Kahler manifold (or more generally to the two generalized complex structures defining the Kahler structure). In order to compare the two N=2 structures of [MSV] with the two in [H] one needs to embed the (anti)holomorphic CDR into the smooth one in a different way which essentially amounts to the orthogonal identifications (here I'm using the standard Kahler structure on $\mathbb{C}$). -$dz \rightarrow dz + \partial_{\bar z}, \qquad \partial_{z} \rightarrow d\bar{z} + \partial_z$ -This is the content of Remark 8 of this article which in turn is an infinite dimensional manifestation of the fact that the p,q decomposition of generalized Kahler manifolds is not the Dolbeaut decomposition, but rather this orthogonal transformation of it as explained in Gualtieri's article. With this different embedding then the two structures coincide (but this is non-trivial since the Kahler form enters the embedding). -Perhaps is just a form of advertizing, but CDR (at least for the purpose of these computations) should be viewed as a Courant-algebroid version of the (N=1 super) affine vertex algebra associated to a Lie algebra with an invariant non-degenerate form (,). The computations in that article with Zabzine mentioned above might look longer due to the generality of the situation (in the Generalized CY situation dilatons are unavoidable) but are coordinate free and explicitly invariant, perhaps taking a look at that article will clarify better your questions than this post. I hope this helped a little.<|endoftext|> -TITLE: What is the protocol for making modifications to someone else's proof to prove something slightly stronger? -QUESTION [12 upvotes]: I have a need to modify Erdős' proof of the Sylvester-Schur Theorem to prove something stronger. See my working document at http://math.rudytoody.us/ or http://math.rudytoody.us/OppermannTheorem.pdf -If I have to modify most of the proof, I will use the entire proof (with proper attribution, of course.) However, I don't believe I will need to do that. So, how much should I show of the original? Could I do a line-by-line comparison of the changes? If I only change a few variables, could I do something along the lines of, "By changing variables a, b, c and relaxing condition x, it's easy to see that Erdős' proof arrives at the same conclusion without breaking the original." -Some suggestions would be appreciated. Thanks. - -REPLY [15 votes]: I'm following Todd Trimble's suggestion and writing my comment as an answer: -I suggest you write your own proof completely, then mention that your proof is a modification of the proof of Erdős, then cite his paper.<|endoftext|> -TITLE: are there soliton solutions for Euler and Navier-Stokes Equation -QUESTION [6 upvotes]: I'm now reading papers about the the well-posedness of Euler and Navier-Stokes Equation, so I wonder if we have soliton solutions for this two equations just like for KdV equation. I'm interested in this because if soliton solutions exist, then we can try larger space for initial data, which includes the soliton, to work in for the well-posedness, and also we can consider the stability for the soliton solutions. -I searched in google, but haven't got any positive result. - -REPLY [13 votes]: There are solitary wave solutions for the Euler equations, but they do not have the "soliton" property of passing through each other without changing shape. Friedrichs and Hyers proved existence of such solutions in the 1950s for the case of zero surface tension. The problem with surface tension was solved in the 1980s and 1990s. -Here is one reference, which will lead you to the earlier ones: -S.M. Sun, Proc. Roy. Soc. London A 455 (1999), 2191-2228.<|endoftext|> -TITLE: Is there a MAGMA function to calculate the absolutely irreducible components of an algebraic curve defined over the rationals? -QUESTION [8 upvotes]: Given a curve defined over the rationals, is it computationaly possible to find all its absolutely irreducible components? -Is there an implementation of this in the MAGMA program? - -REPLY [3 votes]: An algorithm to find irreducible components is described here: -On computing absolutely irreducible components of algebraic varieties with parameters, by Ali Ayad (published in "Computing", 2010) -Abstract: -This paper presents a new algorithm for computing absolutely irreducible components of n-dimensional algebraic varieties defined implicitly by parametric homogeneous polynomial equations over Q, the field of rational numbers. The algorithm computes a finite partition of the parameters space into constructible sets such that the absolutely irreducible components are given uniformly in each constructible set. Each component will be represented by two items: first by a parametric representative system, i.e., the equations that define the component and second by a parametric effective generic point which gives a parametric rational univariate representation of the elements of the component. The number of absolutely irreducible components is constant in each constructible set. The complexity bound of this algorithm is δO(r4)dr4dO(n3), being double exponential in n, where d (resp. δ) is an upper bound on the degrees of the input parametric polynomials w.r.t. the main n variables (resp. w.r.t. r parameters). -I don't know if this, or any other, algorithm is implemented in Magma, but I am guessing that a simple version is not so hard to implement. - -REPLY [3 votes]: Magma (respectively, Sage!) can find the prime and primary components of a scheme $X$ (the irreducible components of a scheme $X$). -Look at the Magma documentation and also the Sage documentation on algebraic schemes. There are examples of usage in the documentation.<|endoftext|> -TITLE: Alternate definition of ordinals ? -QUESTION [8 upvotes]: Hello, recently I came upon some personal notes I'd made several years ago while reviewing some basic set theory (ordinals, transfinite recursion, inaccessible cardinals etc.), and I stumbled upon a loose thread which I obviously had not resolved at the time, and which I would like to lay to rest: -Assuming some standard set theory (say ZF, even though I prefer NBG), without the Axiom of Foundation (preferably), one may define an ordinal $\alpha$ (von Neumann's definition) as a transitive set whose elements are well-ordered with respect to the membership relation $\in$. This is seen to be equivalent to the statement that $\alpha$ is transitive, all its $\beta\in\alpha$ are transitive too, and (as we cannot rely on foundation) for each non-empty $x\subseteq\alpha$ there exists some $\beta\in x$ such that $x\cap\beta=\emptyset$ (except for the last condition, this is as in Schofield's book on Mathematical Logic). One then goes on to prove that the class of all ordinals is well-ordered with respect to membership etc.; along the way a useful intermediate step is to prove that any ordinal $\alpha$ is (ad hoc definition) $\textbf{strange}$ in the sense that one has $x\in\alpha$ for any transitive $x\subsetneq\alpha$. -My question finally (as this would provide an alternate definition of ordinal sets): are elements of strange sets themselves strange, or at least transitive ? -Thanks in advance for any useful comments ! Kind regards, Stephan F. Kroneck. - -REPLY [15 votes]: Theorem. Every strange set is an ordinal. -Proof. Suppose that $\alpha$ is strange. Let $\beta$ be the smallest ordinal such that $\beta\notin\alpha$. Such a $\beta$ exists, because no set can contain all the ordinals, and this does not require the foundation axiom to prove. It follows that $\beta\subset\alpha$ and $\beta$ is transitive. Thus, if $\beta\neq\alpha$, we would have $\beta\in\alpha$, contradicting the choice of $\beta$. Hence $\beta=\alpha$ and $\alpha$ is an ordinal. QED - -REPLY [14 votes]: Strange sets are the same thing as ordinals. Given a strange set $\alpha$, let $\beta$ be the smallest ordinal such that $\beta\notin\alpha$. Then $\beta\subseteq\alpha$. If $\beta\subsetneq\alpha$, then $\beta\in\alpha$ as $\alpha$ is strange, which contradicts the definition of $\beta$. Thus $\beta=\alpha$.<|endoftext|> -TITLE: On sufficient conditions on an analytic map to be algebraic(=regular) -QUESTION [5 upvotes]: Let $X$ and $Y$ be smooth quasi-projective varieties defined over $\mathbf{C}$ and let -$$ -f:X(\mathbf{C})\rightarrow Y(\mathbf{C}) -$$ -be a holomorphic map (not necessarily regular=algebraic). Then it is natural to ask what are additional conditions that one can impose on the data $(f,X,Y)$ in order to force -$f$ to be algebraic. Let me give 3 examples of such conditions: -1) Assume that $f$ is finite, unramified and that $X(\mathbf{C})$ has only one algebraic structure. Then a combination of Grauert-Remmert and GAGA implies that $f$ is algebraic. Note that (a postiori) the finiteness assumption on $f$ is essential since one has for example the exponential map $exp:\mathbf{C}\rightarrow\mathbf{C}^{\times}$ which is not algebraic but satisfy all the other assumptions (except the finiteness). -Moreover, in general, it is also essential to assume that $X(\mathbf{C})$ has only one algebraic structure since there are examples of complex manifolds with at least 2 non-equivalent algebraic structures. -2) If $X$ is compact then from GAGA we ge automatically that $f$ is algebraic -3) Say that $X$ is a curve and $Y=\mathbb{P}^1(\mathbf{C})-\{0,1,\infty\}$. Then Picard's theorem (+removable singularity result) imply that $f$ is meromorphic on the compactification of $X$ and therefore $f$ is algebraic. (If I remember correctly, I think that there is some kind of generalization of Picard's result to higher dimension from the work of Kwack). -So with these 3 examples in mind, here is my question: -Q: what is known in the litterature about additional conditions that one may impose on the data $(f,X,Y)$ in order to force $f$ to be algebraic? - -REPLY [4 votes]: Borel (1972, J. Diffl. Geometry) proved that $f$ is always algebraic if $Y$ is the quotient of a bounded symmetric domain by a torsion-free arithmetic subgroup. This is a super-generalization of your example 3 (the quotient of the complex upper half plane by $\Gamma(2)$ is isomorphic to the projective line minus three points). The proof uses a generalization of work of Kwack plus the resolution of singularities. -Added: Kwack (1969) generalized the big Picard theorem by proving that any -holomorphic map from the punctured unit disk into a hyperbolic complex space -can be extended holomorphically to the whole unit disk. [A reduced complex -space is said to be hyperbolic if the Kobayashi pseudodistance is a distance -(Kobayashi 1967).] -Borel 1972 replaced the punctured disk in Kwack's theorem with a product of -punctured disks and disks. -Resolution of singularities allows you to realize a -smooth algebraic variety as an open subvariety of a smooth projective variety -in such a way that the boundary is a divisor with normal crossings (hence -analytically a product of punctured disks and disks). -These statements sometimes allow you to extend your map to an analytic map of projective varieties, where you can apply Chow's theorem to prove that it is regular. -References: -Borel, Armand. Some metric properties of arithmetic quotients of symmetric -spaces and an extension theorem. J. Differential Geometry 6 (1972), 543--560. -Kwack, Myung H., Generalization of the big Picard theorem. Ann. of Math. (2) -90 1969 9--22. -Kobayashi, Shoshichi, Invariant distances on complex manifolds and holomorphic -mappings. J. Math. Soc. Japan 19 1967 460--480.<|endoftext|> -TITLE: Finite fundamental groups of 3-dimensional Calabi-Yau manifolds -QUESTION [17 upvotes]: Question. Is there an example of a compact $3$-dimensional Calabi-Yau manifold with finite fundamental group $G$ that does not admit a free action on $S^3$? -This question is motivated by the following: it is known that many simply-connected Clabi-Yau 3-folds admit a singular Lagrangian torus fibration over $S^3$. I don't know if there are exceptions. On the other hand, if $\pi_1$ is finite and we still have a lagrangean torus fibration, one can expect that the base is a lens space. But in this case probably $\pi_1$ of the CY-manifold will be equal to $\pi_1$ of the base. -PS. As Tony Pantev explains, the answer to this question is YES -- there are such examples. On the other hand, if we assume that a finite group $G$ is acting freely on a CY 3-manifold preserving the volume form and preserving a Lagrangian torus fibration, this should impose some very strong restrictions on $G$. I wonder if anyone bothered to work out what is the restriction :). - -REPLY [18 votes]: This intuition seems to be only loosely right. There are many smooth compact CY threefolds with large fundamental groups. For instance $\mathbb{Z}/3\times \mathbb{Z}/3$, $\mathbb{Z}/8\times \mathbb{Z}/8$, are allowed fundamental groups and I am pretty sure that those do not act freely on $S^{3}$. -More to the point - the Calabi-Yau threefolds that have these fundamental groups are explicitly constructed and we have a pretty good idea of the shape of (at least one of) their slag torus fibrations. For instance, in the first case the Calabi-Yau fibers by genus one curves over a rational elliptic surface, and the slag fibration is compatible with the genus one fibration. In the second case the Calabi-Yau fibers by abelian surfaces and again the slag fibration is compatible. So guided by the holomorphic picture you can easily imagine a situation where you group acts freely on the CY, preserves the slag torus fibration, and the induced action on the base of the fibration is not free. The only thing you can conclude really is that the action of the group on any fiber sitting over a fixed point in the base is free. This is possible to arrange on a torus by taking action by translations. -So, even if your fundamental group happens to admit some free action on $S^{3}$, this doesn't mean that the action on the base of the slag fibration will be free. And, in general, I don't expect it to be free.<|endoftext|> -TITLE: Is there a smooth $4$-manifold homeomorphic but not diffemorphic to $CP^2$? -QUESTION [11 upvotes]: Is there a smooth $4$-manifold homeomorphic but not diffemorphic to $CP^2$? Are there known non-smooth examples homeomorphic $CP^2$? - -REPLY [22 votes]: This is a notorious open problem. For the moment the simplest compact four-manifold that is announced to admit (infinite number of) exotic smooth structures is $S^2\times S^2$. This result is contained here : http://arxiv.org/abs/1005.3346 -I have to say that I am not at all an expert in the area -(also it seems that the above paper is not yet published). On the other hand there are several published papers showing that $CP^2\sharp 3\overline{CP^2}$ admit exotic smooth structures. -Also, it might be worth to recall that by a theorem of Yau a complex surface homeomrophic to $CP^2$ always has the standard smooth structure (in other words $CP^2$ admits a unique holomorphic structure up to bi-homolorphism). While for $S^2\times S^2$ this is still unknown (is there a surface of general type homeomorphic to $S^2\times S^2$?)<|endoftext|> -TITLE: Are cluster variables prime elements? -QUESTION [10 upvotes]: Cluster algebras introduction -A cluster algebra is a subalgebra $A$ of $k[x_1^{\pm1},...,x_n^{\pm1}]$ generated by a set of cluster variables, which are elements which can be generated from the set $\{x_1,...,x_n\}$ by a certain recursion determined by a skew-symmetrizable matrix $M$ (mutation). -The mutation of the pair (called a seed) -$$((x_1,x_2,...,x_i,...,x_n),M)$$ -at the $i$-th place is the seed -$$((x_1,x_2,...,\mu_i(x_i),...,x_n),\mu_i(M))$$ -$$\mu_i(x_i)=\left[\prod_{j,M_{ij}>0} x_j^{M_{ij}}+\prod_{j,M_{ij}<0}x_j^{-M_{ij}}\right]x_i^{-1}$$ -and $\mu_i(M)$ is a new skew-symmetrizable matrix I cannot get mathjax to display. -Mutation of seeds may be iterated arbitrarily. It is a non-trivial theorem that the functions in the seed are always Laurent polynomials. A cluster variable is any function appearing in a seed obtained by a sequence of mutations. -Ring-theoretic properties of cluster variables -As elements in $A$, each cluster variable $f$ is an irreducible element (it can't be factored). This is not hard to show; it follows from the Laurent embedding for any cluster containing $f$, and the observation that $A$ does not contain any Laurent monomials with negative powers of (non-frozen) variables. -So then I ask, are cluster variables prime elements in $A$? -The analogous argument to the irreducible case doesn't seem to work, since one needs to consider more general Laurent polynomials in a given cluster, and there is no nice criterion for telling when a general Laurent polynomial is in the cluster algebra. It also seems unlikely that $A$ is a UFD in general, which would be a standard trick for deducing primality for irreducibility. Nonetheless, the examples I checked in Macaulay 2 were all prime. - -REPLY [6 votes]: Boo. The answer is no. Consider the Markov cluster algebra, whose corresponding matrix is -$$\left[\begin{array}{ccc} 0 & -2 & 2 \\ 2 & 0 & -2 \\ -2 & 2 & 0 \end{array}\right]$$ -For an initial cluster of $\{x_1,x_2,x_3\}$, the mutation relation at 2 is -$$ x_2x_2'=x_1^2+x_3^2$$ -If the ground field has a square root $i$ of $-1$, then -$$ x_2x_2'=(x_1+ix_3)(x_1-ix_3)$$ -Thus $x_2$ is not prime.<|endoftext|> -TITLE: What is the relationship between the finiteness of the Tate-Shafarevich group and the Tate conjectures? -QUESTION [6 upvotes]: (I asked this on math-stackexchange, but it seems more appropriate to this forum, so I took it off from there and am posting it here) -After the great answer I got for my previous question about the Tate conjectures What is the intuition behind the concept of Tate twists?, I'm ready for my next one: -Let $X$ be an abelian variety defined over a number field $k$. -I am given to believe that there is some relationship between the Tate conjectures and the finiteness of the Tate-Shafarevich group. I imagine that this is because the Tate-Shafarevich group is equal to the Manin obstruction $X(\mathbb{A}_k)^{Br(X)}$ (where $\mathbb{A}_k$ denotes the adeles), and that the Brauer-Grothendieck group of $X$ has something to do with the Tate conjectures. -The relationship between the Tate conjectures and the Brauer-Grothendieck group is not one I understand well. If I understand "Conjectures on Algebraic Cycles in $l$-adic Cohomology" (written by Tate) correctly, the conjecture he calls $T^1(Y)$ (the first Tate conjecture on the variety $Y$) is equivalent to $Br(Y)$ being finite IF $Y$ is a variety over a finite field. -I don't know how to understand this relationship in any way that would be coherent. Is it true that the finiteness of the Tate-Shafarevich group of an abelian variety over a number field is implied by the Tate conjecutres on that abelian variety. Is the reverse true? Why is there even a relationship between these seemingly very different statements? - -REPLY [6 votes]: I believe that Tate was led to formulate his conjecture (first over finite fields, and then, by analogy, over finitely generated fields, in particular number fields) by considering the function field case of BSD. A careful examination of BSD for an elliptic curve over the function field of a curve over a finite field shows that the statement that order of vanishing of the $L$-function equals the Mordell--Weil rank is equivalent to the Tate conjecture (on divisors) for the corresponding elliptic surface over the finite field(the one obtained by taking the minimal regular model of the elliptic curve). -Off the top of my head, I don't remember what happens if one goes further, and tries to understand the leading term of the $L$-function in geometric terms. -You should probably look at the original Bourbaki seminar on this, as well as at the papers mentioned by SGP in their answer.<|endoftext|> -TITLE: Minimum distance between adjacent concentric circles that cross integer lattice points -QUESTION [5 upvotes]: This problem looks simple, but I searched around and couldn't find any similar problems or related resources. Hope someone could provide a clue or at least a hint of what class of prolbems it belongs to. -A = { [$m$, $n$] | $m$ and $n$ are positive integer numbers} is a set of 2-D lattice points -R = { $r$ | $r^2$ = $m^2$ + $n^2$ } is a set whose elements are the distance between the origin and the lattice points in the set A. -The problem is: -If we sort all the $r$ from set R and use these sorted $r$ to creat a series C = {$r{_1}$, $r{_2}$, $r{_3}$..., $r{_N}$}, where $r_i$ $\in$ R and $N\to\infty$, so that $r{_1}$ $\leq$ $r{_2}$ $\leq$ $r{_3}$... $\leq$ $r{_N}$, and we define $\Delta$$r{_i}$ = $r_i$$_+$$_1$ $-$ $r{_i}$, what is the explicit forms of asymptotics of $\Delta$$r_i$, such as $\Delta$$r_i$ $\propto$ $\frac{1}{r_i}$ ? -The problem can be also described as: -What is the explicit forms of asymptotics of the minimum distance between two adjacent concentric circles that cross integer lattice points? -The motivation of this problem is that: integers $m$ and $n$ determine the electromagnetic resonance frequency $r$ in a 2-D rectangular cavity, so $\Delta$$r_i$ is the distance between two adjacent resonance frequencies. -Thank you! - -REPLY [12 votes]: There cannot be an asymptotic answer, because $\Delta r_i$ can be as small as $c/r_i$ (with $c \rightarrow 1/2$ as $i \rightarrow \infty$), but $\Delta r_i$ is of order $(\log r_i)^{1/2} / r_i$ on average. Equivalently, $\Delta(r_i^2)$ can be as small as $1$ but is of order $\sqrt{\log r_i}$ on average. One can also construct gaps of length $\gg \log r_i / \log \log r_i$. -Clearly $\Delta(r_i^2)$ can be no smaller than $1$ because each $r_i^2$ is an integer. A difference of $1$ is easily attained. The proposer of the question asked about positive $m,n$, but the application to eigenmodes of a 2-dimensional cavity — presumably a square cavity — must allow $m=0$ or $n=0$ as well, and then it's clear that $r_i^2 = n^2 + 0^2$ and $r_{i+1}^2 = n^2 + 1^2$ works. Even if you insist on nonzero $m,n$, differences of 1 are easy to construct; e.g. if $n^2 = a^2 + b^2$ then $r_i^2 = a^2 + b^2$ and $r_{i+1}^2 = n^2 + 1^2$ differ by 1. -On the other hand, while the number of solutions of $m^2 + n^2 \leq r_0^2$ in positive $m,n$ is asymptotic to $\pi r_0^2 / 4$ as $r_0 \rightarrow \infty$, the set $R$ of possible $r$ values is not quite as dense as this suggests, because there's a lot of "degeneracy" (repetition) in the values of $m^2+n^2$; equivalently, as $x \rightarrow \infty$, the proportion of integers in $[1,x]$ that can be written as $m^2 + n^2$ goes to zero, so for those that do have such a representation the average number of representations must go to $\infty$. Indeed if $p$ is a prime congruent to $3 \bmod 4$ then no multiple of $p$ can be $m^2+n^2$ unless both $m$ and $n$ are multiples of $p$, which makes $m^2+n^2$ a multiple of $p^2$. This gives $p-1$ forbidden residues $\bmod p^2$. For any finite set $P$ of such primes, we deduce an upper bound $\beta_P := \prod_{p\in P} (1 - p^{-1} + p^{-2})$ on the density of integers representable as sums of two squares; since by Dirichlet the sum of $1/p$ over all such primes diverges, we find that $\log\beta_P$ can be made arbitrarily negative by taking $P$ large enough, which means that the density is less than any positive number, and is thus zero. A more refined analysis shows that the density in $[1,x]$ decays as $1/\sqrt{\log x}$. -Another way to use a finite list $p_1,\ldots,p_k$ of primes congruent to $3 \bmod 4$ is to choose for each of them one of the forbidden residues, i.e. some $r(p_i)$ that's divisible by $p_i$ but not by $p_i^2$, and consider the simultaneous congruences $N+i \equiv r(p_i) \bmod p_i^2$. By Chinese Remainder, there's a solution $N < \prod_{i=1}^k p_i^2$, which gives $k$ consecutive integers none of which is the sum of two squares. If we use all the $3 \bmod 4$ primes up to $x$, we get $k \sim x / 2 \log x$, and $\prod_i p_i^2 < x^{2k}$ so $\log \prod_i p_i^2 < 2k \log x \sim x$, whence we've constructed a "$\log/\log \log$" gap as claimed. There may be enough freedom in the choices we've made (including the order of the $p_i$) to squeeze a bit more out of this argument, but — unless somebody can provide a reference for a paper that already proved such a result — it looks like more effort than one expects to exert on a MathOverflow question... -EDIT A quick Google search turned up a reference: - -Heini Halberstam: Gaps in Integer Sequences, Math. Magazine 56 #3 (May 1983), 131-140 - -The "$\log / \log\log$" bound (with the same Chinese Remainder) argument appears on page 136, followed by an asymptotic formula $A(x) \sim (\pi/\sqrt{12}) (x / \sqrt{\log x})$ for the number $A(x)$ of sums of two squares less than $x$. This formula is attributed to Landau 1908, though there's no paper of Landau in the bibliography. The bibliography does cite Iwaniec's "On the half-dimensional sieve" (Acta Arith, 29 (1976), 69-95) for an alternative approach, and also a two-page paper "On the gaps between numbers that are sums of two squares" by I.Richards in Advances in Math. 46 (1982), 1-2.<|endoftext|> -TITLE: Numerical linear algebra: how to compute $b^TA^{-1}b$ efficiently -QUESTION [7 upvotes]: What is the most efficient way to compute $b^TA^{-1}b$ for a given $A$ and $b$? -Do we have to calculate $A^{-1}b$, or is this not necessary? -edit: I forgot to mention that A is symmetric and positive definite and sparse (so usually you'd use the conjugate gradient method). -What I have is a convex quadratic $x^TAx + b^Tx$. The minimum of this is at $2Ax+b=0$, and if you plug this minimum into the original form, then you get $x^T(-b/2)+b^Tx=b^Tx/2$ and this leads you to have to compute $-1/4\cdot b^TA^{-1}b$. So another way to pose the question is: can you find the height at the minimum faster than the location of the minimum? - -REPLY [6 votes]: This is possibly an answer from a practical point of view: If you use the CG method for solving $x=A^{-1}b$ then $b^T A^{-1}b$ can be obtained along the way. However, it has been shown that computing $b^T A^{-1}b$ during the iteration can converge faster than first solving for $x$ and then multiplying $b^T x$. See "Z. Strakos and P. Tichy, On efficient numerical approximation of the bilinear form c*A-1b , SIAM Journal on Scientific Computing (SISC), 33, 2011, pp. 565-587" and the references therein for the positive definite case.<|endoftext|> -TITLE: Class groups in dihedral extensions - some sort of Spiegelungssatz? -QUESTION [14 upvotes]: Let $p$ be an odd prime and let $F/\mathbb{Q}$ be a Galois extension with Galois group $D_{2p}$, let $K$ be the intermediate quadratic extension of $\mathbb{Q}$, and $L$ an intermediate degree $p$ extension: -$\;\;\;F$ -$\;\;\;\;|\;\;\;\backslash$ -$\;\;\;K$ $L$ -$\;\;\;\;|{\tiny 2}\;\;/{\tiny p}$ -$\;\;\;\mathbb{Q}$ -Then -$$ -\frac{h(F)}{h(K)h(L)^2}\in\left\{1,\frac{1}{p},\frac{1}{p^2}\right\}.\;\;\;\;\;\;\;\;\;\;(1) -$$ -One can also prove a similar statement over any given base field, but let's stick with this simple special case for now. The only way I know of proving this is rather roundabout: - -One first replaces the class numbers by regulators, using the analytic class number formula. -One interprets the resulting quotient of regulators as an invariant of the isomorphism class of the Galois module $\mathcal{O}_F^\times$. -One uses a classification of all $\mathbb{Z}$-free $\mathbb{Z}[G_{2p}]$-modules due to M. Lee from the 60s and computes the corresponding regulator quotient for all the modules that can occur in the above situation. The fact that this module is not $\mathbb{Z}$-free is a subtlety that one has to take care of separately. - -It seems to me that there should be a much easier way, or at least a more direct one, one that actually gives insight into the reasons for the dependence of class numbers on each other. My question is: - -Does anyone know of a direct way of proving (1), one that only uses properties of class groups, but not the analytic class number formula, nor any integral representation theory? - -Here are the few things I know: - -The fact that this class number quotient will have trivial $q$-adic valuation for any $q\neq p$ follows from the formalism of cohomological Mackey functors and is due to R. Boltje, Class group relations from Burnside ring idempotents, J. of Number Theory, 66, (1997). Essentially, the two ingredients are: -(a) class groups form a cohomological Mackey functor, and -(b) for any $q\neq p$, there exists an isomorphism -$$ -\mathbb{Z}_q[G/1]\oplus\mathbb{Z}_q[G/G]^{\oplus 2}\cong\mathbb{Z}_q[G/C_p]\oplus\mathbb{Z}_q[G/C_2]^{\oplus 2}. -$$ -I am happy with that part, so the remaining bit is the p-primary part. -I can immediately see from class field theory that the coprime-to-2 part of the class number of $L$ divides the class number of $F$, because the compositum of $F$ with the coprime-to-2 bit of the Hilbert class field of $L$ gives an abelian unramified extension of $F$ of the same degree. Similarly, if $3^r|h(K)$, then $3^{r-1}|h(F)$. - -But already the simple statement that - -if $p|h(F)$, then either $p|h(K)$ or $p|h(L)$ - -is not clear to me. E.g. the Hilbert class field of $F$ needn't be abelian over $L$, nor even Galois. Am I missing something elementary? A direct proof of this would already be nice. Or in the opposite direction, if the class number of $L$ is divible by $p^{100}$, why must the class number of $F$ compensate that? (Note that in the quotient, $h(L)$ is squared, while $h(F)$ isn't). -The article of Franz Lemmermeyer, Class groups of dihedral extensions gives a pretty extensive overview of the known variants of Spiegelungssätze for dihedral extensions, but as far as I can see, (1) does not follow from any of them (dear Franz, I call upon thee to confirm or to correct my assessment). Some very special cases do follow, but it seems to me that there should be a general direct proof. - -REPLY [4 votes]: I normally don't like to cite my own work on MO, but this time the preprint arXiv:1803.04064 was written, together with L. Caputo, having the OP's question in mind; and so, first of all, let me thank Alex for having asked it. -The main result is a purely algebro-arithmetic proof that for any base field $k$ and for every dihedral extension $F/k$ of degree $2q$ with $q$ odd, it holds -$$ -\frac{h(k)^2h(F)}{h(L)^2h(K)}=\frac{\lvert\widehat{H}^0(D_{2q},\mathcal{O}_F^\times\otimes\mathbb{Z}[1/2])\rvert}{\lvert\widehat{H}^{-1}(D_{2q},\mathcal{O}_F^\times\otimes\mathbb{Z}[1/2])\rvert} -$$ -in the OP's notations (which are different from the ones used in the preprint). When I say that the proof is ``algebro-arithmetic'' I mean that the main ingredient is class field theory and some group cohomology: no $\zeta$- or $L$-functions are involved, neither is the classification of integral representation of dihedral groups. The key point is that, if we call $G_q=\operatorname{Gal}(F/K)\cong\mathbb{Z}/q\subseteq D_{2q}$, then $G_q$-cohomology of a $D_{2q}$-module (typically: units, ad`eles, local units, etc.) has an action of $\operatorname{Gal}(K/k)$ and when the module is uniquely $2$-divisible, this induces identifications $\widehat{H}^i(G_q,-)^+=\widehat{H}^i(D_{2q},-)$ as well as $\widehat{H}^i(G_q,-)^-=\widehat{H}^{i+2}(D_{2q},-)$, where $\pm$ are the eigenspaces with respect to the action of $\operatorname{Gal}(K/k)$. Therefore we can use class field theory for the abelian extension $F/K$ to deduce information about cohomology groups for $D_{2q}$. -As corollary of the above formula we show that, for every prime $\ell$, the following bounds hold -$$ --av_\ell(q)\leq v_\ell\left(\frac{h(k)^2h(F)}{h(L)^2h(K)}\right)\leq bv_\ell(q) -$$ -where $a=\operatorname{rank}_\mathbb{Z}\mathcal{O}_K^\times + \beta_K(q) + 1$, $b=\operatorname{rank}_\mathbb{Z}\mathcal{O}^\times_{k}+\beta_k(q)$ and $v_\ell$ denotes the $\ell$-adic valuation; the "defect" $\beta_M(q)\in\{0,1\}$ (for $M=K,k$) is defined to be $1$ if $\mu_M(q)$ is non-trivial, and $0$ otherwise. From this, we deduce even sharper bounds in case $K$ is either CM (with totally real subfield equal to $k$) or if it is totally real. Since when $k=\mathbb{Q}$ this is always the case, we prove as a special result that in every dihedral extension of $\mathbb{Q}$ of degree $2p$ the formula required by the OP holds, again without resorting to any analytic or ``hard'' representation-theoretic result. Actually, restricting to the prime case $q=p$ has no utility whatsoever, and when $F/\mathbb{Q}$ is any dihedral extension of degree $2q$ ($q$ odd!) we deduce that the ratio of class numbers verifies -$$ -0 \geq v_\ell\left(\frac{h(F)}{h(K)h(L)^2}\right)\geq -\begin{cases} --2&\text{if $K$ is real quadratic}\\ --1&\text{if $K$ is imaginary quadratic} -\end{cases} -$$ -where, again, $v_\ell$ is the $\ell$-adic valuation. It is perhaps interesting to observe that ramification plays little to no role in our proof (ramification indexes only appear as well-controlled orders of cohomology groups of local units) and so assuming particular ramification behaviours wouldn't simplify it.<|endoftext|> -TITLE: Probability of a black path on a random chess board -QUESTION [8 upvotes]: Take a $2n$ by $2n$ chess board (oriented so the grid lines are vertical and horizontal). Usually there are $2n^2$ squares coloured black and $2n^2$ squares coloured white so that a black square is only adjacent to white squares. (Here, two squares are adjacent if they have a common edge.) -Suppose instead we start with a blank $2n$ by $2n$ chess board. We pick $2n^2$ squares at random and assign them black. The other half of the squares are assigned white. - -What is the probability the resulting chessboard has a monotonic black path? (Here, a monotonic black path is one which starts in the South-West corner and finishes in the North-East corner, and consists entirely of black squares adjacent along their North or East edge. -What is the probability that the resulting chessboard has a black path from the South-West corner to the North-East corner? (Here, a black path is a sequence of adjacent black squares) - -REPLY [6 votes]: James quickly gave the right answer in the comments, since $p_c \approx .5927$ for site percolation on the square lattice. -These crossing questions often have elementary answers, but neither the proofs nor the applications are trivial. For example, in critical percolation, the Russo-Seymour-Welsh theorem states that there is a uniform lower bound in the crossing probability. i.e., there is a uniform constant $c$ such that $\mathbb P_n(\mbox{there is a black crossing}) \ge c$, independently of $n$. -For a nice proof of the RSW theorem (with illustrative pictures!), see pages 33-44 of Pierre Nolin's lecture notes. (After deriving RSW, Pierre uses this formula to prove Kesten's theorem: $p_c = 1/2$ for bond percolation on the square lattice) -Another place to look is Section 1.3 of Wendelin Werner's lecture notes on percolation. Werner uses this to prove the Cardy-Smirnov formula, and then that site percolation on triangular lattice converges to $\operatorname{SLE}(6)$. -Cardy's formula is just one of the many elegant results in mathematical conformal field theory. Define $$f(x) = \mathbb P( \mbox{crossing starting from the point $x$ on side $1$ to side $2$} )$$ for site percolation on the unit triangle with spacing $1/n$. Cardy's formula is that $$f(x) = x.$$ (Peter Jones has described Cardy's formula as "the most difficult theorem about the identity function.")<|endoftext|> -TITLE: How can I see the "missing" Poisson center when the rank of the Poisson structure drops? -QUESTION [6 upvotes]: Recall that a Poisson algebra is a commutative algebra $A$ along with a bracket $\lbrace,\rbrace: A^{\otimes 2} \to A$ which is a Lie bracket and which is also a derivation in each variable. The Poisson center of $A$ is the subalgebra of those $f\in A$ such that $\lbrace f,\rbrace : A \to A$ is the $0$ derivation. Elements of the Poisson center are generally called Casimirs. Let me write $Z(A)$ for the Poisson center of $A$. -The most important case is when $A = \mathcal C^\infty(M)$ for a Poisson manifold $M$. Then $M$ is foliated by symplectic leaves, which are the orbits for the Lie algebra action of $A$ on $M$. It follows that every Casimir is constant on each symplectic leaf. Symplectic leaves can have interesting macroscopic topology — for example, they can wrap around $M$ "irrationally" — so I would rather think locally on $M$, and replace $A$ by the corresponding sheaf of Poisson structures. Then there is a sheaf whose sections are local Casimirs; this sheaf might not have very many global sections. Anyway, one generally tries to believe that at least locally the symplectic leaves are precisely the common level sets of the Casimirs. -But even this fails locally when the Poisson bivector drops in rank. As an easy example, consider $\mathbb R^2$ with coordinates $x$ and $y$ and Poisson bivector $x \frac{\partial}{\partial x} \wedge \frac{\partial}{\partial y}$. There are two 2-dimensional symplectic leaves, namely $\lbrace (x,y) \text{ s.t. } x>0\rbrace$ and $\lbrace (x,y) \text{ s.t. } x<0\rbrace$, and uncountably many $0$-dimensional symplectic leaves, namely the points $\lbrace (0,y)\rbrace$ for each $y$. But any Casimir is locally constant on each of the 2-dimensional leaves, and if it extends to the $y$-axis it must continue to be constant. -So, far from $\operatorname{spec}(Z(A))$ being the set of symplectic leaves, $\operatorname{spec}(Z(A))$ is more accurately thought of as the "GIT quotient" of $\operatorname{spec}(A) = M$ under the Poisson action. -But there is a qualitative difference between, say, $\mathcal C^\infty(\mathbb R^2)$ with bracket $x \frac{\partial}{\partial x} \wedge \frac{\partial}{\partial y}$ and the same algebra with the nondegenerate bracket $\frac{\partial}{\partial x} \wedge \frac{\partial}{\partial y}$. This difference is not detected by the algebraic Poisson center, but in some sense this is because $\mathcal C^\infty$ isn't quite the right type of function. Indeed, suppose that I had some type of "delta functions"; then there would be functions of the form $\delta(x)f(y)$ in $Z\bigl( \mathcal C^\infty(\mathbb R^2), x \frac{\partial}{\partial x} \wedge \frac{\partial}{\partial y}\bigr)$, and the collection of all such functions would correctly cut out the symplectic leaves as the common level sets. -Hence, my question: - -Can I always find the symplectic leaves of a Poisson manifold as the common level sets of the Casimirs if I allow some sort of "generalized function"? If so, how precisely should I define such generalized functions? - -As a step towards the second question (assuming the answer to the first is "yes"), let me describe the approach I've been imagining. I do not have $\delta$-functions in $\mathcal C^\infty$, but I do have sequences of functions that approach $\delta$-functions for some norm. I'm not very good at analysis, so I don't have intuition of which is the correct norm to use, but let's suppose I've picked such a norm. Then I might say that $f\in \mathcal C^\infty$ is almost central or an almost Casimir if the operator norm of $\lbrace f,\rbrace$ is small. One should then expect that functions approaching $\delta(x)f(y)$ are almost Casimirs for $x \frac{\partial}{\partial x} \wedge \frac{\partial}{\partial y}$, whereas some version of Heisenberg Uncertainty says that there are no almost Casimirs for the nondegnerate bracket $\frac{\partial}{\partial x} \wedge \frac{\partial}{\partial y}$. So perhaps every Poisson manifold does have (locally) an "almost Poisson center", and this almost center is enough to detect all the symplectic leaves? - -REPLY [2 votes]: I do not have a definite answer but rather some reflections I've being doing myself and with a colleague I'll mention later, recently, on the subject. -Sure Poisson cohomology can help you in detecting the"missing" leaves, my favourite example being the triple of bivectors $\partial_x\wedge\partial_y$ (trivial 1-Poisson cohomology), -$(x^2+y^2)\partial_x\wedge\partial_y$ (1-dimensional 1-Poisson cohomology), $(x^2+y^2)^2\partial_x\wedge\partial_y$ (infinite-dimensional 1-Poisson cohomology). In a sense $1$-dimensional Poisson cohomology represents the tangent space to the set of leaves (which is a badly behaved non Hausdorff space and may therefore have no nonconstant functions defined on it) and is therefore slightly more sensible. -Now, you may know that Poisson cohmology may be defined also with values in a Poisson module. If $T$ is a distribution (I mean a linear continuous functional on $\cal C_0^\infty(M)$ - compactly supported smooth functions) then letting $\left\{f,T\right\} (g)=T(\{f,g\})$ one gets that $\{f,T\}$ is still a distribution. One can show that in this way distributions form a Poisson module over $\cal C_0^\infty(M)$. One can of course look for the annihilator of such module, i.e. Casimir distributions defined by $\left\{f,T\right\}=0$ for every $f$. This space enlarges naturally the space of Casimir functions. On the quadratic singular Poisson structure I was referring to above it is possible to show that the vector space of Casimir distributions is $5$-dimensional generated by the $\delta$ function of the origin, its first derivatives and its mixed second derivative. -One could consider this space of Casimir distributions, or more generally the Poisson cohomology with coefficients in this Poisson module. This I've never attempted to compute, even in easy examples. -I've basically learned this by Paolo Caressa who wrote a couple of notes about this: -Examples of Poisson Modules, I, Rendiconti del Circolo Matematico di Palermo(2) 52 (2003), 419-452 -Examples of Poisson Modules, II, Rendiconti del Circolo Matematico di Palermo(2) 53 (2004), 23-60. -The examples above are all $2$-dimensional. It is important to recall that every $2$-dimensional bivector is Poisson and determined by a smooth functions on the plane. We are therefore looking for something that should be, in this specific case, distinguish singularities of smooth functions, quite subtle, therefore. -Something easier, though not so capable of detailed analysis is to consider a maximal subalgebra of continous functions to which the Poisson bracket may be extended. This was done, for example, in an old paper by Albert Sheu on the quantization of the Podles sphere (the one with an appendix by Lu-Weinstein, I do not have a chance to get the exact reference now) where he quantizes the algebra of all smooth functions on the sphere minus the North Pole that can be continously extended to the North Pole. The reason why this is much less sensible is that as long as you have quadratic singularities in the bivector you can had square root singularities in the functions, which immediatly throws in all continous functions. -Hope this may help.<|endoftext|> -TITLE: Machine model for primitive recursion? -QUESTION [11 upvotes]: General computable functions can be described either functionally (in terms of closure of -the coordinate functions, constant functions, composition, primitive recursion, and $\mu$-recursion), or in terms of a Turing machine. -I have only seen primitive recursion defined in the functional language, i.e. functions obtained by coordinates, constants, composition, primitive recursion. -Is there a similar type of machine model for primitive recursion? -I am aware of some (pedagogical) programming languages, such as Hofstadter's BLOOP, that are PR-complete, but this approach doesn't really look like a Turing machine to me. - -REPLY [5 votes]: If you're willing to accept register machines (rather than just tape machines), you can get what you want via the Grzegorczyk hierarchy, which generates the class of primitive recursive functions in stages. -The ${n+1}^{th}$ stage of the hierarchy ${\mathscr E}_{n+1}$ is the closure of the zero, successor, projection, and hyper operation $H_n$ function under composition and bounded recursion. -Bounded recursion is defined just like primitive recursion, except that when defining a function $f$ at the ${n+1}^{th}$ level, the definitions of the base case and inductive case for $f(m,\bar x)$ must take the form $\text{min}(g(m,\bar x),...)$ where $g$ is a function from the $n^{th}$ level. -Every primitive recursive function belongs to ${\mathscr E}_n$ for some $n$, and every function in the hierarchy is primitive recursive. -Beltiukov's stack register machines (also here) give a "machine-oriented" characterization of the Grzegorczyk hierarchy -- and therefore of the primitive recursive functions. There is a slightly more accessible description of stack register machines here, starting on page 108.<|endoftext|> -TITLE: A recommended roadmap into inner models -QUESTION [24 upvotes]: A friend of mine and myself (both grad students with a relatively decent set theoretic background) want to venture into the universe of inner models. [pun intended :-)] -I would very much like to get some recommendations on not only material to read from, but also on the order of which these should be approached and points which may be important to stop and study more extensively. -We both have studied large cardinals (weak compactness, measurability, $0^\sharp$, some supercompactness. Iterations are missing so is $L[D]$), we have also background in forcing and seen proofs for the covering lemma for $L$ (both Jensen's and Magidor's covering lemmas, although no fine structure was involved). -One of the reason I ask is that there resources are relatively abundant, The Handbook, Jech, Kanamori's The Higher Infinite, etc. and while it is clear to me that some topics should be covered first (iterations, for example) I'd much rather have a general roadmap in mind when approaching this. -Many thanks. - -REPLY [28 votes]: Hi Asaf. Here is a quick answer, I'll try to expand once I have some time. I once prepared a short list to a similar question somebody asked me by email. What follows is based closely on that list: - -Let's see... (It is a long road.) It is useful to have a good understanding of the basics of fine structure before venturing too much into inner model theory proper, so I think one should begin with Jensen's paper, perhaps having Devlin's "Constructibility" book nearby. I would suggest after having some understanding of the basic notions as discussed in Jensen's paper, at least start reading the articles in the Handbook on fine structure, by Schindler-Zeman and Welch. There is a bit of hard work involved in going through these three papers, especially since it takes some time to reach applications, but one must first master the language. [On the side, you may want to read about the Mitchell order, strong cardinals and Woodin cardinals. As we climb up the large cardinal hierarchy, the associated models (the pre-mice) become more complicated, and their iterations become harder to describe, so it is a good idea to at least have the large cardinal notions clear before studying their associated pre-mice.] -Then Mitchell's "Beginning inner model theory", also in the Handbook. And perhaps Schimmerling's "The ABC's of mice", in the Bulletin of Symbolic Logic. Steel has two excellent introductory papers (listed below), but these two papers are a good starting point. -If one is interested in organizing the reading by strength of the assumptions studied, it is then time to look at Zeman's book "Inner models and large cardinals", which is also a useful reference to have. Continue with Schindler's "The core model for almost linear iterations", Annals of Pure and Applied Logic, 116:205-272, 2002 (leaving for later the proofs of iterability and covering). -Schindler's paper appeared after the references below, but its setting is more restrictive, so some of the arguments are simpler. By now it may be a good idea to be reading on the side on iteration trees, which in itself is a demanding project. There is the original article by Martin and Steel, which may be a reasonable place to first find the notion. Its setting is not fine structural, so it lacks some complications. [You may want to look at Neeman's Handbook article as well, to get a good feeling on how iteration trees are used, why we care about them and about Woodin cardinals (Steel's Handbook article mentioned below also treats these topics in detail).] -Then one cannot postpone it anymore, and it is time to read "Fine structure and iteration trees" by Mitchell-Steel and "The core model iterability problem" by Steel, together with Schindler-Steel-Zeman "Deconstructing inner model theory", Journal of Symbolic Logic, 67(2) (2002) 721-736, and Steel's "An outline of inner model theory", in the Handbook, and Löwe-Steel "An Introduction to Core Model Theory", in "Sets and Proofs, Logic Colloquium 1997 , volume 1", London Mathematical Society Lecture Notes 258, Cambridge University Press, Cambridge, 1999. Yes, they must be read more or less concurrently. Yes, this is overly ambitious and almost impossible. - -This plan, unfortunately, takes a huge amount of time. A shorter version would be to do Jensen's paper, with Devlin on the side in case there are details that are not clear in Jensen's paper, and then jump to "Fine structure and iteration trees" and the papers in the paragraph just above. One then revisits the other papers/books as needs demand. - -[That is how Steel introduced me to the subject. "Read Jensen's paper, and the first 3 chapters of FSIT, and next week we can begin with chapter 4."] -The above requires some explanation, I believe. The problem is that "Fine structure ..." is, let's say, not as nicely written as one would like. Steel's Handbook paper covers a lot of the same ground (and much more) and it is very nice to read, but there are details missing, so having "Deconstructing ..." on the side may help (that paper fixes a gap in the definitions in "Fine structure...", so it is essential). Once that's done, it is easier to continue with "The core model iterability ...", which is also a nice read (perhaps skip the last chapter on a first reading), and read Löwe-Steel simultaneously, as it is a gentler, less technical introduction. -The end of Steel's Handbook paper refers to more recent and technical work. There are some companion papers, "HOD^{L(R)} is a core model below theta", Bulletin of Symbolic Logic, vol 1 (1995), 75-84, and "Woodin's analysis of HOD^{L(R)}", an unpublished note available at Steel's page. -Now, this road is intended to take you from the beginning to what three years ago was nearly the top. If you want to see applications to consistency strength questions, then specific details of the theory at levels below Woodin cardinals may be important, while those details are not so relevant higher up. -I much recommend that you bookmark -http://wwwmath.uni-muenster.de/logik/Personen/rds/bibliography.html -although it is in terrible need of an update. There are many papers listed there that I haven't mentioned here. -After the papers above, which form the core of the theory, I guess one could follow with Mitchell's paper on the covering lemma (in the Handbook), Steel's notes on "A theorem of Woodin on mouse sets", and the draft of his book with Schindler on "The core model induction". Past that, Sargsyan's thesis essentially takes you to the boundary of what is known. There is a recent paper by Jensen and Steel, not yet published, "$K$ without the measurable", which explains how to eliminate a technical assumption we needed for many years. An update of Grigor's thesis will appear in the Memoirs, and it is available from his webpage, together with a gentler introduction, that he wrote for the Bulletin. -[At this stage, it should be clear that learning determinacy is indispensable. Again, this is in itself a demanding task, as part of what one now needs is to work through the Cabal volumes.] -And you may want to take a look at http://wwwmath.uni-muenster.de/logik/Personen/rds/core_model_induction_and_hod_mice.html and http://wwwmath.uni-muenster.de/logik/Personen/rds/core_model_induction_and_hod_mice_2.html for papers and references on the state of the art.<|endoftext|> -TITLE: Higman embedding theorem -QUESTION [12 upvotes]: The Higman Embedding theorem says that any finitely generated and recursively presented group can be embedded in a finitely presented group. -My question is if one can embed such a group as a normal subgroup into a finitely presented group? - -REPLY [22 votes]: No. Take a f.g. non-finitely presented group $G$ with trivial $Out(G)$ and trivial center (such groups clearly exist; in fact one can even assume that $Out(G)$ is locally finite, say $G$ is the Grigorchuk group of intermediate growth, by the result of Grigorchuk and Sidki). Suppose that $G$ is a normal subgroup of $H$. Then every $h\in H$ acts on $G$ by conjugation $x\to x^h$. Since $Aut(G)=Inn(G)$, there exists $g\in G$ such that $x^g=x^h$ for every $x\in G$. Hence $g^{-1}h$ centralizes $G$. Therefore $H=G Z_H(G)$ (if $Out(G)$ is locally finite, then instead of $G$ here you will get a finite extension of $G$, the image of $H$ in $Aut(G)$ under the natural homomorphism) where $Z_H(G)$, the centralizer of $G$ in $H$, intersects $G$ trivially (since the center of $G$ is trivial) and is a normal subgroup. Hence $H$ is a direct product of $G$ and $Z=Z_H(G)$. Since $H$ is finitely generated, $Z$ is finitely generated (being a homomorphic image of $H$). Hence if $H$ is finitely presented, then $G$ is presented by the presentation of $H$ plus finitely many relations killing the generators of $Z$, so $G$ must be finitely presented, a contradiction.<|endoftext|> -TITLE: Motivation for the proof of Hilbert's Theorem 90 -QUESTION [9 upvotes]: The proof of Hilbert's Theorem 90 about cyclic extensions goes like this: Let $\sigma$ be the generator of the Galois group of order $n$ and let $b$ have norm $1$, i.e. $b \sigma(b) \cdots \sigma^{n-1}(b) = 1$. For an element $c$ consider -$a := c + b \sigma(c) + \dotsc + b \sigma(b) \cdots \sigma^{n-2}(b) \sigma^{n-1}(c)$. -Then one verifies $b \sigma(a) = a$. We may choose $c$ such that $a \neq 0$ since characters are linearly independent. QED -Question 1. Is there a motivation for the choice of $a$? It is clear that it works, but why did Hilbert or rather Kummer (feel free to add historical details) came up with this sum? -Was he influenced by Lagrange's resolvents? But if this is the source, I would have to ask the same question which idea underlies the definition these resolvents; again here I only know that they solve the cubic. After a quick glance at the original papers I could not find any explanation. -Feel free to forget about history and give another, perhaps more modern motivation of the choice of $a$. I already tried to "get" $a$ starting with -$a = b \sigma(a) = b \sigma(b \sigma(a)) = b \sigma(a) \sigma^2(a) = \dotsc = b \sigma(b) \cdots \sigma^k(b) \sigma^{k+1}(a)$. -Perhaps also the following interpretation helps: We look for a fixed point of the function $b \sigma(-)$. Now perhaps there is a connection with others fixed point theorems, in case I already asked about the motiviation of their proofs. -Question 2. Linear independence of characters is proven indirectly, so the choice of $c$ is above is not canonical. Is there any chance to get an explicit choice of $c$? This would be great because then for every solvable separable polynomial with known Galois group one would get explicit generators of the Galois extension. This is because Hilbert's Theorem 90 makes it possible to classify cyclic extensions, which are the intermediate steps; actually there $b$ is in the ground field, perhaps this simplifies both questions. - -REPLY [12 votes]: The map $T : a \mapsto b \sigma(a)$ is linear and has order $n$. It follows straightforwardly that $c + T c + ... + T^{n-1} c$ is a fixed point of $T$. -More generally, let $V$ be a representation of a finite group $G$ over a field of characteristic not dividing $|G|$ containing the values of every character of $G$ over the algebraic closure. Let $\chi$ be the character of an irreducible representation of $G$. Then -$$v \mapsto \frac{1}{|G|} \sum_{g \in G} \overline{\chi(g)} gv$$ -is the projection from $V$ to the isotypic component $V_{\chi}$ of $V$. When $G$ is a cyclic group we recover Lagrange resolvents. In particular when $\chi$ is the trivial representation, the above is the projection from $V$ to its $G$-invariant subspace.<|endoftext|> -TITLE: univalent axiom as a property of a model category? -QUESTION [5 upvotes]: I am interested to understand the univalence axiom of Voevodsky; however, I know -very little type theory. Thanks to response below, I now understand what is being univalent -means for a morphism. A couple more questions, though: - -Do I understand correctly that tthe Univalence Axiom makes sense for an arbitrary locally cartesian closed model category ? And what is 'the universe of small fibrations'? If I understand correctly, it means that for each small fibration g:Y→X there are morphisms h:X→U and h˜:Y→U˜ such that the corresponding square is a pull-back square. What does exactly 'small' mean here? - Is there a -reformulation of the univalence axiom -stated fully in terms of a model -category, perhaps with an -distinguished fibration ? What is the best reference giving full detail? - -I was not able to find anything in the literature. -The NSF proposal of Voevodsky seem to come quite close to giving such a formulation, -but it does not have full details. - -REPLY [2 votes]: This note in arxiv:1111.3489 gives an formal interpretation of the univalence axiom in an arbitrary (locally cartesian closed) model category: basically, interpret word-by-word a passage of Voevodsky -describing the univalence for sSets. - The authors are non-specialist, though, so various blunders possible.<|endoftext|> -TITLE: The fundamental group of an algebraic surface -QUESTION [6 upvotes]: Question: Let $X$ be a non-singular algebraic surface of general type. Suppose that the canonical class $K_X$ is an integer multiple of another class $L$. Let $\Sigma_k$ be a smooth curve of genus $k$ that represents the class $L$ in $X$ (here I assume such $\Sigma_k$ exists). Is it true that $\pi_1(\Sigma_k)$ surjects into $\pi_1(X)$? Do you know any relationships between them? -Thanks - -REPLY [2 votes]: Yes, $L$ is an ample line bundle and hence $\Sigma_k$ is a smooth ample divisor. The Lefschetz hyperplane theorem now implies that $\pi_1(\Sigma_k)\to\pi_1(X)$ is surjective.<|endoftext|> -TITLE: Negative values of Riemann zeta function on the critical line. -QUESTION [8 upvotes]: From parametric plots of $\zeta \left( \frac{1}{2} + it \right)$ it seems to be the case that: -(1) except for $\zeta(\frac{1}{2})$ the Riemann zeta function does not attain any negative real value on the critical line. -(2) the curve $(t, \zeta(1/2+it))$ is dense in the complex plane. -Are these statements known to be false, if not, is there any proof affirming them? - -REPLY [7 votes]: Update, some recent information on (1): -Kalpokas, Korolev, Steuding recently released a preprint showing that $\zeta(1/2 + it)$ takes aribtrarily large positive and negative (real) values; and also show analog statements for the other lines through the origin, that is positive and negative (real) values of arbitary says of $e^{-i \phi} \zeta(1/2 + it)$ for any $\phi$. The paper contains also more quantitative results along these lines (cf. in particular Corollary 3 and the preceeding discussion). - -Since (1) already received several answers, I expand and upgrade the comments on (2): -Yes, indeed it is conjectured, but unproved, that $\zeta(1/2 + i t)$ for $t \in \mathbb{R}$ is dense in the complex plane. [Side note: It is well-known that this is so for the lines $\sigma +it$ with $1/2 < \sigma < 1$.] -It seems that this conjecture was first formulated by Ramachandra (Durham, 1979), however only appeared in print in the second edition of Titchmarsh's book (note's by Heath-Brown), see the articles below for details. -There is very recent work on this problem due to Delbaen, Kowalski, and Nikeghbali. -See in particular this preprint by the latter two and this by all three. -Among others: in the former, they show how this result would follow "from a suitable version of the Keating--Snaith moment conjectures"; -in the latter, they propose a refinement of the density conjecture, a quantitative version (see Conj. 1, in Sec. 3.9). - -REPLY [7 votes]: The zeta function is real on the critical line only at the zeros and at Gram points, this is because zeta(1/2+it)=exp(-ivartheta(t)) Z(t). -At the Gram point g_k we have by definition vartheta(g_k)=pi k. so that -zeta(1/2+ig_k) =(-1)^k Z(g_k). -Now a Gram point g_k is said a good Gram point if (-1)^k Z(g_k) >0. In other case it is said a bad Gram point. -Since it appear improbable a zero just at a Gram point. You are asking if there exists bad Gram points, there are plenty. The first few bad Gram points are -g_126, g_134, g_195, g_211, ... -g_126 = 282.45472082346217461077 -In fact it is proved there are infinite bad Gram points. -Also we may easily obtain large negative values. For example using data -of T. Kotnik "Computational estimation of the order of zeta(1/2+it) Math of Comp. -(2003) we easily locate the point -t = grampoint(2601005843707) were we have -zeta(0.5+i t) = -119.6304321077241661374 -This is easily confirmed in mpmath (or Mathematica) -( grampoint(2601005843707) = 669980906189.53552206792 ).<|endoftext|> -TITLE: What might the (normalized) pair correlation function of prime numbers look like? -QUESTION [11 upvotes]: Cross-posting from Math.Stackexchange. - -You might have read about the fortuitous meeting between Montgomery and Dyson. The background is that the nontrivial zeros of the Riemann zeta function, when normalized to have unit spacing on average, (seem to) have the pair correlation function $1-\mathrm{sinc}^2(x)$, where $\mathrm{sinc}$ is the normalized function $\sin(\pi x)/ (\pi x)$. It's still a conjecture but it has good numerical support. -So what about prime numbers? Let $\Sigma(x,u)$ be the number of pairs of primes $p,q\le x$ which satisfy the inequality $0\le p-q\le \frac{x}{\pi(x)}u$, where $\pi(x)$ is the prime counting function. This inequality is chosen because multiplying primes by $\frac{\pi(x)}{x}$ will ensure the gaps between consecutives is exactly unity (hence they are normalized). Then what might -$$g(u)=\frac{d}{du}\left(\lim_{x\to\infty}\frac{\Sigma(x,u)}{\pi(x)}\right)$$ -end up looking like? This basically asks, "what is the density of normalized primes around so-and-so apart from each other?" (You can see the original Montgomery conjecture as equation 12 here. I've adapted it to prime numbers by essentially changing the asymptotic number of zeta zeros to the prime counting function instead. I might be acting presumptuous in assuming the naive replacement will still afford a meaningful answer.) - -REPLY [11 votes]: you are asking for the two-point correlation function of a Poisson process with unit density, which is just unity: $g(u)=1$. -the support for this is about as strong as for the Riemann zeroes: there is extensive numerical evidence but no conclusive theorem; see Soundarajan's 2006 paper cited above, or more recent papers on arXiv:0708.2567 and arXiv:1102.3648 -the esssential difference between the function $g_R(u)=1-sinc^2(u)$ for the Riemann zeroes and $g(u)=1$ for the prime numbers, is that the former vanishes $\propto u^2$ for small $u$ ("level repulsion"), while the latter remains constant. This is the difference between the (conjectured) Gaussian unitary ensemble of Riemann zeroes and the Poisson ensemble of prime numbers. For large $u$ all correlations decay and $g_R(u)\rightarrow g(u)$.<|endoftext|> -TITLE: Query on comment in Deligne-Mumford (1969) -QUESTION [5 upvotes]: In Deligne and Mumford's famous 1969 paper, The irreducibility of the space of curves of given genus, definition 4.6 (that of algebraic stacks) has the following footnote: - -This definition is the "right" one only for quasi-separated stacks. It will however be - sufficient for our purposes. - -Note that their definition of an algebraic stack is: - -A stack on $Sch$ with the etale topology such that -The diagonal is representable, and -there is a representable etale surjection from a scheme. - -and it immediately follows the definition of what it means for a stack to be quasi-separated, but they clearly do not mention quasi-separability here. -What do they mean by the footnote? What is the "right" definition for stacks which are not quasi-separated? - -REPLY [10 votes]: I think that what they might have in mind is that for non-quasi-separated Deligne-Mumford algebraic stacks one should not assume that the diagonal is represented by schemes, but by algebraic spaces. For quasi-separated Deligne-Mumford stacks this implies representability by schemes, but this is not true in general. -At least, this was my guess when I read Deligne-Mumford for the first time, as a graduate student. The first articles by Artin on algebraic spaces appeared in 1969 (a momentous year in algebraic geometry), and Deligne and Mumford must have been aware of his ideas.<|endoftext|> -TITLE: Recent claim that inaccessibles are inconsistent with ZF -QUESTION [25 upvotes]: Here it is mentioned that someone claims to have proven that there are no weakly inaccessibles in ZF. -Question 1: What reasons are there to believe that weakly inaccessibles exist? -Question(s) 2: Since all large cardinals are weakly inaccessible, this would have a profound effect on set theory. What are some of the most significant results whose only known proof assumes the existence of weakly inaccessibles? Might any of the arguments go through without their existence? For example, I've heard that the original proof of Fermat's Last Theorem (FLT) assumed (something equivalen to) a large cardinal, but it was then shown that the argument went through without such an assumption. -Edit. I just added the phrase "whose only known proof" to Question 2 above, which is what I intended originally. The point of course, is that I want to know which results, if any, would be "lost" if weakly inaccessibles were lost. FLT is not an example of that, but would have been before it was known that weakly inaccessibles are not necessary in its proof. - -REPLY [31 votes]: François has excellently addressed your question 1; allow -me to address question 2. I understand the question to be: -what will be the mathematical effects if someone were to -show that there are no (weakly) inaccessible cardinals? A -similar question would apply to any of several large -cardinals. So let me list some consequences. -First, let me note that the existence of a weakly -inaccessible cardinal is provably equiconsistent with the -existence of a (strongly) inaccessible cardinal, since any -weakly inaccessible cardinal is strongly inaccessible in -$L$, and so the issue about weakly or strongly inaccessible -is entirely irrelevant when it comes to consistency. -Second, let me note that set theorists are not generally -satisfied by claims of the sort "the only known proof uses -such-and-such," but rather they use the concepts of -consistency strength and equiconsistency, which allow for -precise claims to be proved about exactly which large -cardinals are required to prove which statements. The -situation is that for many mathematical assertions, we can -prove that any proof must use a certain type of large -cardinal or something just as strong, in the sense that the -consistency of the statement itself implies the consistency -of the large cardinal in question. In this way, we avoid -any problematic issue about knowledge concerning whether a better proof is -simply not yet discovered. -As a result, if inaccessible cardinals should be refuted, -then using the known results we immediately gain an -enormous number of positive theorems. So it isn't really a -case of losing theorems, but rather gaining. -Theorem. If inaccessible cardinals are inconsistent, -then (we can prove that) we can construct a non-Lebesgue -measurable set of reals without using the axiom of choice. -This follows from the fact that Solovay and Shelah have -proved that the possibility of constructing a non-Lebesgue -measurable set of reals (in the context of ZF+DC) without -using AC is exactly equivalent to the inconsistency of -inaccessible cardinals. -Most people believe that one must use AC in any Vitali-type -construction of a non-Lebesgue measurable set, and the -theorem above shows that this belief is provably equivalent -to the consistency of inaccessible cardinals. Perhaps many -mathematicians would find their confidence in the -consistency of inaccessible cardinals to increase upon -learning of this, and in this sense, this is also an answer to question 1. -In any case, many well-known set -theorists have emphasized enormous confidence in the -consistency of large cardinals, and have stated quite -explicitly that if inaccessible cardinals should become -known to be inconsistent, then we should expect further -inconsistency much lower in ZFC itself or in the low levels -of PA. -Theorem. If inaccessible cardinals are inconsistent -(and even merely if we can refute infinitely many Woodin -cardinals), then (we can prove that) there is a projective -set of reals $A\subset\mathbb{R}$ whose corresponding -two-person game of perfect information has no winning -strategy for either player. In other words, the infinitary -de Morgan law -$$\neg\forall n_0\exists n_1\forall n_2\exists n_3\cdots A(\vec -n)\iff\exists n_0\forall n_1\exists n_2\forall -n_3\cdots\neg A(\vec n)$$ will fail for some projective set -$A$. -The projective sets of reals are those reals that are -definable by a property involving quantification only over -real numbers and integers. The reason for the theorem is -that projective determinacy is equiconsistent over ZFC with -infinitely many Woodin cardinals, and so if we refute the -large cardinals in ZFC, then we similarly refute projective -determinacy. -Theorem. If inaccessible cardinals are inconsistent -(and even if merely measurable cardinals are inconsistent), -then (we can prove that) there is an analytic set (a -continuous image of a Borel set) that is not determined. -Theorem. If inaccessible cardinals are inconsistent, -then we can prove that the full set-theoretic universe is -very close to the constructible universe in the sense of -covering. In particular, $L$ computes the successors of -singular cardinals correctly. -This shocking conclusion follows in this case from Jensen's -covering lemma, since refuting inaccessible cardinals -implies a refutation of $0^\sharp$. -Theorem. If inaccessible cardinals are inconsistent, -then on no set is there a countably complete real-valued -measure measuring all subsets of the set and giving points no mass. -This is simply because any real-valued measurable cardinal -is measurable and hence inaccessible in an inner model. -Theorem. If inaccessible cardinals are inconsistent, -then (we can prove that) there are no uncountable -Grothendieck universes and the axiom of universes in -category theory is false. -An uncountable Grothendieck universe is exactly $H_\kappa$ -for an inaccessible cardinal $\kappa$, and the axiom of -universes asserts that every set is in such a universe. -There are many more examples. (I invite any knowledgeable -person to edit the answer with additional examples.)<|endoftext|> -TITLE: Line bundles on fibrations -QUESTION [6 upvotes]: Let $f:Y \to X$ be a flat morphism with positive dimensional fibers. Is it always true that line bundles that are trivial along each fiber are of type $f^*L$ for $L$ a line bundle on $X$? - -REPLY [11 votes]: It is true with some extra assumptions. If $f$ is projective (EDIT: in fact proper is enough) and has connected and (EDIT) reduced fibers and $M$ is a line bundle that is trivial on every fiber, then $h^0(X_y, M)=1$ for every $y\in Y$. If $Y$ is integral, then it follows that $L:=f_*M$ is a line bundle and the natural map $f^*L\to M$ is an isomorphism. -EDIT: the argument works provided $h^0(X_y, {\mathcal O})=1$ for every $y$. So in some cases one can remove the assumption that all the fibers are reduced. For instance if $X$ is a smooth complex surface and $Y$ is a smooth curve, then by Zariski's lemma every fiber $X_y$ is either $1$-connected or $X_y=mD$, where $D$ a $1$-connected divisor and $D|_D$ is torsion of order $m$. Using the fact that $h^0({\mathcal O}_D)=1$ if $D$ is $1$-connected and applying induction, one gets $h^0(X_y, {\mathcal O})=1$ for every $y$. - -REPLY [7 votes]: No. -Take $X=Spec(k[x^2,x^3])$, the cusp over the field $k$ , the trivial bundle $Y=X\times_k\mathbb A^1_k $ and the first projection $f=pr_X:Y\to X$ . -Every line bundle$M$ on $Y$ is trivial on the fibers, since said fibers are affine lines over the field $k$. However not every line bundle $M$ on $Y$ can be written $f^*L$ with $L$ a line bundle on $X$. -Here is why: -A ring $R$ is called semi-normal if whenever elements $a,b \in R$ satisfy $a^3=b^2$, you can conclude that there exists $r\in R$ with $a=r^2, b=r^3$ . The ring is then automatically reduced (Costa). This notion is due to Traverso and Swan. -[Strange condition, eh? For example, a normal domain $R$ is semi-normal: take $r=b/a\in Frac(R)$, which by normality must be in $R$ since it satisfies the integrality equation $r^2-a=0$. ] -A theorem of Swan then states that given a ring $R$, the map from $R$ to its polynomial ring $j:R\to R[T]$ induces a surjection $j^*:Pic(R)\to Pic(R[T])$ if and only if the reduced ring $R_{red}=R/Nil(R)$ is semi-normal. This proves the above claim about the cusp (and much more). -Bibliography: -a) Here is a nice, completely self-contained survey by Lombardi and Quitté on semi-normal rings. Its bibliography will lead you to the original articles by Traverso and Swan. -b) And there is another very nice survey by Vitulli. - -REPLY [6 votes]: No. Let $X$ be an elliptic curve, $p\colon Y=X\times\mathbf P^1\to X$ the projection, $g\colon X\to X$ multiplication by $2$, and $f=gp$. Take a line bundle $M$ of degree $1$ on $X$. Then, $p^*M$ is trivial on the fibres of $f$. Suppose $p^*M\cong f^*L$ for some line bundle $L$ on $X$. Since $p^*$ is injective on $\mathrm{Pic}$ (see Hartshorne, Ex. III, 12.5), we have $M\cong g^*L$, but the degree of $g^*L$ is even, contradiction.<|endoftext|> -TITLE: Interpolating between piecewise linear functions, with a family of smooth functions -QUESTION [15 upvotes]: Let $[a,b)\subset\mathbb R$, and $F,G:[a,b)\to\mathbb R$ two decreasing piecewise linear functions so that $F(x)\leq G(x)$ for any $x\in[a,b)$. We assume that: - -there is a number $k\in\mathbb N-\{0\}$ and a set of $k+1$ numbers $a=x_0\lt\ldots\lt x_k=b$ partitioning $[a,b)$ in $k$ intervals $[x_{j-1},x_j)$ on which the restrictions of $F$ and $G$ are linear. -the equalizer set $E=Eq(F,G):=\{x\in[a,b)|F(x)=G(x)\}$ is included in the set $\{x_0,x_1,\ldots,x_k\}$. -the restriction of $F$ to the set $E$ is strictly decreasing -for any $j$, $0\lt j\lt k$, there is an open neighborhood $(x_j-\epsilon,x_j+\epsilon)$, and a linear function $L:(x_j-\epsilon,x_j+\epsilon)\to\mathbb R$, so that $G\geq L\geq F$ on $(x_j-\epsilon,x_j+\epsilon)$. - -(conditions 2-4 forbid some cases when the problem has no solution) -Problem: Find a continuous family of functions $f_t:[a,b)\to\mathbb R$, $t\in[0,1]$ satisfying the conditions: - -$f_t$ is smooth and strictly decreasing for any $t\in(0,1)$. -For any fixed $x\in[a,b)-E$, the application $\vartheta_x:[0,1]\to [F(x),G(x)]$, $\vartheta_x(t)=f_t(x)$ is a strictly increasing and bijective smooth function. - -It is easy to see that from the condition 2 it follows that: - -if $0\leq s\lt t\leq 1$, then $f(s)\lt f(t)$ on $[a,b)-E$ (and of course $f(s)=f(t)=F=G$ on $E$). -$f_0=F$ and $f_1=G$. - -If possible, please also provide some references which can help solving this problem. - -Update 1: -So far I tried to construct the functions from known analytical functions, splines and sigmoids, and to use Schwarz-Christoffel to map the region between $F$ and $G$ to a rectangle in the complex plane. While these methods appeared to have some advantages, it seems difficult to show that they really satisfy the required conditions. Anyway, I don't want to reinvent the wheel. - -REPLY [6 votes]: As noted by Dejan Govc, the set $E$ should also contain those points where the right limit of $F(x)$ equals the left limit of $G(x)$, because any continuous function bounded between $F$ and $G$ will be forced to the unique limit in these points. - -Let's first describe a construction for a family of continuous functions, which will later be refined to give a family of smooth functions. In the picture below, the parts where the smooth construction is identical to the continuous construction are drawn with bold lines. The drawn parts are $f_t(x)$ for $t=\frac{1}{4}$, $t=\frac{1}{2}$ and $t=\frac{3}{4}$. The construction separates an interval into four different regions, either using the diagonal if $F$ and $G$ don't meet each other in the interval, or cutting the interval in half an using the diagonals of the subinterval where $F$ and $G$ don't meet. - -The piecewise linear family $p_t(x)=(1-t)F(x)+tG(x)$ corresponds to a family of straight lines inside of each interval. This family either consists of parallel lines, or these lines meet in a common point which is not inside the interval. The bold lines in the image are the parts where we have $f_t(x)=p_t(x)$. -Let $y_j:=\lim_{x\to x_j^-}F(x)$ be the relevant limit of $F$ at $x_j$ and $Y_j:=\lim_{x\to x_j^+}G(x)$ be the relevant limit of $G$. We have $y_j \leq Y_j$. We use $f_t(x_j)=(1-t)y_j+tY_j$. For the continuous construction, these two different parts of the construction are simply connected by straight lines. Let's denote the resulting family by $f_t^{cont}$. It's easy to check that this construction satisfies to requirements of the question, except that $f_t^{cont}$ (and the corresponding $\vartheta_x^{cont})$ is only continuous instead of smooth, and that the set $E$ had to be enlarged as described above. - -If we want to smooth out the continuous family $f_t^{cont}$ to get a smooth family $f_t$, we need to "specify" how $f_t$ should behave near the non-smooth points of $f_t^{cont}$. It's clear how it should behave near the parts with $f_t(x)=p_t(x)=f_t^{cont}(x)$, so let's focus on the regions around $x_j$. If $y_j < Y_j$, we can use the family of straight lines generated by the straight line given by $F$ on the left side of $x_j$ and by the straight line given by $G$ on the right side of $x_j$. If $y_j=Y_j$, we can use the linear function $L$ from assumption 4 to define how $f_{1/2}$ should behave near $x_j$. Because $f_t$ should not cross $f_{1/2}$, we use $L(x)+(t-\frac{1}{2})x^2$ to define how - $f_t$ should behave near $x_j$. - -At the left side, we see that line $L$ for the case $y_j=Y_j$ can lead to problems with how we separated the interval into different regions. The image also suggests how these problems can be fixed. -Regarding the missing details of this construction, I had started to explicitly construct the smoothing for the case $y_j < Y_j$, but inadvertently ignored the condition that $\vartheta_x(t)$ should be a smooth function of $t$ instead of just a continuous function. For the case with $y_j=Y_j$, I haven't started to think about an explicit smoothing, and I also don't know for sure whether the behavior of $f_t$ around $x_j$ prescribed above will always allow such a smoothing (but a more "generic" linear function $L$ with strict inequalities should fix this, in case it really is a problem). - -Regarding references, this is one of the reasons why I decided to work out a detailed solution. A long time ago, I wrote a German text that starts by construction smooth and differential curves satisfying various conditions (these curves are used as "test curves" later in the text). That text cites normal introductory and slightly advanced analysis texts as main references. I really think this is the place where one gets taught how to construct curves adapted to various purposes. It is laborious to do so, but I fear no theory will save us from that.<|endoftext|> -TITLE: Does a fixed-point free "homotopy involution" imply that a manifold bounds? -QUESTION [6 upvotes]: Let $M^n$ be a closed (compact, connected, without boundary) smooth manifold. It is known that if there exists a fixed point free involution $\tau:M \rightarrow M$, then M bounds. That is, there exists a compact manifold $W^{n+1}$ such that $\partial W = M$. -But now suppose $\tau$ is only a "homotopy involution". That is $\tau^2$ is only homotopic to the identity on $M$ rather than equal to the identity. Can we say that $M$ bounds? -For some reason I feel this statement is not true..., but I have not been able to construct a counterexample yet. For a counterexample, maybe an aspherical, nonbounding manifold would be the best candidate? -On a related question, what if we say that $\tau^2$ is isotopic to the identity on M. Then does M bound? -Thanks, I appreciate any responses. - -REPLY [11 votes]: A manifold with zero Euler characteristic admits a nowhere-vanishing vector field, which generates a one-parameter group of diffeomorphisms that are (smoothly) isotopic to the identity. A sufficiently small element $\tau$ is fixed-point free since the vector field does not vanish and the manifold is compact. -There are manifolds with zero Euler characteristic that do not bound, for instance the unoriented cobordism group in dimension 5 is not trivial, see the Wikipedia page on cobordisms.<|endoftext|> -TITLE: Corvallis 1979 proceedings -QUESTION [6 upvotes]: These proceedings have long been freely available on the AMS website, but now it seems we can't even find them anymore (e.g. http://www.ams.org/publications/online-books/pspum331-index and http://www.ams.org/publications/online-books/pspum332-index). Unfortunately, I don't own a copy, nor have I ever downloaded it on my computer. My question, then: Does anyone know if there exists another website where I could get it? - -REPLY [6 votes]: Given the importance of the proceedings, I leave a link to the copy at Library Genesis: -A. Borel and W. Casselman (Editors) Automorphic Forms, Representations and L-functions (1979)<|endoftext|> -TITLE: Do Richardson varieties have rational singularities in arbitrary characteristic? -QUESTION [11 upvotes]: The title basically asks the question. I'll review the relevant terminology and explain what I have and haven't found in the literature. -Let $G$ be a reductive group. Let $v \leq w$ be elements of the Weyl group, with $X_v$ and $X^w$ the corresponding Schubert and opposite Schubert. Let $R_v^w = X_v \cap X^w$. This is called a Richardson variety. A variety $R$ is said to have rational singularities if there is a resolution of singularities $Z \to R$ which has certain nice cohomological properties. -I know of two references in the literature for the fact that Richardson varieties have rational singularities: Theorem 4.2.1 in Michel Brion's "Lectures on the Geometry of Flag Varieties" and Lemma 2 in his "Positivity in the Grothendieck Group of Flag Varieties". Both of these essentially give the same proof, though in different language. In the notation of "Lectures", they construct a variety $Z_{\underline{v}}^{\underline{w}}$ with a map to $R_v^w$ and proof that it has the correct cohomological properties. This part of the proof works in any characteristic. -However, in order to show that $Z_{\underline{v}}^{\underline{w}}$ is smooth, they appeal to Kleiman transversality, or to generic smoothness. These arguments only work in characteristic zero. -Does anyone know a reference which addresses this? (I have e-mailed Brion, and also Kumar, and am waiting to hear back, but I figured someone here might know this.) - -REPLY [4 votes]: Allen Knutson, Thomas Lam and I prove this in an Appendix we have recently added to our paper Projections of Richardson Varieties. Many thanks to Michel Brion and Shrawan Kumar for helping us by e-mail.<|endoftext|> -TITLE: Stable homotopy category and the moduli space of formal groups -QUESTION [15 upvotes]: The usual disclaimer applies: I'm new to all this stuff, so be gentle. -It seems like the spectrum, as defined by Balmer, of the stable homotopy category of finite complexes is something like $M_{FG}$, the stack of formal groups (that is, $Spec L/ G$ where $L$ is the Lazard ring and $G$ acts by coordinate changes). I'm not actually sure if that's true, I don't think I've seen it written quite like that, but the picture of the spectrum in Balmer's paper looks an awful lot like how I'd imagine $M_{FG}$ looking. -If the above is right, then there's another tensor triangulated category with the same spectrum, namely the derived category of perfect complexes on $M_{FG}$ (whatever that means for stacks...). -So my question is: - -Just how far away is the stable homotopy category from actually being equivalent to this derived category? Is there a theorem to the effect that it can't be equivalent to such a thing? Do we even know that it's not equivalent? - -I've heard that chromatic homotopy theory is about setting up a rough dictionary between algebro-geometric terminology regarding $M_{FG}$ and the stable homotopy category, so I guess the question is about whether or not we can make the dictionary into a proper functor. - -REPLY [22 votes]: It's definitely known that the derived category of ${\cal M}_{FG}$ and the stable homotopy category are not equivalent. This is an instance of - -The Mahowald Uncertainty Principle: - Any spectral sequence converging to - the homotopy groups of spheres with an - $E_2$-term that can be named using - homological algebra will be infinitely - far from the actual answer. - -(The naming is due to Ravenel; this quote is from Paul Goerss' "The Adams-Novikov Spectral Sequence and the Homotopy Groups of Spheres".) There is often a feeling that stable homotopy theory always deviates from algebra as soon as is possible. -As Neil said, the Adams-Novikov spectral sequence starts with morphisms in the derived category and computes stable homotopy groups of spheres. Every place where this spectral sequence does not degenerate indicates a point where the stable homotopy category deviates from the algebraic approximation. This includes the following phenomena. - -Hidden additive extensions, such as the hidden additive extension making $\pi_3^s$ into $\mathbb{Z}/24$ rather than $\mathbb{Z}/12 \times \mathbb{Z}/2$. -Hidden multiplicative extensions. In the (2-local) stable homotopy groups there are elements $\eta \in \pi_1^s$, $\nu \in \pi_3^s$, and $\sigma \in \pi_7^s$. My recollection is that such that $\eta^2 \sigma = \nu^3$ on the $E_2$-term, but Toda showed that this relationship doesn't hold on-the-nose in stable homotopy groups of spheres. -Differentials. For any prime $p$, there is always a nontrivial differential in the Adams-Novikov spectral sequence, and the first differential is called the Toda differential.<|endoftext|> -TITLE: Hyperbolicity on Riemann Surfaces -QUESTION [9 upvotes]: For Riemann surfaces there are at least to possible notions of hyperbolicity. The classical one given by the Uniformization Theorem, or equivalently the type problem, which essentially says that a simply connected Riemann surfaces is conformally equivalent to one of the following: - -Riemann Sphere $\mathbb{C}\cup\{\infty\}$ (elliptic type). -Complex plane (parabolic type). -Open unit disk (hyperbolic type). - -On the other hand, given a Riemann surface one can asks if it is hyperbolic in the Gromov's sense. In other words, does there exists $\delta>0$ such that all the geodesic triangles in the surface are $\delta$-thin? -It seems to me that this two notions of hyperbolicity are not equivalent and one can have counterexamples in both directions. For instance, the two dimensional torus $\mathbb{T}^2$ is hyperbolic in Gromov's sense (since it is compact), but it's also a quotient of the Euclidean plane by a free action of a discrete group of isometries and therefore, of parabolic type. -My questions are: what is a sufficient condition for a surface of hyperbolic type to be Gromov's hyperbolic? what is known about the relation of these two notions? -Related Question: Let $G$ be an infinite planar graph with uniformly bounded degree and assume that the simple random walk is transient. Is the graph necessarily Gromov's hyperbolic? - -REPLY [2 votes]: About the related question: it is a result of Babai that a (connected, locally finite) vertex-transitive, planar graph is isomorphic to the 1-skeleton of an Archimedean tiling of the sphere, or the Euclidean plane, or the hyperbolic plane. So assuming transience singles out the hyperbolic plane, and implies Gromov-hyperbolicity for the graph. See -http://www.cs.uchicago.edu/files/tr_authentic/TR-2001-04.ps<|endoftext|> -TITLE: Gauss sum (with sign) through algebra -QUESTION [11 upvotes]: Let $p$ be an odd prime, and $\zeta$ a primitive $p$-th root of unity over a field of characteristic $0$. -Let $G = \sum\limits_{j=0}^{p-1} \zeta^{j\left(j-1\right)/2}$ be the standard Gauss sum for $p$. (An alternative definition for $G$ is $G = \sum\limits_{j=1}^{p-1}\left(\frac{j}{p}\right)\zeta^j$, where the bracketed fraction denotes the Legendre symbol.) Denote by $k$ the element of $\left\lbrace 0,1,...,p-1\right\rbrace$ satisfying $16k\equiv -1\mod p$. -Then, $\prod\limits_{i=1}^{\left(p-1\right) /2}\left(1-\zeta^i\right) = \zeta^k G$. -Question: Can we prove this identity purely algebraically, with no recourse to geometry and analysis? -If we can, then we obtain an easy algebraic proof for the value - including the sign - of the Gauss sum $G$, since both the modulus and the argument of $\prod\limits_{i=1}^{\left(p-1\right) /2}\left(1-\zeta^i\right)$ are easy to find (mainly the argument - it's a matter of elementary geometry). -Note that my "algebraically" allows combinatorics, but I am somewhat skeptical in how far combinatorics alone can solve this. Of course, we can formulate the question so that it asks for the number of subsets of $\left\lbrace 1,2,...,\frac{p-1}{2}\right\rbrace$ whose sum has a particular residue $\mod p$, but whether this will bring us far... On the other hand, $q$-binomial identities might be of help, since $\prod\limits_{i=1}^{\left(p-1\right) /2}\left(1-\zeta^i\right)$ is a $\zeta$-factorial. -I am pretty sure things like this must have been done some 100 years ago. - -REPLY [8 votes]: A few historical remarks about algebraic determinations of the sign of the quadratic gaussian sum might not be out of order. -The proof in David's post was first given by Kronecker, according to Hasse's Vorlesungen. The only analytic ingredient is the determination of the sign of the sin function. This proof is reproduced in Fröhlich and Taylor, Algebraic Number Theory, pp. 228--231. -A different algebraic proof, using the same analytic ingredient, was given by Schur and can be found in Borevich and Shafarevich, Number Theory, pp. 349--353. -Hasse's Vorlesungen also contain a proof by Mordell in which the analytic ingredient is replaced by the fact that if a polynomidal $f\in{\mathbf Z}[T]$ has opposite signs at $a,b\in{\mathbf R}$, then it has a root between $a$ and $b$. This can be proved using the purely algebraic theory of Artin and Schreier. -If you are looking for a proof using more analysis, not less, see Rohrlich's survey on Root Numbers in Arithmetic of L-functions, pp. 353--448. -Addendum. A nice (if somewhat dated) survey on The determination of Gauss sums can be found in the BAMS 5 (1981), 107-129. I learnt there that the proof attributed by Hasse to Kronecker actually goes back to Cauchy. New proofs are still being given; see for example Gurevich, Hadani, and Howe, Quadratic reciprocity and the sign of the Gauss sum via the finite Weil representation. -Int. Math. Res. Not. IMRN 2010, no. 19, 3729–3745, available here.<|endoftext|> -TITLE: Periodic matrices -QUESTION [6 upvotes]: A square matrix $M$ such that $M^{k+1}=M$, for some positive integer $k$, is called a periodic matrix. - -Can we characterize the periodic matrices in $\mathcal{M}_n(\mathbb{Z})$? -If we replace $\mathbb{Z}$ by an Euclidean domain? -If we replace $\mathbb{Z}$ by a PID? - -REPLY [11 votes]: Geoff already gave a description. Here is a semigroup theory approach. $M^{k+1}=M$ means that $E=M^k$ is an idempotent, $E^2=E$, and $EM=M=ME$. All idempotents in the matrix semigroup over $Z$ are easily described as matrices similar to diag$(0,...,0,1,1,...,1)$ (several 0's followed by several $1$'s) with unimodular conjugator. Hence we can assume that $E$ has that form. Therefore $M=EM=ME$ must have the form described in Geoff's answer. The same description holds for matrices over any ring if the structure of idempotents is similar to the above. - Edit. As Geoff pointed out below, in fact since $EM=ME=M$, we get that the block $A$ in $M$ is 0, so $M$ looks like $$\left(\begin{array}{ll} 0&0\\\ 0 & B\end{array}\right)$$ where $B$ is an integer matrix with $B^k=1$. This is of course an "if and only if" description. I am pretty sure this has been known since the 50s, but I do not have time to dig it up. It should follow from the description of Green relations in the matrix semigroups.<|endoftext|> -TITLE: Product of ultrafilters, is it an ultrafilter? -QUESTION [6 upvotes]: Let $a$ and $b$ are filters. The product $a\times b$ is defined as the filter (on the set of pairs) induced by the base $\{ A\times B | A\in a, B\in b \}$. -It is simple to show that product of a non-trivial ultrafilter with itself is not an ultrafilter (as it is not finer than the principal filter corresponding to the identity relation). -My question: Is product of every two (different) non-trivial ultrafilters always not an ultrafilter? -Note: non-trivial ultrafilter is the same as non-principal ultrafilter. - -REPLY [16 votes]: The product $a\times b$ of two ultrafilters is an ultrafilter if and only if, for every function $f$ from the underlying set of $a$ into $b$ (that's not a typo for "into the underlying set of $b$"), there is a set $A\in a$ such that $\bigcap_{x\in A}f(x)\in b$. One way for this to happen is for the underlying set of $a$ to be small enough and $b$ to be complete enough, as in Joel's answer. Notice, though, that the condition is, despite its appearance, symmetrical between $a$ and $b$. In particular, if $b$ lives on $\omega$ while $a$ is countably complete, then the condition is satisfied because $f$ will be constant on some set in $a$ (because countably complete ultrafilters are closed under intersection of continuum many sets). [Archaeologists may be interested to know that this characterization of ultrafilters whose product is ultra occurs on page 22 of my 1970 Ph.D. thesis, which is, thanks to patient scanning, available from my web page.] - -REPLY [12 votes]: No. If $a$ and $b$ are principal ultrafilters, then so is the product filter as you have defined it. If $a$ and $b$ contain $\{x\}$ and $\{y\}$, respectively, then the base of your product includes the singleton $\{(x,y)\}$, and hence it is the principal ultrafilter. -But perhaps by "nontrivial" you meant nonprincipal. In this case, here is another example. Let $\mu$ be any ultrafilter on $\omega$ and let $\nu$ be a $\kappa$-complete ultrafilter on a measurable cardinal $\kappa$. If we consider the product filter $\mu\times\nu$ on $\omega\times\kappa$, as you have defined it, then it is an ultrafilter, since for any $X\subset \omega\times\kappa$, there are fewer than $\kappa$ many possible horizontal slices $X_\alpha=\{n\mid (n,\alpha)\in X\}$, and so there is some $A\subset \omega$ such that $\{\alpha\lt\kappa\mid X_\alpha=A\}\in \nu$. If $A\in\mu$, then $X$ is in the product filter, and if $A\notin\mu$, then the complement of $X$ is in the product filter. So $\mu\times\nu$ as you have defined it is an ultrafilter.<|endoftext|> -TITLE: What does $L[\mathcal P(Ord)]$ look like? -QUESTION [5 upvotes]: The construction of $L[A]$ can be considered for the case when $A$ is a class of ordinals. We simply consider things which are definable over the language $\{\in, A\}$ instead of just $\{\in\}$. -Originally I thought that $L[Ord]=HOD$, but was told that actually $L[Ord]=L$. -Denote $\mathcal P(Ord) = \bigcup_{\alpha\in Ord}\mathcal P(\alpha)$, all the sets of ordinals in $V$. I then figured that perhaps $L[\mathcal P(Ord)]=HOD$, but that too I was told is not true. -In fact, as the conversation continued, $HOD$ is the second-order $L$, that is we construct $L$ as usual (in $V$), only defining it with second-order logic instead. -If this is the case, what is $L[\mathcal P(Ord)]$? - -REPLY [9 votes]: One should make a distinction between two kinds of relative constructibility. Traditionally, one uses the square bracket notation $L[A]$ to indicate the result of constructing as you said where one allows $A$ as a predicate, so that $L_{\alpha+1}[A]$ consists of the definable subsets of the structure $\langle L_\alpha[A],{\in},A\rangle$. The alternative round-bracket structure $L(A)$ is obtained by throwing in (the transitive closure of) $A$ wholesale, and then constructing just in the language of $\in$. (Some accounts add only $\text{TC}(A)\cap V_\alpha$ at stage $\alpha$, rather than all of it at the beginning, and this is more sensible when $A$ is a class.) -The difference is that in $L[A]$, you are only able to ask queries about membership-in-$A$ for objects that you can construct, whereas in $L(A)$ you are given all of $A$ and its transitive closure at the outset. In particular, a set $A$ is always in $L(A)$, but it may not actually be an element of $L[A]$. -For example, $L[\mathbb{R}]=L$, even when there are non-constructible reals, since having a predicate for $\mathbb{R}$ is not helpful; we can already recognize when an object is a real number without having a predicate for $\mathbb{R}$. But in most of the interesting cases, $L(\mathbb{R})\neq L$. -The same analysis applies to $L[\text{Ord}]=L=L(\text{Ord})$, where the former equality holds since we don't need a special predicate to recognize an ordinal, and the latter equality holds since $L$ already has all the ordinals. -In the case of $P(\text{Ord})$ as you have defined it, Andreas has given you the answer actually for $L(P(\text{Ord}))$, which is all of $V$ precisely because every set in $V$ is coded by a set of ordinals, which is available in $L(P(\text{Ord}))$. And probably this is the case that you intended to ask about. -But meanwhile, for the case you actually asked about, $L[P(\text{Ord})]=L$, since having a predicate for $P(\text{Ord})$ is not so helpful, as we are already able to recognize whether a set is a set of ordinals without having a predicate for $P(\text{Ord})$. So this case is just like $L[\mathbb{R}]=L$. -Finally, I should mention that although the square-bracket and round-bracket notation is fairly standard in most of the set-theoretic community, nevertheless I have heard that some quarters use the notation with exactly the opposite meaning. -Finally, I point you to the Chang model, which is the result of doing the $L$ construction in the infinitary logic $L_{\omega_1,\omega}$, and there are analogous higher versions.<|endoftext|> -TITLE: Why is $\mathbb{R}^{\infty}$ defined the way it is? -QUESTION [9 upvotes]: I've been thinking about Grassmannians recently. Think of $\mathbb{R}^k$ as a $k$-dimensional vector space. Let $\text{Gr}_n(\mathbb{R}^k)$ denote the Grassmannian of all $n$-dimensional vector subspaces of $\mathbb{R}^k.$ (This is a compact, Hausdorf topological manifold of dimension $n(k-n)$.) Let -$ \Gamma^n(\mathbb{R}^k) := \{ (X,v) : X \in \text{Gr}_n(\mathbb{R}^k) \text{ and } v \in X \} . $ -There's a standard idea of a vector bundle $\pi : \Gamma^n(\mathbb{R}^k) \twoheadrightarrow \text{Gr}_n(\mathbb{R}^k)$ given by $\pi(X,v) := X.$ This bundle has the nice property that lots of other bundles can be realised as sub-bundles of it. There is a more general definition, where we use $\mathbb{R}^{\infty}$ in place of $\mathbb{R}^k$. My question is about why we define $\mathbb{R}^{\infty}$ the way we do. -We define $\mathbb{R}^{\infty}$ as the set of infinite sequences $(x_1,x_2,x_3,\ldots)$ where each $x_i \in \mathbb{R}$ and only finitely many of the $x_i$ are non-zero. We identify $\mathbb{R}^k$ with the sequences of the form $(x_1,\ldots,x_k,0,0,\ldots),$ and then topologize $\mathbb{R}^{\infty}$ as the direct limit of the sequence $ \mathbb{R}^1 \subset \mathbb{R}^2 \subset \mathbb{R}^3 \subset \ldots$ Then we get the universal bundle $\pi : \Gamma^n(\mathbb{R}^{\infty}) \twoheadrightarrow \text{Gr}_n(\mathbb{R}^{\infty}).$ -My question is why do we insist that only finitely many of the $x_i$ are non-zero for each $(x_1,x_2,x_3,\ldots) \in \mathbb{R}^{\infty}$? I understand that it gives a countably infinite dimensional vector space, but that's a result of the definition; it doesn't explain why we define it the way we do. I suspect that it's related to the topology, but I don't really know. -Edit: The context is the OP is reading Milnor and Stasheff. - -REPLY [2 votes]: Obviously tastes/opinions vary, but I think some ambiguous, or insufficiently localized, or not publishable-research-y enough, but nevertheless valuable to (a significant demographic of) research mathematicians... In fact, sometimes these questions are exactly the "dumb, non-research" questions that "everyone" (anyway, many people) have asked themselves... and not received a cogent answer. -One cliched-but-important (in my opinion) point is the "naive category-theory" explanation of why $\mathbb R^\infty$ is "defined" to be what it is. This does raise the entirely legitimate meta-meta-question of why we "have definitions", and "who is authorized to make them"... to which the easy answer (in my opinion, with some hindsight) is that, not merely must definitions capture the phenomena of important examples (or else the definitions are dumb), but, actually, as it seems to happen very often, re-ordering the "definition...theorem" sequence to "(mapping-)characterization..." rewrites the narrative so that the required/desired property is written in mildly category-theoretic terms, and the technical bit is perhaps proof-of-existence. -That is, a (typically, set-theoretic) "definition" is actually just _one_specific_construction_ of an object whose important features (if it exists at all) are completely determined by its interactions with other objects. That is, its characterization is "category theoretic" rather than "set theoretic". -(Yes, this is an advertisement for a certain little bit of category theory, though it is not a paid advertisement, insofar as I do not at all advocate formal category theory, nor would I advocate allocating one's personal resources to fretting over reconciliation of set theory and category theory... e.g., Grothendieck's "universes" and large cardinals? Fun, but likely not refering directly to one's original issue...) -So, rewriting the question about "why is the definition of $\mathbb R^\infty$ what it is?", we are required to ask what function this thing should have. Well, it is almost immediate that it should be the _ascending_union_ of the $\mathbb R^n$'s. That is, (upon reflection!) it is a (filtered) colimit (a.k.a. "inductive limit"). That is, it should/must/does have certain diagrammatic/mapping properties... as opposed to goofy set-theoretic constructional details. -Issues about infinite-dimensional Grassmannians... infinite-dimensional simplicial complexes... do share that basic feature, namely, that there is a mapping property (if only ascending-union sorts of (filtered) colimit properties) that are relevant. -Truly, the above viewpoint seems to me to be extraordinarily efficient/effective as explanatory device... -(And, one more time, questions that fail to be "documentable research" sometimes are far more interesting and useful to "us" than more focused ones... Of course, this is not a general rule...)<|endoftext|> -TITLE: Applications of and motivation for von Neumann's mean ergodic theorem -QUESTION [18 upvotes]: I stated von Neumann's mean ergodic theorem (VNMET) in a talk recently and someone in the audience asked what it was good for. The only application I knew of VNMET was to prove Birkhoff's ergodic theorem (BET), which is why I'd stated VNMET in the first place. But I'm pretty sure that VNMET came first, so I doubt it was originally proven with that in mind. -Question 1. How did the theorem (or conjecture) arise in the first place? E.g. was it intended as as mere stepping-stone to BET? -The only applications I've seen (and can find) of ergodic theory to other branches of math (or physics) use BET. -Question 2. What are some applications of VNMET? I'm particularly interested in applications to other branches of mathematics; so I'm looking for something other than the "application" of it to prove BET. -Edit. What I'm really trying to glean with the above questions is an answer to the following question: -Question 3. Is VNMET important/useful beyond proving BET? - -REPLY [10 votes]: von Neumann long argued that for physics, his result suffices (see, e.g., Proc. Nat. -Acad. Sci. U.S.A. 18 (1932), 263–266,). There is not only truth to that but also to the fact that his result suffices for some of the mathematical applications. Moreover, as von Neumann -emphasized [in the above], there is one aspect of his result that is stronger than -Birkhoff’s. If one defines - $$Av(n,L) (\omega; f) = \frac{1}{n} \sum_{j=L}^{n+L-1} f(T_j(\omega))$$ -then as $n \rightarrow\infty$, in $L^2$, $Av(n,L) ( · ; f)$ converges uniformly in $L$ (as can be -seen by looking at either the von Neumann or Hopf proofs), but the pointwise convergence need not be uniform in $L$.<|endoftext|> -TITLE: Characterizing elementary embeddings of $L$ and $L_\alpha$ under 0# -QUESTION [6 upvotes]: Suppose 0# exists. -It is clear that every order preserving map from the indiscernibles to the indiscernibles gives an elementary embedding from $L$ to $L$. Furthermore, following lemmas 18.7 and 18.8 of Jech, if $\alpha$ is an infinite infinite limit ordinal, an increasing map from alpha to beta gives an elementary embedding from $L_{i_\alpha}$ to $L_{i_\beta}$, where $i_\alpha$ is the $\alpha$-th indiscernible. This is because $L_{i_\alpha}$ equals the Skolem hull in itself of the first $\alpha$ indiscernibles. However, I am not clear on the following points. -1) Is it the case that for a finite successor ordinal, n, $L_{i_n}$ is necessarily equal to the Skolem hull in $L_{i_n}$ of the first n indiscernibles? Jech only proves this result for infinite ordinals. -2) Is it possible that there could be an elementary embedding from $L$ to $L$, or from $L_{i_\alpha}$ to $L_{i_\beta}$ ($\alpha, \beta$ may be finite or infinite), that does not always map indiscernibles to indiscernibles? -This sounds weird, but I'm not convinced it's impossible. As far as I know, there's no formula in $L$ that defines "$\alpha$ is a Silver indiscernible." (In fact there is no such formula -- see Andreas Blass's comment below.) - -REPLY [5 votes]: The answer to Q2 is 'No'. Suppose $j:L\rightarrow L$ is a non-trivial elementary embedding. We use the following fact: -$\bullet$ $cp(j)$ (the first ordinal moved by $j$) is always a Silver indiscernible. -Now let $I$ be the class of Silver indiscernibles, and $\delta \in I$ but $j(\delta)\notin I$ for a contradiction. Let $H$ be the Skolem hull in $L$ of $j(\delta)\cup j$''$I\backslash (\delta +1)$. $H$ is isomorphic to $L$. If $j(\delta)\notin H$ but $\pi:H \rightarrow L$ is the transitive collapse, then $\pi^{-1}:L\rightarrow L$ is non-trivial with critical point $j(\delta)$. Hence, by the Fact, $j(\delta)$ must be in $H$. Then we see that for some $\vec \xi j(\delta)$ with $\vec \zeta \in I\backslash (\delta +1)$ that -$L \models $ ''$\exists \vec \xi < j(\delta)( j(\delta) = t(\vec \xi ,\overrightarrow{j(\zeta)}))$''. -for some term $t$. But then: -$L \models $ ''$\exists \vec \xi < \delta( \delta = t(\vec \xi, \overrightarrow{\zeta}))$'' -is a definition of the indiscernible $\delta$ from larger indiscernibles and smaller ordinals, which is impossible. (This works for the variant of the question, taking embeddings between sets, if $\alpha, \beta$ are limit ordinals.)<|endoftext|> -TITLE: A simple minded Poincare duality for orbifolds? -QUESTION [6 upvotes]: Suppose $X^n$ is an orientable compact orbifold (without boundary) with stabilisers in codimension 2, and $\bar X^n$ is the underlying topological space. We can assume, moreover, that $X^n$ is a quotient of a manifold $X'^n$ by an action of finite group $G$. -Is it true that, for simplicial homologies of $\bar X^n$, we have $H_{n-k}(\bar X^n, \mathbb R)$ is dual to $H_k(\bar X^n,\mathbb R)$? -If not, what is a simplest counterexample, and what is the correct statement? If yes, what would be a reference? -PS. It seems to me that this should be true in the case when $X^n$ is a global quotient of a manifold by a finite group, because I guess in this case the simplicial homology of $X^n$ should be equal to the invariants of the action of $G$ on $H_k(X'^n)$. At the same time, actions on $H_k(X'^n)$ and $H_{n-k}(X'^n)$ are dual. Is this reasoning correct? - -REPLY [2 votes]: Chapter 5 of "Orbifolds and stringy topology" by Adem, Leida, Ruan have a version of Poincare duality in the groupoid setting. This is probably a very general result on this.<|endoftext|> -TITLE: Hopf Algebras and Quantum Groups -QUESTION [10 upvotes]: I have studied graduate abstract algebra and would like to learn about Hopf algebras and quantum groups. What book or books would you recommend? Are there other subjects that I should learn first before I begin studying Hopf algebras and quantum groups? - -REPLY [4 votes]: Thomas Timmermann's Invitation to Quantum Groups and Duality.<|endoftext|> -TITLE: Given a curve, under which condition is the set of gonal morphisms finite -QUESTION [9 upvotes]: Recently, in my research I bumped onto gonal morphisms. At the moment, my knowledge is based upon some things I read on the internet. Before stating my questions, I added some definitions/facts that might motivate the questions below. -By a curve, I mean a smooth projective connected curve over $\mathbf{C}$. A non-constant morphism $\pi:X\longrightarrow \mathbf{P}^1$ is gonal if $\deg \pi$ is minimal. The gonality of a curve $X$, denoted by $\gamma_X$, is the degree of a gonal morphism $\pi:X\longrightarrow \mathbf{P}^1$. Thus, for example, a curve of genus $g\geq 2$ is hyperelliptic iff it is $2$-gonal. -The hyperelliptic map of a hyperelliptic curve is unique. (Of course, here by unique we mean unique up to composition with an isomorphism of the projective line.) -Edit: In the questions below, we consider the set of gonal morphisms of a curve modulo the action of Aut$(\mathbf{P}^1)$. -Fact 1. For any curve $X$ of genus $g\geq 2$, we have that $\gamma_X \leq [\frac{g+3}{2}]$. -Fact 2. For any integer $\gamma \geq 2$, the closure of the locus of $\gamma$-gonal curves in the moduli space $\mathcal{M}_g$ of smooth curves of genus $g\geq 2$ is -irreducible of dimension $2g-5+2\gamma$. -Fact 3. For any prime number $p$ and integer $g\geq 2$ such that $g\geq (p-1)^2$, Accola showed that any $p$-gonal curve of genus $g$ has a unique gonal morphism. -I can't prove these facts, but I do remember where I got them from. So if necessary I could give the references. -Question 1. Let $X$ be a $\gamma$-gonal curve of genus $g\geq 2$. Is the set of gonal morphisms for $X$ modulo the action of Aut$(\mathbf{P}^1)$ finite? -I expect the answer to this question to be negative if $g-\gamma$ is small. In view of Fact 3, I would like to propose the following question. -Question 2a. Fix $\gamma\geq 3$. Does there exist an integer $g_\gamma$ such that for any $g\geq g_\gamma$ and any $\gamma$-gonal curve $X$ of genus $g$, the gonal morphism for $X$ is unique? -Question 2b. Fix $\gamma\geq 3$. Does there exist an integer $g_\gamma$ such that for any $g\geq g_\gamma$ and any $\gamma$-gonal curve X of genus $g$, the set of gonal morphisms for $X$ is finite? -Question 3. Does there exist a positive integer $g_0$ with the following property? For any $g\geq g_0$ and curve $X$ of genus $g$, the set of gonal morphisms of $X$ is finite? -Question 4. Do there exist curves with infinitely many gonal morphisms? (Edit: In hindsight, this question is the same as Question 1.) -I think it's clear that these questions aren't unrelated. They are all related to the set of gonal morphisms associated to a curve. It would be wonderful to know when this set is finite. - -REPLY [7 votes]: Extending Rita's example, if $X$ is, say, a double cover of a curve of genus $3$, then $X$ can have arbitrarily large genus and it has gonality (at most) $6$. Moreover it has infinitely many $g^1_6$ (BTW $g^r_d$ means a linear system of degree $d$ and dimension $r$, so a $g^1_d$ is a map to $\mathbb{P}^1$ of degree $d$). So the answer to all of your questions is no. -Here is something that can be done. If you have two maps of degree $d$ from $X$ to $\mathbb{P}^1$, then you get a map from $X$ to $\mathbb{P}^1 \times \mathbb{P}^1$. If this map is injective, then the genus of $X$ is at most $(d-1)^2$ (or something like that). So if the genus is large, there must be a map $X \to Y$ such that any map of degree $d$ from $X$ to $\mathbb{P}^1$ factors through $Y$. If $d$ is prime, this cannot happen, hence your fact 3.<|endoftext|> -TITLE: Tate conjecture for elliptic curves local fields -QUESTION [6 upvotes]: Let $E_1$ and $E_2$ be elliptic curves over a field $k$, and let $l$ be a prime coprime to the characteristic of $k$ (if $char(k) \ne 0$). Let $\varphi$ denote the canonical map -$Hom(E_1,E_2)\otimes_{\mathbb{Z}} \mathbb{Z}_l \rightarrow Hom_{G_k}(T_lE_1,T_lE_2)$ `. -For any field $k$, $\varphi$ is easily shown to be injective, and (a case of) the Tate conjecture says that if $k$ is a finite field or a number field then $\varphi$ is surjective (though the proof is hard). What can we say if instead $k$ is a local field? -More precisely, my questions are: -1)Is there a reason/counterexample explaining why $\varphi$ will not be surjective for $k$ a local field? -2)If the answer to (1) is `yes', is there a weaker statement along the lines of the above which is (or is expected to be) true for $k$ a local field? (Sorry this is rather vague). -Thanks, -David - -REPLY [6 votes]: Hi David, -A nice question. The map $\varphi$ can be very far from surjective! One way to see this is as follows. Let us work over $Q_p$, and suppose first that $l\neq p$. Working with elliptic curves with good reduction, the corresponding Galois representation is determined by $a_p$ (as this determines the characteristic polynomial of Frobenius). -Now suppose instead that $l=p$, and consider elliptic curves with good ordinary reduction. Then the Galois representation is again determined (up to a finite number of possibilities) by $a_p$. More generally, the classification of crystalline Galois representations by weakly admissible modules shows that the local Galois representation is always determined up to a finite number of possibilities by $a_p$. -Now thinking about cardinality shows that there must be many pairs of elliptic curves over $Q_p$ which are not isogenous, but have isomorphic $l$-adic Galois representations.<|endoftext|> -TITLE: Need to bound a trigonometric sum -QUESTION [9 upvotes]: Let $\boldsymbol{\theta}=(\theta_1,\ldots,\theta_m)$ be a vector of real numbers in $[-\pi,\pi]$. For $t\ge 0$, define -$$ f(t,\boldsymbol{\theta}) = \binom{m+t-1}{t}^{-1} - \sum_{j_1+\cdots+j_m=t} \exp(ij_1\theta_1+\cdots+ij_m\theta_m),$$ -where the sum is over non-negative integers $j_1,\ldots,j_m$ with sum $t$. -Note that the number of terms in the sum is $\binom{m+t-1}{t}$, so -$|f(t,\boldsymbol{\theta})|\le 1$ with equality occurring when all the $\theta_j$s -are equal. -For a problem in asymptotic combinatorics, we need a bound on -$|f(t,\boldsymbol{\theta})|$ that decreases rapidly as the $\theta_j$s move apart and -is valid for all $\boldsymbol{\theta}$. -Surely this problem has been studied before? -Note that $\binom{m+t-1}{t}f(t,\boldsymbol{\theta})$ is the coefficient of $x^t$ in -$$\prod_{j=1}^m (1-xe^{i\theta_j})^{-1},$$ -which suggests some sort of contour integral approach. - -REPLY [2 votes]: We shall prove that -$$ -f(t,{\theta}) \le \frac{m-1}{t+m-1} \min_{1\le j,k \le m} \frac{1}{|\sin((\theta_j-\theta_k)/2)|}. -$$ -This shows that if the angles are not too close to each other, then the sum does get -small. -Suppose without loss of generality that the minimum in our bound occurs for $\theta_1$ and $\theta_2$ -(so these are the angles that are furthest apart). Then writing $j_1+j_2 =\ell$ we have -$$ -|f(t,{ \theta})| \le \binom{m+t-1}{t}^{-1} \sum_{\ell=0}^{t} \sum_{j_3+\ldots+j_m=t-\ell} -\Big| \sum_{j_1+j_2 =\ell} \exp(ij_1 \theta_1 + ij_2 \theta_2)\Big|. -$$ -Now the inner sum over $j_1$ (and $j_2=\ell -j_1$) is simply a geometric progression, and -so -\begin{align*} -\Big| \sum_{j_1+j_2=\ell} \exp(ij_1 + i j_2 \theta_2) \Big| &= -\Big| \sum_{j=0}^{\ell} \exp(ij (\theta_1-\theta_2))\Big| = -\Big|\frac{\exp(i(\ell+1)(\theta_1-\theta_2))-1}{\exp(i(\theta_1-\theta_2))-1}\Big| -\\ -&\le \frac{2}{|\exp(i(\theta_1-\theta_2))-1|} = \frac{1}{|\sin((\theta_1-\theta_2)/2)|}. -\end{align*} -Therefore -$$ -|f(t,{\theta})| \le \frac{1}{|\sin((\theta_1-\theta_2)/2)|}\binom{m+t-1}{t}^{-1} \sum_{\ell+j_3+\ldots+j_m=t} 1, -$$ -which proves our claimed bound. -As the original question suggests, one would be able to prove better bounds -using contour integrals. The key would be to integrate on the circle centered at -the origin and with radius $r=t/(m+t)$. One should be able to get good bounds -in terms of -$$ -\sum_{j,k} \sin^2 \Big(\frac{\theta_j-\theta_k}{2}\Big), -$$ -but I have not worked this out carefully.<|endoftext|> -TITLE: When is a Form a Kähler Form? -QUESTION [15 upvotes]: Let $M$ be a complex manifold, and $\omega$ a closed $2$-form. When is $\omega$ a Kähler form? I mean, when does there exist a Kähler metric for which $\omega$ is the corresponding form. -I would (wildly) guess that necessary and sufficient conditions might be got from the Kähler identities. - -REPLY [8 votes]: It's possible to approach the question in a slightly different way if $X$ is compact. Donu and Spiro are of course right in that the condition for a smooth closed $(1,1)$-form $\omega$ to be a Kahler metric is that $\omega$ be positive. This is a pointwise condition on the manifold $X$, which can be difficult to check unless we have quite explicit expressions for $\omega$. -Suppose for a minute that $\omega$ is a Kahler metric. If $Y \subset X$ is a $p$-dimensional closed subspace, then we see that -$$ -\int_Y \omega^p = \int_Y [\omega]^p > 0, -$$ -where $[\omega]$ is the cohomology class that $\omega$ defines. This holds because $\omega$ induces a Kahler metric on (the smooth part of) $Y$ and $\omega^p$ is $p!$ times the volume form of the induced metric. (I'm skipping over some subtle things that involve integration over singular subspaces, but one can make sense of all of this.) -Demailly and Paun proved that the converse holds. That is, if we have a cohomology class $[\omega]$ such that its integral over any subspace of $X$ (including $X$ itself) is positive, then $[\omega]$ contains a Kahler metric. (This is a big generalization of the Nakai-Moishezon criterion in algebraic geometry.) -This doesn't help if what you're interested in is knowing that the precise form you have is Kahler, because given a Kahler form we can always take a nonconstant real smooth function $f$ and add $i\partial\bar\partial f$ to the form; this doesn't change the cohomology class and adding (or subtracting) giant such multiples will break the positivity of the resulting form. But for some questions, all that matters is the cohomology class of the Kahler form, and for those you can check positivity of the class by integration.<|endoftext|> -TITLE: What's so "schematic" about schemes? -QUESTION [11 upvotes]: Well, the title clearly follows the title of this question. -Why the objects so successfully defined by Grothendieck have been called "schemes"? In my opinion the original French word (schéma) doesn't help, by itself, to understand the motivations behind such a choice of nomenclature... - -REPLY [10 votes]: In this context, the introduction of the word schéma is due to Claude Chevalley. -According to Dieudonné (The historical development of algebraic geometry, Amer. Math. Monthly, vol 1979, 1972, p. 827-866), nobody had ever given "an intrinsic definition of an affine variety'' until the 1950s, ie "independent of any imbedding". -For general varieties, Weil's definition used local charts, as in differential geometry and -Chevalley had asked himself what was invariant in Weil's definition. -Cartier (Grothendieck et les motifs. Notes sur l’histoire et la -philosophie des mathématiques IV, 2000), himself quoted by Ralf Krömer (Tool and object: a history and philosophy of category theory, §4.1.1.1, footnote 319, page 164) explains that - -La réponse, inspirée des travaux antérieurs de Zariski, était simple et élégante: le schéma de la variété algébrique est la collection des anneaux locaux des sous-variétés, à l'intérieur du corps des fonctions rationnelles. (Krömer's translation: The answer, inspired by previous work by Zariski, was simple - and elegant: the scheme of the algebraic variety is the collection of local - rings of the subvarieties inside the field of the rational functions.) - -Soon after, Grothendieck's schémas pushed this idea even further.<|endoftext|> -TITLE: Metric spheres in CAT(0) manifolds -QUESTION [9 upvotes]: Let $X$ be a topological manifold of dimension $n$, equipped with a compatible CAT(0) metric. -Are sufficiently small metric spheres in $X$ homeomorphic to metric spheres in Euclidean space $\mathbb{E}^n$? -[In "Ideal boundary of CAT(0) spaces" (1998) by Myung-Jin Jeon, this was unclear to the author; see the bottom of Page 104. That paper examined geodesic completeness for CAT(0) manifolds.] -EDIT: It is well-known that sufficiently small metric balls in any CAT(0) space are contractible. Using the manifold property, I believe that my question reduces to a very basic one: -If $U$ is a contractible open subset of $\mathbb{R}^n$, then is $U$ homeomorphic to $\mathbb{R}^n$? - -REPLY [16 votes]: The answer is no. By the double suspension theorem of Cannon and Edwards, if $X^n$ is a homology $n$-sphere, then the double suspension $S^2X$ is a sphere. In particular, the cone $CSX$ on $SX$ will be a simply connected manifold homeomorphic to $R^{n+2}$, since it is obtained from the double suspension by deleting a point. If there is a simplicial complex structure on $SX$ which is flag (e.g. by barycentric subdivision), then the cone $CSX$ will have a CAT(0) metric. The point is that one may make each of the simplices of the triangulation of $SX$ into a right-angled $n+1$ spherical simplex, then take the metric cone on this to get a quadrant of $R^{n+2}$, and glue these quadrants together in the manner prescribed by the simplicial complex. The metric ball about the cone point of $CSX$ will not be homeomorphic to a ball, since its boundary will be $SX$ which is not a manifold. Check out the papers of Davis-Januskiewicz for constructions of CAT(0) complexes.<|endoftext|> -TITLE: A 'generalized Four Squares Theorem'? -QUESTION [5 upvotes]: The $4$-dimensional lattice $\mathbb{Z}^{4}$ has vectors of length $\sqrt{n}$ for any positive integer $n$ by the Four Squares Theorem, but this need not be true for higher-dimensional integral, unimodular lattices for two reasons: -(i) Some small positive integers could be skipped as squared lengths of lattice vectors. For example, the odd Leech lattice has no $v$ with $v \cdot v = 1$ or $2$. -(ii) The lattice may be even. -Therefore, the way to word the question to recognize these possibilities is: -Let $\Lambda$ be an integral unimodular lattice of dimension $d$, where $d \geq 4$. -(i) If $\Lambda$ is odd, then is it true that every sufficiently large positive integer arises as the squared length of a vector in $\Lambda$? -(ii) If $\Lambda$ is even, then is it true that every sufficiently large even positive integer arises as the squared length of a vector in $\Lambda$? - -REPLY [6 votes]: Yes, both are true. For example, see Theorem 1.6 of Chapter 11 of Cassels's "Rational Quadratic Forms", which says that if $Q$ is a positive definite integral quadratic form, then there is an integer $N$ depending on $Q$ such that if $a > N$ and $a$ is represented primitively by $f$ over all $\mathbb{Z}_p$ then $a$ is represented by $Q$. The local primitive representability is easy to show using the fact that $f$ is unimodular, and the classification of forms over $\mathbb{Z}_p$ by invariants. -For instance, if $p$ is odd and $p \nmid a$, then $f$ is equivalent over $\mathbb{Z}_p$ to $(a, a \det(f), 1, 1, \dots, 1)$, which obviously represents $a$ primitively. If $p | a$ you could use $((a-1), (a-1) \det(f), 1, 1, \dots, 1)$ which represents $a$ primitively. I won't do the analysis for $p = 2$, but see section 4 of chapter $8$ of Cassels.<|endoftext|> -TITLE: Closed but not rational points of a real cubic -QUESTION [5 upvotes]: In Mumford's Red Book of Varieties and Schemes, page 102, he gave the example of the closed but not rational points (that is to say points having residue field the complex field and not the real field) of the cubic $y^2=x^3−x$ on the real field : I have some difficulty to recover by elementary methods the figure he traced. -Especially, he seems to imply that these closed points formed the region $y^2>x^3−x$ in the real plane (which looks like the cylinder he pictured in the projective plane). Can somebody give me a simple explanation ? (I suppose the maximal ideals of the spectrum of the algebra defined by the cubic have to be parametrized the right way ?) - -REPLY [12 votes]: I don't know what Mumford had in mind, but here (in some detail) is a down-to-earth way to topologically identify this space with a cylinder. -Let $C$ be our projective cubic curve with affine equation $y^2=x^3-x$. We're considering complex conjugate pairs of points on $C$, that is, pairs $\{(x,y), (\bar x,\bar y)\}$ of solutions of $y^2=x^3-x$. While those points are not real, the line joining them is real: there are real numbers $a,b,c$, not all zero, such that the line $l_{a,b,c}: aX + bY + c = 0$ passes through $(x,y)$ and $(\bar x, \bar y)$, and the coefficient vector $(a,b,c)$ is determined uniquely up to multiplication by a nonzero scalar. That is, $(a:b:c)$ is a well-defined point in the "dual projective plane" ${\bf P}^*$ of lines on the projective plane with coordinates $(x:y:1)$ where $C$ lives. Now these points $(x,y)$ and $(\bar x, \bar y)$ are on $l_{a,b,c} \cap C$, which contains three points in all, so there is a third point $(x_0,y_0) =: p_0$, necessarily real. Conversely, any line $l$ meets $C$ in at least one real point, and if there is only one such point (and $l$ is not tangent to $C$ at that point) then the other two points of $l \cap C$ constitute a closed-but-not-rational point of $C$. -That is, - -the space we're looking for is homeomorphic with the subset, call it $S$, of ${\bf P}^*$ consisting of lines whose real intersection with $C$, with multiplicity, has size $1$ - -as opposed to size $3$. -One way to describe $S$ is to start from $p_0 = (x_0,y_0)$. It is geometrically clear that this point must be on the infinite component of $C$, call it $C_0$: the other component $C_1$ is a closed curve in the affine plane ${\bf R}^2$, so any line meets it with even total multiplicity. Given $p_0$, the lines through $p_0$ constitute a real projective line, which is topologically a circle and the lines through $p_0$ that meet $C_0$ in two other points $q,q'$ constitute the union of two closed arcs, one for lines where $q,q' \in C_0$ and the other for lines where $q,q' \in C_1$. [The boundary points correspond to the four points $q$ whose tangent passes through $p$, which are the solutions of $2q=-p$ in the group law of $C$.] So the lines through $p_0$ in $S$ constitute two open intervals. Now the subtlety is that when $p_0$ goes around the closed curve $C_0$, these two intervals switch as each of the boundary points makes a complete cycle around $C_0$ or $C_1$, so we must traverse $C_0$ twice to traverse our cylinder once. In effect we're getting a Möbius band cut down the middle, which is indeed a cylinder (with a "full twist", true, but that is an artifact of the embedding in three-dimensional space that we use to visualize $S$). -For a different kind of explicit picture of $S$, note that a real cubic polynomial has one real root (with multiplicity) if and only if its discriminant is negative. So we can describe $S$ by eliminating of the variables from $aX+bY+c=0$, substituting into $Y^2=X^3-X$, computing the discriminant $\Delta$ of the resulting cubic, and plotting the region $\Delta < 0$. For example, in the affine piece $b \neq 0$ of ${\bf P}^*$, we may set $b=1$, compute $Y = -(aX+c)$, find that -$$ -\Delta = -27c^4 - 4(ac)^3 + 30(ac)^2 + 4 a^5 c + 24 ac + a^4 + 4 -$$ -(I didn't promise it would be pretty), and ask www.wolframalpha.com -plot(-27*c^4-4*a^3*c^3+30*a^2*c^2+4*a^5*c+24*a*c+a^4+4 < 0) - -to get a picture with two blue components that join up at infinity to form a topological cylinder: - (source) -[The two visible cusps come from the inflection points where $p=q=q'$, which are real 3-torsion points on $C$; there's a third such singularity at infinity. This means that of the two boundary components of $S$ (it looks like four but they pair up at infinity) the one containing the cusps is $C_0$, and the other is $C_1$.] Try also -plot(-27*c^4-4*a^3*c^3-30*a^2*c^2+(24*a-4*a^5)*c+a^4-4 < 0) - -for the picture arising from the curve $y^2=x^3+x$ with only one real component; this time it is a Möbius band embedded in ${\bf P}^*$ so that there boundary and the complement have only one component each: - (source) -To connect this with the usual (but less elementary) picture of an elliptic curve over ${\bf C}$ as a complex torus: as Lubin noted, the complex locus of $C$ is isomorphic as a Riemann surface with ${\bf C} / L$ where $L$ is the Gaussian lattice ${\bf Z} + {\bf Z} i$; this is consistent with complex conjugation, and the real locus consists of the cosets mod $L$ of the complex numbers of integral or half-integral imaginary part, constituting the components $C_0$ and $C_1$ respectively. We're looking to identify the conjugate pairs $\{(z,\bar z)\} \bmod L$ with a cylinder; in terms of the group law the real point $p_0$ associated above to $\{(z,\bar z)\}$ is $-2 \phantom. {\rm Re}(z)$, which as before can only be on $C_0$ and goes around $C_0$ twice (and in the opposite direction, as it happens) as $z$ goes around the cylinder once.<|endoftext|> -TITLE: Intersection of non transverse submanifolds -QUESTION [5 upvotes]: Hi! -Probably this is an easy question, but i can't see the answer. -Let $X$ be a a smooth real manifold with $\dim(X)=d$ and $M,N\subset X$ two smooth submanifolds -with $\dim(M)=m$ and $\dim(N)=n$. The submanifolds $M,N$ intersect but not transversely. -What can i say about connected components of $M\cap N$? More precisely, is it possible to find three manifolds $X,M,N$ as above such that a connected component of $M\cap N$ is not a manifold? Or a connected component that is not smooth? (In all the examples i thought, connected components of $M\cap N$ were smooth) -Thank you in advance. - -REPLY [7 votes]: Let $M$ be any manifold, and let $Z$ be a closed subset of $M$. Suppose there exists a smooth function $f:M\to\mathbb{R}$ with $f^{-1}\{0\}=Z$. We can then take $X=M\times\mathbb{R}$ and identify $M$ with $M\times\{0\}$ and put $N=\{(m,f(m)):m\in M\}$. Then $M$ and $N$ are embedded submanifolds of $X$ with $M\cap N=Z$. -Moreover, I think it is true that such a function $f$ exists for every closed subset $Z$, no matter how wild or fractal. I don't remember the argument in detail, but if I recall correctly it is not too hard. One issue is to patch together things done locally using a partition of unity, and another is to express $f$ as a countable sum of nonnegative smooth functions $f_n$ which need to be rescaled aggressively to force the higher derivatives of the sum to converge. - -REPLY [6 votes]: You could try the real quadric surface $x^2+y^2-z^2=1$ and the hyperplane $x=1$. They intersect along a union of two (intersecting) lines ($z=\pm y$, $x=1$).<|endoftext|> -TITLE: An exponential polynomial with at least one bounded positivity component -QUESTION [10 upvotes]: In a forthcoming paper on nodal domains of Gaussian random functions, we (I and Misha Sodin) have a statement that is, roughly speaking, the following: if bounded nodal domains are possible at all, they have certain positive density. This sounds great until one asks a naive question "When are they possible at all?". Stripped of all irrelevant high tech terminology, this boils down to the following: -Let $K$ be an origin symmetric compact set in $\mathbb R^n$ having no isolated points and not contained in a hyperplane. Can one always construct a real-valued trigonometric polynomial $f(x)=\sum_{y\in K}\;c_y\; e^{i\,y\cdot x}$ (where all but finitely many $c_y$ vanish and $c_{-y}=\bar c_y$) such that the set $f\ge 0$ has at least one bounded connected component? If not, how to describe $K$ for which it is possible? - -REPLY [3 votes]: Let's denote $F_K$ the family of real-valued trigonometric polynomials corresponding to $K$, and assume that $K$ has a point in the interior of its convex envelope. Then, there is a function $f$ in $F_K$ for which $\{f\ge 0\}$ has a bounded component. -To show this we can freely apply a linear transformation to $K$, for $F_{LK}=\{f\circ L^T\, :\, f\in F_K \}$. In particular we can assume that $K$ includes the standard basis $ \{ e_1,\dots, e_n\}$, and there is in $K$ one more point $y $ with $y_j\ge0$ and $\|y\|_1:=\sum_{j=1}^n y_j <1$. Consider a trigonometric polynomial -$$f(x)= \sum_{j=1}^n \lambda_j \cos(x_j) -\cos(y\cdot x)\, .$$ -It belongs to $F_K$ and has a second-order expansion at $0$ -$$f(x)= \sum_{j=1}^n\lambda_j - 1 - \frac{1}{2}\sum_{j=1}^n \lambda_j x_j^2 + \frac{1}{2}(y\cdot x)^2+o(\|x\|^2)$$ -$$\le \Big(\sum_{j=1}^n\lambda_j - 1\Big) -\frac{1}{2}\sum_{j=1}^n (\lambda_j -\|y\|_1y_j)x_j^2 +o(\|x\|^2) $$ -because by Cauchy-Schwarz, $(y\cdot x)^2 = \big( \sum_{j=1}^n y_j ^{1/2} y_j ^{1/2} x_j\big)^2\le \|y\|_1\sum_{j=1}^n y_j x_j^2 $. -We can now take e.g. $\lambda_j= \|y\|_1y_j +\frac{1}{ n}(1-\|y\|_1^2+\epsilon)$ with $\epsilon>0$ so that $f(0)=\epsilon$ and $f(x)\le\epsilon-\frac{1}{2n}(1-\|y\|_1^2)\|x\|^2+o(\|x\|^2)$ (unif. on $\epsilon$). So for $\epsilon$ small enough $f(x)<0$ on the boundary of a ball around $0$, meaning that the connected component of $0$ in $\{f\ge0\}$ is contained in the ball.<|endoftext|> -TITLE: Sequences of Squares with all square differences -QUESTION [12 upvotes]: Background -The following question was first asked by Alex Rice, who was thinking about small subsets $A\subset [1,\ldots , N]$ with lots of square differences. Certainly for any set $A$ the maximum number of square differences is going to be $\binom{|A|}{2}$. From the point of view of someone working in additive combinatorics, an infinite set of positive integers can't get much less substantial than the squares, and so it's natural to wonder if there are arbitrarily large sets $A$ inside the squares, all of whose differences are squares [edit: I apparently misunderstood the original motivation, see Alex's answer/comment below]. This question was asked of a few others, including Adrian Brunyate, Jacob Hicks and Nathan Walters before it was asked of me by Adrian in this form: - -Definition: We say that a sequence $(a_1, \ldots, a_n) \in \mathbf{Z}^n_{\ge 1}$ is a Super-$n$ if for all $1 \le i \le n$, $a_i$ is an integer square and for all $1 \le i < j \le n$, $a_j - a_i > 0$ is also an integer square. - -Clearly a Super-2 defines a Pythagorean triple. -Perhaps less clearly, a Super-3 defines an Euler Brick, and is strongly related the the question of whether there is a perfect rational cuboid. - -Question 1: For which positive integers $n$ does there exist a Super-$n$ ? - -If the answer is yes to the above question, we may also ask the following: - -Question 2: For which positive integers $n$ do there exist infinitely many Super-$n$'s? - -One may note that the following problems are related to some problems already asked on MO about rational polytopes and sequences of squares -How many sequences of rational squares are there, all of whose differences are also rational squares? -Totally rational polytopes -What seems to be known already -It has been known for millenia that there are infinitely many Pythagorean Triples. -Euler discovered in 1772 that there are infinitely many Super-$3$'s, and in fact he gave a parametrized family of them. -None of us have been able to find a Super 4 (although I haven't been searching myself). -The connection to algebraic geometry - -Definition: The Super-$n$-variety is the intersection of the following $\binom{n}{2}$ quadratic polynomials in projective space over $\mathbf{Q}$. $$d_1^2 = c_2^2 - c_1^2$$ $$\vdots$$ $$d_{\binom{n}{2}}^2 = c_{n}^2 - c_{n-1}^2$$ - -Clearly the Super-2 variety is a copy of $\mathbb{P}^1_{\mathbf{Q}}$. -In Section 8 of the link given above for Euler's family of "Euler Bricks" we see that the Super-3 variety is birational to a singular K3 surface of Mordell-Weil rank 2. In this setting, one could say that Euler found a rational curve on this variety. It is also noted in the article that Narumiya and Shiga found a different rational curve on this variety. - -Question 2': Could there be rational curves on the Super-$n$ variety for all $n$? - -But perhaps (probably) this is way too much to ask. More generally, I'd like to know: - -Question 3: Is there any interesting geometry to the Super-$n$ variety for $n\ge 4$? - -In general this seems like an interesting problem, and one that people may have studied before, but perhaps in some guise that I'm not familiar with, so any input is appreciated. - -REPLY [3 votes]: This is more of a comment. -If I understand correctly, a perfect cuboid will give a Super-4 according to On Perfect Cuboids - -Are there four squares all pairs of which have square differences? For a perfect cuboid we - could take the squares of $y_3 z$, $y_2 y_3$ , $x_1 z$ and $x_1 y_3$. - -While wasting my time with perfect cuboids, I found 2 surfaces on which they are nontrivial rational points. The first might be for all perfect cuboids, the second is not for all. The surfaces are: -$$ x^{4} y^{2} z^{4} + x^{2} y^{4} z^{4} - 2 x^{4} y^{2} z^{2} - 2 x^{2} y^{4} z^{2} - 4 x^{2} y^{2} z^{4} + x^{4} y^{2} + x^{2} y^{4} - 8 x^{2} y^{2} z^{2} + x^{2} z^{4} + y^{2} z^{4} - 4 x^{2} y^{2} - 2 x^{2} z^{2} - 2 y^{2} z^{2} + x^{2} + y^{2} = 0$$ -and -$$ x^{4} y^{3} z^{3} - x^{4} y^{2} z^{4} + x^{2} y^{4} z^{4} + 2 x^{4} y^{3} z + 2 x^{4} y^{2} z^{2} - 2 x^{2} y^{4} z^{2} - 2 x^{4} y z^{3} + 4 x^{2} y^{3} z^{3} - x^{4} y^{2} + x^{2} y^{4} - 2 x^{4} y z + 4 x^{2} y^{3} z - 4 x^{2} y z^{3} + 2 y^{3} z^{3} + x^{2} z^{4} - y^{2} z^{4} - 4 x^{2} y z + 2 y^{3} z - 2 x^{2} z^{2} + 2 y^{2} z^{2} - 2 y z^{3} + x^{2} - y^{2} - 2 y z =0$$ -In machine readable form: -x^4*y^2*z^4 + x^2*y^4*z^4 - 2*x^4*y^2*z^2 - 2*x^2*y^4*z^2 - 4*x^2*y^2*z^4 + x^4*y^2 + x^2*y^4 - 8*x^2*y^2*z^2 + x^2*z^4 + y^2*z^4 - 4*x^2*y^2 - 2*x^2*z^2 - 2*y^2*z^2 + x^2 + y^2 = 0 - -and -2*x^4*y^3*z^3 - x^4*y^2*z^4 + x^2*y^4*z^4 + 2*x^4*y^3*z + 2*x^4*y^2*z^2 - 2*x^2*y^4*z^2 - 2*x^4*y*z^3 + 4*x^2*y^3*z^3 - x^4*y^2 + x^2*y^4 - 2*x^4*y*z + 4*x^2*y^3*z - 4*x^2*y*z^3 + 2*y^3*z^3 + x^2*z^4 - y^2*z^4 - 4*x^2*y*z + 2*y^3*z - 2*x^2*z^2 + 2*y^2*z^2 - 2*y*z^3 + x^2 - y^2 - 2*y*z = 0<|endoftext|> -TITLE: Zeroes of the random Fibonacci sequence -QUESTION [23 upvotes]: Let $X_n$ be the "random Fibonacci sequence," defined as follows: -$X_0 = 0, X_1 = 1$; -$X_n = \pm X_{n-1} \pm X_{n-2}$, where the signs are chosen by independent 50/50 coinflips. -It is known that $|X_n|$ almost surely grows exponentially by a (much more general) theorem of Furstenberg and Kesten about random matrix products: the base of the exponent was determined explicitly by Viswanath to be $1.13\ldots$ -I am not too proud to say that I learned all this from Wikipedia: -http://en.wikipedia.org/wiki/Random_Fibonacci_sequence -What I did not learn from Wikipedia, or any of the references I gathered therefrom, was: -Question: what, if anything, do we know about the probability that $X_n = 0$, as a function of $n$? - -REPLY [12 votes]: First of all, it is somewhat misleading to attribute the claim concerning exponential growth to Furstenberg and Kesten. What they proved (in 1960) is that for partial products of a random stationary sequence of matrices there exists almost surely an exponential rate of growth. However, it does not exclude possibility that this growth rate is zero. Its positivity for i.i.d. sequences was proved (under reasonable irreducubility conditions) by Furstenberg in 1963. -Now to your question. The reduction to matrix products consists in observing that -$$ -\left( X_{n+1} \atop X_n \right) = \left( 1 \;\pm 1 \atop1\;\;\;\;\;\ 0 \right) \left( X_n \atop X_{n-1} \right) \;, -$$ -whence -$$ -\left( X_{n+1} \atop X_n \right) = B_n \left( 1 \atop 0 \right) \;, -$$ -where -$$ -B_n = A_n A_{n-1} \dots A_1 -$$ -is the product of i.i.d. matrices $A_i$ which are either $M=\left( 1 \;1 \atop1\; 0 \right)$ or -$M'=\left( 1 \;-1 \atop1\;\;\; 0 \right)$ with equal probabilities 1/2 (as explained by JSE, one can omit the first $\pm$ in the recurrence relation). Therefore, $X_n=0$ iff $B_n$ is an upper triangular matrix with $\pm 1$ on the diagonal. The Schreier graph of the quotient of $GL(2,\mathbb Z)$ by the group of such triangular matrices is non-amenable, whence one can deduce exponential decay of the return probabilities. -PS The answer to Mark's question is NO. The reason is that there exist amenable groups (e.g., the higher dimensional lamplighter groups), for which the linear rate of escape of the simple random walk is strictly positive (which is analogous to positivity of the exponent for matrix products), but nonetheless the probability of returning to zero decays subexponentially (because of amenability).<|endoftext|> -TITLE: Surface groups and non separating loops -QUESTION [8 upvotes]: QUESTION: Let $g \geq 4$, $S(g)$ be the fundamental group of the genus $g$ surface, and $G$ be finitely generated (the number of generators $\leq 3$) group with abelianization of rank less than equal $2$. Assume that there exist a surjection $\phi: S(g) \rightarrow G$. Is it true that the kernel of $\phi$ contains at least one non separating loop of the surface? If it is any helpful, you can assume $G$ is a perfect group. - -REPLY [6 votes]: The Simple Loop Conjecture is as follows. - -Every non-injective map from a surface group to a 3-manifold group kills a simple closed curve. - -As there are all sorts of 3-manifolds with abelianisation of rank two, I think the answer to your question is unknown. -UPDATE: Sorry, I wrote the above too hastily. I should have said 'I think that the kernel is not known to contain such a loop.' On the other hand, there may well be examples of such maps with no simple loops in the kernel. You could try looking at Louder's recent preprint on the Simple loop conjecture for limit groups, for instance.<|endoftext|> -TITLE: Sheaves with constant fibre dimension one -QUESTION [7 upvotes]: Hi, -I have a simple question about coherent sheaves and line bundles: -if I have a coherent sheaf $F$ on a good scheme $X$ and I know that $F_x \otimes k(x) = k(x)$ for alle points $x$ on $X$ (where $k(x)$ is the residue field of $x$ and the tensor product goes over the local ring $\mathcal O_{X,x}$), can I then say, that $F$ is already a line bundle? -My strategy would be: -the stalk of $F$ is finitely generated over the local ring; by Nakayama you can choose a generating element $s_x$ of $F_x$ for every $x$, so you have $F_x \simeq \mathcal O_{X,x}$ and now use the statement that if a coherent sheaf is free in a point, then it is in a neighborhood. -Is this the right way to see it? - -REPLY [7 votes]: No. You need that the scheme is reduced. -It is certainly true that if $F_x$ is a free $\mathcal{O}_{X,x}$-module of rank $n$, then there exist an open neighborhood $U$ of $x$ such that $F \vert_U$ is a free $\mathcal{O}_U$- module of rank $n$. -But from $\dim_{k(x)} F_x \otimes_{\mathcal{O}_{X,x}} k(x) = 1$ you can deduce that $F_x$ is a cyclic module and not that $F_x$ is free of rank $1$. -However on a reduced scheme the statement is true: exercise II.5.8 of Hartshorne.<|endoftext|> -TITLE: The Matrix-Tree Theorem without the matrix -QUESTION [25 upvotes]: I'm teaching an introductory graph theory course in the Fall, which I'm excited about because it gives me the chance to improve my understanding of graphs (my work is in topology). A highlight for me will be to teach the Matrix-Tree Theorem, which I think is the only place that linear algebra is used in the course. -Let κ(G) denote the number of spanning trees of G (the complexity of G), and let L(G) denote the Laplacian matrix of G. Finally, for a vertex v of G, let L(v|v) denote the Laplacian matrix with the row and column corresponding to v deleted. - -Matrix-Tree Theorem: κ(G)= det L(v|v). - -It seems a shame for Linear Algebra to be a prerequisite for my course. Anyway, I don't expect most of my students to be great Linear Algebra experts. Because my students might not remember things like Cauchy-Binet, but mainly so that I myself feel that I can really understand what I'm teaching, I wonder how the Matrix-Tree Theorem could be proven without ever mentioning matrices. On a planet with strong combinatorics and where linear algebra had not been discovered, how would they prove the Matrix-Tree Theorem? -The RHS of the Matrix-Tree Theorem makes sense without ever mentioning matrices, via the Lindström-Gessel-Viennot(-Karlin-MacGregor) Lemma. Construct a graph H, with a source and a sink corresponding to each vertex of G, so that the signed sum of edge weights gives the entries of the Lagrangian matrix for G (surely there's a clever standard way to do this!) and define the determinant of H to be the signed sum of all n-tuples of non-intersecting paths from the sources to the sinks. This interpretation of the determinant seems a nice topic to teach. Maybe there's even an easier way. -Cauchy-Binet becomes an elementary property of sets of non-intersecting paths in H, but I can't see how to free the rest of the proof of the Matrix-Tree Theorem from linear algebra. - -Question: Is there a proof of the Matrix-Tree Theorem which makes no mention of matrices? Is it simple enough that one could teach it in an introductory graph theory course? - - -Note 1: If the purpose of the Matrix-Tree Theorem is to efficiently calculate the complexity of a graph, then dodging linear algebra is surely counterproductive, because a determinant can efficiently be calculated in polynomial time. But, quite beside from my interest in "correctly-tooled" proofs and my teaching goals, I have vague dreams which are almost certainly nonsense about better understanding the Alexander polynomial as a quantum invariant along the lines of this question which might conceivably factor through the Matrix-Tree Theorem (or rather some version of Kirchhoff's formula), and I think I clearly want to stay in the diagrammatic world. -Note 2: This excellent post by Qiaochu Yuan suggests that the answer to my question is in Aigner. I'm travelling right now, and the relevant section isn't on Google Books, so I don't have access to Aigner for the next few weeks. If memory serves, Aigner still uses linear algebra, but maybe I'm wrong... anyway, if the answer turns out to be "look in Aigner" then please downvote this question, or if you could summarize how it's done, I would be happier. The answer will surely turn out to be "look in [somewhere]" anyway, because surely this is easy and well-known (just I don't know it)... - -REPLY [5 votes]: The algebraic proof goes like this. Fix a number of vertices $n$, and consider the "formal Laplacian" $n \times n$ matrix defined by -$$ -\begin{align*} -L(i,i) &= \sum_{j \neq i} X(i,j) \\ -L(i,j) &= - X(i,j) \qquad \text {for } i\neq j. -\end{align*} -$$ -Here the $X(i,j)$ are commuting indeterminates. For every spanning tree $T$ of the complete graph $K_n$ we can associate a degree $n-1$ monomial $X(T)$ as follows. Root the tree at the vertex $n$. For every non-root vertex $i$ with parent $j$, we include the variable $X(i,j)$ in the monomial. In total $X(T)$ is a product of $n-1$ indeterminates, one for every edge in the tree. -Let $M$ be the $(n,n)$ minor of $L$. We will show that $\det M = \sum_T X(T)$. You can derive the matrix-tree theorem from this statement by substituting the actual graph for the indeterminates $X(i,j)$. If you wish, you can run the entire proof after doing the substitution, and then you don't have to talk about indeterminates. -Recall that -$$\det M = \sum_{\pi \in S_{n-1}} (-1)^\pi \prod_{i=1}^{n-1} M(i,\pi(i)).$$ -We can write every permutation $\pi$ as a product of nontrivial cycles $\pi_1,\ldots,\pi_{\ell(\pi)}$ (where $\ell (\pi) $ is the number of nontrivial cycles) and of fixed points $F(\pi)$, and then we get -$$ -\begin{align*} -\det M &= \sum_\pi (-1)^\pi \prod_{t=1}^{\ell(\pi)} \prod_{i \in \pi_t} (-X(i,\pi(i))) \prod_{i \in F(\pi)} \left(\sum_{j \neq i} X(i,j)\right) \\ &= -\sum_\pi (-1)^{\ell(\pi)} \prod_{t=1}^{\ell(\pi)} \prod_{i \in \pi_t} X(i,\pi(i)) \prod_{i \in F(\pi)} \left(\sum_{j \neq i} X(i,j)\right), -\end{align*} -$$ -using the formula $(-1)^\pi = \prod_{t=1}^{\ell(\pi)} (-1)^{|\pi_t|+1}$. We can rewrite this as -$$ \det M = \sum_\pi D(\pi), $$ -where $D(\pi)$ is the expression appearing above. -If we open up all the sums, we find that $\det M$ is a (weighted) sum of monomials of the form $X(\alpha) = \prod_{i=1}^{n-1} X(i,\alpha(i))$. For each $\alpha$, we can look at the directed graph in which $i$ points at $\alpha(i)$. It's not hard to see that each such graph consists of a (possibly empty) collection of cycles $\alpha_1,\ldots,\alpha_{\ell(\alpha)}$ and a tree $T(\alpha)$ rooted at $n$. We will show that the coefficient of $X(\alpha)$ equals $1$ if $\ell(\alpha) = 0$, and vanishes otherwise. This will complete the proof. -Consider first the case that $\ell(\alpha) = 0$. In this case the only permutation $\pi$ for which $X(\alpha)$ appears in $D(\pi)$ is the identity permutation (since every $X(\beta)$ appearing in any other $D(\pi)$ contains some cycle), and so the coefficient of $X(\alpha)$ in $\det M$ is indeed $1$. -Next, consider the case in which $\ell = \ell(\alpha) \neq 0$. In this case, for every subset $S$ of the cycles of the directed graph corresponding to $\alpha$, $X(\alpha)$ appears in $D(\pi)$ for $\pi = \pi(S)$ which consists of the cycles in $S$, and it only appears in these $D(\pi)$. For every $S$, the coefficient of $X(\alpha)$ in $D(\pi(S))$ is $(-1)^{|S|}$. Therefore the total coefficient of $X(\alpha)$ in $\det M$ is -$$ -\sum_{S \subseteq \{\alpha_1,\ldots,\alpha_\ell\}} (-1)^{|S|} = -\sum_{S_1 \subseteq \{\alpha_1\}} \cdots \sum_{S_\ell \subseteq \{\alpha_\ell\}} (-1)^{|S_1| + \cdots + |S_\ell|} = \\ -\sum_{S_1 \subseteq \{\alpha_1\}} (-1)^{|S_1|} \cdots \sum_{S_\ell \subseteq \{\alpha_\ell\}} (-1)^{|S_\ell|} = (1-1)^\ell = 0. -$$ -(This is just inclusion-exclusion. The previous case is the case $\ell=0$.) -This completes the proof. -This proof is taken from lecture notes by Jacques Verstraete. See also Mark Muldoon's lecture notes, which presents this argument as a version of inclusion-exclusion.<|endoftext|> -TITLE: Analytic implicit function theorem -QUESTION [13 upvotes]: I'm looking for a proof of the analytic implicit function theorem (IFT). The only related proof I could find was the holomorphic inverse function theorem (by Henri Cartan). On Wikipedia, the analytic IFT is mentioned casually in the general article "Implicit function theorem", saying that "Similarly, if f is analytic inside U×V, then the same holds true for the explicit function g inside U. This generalization is called the analytic implicit function theorem." Mmmh, that's fast... -A sketch of the proof may be the following : - -use analytic continuation to transform f into a holomorphic function -use the holomorphic inverse function theorem (Cartan) to prove a holomorphic IFT -restriction : g is holomorphic on $\mathbb{C}$, therefore analytic on $\mathbb{R}$. - -But it seems weird and I don't think it would work (I have no idea whether a so-called holomorphic IFT exists or not). What would be an efficient proof of the theorem ? Thanks a lot by advance. - -REPLY [5 votes]: The implicit function theorem is deduced from the inverse function theorem in most standard texts, such as Spivak's "Calculus on Manifolds", and Guillemin and Pollack's "Differential Topology". Basically you just add coordinate functions until the hypotheses of the inverse function theorem hold.<|endoftext|> -TITLE: Etale endomorphisms of abelian varieties in positive characteristic -QUESTION [15 upvotes]: Let $K$ be the function field of a smooth curve over the algebraic closure $k$ of the finite field ${\bf F}_p$ (where $p>0$ is a prime number). -My question is : does there exist an abelian variety $A$ over $K$, with the following properties : -(a) the $K|k$-image of $A$ is trivial; -(b) there exists an étale $K$-endomorphism of $A$, whose degree is a power of $p$ ? -The condition on the $K|k$-image can be rephrased as : there are no non-zero $K$-homomorphisms -$A\to C_K$, where $C$ is an abelian variety over $k$. -For examples of abelian varieties satisfying condition (b) only, look at abelian varieties $C_K$, where $C$ -is an ordinary abelian variety over ${\bf F}_p$. The abelian variety $C$ is endowed with the étale endomorphism -given by the Verschiebung morphism. -Also, notice that if there is an abelian variety over $K$ satisfying (a) and (b), then the dimension of $A$ is -larger than one (ie it is not an elliptic curve). Indeed, if an elliptic curve $E$ satisfies the above conditions, then -$E$ has an endomorphism, which is not a multiplication by a scalar and thus it has complex multiplications; this implies that it is isogenous to an elliptic curve defined over $k$, by a theorem of Grothendieck (or by more direct arguments). -Finally, I would like to point out that -if $A$ is an ordinary abelian variety over $K$, which has maximal Kodaira-Spencer rank, then $A[p]$$(K^{\rm sep})=0$ by a theorem -of J-F Voloch (see p. 1093 in "Diophantine Approximation on Abelian Varieties in Characteristic $p$", -Amer. J. Math., Vol. 117, No. 4., pp. 1089-1095); this shows that such an abelian variety -cannot have an endomorphism as in (b). - -REPLY [6 votes]: If $p-1$ is divisible by $24$ then there is an explicit example of an ordinary $7$-dimensional abelian variety $X$, whose endomorphism algebra is the imaginary quadratic field $Q(\sqrt{-3})$; in particular, $X$ is absolutely simple and its $K/k$-trace is trivial. Namely, $K$ is the field of rational functions $k(t)$ and $X$ is the jacobian of the $K$-curve $y^3=x^9-x-t$. See Example 4.3 of arXiv:math/0606422 [math.NT] [MR2289628 (2007j:11077)] for details. -If $p>2$ and $g>1$ is an odd integer then there exists a $g$-dimensional ordinary abelian variety $X$ over a suitable $K$, whose endomorphism algebra is an imaginary quadratic field; in particular, $X$ is absolutely simple and its $K/k$-trace is trivial. See Theorem 1.5(i,ii) of the same paper that is based on a construction of Oort and van der Put. (Actually, the condition $p>2$ could be dropped.)<|endoftext|> -TITLE: Embeddings of Sobolev-Orlicz spaces -QUESTION [8 upvotes]: The Birnbaum--Orlicz spaces generalize the Lebesgue spaces (see http://en.wikipedia.org/wiki/Birnbaum-Orlicz_space for a precise definition). The space $L_\Phi(\Omega)$ is defined for convex functions $\Phi:(0,\infty)\rightarrow(0,\infty)$ with $\Phi(0)=0$ and $\Phi(\infty)=\infty$. The norm in $L_\Phi$ is denoted $\|\cdot\|_\Phi$. -When $\Phi(t)=t^p$, then $L_\Phi=L^p$ and $\|\cdot\|_\Phi=\|\cdot\|_p$. -Define the Sobolev space $W^{1,\Phi}_0(\Omega)$ -to be the closure of ${\mathcal D}(\Omega)$ under the norm -$$f\mapsto\|\nabla f\|_\Phi.$$ -Let me recall some of the Sobolev embeddings, when $\Omega$ is bounded. -If $1 -TITLE: Positive definite function zoo -QUESTION [14 upvotes]: I've asked the following question on math.stackexchange but there has been no response so I'll ask it again here: -A positive definite function $\varphi: G \rightarrow \mathbb{C}$ on a group $G$ is a function that arises as a "diagonal" coefficient of a unitary representation of $G$. -For a definition and discussion of positive definite function see here. -I've often wished I had a collection of diverse examples of positive definite functions on groups, for the purpose of testing various conjectures. I hope the diverse experience of the participants of this forum can help me collect a list of such examples. -To clarify what I'd like to see: - -What is an example of a positive - definite function on a group $G$ that - is not easily seen to be a coefficient - of a unitary representation of $G$? - What are some positive definite - functions that arise in contexts - sufficiently removed from studying the - coefficients of unitary - representations? - -Also, the weirder the group $G$ the better. I'd like a collection of quirky beasts... - -REPLY [6 votes]: Perhaps you are already aware of this, but I thought I'd mention it for other interested google-enabled readers. - -Infinitely divisible distributions are one place where positive-definite functions come up (Lévy processes, Lévy-Khintchine formula, etc., are also relevant keywords) -Infinite divisibility in Free Probability is another related place.<|endoftext|> -TITLE: On the definition of regularity -QUESTION [13 upvotes]: In the literature on D-modules, there are many definitions of regularity of holonomic D-modules. -(1) Bernstein first defines regularity on a curve then says a holonomic D-module is regular if its restriction to any curve is regular -(2) Mebkhout defines the irregularity complexes of a complex of D-modules along an hypersurface. The complex is then regular if its irregularity complexes are 0 along any hypersurface. -(3) Kashiwara defines a D-module as regular if it admits a good filtration $F_*M$ such that $\operatorname{Ann}(Gr^F M)$ is a radical ideal of $Gr^F D_X = \pi_*O_{T^*X}$. -I think there are other definitions (in Deligne for example) -Where can I find proofs that all these definitions are equivalent? Thanks. - -REPLY [5 votes]: There are some comparison results in Chapter 5 of Bjork's `Analytic D-modules and Applications'. Also see Chapter 8. In particular, I think Thm. 8.7.3 combined with Thm. 5.6.5 (almost) gives (1) iff (3). Further, I think Prop. 5.6.22 gives the equivalence with (2). There are also results in there comparing Deligne's description. -I must admit though that I find Bjork quite notationally dense and am not particularly familiar with it, so I may be quite off with the references above. I am interested in this question also, so please comment/post if you find better references.<|endoftext|> -TITLE: When is a given matrix of two forms a curvature form? -QUESTION [14 upvotes]: Let's assume we are working over $\mathbb{R}^n$ (but feel free to change to domain to answer the question). I wish to know if the equation $F = dA + A \wedge A$ can be solved for a matrix of 1-forms $A$, given a (smooth) matrix of 2-forms $F$ which satisfies the condition $dF =B \wedge F - F \wedge B$ for some smooth matrix of 1-forms $B$ (i.e. the Bianchi identity is satisfied). Notice that this is true for line-bundles (in fact over any convex open set). - -REPLY [25 votes]: The answer is generally 'no'; for most $F$ that satisfy your condition, there will not exist an $A$ that satisfies $F = dA + A\wedge A$. -The easiest counterexample I know of is when $n=4$ and the matrix $F$ is $2$-by-$2$. To begin, note that you can reduce to the case when both $F$ and the $A$ you seek have trace zero, i.e., they take values in ${\frak{sl}}(2,\mathbb{R})$. (The reason is that the problem breaks into the part of $F$ that is a multiple of the identity matrix and the trace-free part. I'll leave the details to you.) -One can easily check that, for the generic ${\frak{sl}}(2,\mathbb{R})$-valued matrix $F$ of $2$-forms on $\mathbb{R}^4$, the kernel of the mapping $C\mapsto F\wedge C - C\wedge F$ from ${\frak{sl}}(2,\mathbb{R})$-valued matrices $C$ of $1$-forms on $\mathbb{R}^4$ to ${\frak{sl}}(2,\mathbb{R})$-valued matrices of $3$-forms on $\mathbb{R}^4$ is zero. By dimension count, it follows that this mapping is surjective as well. -Thus, for a candidate ${\frak{sl}}(2,\mathbb{R})$-valued $2$-form $F$ that satisfies this open genericity condition, the equation $dF = F\wedge B - B\wedge F$ is always solvable for $B$, and, moreover, the solution is unique. Thus, this $B$ is the only possible candidate for $A$. Heuristically, this makes it almost immediate that, for the generic such $F$, the $B$ that you find will not satisfy $F = dB + B\wedge B$. The reason is that ${\frak{sl}}(2,\mathbb{R})$-valued $1$-forms on $\mathbb{R}^4$ depend on only $3\times 4 = 12$ arbitrary functions of $4$ variables while the generic ${\frak{sl}}(2,\mathbb{R})$-valued $2$-form $F$ depends on $3\times 6 = 18$ arbitrary functions of $4$ variables. There is no chance that you could hit each such $F$ with an $A$. -To construct an explicit example, choose an ${\frak{sl}}(2,\mathbb{R})$-valued $1$-form $A$ such that $F = dA + A\wedge A$ satisfies the genericity condition. Then, of course, $2F$ will satisfy this genericity condition as well, and, since it satisfies $d(2F) = (2F)\wedge A - A \wedge (2F)$, it follows that the only possible $1$-form whose curvature could be $2F$ is $A$. However, the curvature of $A$ is $F\not=2F$. Thus, $2F$ satisfies your condition, but it is not a curvature form.<|endoftext|> -TITLE: Central extensions of group schemes -QUESTION [9 upvotes]: In the category of groups, it is elementary that all central extensions of a cyclic group are abelian. Is the same true, in the category of (finite?) group schemes over a field $k$, for central extensions of the group $\mu_n$ of $n$th roots of unity? - -REPLY [14 votes]: If we have a central extension of group schemes $1\rightarrow B \rightarrow C\rightarrow -A\rightarrow1$ with $A$ abelian, then we get a commutator mapping -$\Lambda^2A\rightarrow B$ (of sheaves as $\Lambda^2A$ in general is not a group -scheme) and the extension is abelian precisely when this map is zero. Hence for -an non-abelian extension to exist there must be a non-zero map -$\Lambda^2A\rightarrow B$. Let us now assume that $A=\mu_n$ and consider first -the case when $n=p$, the characteristic of the field $k$ (which we may assume is -algebraically closed). A non-zero map $\Lambda^2A\rightarrow B$ would give a -non-zero map $A\rightarrow\mathrm{Hom}(A,B)$, where the right hand side is the -sheaf of group homomorphisms. As the Frobenius map is zero on $\mu_p$ we may -replace $B$ by its Frobenius kernel so we may assume that $B$ is either $\mu_p$ -or the Cartier dual of $\alpha_{p^m}$. Now, as sheaves $\mathrm{Hom}(A,B)$ is -isomorphic to $\mathrm{Hom}(D(B),D(A))$, where $D(-)$ denotes the Cartier -dual. However $D(\mu_p)=\mathbb Z/p$ so when $A=\mu_p$ we get that -$\mathrm{Hom}(D(B),D(A))=\mathbb Z/p$ and there is only the zero map from -$A=\mu_p$ into it. In the other case $D(A)=\alpha_{p^m}$ and -$\mathrm{Hom}(\alpha_{p^m},\mathbb Z/p)$ is zero. If instead $n=p^k$, the -argument is the same. The case when $n=\ell^k$ is even simpler so in all cases -all possible commutator maps are zero and the extension is commutative. -(When $A=B=\mathbb G_a$ then there are candidates for commutator maps and in -fact $(a,b)(a',b')=(a+a',b+b'+a^pa')$ gives a non-commutative central extension -which I imagine is the fake Heisenberg group.) -Addendum: A general comment is that it is more convenient to work with sheaves (in the fppf topology say) as that means that we essentially can pretend that we work with set-theoretic groups. It is however also necessary if we want to see the commutator map as a map $\Lambda^2A\to B$ as the sheaf $\Lambda^2A$ (of $A$ considered as an abelian sheaf) is in general not reprsentable. The $\langle-,-\rangle\colon\Lambda^2A\to B$ view point is convenient as it allows us to do what one usually does when having a pairing: We get for instance a map $A\to\mathrm{Hom}(A,B)$ given by $a\mapsto (a'\mapsto \langle a,a'\rangle)$ just from the fact that $\langle-,-\rangle$ is biadditive. -I have implicitly assumed that $B$ is of finite type (as I claim that its Frobenius kernel is finite) even though it may not be necessary (a limit argument anyone?).<|endoftext|> -TITLE: Top Chern Class = Euler Characteristic -QUESTION [13 upvotes]: Let $X$ be a (quasi-)projective, nonsingular, complex variety. Denote by $\mathcal{T}_X$ its tangent sheaf and by $X^{\mathrm{an}}$ its analytification. I am looking for a proof for the equality -        $\displaystyle \int_X c_n(\mathcal{T}_X) = \chi(X^{\mathrm{an}})$, -i.e. the degree of the top chern class is equal to the topological Euler characteristic of $X$. There's Example 3.2.13 in Fulton's book on intersection theory which briefly mentions this, but it does not give a reference. Can someone help me out with one? Thanks in advance. - -REPLY [21 votes]: As an alternative to R. Budney's answer, one might also notice that the Gauss-Bonnet formula (the one you mention - mind that you must assume that $X$ is projective, otherwise the integral might not even make sense) is a consequence of the Hirzebruch-Riemann-Roch theorem. Indeed, the HRR theorem says -$$ -\chi(V)=\int_{X}{\rm Td}({\rm T}X){\rm ch}(V) -$$ -where $$\chi(V):=\sum_{l}{(-1)}^l{\rm rk}(H^l(X,V))$$ is the Euler characteristic of coherent sheaves. Now there is an universal identity of Chern classes -$$ -{\rm ch}(\sum_{r}(-1)^r\Omega_X^r){\rm Td}(\Omega^\vee_X)=c^{\rm top}(\Omega^\vee_X) -$$ -(called the Borel-Serre identity). Here $\Omega_X$ is the sheaf of differential of $X$ and thus $\Omega^\vee_X={\rm T}X$. Plugging the element $\sum_{r}(-1)^r\Omega_{X}^r$ into the HRR theorem, one gets -$$ -\sum_{k,l}(-1)^{l+k}{\rm rk}(H^k(X,\Omega^l))=\int_{X}c^{\rm top}(TX) -$$ -and by the Hodge decomposition theorem -$$ -\sum_{k,l}(-1)^{l+k}{\rm rk}(H^k(X,\Omega^l))=\sum_{r}{(-1)}^r{\rm rk}(H^r(X({\bf C}),{\bf C})) -$$ -where $H^r(X({\bf C}),{\bf C})$ is the $r$-th singular cohomology group. -The quantity $\sum_{r}{(-1)}^r{\rm rk}(H^r(X({\bf C}),{\bf C}))$ is the topological Euler characteristic, so this proves what you want. -The HRR theorem is proved in chap. 15 of Fulton's book (or in Hirzebruch's book "Topological methods...") and the Borel-Serre identity is Ex. 3.2.5, p. 57 of the same book.<|endoftext|> -TITLE: Uniform lattices in semisimple Lie groups -QUESTION [6 upvotes]: Let $\Gamma$ be a uniform lattice in a semisimple Lie group $G$. - -Must $\Gamma$ be virtually torsion-free? -If (1) is true, then does this work more generally if $G$ is reductive? - -I am motivated by a prima facie knowledge of Theorem B of Armand Borel's paper, "Compact Clifford--Klein forms of symmetric spaces" (1963). - -REPLY [9 votes]: If $G$ is linear (which would be the case, for instance, if it is centerless) then it is special case of the more general fact that any finitely generated subgroup of $GL_n(F)$ for a field $F$ of characteristic zero is virtually torsion-free. -So all you need to know is that the lattices are finitely generated. This, for cocompact lattices, is easy, and for instance follows from what is usually called Milnor-Schwarz lemma, which is a very general lemma about cocompact isometric actions of groups on spaces. You can find a version of it in Pierre de la Harpe's book. -For general lattices, in higher rank, this follows from property T and in rank 1 by some more geometric methods. -If not, this may fail. See this for instance: -http://people.uleth.ca/~dave.morris/talks/deligne-torsion.pdf<|endoftext|> -TITLE: Disconnecting sets -QUESTION [7 upvotes]: If E is a metric space, I call a subset C of E a cut if E-C is not connected and if C is minimal for this property (which is obviously equivalent to "for every p in C, E-C union p is connected". The empty set is a cut of any disconnected space, so the obvious conjecture is : every space of more than one point possesses a cut (this is almost obviously true for manifolds, for instance)... but this conjecture is in fact probably wrong, as it is not obvious at all to find cuts in convoluted spaces like, for instance, the common boundary of three open sets in the plane ; on the other hand, i was not able to prove it in this case either. Any hint on such a proof ? (by the way, Zorn's lemma dont apply here ; I wonder is there is any interesting example of a situation where the hypothesis of the lemma fail (ie, like here, the intersection of a decreasing family of sets having the property P does not necessarily have it), but the conclusion (the existence of a minimal such set) is still valid ? - -REPLY [6 votes]: Assume $E$ is what remains after the infinite number of the iterations above. -This is an example of indecomposable continuum, so it can not be presented as a union of any two proper subcontinua. -Assume there is a cut $C\subset E$. -Then each connected component of $E\backslash C$ has $C$ as the boundary. -In particular, $C$ is closed. -Let us present $E\backslash C$ as a union of two disjoint open sets $A$ and $B$. -Clearly both $\bar A =C\cup A$ and $\bar B=C\cup B$ are proper subcontinua of $E$, -a contradiction.<|endoftext|> -TITLE: When is an algebraic variety $\mathbb{Q}$-factorial? -QUESTION [12 upvotes]: A variety is $\mathbb{Q}$-factorial if every global Weil divisor is $\mathbb{Q}$-Cartier. How bad singularities are allowed so that the algebraic variety is still $\mathbb{Q}$-factorial? Is a singular curve $\mathbb{Q}$-factorial? For example is a nodal-cuspidal plane curve $\mathbb{Q}$-factorial? - -REPLY [9 votes]: A complete intersection $X$ in $\mathbb{P}^n$ is $\mathbb{Q}$ factorial if $\dim X_{sing}<\dim X-3$. -In general being $\mathbb{Q}$-factorial depends on the type of singularity you have, but also on the position of the singularities. -E.g., a degree $d$ threefold $X$ in $\mathbb{P}^4$ with only ordinary double points at $p_1,\dots,p_k$ is $\mathbb{Q}$-factorial if and only if the linear system of homogeneous polynomials of degree $2d-4$ polynomials through the singular points of $X$ has no defect (i.e., the codimension of this linear system is precisely $k$). -An example of a non-$\mathbb{Q}$-factorial threefold with only nodes is $f_1f_2+f_3f_4=0$, where $\deg(f_1)+\deg(f_2)=\deg(f_3)+\deg(f_4)$, and the $f_i$ are chosen sufficiently general.<|endoftext|> -TITLE: Exact sequences of bundles on Grassmannians -QUESTION [9 upvotes]: We're looking for a large set of exact sequences of vector bundles on Grassmannians. Here's the set up: -$V$ and $Q$ are complex vector spaces of dimensions $d$ and $r$ respectively $(d\geq r)$, and we're working on the Grassmannian $Gr(V,Q)$. For simplicity let's fix a trivialization of $det(V)$. -Now let $\alpha$ be a partition/Young diagram with at most $(r-1)$ rows and at most $(d-r)$ columns. Let $\beta$ be the Young diagram obtained from $\alpha$ by adding an extra row of length $(d-r)$ at the beginning. What we want is an exact sequence of vector bundles that goes -$$\mathbb{S}_\alpha(Q)\otimes det(Q)^{-1} \rightarrow\;\; ... \;\;\rightarrow \mathbb{S}_\beta(Q) $$ -($\mathbb{S}$ denotes a Schur functor). For $r=1$ there's only one choice for $\alpha$, and the Koszul complex is the required sequence. For $d-r=1$ we have the short exact sequences -$$\wedge^k Q \otimes det(Q)^{-1}\rightarrow \wedge^{k+1}V\ \rightarrow \wedge^{k+1} Q$$ -We can also solve $r=2$ using Eagon-Northcott complexes. These known cases suggest that the exact sequence should have $(d-r+2)$ terms. -Does anyone know a general construction? -Update: we have a precise conjecture for the terms in the sequence. Let $\beta_0$ be the partition obtained from $\alpha$ by deleting the first column. Now define $\beta_i$ recursively as the partition obtained from $\beta_{i-1}$ by adding boxes to the $i$th column until it agrees with the $i$th column of $\beta$. In particular $\beta_{d-r}=\beta$. Then the terms in middle of the exact sequence should be -$$...\rightarrow \wedge^{(|\beta| - |\beta_i|)} V \otimes \mathbb{S}_{\beta_i} Q \rightarrow... $$ -If we fix a single point on the Grassmannian and split the tautological short exact sequence there then we can show that this works, which is pretty good evidence. Surely this isn't a new discovery? - -REPLY [3 votes]: Look at Fonarev's On minimal Lefschetz decompositions for Grassmannians, specifically Proposition 5.3 (link to proposition in the PDF). I guess this exact sequence is what you need.<|endoftext|> -TITLE: How misleading is it to regard $\frac{dy}{dx}$ as a fraction? -QUESTION [96 upvotes]: I am teaching Calc I, for the first time, and I haven't seriously revisited the subject in quite some time. An interesting pedagogy question came up: How misleading is it to regard $\frac{dy}{dx}$ as a fraction? -There is one strong argument against this: We tell students that $dy$ and $dx$ mean "a really small change in $y$" and "a really small change in $x$", respectively, but these notions aren't at all rigorous, and until you start talking about nonstandard analysis or cotangent bundles, the symbols $dy$ and $dx$ don't actually mean anything. -But it gives the right intuition! For example, the Chain Rule says $\frac{dy}{du} \cdot \frac{du}{dx}$ (under appropriate conditions), and it looks like you just "cancel the $du$". You can't literally do this, but it is this intuition that one turns into a proof, and indeed if one assumes that $\frac{du}{dx} \neq 0$ this intuition gets you pretty close. -The debate about how rigorous to be when teaching calculus is old, and I want to steer clear of it. But this leaves an honest mathematical question: Is treating $\frac{dy}{dx}$ as a fraction the road to perdition, for reasons beyond the above, and which have not occurred to me?For example, what (if any) false statements and wrong formulas will it lead to? -(Note: Please don't worry, I have no intention of telling students that $\frac{dy}{dx}$ is a fraction; only, perhaps, that it can usually be treated as one.) - -REPLY [6 votes]: A note from a publication (my own) that occurred several years after this question was asked. $\frac{dy}{dx}$ can be considered a fraction of differentials. -You can think of differentials as infinitesimal values that are related to each other. Non-standard analysis showed that although 19th century mathematics viewed infinitesimals as problematic, they can be easily treated as ordinary mathematical objects, capable of division, multiplication, etc. -There is no problem treating $\frac{dy}{dx}$ as a fraction, but there is a problem in higher-order derivatives and differentials, but that is because we are using a notation that doesn't support it. If you take the idea of $\frac{dy}{dx}$ being a fraction seriously, then, to find the second derivative, you are taking the derivative of a fraction. Therefore, you have to apply the quotient rule. If you apply the quotient rule to $\frac{dy}{dx}$ you do not get the typical result of $\frac{d^2y}{dx^2}$. Instead, you get: -$$\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$$ -Or, written less ambiguously: -$$\frac{d(d(y))}{(d(x))^2} - \frac{d(y)}{d(x)}\frac{d(d(x))}{(d(x))^2}$$ -When written this way, the second derivative can be considered actual fractions just like the first derivative. Third and higher derivatives are even uglier, because you are taking the derivative of that. -You can see more details of this in "Extending the Algebraic Manipulability of Differentials", Dynamics of Continuous, Discrete and Impulsive Systems, Series A: Mathematical Analysis 26(3):217-230, 2019. And, if anyone is concerned for its validity, it had a further review in Mathematics Magazine 92(5), pp. 396–397 in their "Reviews" section.<|endoftext|> -TITLE: construction of the Jacobian of a curve -QUESTION [7 upvotes]: I am trying to understand the construction of the Jacobian of a curve following the notes of J. S. Milne -The question is going to be about a particular step in the proof of Proposition 4.2b in Chapter III, but I will first briefly recall the setup. -Let $X$ be a scheme flat over $T$, a divisor $D$ on $X$ is called relative effective divisor on $X/T$ if it is effective and flat over $T$ as a subscheme of $X$ (definition 3.4). There is a one-to-one correspondence benween relative effective divisors and sheaves $\mathcal L$ over $X$ with a global section $s$ such that $\mathcal{L}/s\mathcal{O}_X$ is flat over $T$. -We are working over a field. Let $C$ be a non-singular curve of genus $\geq 2$. -We are trying to construct a section of the natural map of functors $Div^r_C \to P^r_C$ where the first functor is the functor of relative effective divisors on $C\times T/T$ of degree $r$, and is represented by the $r$-fold symmetric product of $C$, and the second functor is the functor of families of degree $r$ invertible sheaves on $C$ parametrised by $T$, modulo trivial families: -$$ -P^r_C(T) = \{ \mathcal{L} \in Pic(C \times T) \mid deg\ \mathcal{L}_t=r \textrm{ for all }\ t \in T\} / q^* Pic(T) -$$ -(the natural projections are denoted $p: C \times T \to C$, $q: C\times T \to T$.) -Proposition 4.2 deals with subfunctors of $Div^r_C$ and $P^r_C$. We pick an effective degree $(r-g)$ divisor $D_\gamma$ and define -$$ -C^\gamma(T) = \{ D \in Div^r_C(T) \mid h^0(D_t-D_\gamma)=1\ \textrm{ for all }\ t \in T\} -$$ -$$ -P^\gamma(T) = \{ \mathcal{L} \in Pic^r_C(T) \mid h^0(\mathcal{L}_t \otimes \mathcal{L} _\gamma^{-1})=1\ \textrm{ for all }\ t \in T\} -$$ -part b) constructs a section $P^\gamma \to C^\gamma$. Take $\mathcal{L} \in P^\gamma(T)$. By definition of $P^\gamma$ and by Riemann-Roch, $h^1 (\mathcal{L}_t \otimes \mathcal{L}^{-1}_\gamma)=0$. This allows us to apply a base change theorem and coclude that $q_*(\mathcal{L} \otimes p^* \mathcal{L}^{-1} _\gamma)$ is locally free and thus an invertible sheaf on $T$ (call it $\mathcal{M}$). The proof then proceeds to construct a section of $\mathcal{L} \otimes (q^* q_*(\mathcal{L} \otimes p^* \mathcal{L} _\gamma^{-1}))^{-1}$. -In particular, as there is a natural map $q^* q_*(\mathcal{L} \otimes p^* \mathcal{L} _\gamma^{-1}) \to \mathcal{L} \otimes p^* \mathcal{L} _\gamma^{-1}$, one has a canonical global section of $\mathcal{L} \otimes p^* \mathcal{L} _\gamma^{-1} \otimes (q^*\mathcal{M})^{-1}$, and by composing it with the natural map $p^* \mathcal{L} _\gamma^{-1} \to \mathcal{O}_{C\times T}$ one gets the desired. -We did obtain a section $s_\gamma$ of a sheaf equivalent to $\mathcal{L}$ (in the sense of the definition of $Pic^r_C$) -My question is: why does this section give rise to a relative effective divisor, that is, why would $\mathcal{L} \otimes (q^* \mathcal{M})^{-1}/s_\gamma \mathcal{O}_{C\times T}$ be flat over $T$? - -REPLY [2 votes]: Let $N = L \otimes (q^\ast q_\ast (L\otimes L_\gamma^{-1}))^{-1}$. It suffices to show that the zero locus $D \subset C \times T$ of $s \in \Gamma(N)$ is flat over $T$. If $T$ is nice (Noetherian, blah, blah), it then suffices to show that the fiberwise degree of $D$ is constant. Note that the restriction of $N$ to $C \times \{ t \}$ is isomorphic to the restriction of $L$ to $C \times \{ t \}$. Since $L$ is fiberwise degree $r$ by assumption, it follows that $D$ is fiberwise degree $r$. -Oh and you need to check that the section $s$ is fiberwise nonzero. -Well, here's how you do that --- first look at the map $\phi : q^\ast q_\ast (L \otimes p^\ast L_\gamma^{-1}) \to L \otimes p^\ast L_\gamma^{-1}$. What does this map look like on fibers? Well, note that $H^0 (C \times \{ t \} , (L \otimes p^\ast L_\gamma^{-1})|\_{C \times \{ t \}})$ is by assumption 1 dimensional. Hence the restriction of $q^\ast q_\ast (L \otimes p^\ast L_\gamma^{-1})$ to $C \times \{ t \}$ is a trivial line bundle. So, the map $\phi$ on the fiber $C \times \{ t \}$ looks like the map $\mathcal{O}\_{C \times \{ t \}} \to (L \otimes p^\ast L_\gamma^{-1})|\_{C \times \{ t \}}$ corresponding to the one nonzero global section in $H^0 (C \times \{ t \} , (L \otimes p^\ast L_\gamma^{-1})|\_{C \times \{ t \}})$. Clearly this map is not zero. -I'll let you do the rest...<|endoftext|> -TITLE: how to use arxiv? -QUESTION [50 upvotes]: This is a soft question. How do people usually use arxiv to put their papers? At which stage does one usually put his/her paper/report there? Someone suggests me to submit a paper while putting it on arxiv. Is that the convention that people follow? -Thank you! -Anand - -REPLY [10 votes]: Yet another way in which people could use arXiv is as a repository for material which otherwise cannot find a home in a journal. Sometimes in the course of working on a project I wind up with some material which did not make it into the published article -- or perhaps some notes which record my growing understanding of articles already published by others -- which look like they could be useful to the community as expository or supplemental information, but which in my opinion are not otherwise significant enough to warrant submitting to a journal. I have sometimes wondered whether it would be appropriate to post such material on the arXiv. (I have not done so yet.) For that matter, I wonder whether others have done this very thing. -One drawback of this use of the arXiv is that everyone knows that most arXiv articles have not been peer-reviewed at the time of first posting, so they must be taken with a grain of salt. If an article is never published in a journal, you must read the arXiv article with a more critical eye. But as I said, I do believe that there exist some notes that are perhaps worth sharing but not worth wasting the effort to peer review.<|endoftext|> -TITLE: How can one compute the canonical class of the projective completion of the tautological bundle over $P^1\times P^1$? -QUESTION [7 upvotes]: I am interested in computing the (anti)-canonical class of the (total space of the) projective completion of the tautological bundle over $P^1\times P^1$. That is, the canonical class of $\mathbb P_{P^1\times P^1}(J \oplus \mathscr O)$, where $J$ is the tautological line bundle on $P^1\times P^1$. -I believe this can be done by computing the fan of the toric variety and summing the classes of the orbit closures of the one-skeleton? I was hoping for insight into perhaps a slicker/less cumbersome way of approaching this computation. -Thanks in advance. - -REPLY [4 votes]: I like the way you asked to avoid. Forgive me if I describe it in -polytope rather than fan language. -Step 1: ${\mathbb P}^1 \times {\mathbb P}^1$'s polytope is a square (or -any rectangle). The four edges, taken clockwise, correspond to the -${\mathbb P}^1$s giving the classes $h_1,h_2,h_1,h_2$ Michael mentions. -(EDIT: I had signs there before, by overthinking the Danilov relations.) -I can only guess that by "tautological line bundle on -${\mathbb P}^1 \times {\mathbb P}^1$ -you mean ${\mathcal O}(-1) \boxtimes {\mathcal O}(-1)$. -If we blow down that ${\mathbb P}^1 \times {\mathbb P}^1$, we get the -affine cone over the Segre embedding of ${\mathbb P}^1 \times {\mathbb P}^1$. -The polyhedron of that is also a cone, on a square. -Step 2: Blow the singular point back up, which corresponds to cutting the -corner off that cone, leaving a square. So far we have an unbounded -polytope that retracts to the square, just as the line bundle retracts -to ${\mathbb P}^1 \times {\mathbb P}^1$. -Step 3: Projectively complete. This corresponds to bounding the cone. -Combinatorially, we now have a square-based pyramid with the top corner -cut off, so there's a big square on the bottom (whose class is Michael's -$h$) and a little square on the top. -Step 4: Take the anticanonical class. On any toric variety, the boundary -of the polytope defines an anticanonical divisor. -So far our anticanonical class is the bottom square $h$ plus the top -square plus the other four faces. To calculate the linear relations -between them, one needs to be precise about the locations of the vertices. -I have the bottom square at $(0,0), (2,0), (0,2), (2,2)$ with $z=0$ -and the top one at $(0,0), (1,0), (0,1), (1,1)$ with $z=1$. -The Danilov relations from the $z$-axis vector says -$$ (-1) \text{bottom} + (+1) \text{top} -+ 0 \text{west} + 0 \text{south} + (+1) \text{north} + (+1)\text{east} = 0 $$ -so the total of the faces is $2\text{bottom} + \text{south} + \text{west}$, -matching Michael's $2h+h_1+h_2$. -(As it ought, since I learned at least some of this from him.)<|endoftext|> -TITLE: extensions of lebesgue measure -QUESTION [10 upvotes]: The Hahn-Banach theorem implies that Lebesgue measure can be extended give a "measure" on all subsets of [0,1], but this measure is only guaranteed to be finitely additive. It might magically turn out that this measure is countably additive, but this can only happen if the continuum is a real-valued measurable cardinal, a strong set-theoretic assumption. My question is: if it turns out that measure is countably additive on the measure zero sets, does this imply that the measure is countably additive everywhere? - -REPLY [3 votes]: The answer is no. A proof can be found here.<|endoftext|> -TITLE: Blowing up a subvariety - what can happen to the singular locus? -QUESTION [5 upvotes]: Let $X$ be a variety defined over a number field $k$. If I blow-up along some arbitrary subvariety of $X$, what are the possible outcomes for the dimension of the singular locus of the variety? If the subvariety lies outside the singular locus of $X$, then it stays the same, if it is carefully chosen, it might go down. Can it go up? -To be more specific, my variety is a high dimensional hypersurface, and the subvariety I am blowing up is a linear space of much smaller dimension than the singular locus. I don't know if this changes the situation. -I have a feeling this question might be more suited to stackexchange, but it didn't spark much interest over there https://math.stackexchange.com/questions/53676/blowing-up-a-subvariety-what-can-happen-to-the-singular-locus. Apologies for wasting time if so. - -REPLY [4 votes]: Any birational map $\pi:X'\to X$ is the blow-up of some ideal sheaf on $X$, so in general one must expect singularities on $X'$, even if the ideal is reduced (as you assume). -As a concrete example, let $X=\mathbb{A}^n$ and blow-up the complete intersection subvariety given by the ideal $I=(f,g)\subset k[x_1,\ldots,x_n]$. Then the blow-up of $X$ is the Proj of the Rees algebra $R[It]$ which is given by $k[x_1,\ldots,x_n,S,T]/(fS-gT)$. By choosing $f$ and $g$ appropriately one can produce varieties with singular locus of high dimension. -For your specific example, when $Y$ is a linear space of small dimension, I don't know if the above can happen, but there are certainly cases where the dimension of the singular locus will be unchanged after the blow-up, (e.g when $Y$ a point on a singular surface).<|endoftext|> -TITLE: Analytic functions with algebraic Taylor coefficients at some point. -QUESTION [6 upvotes]: This question just came to my mind when reading the question -When may Function (meromorphic) be expanded as power series with coefficients of integers -Suppose $f$ is an analytic function on some open subset $U \subseteq \mathbb{C}$. -Are there sufficient or necessary conditions to be put on $f$ that $f$ has the following property. -There exist a point $\zeta \in U$ such that the Taylor expansion of $f$ around $\zeta$ has only rational (algebraic) coefficients. -More generally, let us call the set of functions with this property $\mathcal{R}(U)$. It is easy to see that -$\mathcal{R}(U)$ is dense in the set of all analytic functions with respect to the topology -given by locally uniform convergence, since all polynomials with rational coefficients are -contained in $\mathcal{R}(U)$ and are dense. But of course $\mathcal{R}(U)$ does not contain -all analytic functions since for example $z^2+\pi$ or constant functions are not all contained. Also $\mathcal{R}(U)$ is uncountable, since $z-\zeta$ for any $\zeta \in \mathbb{C}$ satisfies the property. -So what can be said about $\mathcal{R}(U)$? Is it in some sense interesting? - -REPLY [3 votes]: Suppose you have a polynomial $p(z)=a_nz^n+a_{n-1}z^{n-1}+...+a_0$. Suppose that $p$ has the property you desire, which is equivalent to all of the derivatives of $p$ being simultaneously rational at some point $c$. In particular, this means $a_n$ must be rational, and so we may assume wlog that $a_n=1$. -If $a_{n-1}\in\mathbb{Q}$, then clearly $c$ must be rational, in which case it is immediate that all the $a_i$ are rational and we are done. -So, suppose that $a_{n-1}$ is not rational. Then by looking at the $(n-1)$th derivative at $c$, it must be that $a_{n-1}+nc$ is a rational number. Thus $a_{n-1}$ and $c$ must either both be algebraic or both be in the algebraic numbers adjoin $a_{n-1}$. Looking at the higher derivatives, one sees that all the terms are polynomially related, and so all the $a_i$ must all lie in the algebraic closure of the algebraic numbers adjoin $a_{n-1}$. -This necessary condition guarantees that almost none of the polynomials are in $\mathcal{R}(\mathbb{C})$ and, as a consequence, almost no analytic function is in there either. This is despite them being dense with respect to the topology you mentioned. -For a necessary condition, one can again see that it comes down to the value of $a_{n-1}$. One can "easily" iteratively generate a list of exact polynomial relations between $a_{n-1}$ and each of $a_i$ with $i\le n-2$. This isn't exactly a pleasant relation, however, since the relations are different for different degrees and also one ends up with complicated polynomials. For fourth- and lower-degree polynomials, this allows you to get exact relations in terms of radicals. However, for higher degree polynomials, this problem is decidedly more complicated because finding roots even for 5th degree polynomials is not universally possible. It might be that the polynomials in question are all of a particularly nice flavor which happen to be solvable nicely by radicals, but I don't know enough about that kind of thing. My guess is it is likely a hopeless venture.<|endoftext|> -TITLE: when can we lift an action of Lie algebra? -QUESTION [5 upvotes]: Suppose $G$ is a Lie group, $\mathfrak{g}$ its Lie algebra, if we have a smooth representation $(\pi,V)$, then it induces an action of $\mathfrak{g}$ on $V$. Now conversely, if we have a nice (with properties you may assume) action of $\mathfrak{g}$ on $V$, can we say such action arises from some unique smooth action of $G$? -Here we may assume $G$ to be simply connected if needed. Thank you. - -REPLY [7 votes]: Let $\pi$ represent a finite dimensional real Lie algebra $\mathfrak g$ on a Hilbert space -$\mathcal H$ by skew-adjoint operators. Then $\pi$ integrates to the connected simply connected Lie group $G$ with Lie algebra $\mathfrak g$ if, and only if, the elements of -$\pi(\mathfrak g)$ have a common invariant dense domain. This is an old result of Moshe Flato, Daniel Sternheimer and others. My apologies to the mathematical physicists whose names I have omitted.<|endoftext|> -TITLE: Literature on behaviour of eigenfunctions under multiplication? -QUESTION [6 upvotes]: Dear community, -I would be happy about any literature or comments on the behaviour of the pointwise product of eigenfunctions of a self-adjoint operator with discrete spectrum, acting on a separable Hilbert space which is closed under pointwise multiplication. The operator I'm actually looking at is a symmetric Markov operator acting on $L^2(\mathcal{A},\mu)$, where $\mathcal{A}$ is some function algebra and $\mu$ the invariant measure. -Some questions I'm especially interested in are: - -If you multiply two eigenfunctions, can it happen that the product has an infinite eigenfunction expansion? By "infinite eigenfunction expansion" I mean that it can not be expressed as a finite sum of eigenfunctions. -Somewhat related: If the squares of two eigenfunctions have a finite expansion, respectively, can it happen that the square of the sum of these two eigenfunctions has an infinite expansion? -In the above Markov setting: Is the following "projected Cauchy-Schwarz inequality" always true? -$$ -\int \operatorname{proj}(fg \mid E)^2 \ \text{d} \mu \leq -\sqrt{\int \operatorname{proj}(f^2 \mid E)^2 \ \text{d} \mu} -\sqrt{\int \operatorname{proj}(g^2 \mid E)^2 \ \text{d} \mu} -$$ -Here, $f$ and $g$ are eigenfunctions lying in some common eigenspace, $E$ is another eigenspace and $\operatorname{proj}(f \mid E)$ denotes the projection of $f$ on $E$. - -Note that the answer to questions 1 and 2 is no, if the eigenfunctions are orthogonal polynomials. -Thanks for your help, -Simon - -REPLY [3 votes]: For question 1, one example of interest comes from the energy eigenfunctions of the one dimensional quantum harmonic oscillator. The Hilbert space is separable, and the Hamiltonian satisfies your conditions. Under suitable normalization, the eigenfunctions have the form $p(x)e^{-\pi x^2}$ for $p$ a Hermite polynomial, and the product of any two then has the form $q(x)e^{-2\pi x^2}$ for $q$ a nonzero polynomial. These products are not given by finite linear combinations of eigenfunctions.<|endoftext|> -TITLE: Embedding in f.p. simple groups -QUESTION [10 upvotes]: Dear All! -At the time when Lyndon and Schupp wrote their book there was an open question: -Question: Does every finitely presented group with soluble word problem embed in a finitely presented simple group? -Is it still open? Could you hint at some useful references about this? Thanks! - -REPLY [2 votes]: There is a strengthening of the Boone-Higman result, due to Thompson. He showed that we can take the simple group to be finitely generated. In full, this reads: -"A finitely presented group has solvable word problem if and only if it can be embedded in a finitely generated simple group that can be embedded in a finitely presented group." -You can find the full details in: -R. J. Thompson, "Embeddings into finitely generated simple groups which preserve the word problem", Word Problems II: The Oxford Book, Studies in Logic and the Foundations of Mathematics, Volume 95, (1980). -As far as I am aware, your original question "Does every finitely presented group with soluble word problem embed in a finitely presented simple group?" is still an open problem. --Maurice<|endoftext|> -TITLE: Understanding the analytic index map of the Atiyah-Singer index theorem -QUESTION [8 upvotes]: Hi, -I'm currently trying to understand the Atiyah-Singer index theorem and its proof as presented in the book "Spin Geometry" by Lawson and Michelsohn. -I do not understand why the analytic index map $\operatorname{ind}\colon K_{cpt}(T^\ast X) \to Z$, as defined in chapter III.$13 in equation (13.8), agrees with the Fredholm index of an elliptic pseudo-differential operator. -Recall how the analytic index map is constructed (this is chapter III.§13 in the book): -Given an element $u \in K_{cpt}(T^\ast X) \cong K(DX, \partial DX)$ we can represent it by Lemma III.13.3 via a triple $(\pi^\ast E, \pi^\ast F; \sigma)$, where $E$ and $F$ are vector bundles over $X$, $\pi\colon T^\ast X \to X$ is the bundle projection and $\sigma\colon \pi^\ast E \to \pi^\ast F$ is homogeneous of degree 0 on the fibres of $T^\ast X$ (i.e. $\sigma$ is constant on the fibres). Then for any $m$ there exists an elliptic, classical pseudo-differential operator $P \in \Psi CO_m(E,F)$ whose asymptotic principal symbol is $\sigma$ (in particular, the symbol class $[\sigma(P)] \in K_{cpt}(T^\ast X)$ equals u). Then we set $\operatorname{ind}(u) := \operatorname{Fredholm-ind}(P)$. Then it is proven in the book, that this is well-defined, i.e. independent of all choices. -Now given an elliptic pseudo-differential operator $D \in \Psi DO_m(E,F)$, we can construct its symbol class $[\sigma(D)] \in K_{cpt}(T^\ast X)$ as in chapter III.§1 in equation (1.7). Now I expect that the Fredholm index of $D$ coincides with the analytic index of $[\sigma(D)]$, but I do not see that this is proven in the book. I also can't prove it on my own. Going through the construction above we get an elliptic, classical pseudo-differential operator $P \in \Psi CO_m(E,F)$ with $[\sigma(D)] = [\sigma(P)] \in K_{cpt}(T^\ast X)$. But why do $D$ and $P$ have the same Fredholm index? - -Why is $\operatorname{Fredholm-ind}(D) = \operatorname{ind}([\sigma(D)])$ for an elliptic, pseudo-differential operator? - -REPLY [6 votes]: The Fredholm index of an elliptic operator only depends on the symbol class. Here is the proof (which I memorize from Lawson-Michelsohn and Atiyah-Singer). -If $D: \Gamma(E_0) \to \Gamma(E_1)$ has order $k \neq 0$, pick a connection $\nabla$ on $E_0$. Then $A=(1+\nabla^{\ast}\nabla)$ is a self-adjoint invertible operator of order $2$ and $D \circ A^{-k/2}$ has the same Fredholm index as $D$ and the same symbol class; but it has order $0$. So for any operator, there is an order $0$ operator with the same symbol class and the same index; and this reduces the problem to the order $0$ case. -It has been mentioned that the index of an operator only depends on the homotopy class of its symbol (by the way, the proof of this in Lawson-Michelssohn is incomplete. In the proof of Theorem 7.10, they say ''to construct a family of operators $P_t$ with $\sigma(P_t)=\sigma_t$, [...] is evidently possible locally (in coordinates)''. One needs to know that in $R^n$, the operator norm of a pseudo-DO can be estimated by the symbol. You can see this by going through the proof of Prop. III.3.2 of L.-M.). -Recall the (modified) definition of $K^0 (TX,TX-0)$: it is the group of all equivalence classes of $(E_0,E_1,f)$; $E_i \to X$ vector bundles and $f: \pi^{\ast} E_0 \to \pi^{\ast} E_1$ a bundle map that is an isomorphism away from the zero section and homogeneous of order $0$ outside the zero section. The equivalence relation is generated by (1) homotopy, (2) isomorphism and (3) addition of things of the form $(E,E,id)$. (1) and (2) preserve the index. (3) also preserves the index; since a pseudo-DO with symbol $(E,E,id)$ is e.g. the identity operator, which has index $0$.<|endoftext|> -TITLE: Can Thompson's group F be realized as a semigroup of continuous transformations of a tree? -QUESTION [5 upvotes]: I am not very familiar with F, but I know that it can be realized as a group of homeomorphisms of the boundary of the binary tree. I also know that F cannot be realized as a group of graph automorphisms of any regular rooted tree because F is not residually finite. However, if we topologize our trees with the path metric, can F be realized as a group of continuous prefix-preserving transformations of a regular rooted tree (where the transformations need not be injective or surjective)? If you know the answer, could you provide a reference? Thanks! - -REPLY [3 votes]: There is a way to get $F$ to act on the infinite binary tree bijectively, but I doubt it satisfies most of your other requirements. It basically does something sensible with the "missing finite subtree." I have only partly checked this out (meaning it seems to check for one of the two relations needed). -We let $T$ be the set of finite words (including the empty word) on the alphabet $\{0,1\}$. This is a binary tree by letting the left child of $u\in T$ be $u0$ and the right child of $u$ be $u1$. We define two permutations of $T$. -The permutation $x_0$ is determined by the following rules: -$\emptyset\rightarrow 1$; -$0\rightarrow \emptyset$; -$00u\rightarrow 0u$; -$01u\rightarrow 10u$; -$1u\rightarrow 11u$. -The permutation $x_1$ is determined by the following rules: -$\emptyset\rightarrow \emptyset$; -$0u\rightarrow 0u$; -$1\rightarrow 11$; -$10\rightarrow 1$; -$100u\rightarrow 10u$; -$101u\rightarrow 110u$; -$11u\rightarrow 111u$. -These are the usual rules for the action of $x_0$ and $x_1$ in $F$ on infinite words in $\{0,1\}$ restricted to finite words and extended to the few cases that the rules usually omit. -As I said, it checks for the relation: -$(x_1)^{x_0x_0} = (x_1)^{x_0x_1}$. -Here $a^b$ means $b^{-1}ab$ and the actions are to be composed from left to right (they are right actions). -The other relation that defines $F$ with the one above is -$(x_1)^{x_0x_0x_0} = (x_1)^{x_0x_0x_1}$. -If the second fails while the first succeeds, I will be stunned. -Assuming that the second relation checks out (not too hard, I am just too lazy), then these two permutations of $T$ generate a copy of $F$. On "most" of $T$, the action agrees with the usual action. How well this cooperates with what you want is for you to decide. -The definitions can be tinkered with a bit. I doubt that the relations can survive a lot of tinkering though.<|endoftext|> -TITLE: Maximal number of edges and triangular cells for n points in a triangular lattice -QUESTION [7 upvotes]: Consider a subset of $n$ points in an equilateral triangular lattice. Draw all the edges between nearest-neighbor points. -What is the maximum, over all such subsets, of the number of edges? This sequence appears to start 0, 1, 3, 5, 7, 9, 12, 14, 16... -What is the maximum number of triangular lattice cells? (Not the number of all triangles, just the number of smallest possible equilateral triangles in the lattice.) This sequence appears to start 0, 0, 1, 2, 3, 4, 6, 7, 8, 10, 11, 13... -http://oeis.org/A047932 is related to the first sequence but I have no proof it's the same. (There might be some other way of arranging the pennies that yields a higher number of contacts. A047932 is a lower bound on my sequence.) I can't find any OEIS sequences relevant to the second one. - -REPLY [10 votes]: You might have noticed that the difference between your two sequences is $0,1,2,3,4,5,6,7,8,\ldots$ and that the optimal configurations seem to be the same for both problems. -This is true in general, and is a nice application of Euler's formula $V-E+F=2$. (NB this requires showing that in an optimal configuration the edges form a connected component.) Here $V=n$, the number of edges is $E$, and the number of triangles, call it $t$, is $F-1$ because we must count the exterior of the graph as a face. So the difference between the edge and triangle maxima is $(n-1)$ — at least assuming we can prove it's never to our advantage in either problem to have holes bigger than a unit triangle in the picture, which seems clear but may be annoying to prove rigorously. [EDIT see below on this point.] -Another standard graph-theory formula that applies here: the sum of all the faces' edge-counts is $2E$. [Proof: count in two ways the pairs $(e,f)$ where $f$ is a face and $e$ is one of its edges.] In our setting all but one of the faces has $3$ edges, so $2E=3t+p$ where $p$ (for perimeter) is the number of outside edges (counting an edge twice if both sides abut the infinite face, as happens for $n=2$, and again assuming no internal holes). So $t+2 = 2n-p$, and the problem comes down to minimizing $t$. It certainly looks plausible that the "penny spiral" does this, but it's getting late so I'll leave it as an exercise :-) This would identify your first sequence with OEIS A047932, and thus determine the second sequence as well. -UPDATE Denote the sequences in question by $s_1(n)$ and $s_2(n)$ respectively. As pointed out in G.Zaimi's accepted answer, the formula $s_1(n) = \lfloor 3n-\sqrt{12n-3}\rfloor$, consistent with the original proposer's guess/question, follows from a much more general result of Harbroth that this is the maximal number of times that the minimum distance can occur in any configuration of $n$ points in the plane. I looked up the solution (which is freely available online), and it seems to use ultimately the same technique: see the reference to the "Eulerschen Polyedersatz" before equation (3). Using the Euler "Polyedersatz" we also deduce $$s_2(n) = s_1(n) - (n-1) = \lfloor 2n-\sqrt{12n-3}+1\rfloor,$$ answering the second question. Indeed Euler says that $s_2(n) \leq s_1(n) - (n-1)$ with equality iff there's an optimal configuration without holes bigger than a unit triangle; and Harbroth's solution gives such a configuration. - -REPLY [10 votes]: The following was conjectured by D. Reutter in problem 664A, Elemente der mathematik 27 and proved by H. Harborth in Solution to problem 664A, Elemente der mathematik 29, 14-15 - -The maximum number of times the minimum distance can occur among $n$ points in the plane is $\lfloor 3n-\sqrt{12n-3}\rfloor$. - -This is achieved by a hexagonal piece in the triangular lattice, i.e. from points forming a hexagonal spiral. In particular, this gives a formula for your first sequence.<|endoftext|> -TITLE: How to prove a random d-regular graph is an expander with prob >= 0.5? -QUESTION [7 upvotes]: Context: Many resources, like -http://math.mit.edu/~fox/MAT307-lecture22.pdf -state the theorem in the general case, but then prove it only for the bipartite case. -The full case is supposedly proved in Pinsker's 1973 paper. However, I can't dig up a copy. -Anyone know of a proof for the general case (i.e. d-regular, undirected, not-necessarily-bipartitite graph)? -Thanks! - -REPLY [3 votes]: Pinsker's original paper is now available online in the archive of the International Teletraffic Congress: -http://ww.i-teletraffic.org/fileadmin/ITCBibDatabase/1973/pinsker731.pdf<|endoftext|> -TITLE: Discrete version of Nullstellensatz? -QUESTION [12 upvotes]: Hi. I was reading the paper "On the foundations of combinatorial theory (VI): The idea of a generating function" by Doubilet, Rota and Stanley, and there is a relation treated which is very reminiscent of the relation between ideals in a polynomial ring and affine algebraic varieties (i won't go more specific in the definitions). -It goes as follows (everything quoted from the above paper): Let $P$ be a finite poset (can be generalized to locally finite), and consider its incidence algebra $I(P,K)$, consisting of all the functions from the intervals in $P$ to some field $K$ (of characteristic zero). Sum and product by scalars are inherited from $K$, and product of two functions $f,g \in P$ is defined as the convolution: -\begin{equation} -(f*g)(x,y)=\sum_{x\leq z \leq y}f(x,z)g(z,y) -\end{equation} -Some special elements in $I(P,K)$ needed to state the connection are the units: -\begin{equation} -\delta_{x,y}(u,v)=\begin{cases}1&\text{if $u=x$ and $v=y$,}\\\\0&\text{otherwise.}\end{cases} -\end{equation} -Now, the (two sided) ideals in this algebra and the varieties have a very nice relation just very similar to the one from commutative algebra and algebraic geometry. But in this case the relation is tighter, because varieties have an algebraic structure coming from a natural partial ordering. -Define the support of and ideal $J$, $\Delta (J)$, as the set of all the units $\delta_{x,y}$ belonging to $J$. It turns out that every ideal $J$ in $I(P,K)$ consists of all the functions $f$ for which $f(x,y)=0$ whenever $\delta_{x,y}\notin \Delta(J) $. -On the other hand, define $Z(J)$ as the set of all intervals $\[ x,y \]$ such that $f(x,y)=0$ for all $f\in J$ (this would be the variety). $Z(J)$ is an order ideal of the poset of all intervals of $P$ (ordered by inclusion). -Theorem: Let $P$ be a finite poset and $S(P)$ the poset of its intervals, ordered by inclusion. Then there is a natural anti-isomorphism between the lattice of ideals of $I(P,K)$ and the lattice of order ideals of $S(P)$. -(For more details and background, check the paper, or "Enumerative Combinatorics Vol.1" by Stanley) - -My question is: Does anyone know if this ideal-variety duality has been exploited or studied further in the context of posets from an algebraic geometry point of view? (apart from the material in the mentioned paper). - -REPLY [7 votes]: Joyal and Tierney have done Algebraic Geometry with lattices (in topoi) somewhat along the lines you describe in their landmark monograph An extension of the Galois Theory of Grothendieck<|endoftext|> -TITLE: Do the converses of [weak law of large numbers / central limit theorem] hold? -QUESTION [8 upvotes]: Let $\; X_0,X_1,X_2,X_3,...\;$ be independent and identically distributed (real-valued) random variables. -1. -Suppose $\frac1n \cdot\sum\limits_{m=0}^n X_m$ converges in probability. Does it follow that $\operatorname{E}(X_0)$ exists? -2. -Suppose $\operatorname{E}(X_0) = 0$ and that $\frac1{\sqrt n} \cdot\sum\limits_{m=0}^n X_m$ converges in distribution to a normal random variable. -Does it follow that $\operatorname{E}((X_0)^2)$ is finite? -(I already found that the converse of the strong law of large numbers holds.) - -REPLY [8 votes]: (As suggested, I promote my comment to an answer, with pgassiat's complement.) -Necessary and sufficient conditions (in terms close to those you want) for the WLLN and the CLT can be found, e.g., in "Foundations of modern probability" by Kallenberg (Theorems 4.16 and 4.17 in the first edition, Theorems 5.16 and 5.17 in the second edition).<|endoftext|> -TITLE: True by accident (and therefore not amenable to proof) -QUESTION [27 upvotes]: The graph reconstruction conjecture claims that (barring trivial examples) a graph on n vertices is determined (up to isomorphism) by its collection of (n-1)-vertex induced subgraphs (again up to isomorphism). -The way it is phrased ("reconstruction") suggests that a proof of the conjecture would be a procedure, indeed an algorithm, that takes the collection of subgraphs and then ingeniously "builds" the original graph from these. -But based on some experience with a related conjecture (the vertex-switching reconstruction conjecture), I am led to wonder whether this is something that is simply true "by accident". By this I mean that it is something that is just overwhelmingly unlikely to be false ... there would need to be a massive coincidence for two non-isomorphic graphs to have the same "deck" (as the collection of (n-1)-vertex induced subgraphs is usually called). In other words, the only reason for the statement to be true is that it "just happens" to not be false. -Of course, this means that it could never actually be proved.. and therefore it would be a very poor choice of problem to work on! -My question (at last) is whether anyone has either formalized this concept - results that can't be proved or disproved, not because they are formally undecidable, but just because they are "true by accident" - or at least discussed it with more sophistication than I can muster. -EDIT: Apologies for the delay in responding and thanks to everyone who contributed thoughtfully to the rather vague question. I have accepted Gil Kalai's answer because he most accurately guessed my intention in asking the question. -I should probably not have used the words "formally unprovable" mostly because I don't really have a deep understanding of formal logic and while some of the "logical foundations" answers contained interesting ideas, that was not really what I was trying to get at. -What I was really trying to get at is that some assertions / conjectures seem to me to be making a highly non-obvious statement about combinatorial objects, the truth of which depends on some fundamental structural understanding that we currently lack. Other assertions / conjectures seem, again, to me, to just be saying something that we would simply expect to be true "by chance" and that we would really be astonished if it were false. -Here are a few unproved statements all of which I believe to be true: some of them I think should reflect structure and others just seem to be "by chance" (which is which I will answer later, if anyone is still interested in this topic). -(1) Every projective plane has prime power order -(2) Every non-desarguesian projective plane contains a Fano subplane -(3) The graph reconstruction conjecture -(4) Every vertex-transitive cubic graph has a hamilton cycle (except Petersen, Coxeter and two related graphs) -(5) Every 4-regular graph with a hamilton cycle has a second one -Certainly there is a significant chance that I am wrong, and that something that appears accidental will eventually be revealed to be a deep structural theorem when viewed in exactly the right way. However I have to choose what to work on (as do we all) and one of the things I use to decide what NOT to work on is whether I believe the statement says something real or accidental. -Another aspect of Gil's answer that I liked was the idea of considering a "finite version" of each statement: let S(n) be the statement that "all non-desarguesian projective planes of order at most n have a Fano subplane". Then suppose that all the S(n) are true, and that for any particular n, we can find a proof - in the worst case, "simply" enumerate all the projective planes of order n and check each for a Fano subplane. But suppose that the length of the shortest possible proof of S(n) tends to infinity as n tends to infinity - essentially there is NO OTHER proof than checking all the examples. Then we could never make a finite length proof covering all n. This is roughly what I would mean by "true by accident". -More comments welcome and thanks for letting me ramble! - -REPLY [17 votes]: This is a very interesting (yet rather vague) question. Most answers were in the direction of mathematical logic but I am not sure this is the only (or even the most appropriate) way to think about it. The notion of coincidence is by itself very complicated. (See http://en.wikipedia.org/wiki/Coincidence ). One way to put it on rigurous grounds is using probabilistic/statistical framework. Indeed, as Timothy mentioned it is sometimes possible to give a probabilistic heuristic in support of some mathematical statement. But its is notorious statistical problem to try to determine aposteriori if some events represent a coincidence. -I am not sure that (as the OP assumes) if a statement is "true by accident" it implies that it can never be proved. Also I am not sure (as implied by most answers) that "can never be proved" should be interpreted as "does not follow from the axioms". It can also refers to situations where the statement admits a proof, but the proof is also "accidental" as the original statement is, so it is unlikely to be found in the systematic way mathematics is developed. -In a sense (as mentioned in quid's answer), the notion of "true by accident" is related to mathematics psychology. It is more related to the way we precieve mathematical truths than to some objective facts about them. -Regarding the reconstruction conjecture. Note that we can ask if the conjecture is true for graphs with at most million vertices. Here, if true it is certainly provable. So the logic issues disappear but the main issue of the question remains. (We can replace the logic distinctions by computational complexity distinctions. But still I am not sure this will cpature the essence of the question.) There is a weaker form of the conjecture called the edge reconstruction conjecture (same problem but you delete edges rather than vertices) where much is known. There is a very conceptual proof that every graph with n vertices and more than nlogn edges is edge-reconstructible. So this gives some support to the feeling that maybe vertex reconstruction can also be dealt with. -Finally I am not aware of a heuristic argument that "there would need to be a massive coincidence for two non-isomorphic graphs to have the same 'deck'" as the OP suggested. (Coming up with a convincing such heuristic would be intereting.) It is known that various graph invariants must have the same value on such two graphs.<|endoftext|> -TITLE: Properties from Tropical Geometry that do not imply their algebraic counterpart. -QUESTION [9 upvotes]: One of the motivations to study tropical geometry is that there are some hard Algebraic Questions that can be answered by proving them in the Tropical World. For example one can show that tropical Bezout's Theorem implies the Algebraic Bezout. -What properties are there known that are true (or might be) in tropical geometry that don't imply that their algebraic version is true? - -REPLY [7 votes]: There is a simple nice fact which holds in the tropical plane that has no counterpart in algebraic geometry (nor in any kind of standard geometry I might think of): two tropical lines always "intersect" in a single point... even if they coincide! -Of course this property relies on the fact that "intersection" is not defined in the usual way. We define the "intersection" of two tropical curves $C_1$ and $C_2$ as follows: the union $C_1\cup C_2$ has a natural cellularization into vertices and edges, and "$C_1\cap C_2$" is the union of the vertices contained in the set-theoretic intersection $C_1\cap C_2$. One may also define a multiplicity on each intersection point. With this definition, the intersection of a tropical curve with itself is the union of its vertices. -Therefore two (possibly coinciding) tropical lines always intersect in a point. By defining anagously an appropriate (dual) notion of "span", the dual sentence is also true: two (possibly coinciding) points always span a single line.<|endoftext|> -TITLE: What is the theory of polynomials? -QUESTION [16 upvotes]: We all know what polynomials are, along with their elementary properties and many effective algorithms for different representations of polynomials. -The question here is more of a universal algebra question: what is the signature of the theory which best corresponds to polynomials? To illustrate what this question means, it is probably easiest to do this by example: - -In the category of Unital Rings, the integers are the initial algebra. -In the category of semirings, the natural numbers are the initial algebra. - -So what is a small presentation of a category (in the sense of giving signatures and axioms) for which the polynomials are initial? Naturally, for $R[x]$, one expects that this presentation will either include the presentation of the ring $R$, or be parametric in that presentation. But what else is needed to characterize univariate polynomial rings from general rings? -The motivation is that I am looking for a semantic type for univariate polynomials. In most cases, the type for polynomials one encounters in the litterate is the type of its representations. This is like saying that a matrix has semantic type 'square array', rather than to say that a matrix (in linear algebra) is a representation of a linear operator (with linear operator being the correct semantic type). - -EDIT: one note of clarification. After I figure out the 'theory of polynomials', I then wish to be able to write it down as well, so I want a 'presentation', universal-algebra-style, of the 'theory of polynomials' (whether that is plethories or free V-algebras on one generator or ...). With the integers, it is easy to write down a set of axioms that define unital rings. - -REPLY [3 votes]: IMHO the answer to “where polynomials are initial” (not to the title question, which is too broad for me) is already given in “Awodey. Category theory. 9. Adjoints. 9.3. Examples of adjoints. Example 9.10.” -In that example, the adjunction of functors is constructed, where its free (left adjoint) functor $F$ goes from the category of rings (=RingCat) to the category of rings with distinguished element (= pointed rings). If we define this adjunction via unit ($\eta$), then the definition says that - -for every object $R$ in RingCat (= for - every ring $R$) there is an - initial object in the category $select(R)\downarrow U$ (comma - category), - -where $U$ is the forgetful functor. -Furthermore, a chosen initial object for $R$ consists of $F(R)$ (= $R[x]$) and $\eta(R):R\to U(F(R))$ (= a ring homomorphism constructing constant polynomials). The distinguished element in $F(R)$ is “$x$” (the projection polynomial).<|endoftext|> -TITLE: Cohomology of Theta divisor on Jacobian? -QUESTION [6 upvotes]: Let $C$ be a curve of genus $g \geq 1$ and let $J^d$ be its degree $d$ Jacobian. -Inside of $J^{g-1}$ there is the Theta divisor $\Theta$, which can be defined in various ways; the quickest definition is probably: it's the image of the Abel-Jacobi map $C^{(g-1)} \to J^{g-1}$ sending an effective degree $g-1$ divisor to the corresponding line bundle. Picking an isomorphism $J^{g-1} \cong J^d$, we also write $\Theta$ for the corresponding divisor in $J^d$. - -How to compute $H^\ast(J;\Theta)$, or $h^\ast(J;\Theta)$? Or alternatively, what is known about these groups? - -I suspect this is something embarrassingly standard and/or obvious and/or well-known and/or classical, but I haven't been able to figure anything out. The only thing along these lines that I was able to figure out was how to compute the Euler characteristic $\chi(J;\Theta^k)$ where $k$ is an integer: By Hirzebruch-Riemann-Roch and the Poincare formula it's $$\int_J \operatorname{ch}(\Theta^k) = \int_J e^{k\theta} = \int_J k^g \theta^g / g! = k^g.$$ - -REPLY [5 votes]: this seems to be the kodaira vanishing theorem. i.e. any line bundle of form K+A where is ample, has no higher cohomology. for an abelian variety K is trivial, and Theta is ample. qed.<|endoftext|> -TITLE: "Rounding the corners" to get contact boundary -QUESTION [5 upvotes]: Suppose we have symplectic manifolds $(M_1, \omega_1)$ and $(M_2, \omega_2)$ with non-empty boundary of contact . Often we need to deal with the product $M_1 \times M_2$ with the product symplectic structure. Can we round the corners to get a contact manifold as boundary? - -REPLY [6 votes]: In that generality, the answer is no: a symplectic form $\omega$ on $X$ which has contact-type boundary is exact on $\partial X$. Yet $\omega_1 \oplus \omega_2$ need not be exact on $M_1\times \partial M_2$, nor on $\partial M_1 \times M_2$. -It is possible, however, if $M_1$ and $M_2$ are Liouville domains, i.e., if the symplectic form $\omega_i$ is given as $d\theta_i$ for 1-forms $\theta_i$ whose dual vector field $\lambda_i$ points strictly outwards along the boundary. In fact, if you round corners sensibly, $\theta_1 \oplus \theta_2$ will have those same properties on the product. -Here's a relevant article by Alex Oancea: -http://arxiv.org/abs/math/0403376<|endoftext|> -TITLE: An exercise in Jech's Set Theory -QUESTION [6 upvotes]: I had a hard time trying to solve exercise 7.24 in Jech's book (3rd edition, 2003) and finally came to the conclusion that the result there, which should be proved might be wrong. The claim goes like this: -Let $A$ be a subalgebra of a Boolean algebra $B$ and suppose that $u \in B-A$. Then there exist ultrafilters $F,G$ on $B$ such that $u \in F$, $-u \in G$ and $F \cap A= G \cap A$. -A (perhaps flawed, as I believe) proof of this can be found here. -http://onlinelibrary.wiley.com/doi/10.1002/malq.19690150705/abstract -A counterexample to the claim above is the following: -Let $A$ be the algebra of finite unions of (open, closed, half-open) intervals on $[0,1]$ with rational endpoints, and let $B$ be defined as $A$ but with real endpoints. Each ultrafilter $U$ on $A$ converges to a rational or irrational number $r$ and the elements of $U$ are exactly those sets in $A$ that include $r$. Now if $F$ and $G$ are two ultrafilters on the bigger algebra $B$, both extending $U$ then they converge again towards $r$ and for any $u\in B$ we have that $u\in F$ iff $r \in u$ iff $u\in G$, which makes it impossible to have $u \in B$, yet $-u \in G$. -My questions are now: - -Is my counterexample correct? -The claim is used to show that each Boolean algebra of size $\kappa$ has at least $\kappa$ ultrafilters (this is theorem of the paper mentioned above). Does this remain valid ( I suppose not, see the comments) - -REPLY [9 votes]: The exercise is correct. Let $u\in B\setminus A$. We say that $u$ splits an ultrafilter $H$ of $A$ if $\{u\}\cup H$ and $\{-u\}\cup H$ both have the finite intersection property. (If $u$ splits $H$, then there are ultrafilters $F$ and $G$ of $B$ such that -$F\cap A=H=G\cap A$, $u\in F$, and $-u\in G$.) -Suppose no ultrafilter $H$ is split by $u$. -We say that an ultrafilter $H$ of $A$ is compatible with $b\in B$ iff each $a\in H$ has -nonempty intersection (in $B$) with $b$. If no ultrafilter of $A$ is split by $u$, -then each ultrafilter is either compatible with $u$ or compatible with $-u$. -Let $C$ be the set of ultrafilters of $A$ compatible with $u$, and let $D$ be the set of ultrafilters of $A$ compatible with $-u$. -Now the set Ult$(A)$ of all ultrafilters of $A$ is the disjoint union of $C$ and $D$. -Hence an ultrafilter of $A$ is compatible with $u$ iff it is not compatible with $-u$ and vice versa. So, if $H\in C$, then there is $a\in H$ such that $a$ is disjoint from $-u$. -All ultrafilters of $A$ that contain $a$ are incompatible with $-u$ and hence compatible with -$u$. This shows that $C$ is open in the Stone space Ult$(A)$ of $A$. -The same is true for $D$. It follows that the two sets are clopen. -By the Stone representation theorem, there is $a\in A$ such that $C$ is the set of all ultrafilters $H$ of $A$ that contain $a$. $D$ is the set of all ultrafilters of $A$ that contain $-a$. -In other words, an ultrafilter $H$ of $A$ is compatible with $u$ iff $a\in H$. -But this implies that an element $b$ of $A$ has a nonempty intersection with $u$ -iff it has a nonempty intersection with $a$. Hence $-a$ is disjoint from $u$. -In other words, $u\leq a$. -The symmetric argument shows that $-u\leq-a$. -It follows that $a=u$ and hence $u\in A$, a contradiction. -And yes, this exercise implies that every infinite Boolean algebra of size $\kappa$ has at least $\kappa$ ultrafilters.<|endoftext|> -TITLE: Reference request: Spec A_* is the automorphism group of the additive formal group law -QUESTION [7 upvotes]: Dear all, -I'm seeking a reference for a claim made in lecture 8 of Jacob Lurie's chromatic homotopy theory notes (http://www.math.harvard.edu/~lurie/252xnotes/Lecture8.pdf). More particularly, Theorem 6 of this lecture states that (say over $\mathbb{F}_2$, so that things are commutative) the spectrum $\mathbb{G} = \operatorname{Spec} \mathcal{A}_*$ of the dual Steenrod algebra $\mathcal{A}_*$ is the automorphism group of the additive formal group law, in the obvious sense. -Lurie argues convincingly that $\mathbb{G}$ does act on the additive formal group law, but I don't think he attempts to prove that this action gives an isomorphism with the automorphism group. I'd be grateful if someone could give me a reference for this fact. -Cheers, -Saul - -REPLY [7 votes]: If you're looking for a reference in print, it's in Ravenel's book Complex Cobordism and Stable Homotopy Groups of Spheres. See the comments after the proof of Theorem A2.2.18. (This book is available online, and you want Appendix 2.)<|endoftext|> -TITLE: Dehn's solution to Hilbert's 3rd: 1901 or 1902? -QUESTION [5 upvotes]: This is a simple bibliographic request that I have been unable to pin down. Max Dehn's -solution to Hilbert's 3rd problem is: - -Max Dehn, "Über den Rauminhalt." Mathematische Annalen 55 (190x), no. 3, pages 465–478. - -It is variously cited as either 1901 or 1902 (but always volume 55; Hilbert's own footnote -cites volume 55 "soon to appear"). E.g., - -Mathworld cites it as 1902. -The Encyclopedic Dictionary of Mathematics cites it as 1902. -Wikipedia says 1901. -Various papers, e.g., this one, and Tao's book, cite it as 1901. - -I have been unsuccessful in finding the definitive year via the web, because of all -the conflicting citations. The next step is to retrieve -Mathematische Annalen volume 55, but perhaps someone can spare me that trouble...? -Thanks! - -REPLY [6 votes]: Another point to consider is whether "Über den Rauminhalt" -is in fact Dehn's first solution to Hilbert's 3rd Problem. I -believe his first solution was in the paper "Über raumgleiche -Polyeder" in the Nachrichten der Königliche Gesellschaft der -Wissenschaften zu Göttingen of 1900, pp. 345 -- 354.<|endoftext|> -TITLE: when mapping cone is contractible -QUESTION [9 upvotes]: It is quite obvious that if a map is a homotopy equivalence, then its mapping cone is contractible, but is the converse true: mapping cone contractible => the map is a homotopy equivalence? I am thinking about both the topological category and the category of chain complexes. - -REPLY [3 votes]: The accepted answer addresses the version of the question for topological spaces. But there is also a positive result for triangulated categories: a morphism $f \colon X \to Y$ is an isomorphism if and only if any distinguished triangle $X \to Y \to Z \to X[1]$ has $Z \cong 0$. See Tag 05QR. -This in particular applies to the homotopy category $K(\mathcal A)$ of chain complexes with values in an additive category $\mathcal A$, which is a triangulated category by Tag 014S. Applying this to the mapping cone in $K(\mathcal A)$ shows that $C(f)$ is contractible if and only if $f$ is a homotopy equivalence. -This also applies to the stable homotopy category of spectra, where it implies that a map of topological spaces $f \colon X \to Y$ with contractible mapping cone is a stable homotopy equivalence. -In examples like David White's answer, you only need to go up to the first suspension $\Sigma X$, by construction (but see also this answer). I'm not sure if something like this is supposed to be true in general, but I'm guessing it's more complicated.<|endoftext|> -TITLE: What does the "category" of $(\infty,1)$ category look like. -QUESTION [6 upvotes]: One knows that in higher category theory, the category of $(\infty,n-1)$ categories is naturally an $(\infty,n)$ category ,(I use the word category to mean category in the correct weakened sense). When the category of $(\infty,1)$ categories is regarded as a weakened kan complex, we may regard these objects as a full subcategory of simplicial sets. This is a category in the strict sense. One ought to expect that associativity of the maps between the weakened kan complex be some sort of weakened associativity. The question is: is this weakened associativity there, and if so how is it understood? - -REPLY [11 votes]: You can see the collection of $(\infty,1)$-categories as forming themselves an $(\infty,1)$-category, which is sufficient to see where weak associativity shows up: There are many models for the intuiti9ve concept of $(\infty,1)$-category, the simplest is that of a usual 1-category endowed with a class of weak equivalences (see Barwick/Kan's "Relative Categories: Another model for the homotopy theory of homotopy theories" here). -With the weak Kan complexes - together with the notion of equivalence between them - you happen to have found a strictly associative model for the $(\infty,1)$-category of $(\infty,1)$-categories. You can transform it into different other models, e.g. into simplicially enriched categories or quasicategories or Segal categories, as exposed e.g. in Bergner's "Three models for the homotopy theory of homotopy theories" (available here). -The Segal category and the quasicategory of $(\infty,1)$-cats do no longer have strict associativity and the fact that they are equivalent descriptions of the $(\infty,1)$-cat of $(\infty,1)$-cats reflects that the strict associativity in your model was an accident and not an essential feature... -Edit (in response to the comment) About the significance of having a strict model: Well, different models have different advantages. Your strict one is certainly good for computing compositions of functors between $(\infty,1)$-cats. The quasicategory of $(\infty,1)$-cats on the other hand is e.g. a better model to relate the $(\infty,1)$-cat of $(\infty,1)$-cats to other $(\infty,1)$-cats - examples are the relations between the quasicategories of (small) $(\infty,1)$-cats, presentable $(\infty,1)$-cats and stable $(\infty,1)$-cats given in Lurie's "Higher Topoi" and in DAG 1 (now "Higher Algebra"): There are $(\infty,1)$-adjunctions between these - e.g. between $(\infty,1)$-cats and stable $(\infty,1)$-cats given by taking spectra in an $(\infty,1)$-cat, forgetting the stability of a stable $(\infty,1)$-cat, respectively - and these facts would be hard to express using your model.<|endoftext|> -TITLE: countable union of closed subschemes over uncountable field -QUESTION [9 upvotes]: I am looking for a reference for the following well-known fact: -Let $k$ be an uncountable field, and let $X$ be a $k$-variety. Let $Z_1, Z_2, \dots \subseteq X$ be proper closed subschemes. Then $\bigcup Z_i(k) \neq X(k)$. -Thanks! - -REPLY [11 votes]: Suppose $\dim X>0$ and $k$ is algebraically closed and uncountable. Moreover, if a "variety" is not necessarily irreducible, the $Z_i$ are supposed to have positive codimension in $X$ (otherwise one could take the irreducible components of $X$). -As in MP's answer, one can suppose $X$ is affine. By Noether's Normalization Lemma, there exists a finite surjective morphism $p: X\to \mathbb A^m_k$ with $m=\dim X$. Let $Y_i=p(Z_i)$. This is a closed subset -of $\mathbb A^m_k$ of positive codimension. Moreover $\mathbb A^m_k(k)=\cup Y_i(k)$ because $k$ is algebraically closed (which implies that $Y_i(k)=p(Z_i(k))$). As $k$ is uncountable, there exists a hyperplane $H$ in $\mathbb A^m$ not contained in any $Y_i$ (note that $H\subseteq Y_i$ is equivalent to $H=Y_i$). So by induction on $m$ we are reduced to the case $m=1$, and the assertion is obvious. -Without the hypothesis $k$ algebraically closed, one can show similarly that $X\ne \cup_i Z_i$. This is Exercise 2.5.10 in my book. EDIT In fact this statement is trivial because the generic points of $X$ don't belong to any of the $Z_i$'s. But the proof shows that the set of closed points of $X$ is not contained in $\cup_i Z_i$.<|endoftext|> -TITLE: torsion group of the multiplicative group of a field -QUESTION [11 upvotes]: Let $F$ be any field of zero characteristic, $F^{\ast}$ its multiplicative group and $T$ is the torsion group. -Is it true that $T$ is a direct summand for $F^{\ast}$? - -REPLY [14 votes]: This was a problem that was asked by Fuchs in his book "Abelian groups" (1958). It was first solved in negative by P. M. Cohn in "Eine Bemerkung uber die multiplikative Gruppe eines Korpers", Arch. Math. (Basel) 13 (1962) 344-48. (MR0146252). Later W. May gave a counterexample as an algebraic extension of $\mathbb Q$, in "Multiplicative groups of fields", Proc. London Math. Soc. (3) 24 (1972), 295–306. (MR0294490)<|endoftext|> -TITLE: Concrete example of $\infty$-categories -QUESTION [26 upvotes]: I've seen many different notions of $\infty$-categories: actually I've seen the operadic-globular ones of Batanin and Leinster, and the opetopic, and eventually I'll see the simplicial ones too. Although there are so many notions of $\infty$-category, so far I've only seen the following examples: - -$\infty$-groupoids as fundamental groupoids topological spaces; -$(\infty,1)$-categories, mostly via topological example and application in algebraic geometry (in particular in derived algebraic geometry); -strict $(\infty,\infty)$-categories, and their $n$-dimensional versions, for instance the various categories of strict-$n$-categories (here I intend $n \in \omega+\{\infty\}$). - - -There are other examples of $\infty$-categories, especially from algebraic topology or algebraic geometry, but also mathematical physics and computer science and logic? - In particular I am wondering if there's a concrete example, well known, weak $(\infty,\infty)$-category. - -(Edit:) after the a discussion with Mr.Porter I think adding some specifications may help: - -I'm looking for models/presentations of $\infty$-weak-categories for which is possible to give a combinatorial description, in which is possible to make manipulations and explicit calculations, but also $\infty$-categories arising in practice in various mathematical contexts. - -REPLY [6 votes]: A concrete example of a weak $\infty$-category, but not well studied abstractly, is the cubical singular complex of a space, preferably with the connections introduced in our 1981 JPAA paper with Philip Higgins. The clear advantage of the cubical setup is the command of multiple compositions, using an easy matrix notation. Thus one can express for the diagram - (source) -that the big square is the composition of the little squares, by simply writing something like $\alpha=[\alpha_{ij}]$. (How does one do that simplicially, or globularly?) This and higher dimensional versions are useful in expressing algebraic inverses to subdivision, a useful tool in local-to-global problems. Because higher groupoids are nonabelian, unlike higher groups, one can also obtain nonabelian results in algebraic topology. -The book Nonabelian algebraic topology: filtered spaces, crossed complexes, cubical homotopy groupoids (EMS, (2011), pdf available from here) has a large number of uses of algebraic and geometric (e.g. homotopical) conclusions derived from rewriting such multiple arrays. The main results were conjectured and eventually proved by cubical methods.<|endoftext|> -TITLE: Drawing natural numbers without replacement. -QUESTION [24 upvotes]: Suppose we start with an initial probability distribution on $\mathbb{N}$ that gives positive probability to each $n$. Let's call this random variable $X_1$ so we have $P(X_1=n)=p_{1,n}>0$ for all $n\in\mathbb{N}$. $X_1$ wil be the first draw from $\mathbb{N}$. For the next draw $X_2$ we define a new distribution on $\mathbb{N}\setminus\{ X_1 \}$ by rescaling the remaining probabilities so they add up to 1. So $p_{2,X_1}=0$ and $p_{2,n}=\frac{p_{1,n}}{1-p_{1,X_1}}$ for $n\neq X_1$. Continuing in this manner we get a stochastic process (certainly not Markov) that corresponds to drawing from $\mathbb{N}$ without replacement. My question is whether this process has ever been studied in the literature. In particular, I'm wondering if a clever choice of the initial distribution could result in tractable expressions for the distributions of $X_n$ for large $n$. - -REPLY [4 votes]: Here are some preliminary computations. Assume the reference distribution is $(p(n))$. For every finite subset $I$ of $\mathbb N$, introduce the finite number $r(I)\ge1$ such that -$$ -\frac1{r(I)}=1-\sum_{k\in I}p(k). -$$ -Obviously, $P(X_1=n)=p(n)$ for every $n$. Likewise, -$P(X_2=n)=E(p(n)r(X_1);X_1\ne n)$ hence -$$ -P(X_2=n)=p(n)(\alpha-p(n)r(n)),\qquad -\alpha=\sum\limits_kp(k)r(k). -$$ -This shows that $X_1$ and $X_2$ are not equidistributed (if they were, $\alpha-p(n)r(n)$ would not depend on $n$, hence $p(n)$ would not either, but this is impossible since $(p(n))$ is a measure with finite mass on an infinite set). -One can also compute the joint distribution of $(X_1,X_2)$ as -$$ -P(X_1=n,X_2=k)=p(n)r(n)p(k)[k\ne n], -$$ -and this allows to expand -$$ -P(X_3=n)=E(p(n)r(X_1,X_2);X_1\ne n,X_2\ne n), -$$ -as the double sum -$$ -P(X_3=n)=p(n)\sum_{k\ne n}\sum_{i\ne n}[k\ne i]r(k,i)p(k)r(k)p(i), -$$ -but no simpler or really illuminating expression seems to emerge.<|endoftext|> -TITLE: Continuous extensions reals and to p-adic numbers -QUESTION [7 upvotes]: Assume $f\colon \mathbb Q\to \mathbb Q$ is a function which admits continuous extensions - -$f_0\colon\mathbb R\to \mathbb R$ and -$f_p\colon \mathbb Q_p\to \mathbb Q_p$ for each prime $p$. - - -Is it true that $f$ is a polynomial? - -I guess the answer is no, but I do not see a counterexample. - -REPLY [14 votes]: The answer is no, and one can essentially use the same construction as in the answer: -Is a real power series that maps rationals to rationals defined by a rational function? -Specifically, enumerate the non-zero rationals $\{r_1,r_2, \ldots\}$ in some way. Now consider the function: -$$f(x) = \sum_{n=1}^{\infty} c_n x^{n^2} \prod_{i=1}^{n} (x - r_i).$$ -If $c_n \in \mathbf{Q}$, then this is a well defined function from rationals to rationals. -On the other hand, $f(x)$ converges to an analytic function in $\mathbf{Q}_v$ if and only if the coefficients of this power series converge to zero fast enough. -Since the coefficients of the power series in the range $k = n^2$ to $k < (n+1)^2$ -are simply the cofficients of $c_n x^{n^2} \prod_{i=1}^{n} (x - r_i)$, this can be ensured -by forcing these coefficients to be very highly divisible by the first $n$ primes, and small in the archimedean sense (by including a very very large prime in the denominator).<|endoftext|> -TITLE: Relationship between the focal locus and the cut locus -QUESTION [5 upvotes]: I am seeking -clarification of -the relationship between the -focal locus -and the -cut locus -of a curve $C$ in $\mathbb{R}^2$, and -of a surface $S$ in $\mathbb{R}^3$. -Essentially my question is, - -Under what conditions is the focal locus and the cut locus identical, - when do they differ, and when they differ, how do they differ. - -For example, it believe the two coincide for a sphere -(in any dimension): both are simply the center point of the sphere. -It may be that these issues are primarily definitional -rather than substantive. Let me offer the definitions -with which I am working. -Cut Locus. -Generally the -cut locus -is defined on a Riemannian manifold with respect to a point. -But instead I want to define the cut locus of a set in $\mathbb{R}^n$. -Let me follow the definition of -Franz-Erich Wolter, who wrote his Ph.D. thesis on the topic: - -"The cut locus $C_A$ of a closed set $A$ in the Euclidean space $E$ - is defined as the closure of the set containing - all points $p$ which have at least two shortest paths to $A$." - -(This is quoted from reference (1) below.) -This definition is in accord with that of the -medial axis, -extensively explored in computer science. -Focal locus. -I am having more difficulty locating a widely accepted definition -of the focal locus. -Let me follow Thorpe: - -"The focal locus of a plane curve $C$ is the locus of the centers of curvature and is often called the evolute of $C$." ... - "The set of all focal points along all normal lines to an $n$-surface $S$ - in $\mathbb{R}^{n+1}$ is called the focal locus of $S$." - ... - Let $\phi$ be a parametrized $n$-surface, - and let $\beta$ be "a unit-speed parametrization of the line normal to Image $\phi$ at $\phi(p)$. - A point $f$ is said to be a focal point of $\phi$ along $\beta$ - if $f = \beta(s_0)$ where $s_0$ is such that the map ... - $\phi(q) + s_0 N^\phi(q)$ is singular (not regular) at $p$" [where $N^\phi(q)$ is - the normal at $q$]. - -(These quotes are from reference (2) below.) -One aspect of the focal locus that confuses me is that there -is a notion of focal surfaces, which derive from -"the reciprocal of the principal curvatures," -as described in (3). Here there are two surfaces, -as opposed to one focal locus, as depicted in this intriguing figure: - -                   - - -It may be that there are references that -would resolve my definitional confusions, in which case -pointers would be much appreciated. Thanks! - -References. - - -Franz-Erich Wolter. -"Cut Locus and Medial Axis in Global Shape Interrogation and Representation." -MIT Ocean Engineering Design Laboratory Memorandum 92-2. -December 1993. -PDF. - -John A. Thorpe. -Elementary Topics in Differential Geometry. -Springer, 1979. -Google books. Quote from p.137. - - -Jingyi Yu, Xiaotian Yin, Xianfeng Gu, Leonard McMillan, and Steven Gortler. -"Focal Surfaces of Discrete Geometry." 2007. -Proceedings of the 5th Euro -Graphics Symposium on Geometry Processing. - -REPLY [3 votes]: I believe that the focal locus is the same thing as the conjugate locus in Riemannian geometry. Given a codimension 1 submanifold $S \subset \mathbb{R}^n$, the exponential map $S \times \mathbb{R} \rightarrow \mathbb{R}^n$ is given by $(x, t) \mapsto x + t\gamma(x)$, where $\gamma$ is the Gauss map. The cut locus is the closure of all points in $\mathbb{R}^n$ that have more than one pre-image. The focal points is the closure of all points where the map is not a diffeomorphism. -Given any point where the curvature $\kappa$ is nonzero on a smooth curve in the plane, there is a corresponding point on the focal locus at distance $1/|\kappa|$ on the side of the curve that is inwardly curved. -I believe that the focal locus is always contained in the cut locus. -ADDED: (Corrected definition of map above)...If you differentiate the exponential map (the one I define above), then since the differential of the Gauss map is the second fundamental form, call it $A$, then you see that if a point $y$ lies in the focal locus, there exists $x \in S$, $t \in \mathbb{R}$, and a nonzero $v \in \mathbb{R}^n$ such that $v + tAv = 0$. I neglected to say above that the focal locus corresponds to the closest point on either side of $S$ along the geodesic normal to $x$ where this equation holds. Therefore, there is a focal point on the half line where $t > 0$ only if there is a negative principal curvature and the focal point is at distance $-1/\kappa$, where $\kappa$ is the negative principal curvature with largest magnitude. There is a focal point on the other half line only if there is at a positive principal curvature, and it is at distance $1/\kappa$, where $\kappa$ is the largest positive principal curvature.<|endoftext|> -TITLE: Does ZF prove that topological groups are completely regular? -QUESTION [17 upvotes]: Let $\mathbf{G} = \langle G,\cdot,\mathcal{T}\;\rangle$ be a topological group. Let $\mathbf{e}$ be the identity element of $\langle G,\cdot \rangle$. -Assume $\{\mathbf{e}\}$ is closed in $\langle G,\cal{T}\;\rangle$. Then, I have managed to convince myself that: - -ZF proves that $\langle G,\cal{T}\;\rangle$ is regular Haudorff. -ZF + (Dependent Choice) proves that $\langle G,\cal{T}\;\rangle$ is completely regular. - -My questions are: - -Are those right? -Does ZF prove that $\langle G,\cal{T}\;\rangle$ is completely regular? -If no to question 2, does assuming one or more of following suffice for ZF to conclude that $\langle G,\cal{T}\,\rangle$ is completely regular? - -$\mathbf{G}$ is two-sided complete -$\langle G,\cdot \rangle$ is abelian -Countable Choice - -REPLY [3 votes]: I'm not sure about the proof I gave, But as I checked, it didn't use full AC, but as Asaf mentioned in a comment it uses DC. The following theorem is due to Pontrjagin. See Book by Montgomery and Zippin(Page 29). -I will give a sketch of proof. -Note: Maybe it's needed to add some separation axiom to the following theorem. please edit it, if needed. -Theorem: Every $T_{0}$ topological group is completely regular. -Proof. -It's enough to prove that a given topological group $(G,*)$ is completely regular at $e$. Let $F$ be a closed set not containing $e$. Put $O=F^{c}$. Choose symmetric open neighborhoods of $e$, $U_{n}$ By continuity of $*$, such that $U_{0}=O$(w.l.o.g assume $O$ is symmetric) $U_{n+1}^2 \subseteq U_{n} \cap O ~~~~n=0,1,2...$. -Now for rational numbers of the form $r=\frac{k}{2^n}~~k \in \{1,2,3,...2^n \},~~ n \in \{0,1,2,...\}$ inductively define open neighborhood $V_{r}$ of $e$ such that: -1) $V_{\frac{1}{2^n}}=U_{n}~~~~~\forall n$ -2) $V_{\frac{2k}{2^{n+1}}}=V_{\frac{k}{2^n}}$ -3) $V_{\frac{2k+1}{2^{n+1}}}=V_{\frac{1}{2^{n+1}}}V_{\frac{k}{2^n}}$ -The definition does not depend to the representation of $r$ and the family $V_{r}$ has the following properties: -4) $V_{\frac{1}{2^n}}V_{\frac{m}{2^{n}}} \subseteq V_{\frac{m+1}{2^n}}~~~~~~~~~m+1 \leq 2^n$ -5) $V_{r} \subseteq V_{s}~~~~~$if $~r -TITLE: Why "Classification" of 4 manifolds is NOT possible? -QUESTION [15 upvotes]: I know classification of 2 manifolds and geometrization for 3 manifolds. -Why for dimension great or equal to 4, this task become impossible? -edit: Or should I ask "why geometrization won't be possible for 4 or higher dimension?" - -REPLY [34 votes]: As pointed out in a comment by Autumn Kent to Allen Knutson's answer, -the problem is a bit more subtle than it may appear. In order to -prove that the homeomorphism problem for compact 4-manifolds, say in -the topological category, is recursively unsolvable, it is not enough -to know that (1) every finitely presented group can be realized as -the fundamental group of some compact 4-manifold, and (2) the isomorphism problem for finitely presented groups is recursively unsolvable. -Instead, what you do is give a construction which to any finite presentation $< S | P >$ of a group associates a 4-manifold $M(S,P)$ in such a way that $\pi_1(M(S,P))$ is isomorphic to the group defined by -the presentation $< S | P >$, and moreover two such manifolds are homeomorphic if and only if they have isomorphic fundamental groups. -Then you have constructed a class of 4-manifolds for which the homeomorphism problem is equivalent to the isomorphism problem for -finitely presented groups, and therefore unsolvable. -About "geometrization for manifolds of dimension 4 or higher", well -as far as I know there is no theorem which says it is impossible. It -depends on what you mean by `geometric structure', and what you -want those structures to do for you.<|endoftext|> -TITLE: A question about MathSciNet etiquette -QUESTION [60 upvotes]: Hello, -Recently, a colleague of mine pointed me to a MathSciNet review of one of my papers that is completely off the mark - it is not negative or anything like that, but it grossly misrepresents the contents of the paper (when describing the origins of the techniques and questions in the paper, for instance, as well as the "position" of the results within the current litterature and the meaning of the results themselves). -I'm not sure what I should do - actually, I probably won't do anything because the paper seems pretty inconsequential and the quality of the review most likely does not matter much. Still, the same thing could have happened with a paper I truly care about, and this led me to wondering what the proper behaviour is: should I contact the reviewer and ask him/her to retract his/her review (explaining why, of course)? Should I contact MathSciNet and let them know that I believe the review is incorrect? The first option raises some "diplomatic" problems, while the second one seems to me both to be abrupt and to waste several people's time... I think, if pressed to act, I would choose the first course of action, but I'd be grateful for any suggestions (e.g on how to say "you absolutely mangled that review!" without being rude..) -Final note: I'm asking this question anonymously because I don't have that many papers and it would be easy to identify the reviewer by looking at my MathSciNet profile, and I'm not out to embarass anyone. -Thanks for your help - if the question is inappropriate for this site then please close, of course! - -REPLY [11 votes]: Thanks to everyone for the advice. It seems that the question has stopped attracting new reactions, so I'll try to summarize what I took from the discussion. - -There is no way (nor should there be) or hiding the fact that I'm the one complaining about the review's quality - anyway, the reviewer signed his/her review so it would be highly discourteous to try to stay anonymous. -In all probability the reviewer tried to do a fair job and simply failed; it does not seem unlikely that he/she would prefer to be told that the review is off the mark and have a chance to make it right. This helps prevent his/her name to be attached to an erroneous review. -In view of all this, it seems at least courteous to begin by contacting the reviewer and explaining why the review seems wrong to me, and then give him/her a chance to correct it on his/her own (or to explain me why I misunderstood my paper, of course). -In the course of such communication, one should as much as possible avoid being too directly critical of the original review (and remember that the reviewer wrote it on what is essentially his/her own free time!). -If the above option does not produce the desired results, then the way to go (if sufficiently motivated!) is to contact mathrev@ams.org, explain the issues with the review, and leave it to them (at least one contributor thinks that this should be the first step, without any direct contact with the reviewer, but I do not see the downside of contacting the reviewer first and MathSciNet second). - -Also, note that a somewhat similar question may be found at How do I fix someone's published error? , and that some of the advice there is relevant to the problem at hand. -Again, thanks to everyone for the advice! -Note possibly to be removed at a later point: I plan to accept this answer, so that it appears first to anyone interested in a similar problem at a latter date. I'll wait a few days before doing so, that way you can let me know if this clashes with this forum's usual rules.<|endoftext|> -TITLE: How to find Casimir operators? -QUESTION [8 upvotes]: Given a general Lie algebra, is there a general procedure to find all its Casimir operator? - -REPLY [3 votes]: I'm assuming you're thinking of some specific matrix representation $X_i \in \mathfrak{g}$ (let's assume it's the defining representation). Compute the Killing form, $\kappa_{ij} \doteq Tr (X_i\cdot X_j)$ (actually usually this is defined in the adjoint representation, but any faithful rep will do). The quadratic Casimir is then simply $ X_i \kappa_{ij} X_j$ (Einstein convention). -Other Casimirs can be obtained from the characteristic (secular) equation: define $X(\omega) = \omega^i X_i$. The characteristic equation is $\det\left( X(\omega) - \lambda I \right) = \sum\limits_{j} (-\lambda)^{N-j} \phi_j(\omega) \equiv 0 $ ($N$ is the matrix dimension, and/or the dimension of the Lie algebra if you're using the adjoint representation). If you now perform the substitution $\omega^i \to X_i$ in the coefficients $\phi_j (\omega)$, you get Casimir invariants $\phi_j (X)$! -It might seem like the higher the representation, the more invariants you get, but in fact all the invariants can be expressed in terms of the fundamental invariants of the defining representation. I can't think of many references right at the moment, but e.g. Gilmore: Lie groups, physics and geometry pp. 140 has a nice explanation. Also, google the boldface texts above.<|endoftext|> -TITLE: Seeking proof for linear algebra constraint problem. -QUESTION [13 upvotes]: Given a symmetric real matrix with a zero diagonal $M$, I am trying to find a diagonal matrix $D$, such that the matrix $M + D$ is positive definite, and $(M+D)^{-1}$ has a diagonal consisting of all 1's. This problem looks vaguely like a semidefinite programming problem, except that both the matrix $(M+D)$ and it's inverse have linear constraints. Overall, the system has as many constraints as variables. -Based on small scale numerical testing, it strongly appears that there is always a unique solution. I've implemented an algorithm (which is $O(n^6)$, $n$ being the size of the matrix), that works by constructing a second matrix $X$, and minimizing $||(M+D) X - I||$ with respect to $D$ and then with respect to $X$ in an alternating fashion. Note that I am using the Frobenius norm here. Given the proper initialization, such the $(M+D)$ is positive definite, this usually appears to converge, and each step is simply a quadratic minimization. -That said, I have no proof that there is a unique solution for $D$, or that the algorithm above works in the general case, and moreover I have a strong intuition that there is an algorithm with is closer to $O(n^3)$. -What I seeks is proof or a theoretical justification that the solution to $D$ -is unique (or of course a counterexample). Even better would be a provably polynomial time algorithm to find $D$. -My approach for a proof up till this point has been along the lines of finding some error function $f(X)$, $X = M + D$, such as $f(X) = \sum_i ((X^{-1})_{ii} - 1)^2$. This function (and lots of other variants) have a minimum at the desired solution. My hope was to then show that the function is convex over all positive definite matrices $X$. However I have not been able to accomplish this so far. -Edit: For the $f(X)$ given above I have found a number of counterexamples to it's convexity, although perhaps the overall method is still salvageable with a different error function. -Edit: Some additional facts I've been able to show (in part with help from the comments) -The set of all positive definite matrices $X = M + D$ is clearly a convex set (since it is the intersection of two convex sets, the positive definite matrices and the set of all matrices with non-diagonal elements $M$). -Moreover, the set of all $X$ above such that all the diagonals elements of $X^{-1}_{ii} \le 1$ is also a convex set. This follows from the convexity of $e^T X^{-1} e$, and the statement above. The solution in question is clearly on the boundary of this set. - -REPLY [5 votes]: (Edit: my original answer was perhaps not clear enough, let me try to improve it). -First some notation: for a matrix $x$, let me denote by $E(x)$ the diagonal matrix with the same diagonal as $x$: if $x=(x_{i,j})_{i,j\leq n}$, $E(x) = (x_{i,j}\delta_{i,j})_{i,j \leq n}$. Equivalently, $E$ is the orthogonal projection on the diagonal matrices when you consider the usual euclidean structure on $M_n$. If will also write $x>0$ to mean $x$ is symmetric positive definite. -You are asking whether the map $f:x \mapsto x^{-1} - E(x^{-1})$ is a bijection from its domain $D=\{x \in M_n(\mathbb R), x>0\textrm{ and }E(x)=1\}$ to its image $I=\{x \in M_n, x=x^*\textrm{ and }E(x)=0\}$. And the answer is yes. I prove first that $f$ is injective, and then that it is surjective. - -f is injective -In fact let me prove the following fact, which is equivalent to the injectivity of $f$. - -Let $x$ be a positive matrix. If $d$ is a diagonal matrix such that $x+d$ is positive and such that $x^{-1}$ and $(x+d)^{-1}$ have the same diagonals, then $d=0$. - -Proof: Since $d$ is diagonal, the trace of $d\left(x^{-1} - (x+d)^{-1}\right)$ is zero. But one can write this expression as \[dx^{-1/2}\left(1- (1+x^{-1/2}dx^{-1/2})^{-1}\right)x^{-1/2},\] -so that taking the trace and denoting by $a=x^{-1/2}dx^{-1/2}$, we get $0= Tr(a(1-(1+a)^{-1}))$. -If $\lambda_1,\dots,\lambda_n$ are the eigenvalues of $a$, the condition $x+d$ positive becomes $1+\lambda_i >0$, and the last equality becomes $0 = \sum \lambda_i(1-1/(1-\lambda_i)) = \sum \lambda_i^2/(1+\lambda_i)$, which is possible only if the $\lambda_i$ are all zero, i.e. $a=0$, i.e. $d=0$. - -** f is surjective ** -The surjectivity is true just for topological reasons. More precisely, to prove that $f$ is surjective, it is enough to prove that it is continuous, open and proper (because this would imply that the image is an open and closed subset of $I$, and hence everything since $I$ is connected). The continuity is obvious. $f$ is even differentiable, and the differential is explicitely computable and easily seen to be invertible at every point, so that $f$ is indeed open. It remains to check that it is proper. -The proof I have is not completely obvious, maybe I am missing something. Let me only sketch it. Take a sequence $x_k \in D$ that escapes every compact subset of $D$. Since $\|x\|\leq n$ for all $x \in D$, we have that $u_k=\|x_k^{-1}\|\to \infty$ (I consider the operator norm, and the inequality $\|x\|\leq n= Tr(x)$ is because the norm of $x>0$ is its largest eigenvalue, whereas its trace is the sum of its eigenvalues). We want to prove that $\|f(x_k)\|\to \infty$. Assume for contradiction that this is not the case, and that $\|f(x_k)\|\leq C$ for all $k$. We will get a contradiction through a careful study of the spectral decomposition of $x_k$. -Let $\xi_k$ be a sequence of unit eigenvectors of $x_k$ relative to the smallest eigenvalue of $x_k$, i.e. $x_k \xi_k = 1/u_k \xi_k$. Now the key observation: the assumption that $\|f(x_k)\|\leq C$ implies that, for all diagonal matrix $d$ with $1$ or $-1$ on the diagonal, the distance from $d \xi_k$ to the space $F_k$ spanned by the eigenvectors of $x_k^{-1}$ relative to the eigenvalues in an interval $[u_k/2,u_k]$ goes to zero. For a proof, consider the random diagonal matrix $d$ in which the diagonal entries are iid random variables uniform in $\{-1,1\}$, so that $E(x) = \mathbb E (d x d)$ (hoping there will be no confusion between $E$ and $\mathbb E$). Then $\langle f(x_k) \xi_k,\xi_k\rangle = \mathbb E ( u_k - \langle x_k d \xi_k, d \xi_k\rangle)$. The lhs of this equality is by assumption smaller than $C$. On the rhs, $u_k - \langle x_k d \xi_k, d \xi_k\rangle \geq 0$ because $d \xi_k$ is a unit vector. This implies that $u_k - \langle x_k d \xi_k, d \xi_k\rangle \leq 2^n C$ for any diagonal matrix with $\pm 1$ on the diagonal. But now use the fact that, for $x>0$ in $M_n$, if a unit vector $\xi$ in $\mathbb R^n$ satisfies $\langle x \xi,\xi\rangle \geq \|x\|-\delta$, then $\xi$ is at distance less than $\sqrt{2\delta/\|x\|}$ from the space spanned by the eigenvectors of $x$ relative to eigenvalues in the interval $[\|x\|/2,\|x\|]$ (hint for a proof: consider the decompostion of $\xi$ in an orthonormal basis of eigenvectors of $x$). Here if $\epsilon_k = \sqrt{2^{n+1} C/ u_k}$, we have indeed proved that $d \xi_k$ is at distance less than $\epsilon_k$ from $E_k$ for any diagonal matrix with $\pm 1$ on the diagonal. -I now claim that there is a vector $\eta_k$ in the canonical basis of $\mathbb R^n$ at distance less than $\sqrt n \epsilon_k$ from $E_k$. This will conclude the proof since it will in particular imply that $\langle x_k \eta_k,\eta_k\rangle \to 0$, whereas the assumption $E(x_k)=1$ implies that $\langle x_k \eta_k,\eta_k\rangle = 1$, a contradiction. To prove the claim, let $i$ be such that the $i$-th coordinate of $\xi_k$ is larger than $1/\sqrt n$ in absolute value. Observe that $\xi_k(i) e_i$ is the expected value of $d \xi$, where $d$ is the same random matrix as above, but conditionned to $d_i = 1$. This implies that $\xi_k(i) e_i$ is at distance at most $\epsilon_k$ from $E_k$, which proves the claim. -A remark I do not like this proof, since it really relies on finite-dimensional techniques. In particular, it does not extend to general von Neumann algebras (whereas the injectivity part does). I would prefer a more direct proof.<|endoftext|> -TITLE: A necessary condition for S4-completeness? -QUESTION [6 upvotes]: It is well-known that the modal logic S4 is complete with respect to the class of all finite quasi-trees (where we interpret the $\Box$ modality as topological interior, and topologize a quasi-tree with the up-set topology). It is also well-known that p-morphisms (open, continuous surjections) preserve modal validity. Thus, for any space $X$, the existence of p-morphisms from $X$ onto every finite quasi-tree is a sufficient condition for $X$ to be S4-complete. This technique can be used to establish, for example, McKinsey and Tarski's famous result that S4 is the logic of any dense-in-itself, metrizable space. -My question is: - -Is this condition also necessary? Said differently: is there a space $X$ and a finite quasi-tree $Q$ such that $X$ is S4-complete but there exists no p-morphism $\rho: X \to Q$? - -This seems like a natural question to ask, but I haven't had much luck in finding any discussion about it. Even just a pointer to the right body of literature would be very much appreciated. - -Addendum -Here I'll define my terms a little more carefully, and spell out the translation of my question in terms of the more standard Kripke semantics. -Recall that quasi-orders are sets equipped with reflexive, transitive binary relations, which is precisely the class of Kripke frames corresponding to S4. A quasi-order $Q = (Q,\leq)$ is called a quasi-tree if $Q/\sim$ is a tree, where $\sim$ is the equivalence relation on $Q$ defined by -$$x \sim y \iff x \leq y \textrm{ and } y \leq x.$$ -As mentioned in the comments, there is a correspondence between quasi-orders and Alexandrov spaces, one direction of which is given by topologizing quasi-orders with the up-set topology. There is also a notion of a p-morphism between quasi-orders, nicely outlined by Wikipedia. A p-morphism between quasi-order corresponds to an open, continuous map between the corresponding Alexandrov spaces. -I use the phrase "$X$ is S4-complete" (perhaps somewhat idiosyncratically?) to mean that every formula validated by $X$ is provable in S4; equivalently, $X$ refutes all non-theorems of S4. It is known that if $Q$ is any quasi-order and for each finite quasi-tree $Q_{t}$ there exists a surjective p-morphism $\rho_{t}: Q \to Q_{t}$, then $Q$ is S4-complete. One can then ask: - -Is the converse true? Does every S4-complete quasi-order Q admit maps $\rho_{t}$ as above? - -If not, then a counter-example can be "lifted" into the topological setting, thus answering my original question. However, a positive answer to this question does not immediately resolve the topological version since the quantification in the topological version is over all spaces, rather than just the Alexandrov spaces. Nonetheless, I would be interested in an answer (or even a hint at an answer) to either question. - -REPLY [2 votes]: S4 is complete with respect to a Kripke frame or general frame or topological frame $F$ if and only if $F$ is an S4-frame and for every finite rooted S4-frame $G$, there exists a p-morphism of a generated subframe of $F$ onto $G$. -The left-to-right implication follows from the existence of Fine’s frame formulas: there is a formula $\alpha_G$ such that for any K4-frame $H$, $\alpha_G$ is refutable in $H$ if and only if there exists a p-morphism of a generated subframe of $H$ onto $G$. One way of constructing $\alpha_G$ is as follows. Assume that $r$ is a root of $G$, and let $R$ be the accessibility relation of $G$. We put -$$\alpha_G=\Box^+\biggl(\bigwedge_{\substack{i,j\in G\\\\i\ne j}}(p_i\to\neg p_j)\land\bigwedge_{\substack{i,j\in G\\\\i\mathrel Rj}}(p_i\to\Diamond p_j)\land\bigwedge_{i\in G}\Bigl(p_i\to\Box\bigvee_{i\mathrel Rj}p_j\Bigr)\biggr)\to\neg p_r,$$ -where $\Box^+\phi=\phi\land\Box\phi$ (which is equivalent to $\Box\phi$ in S4; the formula above works for K4 as well). Let $\models$ be a valuation in $H$ such that $x\not\models\alpha_G$ for some $x\in H$. Let $H_x$ be the generated subframe of $H$ rooted at $x$. For every $y\in H_x$, there exists a unique $i\in G$ such that $y\models p_i$; put $f(y)=i$. Then $f\colon H_x\to G$ is a p-morphism such that $f(x)=r$. Conversely, given such a p-morphism, one can construct a valuation refuting $\alpha$ by reversing the process. -Now, if $G$ is an S4-frame, it is a p-morphic image of itself, hence $\alpha_G$ is not valid in $G$, and a fortiori it is not an S4-tautology. Thus, any $H$ wrt which S4 is complete must also refute $\alpha_G$, hence there exists a p-morphism from a generated subframe of $H$ onto $G$. -Since every finite rooted S4-model is a p-morphic image of a finite quasi-tree, one can restrict attention to such $G$’s. -I’m not quite familiar with the topological semantics of S4, but I suppose the criterion translates to something to the effect of: S4 is complete wrt a space $X$ iff for every finite quasi-tree $G$, there exists an open subset $U\subseteq X$ and a p-morphism (whatever that means when the space is not ordered) of $U$ onto $G$.<|endoftext|> -TITLE: Torsion points of abelian varieties in the perfect closure of a function field -QUESTION [18 upvotes]: The following is a problem, which was recently brought to my attention by H. Esnault and A. Langer. -Let $K$ be the function field of a smooth curve over the algebraic closure $k$ of the finite field ${\bf F}_p$ (where $p>0$ is a prime number). -Let $A$ be an abelian variety over $K$ and suppose that the $K|k$-image of $A$ is trivial (ie there are no non-vanishing $K$-homomorphisms from $A$ to an abelian variety over $K$, which has a model over $k$). -Question : is it true that $\#{\rm Tor}(A(K^{\rm perf}))<\infty$ ? (*) -Here $K^{\rm perf}$ is the maximal purely inseparable extension of $K$ and -${\rm Tor}(A(K^{\rm perf}))$ is the subgroup of $A(K^{\rm perf})$ consisting of -elements of finite order. -To put things in context, recall that by the Lang-Néron theorem, we have $\#{\rm Tor}(A(K))<\infty$. -Furthermore, one can show using a specialization argument that -$\#{\rm Tor}(A(K^{\rm perf}))<\infty$ if $k$ is replaced by a finite extension of ${\bf F}_p$; in this case, the assumption on the $K|k$-image can actually be dropped. -Notice also that the inequality in question (*) is actually equivalent to the inequality -$\#{\rm Tor}_p(A(K^{\rm perf}))<\infty$, where ${\rm Tor}_p(A(K^{\rm perf}))$ is the subgroup of $A(K^{\rm perf})$ consisting of the elements, whose order is a power of $p$. -This follows from the fact the multiplication by $n$ morphism is étale if $p\not|n$. -Question (*) has a positive answer if $A$ is an elliptic curve by the work of M. Levin, who -proves a much stronger result (see "On the group of rational points...", Amer. J. Math. 90 (1968)). -The question (*) is in part complementary to the following other question in MO : -Etale endomorphisms of abelian varieties in positive characteristic - -REPLY [11 votes]: The answer to question (*) is yes. It is Theorem 1.2.2 in the following preprint.<|endoftext|> -TITLE: The weak equivalences in the covariant model structure -QUESTION [14 upvotes]: Let $S$ be a simplicial set. Recall that there is a model structure, called the covariant model structure (see HTT ch. 2 and this question), on $\mathbf{SSet}/S$ such that: - -The cofibrations are the monomorphisms. -A map $X \to Y$ of simplicial sets over $S$ is a weak equivalence if $X^\vartriangleleft \sqcup_X S \to Y^\vartriangleleft \sqcup_Y S$ is a categorical equivalence (i.e., a weak equivalence in the Joyal model structure -- that is, one such that when applying the simplicial category functor $\mathfrak{C}$ gives an equivalence of simplicial categories). Here the triangle denotes the left cone. -The fibrations are determined; the fibrant objects are the left fibrations $Y \to S$. - -I think I understand the motivation for most of this: as Lurie explains, left fibrations are the $\infty$-categorical version of categories cofibered in groupoids (so the fibrant objects model a reasonable concept), and the cofibrations are as nice as can be. But I fail to understand the motivation for the weak equivalences -- not least because I don't have a particularly good picture of what these "left cones" are supposed to model. Why should the weak equivalences be what they are? - -REPLY [20 votes]: Maybe it would be helpful to think about the analogous situation in ordinary category theory. Suppose you are given a category $\mathcal{E}$ and a functor $F$ from -$\mathcal{E}$ to the category of sets. There are several ways to encode this functor: -$(a)$: Via the Grothendieck construction, $F$ determines a category $\mathcal{C}$ cofibered in sets over $\mathcal{E}$, so that for each object $E \in \mathcal{E}$ you can identify $F(E)$ with the fiber $\mathcal{C}_E$ of the map $\mathcal{C} \rightarrow \mathcal{E}$ over $E$. -$(b)$: Using the functor $F$, you can construct an enlargement $\mathcal{E}_F$ of the category $\mathcal{E}$, adding a single object $v$ with -$$Hom(E,v) = \emptyset \quad \quad Hom(v,E) = F(E) \quad \quad Hom(v,v) = \{ id \} $$ -Now suppose we are given another functor $G$ from $\mathcal{E}$ to the category of sets, -and a natural transformation $F \rightarrow G$. Then $G$ determines a category -$\mathcal{D}$ cofibered in sets over $\mathcal{E}$, and an enlargement $\mathcal{E}_G$ of $\mathcal{E}$. The natural transformation $F \rightarrow G$ determines functors -$$ \alpha: \mathcal{C} \rightarrow \mathcal{D} \quad \quad \beta: \mathcal{E}_F \rightarrow \mathcal{E}_G$$ -In this situation, the following conditions are equivalent: -$(i)$: The natural transformation $F \rightarrow G$ is an isomorphism (that is, for each object $E \in \mathcal{E}$, the induced map $F(E) \rightarrow G(E)$ is bijective. -$(ii)$: The functor $\alpha$ is an equivalence of categories. -$(iii)$: The functor $\beta$ is an equivalence of categories. -Now observe that the category $\mathcal{E}_F$ can be described as the pushout (and also homotopy pushout) of the diagram $$\mathcal{E} \leftarrow \mathcal{C} \rightarrow \mathcal{C}^{\triangleleft},$$ -where $\mathcal{C}^{\triangleleft}$ is the category obtained from $\mathcal{E}$ by adjoining a new initial object. -Let's now forget the original functors $F$ and $G$, and think only about the categories -$\mathcal{C}$ and $\mathcal{D}$ cofibered in sets over $\mathcal{E}$. The equivalence of conditions $(ii)$ and $(iii)$ shows that functor $\alpha: \mathcal{C} \rightarrow \mathcal{D}$ of categories cofibered over $\mathcal{E}$ is an equivalence of categories if and only if the induced map -$$ \mathcal{E} \amalg_{ \mathcal{C} } \mathcal{C}^{\triangleleft} -\rightarrow \mathcal{E} \amalg_{ \mathcal{D} } \mathcal{D}^{\triangleleft}$$ -is an equivalence of categories. -Now go to the setting of quasi-categories. Assume for simplicity that $S$ is a quasi-category, and let $f: X \rightarrow Y$ be a map of simplicial sets over $S$. If -$X$ and $Y$ are left-fibered over $S$, then we would like to say that $f$ is a covariant equivalence if and only if it an equivalence of quasi-categories. However, we would like to formulate this condition in a way that will behave well also when $X$ and $Y$ are not fibrant. -Motivated by the discussion above, we declare that $f$ is a covariant equivalence if and only if it induces a categorical equivalence -$$ S \amalg_{X} X^{\triangleleft} \rightarrow S \amalg_{Y} Y^{\triangleleft}.$$ -You can then prove that this is a good definition (it gives you a model structure with the cofibrations and fibrant objects that you described, and when $X$ and $Y$ are fibrant a map -$f: X \rightarrow Y$ is a covariant equivalence if and only if it induces a homotopy equivalence of fibers $X_s \rightarrow Y_s$ for each vertex $s \in S$).<|endoftext|> -TITLE: Canonical liftings of endomorphisms of ordinary abelian varieties -QUESTION [7 upvotes]: I am looking for a reference to the following ``well known" fact. -Let $k$ be a perfect field of prime characteristic $p$ and $W(k)$ its ring of Witt vectors. Let $A_0$ be an ordinary abelian variety over $k$ and let $A$ be an abelian scheme over $W(k)$ that is the canonical (Serre--Tate) lifting of $A_0$. Then every endomorphism $u_0$ of $A_0$ lifts to an endomorphism of $A$. In other words, the natural map $End(A) \to End(A_0)$ is bijective. -My problem is that I need it for infinite $k$. (I know a couple of references that deal with finite $k$.) - -REPLY [6 votes]: The canonical reference is Messing, LNM 264 1972, Chapter V, 3.4, p174.<|endoftext|> -TITLE: Is there a name for this map induced by bilinear forms? -QUESTION [6 upvotes]: Let $V$ be a real vector space. A bilinear form $\langle \rangle:V\times V\to {\mathbb{R}}$ induces a linear functional $\theta$ on the tensor product $V\otimes V$ given by sending the finite sum $\sum_i v_i\otimes w_i $ to $\sum_i \langle v_i,w_i\rangle$. -Is there a name for this induced linear functional? -In addition, if the bilinear form is symmetric, then this linear functional $\theta$ respects the natural involution on $V\otimes V$. That is $\theta(v\otimes w)=\theta(w\otimes v)$. - -REPLY [2 votes]: A bilinear form on $V$ (if non-degenerate) lets you identify $V$ with $V^\star$: -$v \mapsto \langle v, \cdot \rangle$ -In this case, your map is just a contraction of the identity map on $V^\star \otimes V$ (which, considering our identification, is the same as the identity on $V \otimes V$). -http://en.wikipedia.org/wiki/Tensor_contraction<|endoftext|> -TITLE: $Sq^1$ cohomology of spaces -QUESTION [12 upvotes]: For any space $X$, the first Steenrod square cohomology operation -$$Sq^1\colon H^\ast(X;\mathbb{Z}_2)\to H^{\ast +1}(X;\mathbb{Z}_2)$$ -is a derivation, meaning that $Sq^1\circ Sq^1 = 0$ and $Sq^1(a\cup b) = Sq^1(a)\cup b + a\cup Sq^1(b)$ (there are no signs since we are working in characteristic two). -Hence we may form the $Sq^1$-cohomology of the space, -$$H\left(H^\ast(X;\mathbb{Z}_2),Sq^1\right)$$ -which will be a graded algebra over $\mathbb{Z}_2$. -I am looking for references on this object. From McCleary's "User's guide to spectral sequences", I know that this is related to the Bockstein spectral sequence. More specifically, I would like to know: - - -What is the precise relationship between the $Sq^1$-cohomology of a space $X$ and $2$-torsion of higher order in $H^\ast(X;\mathbb{Z})$? -Is there a reference with specific calculations of the $Sq^1$-cohomology of the Eilenberg-Mac Lane spaces $K(\mathbb{Z}_2,n)$? -Are there any canonical references I should know about (besides McCleary and Mosher-Tangora)? - -REPLY [9 votes]: Several people have addressed question 1 (Torsten Ekedahl and Neil Strickland). Question 2 is interesting, but I don't have a good answer for it. For question 3, as Sean Tilson points out, this is a special case of "Margolis homology", a.k.a. $P^s_t$-homology. Try - -Adams and Margolis, "Modules over the Steenrod algebra", Topology 10 (1971) -Anderson and Davis, "A vanishing theorem in homological algebra", Comment. Math. Helv. 48 (1973) -Margolis, Spectra and the Steenrod algebra (1983) - -I also wonder if there is anything helpful in - -Adams and Priddy, "Uniqueness of BSO". - -You might also search for the phrase "Bockstein acyclic", since $\textrm{Sq}^1$ is the mod 2 Bockstein.<|endoftext|> -TITLE: Optimal 8-vertex isoperimetric polyhedron? -QUESTION [16 upvotes]: I know from Marcel Berger's - -Geometry Revealed: -A Jacob's Ladder to Modern Higher Geometry -(p.531) -that it is not yet established which polyhedron in $\mathbb{R}^3$ on 8 vertices achieves the optimal isoperimetric ratio $A^3/V^2$, where $A$ is the surface area and $V$ the volume. -Berger says "We also know that the cube ... [is] not the best for $v=8$" (where $v$ is the number of vertices). -Many other aspects of isoperimetry for polyhedra are unresolved, but this one especially interests me. It is not even clear to me that it is known that there is an optimal polyhedron for each $v$. -I've been trying to imagine what would be a strong candidate for an optimal 8-vertex polyhedron. I've been unsuccessful in finding information on this, although it seems likely to have been explored computationally. Does anyone have a candidate, or know of one proposed/calculated? A pointer or reference would be greatly appreciated. Thanks! -Addendum. -From the reference Igor provided (Nobuaki Mutoh, "The Polyhedra of Maximal Volume Inscribed in the Unit Sphere and of Minimal Volume Circumscribed about the Unit Sphere," 2009), here is a piece of Mutoh's Fig.1, which computationally verifies the earlier derivation of the max volume inscribed 8-vertex polyhedron by Berman and Haynes ("Volumes of polyhedra inscribed in the unit sphere in $\mathbb{R}^3$," Math. Ann., 188(1): 78-84, 1970, doi: 10.1007/BF01435416, eudml), as mentioned in the comments: - -           - - -This is surely a candidate for achieving the min of $A^3/V^2$! -I thank Jean-Marc, Igor, and Anton for the rapid convergence to what I sought. -...And then a bit later to Henry for showing that this candidate does not in fact achieve the best ratio! -Here is Henry's polyhedron, if I have interpreted him correctly: - -REPLY [17 votes]: An $8$-vertex polyhedron can achieve an isoperimetric ratio of $A^3/V^2 = 159.3243297053\dots$, and based on some quick experiments I'm pretty confident this is optimal (although I wouldn't be shocked if it could be beaten). -To construct it, let $V_\alpha$ denote the squashed tetrahedron with vertices $(\pm \sqrt{1-\alpha^2},0,\alpha)$ and $(0,\pm \sqrt{1-\alpha^2},-\alpha)$. Then the optimal $8$-vertex polyhedron seems to be the union of $V_\alpha$ and $-\beta V_\gamma$, with $\alpha = 0.2272117725\dots$, $\beta = 0.87345300464\dots$, and $\gamma = 0.83792301859\dots$. The optimal values of $\alpha$, $\beta$, and $\gamma$ are algebraic, but they're pretty complicated and I haven't computed their minimal polynomials. -For comparison, the maximum volume polyhedron inscribed in a sphere has a worse isoperimetric ratio, namely $162.248792\dots$. For the cube, it's $216$. -In general there's no reason to expect the optimal polyhedron to be inscribed in a sphere. The $5$-vertex case is a particularly nice example: it consists of an equilateral triangle on the equator of the unit sphere together with $1/\sqrt{2}$ times the north and south poles. This achieves an isoperimetric ratio of $243$, and I'd be very surprised if that's not optimal. Five vertices is few enough that a rigorous proof may be possible, but I can't think of a non-painful way to do it.<|endoftext|> -TITLE: Axiom of choice and non-measurable set -QUESTION [8 upvotes]: We know that existence of a Lebesgue non-measurable set follows from the Axiom Of Choice. Is the converse true? That is, does the existence of a Lebesgue non-measurable set imply the Axiom Of Choice? - -REPLY [20 votes]: No, the existence of a non-Lebesgue measurable set does not imply the axiom of choice. If ZF is consistent, then set-theorists can construct models of ZF having a non-Lebesgue measurable set, but still not satisfying AC. -This is quite reasonable, because the existence of a non-Lebesgue measurable set is a very local assertion, having to do only with sets of reals, and thus can be satisfied with a small example, by set-theoretic standards. The axiom of choice, in contrast, is a global assertion insisting that every set, even a very large set, has a well-order. So we don't expect to turn a mere non-measurable set into well-orderings of enormous sets, such as the power set $P(\mathbb{R})$. -And indeed, one can use forcing to produce a model $L(P(\mathbb{R})^{V[G]})$ which satisfies $ZF+\neg AC$, for similar reasons as in the usual $\neg AC$ models, but since it has the true $P(\mathbb{R})$, it will have all the same non-Lebesgue measurable sets as in the ambient ZFC universe $V[G]$. -(Finally, let me make a minor objection to the question: consistency is a symmetric relation, and so if $A$ is consistent with $B$, then $B$ would be consistent with $A$, and so one wouldn't ordinarily speak of a "converse". You seem instead to be refering to the implication that AC implies there is a non-measurable set, and this is how I took your question.)<|endoftext|> -TITLE: On a remark in Foundations of mechanics, 2nd Edition, by Abraham and Marsden -QUESTION [5 upvotes]: I don't know if this question is appropriate to this site, but I posted here without an answer, so I tried this alternative. -Given a $2$-form $\omega$ on a manifold $M$, let us denote by $N$ the kernel of $\omega$, i.e. $N:=\{u\in TM : \omega(u,\cdot)=0\}$. Their Proposition 5.1.2 shows that if $\omega$ has constant rank (and is closed) then $N$ is a tangent distribution on $M$ (and completely integrable). -In the following remark they say that ``the reader can easily prove the converse of the previous conclusion''. While I understand that $N$ is a tangent distribution if and only if $\omega$ has constant rank. Instead I think that, for $\omega$ of constant rank, $N$ can be completely integrable even if $\omega$ is not closed, (e.g. $\omega=e^z dx\wedge dy$). -Starting from this consideration I have asked myself a question: -Given a $\Omega\in\mathcal{A}^p(M)$, with $p>1$, whose rank is constant, let us define its kernel $N$ as above. Evidently $N$ is a tangent distribution on $M$, and I find it is completely integrable at least when there exists a $1$-form $\phi$ such that $d\Omega=\phi\wedge\Omega$. Clearly, if $\Omega$ is decomposable then the last condition is even necessary. -My question (edited after the comment of Willie Wong): - -Is this last condition (the ``divisibility'' of $d\Omega$ by $\Omega$) necessary for the complete integrability of $N$ even when $\Omega$ is not decomposable? (Using Frobenius' Theorem I understand the case $p=1$, but what about the case $p>1$?.) - -Any suggestion and\or counterexample are welcome. - -REPLY [5 votes]: The answer is 'no'. To see why, just take any nondegenerate $2$-form $\omega$ on, say, $\mathbb{R}^6$, that has the property that $d\omega$ is not a multiple of $\omega$. (This will be true for a generic such $2$-form.) The kernel of this $\omega$ is trivial, but now, you can just regard it as being defined on $\mathbb{R}^8$, say, by pullback. Then the kernel has (constant) positive dimension and is obviously integrable, but $d\omega$ is still not a multiple of $\omega$. -By the way, the theorem about integrability is true in much greater generality than for a single $p$-form: If $\mathcal{I}\subset\Omega^*(M)$ is a graded ideal that is closed under exterior derivative and its `kernel' $N$, which is defined as the set of vectors $v\in T_xM$ such that $\iota(v)\phi$ is in $\mathcal{I}_x$ for all $\phi\in \mathcal{I}_x$, has constant rank, then $N$ is integrable. This is a classic theorem of Cartan on 'Cauchy characteristics', and can, for example, be seen in Exterior Differential Systems by Bryant, et al.<|endoftext|> -TITLE: On a weak choice principle -QUESTION [9 upvotes]: [PLEASE SEE EDITS AT BOTTOM OF QUESTION] -Consider the following set-theoretic axiom: - -For each set $X$ there exists a set-indexed collection $\{C_i \to X\}_{i\in I_X}$ of surjections such that for every surjection $Z\to X$ there is a map $C_i\to Z$ for some $i$ such that the obvious triangle commutes. - -This is known as WISC (Weakly Initial Set of Covers), and can be interpreted as saying Choice fails to hold in at most a 'small' way. It is clearly implied by AC, and I'm willing to bet that it is independent of other usual set-theoretic axioms (ZF, say). WISC is implied by COSHEP (take $I_X$ to be a singleton for all $X$), SVC and AMC. -My questions are these: - -Does anyone know of a weaker choice principle? (Edit: a global choice principle, or at least one for a sizable collection of sets, like all elements $\bigcup_n \mathcal{P}(\mathbb{R}^n)$) - -and - -In which popular/common models of set theory does WISC hold? That is, aside from the ones listed at the linked page above (which are particularly category theory-oriented). - -(As a bonus question: Come up with a model of ZF that violates WISC or prove we can use forcing to construct one) - -Edit: There is also the axiom WISC${}_\kappa$, where we require the set $I_X$ to be bounded by some cardinal $\kappa$ (either less than or at most). This is perhaps more interesting than the unbounded case, especially in topological applications. - -Edit2: Benno van den Berg has now shown that Gitik's model of ZF (Israel J. Math 1980) violates WISC. This model, which relies on the consistency of a large cardinal assumption (that is, the existence of an unbounded collection of strongly compact cardinals), has the property that only $\aleph_0$ is a regular cardinal. What Benno showed was that ZF+WISC implies the existence of an unbounded collection of regular cardinals. Now one can clearly ask (thanks to godelian in the comments) whether weaker large cardinal assumptions suffice. One would only need to find a model of ZF in which there is only a bounded collection of regular cardinals. This to me sounds reasonable. - -REPLY [3 votes]: After Benno van den Berg's proof, there are now two more proofs: -First, building on his answer, Asaf removed the requirements for large cardinals in - -Asaf Karagila, Embedding Orders Into Cardinals With $DC_\kappa$, Fund. Math. 226 (2014), 143-156, doi:10.4064/fm226-2-4, arXiv:1212.4396. - -Second, as suggested by Mike Shulman, a suitable topos of $G$-sets where $G$ is a large topological group violates WISC in its internal language (how to make this construction precise takes a little bit of ingenuity): - -David Michael Roberts, The weak choice principle WISC may fail in the category of sets, Studia Logica Volume 103 (2015) Issue 5, pp 1005-1017, doi:10.1007/s11225-015-9603-6, arXiv:1311.3074.<|endoftext|> -TITLE: Homotopy Fixed Points of SO(2) on Fully Dualizable Algebras -QUESTION [14 upvotes]: Note: by fixed points, I always mean homotopy fixed points. -As explained in Jacob Lurie's paper on the cobordism hypothesis, we have an action of O(2) on the $\infty $-groupoid $X$ given by considering fully dualizable objects and invertable morphisms in some symmetric monoidal $(\infty ,2)$ category $\mathcal C$. I am interested in the case when $\mathcal C$ is the category where objects are algebras, 1-cells are bimodules, 2-cells maps of bimodules etc... By an algebra, I want to include algebra objects in some $\infty$-category (e.g. chain complexes), so that $\mathcal C$ really has non-trivial 3-cells, 4-cells and so on. -Now, I know (from remark 4.2.7) that the fixed points for the induced action of SO(2) on this space correspond to cyclic Frobenius algebras, i.e. smooth, proper algebras $A$, with a non-degenerate and $SO(2)$-equivariant trace -$A \otimes _{A^e} A \to k$. -Here, non-degenerate means that the pairing $A\otimes A \to A\otimes _{A^e} A \to k$ is nondegenerate. Recall also that the Hochschild homology $A \otimes _{A^e} A$ carries and $SO(2)$ action, so it makes sense to talk about $SO(2)$-equivariant map above. -The data of SO(2) equivariance should be equivalent to descending the trace to cyclic homology. -However, the reason why the fixed points are as above is that both can be identified with 2-d oriented TFTs, after theorem 3.1.8 which describes how to extend from a (n-1)-dimensional TFT to an n-dimensional one. -Question: Can we calculate this action and identify the fixed point space directly? -Part of this is not too hard to see (I think): naively, the SO(2) action on $X$ gives a canonical (Morita) automorphism of each (f.d.) algebra $A$. This automorphism is given by the bimodule dual $A^!$ (or the other one $A^\vee$...), so being a fixed point means to give an isomorphism $A^! \to A$, which is the same as a non-degenerate map $A\otimes _{A^e} A \to k$. If the base category where our algebra lives has no higher structure, then I think this is enough, but this does not explain the $SO(2)$ equivariance. -In particular, where does the SO(2)-equivariance data come from? -To see this, I tried to go a little deeper (with help from Takuo Matsuoka): the action is given by a map $B\mathbb Z =SO(2) \to Aut(X)$, which we transformed into a map $\mathbb Z \to \Omega Aut(X)$. The data described above corresponds to picking an algebra $A$ and composing to get a map $\mathbb Z \to \Omega X$ = Invertable $A-A$-bimodules. However, we haven't used that the original map from SO(2) is a group homomorphism - this should correspond to the map $\mathbb Z \to \Omega Aut(X)$ being an $E_2$-map. So we get braiding data attached to each of the $A^!$, which should be trivialized (because the E_2 -structure on $\mathbb Z$ is trivial (?)). Then to give a fixed point we must take into account all this data... -This has only given me a headache so far, but maybe there is an easier way to think about this? I tried (failed?) to be brief, so let me know if anything here is unclear! - -REPLY [10 votes]: I might be confused about your question. Are you asking... - -How is trivializing the $O(n)$-action the same as giving an $O(n)$-equivariant non-degenerate trace? (as per Lurie's theorem 3.1.8). -How can we identify the $SO(2)$-action with the usual $SO(2)$-action on Hochschild homology? - -For the first one, I think this is pretty well described in Jacob's paper. The key is understanding what the condition "non-degenerate" means in that setting. This is the condition that allows you to turn the equivariant trace into the trivialization of an action. -If it is the second thing you are asking, here is one way to see this, though there may be an easier way. Let's first discuss the more direct approach which is giving you a headache. Instead of thinking about $E_2$-structures or (homtopical) group homomorphisms, I prefer to think of the map of delooings: -$$ \alpha: BSO(2) = K(\mathbb{Z}, 2) = \mathbb{CP}^\infty \to BAut(X)$$ -This is equivalent to understanding the $E_2$-structure you mentioned, but for me it is easier to comprehend. A (homotopical) $ SO(2)$-action on $X$ is the same as the map $\alpha$. -I find this easier probably because of my background. Now $BSO(2) = \mathbb{CP}^\infty$ has a well known cell structure which goes: -$$ S^2 \cup e_4 \cup e_6 \cup \dots $$ -So part of the $SO(2)$-action is an element in $\pi_2 BAut(X)$ which gives the invertible $A-A$-bimodule you identified up above. This is also called the Serre bimodule in this context. The element in $\pi_2$ actually contains more information: it is a natural transformation (from the identity functor to itself). -From this point of view, to understand the rest of the $SO(2)$-action, you need to extend this map all the way through the entire CW-structure. You can start doing this explicitly. For example the fact that the Serre is a natural transformation allows you to construct an element in $\pi_3 BAut(X)$ which is essentially the "self braiding" of the Serre with itself. Trivializing this element allows you to extend the map to the 4-skeleton $\mathbb{CP}^2 = S^2 \cup e_4$. This then gives rise to a higher "self braiding" which you then have to trivialize, etc. -This quickly becomes a huge headache, as each of thee trivializations is supposed to be describable in terms of the basic fully-dualizable structure. While possible, it is not entirely obvious how to procede at each step. Also, even if we succeed this doesn't yet do what you want; we still haven't identified this SO(2)-action with the usual one on Hochschild homology. -A different approach is to realize that -$$BSO(2) \simeq |N \Lambda^{op}|$$ - where $\Lambda$ is the cyclic category. This allows us to get a different "filtration" of $K(\mathbb{Z}, 2)$. If we view the map $\alpha$ as classifying an $X$-bundle over $BSO(2)$, then we can use this description, plus the material of Dwyer-Hopkins-Kan "The homotopy theory of cyclic sets" to realize that this is equivalent to the data of a certain cyclic set. -I think that in the universal case we get a cyclic set which essentially comes from the various ways to decompose a circle cyclicly, and in the specific case at hand we recover the Hochschild homology complex. This would identify the two circle actions. I hope this helps.<|endoftext|> -TITLE: Multilinear generalization of Cauchy-Schwarz inequality -QUESTION [9 upvotes]: Let $V$ be a real vector space, and let $(\cdot,\cdot;\cdot,\cdot) : V^4 \to \mathbb{R}$ be a multilinear form with the following properties: - -$(x,y;z,w) = (y,x;z,w) = (x,y;w,z)$ (symmetry in the first and second pairs) -$(x,x;z,z) \ge 0$ (positive semidefiniteness in the first and second pairs). - - -Must such a form satisfy the inequality $$|(x,y;z,w)| \le \sqrt{(x,x;z,z)(y,y;w,w)}?$$ - -The prototype I have in mind is something like $V = C_c^\infty(\mathbb{R}^n)$, with -$$(f,g;h,k) = \int f g \nabla h \cdot \nabla k$$ -in which case the inequality follows by using Cauchy-Schwarz twice (first in $\mathbb{R}^n$, and then in $L^2(\mathbb{R}^n)$). -I'd settle for the inequality -$$|(x,y;z,w)| \le C({\epsilon}(x,x;z,z) + \epsilon^{-1}(y,y;w,w))$$ -which follows from the above by AM-GM (with $C = 1/2$). I'd also settle for the special case $x=w, y=z$ where it reads $|(x,z;x,z)| \le \sqrt{(x,x;z,z)(z,z;x,x)}$. -Simply using Cauchy-Schwarz in each pair gives the inequality -$$|(x,y;z,w)| \le [(x,x;z,z)(x,x;w,w)(y,y;z,z)(y,y;w,w)]^{1/4}$$ which has cross terms that I don't want. Edit: Of course, as Willie Wong points out and zeb's counterexample confirms, this doesn't work. -Thanks! - -REPLY [9 votes]: Even the inequality $(x,z;x,z)^2 \le (x,x;z,z)(z,z;x,x)$ is false: -Let $V = \mathbb{R}^2$, with basis $x,z$. Take $(x,x;x,x) = 100$, $(x,z;x,x)=0$, $(z,z;x,x)=1$, $(x,x;x,z)=0$, $(x,z;x,z)=50$, $(z,z;x,z)=0$, $(x,x;z,z) = 1$, $(x,z;z,z)=0$, $(z,z;z,z)=100$, and extend to all of $V^4$ by symmetry and multilinearity. -To check that positive semi-definiteness holds, note that we just need to check that -$(x+az,x+az;x+bz,x+bz) = 100 + a^2 + 200ab + b^2 + 100a^2b^2 \ge 0$, -which easily follows from AM-GM. -Now note that we have $2500 = (x,z;x,z)^2 > (x,x;z,z)(z,z;x,x) = 1$. -In fact, we even have $6250000 = (x,z;x,z)^4 > (x,x;x,x)(x,x;z,z)(z,z;x,x)(z,z;z,z) = 10000$. -Edit: On the other hand, we can prove the following inequality: -$4(x,y;z,w)^2 \le ((x,x;z,z)+(x,x;w,w))((y,y;z,z)+(y,y;w,w))$. -To see this, note that by positive semi-definiteness we have -$0 \le (x+ay,x+ay;z+w,z+w) + (x-ay,x-ay;z-w,z-w)$ -$ = 2((x,x;z,z)+(x,x;w,w)) + 8a(x,y;z,w) + 2a^2((y,y;z,z)+(y,y;w,w))$ -for any $a$, and plugging in $a = -\frac{2(x,y;z,w)}{(y,y;z,z)+(y,y;w,w)}$ we get the desired inequality.<|endoftext|> -TITLE: Book on mixed Hodge structures? -QUESTION [7 upvotes]: Is there any English textbook about Deligne's mixed Hodge structures? Can you tell me about a reference where they are introduced at least for smooth quasi-projective varieties? - -REPLY [7 votes]: Mixed Hodge Structures, Peters and Steenbrink - -REPLY [7 votes]: Maybe you already know of the notes by Benoît Audoubert and Orsola Tommasi entitled "Mixed Hodge Structures," -notes on Audoubert's seminars on Mixed Hodge structures at the University of Nijmegen in 2002? -These notes certainly discuss quasi-projective algebraic varieties (in Section 3). -It is about 50 pages long, with the following (high-level) Table of Contents: - -Hodge structures -Varieties with normal crossings -Smooth quasi-projective varieties -The theory of Deligne<|endoftext|> -TITLE: Are wild problems related to undecidable ones? -QUESTION [14 upvotes]: In representation theory, there is a well-known notion of a wild classification problem (such problems have been discussed often on this forum, for example, here). In logic, there is a notion of an undecidable problem. - -Is there a theorem which says that there is something undecidable about a wild classification problem? - -A reference where such issues are discussed would be very helpful. - -REPLY [16 votes]: Yes, there is a connection, but I think it is conjectural in its full generality. The mosst general reference could be, where it is proven, that for a subclass of wild algebras, the representation theory is undecidable: -Mike Prest: Wild representation type and undecidability, Comm. Alg. 19 (3), 1991. -It is also well-known (it is stated with references for example in Benson), that the representation theory of the algebra used to define wildness (i.e. $k\langle X,Y\rangle$) is undecidable.<|endoftext|> -TITLE: Wanted: example of a non-algebraic singularity -QUESTION [31 upvotes]: Given a finitely generated $\def\CC{\mathbb C}\CC$-algebra $R$ and a $\CC$-point (maximal ideal) $p\in Spec(R)$, I define the singularity type of $p\in Spec(R)$ to be the isomorphism class of the completed local ring $\hat R_p$, as a $\CC$-algebra. -Do there exist non-algebraic singularity types? That is, does there exist a complete local ring with residue field $\CC$ which is formally finitely generated (i.e. has a surjection from some $\CC[[x_1,\dots, x_n]]$), but is not the complete local ring of a finitely generated $\CC$-algebra at a maximal ideal? -Googling for "non-algebraic singularity" suggests that the answer is yes, but I can't find a specific example. I would expect that it should be possible to write down a power series in two variables $f(x,y)$ so that $\CC[[x,y]]/f(x,y)$ is non-algebraic. - -What is a specific formally finitely generated non-algebraic singularity? - -REPLY [15 votes]: The main question of the PI has been beautifully answered by Moret-Bailly, but not the -secondary question arisen from his expectation: "I would expect that it should be possible to write down a power series in two variables f(x,y) so that ℂ[[x,y]]/f(x,y) is non-algebraic." -though we got quite close in comments. -So for the record: this is not possible. Indeed, such a singularity would be analytic by a result of Michael Artin ( "On the solutions of analytic equations", Invent. Math. 5 1968, 277–291, cf. Angelo's answer to Analytic vs. formal vs. étale singularities) and then algebraic by Ulrich's comment (that is by Corollary 7.7.3 of the book by Casas-Alvero "Singularities of plane curves", London Mathematical Society Lecture Note Series, 278). -This is of course consistent with the fact that the example quoted by Moret-Bailly is in -three variables.<|endoftext|> -TITLE: Intersection of a smooth projective variety and a plane -QUESTION [5 upvotes]: Let $X \subset P^n$ be an irreducible smooth complex projective -variety embedded in the $n$-dimensional projective space. -Let $k$ be the dimension of $X$ and $d$ its degree. -Let $L \subset P^n$ be a linear subspace of dimension $n-k$ -and $Z=L \cap X$. Assume that -(a) $X$ is not contained in any hyperplane of $P^n$ and -(b) $Z$ is finite of cardinality $d$. -Question: Is it true that $Z$ spans $L$? -Comment: I was told that this is true if $X$ is ACM -(arithmetically Cohen-Macaulay). A reference for this -would be appreciated. - -REPLY [8 votes]: It is true that $Z$ spans $L$ — even if $X$ isn't ACM. You can also allow $X$ to be singular (but you do need $X$ irreducible and non-degenerate, of course). To illustrate one of the main ideas it is useful to first look at the case when $X$ -is a curve. -If $X$ is a curve. Let $M$ be the span of $Z$ and suppose that $M\neq L$. (In the curve case, $L$ will be a hyperplane). Let $p$ be any point of $X$ outside of $Z$ and let $H$ be any hyperplane containing $M$ and $p$. Then $H\cap X$ contains at least $d+1$ points, so by Bezout's theorem the intersection cannot be zero dimensional. Since $X$ is irreducible and one dimensional, this means that the intersection must be all of $X$, so $X$ is contained in $H$, contrary to hypothesis. -The general case. -The idea when $k\geqslant 2$ is to show that if $H$ is a general hyperplane containing $L$ then $H \cap X$ is irreducible and non-degenerate (i.e, the intersection $H\cap X$ is not contained in a smaller linear space of $H$). But now all dimensions have been reduced by $1$, and so iterating this procedure reduces us to the curve case, which we've already solved. -To set this up, note that hyperplanes in $\mathbb{P}^n$ containing $L$ are parameterized by a $\mathbb{P}^{k-1}$ (If $V$ is the underlying vector space of $\mathbb{P}^{n}$, $W$ the underlying vector space of $L$, then the hyperplanes are parameterized by the projectivization of $(V/W)^{*}$). We'll use $H$ to refer both to a point of $\mathbb{P}^{k-1}$ and the corresponding hyperplane in $\mathbb{P}^n$ containing $L$. Define $\Gamma\subset \mathbb{P}^{k-1}\times (X\setminus Z)$ to be the set -$$\Gamma = \left\{(H,p) \mid p\in H\right\}$$ -i.e, the pairs $(H,p)$ so that $H$ is a hyperplane containing $L$, and $p$ a point of $H\cap X$ not on $Z$. -If we fix $p$, then the set of possible $H$'s satisfying this condition are simply the hyperplanes $H$ containing the span of $L$ and $p$, and this is parameterized by a $\mathbb{P}^{k-2}$. In other words, $\Gamma$ is a $\mathbb{P}^{k-2}$ bundle over $X\setminus Z$. (This fibration is where we use $k\geqslant 2$.) Since $X\setminus Z$ is irreducible this implies that $\Gamma$ is irreducible. -Let $\overline{\Gamma}$ be the Zariski-closure of $\Gamma$ in $\mathbb{P}^{k-1}\times X$. Then $\overline{\Gamma}$ is irreducible since $\Gamma$ is. For a fixed $H\in \mathbb{P}^{k-1}$ the fibre of the projection $\overline{\Gamma}\longrightarrow \mathbb{P}^{k-1}$ over $H$ is simply the intersection $X\cap H$, of dimension $k-1$. -Now let $q$ be any point of $Z$. Then $q\in X\cap H$ for every $H\in \mathbb{P}^{k-1}$ so $q$ gives a section of $\overline{\Gamma}\longrightarrow\mathbb{P}^{k-1}$. Since $Z$ consists of $d$ distinct points where $d$ is the degree of $X$ we conclude that $q$ is a smooth point of $X$. Finally, since $Z$ is the intersection of all $X\cap H$ for $H\in \mathbb{P}^{k-1}$ this implies that the general intersection $X\cap H$ is smooth at $q$. Summarizing, we have a section of the map which generically lies in the smooth locus of the fibres. Since $\overline{\Gamma}$ is irreducible, this implies that the generic fibre is irreducible, i.e, if $H$ is a generic hyperplane containing $L$, then $H\cap X$ is irreducible. -(The intuitive reason for this implication is that, generically over $\mathbb{P}^{k-1}$ the section lets us pick out precisely one irreducible component of the fibre. The union of these components gives us a subset of $\overline{\Gamma}$ which has the same dimension as $\overline{\Gamma}$, and hence whose closure must be all of $\overline{\Gamma}$ by irreducibility. But if there is more than one component in a general fibre, this is a contradiction, thus the general fibre must be irreducible. To make this intuitive construction rigorous requires passing to the normalization of $\overline{\Gamma}$ and then looking at the Stein factorization of the map from the normalization to $\mathbb{P}^{k-1}$. The section gives a generic section of the finite part of the Stein factorization, and that allows one to construct the ``union of the components containing the section''.) -Finally, the same trick as in the curve case also shows us that for any hyperplane $H$, $H\cap X$ must be non-degenerate. Let $Y=H\cap X$, so that $Y$ is a variety of degree $d$ and dimension $k-1$. Let $M$ be the span of $Y$. If $M\neq H$ then pick any point $p\in X\setminus Y$ and let $H'$ be any hyperplane containing $M$ and $p$. Then $H'\cap X$ can't be all of $X$ (since this would contradict the non-degeneracy of $X$), so $Y'=H'\cap X$ must be a subvariety of dimension $k-1$ (more precisely, all components of $Y'$ have dimension $k-1$) and degree $d$. But $Y$ is therefore a component of $Y'$, and the equality of degrees tells us that $Y'$ can't have any other components so we must have $Y'=Y$. This contradicts the fact that $p\in Y'$ and $p\not\in Y$. -Together this shows the required inductive step: If $H$ is a general hyperplane containing $L$ then $H\cap X$ is irreducible and non-degenerate. -Other remarks. I'm guessing from the setup of the question that you want to apply the result for a particular $L$ that you have chosen. If, in the application, you're allowed to pick a general $L$ then you can say something stronger. The classical uniform position principle (where ''classical'' in this case means ''established by Joe Harris in the 80's'') states that for a general subspace $L$ of dimension $n-k$ the finite set of $d$-points in $Z=L\cap X$ have the property that any subset of $r+1$ of the points (with $r\leqslant n-k$) span a $\mathbb{P}^{r}$. Picking $r=n-k$, this means that any subset of $n-k+1$ points of $Z$ spans all of $L$, and so in particular $Z$ spans $L$. (Note that $d\geqslant n-k+1$; for instance, as a consequence of the argument above: if $d < n-k+1$ then the $d$ points of $Z$ would never span $L$.)<|endoftext|> -TITLE: A combinatorial proof for the property of KM numbers? -QUESTION [8 upvotes]: Kendell-Mann numbers $M(n)$ ( see the sequence A000140 http://oeis.org/A000140 ) have the simple property: $M(n+1) \approx (n-1/2)M(n)$. -The property can be proved by different methods. -For eg. The property of Kendall-Mann numbers -What I am looking for is to find out if a combinatorial proof exists? -For eg. Let us start: Suppose we look at all the permutations of $n-1$ in the maximal grouping, then at all the permutation of $n$ in that maximal grouping; is there any simple way in which each permutation in the first set gives rise to $n$ permutations in the second? Better yet, a simple way in which about half the $n-1$-permutations give rise to $n$ $n$-permutations each, and the other half give rise to $n+1$ $n$-permutations each? -Any hints are higly welcomed. -I hope that the combinatorial proof will makes the reason for the simple property more transparent. - -REPLY [8 votes]: Here is a quick and dirty probabilistic analysis which gets the right answer. For a permutation $w \in S_n$, define -$$I(w) = \sum_{1 \leq i < j \leq n} \begin{cases} -1 & w(i) < w(j) \\ 1 & w(i) >w(j) \\ \end{cases}.$$ -So $I(w) = 2 \# (\mbox{number of inversions of $w$}) - \binom{n}{2}$. If $w$ is chosen uniformly at random, then the expected value of $I(w)$ is $0$. -Now, let's think about the expected value of $I(w)^2$. Squaring the sum, we get terms indexed by $(i_1, j_1, i_2, j_2)$ with $i_1 -TITLE: Is there a good definition of (topological) K-Theory over arbitrary spaces? -QUESTION [13 upvotes]: Hi -(this is my very first question here, so please don't hurt me...) -for some time now i've been looking for a sufficiently aesthetical definition of (topological) K-theory of arbitrary spaces, yet been unable to find or come up with one. The definition I know goes as follows: -For X connected, compact, hausdorff one defines $V(X) = \text{ set of isoclasses of vectorbundles over} X$ which becomes a comm. monoid under direct sum and then $KO(X) = K(V(X))$ where the righthandside just means group completion of a comm monoid. -Here already the "isoclasses" of bundles bothers me, because this "set" is not really a set, is it? (?its elements being proper classes?). I guess this may be salvaged by instead looking at equi. classes of systems of transition functions taking values in $GL(\mathbb R)$, modulo some further restriction and relations !? -Anyway, this is something I might even live with, but -one goes on to show $KO(X) \cong [X, \mathbb Z \times BGL(\mathbb R)]$ for such spaces, and for a CW-complex C sets $KO(C) = [C, \mathbb Z \times BGL]$. This is the only possible defintion (up to nat. iso), when trying to end up with a cohomology-like functor; right? -Finally for a general space Z we pick (for every space simultaneously?!) a CW-substitue say C' and put $KO(Z) = [C', \mathbb Z \times BGL]$. -However, using this as defintion, there really is very little beauty left in K-Theory for me. I know that just putting $KO(X) = K(V(X))$ goes awry (bundles must be allowed to have varying dimension over different components and in turn must allow for a partition of unity and so on...) -So my question is: -Is there a way of altering the definition of V(X) sufficiently so as to give a "correct" definition of K-Theory? Or some other way of producing these groups, nicely? Nicely should in particular mean, without homotopy theory oder cell complexes, so that for instance homotopy invariance is not directly built into the definition. And if so, what about relative groups, or even higher ones? -After all for singular cohomology and bordism there also are descriptions using homotopy theory (via Eilenberg-MacLane resp. Thom-Spaces) just as above, but for an arbtrary space there are entirely different (better?) descpriptions in terms of singular chains and manifolds. -Thanks in advance -Maybe I should add that passing to spectra makes everything even worse in my opinion. - -REPLY [6 votes]: This is a response only to a tangential part of your question, on set theory, and not to the bulk. It continues some of the discussion in the comments to Alain Valette's answer. -I guess the thing to emphasize is that in some technical sense you are correct: standard (e.g. ZFC) approaches to set theory often do not give meaning to sentences like "The set of isomorphism classes of all finite-dimensional vector bundles". Certainly if you think that an "isomorphism class" is itself necessarily a "set" of things, and that the elements of a "set" should themselves be "sets", as is often done, that is. -But from a modern perspective, this is a perverse way to go about doing mathematics. The notion of "the collection of isomorphism classes of finite-dimensional vector bundles" is perfectly good. Moreover, there are many sets-with-structure that represent this notion. So pick one --- for example, choose your favorite copy of $\mathbb R$, your favorite construction of $\mathrm{GL}(\mathbb R,n)$, etc., and form some set of transition functions and so on. At the end of the day, whatever honest set you actually construct is just as good as any other. The most important moral from category theory is that what you should mean by "the X" is "any object satisfying the distinguished properties of X, along with the instructions of how to make it isomorphic to any other object with the same distinguished properties". -Since you should not expect the Axiom of Choice, and other difficult things, to work for very large collections, it is important to worry about the "size" of a collection. Most mathematicians only have two "sizes". One size is called "is a set", and the other is called "is larger than a set". Things like Choice are assumed to work without complication for anything that "is a set", and you may be more suspicious for "things larger than a set". For this notion of "is a set", it certainly is true that the collection of isomorphism classes of finite-dimensional vector bundles "is a set". Proof: represent it by a set in some way, I don't care how. -Anyway, in the comments to Alain's answer, it looks like you understand all of this perfectly well. So don't sweat this part of the language!<|endoftext|> -TITLE: Classical invariants involving exterior powers of standard representation -QUESTION [8 upvotes]: While investigating certain conformal blocks line bundles on $\overline{M}_{0,n}$, I was led to what seems to be an identification between two spaces of invariants, and I am curious if there is a direct way to see this identification. -Statement: for any integers $n\ge 4$ and $r\ge 2$, and any integers $i_1,\ldots,i_n$ such that $1 \le i_j \le r-1$ and $r=\frac{1}{2}\sum_{j=1}^n i_j$, I believe there is a vector space isomorphism $$(\wedge^{i_1}\mathbb{C}^r\otimes\cdots\otimes \wedge^{i_n}\mathbb{C}^r)^{SL(r)} \cong (S^{i_1}\mathbb{C}^2\otimes\cdots\otimes S^{i_n}\mathbb{C}^2)^{SL(2)},$$ where $\mathbb{C}^m$ denotes the standard representation of $SL(m)$. The invariants on the RHS are classical and well-known: a basis is given by all $2\times r$ semi-standard tableaux with entries in $\{1,\ldots,n\}$ such that $j$ occurs exactly $i_j$ times. I wonder if the invariants on the LHS are also known, and if there's a conceptual reason why they might be in bijection with those on the RHS. -Background: -This is not relevant for the question itself, but I am including it in case you are curious how this purported identity arose. It seems likely that for conformal blocks bundles on $\overline{M}_{0,n}$ of level 1 and Lie algebra $\mathfrak{sl}(r)$, the global sections are naturally identified with a space of covariants. Specifically, the conformal blocks line bundle with weights $(\omega_{i_1},\ldots,\omega_{i_n})$, where $\omega_i$ are fundamental weights, should have global sections $(\wedge^{i_1}\mathbb{C}^r\otimes\cdots\otimes\wedge^{i_n}\mathbb{C}^r)_{\mathfrak{sl}(r)}$, since the irreducible representation associated to $\omega_i$ is $\wedge^i\mathbb{C}^r$. This space of $\mathfrak{sl}(r)$-covariants is isomorphic to the corresponding space of $\mathfrak{sl}(r)$-invariants, which in turn is the same as the space of $SL(r)$-invariants for this representation. On the other hand, it is known (by a result of Fakhruddin) that when $\sum_{j=1}^n i_j = 2r$ then this conformal blocks line bundle induces the GIT morphism $\overline{M}_{0,n} \rightarrow (\mathbb{P}^1)^n//_{(i_1,\ldots,i_n)}SL(2)$, so we know that its space of global sections is $H^0((\mathbb{P}^1)^n,\mathcal{O}(i_1,\ldots,i_n))^{SL(2)}$. - -REPLY [12 votes]: To follow up on Sasha's answer, yes there is a natural isomorphism of vector spaces which lifts the combinatorial equality. All isomorphisms in this answer will be natural. -Schur-Weyl duality: -Let $\lambda$ be a partion; set $d = |\lambda|$ and let $m$ be greater than or equal to the number of parts of $\lambda$. Let $V$ be a vector space of dimension $m$, let $V_{\lambda}$ be the irrep of $GL(V)$ with highest weight $\lambda$ an let $M_{\lambda}$ be the $S_d$-irrep (aka Specht module) indexed by $\lambda$. Schur-Weyl duality is the isomorphism of $GL(V)$ representations: -$$\mathrm{Hom}_{S_d}(M_{\lambda}, V^{\otimes d}) \cong V_{\lambda}.$$ -I'll take this to be the definition of $V_{\lambda}$. -Tensor product and restriction: -Let $\mu$, $\lambda_1$, ..., $\lambda_k$ be partitions, with $|\mu| = \sum |\lambda_i|$. Let $m$ be greater than or equal to the number of parts of $\mu$. Set $d_i = |\lambda_i|$. Then the above shows that -\begin{align*} -&\mathrm{Hom}_{GL(V)}\left( V_{\mu}, V_{\lambda_1} \otimes \cdots \otimes V_{\lambda_k} \right)\\ -&\cong \mathrm{Hom}_{GL(V)} \left( \mathrm{Hom}_{S_{\sum d_i}} \left( M_{\mu}, V^{\otimes \sum d_i} \right), \mathrm{Hom}_{S_{d_1} \times \cdots \times S_{d_k}}\left( M_{\lambda_1} \boxtimes \cdots \boxtimes M_{\lambda_k}, V^{\otimes \sum d_i} \right) \right) \\ -&\cong \mathrm{Hom}_{S_{d_1} \times \cdots \times S_{d_k}} \left( M_{\lambda_1} \boxtimes \cdots \boxtimes M_{\lambda_k}, \left( M_{\mu} \right)|_{S_{d_1} \times \cdots \times S_{d_k}} \right). -\end{align*} -In other words, every Hom in the second expression is induced from composition with one of the Hom's in the third expression. Here, if $U$ and $V$ are $G$ and $H$-representations, then $U \boxtimes V$ denotes $U \otimes V$ with $G$ and $H$ acting on the first and second factors respectively. -Relation to transpose: Now let $\lambda_i$ and $\mu$ be as above. Let $\epsilon(d)$ be the sign representation of $S_d$, and let $\lambda^T$ be the transpose of $\lambda$. Then $M_{\lambda} \cong M_{\lambda^T} \otimes \epsilon(|\lambda|)$. How to make this natural depends on exactly how you define $M_{\lambda}$. For example, if you use the Vershik-Okounkov approach, the bases they construct for $M_{\lambda}$ and $M_{\lambda^T}$ correspond to each other. -Now, $\epsilon(d_1) \boxtimes \cdots \boxtimes \epsilon(d_k) \cong \epsilon(\sum d_i)|_{S_{d_1} \times \cdots \times S_{d_k}}$, and is irreducible. So we deduce that $$\mathrm{Hom}_{S_{d_1} \times \cdots \times S_{d_k}} \left( M_{\lambda_1} \boxtimes \cdots \boxtimes M_{\lambda_k}, \left( M_{\mu} \right)|_{S_{d_1} \times \cdots \times S_{d_k}} \right) \cong \mathrm{Hom}_{S_{d_1} \times \cdots \times S_{d_k}} \left( M_{\lambda_1}^T \boxtimes \cdots \boxtimes M_{\lambda_k}^T, \left( M_{\mu}^T \right)|_{S_{d_1} \times \cdots \times S_{d_k}} \right)$$ -and this isomorphism is natural if we define $M_{\lambda} \cong M_{\lambda^T} \otimes \epsilon(|\lambda|)$ in a natural way. -Putting it all together, if $\dim V$ is greater than or equal to the number of parts in $\mu$, and $\dim W$ is greater than or equal to the number of parts in $\mu^T$, then we have a natural isomorphism: -$$\mathrm{Hom}_{GL(V)}\left( V_{\mu}, V_{\lambda_1} \otimes \cdots \otimes V_{\lambda_k} \right) \cong \mathrm{Hom}_{GL(W)}\left( W_{\mu^T}, W_{\lambda^T_1} \otimes \cdots \otimes W_{\lambda^T_k} \right).$$ -Your question: -Take $\lambda_j = 1^{i_j}$ and $\mu = 2^r$. Let $\dim V = r$ and $\dim W=2$. -Since $M_{\lambda_j}$ is the sign rep, we have $V_{\lambda_j} \cong \bigwedge^{i_j} V$ naturally. Similarly, since $M_{\lambda_j^T}$ is the trivial rep, we have $W_{\lambda_j^T} \cong \mathrm{Sym}^{i_j}(W)$. -Finally, one needs to use the fact that $V_{\mu}$ and $W_{\mu^T}$ are trivial one-dimensional $SL(V)$ and $SL(W)$ reps. (More precisely, they are $\det^2$ and $\det^r$, as $GL$-reps.) I'm not sure what the easiest proof of this is, but it can only introduce a scalar factor to your isomorphisms.<|endoftext|> -TITLE: finite codimension implies closed? -QUESTION [5 upvotes]: Let $E$ be a (complete) topological vector space, and $u:E\to E$ be continuous. Is it always true that if ${\rm Im}(u)$ is of finite codimension in $E$, then it is closed in $E$ or do we have to assume something on $E$? (It is OK if $E$ is Frechet by the open mapping theorem applied to ${\rm id}\oplus u:F\oplus E\to E$, where $F$ is a supplementary subspace to $E$.) - -REPLY [3 votes]: Bill: The existence of Hamel basis in (all) vector spaces is equivalent to the axiom of choice (see Blass, Andreas "Existence of bases implies the axiom of choice". Contemporary mathematics 31, 1984). For the existence of a Hamel basis in $l_2$ it is enough to have a well-ordering of the reals. -Also, it is consistent with ZF that all linear functionals on $l_2$ are continuous (for instance, a model in which every set of reals has the property of Baire).<|endoftext|> -TITLE: What does the σ in σ-algebra stand for? -QUESTION [18 upvotes]: I was tutoring someone in analysis and realized I have no idea where this notation comes from (or analogous terms: σ-additive, σ-ring, etc). I would like to know why the letter σ was chosen. I can't think of anything relevant that starts with "S" in either English or French. My German is nearly nonexistent, but I didn't see an explanation while trying to read the German wikipedia page. -Bonus points if you can tell me who introduced this notation and when. -(By the way, I really don't like this notation very much. I think it would be much more reasonable if we just wrote "$\aleph_1$-algebra" instead. Or better yet, replaced "algebra" with a less overloaded word. But I might change my mind, if it turns out there is a good explanation for the σ!) - -REPLY [42 votes]: From Elstrodt's book Maß- und Integrationstheorie, pages 13-14: - -Bei den Wörtern „$\sigma$-Ring", „$\sigma$-Algebra" weist der Vorsatz „$\sigma$-..." darauf hin, daß das betr. - Mengensystem abgeschlossen ist bez. der Bildung abzählbarer Vereinigungen. Dabei soll der - Buchstabe $\sigma$ an „Summe" erinnern; früher bezeichnete man die Vereinigung zweier Mengen als ihre Summe (s. z.B. F. Hausdorff 1, S. 5 und S. 23). - Eine entsprechende Terminologie ist üblich mit dem Vorsatz „$\delta$..." für abzählbare Durchschnitte (z.B.„$\delta$ -Ring"). - -My translation: - -In the words "$\sigma $-ring","$\sigma$-algebra" the prefix "$\sigma$-..." indicates that the system of sets considered is closed with respect to the formation of denumerable unions. Here the letter $\sigma$ is to remind one of "Summe"[sum]; earlier one refered to the union of two sets as their sum (see for example F. Hausdorff 1, p. 5 and p. 23). - A corresponding terminology is usual with the prefix „$\delta$-..." for denumerable intersections [Durchschnitte] (for example "$\delta$ -ring") - -(The reference is to Hausdorff's Grundzüge der Mengenlehre. published in 1914.) -To sum up: the excerpt says that $\sigma$ [=Greek s] and $\delta$[=Greek d] come from the German words Summe and Durchschnitt, whose English translations are respectively sum and intersection.<|endoftext|> -TITLE: What's a magical theorem in logic? -QUESTION [32 upvotes]: Some theorems are magical: their hypotheses are easy to meet, and when invoked (as lemmas) in the midst of an otherwise routine proof, they deliver the desired conclusion more or less straightaway—like pulling a rabbit from a hat. Here are five examples. Some are from outside of logic, but each is often useful within logic. -Baire Category Theorem. In any completely metrizable topological space, each nonempty open set is nonmeager. -Gödel's Diagonal Lemma. If a theory $T$ relatively interprets Robinson's arithmetic, then for each first order formula $\varphi(x,v_1,\dots,v_n)$ in the language of $T$, there is a $\psi(v_1,\dots,v_n)$ such that $T$ proves the sentence $\forall v_1 \dots \forall v_n[\psi(v_1,\dots,v_n) \leftrightarrow \varphi(\overline\psi,v_1,\dots,v_n)]$, where $\overline\psi$ is the code of $\psi$. -König's Tree Lemma. Every finitely splitting tree of infinite height has an infinite branch. -Knaster–Tarski Theorem. Every monotone nondecreasing operator on the powerset of a set has a fixed point. -Mostowski Collapsing Lemma. If $E$ is a well founded, set like, and extensional binary class relation on a class $M$, then there is a unique transitive class $N$ such that $(M, E)$ is isomorphic to $(N, \in)$. -Let's list other magical theorems that every logician can wield. Students among us will thereby learn of useful results that might otherwise escape their attention until much later. (There is related question here. But it and most of its answers don't focus on theorems useful in logic.) Please treat only one theorem per answer, and write as many answers as you like. Don't just link to Wikipedia or whatever; give a pithy statement. If possible, keep it informal. Bonus if the theorem isn't well known, or if you show it in action. - -REPLY [4 votes]: The Feferman-Vaught theorem is a fairly magical result in logic since it allows one to figure out the truth value of a first order formula in a product of structures in terms of the individual structures and the type of product used. More generally, the Feferman-Vaught technique can be used to analyze complex structures based on their components. There are different versions of the Feferman-Vaught theorem, so I should say that the Feferman-Vaught theorem is a class of theorems or a technique rather than an individual result. This technique was developed by Feferman and Vaught in the well-known paper The first Order Properties of Products of Algebraic Systems. The following result is a version of the Feferman-Vaught theorem for reduced products. For this result, we will need the following definition. If $\psi(x_1,...,x_n)$ is a formula and $\prod_{i\in I}\mathcal{A}_{i}/Z$ is a reduced product, then for $[f]_{1},...,[f]_{n}\in\prod_{i\in I}\mathcal{A}_{i}/Z$, let $\|\psi([f]_{1},...,[f]_{n})\|=\{i\in I|\mathcal{A}_{i}\models\psi(f_{1}(i),...,f_{n}(i))\}/Z$. In other words, $\|\psi([f]_{1},...,[f]_{n})\|\in I/Z$ is the Boolean value of $\psi([f]_{1},...,[f]_{n})$. - -For each formula - $\phi(x_{1},...,x_{n})$ there is a - sequence of formulas - $(\theta;\psi_{1},...,\psi_{m})$ such - that - -the formulas $\psi_{i}$ have at most the variables $x_{1},...,x_{n}$ free -$\theta=\theta(y_{1},...,y_{m})$ is a formula in the language of Boolean - algebras and -If $I$ is a set and $Z$ is a filter on $I$ and $\mathcal{A}_{i}$ is a - first order structure for $i\in I$, - then for - $[f_{1}],...,[f_{n}]\in\prod_{i\in -> I}\mathcal{A}_{i}/Z$, we have - $$\prod_{i\in -> I}\mathcal{A}_{i}/Z\models\phi([f_{1}],...,[f_{n}])$$ - if and only if - $$P(I)/Z\models\theta(\|\psi_{1}([f_{1}],...,[f_{n}])\|,...,\|\psi_{m}([f_{1}],...,[f_{n}])\|).$$ - - -The above result also holds for the limit reduced power and the Boolean product (see the paper Sheaf Constructions and Their Elementary Properties for more details on this result). An immediate consequence of this result is that reduced products preserve elementary equivalence. In particular, direct products, direct powers, and reduced powers all preserve elementary equivalence and elementary embeddings. I posted this result since I recently used an application of a version of the Feferman-Vaught theorem to produce a result about ultrapowers and limit reduced powers.<|endoftext|> -TITLE: Parabolic convolution of perverse sheaves in terms of the Hecke algebra -QUESTION [8 upvotes]: It is "well-known" that the Hecke algebra $\mathcal{H}$ can be thought -of as the Grothendieck group for the category of perverse sheaves on -$G/B$, where the product in $\mathcal{H}$ corresponds to convolution -of sheaves by the Borel subgroup. This means, given perverse sheaves -$X$ and $Y$ on $G/B$ and their classes $[X]$ and $[Y]$ in the -Grothendieck group considered as elements of $\mathcal{H}$, the -element $[X][Y]$ is (the class of) the sheaf -$$q_B(\pi^*(X)\boxtimes Y),$$ -where $\pi$ is the quotient map $G\rightarrow G/B$ and $q_B$ is the -operation that takes a sheaf on $G\times G/B$ which is equivariant -under the action of $B$ by $b\cdot (g, hB) = (gb^{-1}, bhB)$ and -quotients out by this action, mapping to a sheaf on $G/B$. -Now I want to think about convolution using a parabolic subgroup -larger than the Borel. Suppose $P$ is a parabolic, $X$ is a perverse -sheaf on $G/B$ which is pulled back from $G/P$, and $Y$ is $P$-equivariant. -What is the class of -$$q_P(\pi^*(X)\boxtimes Y),$$ -where $q_P$ now quotients by the action of $P$ by $p\cdot (g, hB) = -(gp^{-1}, pHB)$ in terms of the Hecke algebra? -Motivation: Actually, I only care about the case where $X$ and $Y$ are -IC sheaves of Schubert varieties, so $[X]$ and $[Y]$ are -Kazhdan--Lusztig basis elements. In this case, by the Decomposition -Theorem the product will give a positive, bar--invariant sum of -Kazhdan--Lusztig basis elements, so if you find good choices of $X$ -and $Y$ and understand this product, you get a good inductive method -for calculating Kazhdan--Lusztig polynomials. In some cases I am -interested in, I have calculated the product geometrically by -localization, but sticking to algebra would make things cleaner and -likely easier to write up in general. There are cases in which this -has been done, most notably Polo's paper showing that any polynomial -with positive integer coefficients and constant term 1 is a -Kazhdan--Lusztig polynomial; an answer to this question should allow -one to dispense with geometry in his paper and formulate his -calculations entirely within the Hecke algebra (given a theorem that -says the algebraic analogue of parabolic convolution produces positive sums of Kazhdan--Lusztig elements). -Although everything makes sense for an arbitrary Kac-Moody group, I am -happy with an answer for the finite dimensional case, or even with -answers for type A. -Pre-emptive requests: (1) I expect parabolic Kazhdan--Lusztig elements -will come up. As there are several variants, please tell us which one -you mean, with reference to a paper using your notation. (2) As you -may have noted from my vague description of $q_B$ and any other -mistakes I might have made above, I don't really understand perverse -sheaves and intersection cohomology. Please feel free to correct me, -and please give me an answer I can understand. - -REPLY [3 votes]: Let $G$ be a connected reductive algebraic group (over $\mathbb{C}$) and fix a Borel subgroup $B \subset G$. One can consider the 2-category with objects parabolic subgroups $P \supset B$ and 1-morphisms $P \to Q$ given by $D^b_{P\times Q}(G)$ (the $P \times Q$-equivariant derived category of $G$ with respect to the action $(p,q) \cdot g = pgq^{-1}$ for $p \in P$, $q \in Q$ and $g \in G$). 2-morphisms are the morphisms of $D^b_{P \times Q}(G)$. Composition of 1-morphisms is given by convolution: -$* : D_{P \times Q}(G) \times D_{Q \times R}(G) \to D_{P \times R}(G)$ -(you can probably guess how this is defined using the description you give above). -Then your question is a special case of the following question: -Q: describe the Grothendieck group of this 2-category (a 1-category). -The answer is that Grothendieck group is* what I call the Schur algebroid (for want of a better name). See the beginning of my paper "Singular Soergel bimodules" on the arXiv. (I am only referring to my paper because it is a convenient reference. Certainly these things have been known to experts since at least the early 80's.) -If you would prefer not to look at this paper here is a direct answer to your question. For any subset $I \subset S$ of the simple reflections let $\underline{H}_I$ denote the Kazhdan-Lusztig basis element indexed by the longest element of the standard parabolic $W_I$. Then, given $h \in \mathcal{H}\underline{H}_I$ and $h' \in \underline{H}_I \mathcal{H}$ one can define -$h*_I h' := \frac{1}{\pi(I)} hh'$ -where $\pi(I)$ denotes the Poincaré polynomial of $W_I$. (Note that this element really lives in the $\mathbb{Z}[v,v^{-1}]$ form of the Hecke algebra.) This is the class in the Hecke algebra you are looking for. -*: Of course this is not true because one has lost the $q$: what I really mean is that one should either consider the split Grothendieck group of semi-simple complexes (or parity sheaves if one prefers) or use an appropriate mixed version. -Final note: you talk about Polo's result about arbitrary polynomials $ \in 1 + q\mathbb{N}[q]$ occuring as KL-polynomials. I recall that there is a purely combinatorial proof of this result in the literature. Unfortunately I can't remember the title or author, but it shouldn't be difficult to find.<|endoftext|> -TITLE: Are the closures of the tori in the decomposition of a torified variety toric varieties? -QUESTION [6 upvotes]: In "Torified varieties and their geometries over $\mathbb{F}_1$", J. L. Pena and O. Lorscheid define a torified variety as a variety $X$ over $\mathbb{Z}$ along with a family of locally closed subvarieties $T_i \cong \mathbb{G}_m^{r_i}$ isomorphic to algebraic tori such that $\bigsqcup_{i \in I} T_i(K) = X(K)$ for every algebraically closed field $K$. I'm trying to prove that $\Omega_{\overline{T_i}}^1(\text{log}D_i)$, the differential forms on $\overline{T_i}$ with logarithmic poles along $D_i = \overline{T_i} \setminus T_i$, is trivial for each $T_i$. I'm actually not sure if this is true but that's my hope. A similar result is true for toric varieties. If $Y$ is a toric variety with open orbit $B$, then $\Omega_{Y}^1(\text{log}(Y \setminus B))$ is trivial (proven in "Toric varieties" by Fulton, pg 87). -My specific question is this: are the $\overline{T_i}$ in $X$ toric varieties? I know this would be true if the action of $T_i$ on itself extended to an action of $T_i$ on $\overline{T_i}$ but I'm not sure if this is possible in general or if not when this is possible. Thanks so much for the help! -Edit: For completeness, the definition of toric variety I'm using is a variety $X$ with an open dense subset $B$ isomorphic to a torus such that the action of $B$ on itself extends to $X$. Thus since each $T_i$ above is locally closed by assumption, then it is an open dense subsets of $\overline{T_i}$ so I only need to guarantee that the action of $T_i$ on itself extends to the closure. -Edit 2: I'll rephrase my question in a more general way so that it maybe easier to approach: if $X$ and $Y$ are varieties, $\overline{Y}$ is a closure of $Y$ such that the inclusion of $Y$ is an open immersion, and $f:X \times Y \to Y$ is a morphism, under what conditions if any does $f$ extend to a morphism $f_* : X \times \overline{Y} \to \overline{Y}$? Furthermore, does the special case of $X$ and $Y$ being a torus $T$, and $f$ being the map $T \times T \to T, \enspace (t,s) \mapsto ts$ as above satisfy these conditions? - -REPLY [3 votes]: Sorry that I reply so late, I haven't been hanging much on MO recently. -The quick answer to your question is no, the closure of tori appearing in the torification are in general not toric varieties. If that was the case, every irreducible torified variety would be toric (as it would coincide with the closure of the unique dense torus in the torification), examples are for instance the Grassmannian $Gr(2,4)$ which is torified but not toric. -Now, what can actually help you with your problem is the following. You can initially restrict yourself to study regular torified varieties, where a torification is called regular if the closure of the tori are torified with (a subset of) the same torification: $\overline{T_i}=\bigsqcup_{j \leq i} T_j$ (where the ordering is simply "being in the closure of"). Under these conditions, the complement $\overline{T_i}\setminus T_i$ is again a torified variety (which might be disconnected, but it will surely be a disjoint union of torified varieties), with strictly smaller dimension than the one of $T_i$. -It is an open problem to decide whether every torified variety can be regularly torified, but so far in the examples we worked with we have been able to find regular torifications. -Your project sounds very interesting, I am looking forward to see what you come up with!<|endoftext|> -TITLE: A nonlinear system with special structure -QUESTION [5 upvotes]: Suppose we have an $n \times n$ uniform grid, covering $[-1,1] \times [-1,1]$ (typically, n $\approx$ 500). We have a smooth, differentiable function $z(x,y)$ that we want to determine on the nodes of this grid. At every point, the function $z(x,y)$ satisfies a relation of the form: -$\displaystyle\frac{az + b}{cz + d} = \displaystyle\frac{z_x}{z_y}$ -where $z_x$ and $z_y$ are the partial derivatives. Note that the values of $a, b, c, d$ are samples from four known smooth functions (so they are not constant at every point on the grid). How can we solve systems of equations with such structure? -[EDIT 1] (Boundary Condition): The ratio $\displaystyle\frac{z_x}{z_y}$ is known at the boundary. Thus, $z$ is known at the boundary. -[EDIT 2] Am I correct in understanding that this is a quasilinear first-order PDE and the method of characteristics will solve it up to level sets of z? Is there a robust way to solve such PDEs in the presence of noise and isolated singularities in a, b, c, d? - -REPLY [2 votes]: It seems unlikely that your problem is going to have a solution in the generality you've described but here goes. There are two approaches you could try. -(1) Discretize the partial derivatives using finite differences on the grid. At every point on the grid your PDE will give you a nonlinear equation. You will have a total of $n^2$ equations in $n^2$ unknowns. But each equation will only contain a small number of unknowns. You can use a variant Newton's method to solve these equations. But you have to work with the restrictions that you probably don't want to code up the Jacobian for the system, and you could not use a direct method to calculate the solution even if you did. I would recommend looking at a Jacobian-Free Newton-Krylov method. -(2) Exploit the method of characteristics. Rewrite your equation as -$$ -(cz+d) z_x - (az+b) z_y = 0. -$$ -This gives characteristic equations -$$ -\frac{dx}{dt}=cz+d, \ \ \ \frac{dy}{dt}=a z+b, \ \ \ \frac{dz}{dt}= 0. -$$ -There are numerical methods for solving nonlinear hyperbolic equations exploiting characteristics. Sethian and Vladimirsky have a nice one. Your problem does not quite fit into their scheme but their paper might help give you ideas. -So, if you problem does have a solution, one of these might work. I would expect (1) to be more robust than (2), but also more expensive.<|endoftext|> -TITLE: Random graphs in $\mathbb R^2$ (or random rays from $\mathbb Z^2$) -QUESTION [15 upvotes]: The model: -Suppose that for each lattice point in $\mathbb Z^2$ we pick a random direction uniformly and independently. At time $t=0$ we start drawing rays starting from each lattice point in the chosen directions with unit velocity. The drawing of a ray will continue until it intersects another ray, at which point both rays stop. -In the end we are left with a graph $G$ (more like a union of segments, but let's pretend it's a graph), and I would like to understand it's properties. -Question: -This seems at first like it might share some properties with percolation in $\mathbb Z^2$, but I'm not so sure. I'd like to know how the connected components are distributed. For example, say we restrict to the lattice points inside the rectangle $[0,n]\times[0,m]$, and perform the process above (with the modification that if a ray hits the boundary rectangle it also stops). What is the expected number of connected components in $G$? (the boundary rectangle is not a part of $G$). -(There used to be a question here about cycles formed, but I removed it since it is more appropriate for motorcycle graphs mentioned in the answer below. For now, I want to focus on connected components. jc's answer seems to indicate that connected components will usually be small.) - -REPLY [4 votes]: edit: -At Tom LaGatta's request, here are a few more very rough graphs relating to the number of components. For $n=3,\dots,15$, I ran just 10 instances of each (took about 15-20 minutes). Hopefully that's enough to give a little bit of the idea of the spread. -Here are the total number of components (with size>1) versus $n$. I noticed that it appeared parabolic, so I plotted it on a log-log scale with the function $n^2/3$ (the factor of 1/3 just happened to be close). Each data point corresponds to a different instance. - -Here on semi-log scales are the distributions of numbers of components (with size>1) for the various grid sizes I considered. The different colors correspond to distributions observed in different instances. They look much better on semi-log as opposed to log-log, though that's fairly meaningless with not even a decade of data and small number effects. - -Code for this section (run with the functions in the notebook linked to below): -(* code to generate data, change parameters as needed *) -minn = 3;maxn = 15;numruns = 10; -data = Table[test = Flatten[makegrid[n, n], 1];Truncate[test], {j, numruns}, {n, minn, maxn}]; -(* code to process and plot data *) -averagecomponents = Table[{n, Length[data[[i, n - minn + 1, 2]]]}, {n, minn, maxn}, {i, -numruns}]; -plot1 = ListLogLogPlot[averagecomponents, AxesLabel -> {"n", "total number of components"}];plot2 = LogLogPlot[x^2/3, {x, 1, maxn}];Show[plot1, plot2] -compdist = Table[Table[Tally[Table[Length[data[[j, n, 2, i]]], {i, averagecomponents[[n, j, 2]]}]], {j, numruns}], {n, maxn-minn+1}]; -ListLogPlot[compdist[[12]], PlotStyle -> PointSize[Large]] -ListLogPlot[compdist[[13]], PlotStyle -> PointSize[Large]] -GraphicsGrid[Partition[Table[ListLogPlot[compdist[[j]], PlotStyle -> PointSize[Medium], PlotLabel -> "n=" <> ToString[j + minn - 1], AxesLabel -> {"size of component", "frequency"}], {j, maxn - minn + 1}], 3, 3, {1, 1}, {}]] -(* not shown: number of infinite rays *) -numinfinities = Table[{n, Count[data[[i, n - minn + 1, 1]], Infinity]}, {n, minn, maxn}, {i, numruns}] - - -original answer: -I wrote a bit of Mathematica code to try to simulate your process (link here). I only partially commented things, so let me know if you have trouble understanding how to use it. More on this after some pictures and graphs. -I apologize in advance, I don't have the time now to do a real systematic study or gather any real statistics on my computer. I'll have to leave it to you or other interested readers... -I didn't bother to add boundary conditions to cut off the rays leaving the grid. Below, components of size 1 are just rays that don't terminate. -5x5: - -8x8: - -15x15 grids: -(1) -(2) -Number of components of a given size for 15x15: -(1) -(2) -Stopping time distribution for 15x15 (excluding infinite rays, of which there are 6 in #1 and 7 in #2): -(1) -(2) -Note: I didn't see any cycles at all, but I had to check this by eye, since I didn't write a routine to extract the graph structure of the connected components. What I have now just records their size. -There is what looks like a cycle in the upper left quadrant of 15x15 #1, but you can easily convince yourself by comparing the lengths of the edges that the rays couldn't actually touch. - -I used pretty much the most naïve algorithm I could think of: - -Generate a pseudorandom angle at each grid point, and thus get a set of lines. -Find the intersection times of all pairs of lines. (Lines are naturally parametrized as $x(t)=x_0+t(\cos\theta,\sin\theta)$, where $x_0$ is the starting point and $\theta$ is the angle.) -From this set of pairs, select pairs of rays that intersect in positive time rather than negative, and sort from smallest time to largest. -Choose the intersection that occurs at the smallest time, mark the two rays involved as truncated, and record the connected component formed by this as a new one, and record the time of intersection as the "ending time" for the two rays. -Check the intersection occurring at the next smallest time to see whether it's been preempted by existing rays (by comparing the intersection times with ending times), and if the intersection passes the test, mark these rays as truncated, and record the connected component and any new ending times as in 4. -Repeat 5. until all intersections or all rays have been eliminated. - - -What are smarter ways of doing this? - -Because of 2., the code runs rather slowly, probably $O(n^4)$ for an $n$ by $n$ grid ($n^2$ grid points, thus $(n^2)^2$ possible intersections). An instance at $n=15$ takes about 40 seconds on my machine to run.<|endoftext|> -TITLE: Closed form or/and asymptotics of a hypergeometric sum -QUESTION [6 upvotes]: Dear mathematicians, -I am a computer scientist wandering in the deep sea of combinatorics and asymptotics to pursue a recent interest in average case analysis of algorithms. In doing so, I designed the following double summation: -$$ -\sum_{h = 1}^{n}\sum_{j=0}^{n-h}h{2j+h-1\choose j}{2n - 2j - h\choose n-j}, -$$ -which, divided by the $n$th Catalan number, yields the statistics I want. Since the variable $j$ occurs in many places, I thought that I could exchange the summations without worrying about the bounds and then work on $h$ instead: -$$ -\sum_{j\geq 0}\sum_{h>0}h{(2j-1)+h\choose j}{2(n-j)-h\choose n -j}. -$$ -Does anyone know if there is a closed form for -$$ -\sum_{h>0}{h}{p+h\choose q}{2r - h \choose r}? -$$ -I read the relevant chapter in Concrete Mathematics, to no avail. Perhaps generating functions would help? Anyhow, I would be interested in the main term of the asymptotic expansion of the original double summation. -Thanks for reading me. - -REPLY [4 votes]: For your sequence -s:=[0, 1, 7, 39, 198, 955, 4458, 20342, 91276, 404307, 1772610, 7707106, 33278292, 142853854, 610170148, 2594956620, 10994256152, 46425048451, 195456931506, 820725032042, 3438011713540] - -I use with(gfun) in Maple and then guessgf(s, x) to find out that the generating function for your sequence is -$$ -\frac{x(1+\sqrt{1-4x})}{2(1-4x)^2}. -$$ -This isn't now hard to establish rigorously by verifying that your double sum satisfies a polynomial recurrence (a multivariate version of the Gosper-Zeilberger creative telescoping).<|endoftext|> -TITLE: Fréchet manifolds vs ILH manifolds -QUESTION [7 upvotes]: What is the precise relation between ILH manifolds and Fréchet manifolds? Specifically: - -Does any ILH manifold has a canonical structure of a Fréchet manifold? -If so, is it true that any ILH submanifold is a Fréchet sumanifold? - - Background. -The notion of an ILH (Inverse Limit Hilbert) manifold was developed by Omori in his studies of diffeomorphism groups. Very roughly it is a manifold modelled on a topological vector space that is the inverse limit of a countable family of Hilbert spaces. This object is easier to deal with than the usual Fréchet manifolds due to the avaiability of the inverse function theorem. -For all I know the inverse limit of Hilbert spaces need not be Fréchet. More precisely, it seems that the inverse limit inherits a countable family of seminorms from the Hilbert spaces, but I see no reason for the inverse limit to be complete (as a uniform space), and I am not sure if the inverse limit is always Hausdorff. -The ILH manifold I am trying to understand is the diffeomorphism group of a compact manifold, so in particular, it is what Omori called a strong ILH manifold, which perhaps makes a difference in answering 1-2. The diffeomorphism group is also a Fréchet manifold (indeed, it is the so called Fréchet-Lie group), but I am not sure how ILH and Fréchet manifold structures interact. - -REPLY [2 votes]: Dear Igor -In order to answer your questions we first need to ensure what we mean by a Fréchet manifold. -Obviously a Fréchet manifold should be modeled on Fréchet spaces, so the transition maps should be smooth maps between Fréchet spaces. Thats where the problem lies. -How is differentiability of maps between Fréchet spaces defined ? -There is no natural (canonical) way to extend the classical notion of -differentiability from Banach spaces to Fréchet spaces. -(because there is no norm in a Fréchet space) -There are many inequivalent ways to define differentiability in -Fréchet spaces, and the choice may depend o the applications in mind. -For some of them there are even Nash-Moser type Inverse Function Theorems. -That was the main reason we considered the ILH structure when studying -diffeomorphism groups, or the groups of pseudo differential operators or Fourier integral operators. -For $Diff^{\infty}(M)$ the Fréchet spaces are given as spaces of smooth sections of certain vector bundles and the Fréchet topology of $Diff^{\infty}(M)$ is known as the smooth compact-open topology. -In other words, we don't start with a sequence of Hilbert spaces to get as -inverse limit our Fréchet space, but we have the Fréchet space to begin with -and consider it as inverse limit of Hilbert spaces by changing the topologies. -This way we have to our proposal the whole powerful theory of Hilbert spaces. -Rudolf Schmid<|endoftext|> -TITLE: How to construct a scalar differential operator having the same spectrum as a non-scalar differential operator exploiting symmetries? -QUESTION [7 upvotes]: I am interested in eigenvalue problems for differential operators acting on one forms on closed two-dimensional manifolds and how they relate to eigenvalue problems of associated operators acting on scalar functions. -For the discussion I assume $M$ to be a compact two-dimensional Riemannian manifold without boundary. Let $\Delta^k=dd^*+d^*d$ be the De Rham Laplacian on the space $\Omega^k(M)$ of real-valued $k$-Forms. Here $d:\Omega^k(M) \to \Omega^{k+1}(M)$ is the exterior derivative and $d^*: \Omega^{k+1}(M) \to \Omega^{k}$ is its adjoint. -For $k=0$ we get the Laplace Beltrami Operator acting on functions: $\Delta^0=d^*d$. -Consider the eigenvalue problem: Find $\lambda \in \mathbb{R}$ and an $1$-Form $\alpha$ such that $\Delta^1\alpha = \lambda \alpha$. Because of the identity -$d^* \Delta^1 = \Delta^0 d^* $ -any eigenform $\alpha$ of $\Delta^1$ yields an eigenfunction $f:= d^*\alpha$ of $\Delta^0$ for the same eigenvalue, provided that $\alpha$ is not co-closed. -Also, since $\Delta^1$ commutes with the Hodge star we have that with any eigenform $\alpha$, the $1$-form $*\alpha$ (imagine $\alpha$ being rotated pointwise by 90 degrees) is also a linearly independent eigenform for the same eigenvalue. Therefore all eigenspaces of $\Delta^1$ have even dimension and are invariant under the symmetry -$$\alpha \mapsto a \alpha + b (*\alpha) \qquad a,b \in \mathbb{R}$$ -To sum up, the spectra of $\Delta^1$ and $\Delta^0$ are closely related: They are essentially the same except for the multiplicities of the eigenvalues. More precisely, any non-zero eigenvalue of the scalar operator $\Delta^0$ with multiplicity $m$ becomes an eigenvalue of the $1$-form operator $\Delta^1$ with multiplicity $2m$. -Now for the question: Does a similar relation hold for other differential operators? For example, I consider the Bochner Laplacian $\widehat{\Delta^1} = \nabla^* \nabla$ on $T^*M$. The Weizenböck identity -$$\Delta^1 \alpha = \widehat{\Delta^1} \alpha + K\cdot \alpha$$ -shows that the Bochner Laplacian and the De Rham Laplacian on $1$-Forms differ only by a an operator that is a pointwise multiplication with the Gaussian curvature $K$. Moreover, the Bochner Laplacian also commutes with the Hodge star $*$, therefore all eigenspaces of $\widehat{\Delta^1}$ have even dimension and are invariant under the symmetry mentioned above. So far, the situation looks similar. Now, is there an operator $\widehat{\Delta^0}$ acting on $\Omega^0(M)$ and an operator -$\widehat{d^*}: \Omega^1(M) \to \Omega^0(M)$ such that the following intertwining relation holds? -$$\widehat{d^*} \ \widehat{\Delta^1} = \widehat{\Delta^0} \ \widehat{ d^* }$$ -In that case the the spectrum of the Bochner Laplacian on $1$-Forms would be essentially equal (up to multiplicity as discussed above) to the spectrum of the unknown scalar operator $\widehat{\Delta^0}$. Is this possible? More generally, for what other -operators (apart from the De Rham Laplacian) is such a "reduction" possible? -Any references would be appreciated. -Thanks. -Update: (in reply to the answer of Robert Bryant below) -I have tried to spell out the calculations leading to some of the results in the answer below in some more detail in order to understand them by myself and for future reference. Unfortunately I am not familiar with the principal symbol calculus, so I apologize for my low-level approach. My goal is to calculate the eigenvalues for the Bochner Laplacian numerically. From the implementation point of view it is easier to deal with scalar valued second order equations than with vector (or in this case 1-form valued equations). I think the special case below is instructive. I wonder if it is possible to avoid the need of putting restrictions on the metric. Of course, that would be great. But it would be also interesting to have an argument that says that it is impossible in the general case. -Anyway, In the special case -$$\widehat{\Delta^1} \alpha := \nabla^*\nabla \alpha + L \alpha = \Delta^1 \alpha + (L-K) \alpha$$ -$$\widehat{\Delta^0f} := \Delta^0 f + H f$$ -$$\widehat{d^*}\alpha :=d^* \alpha + \langle \phi,\alpha \rangle $$ -we get -$$\begin{aligned} E \alpha &:= \widehat{d^*} \widehat{\Delta^1}\alpha - \widehat{\Delta^0} \widehat{d^*} \alpha \\& = \widehat{d^*}(\Delta^1\alpha+(L-K)\alpha)-\widehat{\Delta^0}(d^*\alpha + \langle \phi,\alpha\rangle) \\ -& =d^*\Delta^1\alpha+d^*((L-K)\alpha)+\langle \phi,\Delta^1\alpha \rangle + \langle \phi, (L-K) \alpha \rangle \\ & -\Delta^0 d^*\alpha -Hd^*\alpha -\Delta^0\langle \phi,\alpha \rangle -H \langle \phi,\alpha \rangle\end{aligned}$$ -Because of the identities -$$ \begin{aligned} -d^*( (L-K) \alpha ) &= (L-K)d^*\alpha -\langle d(L-K) ,\alpha \rangle \\ -\Delta^0 \langle \phi,\alpha \rangle & = \langle \nabla^*\nabla -\phi,\alpha \rangle + \langle \phi,\nabla^*\nabla \alpha \rangle - 2 -\langle \nabla \phi, \nabla \alpha \rangle \\ -\nabla^*\nabla &= \Delta^1 - K\\ -d^*\alpha &= -\nabla^i \alpha_i = -g_{ab} g^{ak} g^{bi} \nabla_k\alpha_i = \langle -g, \nabla \alpha \rangle \\ -d^*\Delta^1&=\Delta^0 d^* \end{aligned} -$$ -the third and second order terms in $\alpha$ cancel, yielding the -following first-order operator: -$$E\alpha = \langle -g(L-K-H) + 2\nabla \phi, \nabla \alpha \rangle + -\langle (L-K-H)\phi - d(L-K)- \nabla^*\nabla \phi + K \phi, \alpha \rangle $$ -Now I see that in order for the first-order terms to vanish for all -$\alpha$ we need $\nabla\phi = f g$ for $f := \frac{1}{2}(L-K-H)$. -Taking the covariant derivative of this equation yields $\nabla \nabla \phi = df \otimes g + df \otimes 0$ and taking the trace yields $-\nabla^*\nabla \phi = -df$. Therefore, if we set $df = -K\phi$, the zeroth-order part reduces -to the operator -$$E\alpha = \langle 2f\phi -d(L-K),\alpha \rangle$$ -I have two questions at this point: Why is $df =-K\phi$ necessary for $E$ to vanish? -More important (since that leads to the restrictions on the metric): -Why is $*\phi$ dual to a Killing field? -Maybe this is obvious but i don't see it. - -REPLY [8 votes]: I thought a little bit about your question, which is phrased a little more generally than I like, but I decided to think about it with the restrictions that $\widehat{\Delta^0}$ be a differential operator and $\widehat{d^\ast}$ be $d^\ast$ plus a lower-order (i.e., zeroth order) differential operator. I also decided to think, not of the most general perturbation of $\widehat{\Delta^1}$, but of perturbations of the form $\widehat{\Delta^1} = \nabla^\ast\nabla+L$, where $L$ means scalar multiplication by a function $L$ on $M$. -The first thing to notice is that, since the principal symbol $\sigma^1_\xi$ of $\widehat{\Delta^1}$ is just scalar multiplication by $|\xi|^2$ and the principal symbol of $\widehat{d^\ast}$ is $\alpha\mapsto \xi\cdot\alpha$, it follows from the symbol calculus that the principal symbol of $\widehat{\Delta^0}$ must also be scalar multiplication by $|\xi|^2$, i.e., the differential operator $\widehat{\Delta^0} - \Delta^0$ must be of order at most $1$. However, it must also be self-adjoint, and this implies that it must be of order $0$, i.e., that $\widehat{\Delta^0} = \Delta^0 + H$ (where '$H$' means scalar multiplication by a smooth function $H$ on $M$). -Now, $\widehat{d^\ast}\alpha = d^\ast\alpha + \phi\cdot\alpha$ for some $1$-form $\phi$ on $M$. Under the given assumptions, it is not hard to see that the equation -$$ -\widehat{d^\ast}\widehat{\Delta^1} = \widehat{\Delta^0}\widehat{d^\ast} -$$ -implies that $\widehat{\Delta^0} = \Delta^0 + H$ for some function $H$ on $M$. -(This is what I just explained above.) -Now, the operator $E = \widehat{d^\ast}\widehat{\Delta^1} - \widehat{\Delta^0}\widehat{d^\ast}$, when expanded out, is of first order (not second order, as you might have expected). Looking at the principal symbol of $E$ and setting this equal to zero gives $4$ equations, and these are equivalent to $\nabla\phi = f\ g$, where, for simplicity, I have set $f = \tfrac12(L-K-H)$. Taking the covariant derivative of both sides of $\nabla\phi = f\ g$ and using the definition of $K$, one gets $df = -K\ \phi$. Substituting this back into the expression for $E$, it reduces to a $0$-th order operator which turns out to be $E(\alpha) = (2f\phi +dK - dL)\cdot\alpha$. Setting this equal to zero gives $d(|\phi|^2 + K - L) = 0$, since $d\bigl(|\phi|^2\bigr) = 2f\phi$ (which is a consequence of $\nabla\phi = f\ g$). Thus, $L = K + |\phi|^2 - c$ for some constant $c$. -Thus, one finds that one must have the identities -$$ -\nabla \phi = f\ g\qquad\text{and}\qquad df = -K\ \phi -$$ -for some function $f$ on $M$ while -$$ -L = K + |\phi|^2 - c\qquad\text{and}\qquad H = |\phi|^2 -2f - c -$$ -for some constant $c$. -Obviously, there is always the trivial solution $(\phi,f) = (0,0)$, which gives the well-known intertwining of the Hodge Laplacian on $1$-forms and $0$-forms. More interesting is the case when there are nontrivial solutions $(\phi,f)$ to the above equations. This puts severe restrictions on the metric $g$, but these can be understood. -Let's assume that $M$ is connected and complete. Then the pair of equations $\nabla\phi = f\ g$ and $df = -K\ \phi$ forms a linear total differential system, so if the pair $(\phi,f)$ vanishes anywhere, it vanishes identically. Let's assume that it does not. The equation $\nabla\phi = f\ g$ implies that $d\phi = 0$ and that the vector field $F$ that is $g$-dual to $\ast\phi$ is a Killing field. Thus, $(M,g)$ is, at least locally, a surface of revolution. Any fixed points of $F$ are isolated elliptic points. Write $\phi = du$ for some function $u$ (at the moment locally defined). Note that the critical points of $u$ are nondegenerate and of index 0 or 2. It is then not hard to show that $|\phi|^2 = a(u)$ for some function $a$ and that, in the region where $a>0$, the metric $g$ takes the form -$$ -g = \frac{du^2}{a(u)} + a(u)\ d\psi^2 -$$ -for some (locally defined) function $\psi$. In these coordinates, one has $\phi = du$, $f = \tfrac12a'(u)$, and $K = -\tfrac12a''(u)$. One also has $L = a(u) -\tfrac12a''(u)- c$ and $H = a(u) - a'(u) - c$. -Conversely, if one starts with a function $a$ positive on some domain on the $u$-line, then the above formulae give a solution to the intertwining equation. If $a$ is positive and periodic on the $u$-line, then one obtains a solution on the torus. -One can also obtain solutions on the $2$-sphere: If $a$ is smooth and satisfies $a(u_0) = a(u_1) = 0$ (where $ u_0 < u_1 $) while $a$ is positive between $u_0$ and $u_1$ and satisfies $a'(u_0) = -a'(u_1) = b >0$, then the metric -$$ -g = \frac{du^2}{a(u)} + \frac{4a(u)}{b^2}\ d\theta^2 -$$ -defines a smooth metric on the $2$-sphere with $(u,\theta)$ as 'polar coordinates' (with $\theta$ being $2\pi$-periodic and $u_0\le u\le u_1$, with $u=u_i$ corresponding to the 'poles' of rotation), and this gives a family of solutions to the intertwining equation. -To get $\widehat{\Delta^1}$ to be the Bochner Laplacian, one must have $L = 0$, which is equivalent to $a''(u) = 2\bigl(a(u)-c\bigr)$. This has 'spherical solutions', such as -$$ -a(u) = c - e\ \cosh\bigl(\sqrt{2}\ u\bigr), -$$ -where the constants $c$ and $e$ satisfy $c > e > 0$. Thus, there is a nontrivial $2$-parameter family of metrics on the $2$-sphere such that the Bochner Laplacian has the desired intertwining property. -A similar calculation can be done for the more general case when $\widehat{d^\ast}$ is allowed to be somewhat more general (but still underdetermined elliptic), but I won't go into that unless someone is interested. I'll just say that, aside from the metrics above, the only metrics that admit a nontrivial solution when $\widehat{d^\ast}$ is an underdetermined elliptic first order differential operator are the metrics of the form -$$ -g = \bigl( a\ (x^2{+}y^2) + 2b\ x + c\bigr)\ \bigl(dx^2 + dy^2\bigr) -$$ -where $a$, $b$, and $c$ are constants such that the open set $M$ in the $xy$-plane defined by $a\ (x^2{+}y^2) + 2b\ x + c > 0$ is nonempty. Thus, the metrics that allow this are very restricted.<|endoftext|> -TITLE: Diagonalization of a matrix of differential operators -QUESTION [5 upvotes]: Dear community, -i have a question regarding differential operators acting on vector valued functions and how to "diagonalize" them. -To explain my question i will use an example: -Let $V^k$ be the space of twice differentiable functions $U:[0,2\pi] \to \mathbb{R}^k$ -with periodic boundary conditions. -Consider the differential operator $L:V^2\to V^2$ defined via -$L :=\begin{pmatrix} -\partial^2 & 0 \\ 0 & -\partial^2 \end{pmatrix}$ -That means it acts on $U\in V$, $U(t)=(u(t),v(t))^T$, by mapping it to $LU=(-u''(t),-v''(t))^T$. -$L$ is self adjoined with respect to the scalar product -$(U,V) := \int_0^{2\pi} U^T V \ dt$. So it has real eigenvalues. -It commutes with the operator $J=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. -In fact, the spectrum of $L$ consists of all $\lambda=k^2$ with $k \in \mathbb{Z}$ with multiplicity 4. The spectrum of $-\partial^2:V^1 \to V^1$ is the same, except that the multiplicities of the eigenvalues are halved. This is evident from the diagonal form of $L$. -Now, what happens for other self-adjoined operators on $V^2$ that commute with $J$? -For example consider the operator $M:V^2 \to V^2$ defined by -$M :=\begin{pmatrix} -\partial^2 & -\partial \\ \partial & -\partial^2 \end{pmatrix}$ -It is also self-adjoined with respect to the above mentioned inner product and it also commutes with $J$. Is it possible to "diagonalize" this operator into a form -$\begin{pmatrix} m & 0 \\ 0 & m \end{pmatrix}$ -with a scalar differential operator $m: V^1 \to V^1$ having the same spectrum as $M$ except for the multiplicities? -Any references would be appreciated. Thanks. - -REPLY [5 votes]: Your matrix is a matrix with coefficients in the ring $\mathbb C[\partial]$ of differential operators with constant coefficients, which happens to be a principal ideal domain. Therefore we can take your matrix — let's call it $A$ — to its Smith normal form. In this case, the normal form is $$S=\left( -\begin{array}{cc} - \partial & 0 \\\\ - 0 & \partial ^3+\partial -\end{array} -\right)$$ -Indeed, if we let $$P=\left( -\begin{array}{cc} - -1 & 0 \\\\ - -\partial & 1 -\end{array} -\right)\qquad\text{and}\qquad Q=\left( -\begin{array}{cc} - 0 & 1 \\\\ - 1 & -\partial -\end{array} -\right),$$ which are invertible, we have $$PAQ=S.$$ This means that you can change coordinates to get a diagonal matrix. -(Of course, in your situation you want to consider $A$ as defining an endomorphism, so you probably want to restrict changes of coordinates which coincide in the domain and the codomain of the map...)<|endoftext|> -TITLE: Maiden Names vs. Married Names -QUESTION [18 upvotes]: Is there a set convention for which name (maiden name or married name) a female married mathematician should use? -While this question addresses women's maiden name it applies equally to men's maiden name when it differs from their married name. The question seeks for an advice for the dilemma: whether to use the maiden name or the new married name. -For example, Fan Chung is married to Ron Graham, but she publishes under "Fan Chung." Vera T. Sós is another married woman who continued to use her maiden name, but the T. stands for Turán. Yet, I'm pretty sure that Emma Lehmer (née Trotskaia) published under her married name. -Does it have something to do with the name under which the woman first publishes or the name under which name she receives her Ph.D.? - -REPLY [12 votes]: There are numerous different circumstances when people marry and/or change their names. There is clearly no general rule on this, as the question hints already. Let me describe the ways. -First, conventions. There are not two but really four types of surnames we are talking about: -a) full legal name (usually found in passports) -b) professional name (university websites, wikipedia) -c) pen name (used to author papers, see arXiv, mathscinet) -d) maiden names and other former names -I believe neither two have to be the same, some people have more than one version of at least one of these items (say, have two passports from different countries, or publish under two different names, whatever), and occasionally people have different all four. -EXAMPLES - -a) Julia Hall Bowman Robinson, b),c) Julia Robinson, d) Julia Bowman -b) Sofia Kovalevskaya, c) Sophie Kowalevski, d) Sofia Vasilyevna Korvin-Krukovsky -as for a), I am not sure if there was a passport back then; if issued today she would have a passport in the name Sofia Vasilyevna Kovalevskaya (modern transliteration, changed several times) -a),b) Mary Ellen Rudin, d) Mary Ellen Estill, c) published first as Estill, then as Rudin -a),b),c) Cathleen Synge Morawetz, d) Cathleen Synge (published also with S. as an initial) -a) Ruth Elke Lawrence-Naimark, b),d) Ruth Elke Lawrence, c) Ruth Lawrence -a),b) Dmitry Feichtner-Kozlov, c),d) Dmitry N. Kozlov (although Russian passport is notoriously difficult to change, and usually continues to have maiden name). -a),b),c) Lane A. Hemaspaandra, d) Lane A. Hemachandra (this is an example of this trend sometimes considered bogus) - -Browsing here will give you many more different examples. For legal background in the US, see here. -P.S. To further appreciate complexity arising sometimes, consider e.g. this explanation by Paco Santos (ht Gil Kalai).<|endoftext|> -TITLE: rational function identity -QUESTION [17 upvotes]: I just had to make use of an elementary rational function identity (below). The proof is a straightforward exercise, but that isn't the point. First, "my" identity is almost surely -not original, but I don't have a reference for it. Perhaps someone knows it (like a lost cat without a collar) or, more likely, could spot this as a special case of a more general identity. Second, the obvious proof is not much of an explanation: a combinatorial identity often arises for a conceptual reason, and I'd be happy to hear if anyone sees mathematics behind this one. -Let $f(x_1,\ldots,x_n)=\prod_{p=1}^n\big(\sum_{i=p}^n x_i\big)^{-1}$. Then -$$ -f(x_1,\ldots,x_n)+f(x_2,x_1,x_3,\ldots,x_n)+\cdots+f(x_2,\ldots,x_n,x_1)=\big(\sum_{i=1}^n x_i\big)/x_1\cdot f(x_1,\ldots,x_n), -$$ -where $x_1$ appears as the $i$th argument to $f$ in the $i$th summand on the left side, for $1\leq i\leq n$. But why? - -REPLY [10 votes]: A simple proof of the Sh(a,b) cat, using iterated integrals, is as follows. -Note that -$$ -f(x_1,\ldots,x_n)=\int_{1>t_1>\cdots>t_n>0} dt_1\cdots dt_n -\ t_1^{x_1-1}\cdots t_n^{x_n-1}\ . -$$ -Littlewood's identity follows from changing variables using the permutation so as to keep the -integrand fixed. Then one has a sum of simplices (corresponding to all possible relative orderings of the variables) which recombines into a cube of integration -$[0,1]^n$. -The proof of the Sh(a,b) identity follows the same idea. Here the total volume of integration is a product of simplices which is broken into a union of simplices. -This is probably well known to people working with moulds, operads, etc. -An additional remark: -Littlewood's identity follows from Lemma II.2 in my article -"Trees forests and jungles: a botanical garden for cluster expansions" -with V. Rivasseau. -To see this, extract the coefficient of the highest degree monomial in the v variables (notations of that article), then specialize the u variables to the case where -$u_{i, i+1}=x_i$ and all other pair variables are zero (killing all edges of the complete graph which are not in a `spanning chain'). -The Lemma in our article is related to many other topics in mathematical physics such as the Wilson-Polchinski renormalization group equation, -see e.g. these slides.<|endoftext|> -TITLE: Crystalline analogue of perverse sheaves -QUESTION [10 upvotes]: Consider a variety $X$ over a field $k$ and let $\ell$ be a prime different from the characteristic of $k$. One has the derived category $D(X, Q_{\ell})$ of $\ell$-adic sheaves. There are very important abelian subcategories (of perverse sheaves) corresponding to $t$-structures given by various perversity conditions. -Question: -What is the precise crystalline analogue of $D(X, Q_{\ell})$ and the subcategories of perverse sheaves? Here I am thinking of crystalline as being some sort of "$p$-adic sheaves" where $p$ is the characteristic of $k$. Perhaps one needs to assume $k$ is perfect.. -Thank you in advance for your help. - -REPLY [3 votes]: Your question seems to contain an implicit assumption that analogues are unique. It is not clear to me that for any choice of formalism you will get precise analogues of all the theorems you see in the $\ell$-adic world. -If $k$ is the complex field, you can apply the Riemann-Hilbert correspondence to get an equivalence between regular holonomic $\mathcal{D}_X$-modules and middle-perverse sheaves in the analytic topology with complex coefficients. Then, you can look for a $p$-adic analogue of regular holonomic $\mathcal{D}_X$-modules, and as Emerton commented, one possible answer is provided by Berthelot's theory of $\mathcal{D}^\dagger$-modules. Caro has some work on the ArXiv on overholonomic modules that produces the same sort of finiteness under smooth proper maps you get from the $\ell$-adic setting, but I do not know enough about this field to commment further.<|endoftext|> -TITLE: Any good reference for Tits Building? -QUESTION [17 upvotes]: For beginers, any suggestions? - -REPLY [8 votes]: Since this question was asked, Brent Everitt wrote A (very short) introduction to buildings. It comes with many pictures to help you visualize them, and a long list of pointers to where to find out more about each aspect in the literature.<|endoftext|> -TITLE: Magnus hierarchy of 1-related groups -QUESTION [8 upvotes]: By Magnus-Moldavansky theorem (see Wilhelm Magnus, Abraham Karrass, Donald Solitar, Combinatorial group theory. Presentations of groups in terms of generators and relations, Reprint of the 1976 second edition, Dover Publications, Inc., Mineola, NY, 2004 and http://arxiv.org/PS_cache/math/pdf/0608/0608635v3.pdf), every 1-related group is an HNN extensions of a "smaller" 1-related group. Say, if one letter occurs in the relator with total exponent 0, one can take this letter as the free letter in the HNN extension. - Question. Is it true that in this HNN extension one of the associated subgroups is always undistorted? Is it possible that one can always represent the group as an HNN extension satisfying that property (one of the associated subgroups is undistorted)? -For example, in some sense the "worst" 1-related group is the Baumslag group $$\langle a,t\mid a^{a^t}=a^2\rangle=\langle a,b,t\mid a^b=a^2, a^t=b\rangle.$$ In that case $t$ is a free letter, the base group is $\langle a,b \mid a^b=a^2\rangle$, the associated subgroups are $\langle a\rangle$ (exponentially distorted in the base group) and $\langle b \rangle$ (undistorted in the base group). - -REPLY [6 votes]: You can reverse engineer distortion. Consider -the group $\langle a_0, a_1, a_2 |a_1^{a_2a_0}= a_1^2\rangle$. By the Freiheitssatz, the subgroups generated by $\langle a_0, a_1\rangle$ and $\langle a_1, a_2\rangle$ are free. But both are distorted, since $a_1$ is conjugate to $a_1^2$ (if I understand the notion of distortion correctly). -Now, take the HNN extension of these two free subgroups to get a 1-relator group -$$\langle a_0, a_1, a_2, t | a_1^{a_2a_0}= a_1^2, a_0^t = a_1, a_1^t=a_2\rangle=\langle a, t | (a^t)^{a^{t^2}a} = (a^t)^2\rangle.$$<|endoftext|> -TITLE: Computational methods for dealing with geometrically complicated solid boundaries in fluid-air interface problems -QUESTION [8 upvotes]: Hello, -I am a PhD student who does not have extensive computational experience seeking advice from those experienced with computational modelling as to which method would be most appropriate for solving my particular problem. -Background -Physical Scenario -The Salvinia is a small floating fern. Its leaves have upon them a forest-like structure of fronds, with a particular shape and particular regions of hydrophobia and hydrophilia. This structure, also found elsewhere in nature, allows the Salvinia to maintain a persistent, stable air layer on its surface due to the phenomenon of surface tension. -For those familiar with the phenomenon of capillary action, this physical scenario is closely related, and involves many of the same considerations. -Mathematics -In brief, the interface between the air and the water is often constructed according to the method of Gauss. Operating on a variational principle, this method involves writing the free surface energy, the "wetting energy" due to contact with the solid boundaries (fronds) and gravitational potential (if desired) as action functionals to be minimised. It is also usually desirable to impose a condition on the volume bounded by the surface, specifying that it should not change under the variation, in order to fix a unique surface with respect to translations. -Classical formulations have viewed the surface as a height function over a euclidean domain. This can cause problems when the surface curves back on itself, as in the case of some sessile drops, for example. Thus, I have written the action functionals in terms of functions $X^A$ (where $A=1,2,3$) which define the embedding of the surface into three-dimensional euclidean space. -Why is this a problem? -The difficulty I have encountered is that the solid boundaries (fronds) do not have a simple geometry. Take, for example, the case where the fronds are cylinders. It is then possible to define the surface as a function $u(x,y)$ over a domain $\Omega$. The functional is can then be decomposed into an interior term and a boundary term using the divergence theorem, and solved using a finite element method. -Now consider the (still relatively simple) case where the frond is a cone, rather than a cylinder. Now, as the surface moves up and down vertically, the location and shape of the boundary as viewed in the $(x,y)$ plane changes, depending on the height and curvature of the surface. -What I have already tried -I have produced solutions for the cylinder case using FEMLAB, and replicated those results with COMSOL. However, I was unable to think of a way to incorporate more complicated boundary structures (even simple ones such as a cone). -I have had slightly more success with the Surface Evolver, developed by Ken Brakke. This is also a finite-element-style scheme, which works by evolving an initial surface using a gradient descent method. The software is stable and well-written, and I have been able to produce results for a cylinder, a cone and hyperboloid. However, as the solid boundaries must be defined as level-set constraints, I assume that building more complex solid boundaries would require overlapping level-set constraints and some criteria for switching between them appropriately. -Notes -I am aware of several different methods which may be applicable, including: Volume-of-Fluid methods, Level-Set Methods (Osher & Sethian), Finite Element Methods for PDEs and the Dorfmeister-Pedit-Wu algorithm. I have been endeavouring to determine for myself whether any or all of these might be appropriate, but due to my limited computational experience, I am quite unsure as to what method might be appropriate. -Important Comment -I am not attempting, in any way, to avoid the long and possibly laborious process of learning the ins and outs of a computational method. If referred by consensus or expert advice to an appropriate method, I will most happily plow into every piece of material I can find on the subject until I am able to address my problem. At this stage, I simply do not have the breadth of knowledge necessary to investigate every possible method and assess each for its strengths and weaknesses with respect to my problem. -Summary -Is there a computational method, or already-existing software package, which is appropriate for modelling fluid-air interfaces with solid boundaries of complicated geometry? -With thanks in anticipation, -Christopher Laing - -REPLY [5 votes]: You may also consider the immersed boundary method, and its variants. It was specifically developed for situations involving complex fluid/structure interactions. The method is quite successful for solid structures (complexity is not a problem); it's a bit trickier for porous media. In essence one write down the interaction forces experienced by the particles in the solid body, and integrates. -The methods do indeed rely on the computation of integrals, but fast quadratures work quite well here. -A really great starting point for this field is the Acta Numerica paper by Charles Peskin (2002). -He also has some nice course notes, and code, online: -http://math.nyu.edu/faculty/peskin/ib_lecture_notes/index.html<|endoftext|> -TITLE: History of fundamental solutions -QUESTION [21 upvotes]: I have a few questions on the history of PDE. - -Who first wrote down the formula for the solution of the Cauchy problem for the heat equation involving the heat kernel? I have seen it called Poisson's formula. If it is true Poisson has a formula for each of the heat, wave, and Laplace equations. -Who is the discoverer of the analogous formula for the wave equation in 2 and 3 dimensions? I have seen they were called both Kirchhoff's formula and Poisson's formula. -Is there a book to look up such questions? I have Dieudonne's History of functional analysis, but it does not have much on PDEs other than the Laplace equation. - -REPLY [10 votes]: This is an update on Question 1. As Willie observed, in his 1819 memoir Poisson studies not only the wave equation but also the heat equation from page 143 on, and reaches the heat kernel on page 145. However, amazingly, in Fourier's original memoir where he derived the heat equation and gave a convincing case for the importance of trigonometric series, the heat kernel appears on page 454 for 1D, on page 475 for 1D in the usual form as presented today, and on page 479 for 3D. Fourier's memoir was published in 1822 after a long delay, and it is said that the memoir is essentially Fourier's 1811 work that won a mathematical prize, which was in turn a continuation of his work presented in 1807, and summarized by Poisson in 1808. That said, even more amazingly, a new player appears in the story. After giving the 1D heat kernel on page 454, Fourier says something like - -This integral, which contains an arbitrary function, was not known when we started our research on the theory of heat, which were presented at the Institute of France in December 1807. It was given by Mr. Laplace, in volume VI of des Mémoires de l'école polytechnique, and we have only applied his results here. - -Poisson also mentions Laplace on page 148, and says that his 3D result was a straightforward extension of Laplace's formula. I found volume 6 of Journal de l'école polytechnique but there is nothing by Laplace, and moreover the journal is from 1806. I wondered if des Mémoires is different than Journal, but skimmed through Fourier's book to find that on page 513 he cites Laplace again, but now says volume 8. Then volume 8 it is! It is published in 1809, and the heat kernel appears on its page 241!<|endoftext|> -TITLE: Formally undecidable problems on finitely presented quandles -QUESTION [15 upvotes]: In the literature, one sometimes sees the claim that finitely presented quandles (in particular, knot quandles) are "hard to deal with". Hence, a great deal of effort has gone into studying finite quandles and counting homomorphisms onto them, and so on. However, I have not yet come across any theorems that state formal undecidability results for finitely presented quandles similar to those for finitely presented groups. In fact, I have yet to see any formulation of such problems. (For instance, a theorem stating that the isomorphism problem is undecidable for finitely presented quandles.) -Do such results exist in the literature and, if so, could someone please provide references? -(Asked previously here on math.stackexchange, without response.) - -REPLY [9 votes]: I do not believe that the word problem for quandles was first formulated by the team of Rena Levitt and Sam Nelson, if this formulation was made while Levitt was at Pomona. I had been working on this question well before 2011, which is when I assume Levitt arrived in Pomona and I am certain that it had been asked before. This is a standard question that people working in universal algebra or computation would ask of such a domain. I would not be surprised at all if Sam Nelson had encountered the question well before then, given the impressive array of work he has done on quandles. -If my chronology of this problem is missing some important historical details, I welcome any and all information. -A proof for the undecidability of the word problem for finitely presented quandles was discovered by James Belk and me at Bard College in June of 2009. This result was presented to the Bard College Mathematics Seminar in September of 2009, the SUNY New Paltz Mathematics Seminar in November of 2010, and the Algebra without Borders workshop at Yeshiva University in August of 2011. There might be some earlier proof, but it was and remains unknown to us. -Moreover, I shared a manuscript of the proof in the summer of 2012 with Benjamin Fish, then a Bard MATH REU student of mine and a Pomona undergraduate and now a graduate student in the mathematics department at the University of Illinois at Chicago. Fish then completed his senior research project in 2012-2013 on a related topic under Levitt at Pomona. I assume that this manuscript, or at least its contents were shared with Levitt. -Apropos the theorem at the end of the answer above… Joyce showed in his original paper that the free quandle FQ(A) over a generating set A can be embedded in the group quandle FG(A) over the free group on A. Hence the decidability of the word problem for FQ(A) reduces to the decidability of the word problem for FG(A), and the word problem for FG(A) is certainly decidable if A is finite. Hence the first part of the theorem is a straightforward consequence of Joyce's paper in 1982. -However, one can ask whether there is a more direct proof of the decidability of the word problem for free quandles over a finite generating set that does not appeal to group theory. There is such. Fix the following theory for the theory of quandles. -x*x = x -(x*y)/y = x -(x/y)*y = x -(x*y)z = (xz)(yz) -I discovered the following term rewriting system also in 2009: -x*x -> x -x/x -> x -(x*y)/y -> x -(x/y)*y -> x -x*(y*z) -> ((x/z)*y)*z -x/(y*z) -> ((x/z)/y)*z -x*(y/z) -> ((x*z)*y)/z -x/(y/z) -> ((x*z)/y)/z -This system is both terminating and confluent and hence has unique normal forms. In particular, you can use it to decide whether two words over FQ(A) are provably the same. -Extensions of this result were the subject of a Bard senior research project with Hannah Quay-de la Vallee in 2009-2010 and my summer 2011 REU at Bard with then Drew University undergraduate Gregory Hunt, who is now in the Ph.D. program in Statistics at Michigan. The term rewriting system along with a proof of confluence and termination was also shared with Ben Fish and his REU colleagues in 2012. -That is not to say that the latter half of the Theorem stated above is not original. That appears to be new work. In fact, I had thought about this problem myself to no avail. However, the manner in which the answer above is presented might mislead the reader about the origin of some of the claims. I assume that this was an oversight and not intentional.<|endoftext|> -TITLE: Applications of PDE in mathematical subjects other than geometry & topology -QUESTION [24 upvotes]: Partial differential equations have been used to establish fundamental results in mathematics such as the uniformization theorem, Hodge-deRham theory, the Nash embedding theorem, the Calabi-Yau theorem, the positive mass theorem, the Yamabe theorem, Donaldson's theory of smooth 4-manifolds, nonlinear stability of the Minkowski space-time, the Riemannian Penrose inequality, the Poincaré conjecture in 3D, and the differentiable sphere theorem. These examples all come from geometry and topology, and I was trying to find similar examples in other branches of mathematics without luck. I can sort of imagine why geometry and topology maybe amenable to PDE but this does not mean PDE cannot find applications in other branches. I asked probabilists and was told that most of the examples they think of seem to be the other way around, i.e., using probability theory to say something about PDE. Can you provide an example, or give a reason why such examples must be confined to geometry and topology. -The reason I am asking this question is that majority of "pure math" students don't seem to like PDE courses, thinking it as an "applied" subject so it has nothing to do with them. My impression is that for instance students in algebraic or differential geometry somehow get their "own version" of PDE theory from specialized books in their subject, specifically tailored for the problem at hand. It would be much easier and methodical if the student had taken a general PDE course before. So I thought this kind of list maybe helpful in convincing the beginning student to take PDE classes. As the list stands now, we have enough for geometry/topology and perhaps mathematical physics students, but it would be great for instance to have something for probability, number theory, analysis, and algebra students. - -REPLY [3 votes]: Probability: PDEs are all over the place in problems related to optimal filtering problems. For example, the Kushner-Stratanovich equation of nonlinear filtering. -Several of the optimal filtering type results are approached by forming a cost functional, and reducing the problem to a Euler-Lagrange PDE. -My impression is that the Euler-Lagrange PDE is pervasive in lots of areas of math, although my exposure is mostly with Hamiltonian dynamics, dynamical systems and estimation theory.<|endoftext|> -TITLE: Classification of holomorphic disc bundles -QUESTION [10 upvotes]: I've had difficulty finding sources which treat the classification of holomorphic disc bundles over (compact and noncompact) Riemann surfaces. Note that by "bundle", I mean a holomorphic fiber bundle, which means it is locally holomorphically trivial. -I'm really just looking for information about holomorphic disc bundles, but if you want a specific question: -What is the classification of holomorphic disc bundles over a Riemann surface? - -REPLY [3 votes]: If I don't misunderstand the question, I don't believe that any kind of such classification can exists in the case when the normal bundle of the surface has positive degree. This moduli is infinite-dimensional. -Even in the case of negative degree there are problems. If we take a disk bundle over $\mathbb CP^1$ of degree -1, and contract $\mathbb CP^1$ we will get in particular plenty of complex balls in $\mathbb C^2$. Can one classify them up to biholomorphism?<|endoftext|> -TITLE: What is the degree of a symmetric boolean function? -QUESTION [5 upvotes]: (previous title " Zero sum of binomials coefficients - a stronger version ") -This is a stronger version of another question. -Is there an $N\in \mathbb N$ and a sequence of non-constant functions $ \left\{ p_n:[n] \to \{ 1,-1 \} \right\}_{n=N} ^{\infty}$ such that for all $n>N$ we have: -$$ \sum _{i=0} ^{n} (-1)^{i} p_n(i) \cdot \binom {n} {i} = 0$$ -For instance, for all odd values of n, we may choose -$p_{n}(i)=\begin{cases} -(-1)^{i} & i\leq\frac{n-1}{2}\\\ -(-1)^{i+1} & i\geq\frac{n+1}{2}\end{cases}$. This simply means we sum the first half of the binomial coefficients and subtract the second half. The fact that for odd values we can partition the set of binomial coefficients evenly allows us to do that, so I don't see how the same trick may be applied for even values. -To my understanding, the methods that solved the previous question (for which I thank darij grinberg and Mikael de la Salle) are not applicable here. -My guess, as before, is that there is no such sequence (in which all functions are non-constant), any ideas on how to prove it? -(a counter example would surprise me, but is of interest as well) - -REPLY [5 votes]: It was already pointed out in the comments that determining for which $n$ one can find a non-constant $p_n$ is an open problem. I thought I'd give a bit of context and my understanding on what is known so far. The problem as stated has a negative answer because when $n+1$ is prime, $p_n$ must be constant. -The sums $\sum_{i=0}^n \varepsilon_i \binom{n}{i}$ are the leading coefficients of the polynomials we get from Lagrange interpolation on points $(i,\alpha(i))$ where $0\le i\le n$ and $\alpha(i)\in \lbrace 0,1\rbrace$. So the question is equivalent to - -Is there a polynomial that sends $\lbrace 0,1,\dots,n\rbrace$ to $\lbrace 0,1\rbrace$, that is not constant but has degree $\le n-1$? - -Let us denote the number of such polynomials by $\mathcal B(n)$. Some examples are given by $\varepsilon_i=(-1)^i$ when $n$ is even and $\varepsilon_i=-\varepsilon_{n-i}$ for odd $n$. This implies that $\mathcal B(n)\geq 2$ when $n$ is even and $\mathcal B(n)\geq 2^{\frac{n+1}{2}}$ when $n$ is odd. Finding non-trivial solutions to the problem implies improving on these lower bounds. -Here is a simple argument, that when $p$ is an odd prime $\mathcal B(p-1)=2$, so there are no non-trivial solutions. This is because $\binom{p-1}{i}\equiv (-1)^i\pmod{p}$ so the only way for the sum to be divisible by $p$ is if the sequence $(-1)^i\varepsilon_i$ is constant. This includes the examples $n=16,18$ that you confirmed with a computer search. However there are even values of $n$ for which $k(n)\geq 3$. The first example is -$$\binom{8}{0}-\binom{8}{1}-\binom{8}{2}+\binom{8}{3}+\binom{8}{4}-\binom{8}{5}-\binom{8}{6}-\binom{8}{7}+\binom{8}{8}=0$$ -and the next one is the one given by Darij in the comments for $n=14$. The even values of $n$ for which $\mathcal B(n)\geq 3$ and $n\le 128$ were found in J. von zur Gathen and J. Roche, “Polynomials with two values”, Combinatorica 17, no. 3 (1997), 345–362. The sequence is $\lbrace 24,34,48,54\rbrace$ and numbers $2\pmod{6}$. -Your question is really about proving that $\mathcal B(n)=2$ for infinitely many $n$, and it is an open to determine such $n$ besides the values found in the von Zur Gathen-Roche paper. As I mentioned above it is equivalent to proving for such $n$ that the minimum degree of a non-constant polynomial sending $\lbrace 0,1,\dots,n\rbrace\to \lbrace 0,1\rbrace$ is $n$. The best results known so far are that the degree is $n-o(n)$, where the $o(n)$ comes from the gaps in consecutive primes (so one can take $O(n^{.525})$), but conjecturally this can be improved to $n-O(1)$. -One thing that is surprising is the following threshold phenomenon. If we look at non-constant polynomials sending $\lbrace 0,1,\dots,n\rbrace\to \lbrace 0,1,\dots,n\rbrace$, the minimum degree is $1$ (for instance $f(x)=x$), but as soon as we look at $\lbrace 0,1,\dots,n\rbrace\to \lbrace 0,1,\dots,n-1\rbrace$ the degree is at least $n-o(n)$. The current methods don't seem to make use of the fact that in the boolean case the range is simply $\lbrace 0,1\rbrace$, as they give the same bound for larger ranges. A recent paper on the topic is "On the Degree of Univariate Polynomials Over the Integers" by G. Cohen, A. Shpilka and A. Tal.<|endoftext|> -TITLE: Meaning of 'alternating' group ? -QUESTION [18 upvotes]: What's the meaning of the adjective 'alternating' in the name of the 'alternating group' ? - -REPLY [21 votes]: Evidence for Brendan McKay's suggestion that it comes from the older concept of an alternating polynomial may be found in Burnside's classic book, Theory of Groups of Finite Order. When defining the symmetric and alternating groups, he puts an asterisk next to the word "alternating" and adds the following footnote: - -The symmetric group has been so called because the only functions of the $n$ symbols which are unaltered by all the substitutions of the group are the symmetric functions. All the substitutions of the alternating group leave the square root of the discriminant unaltered. - -By "the square root of the discriminant," Burnside means the polynomial -$$\prod_{r=1}^{n-1}\prod_{s=r+1}^n (a_r-a_s),$$ -which of course is an alternating polynomial.<|endoftext|> -TITLE: Have you ever seen this product? -QUESTION [12 upvotes]: Given $k$, what is the value of the following product? -$$\prod_{p\textrm{ prime}}\left(1-\frac{k}{p^2}\right).$$ -Clearly for $k=1$, we have $\zeta(2)^{-1}$ (where $\zeta(s)$ is the Riemann Zeta function). - -REPLY [5 votes]: Elementary comment. For a fixed real number $x$, define a multiplicative function $f(x;n)$ as follows. Factor $n=p_1^{e_1}\cdots p_r^{e_r}$ and set $f(x;n)=x^{e_1+\cdots+e_r}$. Then -$$ - \prod \left(1-\frac{x}{p^s}\right)^{-1} = \sum_{n=1}^\infty \frac{f(x;n)}{n^s}. -$$ -The function $n\to e_1+\cdots+e_r$ that counts the number of prime divisors of $n$ with multiplicity must come up in lots of places, but I don't recall seeing it appear like this in a Dirichlet series.<|endoftext|> -TITLE: Is it decidable whether or not a collection of integer matrices generates a free group? -QUESTION [29 upvotes]: Suppose we have integer matrices $A_1,\ldots,A_n\in\operatorname{GL}(n,\mathbb Z)$. Define $\varphi:F_n\to\operatorname{GL}(n,\mathbb Z)$ by $x_i\mapsto A_i$. Is there an algorithm to decide whether or not $\varphi$ is injective? - -REPLY [5 votes]: Some comments on the question 'Is it decidable whether or not a -collection of integer matrices generates a free group?' -Given a finite set of matrices S over a field, the problem of -testing whether the group H generated by S contains a free -non-abelian subgroup is decidable. An algorithm solving the -problem as well as its implementation (in Magma) available. Notice -that the algorithms does not construct a free non-abelian subgroup -in H but justifies its existence. As to testing freeness of -finitely generated linear groups then the problem has quite a long -history. I may recommend as a start point the paper by John Dixon -Can.J Math, v. 37, n. 2, 1985, 238-259 (see p. 240 there, and then -follow the references).<|endoftext|> -TITLE: Ascending Chain Condition for finite normalizers -QUESTION [7 upvotes]: Let $G$ be a group and $H$ a subgroup. Consider the ascending chain of iterated normalizers: -$$ -H \trianglelefteq N_G(H) \trianglelefteq N_G(N_G H) \trianglelefteq \cdots \trianglelefteq N^{(k)}_G(H) \trianglelefteq \cdots. -$$ -Is there an example where all the terms are finite, but the chain fails to stabilize? - -REPLY [7 votes]: Can't we just take $G$ to be the union of the chain $H_1 -TITLE: The vanishing of non-connective K-theory in negative degrees -QUESTION [8 upvotes]: In the works of Cisinski, Tabuada, and Schlichting certain non-connective K-theory groups for a differential graded category $C$ are defined. As far as I understand, $K_i(C)$ is not necessarily zero for $i<0$. Yet are there any sufficient conditions on $C$ that ensure that this condition is fulfilled? Are $K_i(C)$ just the 'standard' $K$-groups of a smooth variety $X$ if $C$ is an enhancement of the (derived) category of perfect complexes of sheaves over $X$? -I would be deeply grateful for any explanations and/or precise references here. - -REPLY [11 votes]: This abstract non-connective $K$-theory, when restricted to schemes, is known from a long time: this is the Bass $K$-theory functor $K^B$ considered by Thomason and Trobaugh (the article of Thomason and Trobaugh is a classic on the subject which must be read anyway). The comparison of the abstract construction of non-connective $K$-theory with the Thomason-Trobaugh construction follows straight away from Theorem 5 (Section 8) and Theorem 8 (Section 12) in Schlichting's paper -M. Schlichting, Negative K -theory of derived categories, Math. Z. 253 (2006), 97–134. -The only general way to see negative K-groups vanishing is Theorem 7 of loc. cit: for any noetherian abelian category $A$, $K_i(D^b(A))=0$ for $i<0$ (where $D^b(A)$ stands for a dg version of the bounded derived category of $A$). In particular, if $X$ is a noetherian regular scheme, then the equivalence $Perf(X)\simeq D^b(Coh(X))$ implies that $K_i(X)=K_i(Perf(X))\simeq K_i(D^b(Coh(X)))=0$ for $i<0$. -Weibel's conjecture predicts that, for any noetherian scheme of Krull dimension $\leq d$, $K_{-i}(X)=0$ for $i>d$. This conjecture is proved in characteristic zero in this paper: -G. Cortiñas, C. Häsemeyer, M. Schlichting and C. A. Weibel, Cyclic homology, cdh-cohomology and negative K-theory, Ann. of Math. 167 (2008), 549 - 573. -Assuming a rather strong version of resolution of singularities, the conjecture has been proved in positive characteristic as well in this paper: -T. Geisser, L. Hesselholt, On the vanishing of negative K-groups, Math. Ann. 348 (2010), 707-736. -Finally, we conjecture that $K_i(A)=0$ for $i<0$ for any saturated dg algebra $A$ (possibly such that $A^n=0$ for $n<0$). Such a vanishing would imply the existence of a weight structure à la Bondarko on (an adequate version of) Kontsevich's triangulated category of "non-commutative motives". Evidence for this is given by the fact that this is known if $A$ is Morita equivalent to $Perf(X)$ (for a smooth and projective variety $X$), as well as by the degeneration of the non-commutative version of the Hodge-to-de Rham spectral sequence, see -D. Kaledin, Non-commutative Hodge-to-de Rham degeneration via the method of Deligne-Illusie, Pure Appl. Math. Q. 4 (2008), 785–875.<|endoftext|> -TITLE: embedding of local tempered representation into cuspidal automorphic representation -QUESTION [9 upvotes]: Let v be a finite place of a number field F. Let $\pi_{v}$ be an irreducible tempered representation of $ GL_{n}(F_v)$. Is it true that we can find some irreducible cuspidal automorphic representation $\pi$ of $GL_{n}(\mathbb{A_{F}})$ with $v$-component isomorphic to $\pi_{v}$ ? - -REPLY [7 votes]: As Joel points out, this is not possible. Even in the case of a discrete series rep'n, one can't control the value of the central character on the non-compact part of $F_v^*$. What one can do (in some generality; for precise references recent work of Sug Woo Shin might be relevant) is show that the set of tempered representations which are local components of automorphic forms are "large" in the space of all tempered rep'ns, where "large" can be taken to mean something like "equidistributed according to Plancherel measure", or also "Zariski dense in the Spec of the Bernstein centre". (There may well be caveats to both these statements, though, so I would look at Shin's work and whatever references are contained therein.)<|endoftext|> -TITLE: Things that should be positive integers...really? -QUESTION [19 upvotes]: Kronecker. Nuff said. Even the numbers themselves historically started -as positive integers and were subsequently generalized to hell and back. -Here are some other well known concepts that "should" involve $\mathbb{N}$ -but were generalized to $\mathbb{Q}$, $\mathbb{R}$ or even $\mathbb{C}$: - -Dimension $\rightarrow$ Hausdorff dimension. -Factorial $\rightarrow$ gamma function. -Differentation $\rightarrow$ half-differentation (etc.) - -So, can you extend this small to a big list? -(Motivation: Some hypothetic knot polynomial I calculated with -demanded a dimension of its associated group representation -- thus the "rt" tag - of 60/11. That is noooooot boding well -for its existence. :-) - -REPLY [8 votes]: Let $f:\mathbb{Z}_p\to\mathbb{Z}_p$ be a "nice" map on the $p$-adic integers (or a map on some more general space with a $p$-adic topology). People who study $p$-adic dynamcis investigate what the iterates of $f$ do to points of the space. So if we fix a point $\alpha\in\mathbb{Z}_p$, we can define an iteration map -$$ - I : \mathbb{N} \longrightarrow \mathbb{Z}_p,\qquad - I(n) = f^n(\alpha). -$$ -The map $I$ is naturally defined on $\mathbb{N}$, and if $f$ is invertible, then it clearly extends to $\mathbb{Z}$. But for various applications, one would like to evaluate $I(n)$ for $n\in\mathbb{Z}_p$. So the example is - -iteration an integral number of times $\to$ iteration a $p$-adic number of times. - -A very pretty application of this idea is in the paper: -Bell, J. P. ; Ghioca, D. ; Tucker, T. J. The dynamical Mordell-Lang problem for étale maps. - Amer. J. Math. 132 (2010), no. 6, 1655--1675.<|endoftext|> -TITLE: Example of an infinite abelian but non-cyclic group whose automorphism group is cyclic -QUESTION [6 upvotes]: Can anyone give me an example of: - - -An infinite abelian but non-cyclic group whose automorphism group is cyclic. - -REPLY [6 votes]: One has a very general statement in this direction: -For any cotorsion-free ring $A$ there exist abelian groups $G$ of arbitrarily large cardinalities such that -$$A\cong\mathop{\rm End}(G)$$ -In particular if you need $\mathop{\rm Aut}(G)$ to be cyclic then you take $A$ such that the units of $A$ form a cylic group (e.g. $A=\mathbb{Z}$). Even more generally, you can replace abelian groups with $R$-modules, where $R$ is any commutative, cotorsion-free ring. Then you obtain $A$ - any cotorsion-free $R$-algebra. -You will find various constructions of such groups in Goebel and Trlifaj "Approximations and Endomorphism Algebras of Modules" (2006 and much expanded edition 2012). -cotorsion-free above means no nontrivial homomorphisms of the underlying groups: $\mathbb{Z}^\wedge_p\to A$ for all primes $p$.<|endoftext|> -TITLE: Conformal Field Theory and Langlands -QUESTION [7 upvotes]: I'm a Mathematics masters student currently -studying some aspects of TQFT. I'm interested in Langlands, mainly -because it sounds oppressive! Is anyone familiar with any links between -CFT and Langlands, or more importantly any readable intros to these -sorts of areas. -How do Hecke Eigensheaves come into things? What does one need to know before hand. Are there any lists of requisite knowledge before tackling such concepts? Sometimes knowing where to start is a hard problem! - -REPLY [7 votes]: There's a review article by Edward Frenkel which exactly fits your need: see "Lectures on the Langlands Program and Conformal Field Theory". It's posted on hep-th, but as I (as a string theorist) had difficulties reading it, it should be written for mathematicians :)<|endoftext|> -TITLE: reference help needed on a fact about decomposition of representations -QUESTION [5 upvotes]: For simplicity, let $G_n$ be $GL_n(\mathbb{R})$, $\mathfrak{g}_n$ be its Lie algebra. $K_n$ be $O(n)$. I want to know any reference about the following statement. -For any irreducible admissible $(\mathfrak{g}_n\oplus \mathfrak{g}_m,K_n\times K_m)$ module $V$, there exist an irreducible admissible $(\mathfrak{g}_n, K_n)$ module $U$, and an irreducible admissible $(\mathfrak{g}_m, K_m)$ module $W$, such that $V$ is isomorphic to $U\otimes W$. Both $U$ and $W$ are uniquely determined by $V$ up to isomorphism. -Similar of this is true for representations of finite groups, smooth representations of p-adic groups. It seems to me that this is also true for real reductive groups, and I can't find it in any standard reference. I appreciate a lot if anyone would provide some related paper or book. - -REPLY [2 votes]: Well it's slightly later than you probably would've liked, but here's a reference for exactly what you're looking for: - -Gourevitch, D. and Kemarsky, A., 2012. Irreducible representations of product of real reductive groups. (arXiv link: https://arxiv.org/abs/1212.6004) - -See Theorem 1.1 and its proof. There the authors show that each irreducible Harish-Chandra $(\mathfrak{g}_1 \times \mathfrak{g}_2, K_1 \times K_2)$-module has a unique decomposition into a tensor product of irreducible Harish-Chandra $(\mathfrak{g}_i, K_i)$-modules. (Here by a Harish-Chandra $(\mathfrak{g},K)$-module we mean a $(\mathfrak{g},K)$-module that is admissible and finitely generated as a $U(\mathfrak{g}_\mathbb{C})$-module.) -They also reference Aizenbud and the first author's 2009 paper wherein a converse to this is stated and proved in Appendix A. (That direction is easier, and IIRC their proof is maybe a few paragraphs in length.)<|endoftext|> -TITLE: Interesting mathematical topics arising from biology -QUESTION [36 upvotes]: I've heard that there's a relatively new field of science called mathematical biology. -It will certainly apply well known and less known mathematical techniques to the understanding of some biological phenomena such as strings of DNA knotting together, proteins bending, cell membranes, population dynamics,... - - -Which interesting advances from the point of view of a mathematician are there that have been inspired by biology and related areas? - - -For example, I heard of "membrane computation", but I don't know if it's genuinely inspired by biology in some nontrivial way or if it just bears that name by a loose analogy... - - -Which fields of mathematics are currently used in the discipline called mathematical biology? - - -If I were asked about economy and mathematical Finance I would loosely quote probability theory, stochastic differential equations, Ito integrals, ... So, what about biology? - -REPLY [2 votes]: I think this is quite relevant, -https://liorpachter.wordpress.com/2014/12/30/the-two-cultures-of-mathematics-and-biology/ -especially the references to Haussler's work, -[Mapping to a Reference Genome Structure][https://arxiv.org/ftp/arxiv/papers/1404/1404.5010.pdf] and -[Efficient haplotype matching and storage using the positional Burrows–Wheeler transform (PBWT)][http://bioinformatics.oxfordjournals.org/content/30/9/1266.short]. -(Incidentally, I'd say "David Haussler ... studied mathematics" is quite an understatement, as he's made seminal contributions to computational geometry and combinatorics.)<|endoftext|> -TITLE: Intervals in posets: how to extend interval orders, Allen's algebra, and interval graphs to intervals of posets? -QUESTION [5 upvotes]: BACKGROUND -Assume a poset $\langle P, \le \rangle$. For two points $a,b \in P$ -with $a \le b$, then $I = [a,b] = \{ x : a \le x \le b \}$ is the -interval between $a$ and $b$. -When $P$ is a chain (e.g. ${\mathbb Z}, {\mathbb R}$), then the $I$ -are just standard intervals. Two real intervals $I=[a,b],J=[c,d] -\subseteq {\mathbb R}$ are ordered usually to mean that $I \le J$ iff -$b \le c$. Call this the "strong order", which isn't actually a proper -order (it needs to be "reflexivized" to require that $I \le I$). Two -other true orders are also available, namely that $a \le c$ and $b \le -d$ (the product order of the endpoints), or that $a \le c$ and $b \ge -d$ (subset order). These last two are conjugate orders. All of these -are defined in the context of Allen's alegbra, enumerating all the -possible relations between $I,J$ given combinations of both equal and -unequal endpoints. -Additionally, the intersection graphs of sets of real intervals are -interval graphs, which are well studied. -MOTIVATION -We work with data objects represented as finite, bounded posets. -Analyzing the intervals therein, and their orderings and -intersections, is very useful in a range of applications in layout and -display. -QUESTION -We are thus seeking extensions from real intervals to poset intervals -for the concepts of interval order, Allen's algebra, and inteveral -graphs. Our preliminary literature reviews haven't turned up anything, -and we're preparing to start the development from first -principles. Pointers appreciated, thanks! - -REPLY [4 votes]: The subset order on intervals of a finite poset has received some -attention. See for example Exercises 3.10, 3.76(b), 3.138, and 3.158(c) -of http://math.mit.edu/~rstan/ec/ec1.pdf.<|endoftext|> -TITLE: A book on Banach Manifold for a Dynamicist -QUESTION [7 upvotes]: Hi all, -Could you give me a suggestion of suitable book about Banach Manifolds for someone that have background in functional analysis at the level of Conway's book and Do Carmo's book on Riemannian Geometry ? -To help the indication the problems I am facing in my research are for example, what are the usual conditions required to a Banach manifold to be metrizable. Is there a standard way to construct a metric on $C^r(M,M)$, where $M$ is a two-dimensional manifold ? - -REPLY [3 votes]: The first textbook I thought is: Palais, The Foundations of Global Non-linear Analysis, Benjamin-Cummings, 1968. -There is also: Marsden, Applications of Global Analysis in Mathematical Physics, Publish or Perish, 1974. -Finally you could look here at Graff's review of ``The Metric theory of Banach Manifolds'' by Ethan Atkin, for many other references.<|endoftext|> -TITLE: Symmetric extensions and class forcing -QUESTION [7 upvotes]: Suppose $V\models ZFC$ and $P\in V$ is a poset of forcing conditions. - -It is a basic theorem in forcing that $V[G]\models ZFC$ for any generic extension by a $V$-generic filter $G$. -It is also known that if $V\subseteq M\subseteq V[G]$, and $M\models ZFC$ then $M$ is a forcing extension of $V$ and $V[G]$ is a forcing extension of $M$. - -Now, consider $V[G]$ to be some generic extension. We can define in $V[G]$ a transitive subclass which is a model of $ZF$, but often not of $AC$. This is done by considering some permutation group of the forcing conditions and taking only names which obey some condition. The interpretation of this class of special names is called a symmetric extension of $V$. -In most interesting cases the symmetric extension negates the axiom of choice one way or another. This doesn't sit right with the two theorems mentioned above, since if $M$ is actually a generic extension of $V$ then the axiom of choice should hold in $M$, but it does not. (This is nullified by the correction points by Andreas Blass and Amit Kumar Gupta in the comments). -Edit: Instead of the above, then, how much choice is needed to have the second theorem stated in $ZF$ alone? does that depend on the forcing $P$ at hand? If the answer is that the theorem is nontransferable? -The way I see it negative answers would mean at least one of two possible things: - -We consider $V$ as a model of $ZF$ and by some forcing which preserves only $ZF$ we obtain $M$ as the generic extension, this is similar to the way we may violate the continuum hypothesis or collapse the continuum to be $\aleph_1$, and so change the truth value of an unprovable statement (very much like $AC$ is unprovable from $ZF$). -$M$ can be achieved by class forcing. I have very little intuition on that topic, so I cannot see any reason why this may be either true or false. - -Is my intuition correct? -While we're on the topic, I do recall forcing is indeed possible without choice, but that should require extra assumptions or different methods to handle the genericity. Is there a good introduction to the topic available? - -REPLY [7 votes]: These intermediate model questions were thoroughly investigated by Serge Grigorieff in Intermediate Submodels and Generic Extensions in Set Theory [Annals of Mathematics 101 (1975), 447-490]. -The basic result of the kind you are looking for is due to Solovay (according to Grigorieff): - -Let $P$ be a poset in $M$, a model of ZF, and let $G$ be $P$-generic over $M$. If $a \in M[G]$ contains only elements of $M$, then $M[a]$ is a generic extension of $M$ and $M[G]$ is a generic extension of $M[a]$. - -This is similar to what you state, but note the special form of the intermediate model. In fact, the intermediate models of ZF between $M$ and $M[G]$ that are generic extensions of $M$ are precisely those of the form $M[a]$ as described above. -When $M$ satisfies ZFC, then any intermediate model of ZFC between $M$ and $M[G]$ is of the form $M[a]$ for some set of ordinals $a \in M[G]$. Thus, your stated result does hold provided that all models involved satisfy ZFC. -Since you mentioned symmetric models in the questions, note that Grigorieff goes on to classify all of those too: - -The symmetric submodels of $M[G]$, are all intermediate models of the form $(HOD\ M[a])^{M[G]}$ as $a$ varies over $M[G]$.<|endoftext|> -TITLE: Multiplicativity in the descent spectral sequence -QUESTION [17 upvotes]: For a homotopy sheaf $\mathcal{F}$ of ring spectra over some space (/ site / whatever) $X$ with a cover $U_i$, we can build a "descent spectral sequence" with signature $$E^1_{p, q} = \pi_{p+q} \mathcal{F}\left(\coprod_{|I| = q} U_I \right) \Rightarrow \pi_{p+q} \mathcal{F}(X).$$ This comes about by building the simplicial object associated to the cover, applying the sheaf to get a cosimplicial object, and then instantiating a filtration-type spectral sequence given by looking at varying truncations of totalizations --- the standard spectral sequence for a cosimplicial object. The $E^2$-page of the spectral sequence can be identified with Cech cohomology, and so the spectral sequence is meant to provide an intermediary between that homological object and the homotopy-sensitive information in the sheaf of spectra. This construction is natural enough that... - -...a refinement of the covering induces a map of spectral sequences. Just like limiting Cech cohomologies over cover refinements gives sheaf cohomology, limiting Cech descent spectral sequences gives a descent spectral sequence with $E^2$-page described by sheaf cohomology. -...a map of sheaves induces a map of spectral sequences. - -This construction doesn't really use the fact that $\mathcal{F}$ takes values in rings. I feel that this must appear in the spectral sequence, that we should expect some kind of multiplicativity --- maybe one that mixes the products of the ring spectra and the Cech product. - -So, my question is: Is there a multiplicative structure in any of these spectral sequences? What is its signature? How might I compute with it? Better yet, are there examples of other people computing with it that I can read about? - -For what it's worth, I'm interested most in starting with a sheaf of ring spectra $\mathcal{F}$ and computing $\mathcal{F}(X)^*(Y)$ for my favorite space $Y$ by augmenting $\mathcal{F}$ to $F(\Sigma^\infty_+ Y, \mathcal{F})(U) := F(\Sigma^\infty_+ Y, \mathcal{F}(U))$ and working with that. This comes with a map $F(\Sigma^\infty_+ (Y \times Y), \mathcal{F}) \to F(\Sigma^\infty_+ Y, \mathcal{F})$, which is in the vein of the usual construction of the cup product. -(There are a lot of details I'm eliding past, like conditional convergence, $\lim^1$ problems, when my augmented guy is actually a homotopy sheaf, when the analogy with sheaf cohomology can be made, so on and so forth. I don't think I've said anything not true in the nicest of settings, which is where I'd like to start learning, at least. Sorry if this is off-putting.) - -REPLY [9 votes]: Using your $X$ and $Y$, you get an augmented simplicial object as follows: -$$ -\cdots Y \times_X Y \times_X Y \Rrightarrow Y \times_X Y \Rightarrow Y -$$ -Applying $\cal F$ to this diagram, you get a coaugmented cosimplicial ring spectrum. The spectral sequence for the homotopy groups of Tot of this which realizes your Cech cohomology. -So this reduces you to a question: Given a cosimplicial object in ring spectra, do you get a multiplication on the associated spectral sequence converging to the multiplication on Tot? -This is true; most types of multiplicative structure carry over like this (although it may turn into "coherent" multiplicative structure). However, I've had a little trouble chasing this through the literature this morning. -One method you could use is a method of universal example, where $E_r$-cycles are carried by certain maps of cosimplicial objects. In Bousfield and Kan's "A second quadrant homotopy spectral sequence," they do this for the smash product in the homotopy spectral sequence of cosimplicial spaces (which is harder because you have to worry about basepoints!), and other authors (James Turner, and recently Philip Hackney) have studied operations arising on cosimplicial chain complexes by methods that should translate to the context of ring spectra. -Wish I had a more definitive reference for you.<|endoftext|> -TITLE: Helmholtz-Decomposition on compact Riemannian manifolds -QUESTION [6 upvotes]: For smooth domains $\Omega$ in $\mathbb{R}^n$ it is known that one can decompose vector fields in $L^p(\Omega)^n$, $1 < p <\infty $ into a "gradient"- and a "divergence-free"-part such that -$L^p(\Omega)^n=G^p(\Omega) \oplus D^p(\Omega)$, -where $G^p(\Omega)= \{ w\in L^p(\Omega)^n; w= \nabla p$ for some $p\in W^{1,p}(\Omega)\}$, and $D^p(\Omega)$ is the completion of $\{ u\in \mathcal{C}^\infty_0(\Omega)^n; \nabla \cdot u=0 \}$ in $L^p$. -Is such a decomposition also available on a compact Riemannian manifold (with boundary) $M$ with respect to the gradient- and divergence-operator induced by the Riemannian metric? Does one additionally have a "annihilator"-property in the spirit of $D^p(\Omega)^\perp = G^q(\Omega)$ (with dual exponent $q$)? - -REPLY [3 votes]: Günter Schwarz, Hodge Decomposition - A Method for Solving Boundary Value Problems, Lecture Notes in Maths 1607 (1995)<|endoftext|> -TITLE: Recovering a polyhedron from its tumble-density profile -QUESTION [6 upvotes]: Imagine a white convex polyhedron $P$ tumbling randomly about its fixed center of gravity (c.g.) -$c$ against a blue background. -A long-exposure photo would show pure white in a neighborhood of $c$ -(because an opaque ball about $c$ is interior to $P$), -and diminished white and increasing blue in circles of larger radius $r$ -about $c$, for a line of sight at a given $r$ only hits $P$ a fraction of the tumbling time. -My question is to what extent this tumble-density profile uniquely determines $P$. -Henceforth I will specialize to convex polygons $P$ spinning about their c.g. $c$ -in $\mathbb{R}^2$, although all questions generalize to $\mathbb{R}^d$. -Define the profile function $\rho(r)$ at radius $r$ to be the fraction of -the circumference of a circle of radius $r$ centered on $c$ that is interior to $P$. -An example for an isosceles right triangle $P$ (edge lengths 1, 1, $\sqrt{2}$) -is shown below. -Up to $r=\frac{\sqrt{2}}{6} \approx 0.24$, $\rho(r)=1$. Beyond that, $\rho(r)$ -diminishes as illustrated as larger circles have less of their circumference -inside $P$. Derivative discontinuities occur where circles pass through vertices or are -tangent to edges, in this case at $r = \frac{1}{3}$ and $r= \frac{\sqrt{2}}{3} \approx 0.47$. - - - -Spinning the profile function around $r=0$ shows the density at any point around $c$: - -           - - -Q1. - Does $\rho(r)$ uniquely determine $P$ if it is known that $P$ is a triangle? - -The concave sections of $\rho(r)$ seem to be functions specific enough -(sums of inverse trig functions) to -perhaps determine the geometry. - -Q2. For arbitrary convex polygons $P$, are almost all - uniquely determined by their profiles $\rho(r)$? - -Certainly there are pairs of incongruent polygons that have the same profile, e.g., -this pair of augmented regular octagons: - -      - -However, it seems there need be special relationships between these polygons, -so that in some appropriate sense, these are density-zero coincidences, -and generic polygons have unique profiles. - -Q3. - Has this notion of density profile been studied before? - -My questions are related in spirit to those explored in -Richard Gardner's -Geometric Tomography -(Cambridge University Press, Cambridge, 2nd ed., 2006), -but his natural focus on X-rays along lines seems a -different flavor than the -integration around circles in my profiles. Thanks for ideas and pointers! - -REPLY [4 votes]: Q1: Yes, the first three non-differentiable points in the graph of $\rho(r)$ determine the lengths of $GP_A,GP_B,GP_C$, where $G$ is the centroid of $\triangle ABC$, and $P_A$ is the projection of $G$ on $BC$, etc. On the other hand $|GP_A|=\frac{1}{3}h_a$, the altitude from vertex $A$. -It is well known that one can reconstruct the triangle knowing the lengths of the three altitudes (hint: $h_a:h_b:h_c=a^{-1}:b^{-1}:c^{-1}$) - -(Added) Q2: The answer here is also intuitively yes, but it may be a bit painful to write down a rigorous proof. The following holds: - -If $O$ is a point inside the polygon $P=P_1\cdots P_n$ so that all distances $|OP_i|$ and all distances from $O$ to the sides of $P$ are distinct, then we can recover $P$ from the profile $\rho_{O,P}(r)$. - -The proof doesn't use more than just some basic combinatorial geometry and a little analysis. Let's assume that $O$ is the origin. For every convex cone $C$ (the interior of some angle) we can define it's profile -$$\chi_C(r)=Vol(B(r)\cap C).$$ -If the cone $C$ is so that all rays from $O$ intersect it only once, then $\chi_C$ has only one non-differentiable point, it is constant before that point and real analytic after that. With a little work one can show that if $\chi_C=\chi_{C'}$ on some interval then one can obtain $C'$ from $C$ by a rotation and/or a reflection. -Coming back to our problem by integrating $\rho(r)$ we can write $\chi_P(r)=\int_0^r \rho(x) dx=Vol(B(r)\cap P)$. Now $\chi_P$ can be written as a linear combination of $\chi_C$'s by taking $C$'s to be (1) the outer halfplanes determined by the edges of the polygon and (2) the exterior angles at it's vertices. (Technically there is some case work here, but for the sake of brevity I'm assuming $O$ is so that the perpendicular to each edge of the polygon falls on that edge.) -Since we are working with a generic polygon we can obtain all $\chi_C$'s from $\chi_P$ by looking at the intervals between non-differentiable points and taking the successive differences of their analytic continuation. Therefore we can learn the angles of the polygon, the direction of it's edges and the distances between the edges and $O$. This is enough information to reconstruct $P$ up to a rotation and reflection.<|endoftext|> -TITLE: Non isomorphic finite rings with isomorphic additive and multiplicative structure -QUESTION [45 upvotes]: About a year ago, a colleague asked me the following question: - -Suppose $(R,+,\cdot)$ and $(S,\oplus,\odot)$ are two rings such that $(R,+)$ is isomorphic, as an abelian group, to $(S,\oplus)$, and $(R,\cdot)$ is isomorphic (as a semigroup/monoid) to $(S,\odot)$. Does it follow that $R$ and $S$ are isomorphic as rings? - -I gave him the following counterexample: take your favorite field $F$, and let $R=F[x]$ and $S=F[x,y]$, the rings of polynomials in one and two (commuting) variables. They are not isomorphic as rings, yet $(R,+)$ and $(S,+)$ are both isomorphic to the direct sum of countably many copies of $F$, and $(R-\{0\},\cdot)$ and $(S-\{0\},\cdot)$ are both isomorphic to the direct product of $F-\{0\}$ and a direct sum of $\aleph_0|F|$ copies of the free monoid in one letter (and we can add a zero to both and maintain the isomorphism). -He mentioned this example in a colloquium yesterday, which got me to thinking: - -Question. Is there a counterexample with $R$ and $S$ finite? - -REPLY [28 votes]: There do exist pairs of finite unital rings whose additive structures -are isomorphic and whose multiplicative structures are isomorphic, -yet the rings themselves are not isomorphic. -To see this, let $\mathbb F$ be a field and let $X = \{x_1,\ldots, x_n\}$ -be a set of variables. The polynomial ring $\mathbb F[X]$ -is graded by degree -$$ -\mathbb F[X] = H_0\oplus H_1\oplus H_2\oplus\cdots. -$$ -Let $Q(x_1,\ldots,x_n)$ be a quadratic form over $\mathbb F$. -Let -$$I = \mathbb F\cdot Q(X)\oplus H_3\oplus H_4\oplus\cdots$$ -be the -ideal generated by $Q(X)$ and the homogeneous components -of degree at least $3$. -Let $S_{\mathbb F,Q}$ denote the $\mathbb F$-algebra -$\mathbb F[X]/I$. It is a commutative, local ring, which encodes -properties of the quadratic form $Q$. -Two quadratic forms $Q_1$ and $Q_2$ are equivalent -if they differ by an invertible linear change of variables. -Claim. - Let $\mathbb F$ be a finite field of odd characteristic $p$. -Let $Q_1(x_1,\ldots,x_n)$ and $Q_2(x_1,\ldots,x_n)$ be -nonzero quadratic forms over $\mathbb F$. - -$S_{\mathbb F,Q_1}$ and $S_{\mathbb F,Q_2}$ have isomorphic -$\mathbb F$-space structures. -If $n>4$ and $Q_1$ and $Q_2$ are nondegenerate, then -$S_{\mathbb F,Q_1}$ and $S_{\mathbb F,Q_2}$ have -isomorphic multiplicative monoids. -$S_{\mathbb F,Q_1}\not\cong S_{\mathbb F,Q_2}$ as $\mathbb F$-algebras, -unless $Q_1$ is equivalent to a nonzero scalar multiple of $Q_2$. - -Proof. Exercise! \\ -So let $\mathbb F = \mathbb F_3$ be the $3$-element field. -It is known that over a finite field of odd characteristic -the quadratic forms are classified by the dimension -and by the determinant of the form modulo squares. -The determinant of -$$ -Q(x_1,\ldots,x_n)=a_1x_1^2+a_2x_2^2+\cdots+a_nx_n^2 -$$ -is $a_1\cdots a_n$. If -$\alpha\in \mathbb F_3^{\times}=\{\pm 1\}$, -then $\alpha\cdot Q$ has determinant -$\alpha^n a_1\cdots a_n=(\pm 1)^n a_1\cdots a_n$. -If $n$ is even, then the determinants of $Q$ and $\alpha\cdot Q$ will be equal, -so $Q$ will be equivalent -to $\alpha\cdot Q$ for every $\alpha\in \mathbb F_3^{\times}$. -This implies that, when working over $\mathbb F_3$ in an even dimension, -if $Q_1$ is not equivalent to $Q_2$, $Q_1$ will also -not be equivalent to any nonzero scalar -multiple of $Q_2$. -In particular, no scalar multiple of -$$ -Q_1 = x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2, -$$ -is equivalent to -$$ -Q_2 = x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 - x_6^2 -$$ -over $\mathbb F_3$. For these forms we have -that $S_{\mathbb F,Q_1}$ and $S_{\mathbb F,Q_2}$ are nonisomorphic -finite unital rings with isomorphic additive and multiplicative structures. -(These rings have size $3^{27}$.) -Minor side comment 1: If you allow nonunital rings, there is a pair of nonisomorphic $8$-element rings whose additive and multiplicative structures are isomorphic. -Minor side comment 2: -The solution to the exercise above (that is, the proof of the Claim) -can be found here.<|endoftext|> -TITLE: How should I think of the $\infty$-category of spectra? -QUESTION [10 upvotes]: I've seen a bunch of definitions of spectra in the literature, and the fanciest seems to be the $(\infty, 1)$-category of spectra obtaining by "stablizing" the higher category of spaces, as in DAG I. I don't really understand this stabilization procedure yet and would like to connect this idea to the more concrete notions that I've heard about, such as: - -The Boardman category of spectra: here a spectrum is a bunch of spaces (say, CW complexes) $E_n$ with closed cellular imbeddings $SE_n \to E_{n+1}$, and morphisms are defined via cofinal subspectra. -Symmetric or orthogonal spectra, where one just has spaces and morphisms $SE_n \to E_{n+1}$, but there is some additional equivariance condition (and this way we get an honest symmetric monoidal category). - -Presumably from one of these other constructions one can still recover the $\infty$-category of spectra. -For concreteness, I still like to think of a higher category as a topologically (or simplicially) enriched category, or even more concretely a set of objects together with 1-morphisms, 2-morphisms, etc. and various ways of composing them. The 1-morphisms in all these concrete categories are spectra are known (e.g. they're equivariant morphisms in the symmetric or orthogonal case). How should I think of the higher morphisms? - -REPLY [9 votes]: Basically you want to know what the space of maps between two spectra $X$ and $Y$ is. Well, each map from $X$ to $Y$ is a sequence of maps from $X_n$ to $Y_n$ and thus $\mathrm{map}(X,Y)$ is a subspace of the product of the $Y_n^{X_n}$. And you can build an entire spectrum of maps from $X$ to $Y$, whose nth space is just maps from $X$ to the nth suspension of $Y$. -For modern constructions (S-modules, orthogonal spectra, symmetric spectra) of spectra which give symmetric monoidal categories this internal hom is adjoint to the smash product. These modern categories have model structures and between cofibrant and fibrant objects the mapping space described above is homotopically correct. (And, in Boardman's construction --which is not monoidal before passing to the homotopy category--, you still get that this mapping spectrum is the right one for maps between a CW-spectrum and an $\Omega$-spectrum). So one version the stable $(\infty,1)$-category of spectra is the topologically enriched category of fibrant-cofibrant spectra in any of these modern categories, with mapping spaces as above. -Of course, for these model categories of spectra, any of the other ways of getting at the $(\infty,1)$-category they represent, such as Dwyer-Kan localization, will give an equivalent stable $(\infty,1)$-category of spectra.<|endoftext|> -TITLE: Do invariant measures maximize the integral? -QUESTION [32 upvotes]: Update: The negative answer to the following question has been provided by Matthew Daws, who won, but also rejected, the bounty of 100 euro that I set over the question. -Let $\mathcal M(\mathbb Z)$ be the set of all finitely additive probability measures on the power set of $\mathbb Z$. Let $\phi:\mathbb Z\rightarrow\mathbb R$ be nonnegative and bounded. Observe that $\phi$ is integrable with respect to any $\mu\in\mathcal M(\mathbb Z)$. Let me say that $\mu$ is $\phi$-translation invariant if for all $y\in\mathbb Z$ one has -$$ -\int\phi(x+y)d\mu(x)=\int\phi(x)d\mu(x) -$$ -Let $I_\phi(\mathbb Z)$ be the class of $\phi$-invariant measures in $\mathcal M(\mathbb Z)$. Let $\mu\in I_\phi(\mathbb Z)$ be fixed. - -Question: Is it true that the mapping $F:\nu\in\mathcal M(\mathbb Z)\rightarrow\int\int\phi(x+y)d\nu(x)d\mu(y)$ attains its maximum on a measure $\nu\in I_\phi(\mathbb Z)$? - -Thanks in advance, -Valerio - -REPLY [23 votes]: Edit: Here is what I think is a counter-example. -Let $\phi$ be the indication function of the even natural numbers, let $\mathcal U$ be an ultrafilter supported on the even naturals, and let $\mathcal V$ be an ultrafilter defined on the even negative integers. Define $\mu\in M(\mathbb Z)$ by -$$\int f(x) \ d\mu(x) = \lim_{x\in\mathcal V} f(x).$$ -Then $\mu\in I_\phi$ (as, in fact, $\int \phi(x+y) \ d\mu(y)=0$ for all $x$). If $\nu\in I_\phi$ then $\nu$ must assign the same measure to $2\mathbb N$ and $2\mathbb N+1$, say $\alpha\leq 1/2$. You also need to argue that $\nu$ must assign zero measure to any finite set (else it won't be $\phi$-invariant). So for any $y\in\mathbb Z$, -$$\int \phi(x+y) \ d\nu(x) = \nu(2\mathbb N-y) = \begin{cases} -\nu(2\mathbb N) &: y\in 2\mathbb Z, \\ \nu(2\mathbb N+1) &: y\in 2\mathbb Z+1, \end{cases} -= \alpha.$$ -Thus -$$\int \int \phi(x+y) \ d\nu(x) \ d\mu(y) = \int \alpha \ d\mu(y) = \alpha,$$ -as $\mu$ is a probability measure. By contrast, let $\nu$ be defined by -$$\int f(x) \ d\nu(x) = \lim_{y\in\mathcal U} f(y).$$ -Then -$$\int \phi(x+y) \ d\nu(x) = \begin{cases} 1 &: y\in 2\mathbb Z, \\ -0 &: y \in 2\mathbb Z+1, \end{cases}$$ -and so -$$\int \int \phi(x+y) \ d\nu(x) \ d\mu(y) = \int \chi_{2\mathbb Z}(y) \ d\mu(y) -= 1.$$ -So $F$ is not maximised on $I_\phi$. -In fact, by replacing $2\mathbb N$ by $k\mathbb N$, I think you get that $F$ has norm one, but $F(\nu)\leq 1/k$ for any $\nu\in I_\phi$. -But somehow, to my mind, what's wrong is that the $\mu\in I_\phi$ you choose is very poor. So here's a revised conjecture: - -Let $\mu\in I_\phi$ maximise the integral $\int \phi(x) \ d\mu(x)$. Then $F$ attains its maximum on $I_\phi$. - -Old post: (Explains my thinking). -I think of these questions using the Arens products, from abstract Banach algebra theory. So I work over the complex numbers; but this is not a problem. -Consider $A=\ell^1(\mathbb Z)$ with the convolution product, so $A$ is commutative. Then $A^*=\ell^\infty(\mathbb Z) = C(\beta\mathbb Z)$ is an $A$-module: $(a\cdot f)(b) = f(ba)$ for $a,b\in A,f\in A^*$. Then $A^{**}=M(\beta\mathbb Z)$ the space of finite Borel measures on the Stone-Cech compactification $\beta\mathbb Z$. Your space $M(\mathbb Z)$ is just the positive measures $\mu\in A^{**}$ with $\mu(1)=1$. -We try to extend the product of $A$ to $A^{**}$. Firstly we define a bilinear map $A^{**}\times A^*\rightarrow A^*$ by -$$(\mu\cdot f)(a) = \mu(a\cdot f) \qquad (\mu\in A^{**}, f\in A^*, a\in A).$$ -But then we have two choices for the product on $A^{**}$: -$$(\mu \Box \lambda)(f) = \mu(\lambda\cdot f), \quad -(\mu\diamond\lambda)(f) = \lambda(\mu\cdot f) -\qquad (\mu,\lambda\in A^{**}, f\in A^*).$$ -A little thought shows that $\mu\diamond\lambda = \lambda\Box\mu$. -So if $\phi\in A^*$ if positive then $\mu\in I_\phi$ if and only if $\mu\cdot\phi -= \mu(\phi) 1$. This follows, as writing $\delta_x\in A=\ell^1(\mathbb Z)$ for the point mass at $x\in\mathbb Z$, we have -$$(\phi\cdot\delta_x)(\delta_y) = \phi(\delta_{x+y}) \implies -(\mu\cdot\phi)(\delta_x) = \mu(\phi\cdot\delta_x) -= \int \phi(x+y) \ d\mu(y).$$ -So the condition that $\mu\in I_\phi$ becomes that $(\mu\cdot\phi)(\delta_x)$ is constant in $x$, which is seen to be equivalent to $\mu\cdot\phi = \mu(\phi) 1$. -Similarly, your map $F$ is just $F(\nu) = (\mu\Box\nu)(\phi)$. -As you allude to, it's known that $\lambda\Box\mu \not= \mu\Box\lambda$ for arbitrary $\lambda,\mu$. However, we say that $f\in A^*$ is "weakly almost periodic" (WAP) if $(\lambda\Box\mu)(f) = (\mu\Box\lambda)(f)$ for all $\mu,\lambda\in A^{**}$. So if $\phi$ is WAP and $\mu\in I_\phi$ then for any $\nu\in M(\mathbb Z)$, -$$F(\nu) = (\mu\Box\nu)(\phi) = (\nu\Box\mu)(\phi) = \nu(\mu\cdot\phi) -= \nu(1) \mu(\phi) = \mu(\phi),$$ -as $\nu$ is a probability measure. So actually $F$ is constant on $M(\mathbb Z)$ and so certainly attains its maximum at a point of $I_\phi$. -So, to be interesting, we need to ask the question for $\phi$ which are not WAP. An alternative characterisation of $\phi$ being in WAP is that the set of translates of $\phi$ in $\ell^\infty(\mathbb Z)$ forms a relatively weakly compact set. A nice characterisation of Grothendieck shows that this is equivalent to -$$\lim_n \lim_m \phi(x_n+y_m) = \lim_m \lim_n \phi(x_n+y_m)$$ -whenever all the limits exist, for sequences $(x_n),(y_m)$ in $\mathbb Z$. If $\phi$ is the indicator function of $\mathbb N$, then it's not in WAP. -We may as well assume that $\|\phi\|_\infty=1$. -Another "easy" case is when we can find $\nu\in I_\phi$ with $\nu(\phi)=1$. Then $F(\nu) = \mu(\nu\cdot\phi) = \mu(1) \nu(\phi) = 1$; while for any $\lambda\in M(\mathbb Z)$, clearly $|F(\lambda)| = |\mu(\lambda\cdot\phi)| \leq 1$ as $\mu$ is a probability measure, and $\lambda\cdot\phi$ is bounded by $1$ (again, as $\lambda$ is a probability measure and $\phi$ is bounded by $1$). Notice that this case covers your example of when $\phi$ is the indicator function of $\mathbb N$. - -So a test case is to find $\phi$ not in WAP and with $\nu(\phi)<\|\phi\|_\infty$ for all $\nu\in I_\phi$ (notice that $I_\phi$ is always non-empty, as $\mathbb Z$ is amenable). Do you have an example of such a $\phi$? - -Actually, if $\phi$ is the indicator function of the even natural numbers, then that's an example. And that leads to my (hopeful) counter-example.<|endoftext|> -TITLE: Who introduced the notion of "stability" in numerical analysis? -QUESTION [17 upvotes]: I am preparing a lecture course on the applications of operator theory where I intended to make some numerical analysis application. I was wondering about this question while browsing the literature I can access. -Lax and Richtmyer (1955) start with a vague reference to Courant, Friedrichs, and Lewy (1928). -Trotter (1958) on the other hand refers to von Neumann, without any specific reference. -Since I have no deeper background in numerical analysis, I might be missing some obvious point. So can someone tell me (with a reference) who introduced the notion of stability in numerical analysis? - -REPLY [17 votes]: John von Neumann is credited as having pioneered the stability analysis of finite difference schemes. Crank and Nicholson [1] acknowledge Von Neumann when they demonstrate the stability of their scheme in 1947, and a few years later the method was applied in a meteorological context in a paper co-authored by Von Neumann [2]. That 1950 paper introduces the stability analysis as being "patterned after the rigorous method of Courant, Friedrichs, and Lewy." -[1] J. Crank and P. Nicolson, Proc. Camb. Phil. Soc. 43, 50–67 (1947). -[2] J.G. Charney, R. Fjörtoft, and J. von Neumann, Tellus 2, 237-254 (1950). Online at -http://mathsci.ucd.ie/~plynch/eniac/CFvN-1950.pdf -Further reading: -http://en.wikipedia.org/wiki/Von_Neumann_stability_analysis<|endoftext|> -TITLE: Poisson algebras as deformations vs. Poisson algebras in algebraic topology -QUESTION [14 upvotes]: Commutative Poisson algebras $A$ can be thought of as commutative algebras equipped with a first-order deformation into a noncommutative algebra given by the Poisson bracket. A simple example is the symmetric algebra $S(\mathfrak{g})$ of a Lie algebra, which can be deformed into the universal enveloping algebra $U(\mathfrak{g})$. -Today I learned that Poisson algebras also appear in algebraic topology as follows: - -If $X$ is a pointed space, the iterated loop space $\Omega^d(X)$ is an algebra over the little $d$-disks operad. -Hence the homology $H_{\ast}(\Omega^d(X))$ is an algebra over the homology of the little $d$-disks operad. -The homology of the little $d$-disks operad is an operad $\text{Pois}^d$ whose algebras are (graded commutative) Poisson algebras where the bracket has degree $1 - d$. - -(I may have that last statement slightly wrong.) - -Can these two points of view be related? - -For example, is it known whether $H_{\ast}(\Omega^d(X))$ is naturally the associated graded of some filtered algebra $F(X)$ such that the Poisson bracket arises from the (super) commutator in $F(X)$? - -REPLY [7 votes]: Let me start by rephrasing what is already in the answers of David Ben-Zvi and Theo Johnson-Freyd. The DG $\mathbb{Q}$-linear operad $\mathbb{E}_n:=C_{-\bullet}(E_n,\mathbb{Q})$ is filtered. For $n\geq2$ the filtration is the degree filtration, and thus $gr(\mathbb{E}_n)=H_{-\bullet}(E_n,\mathbb{Q})={\rm Pois}^n$. -The situation for $n=1$ is a bit different. We know that $\mathbb{E}_1\cong {\rm As}$ (this is the formality theorem for $E_1$ which, contrary to the case when $n\geq2$, is easy to prove). The operad ${\rm As}$ of associative algebras is also filtered, but in a less obvious way. To be short, one assigns the following two-step filtration onto ${\rm As}(2)=\mathbb{Q}[\Sigma_2]$ (which generates ${\rm As}$): -$$ -F^0{\rm As}(2)=\mathbb{Q}(1-\sigma)\subset F^1{\rm As}(2)={\rm As}(2). -$$ -It then an exercise to check that $gr({\rm As})={\rm Pois}^1$. -Then, in order to relate the two stories, I have the feeling that one does not need to invoke the formality of $E_n$ for $n\geq2$. Given a filtered $\mathbb{E}_n$-algebra $A$ (i.e. a filtered DG $\mathbb{Q}$-vector space equipped with an action of $\mathbb{E}_n$ that is compatible with the above filtration), then $gr(A)$ is a ${\rm Pois}^n$-algebra. -Concerning the last example in the question, one has to take $A=C_{-\bullet}(\Omega^d(X),\mathbb{Q})$ equipped with the degree filtration. Then $gr(A)=H_{-\bullet}(\Omega^d(X),\mathbb{Q})$ is going to be a ${\rm Pois}^d$-algebra. -Side remark: Observe that the story for $E_0$ is even more degerated. Nevertheless,deformation theory of $E_0$-algebras is still very interesting (for a discussion about this issue and its relation to the BV formalisms, see Costello-Gwilliam work-in-progress http://math.northwestern.edu/~costello/factorization_public.html - especially 5b and 5c).<|endoftext|> -TITLE: Is the preimage of the closure the closure of the preimage under a quotient map? -QUESTION [11 upvotes]: Let $f : X \to X/\sim$ be a quotient map from a topological space $X$ to the quotient space $X/\sim$ for $\sim$ some equivalence relation. Let $S \subseteq X/\sim$. Is it true that $f^{-1}(\overline{S}) = \overline{f^{-1}(S)}$? -The specific case I have in mind is a Borel subgroup $B$ of a Chevalley group $G$ acting on $G$. The Bruhat decomposition decomposes $G$ into the disjoint union of Bruhat cells $BwB$ over representatives $w$ of the elements of the Weyl group $N(T)/T$ where $T$ is a maximal torus in $B$ in $N(T)$ is it's normalizer in $G$. Taking the quotient under the action of $B$ on $G$ induces a decomposition of $G/B$ into the disjoint union of $Bw.B$ where $Bw.B$ is the set of cosets of the form $bwB$. The closures of the Bruhat cells $Bw.B$ in $G/B$ are unions of Bruhat cells and I want to know if this therefore implies that the closures of the corresponding Bruhat cells in $G$ are also unions of Bruhat cells. - -REPLY [4 votes]: Since the question has been bumped to the front page, I won't feel guilty about adding another answer. It's a bit more complicated that mathmatrucker's answer, but it involves nicer spaces (metric, whereas mathematrucker's quotient isn't even $\text{T}_1$); it may also be related to the "fairly simple semilinear set" mentioned by Thierry Zell in his comment on his answer. -Let $X$ be the subspace $\{\frac1n:n=1,2,\dots\}\cup\{0\}$ of $\mathbb R$, i.e., a convergent sequence together with its limit. Form a quotient by identifying $1$ with $0$, call the resulting point $p$, and let $S$ be the set of all points except $p$ in the quotient. (So $S$ consists of the images of the points $\frac12,\frac13,\dots$ of $X$.) Then the closure of $S$ is the entire quotient space, so $f^{-1}(\overline S)$ is all of $X$. But $f^{-1}(S)$ is $\{\frac12,\frac13,\dots\}$, whose closure in $X$ does not contain the point $1$.<|endoftext|> -TITLE: A question on the sum of element orders of a finite group -QUESTION [9 upvotes]: Let G be a nontrivial finite group. Is it true that the sum of the orders of all elements of G is not divisible by the order of G? - -REPLY [20 votes]: It is false in general, for instance there's a group of order $3\cdot 5\cdot 7=105$ with sum of orders equal to $1785=3\cdot 5\cdot 7\cdot 17$. (In Magma, it is the first of the two groups of order 105 in the "small groups" database). -However it is true for all groups of even order, because the sum of orders of elements is always odd (this is shown by partitioning $G$ according to the equivalence relation $x\sim y$ if $x$ and $y$ generate the same cyclic subgroup, and using the fact that, for a positive integer $n\geq 1$, $n\varphi(n)$ is odd only if $n=1$.)<|endoftext|> -TITLE: Does the name divisor in algebraic geometry relate to divisor in the basic arithmetic or ring theory sense? -QUESTION [13 upvotes]: This is just a random question I was thinking of. There are lot of cases of things in algebraic geometry unifying and generalizing geometric and arithmetic ideas. For example, the etale fundamental group putting together both Galois theory and covering spaces, so that the etale covers are just field extensions. -I was wondering if there was any analagous reason as to why divisors are called divisors in algebraic geometry. Does their name have any relationship to a divisor as in an element that divides another element in arithmetic or ring theory? Does it reduce to something like that in any special case like the fundamental group reducing to the galois group? Is there any relation at all or is does the name divisor mean something totally different and/or is coincidental or random? -Edit: To clarify I mean divisor as in formal linear combinations of irreducible subvarieties of codimension 1. - -REPLY [12 votes]: This answers address the part of the question asking what the divisors above correspond to in number theory and ring theory. I also include some vague remarks on the naming. -The relevance of Dedekind domains in the geometric context was already mentioned in a comment. And, the notion of Dedekind domains in some sense 'comes from' number theory. -The rings of algebraic integers of algebraic number fields are important examples of Dedekind domains and the historical root for the development. -Rings of algebraic integers are not necessarily unique factorization domains, and one tried to remedy this lack of uniqueness of factorization into irreducible elements by passing to a different (larger) structure. A structure consisting of 'ideal elements' that would not lack the nice property of having unique factorizations into irreducibles. -Now, todays ideals are in some sense these 'ideal elements' for rings of algebraic integers and more generally Dedekind domains. -(One way to define the notion Dedekind domain is: every non-zero ideal is a product of prime ideals.) -Yet, besides the approach via ideals there was also a different/competing approach, namely that of establishing a divisor theory (orig. Divisorentheorie) promoted by Kronecker. -(When I searched for a reference to link to, almost first thing I found was a related MO question, how practical!) -In the number theoretic context today this approach and name is not very wide spread, but might explain the name. And, it is still preserved in more general contexts (cf. below). -But, it is essentially equivalent to considering ideals. And, as mentioned in one of the comments (in suitable circumstances) the fractional ideals correspond to the divisors (in the geometric context). -As said above for a Dedekind domain all non-zero ideals are in a unique way products of prime ideals, so $I= \prod_P P^{v_p}$ where $P$ runs through the non-zero prime ideals and $v_p$ are nonnegative integers all but finitely many $0$. If one allows $v_p$ to be negative too, one gets the fractional ideals. -The link is perhaps even better seen when generalizing a bit, and here the standard terminology even includes words resembling 'divisor'. -Another way to define Dedekind domain would be to say it is a noetherian, integrally closed domain of dimension $1$. -Now, here is a generalization of Dedekind domains: Krull domain. -A way to define what a Krull domain is, is to say that it is a completely integrally closed domain and a v-noetherian domain; where v-noetherian means ACC only on divisorial ideals that is those for which $(I^{-1})^{-1} = I$, in particular v-noetherian is weaker than noetherian; and where completely integerally closed is a condition that is in principle stronger than integrally closed but for noetherian domains it is the same. -So, since Dedekind domains are noetherian and integrally closed they are completely integrally closed, and since they are noetherian they are v-noetherian. Thus, they are Krull domains. -More generally, every integrally closed and noetherian domain is Krull (without a condition on dimension!). -In particular, Krull domains have the property that the localization of the domain at every height one prime ideal is a DVR, and with an extra condition one could take this property as definition, too. -Now, the problem arises that also in Krull domains one would like to have 'ideal elements' or a 'divisor theory'. Yet, the usual ring ideals are in this case not the 'ideal elements'. One needs to consider a different type of ideals namely the divisorial ideals. -It is then true that every divisorial ideal can be uniquely factorized as a product of prime divisorial ideals (in perhaps more classical terms these are the prime ideals of height one). A key-point is that principal ideals are divisorial ideals, and what one cares about mainly is to have a substitute of unique factorization for the ring elements, so principal ideals. -The product used for this factorization is not the usual product of ideals, but rather the product as divisorial ideals that is one forms $((IJ)^{-1})^{-1}$ to get the product of $I$ and $J$. -One can still generalize this more to commutative semigroups with identity and cancellation law. Namely, a semigroup $S$ with a divisor theory is one for which there exists a free commutative semigroup $F$ and a homomorphism $f:S \to F$ such that $a|b$ if and only if $f(a)|f(b)$ for all $a,b \in S$ [one implication is trivial as $f$ is a homomorphism, but the other one is important]. In other words, the arithmetic (or divisibility relations in $S$) are governed directly by the ones in a free semigroup (so a factorial/unique factorization one). -Side note: the $f$ above is not strictly speaking a divisor theory as there is a minimality condition omitted, but still one can rigorously take this as definition; let us assume for simplicity we have in addition this suitable minimality condition (whose definition we omit). -Then we have an $S$ which we care about and an $F$ which is easy to understand (consisting of the 'divisors'). So, the divisors of $S$ control or govern the divisibility relations in $S$. To be a bit more precise, in $F$ itself we only have what corresponds to the positive divisors (that is no negative 'coefficient'). However, to make the link more direct, one can (and does, e.g., for defining the class group) instead consider the quotient group, which consists precisely of all (formal) products $\prod_p p^{s_p}$ where $p$ runs through the prime elements of $F$ the $s_p$ are integers (all but finitely many $0$). [The sole difference is that here multiplicative instead of additive notation is used.] -What is the link of these semigroups to domains: a domain is Krull if and only if its multiplicative semigroup is a semigroup with divisor theory. -In fact, one can take the semigroup of fractional ideals for the $F$. -And, since a Dedekind domain is Krull its multiplicative monoid is also a semigroup with divisor theory and in this case one can take $F$ to be the (usual) ideals. - -In view of the above, if one completely reduces down to the most classical number theory situation of the integers or natural numbers then everything 'collapses' and the positive divisors (so no negative 'coefficient') 'are' just the elements of the original structure, except for zero and 'forgetting the sign' for the integers. [Omitting the positivity condition, the totality of divisors would correspond to the respective quotient group, i.e., the positive rationals with multiplication (one allows positive and negative exponents in the 'prime factorization')] -In the classical situation of algebraic number theory (ring of integers of an algebraic number field) and more generally Dedekind domains the positive divisors correspond to the ideals and the divisors to the fractional ideals. -And, one could say, these divisors control or govern the divisibility relations (in an efficient way). This is also true for the integers; what one omits is the sign which is in fact irrelevant for divisibility. -So, the divisors (in the geometric context) are closely linked (in certain cases the same) as objects that in number and ring theoretic contexts describe or govern the arithmetic, i.e. divisibility relations.<|endoftext|> -TITLE: Definition of a Grothendieck ring -QUESTION [10 upvotes]: I've been looking at some definitions of Grothendieck rings. However I've not found a good definition that I've understood. Any recommendations? - I'm referring to the definition in tensor categories, more specifically I've discovered there are some structure coefficients in a Grothendieck Ring, I understand the mathematics from a physics perspective. I learned this first from Physics papers, but I want to understand more deeply the mathematical structure. -http://www.math.sunysb.edu/~kirillov/tensor/tensor.html is the set of lecture notes I'm referring to, and I'd love if someone could help me tag this post more effectively. - -REPLY [15 votes]: I'll expand my comments into an answer. Since I'm not quite sure what parts bother you, -I'll assume it's everything! The Grothendieck construction is actually -a family of related constructions, which is brilliant in its simplicity. -Whenever, you have a collection of things (e.g. finite sets) that can split -into parts, you can force it be an abelian group by requiring that the sum of parts -correspond to addition in the group. If your things have more structure, then the Grothendieck group can be expected to inherit this as well. -To get closer to what you seem to be interested in, suppose that $G$ is a discrete or Lie group, and $C$ is the category -of finite dimensional complex representations (as a Lie group). Then $C$ is a good example -of a tensor category: It's abelian, so we can speak of exact sequences, there are also tensor products (usual product with $g(v\otimes w) =gv\otimes gw$), and various compatibilities hold. -Given $\rho:G\to V$, one can attach a character $\chi_V= g\mapsto trace(\rho(g))$ -which sometimes determines $V$ when $G$ is compact, but not in general. -Nevertheless, the standard relations - -$\chi_V = \chi_{U}+\chi_{W}$ when $V$ is extension of $W$ by $U$ -$\chi_{V\otimes W} = \chi_{V}\chi_{W}$ - -always hold, which makes this notion quite useful. -Now suppose that $C$ is a more general tensor category. What would be the analogue of the ring of characters on $G$? It would be the Grothendieck ring, generated by symbols $\chi_V$ -where we simply impose the above relations.<|endoftext|> -TITLE: When is the diagonal inclusion a Hurewicz cofibration -QUESTION [10 upvotes]: Given a space $X$, what conditions on $X$ can you give to ensure that the diagonal map $X\to X\times X$ is a Hurewicz cofibration? (I am happy to assume that $X$ is compactly generated weak Hausdorff, or even just Hausdorf.) -More generally, given a map $f:X\to Y$, when is $X\to X\times_YX$ a Hurewicz cofibration? - -REPLY [8 votes]: The diagonal condition was used crucially in Milnor's classical paper. -Milnor, John. On spaces having the homotopy type of a CW-complex. -Trans. Amer. Math. Soc. 90 1959 272–280. -He gives earlier references to Fox and Serre. In the parametrized -generalization, there is a general fiberwise NDR pair characterization -of cofibrations that applies (e.g. Lemma 5.2.4 in May and Sigurdsson, -Parametrized homotopy theory).<|endoftext|> -TITLE: Are commutative alternative rings associative? -QUESTION [8 upvotes]: An alternative ring is an algebraic structure where all the field axioms are true except for the commutativity and associativity of multiplication, but it is alternative, i.e. for all a,b $a(ba)=(ab)a$ and $(aa)b=a(ab)$. If I prove in such a structure that for all a,b $ab=ba$ holds, does it follow that $a(bc)=(ab)c$? -I know that every alternative ring is associative or a Cayley-Dickson ring. So it is enough to decide, wether it is possible that a Cayley-Dickson ring is commutative but not associative. - -REPLY [13 votes]: No. There are commutative alternative rings that are not associative. An example due to Kaplansky is a commutative alternative algebra over the $\mathbb F_3$ with basis $\lbrace x,y,z,u,v,w\rbrace$ and relations $xy=u, yz=v, xv=w, uz=-w$ (the other products are zero). - -It turns out you are interested in commutative alternative division rings. These must be associative here are the steps of the proof (due to R.H. Bruck): - - Prove that for any three elements $x,y,z$ we have $$(x(yz)x=(xy)(zx)$$ $$x(y(zy))=((xy)z)y$$ - Using these identities show that the associator $(x^3,y,z)=0$, where $(a,b,c)=a(bc)-(ab)c$ - Assume that for some three elements $a,b,c$ we have $(ab)c=t(a(bc))$, using the previous identity show that $a^3b^3c^3=t^3a^3b^3c^3$ - Show that $3(a,b,c)=0$. Thus so far we have $3(t-1)=0$ and $t^3=1$. This implies $(t-1)^3=0$, and because there are no zero-divisors we must have $t=1$.<|endoftext|> -TITLE: The fundamental groupoid and a pushout in the category of groupoids. -QUESTION [8 upvotes]: Hi, -Recently I've been looking at the more general version of Van Kampen's theorem, or R. Browns version of it, for the fundamental groupoid. It mentions that if a space X is the union of the interiors of $X_1$ and $X_2$ , then: - -is a pushout in the category of groupoids. In the category of groups, we have the concrete description that the pushout is just the free product with amalgation. Does something similar hold here? Is there any explicit description, like free product with amalgation? - -REPLY [4 votes]: I think the relevant formula is 8.4.1 in T&G. This is applied in section 9.2 to the Phragmen-Brouwer property and the Jordan Curve Theorem. -My original motivation for the investigation was to avoid a detour to compute the fundamental group of the circle: a basic theorem should compute THE basic example! I like the view of the integers (an infinite set) as an identification of a groupoid $\mathbf I $ with 4 arrows, identifying 0 and 1. -Also I tend to see covering spaces in terms of covering morphisms of groupoids, since then a covering map is algebraically modelled by a covering morphism, whereas an action is one step further. -In the new book `Nonabelian algebraic topology', published by the EMS, the van Kampen style arguments are used to compute relative homotopy groups as modules, and second relative homotopy groups as crossed modules, using colimit calculations. -January 25,2016 There is a small correction to the Phragmen-Brouwer proof. -October29, 2020 -It may be useful to think about forming the coequaliser of two morphisms $a,b: G \to H$ of groupoids. If $a,b$ are the identity on objects, the formula is just as you would expect: factorise by the normal subgroupoid of $H$ generated by ....If they are not the identity on objects, you first have to coequalise the objects. This gives a function $f: Ob(G) \to Ob(H)$. So you now need Higgins' "Universal morphism" say $U_f : G \to f_*(G)$. The construction of this generalises free groups, free products of groups, free groupoids ... and is well explained in Higgins' book. "Categories and Groupoids" available as a TAC reprint. -Nov 3, 2020 In the NAT book, Appendix B, this is put in the context of the functor $Ob: Gpds \to Sets$ being a bifibration of categories.<|endoftext|> -TITLE: Multiplying functions on the unit square as generalized matrices -QUESTION [8 upvotes]: Consider the $\mathbb{R}$-vector space of sufficiently nice real-valued functions on the unit square $I^2$, where "sufficiently nice" could be taken to mean any one of a number of things - say continuous for now. -In analogy with matrix multiplication, we can define the product of two such functions $F$ and $G$ as -$$(F\times G)(i,j) = \int_0^1F(i,t)G(t,j)dt.$$ -We can check immediately that this operation is associative (the proof is exactly the same): -$$((F\times G)\times H)(i,j) = \int_0^1(F\times G)(i,t)H(t,j)dt$$ -$$=\int_0^1\left(\int_0^1F(i,s)G(s,t)ds \right)H(t,j)dt$$ -$$=\int_0^1F(i,s)\left(\int_0^1 G(s,t)H(t,j)dt \right)ds$$ -$$=\int_0^1F(i,s)(G\times H)(s,j)ds = (F\times (G\times H))(i,j)$$ -Also, $\times$ is obviously bilinear with respect to usual addition of real-valued functions, and hence defines a ring structure on $C(I^2)$ which is considerably different from the usual ring structure (but addition is the same). -Extending the matrix analogy, we see that each $F$ also defines a linear operator $C(I) \to C(I)$ in the usual way, as -$$F(f)(i)=\int_0^1 F(i,t)f(t) dt$$ -for each $f : I \to \mathbb{R}$. -Also, all of this actually generalizes usual matrix multiplication if we subdivide the square $I^2$ into a bunch of small rectangles and let $F$ be constant on each subrectangle, being more or less careful on boundaries. -The only candidate for a unit element for $\times$ is the distribution which has a weight $1$ dirac delta on the diagonal, and is $0$ everywhere else (in other words, the product of the dirac delta with the Kronecker delta! :)) -Now my question is: what is this? Is it of any interest, or a mere curiosity? For example, could a notion of "determinant" be assigned to these objects? - -REPLY [3 votes]: This is a partial answer concerning the determinants of such objects. While Will Jagy is right that such objects will have an uncountable spectrum, there still may be some hope for assigning a determinant in some cases as follows. First there is one obvious candidate of a trace for such objects: $Tr(F) = \int_0^1 F(t,t)dt$ (Note that the Identity does not have trace equal the dimension of the matrix under this definition, but this definition parallels your approach of replacing all finite summations with integrals). Then in analogy with finite-dimensional matrix theory, the $k^{th}$ powersum of the spectrum of $F$ should be $Tr(F^k)$. Now using the Newton identities, one can recursively construct the sequence of elementary symmetric functions of the spectrum of $F$. -On the one hand, for ordinary matrices $M$ acting on a vector space $V$ of dimension $d$, the elementary symmetric function $e_i$ evaluated on the eigenvalues of $M$ is equal to the trace of the operator defined by the action of $M$ on $\Lambda^i V$. In particular, $\Lambda^d V$ is one-dimensional and therefore the action of $M$ on this space is simply a scalar, this scalar is exactly $e_d = det(M)$. -In your case, your generalized matrices act on an infinite-dimensional function space $V$ and $\Lambda^\infty V$ does not make sense. Nevertheless, combining the earlier definition of the trace of $F$ with the interpretation of the elementary symmetric functions above, we therefore have the following candidate for a determinant of such a generalized matrix $F$ (provided the limit exists): -$Det(F) = \lim_{i\rightarrow\infty}e_i(F)$ -Where: -$e_1(F) = Tr(F)$ -$e_2(F) = \frac{1}{2}\left(e_1(F)\cdot Tr(F)-Tr(F^2)\right)$ -$e_3(F) = \frac{1}{3}\left(e_2(F)\cdot Tr(F)-e_1(F)\cdot Tr(F^2)+Tr(F^3)\right)$ -$\ldots$ -Of course the difficulty in this approach lies in efficiently calculating the $e_i(F)$ and showing the above limit exists. Thus you might need to assume some additional constraints to make this approach work.<|endoftext|> -TITLE: What makes Langlands for n=2 easier than Langlands for n>2? -QUESTION [13 upvotes]: I must confess a priori that I haven't read the proof of Taniyama-Shimura, and that my familiarity with Langlands is at best tangential. -As I understand it Langlands for $n=1$ is class field theory. Not an easy theory, but one that was known for a long time. -Langlands for $n=2$ is the Taniyama-Shimura conjecture, proven recently by Andrew Wiles and others (some of whom participate in this forum). -Clearly Taniyama-Shimura required new ideas. What special property of the $n=2$ case made the proof of Taniyama-Shimura possible, that doesn't exist for Langlands with $n\geq 3$? - -REPLY [3 votes]: One could (and sometimes would) argue the opposite of your claim/question. Namely, that "Langlands for $n = 2$" is more difficult than "Langlands for $n > 2$". Or that more specifically (since I really can't agree with my previous sentence in full generality), "Langlands for $n = 2$" is more difficult than "Langlands for $n > 2$ an odd prime". Here are two key reasons : -1) Suppose $F$ is a $p$-adic field, $n$ is prime, and $p$ doesn't divide $n$. Then the part of the local Langlands correspondence of $GL(n,F)$ dealing with supercuspidal representations (proven by Bushnell/Henniart) is given by $$Ind_{W_E}^{W_F}(\chi) \mapsto \pi(\chi \Delta_{\chi})$$ -where $W_F$ is the Weil group of $F$, $E/F$ is a degree $n$ separable extension, $\chi : W_E \rightarrow \mathbb{C}^*$ is a certain type of character ("admissible" to be precise, see Bushnell/Henniart papers/book), $\Delta_{\chi} : W_E \rightarrow \mathbb{C}^*$ is a "twisting character" associated to $\chi$ (again see Bushnell/Henniart), and $\pi(\chi \Delta_{\chi})$ is the supercuspidal representation of $GL(n,F)$ attached to $\chi \Delta_{\chi}$ via the "Howe construction". If $n$ is an odd prime, then $\Delta_{\chi}$ is either trivial or the unramified quadratic character of $W_E^{ab} \cong E^*$ (note that any character of $W_E$ factors through the abelianization $W_E^{ab}$). If $n = 2$, then $\Delta_{\chi}$ is much more complicated (see page 217 of Bushnell/Henniart's recent book on $GL(2)$). -2) Let $\pi$ be a supercuspidal representation of $GL(n,F)$ (same assumptions as in the beginning of 1) above). If $n$ is an odd prime, then the distribution character $\theta_{\pi}$ of $\pi$ is much simpler to write down on elliptic tori than if $n = 2$. In particular, if $n = 2$, a non-trivial Gauss sum arises in $\theta_{\pi}$, whereas no such term arises if $n$ is an odd prime. So the theory for $GL(2,F)$ contains some nontrivial arithmetic information that just doesn't arise for $GL(n,F)$, $n$ an odd prime.<|endoftext|> -TITLE: Notable math from those without math PhDs -QUESTION [8 upvotes]: Possible Duplicate: -What recent discoveries have amateur mathematicians made? - -Can anyone give historical examples of people outside of mathematics that have contributed to the literature? I'm particularly interested in academics from distant fields (humanities, social sciences) who have developed their hobby into a noteworthy contribution, but would also be interested in those from nearer fields (sciences), where the math was not part of their area of research, and those not affiliated with academia at all. -I'm a math graduate student with hopes to contribute to philosophy or humanities someday. I'm just curious how impenetrable different disciplines are, and realised I know of no examples like this. -Please forgive me if this is a duplicate, I couldn't find another like it, and think others may find it interesting as well. - -REPLY [2 votes]: Hilary Putnam is a philosopher who played a key role in the solution of Hilbert's 10th problem. -http://en.wikipedia.org/wiki/Hilary_Putnam<|endoftext|> -TITLE: More questions involving characteristic 2 theta series identities -QUESTION [5 upvotes]: In my answer to my earlier question, "Existence of certain identities involving characteristic 2 thetas", I established some curious identities when the thetas have prime "level" congruent to 1 mod 4 or to 3 mod 8. This question concerns the case when the level is 7 mod 8. -I reprise notation from earlier questions. l is an odd prime and [j] is the sum of the x^(n^2), where n runs over the integers congruent to j mod l; we view the "theta series" [j] as elements of Z/2[[x]]. F is the power series x+x^9+x^25+x^49+x^81..., G=F(x^l) and H=G(x^l). My identities involve G,H and the various [j]. -There is evidently a unique C in Z/2[[x]], having constant term 0, with C^2+C=G+H. I showed that when l is 1 mod 4 or 3 mod 8 (or when l=7), then C can be written explicitly as a -polynomial in the [j]. Here is what the computer suggests when l=7 mod 8 and is < 50. First some notation. If (r,s,t) is a triple of integers, we define C(r,s,t) to be the sum of the power series [rj][sj][tj] where j runs from 1 to (l-1)/2. Define C(r,s,t,u) similarly. (When l is 3 mod 8, I showed that C is C(1,1,t) where t^2 is congruent to -2 mod l). -(1) When l=7, I can show that C=C(1,1,1,2)+C(1,2,3) -(2) When l=23 I think that C=C(3,3,1,2)+C(1,3,6) -(3) When l=31 I think that C=C(3,3,2,3)+C(2,3,7) (In my original post I wrote C(2,5,8), but C(2,3,7)=C(2,5,8)) -(4) When l=47 I think that C=C(3,3,2,5)+C(2,3,9) -(Note that the sum of the squares of 3,3,2 and 5 is 47, etc.) -QUESTION 1: Can one establish the truth of (2),(3) and (4)? Kevin Buzzard explained to me that it's enough to show that the power series expansions agree up to a certain exponent, but I'm not sure what that exponent is, and I doubt that I have the computer power. -QUESTION 2: Are there identities like those above for l>50? And if so, what are these -identities explicitly? -EDIT: Let V be the space spanned by the C(r1,r2,r3,r4) with r1=r2 and l dividing the sum of the squares of r1,r2,r3 and r4, together with the C(s1,s2,s3) with l dividing the sum of the squares of s1,s2 and s3. When l=7 mod 16 I can use Jacobi's 4-square theorem to show that C is in V. It's then possible to prove identities like those of (2) above by exploiting the -geometry of of Spec R where R is the subring of Z/2[[x]] generated by the theta series [j]. ------One can show that an element of V has at most l(l-1)(l+1)/6 poles, counted with -multiplicity, on the obvious projective completion of this curve. So if it has a zero of large enough order at the origin, it vanishes. I applied this technique for various l congruent to 7 mod 16; the results boggled my mind. It's only necessary to use 2 terms in -the power series expansion of each theta series. When l=23, I got (2) above. -When l=71, I found that C=C(3,3,2,7)+C(5,6,9) -When l=103, I got 5 different expressions for C! Explicitly: -a----C(3,3,6,7)+C(2,9,11) -b----C(7,7,1,2)+C(5,9,10) -c----C(5,5,2,7)+C(1,3,14) -d----C(3,3,2,9)+C(6,7,11) -e----C(1,1,1,10)+C(1,6,13) -It seems possible to me that in general, for l=7 mod 8, one gets h/4 formulae of this -sort where h is the class-number of Q(Root(-2l)). I've discussed the case l=31 in the comment to ARupinski. When l=47, I can show that C(3,3,2,5)+C(2,3,9)=C(1,1,3,6)+C(3,6,7). -So if (4) above holds, there's a second formula for C in this case, just as in the case l=31. But I can't prove that C is in V when l=15 mod 16. -UPDATE__ Suppose l is 7 mod 8; consider the vectors W in Z^3 with(W,W)=2l. There is a group of order 48 operating on the set of such W by permutation and sign change of co-ordinates; the group operates without fixed points. So if there are 12h such W there are h/4 orbits under the group action. -----Ira Gessel's calculations, carried out for l<1500, indicate that there is an involution, O-->O' on the set of orbits, which has the following property. Let O be any of the (h/4) orbits and (r1,r2,r3) be a representative of O with r1 even (so that r2 and r3 are odd). -Then if (s1,s2,s3) is a representative of O', we have the explicit identity C((r1)/2,(r1)/2,(r2+r3)/2,(r2-r3)/2)+C(s1,s2,s3)=C. -----But to know what these conjectured(but true beyond possibility of doubt) equations are for l>1500, we need to describe the involution. Franz Lemmermeyer suggested that the involution comes from an involution on a set of equivalence classes of quadratic forms of -discriminant -8l. This is surely the case; I'll explain what the involution on the forms is, and how to transfer it to the orbits. -----Consider positive quadratic forms rx^2+2sxy+ty^2 with s^2-rt=-2l. Gauss showed that these fall in exactly h equivalence classes under the action of SL_2(Z), where 12h is the number of W with (W,W)=2l; we'll be interested in GL_2 equivalence however. Since rt=2l+s^2, -we find that mod 16, rt is 2,7,14 or 15. This can be used to show that one of the following possibilities must occur: -a.--- Every non-zero n represented by the form is the product of an integer that is 1 or 7 mod 8 by a power of 2. -b.---Every non-zero n represented by the form is the product of an integer that is 3 or 5 mod 8 by a power of 2. -----In the first case we say that the form is in the principal genus, while in the second that it is in the non-principal genus. There are (h/4) GL_2 classes in the non-principal genus. Furthermore there is an involution on this set of classes taking the class of rx^2+2sxy+2ty^2 to the class of 2rx^2+2sxy+ty^2. I'll call this involution "composition with 2x^2+ly^2". -----I now describe a map from the set of (h/4) orbits to the set of (h/4) classes. The map can be shown to be onto, and so is bijective. When we transfer composition with 2x^2+ly^2 to the set of orbits, we get our desired involution; one which is in complete accord with Gessel's calculations. Suppose (W,W)=2l. Let W# consist of all elements of Z^3 orthogonal to W. We attach to W the class of the form (xU+yV,xU+yV), where U and V are a basis of W#. This class is evidently independent of the choice of basis; one can show that it consists of forms of discriminant -8l and lies in the non-principal genus. This gives the desired map from orbits to classes of forms; as I've indicated it is bijective. -EXAMPLE____Take l=1567, and W=(3,25,50) so that (W,W)=2l. Let O be the orbit of W. I'll calculate O', and write down the conjectured equations coming from O and O'. A basis for W# consists of U=(0,2,-1) and V=(25,1,-2). Then (U,U)=5, (U,V)=4, (V,V)=630, and a form attached to O is 5x^2+8xy+630y^2. Composition with 2x^2+1567y^2 takes this to 10x^2+8xy+315y^2. So we seek U' and V' with (U',U')=10, (U',V')=4, and (V',V')=315. Take U'=(3,1,0). A little experimenting, writing 315 as a sum of 3 squares, shows that we should take V'=(5,-11,13). Then W' which is orthogonal to U' and V' can be taken to be their vector product (13,-39,-38). So O' is the orbit of (13,38,39). And one of our predicted expressions for C is C(25,25,11,14)+C(13,38,39), while another is C(19,19,13,26)+C(3,25,50). - -REPLY [2 votes]: I can now, with less computer calculation than I'd feared, answer Question 1. (I'll say more about Question 2 later). -Lemma:__ Let V be the vector space over Z/2 spanned by the C(r1,r2,r3) and the C(s1,s2,s3,s4). If an element of V has its power series expansion divisible by x^(l^2), it is 0. -To see this, let K be an algebraic closure of Z/2, S' be the subring of K[[x]] generated over K by the [j] and L be the field of fractions of S'. I can show that L/K is the function field of a curve, that there are exactly l(l-1)(l+1)/24 valuation rings in L/K that don't contain S', and that each of [1],...,[l-1] has a simple pole at each of them. So an element of V has at most l(l-1)(l+1)/6 poles in L/K, counted with multiplicity. -Also, the localization of S' at the maximal ideal generated by [1],...,[l-1] is dominated by exactly (l-1)/2 valuation rings in L/K. I can show that for each r prime to l there is an automorphism of L/K taking [j] to [rj] for each j, and that these automorphisms act transitively on this set of valuation rings. Now the elements of V are fixed by these automorphisms. So an element of V whose power series expansion is divisible by x^(l^2) has zeros of order at least l^2 at each of these valuation rings. Since (l^2)(l-1)/2 > l(l-1)(l+1)/6, such an element must vanish. -Suppose now that l=23. Since l=7 mod 16, my answer to my question "Existence of certain identities..." shows that C is in V. The lemma then shows that to prove 2. it's enough to show that C(3,3,1,2)+C(1,3,6)+C is divisible by x^529 in Z/2[[x]]. Now C(3,3,1,2)+C(1,3,6) is a sum of monomials in the [j]. Replacing each [j] by the sum of the first 2 terms in its power series expansion only modifies the sum by something divisible by x^529, and we're reduced to an easy computer calculation. -Establishing 3. and 4. is harder since we don't know in advance that C is in V. I'll use: -Theorem____Suppose there is an element R of V such that R+C is divisible by x^(d+1) where d=l(l+1)(l+1). Then R=C. -To see this, recall that F=x+x^9+x^25+..., that G=F(x^l), that H=G(x^l) and that C^2+C=G+H. Now there is a symmetric degree l+1 2-variable polynomial P over Z/2 with P(F,G)=0; furthermore P(z,H) is monic in z of degree l+1. This is discussed in my question "What's known about the mod 2 reduction...". Suppose a^l=1. Replacing x by ax^l in the identity P(F,G)=0, we get a root of P(z,H)=0 of the form ax^l+... This gives us l distinct roots, with G among them. By symmetry P(H,G)=0. So H(x^l) is still another root, and P(z,H) factors into linear factors over K[[x]]. -Now R^2+R+G+H=(R+C)^2+ (R+C), and so is divisible by x^(d+1). Since P(G,H)=0, P(R^2+R+H,H) is divisible by x^(d+1). Also R^2+R has poles of order at most 8 at each valuation ring in L/K that doesn't contain S', while H has poles of order at most 12. It follows that P(R^2+R+H,H) has at most (l+1)l(l-1)(l+1)/2 =d(l-1)/2 poles, counted with multiplicity, in L/K. Arguing as in the proof of the lemma we see that it has more than d(l-1)/2 zeros, counted with multiplicity. So it vanishes, and R^2+R+H is a root of P(z,H)=0. Examining the roots of this equation we see that R^2+R+H can only be G. So R^2+R=G+H, and R=C. -Suppose now that l=31. To prove 3. we now see that it suffices to show that C(3,3,2,3)+C(2,3,7)+C is divisible by x^(186^2), since 186^2 >(31)(32)(32). This is carried out as in the case l=23, but now we have to use the first 12 terms in the power series expansion of each [j] rather than just the first 2. The treatment of 4. is similar, but now we show divisibility by x^(329^2) using the first 14 terms in the power series expansion of each [j]. -EDIT___(UPDATE ON QUESTION 2) -Ira Gessel has now carried out computer calculations for all l<1500; here's a summary of his remarkable results. Consider the triples (r1,r2,r3) where the squares of r1,r1,r2 and r3 sum to l. To avoid duplicates, normalize each such triple so that r1,r2 and r3 are positive with r2>r3. Ira finds that for each such triple there is a unique second (normalized) triple (s1,s2,s3) such that the power series C(r1,r1,r2,r3)+C(2s1,s2+s3,s2-s3)+C is divisible by x^(l^2). Furthermore (r1,r2,r3)-->(s1,s2,s3) is an involution on the set of normalized triples with at most 1 fixed point. -When l=7 mod 16, the argument I gave above when l=23 then shows that for each normalized -triple (r1,r2,r3) and the corresponding (s1,s2,s3) we have the identity C=C(r1,r1,r2,r3)+C(2s1,s2+s3,s2-s3); note that the squares of 2s1, s2+s3, and s2-s3 sum to 2l. So, for example, when l=1447 we get 17 distinct formulae for C. -When l=15 mod 16, it seems certain that once again C(r1,r1,r2,r3) and C(s1,s2,s3) sum to C. -This could be proved by extending Ira's calculations to prove divisibility by x^(d+1) where d=l(l+1)(l+1), as in my treatment of l=31 and l=47. But I think this extension is unnecessary, and that one may deduce the identities simply from the divisibility by x^(l^2) -established by Ira. If my idea for demonstrating this works out I'll post it as a comment. -But great mysteries remain. Why should this all be true? And can one describe the mysterious involution (r1,r2,r3)-->(s1,s2,s3) explicitly? By the way, it's known that the number of normalized triples is odd or even according as n is 7 or 15 mod 16. The proof of this goes -back to Hasse; one shows that the number of triples is h/4 where h is the class number of Q(Root(-2l)) and uses results of Gauss on representations by sums of 3 squares, together with some genus theory for binary quadratic forms.<|endoftext|> -TITLE: Examples of results first proved using geometrical methods? -QUESTION [6 upvotes]: Hi all, -I am beginning to learn about geometric group theory. I would like to write a little exposé intended to be read by the uninitated, so it would be nice to talk about (preferably simple) results which were inaccessible until geometric methods had been applied. Do you have suggestions? -Edit: I would also be interested to hear about results which aren't necessarily inaccessible otherwise, but admit simpler proofs within a geometric approach. -Best to all - -REPLY [2 votes]: The proof of Stallings's Ends Theorem is topological. Note that the set of ends of a group $\Gamma$ can be identified with $H^1(\Gamma,\mathbb{Z}_2\Gamma)$, so you don't have to define ends geometrically.<|endoftext|> -TITLE: Jumping in the integers -QUESTION [23 upvotes]: Consider infinite digraphs whose vertices are the integers $\mathbb Z$, with the property that there are exactly two arcs coming out of each vertex. (There is no restriction on the number of in-coming arcs.) -The question: is there such a digraph such that for $m,n\in\mathbb Z$, with $m0$ then -$$ -x \rightarrow x \pm 1, \phantom+ x \pm 2^{e-1}, \phantom+ x \pm 2^e, \phantom+ x \pm 3\cdot 2^e. -$$ -If $x$ is odd, omit $x \pm 2^{e-1}$ so $x$ goes only to $x \pm 1$ and $x\pm 3$. If $x=0$ then $x\mapsto \pm 1$ only. -Then: -(i) For each $e \geq 0$, the integers of valuation $e$ constitute an arithmetic progression of common difference $2^{e+1}$. We can get from such $x$ to $x \pm 2^{e+1}$ in two steps: if $e=0$ then add or subtract $1$ twice; if $e>0$ then add or subtract $2^{e-1}$ and then $3 \cdot 2^{e-1}$. -(ii) If $v(x) = e$ and $|n-m| \leq 2^e$ we can get from $m$ to $n$ in at most $2 \phantom. \log_2 |n-m| + 1$ steps. Indeed if $|n-m| = 2^e$ one step suffices, so assume $|n-m| < 2^e$ and then $|n-m| \in [2^d, 2^{d+1})$ with $d < e$. By symmetry we may assume $n>m$. Then -$$ -m \rightarrow m + 1 \rightarrow m + 2 \rightarrow m+4 \rightarrow \cdots \rightarrow m+2^d -$$ -in $d+1$ steps, and then in a further $d - v(n)$ steps we reach $n$ by writing -$$ -n = m + 2^d + 2^{d-1} \pm 2^{d-2} \pm 2^{d-3} \pm \cdots \pm 2^{d-v(n)}. -$$ -(iii) If $v(m) = e < f$, and $n$ is the integer of valuation exactly $f$ that's nearest to $m$, then we can get from $m$ to $n$ in $f-e$ steps. At the $i$-th step, add either $\pm 2^{e+i-1}$ or $\pm 3 \cdot 2^{e+i-1}$ to reach a number of valuation exactly $e+i$; there are two choices, so take the one closer to $n$. The distance to $n$ always stays below $2^f$ so at step $f-e$ we hit $n$. -(iv) Finally, suppose $m$ is odd and $|n-m| \in [2^k, 2^{k+1})$. Find $n'$ of valuation $k$ such that $|n-n'| \leq 2^k$. Thus $|m-n'| < 3 \cdot 2^k$. By (iii), in $k$ steps we get from $m$ to some $n_1$ such that $|m-n_1| < 2^k$ and $v(n_1) = k$. So either $n_1=n'$ or $n_1 = n' \pm 2^{k+1}$, and in the latter case we reach $n'$ in a further two steps using (i). Then in at most $2k+1$ further steps we reach $n$ using (ii). -We are done, in a total of $3 \phantom. \log_2 |n-m| + O(1)$ steps, because if $m$ is not odd then we can use our first step to move to the odd number $m \pm 1$ closer to $n$. -Now to get from out-degree $8$ down to $2$: for each $x \in {\bf Z}$ choose $f_i(x)$ for $i=0,1,2,\ldots,7$ so that $\{f_0(x),f_1(x),f_2(x),\ldots,f_7(x)\}$ is the set of integers reachable from $x$ (with repetitions if $x$ is odd or zero). Define a new graph of out-degree $2$ as follows: write every integer as $8x+i$ for some $x \in \bf Z$ and $i\in\{0,1,2,\ldots,7\}$, and then $8x+i \rightarrow 8f_i(x)$ and $8x+i \rightarrow 8x+i'$ where $i'=0$ if $i=7$ and $i'=i+1$ otherwise. That is, we partition $\bf Z$ into directed $8$-cycles, which consumes only one outgoing edge per integer and leaves us with $8$ further edges per cycle to assign as we wish, letting us use our graph of out-degree $8$. Each step of our algorithm then expands to at most $8$, because we might have to take as many as $7$ steps cycling around until we reach the correct $i$ to make the next step in our original graph, and then at the end we might still have to spend an extra $7$ steps to reach the target $n$ within its cycle. But that's still at most $24 \phantom. \log_2 |n-m| + O(1)$ steps, and still with an explicit and "quick" algorithm. -[EDIT we can reduce $3 \phantom. \log_2 |n-m| + O(1)$ to $\frac52 \log_2 |n-m| + O(1)$ by compressing -$$m \rightarrow m + 1 \rightarrow m + 2 \rightarrow m+4 \rightarrow \cdots \rightarrow m+2^d$$ -to -$$m \rightarrow m + 1 \rightarrow m + 4 \rightarrow m+16 \rightarrow \cdots \rightarrow m+2^d,$$ -and there's probably another multiple of $\log_2|n-m|$ to be saved by permuting -$f_0(x),f_1(x),f_2(x),\ldots,f_7(x)$ to reduce the number of steps spent within directed $8$-cycles.]<|endoftext|> -TITLE: Different uses of the word "ergodic" -QUESTION [13 upvotes]: There appear to be two definitions of the word ergodic. -The dynamical systems definition says that a measure space $(X,\mathit B, \mu)$ and measure preserving transformation $T: X \mapsto X$ is ergodic if - -the only $T$-invariant sets have measure 0 or 1. - -However, a Markov chain is ergodic if - -there exists $t$ such that for all $x,y \in \Omega, P^t(x,y) >0$ - -I've used the Markov chain notation and definition found here -I would like to know if these definitions are equivalent. -Of course, I am asking here because it seems to me that they are not. For example, if $X=\{0,1\}$, $\mathit B = \{\emptyset, \{0\},\{1\},X\}$, $\mu(\{0\})=0,\mu(\{1\})=1$ and $T(x) = 1$ for all $x\in X$, then $(X,\mathit B, \mu, T)$ is ergodic as a dynamical system, but the equivalent Markov chain is not ergodic, since the probability of traveling from $0$ to $1$ is zero. - -REPLY [15 votes]: Unfortunately, the way the term "ergodic" is used in the theory of (finite) Markov chains is completely misleading from the point of view of general ergodic theory. To be consistent, one should have called "ergodic" the chains whose state space does not admit a decomposition into non-trivial non-communicating subsets. The notion of ergodicity you are referring to would rather correspond to what is called "mixing" in ergodic theory. -More precisely, an initial distribution $m$ of a Markov chain on a state space $X$ determines the associated measure $\mathbf P_m$ on the space of sample paths $X^{\mathbb Z_+}$. The measure $\mathbf P_m$ is shift invariant iff the measure $m$ is stationary. Now, if $m$ is finite (this condition is important; otherwise the following claim is false), then ergodicity of the time shift is equivalent to absence of non-trivial partitions of $X$ into non-communicating subsets. -By the way, your example is really too degenerate: the standard example for difference between ergodicity and mixing for Markov chains is presence of so-called periodic classes $A_1\to A_2\to\dots\to A_k\to A_1$ (the only allowed transitions are from $A_i$ to $A_{i+1}$ mod k). For finite chains this is actually the only reason for difference between ergodicity and mixing, but for general state spaces the situation is more complicated.<|endoftext|> -TITLE: Finite order automorphisms of complex projective manifolds isotopic to identity -QUESTION [10 upvotes]: Question. Let $V$ be a complex projective manifold of general type (we can even assume that the canonical bundle of $V$ is ample). Suppose $\varphi: V\to V$ is a non-identical automorphism. Can $\varphi$ be isotopic to the identity map (i.e. $\varphi\in Diff_0(V)$)? -I hope the answer is no, and this can be easily proven when $K_V$ is very ample. -More generally what restrictions are known on smooth manifolds that admit self-diffeos of finite order that are isotopic to identity? - -REPLY [13 votes]: The answer to your question is unknown already for surfaces $S$ of general type. -Note that, if $S$ is simply connected, by a result of Quinn (see "Isotopy of 4-manifolds", Journal of Differential Geometry 1986) every automorphism acting trivially on rational cohomology must be topologically isotopic to the identity. -At any rate, it seems that people conjecture that the answer to your question is no for simply connected surfaces of general type. See Catanese's paper "A Superficial Working Guide to Deformations and Moduli" (arXiv:1106.1368), Section 1.4 for further details. In this paper, complex manifolds which do not admit non-trivial automorphisms isotopic to the identity are called rigidified.<|endoftext|> -TITLE: Have commuting functions a common value ? -QUESTION [7 upvotes]: Let $f,g: I \to I := [0,1]$ be continous functions satisfying $f \circ g = g \circ f$. Does -there exist $x_0 \in I$ such that $f(x_0) = g(x_0)$ ? -Background: In a homework the problem was posed with $g=\operatorname{id}$ (where -it can easily be solved with the help of the intermediate value theorem). The -lecturer said the stronger statement above is true, but he didn't know a proof. -I googled a little around, but could only find something about the "commuting -function problem" (existence of a common fixed point of $f$ and $g$) that is -known to be false. - -REPLY [13 votes]: Yes. The set of fixed points of $g$ is closed, nonempty, and is mapped into itself by $f$. Letting $a\le b$ be, respectively, the minimum and maximum fixed points of $g$, we have $f(a)\ge a=g(a)$ and $f(b)\le b=g(b)$. So, by the intermediate value theorem, there is an $x\in[a,b]$ with $f(x)=g(x)$. -Also, to reiterate the points made in the comments, this is a difficult problem for more general domains. The case of commuting maps on the closed disc has been asked before, and is still open. In fact, even the case of commuting maps on the simple triod (i.e., a capital 'T') appears to be an open problem, according to the contributed problem from Jeff Norden here (Commuting, coincidence-point-free maps on a triod).<|endoftext|> -TITLE: Are all countable, nonstandard models of arithmetic given by ultrapowers? -QUESTION [7 upvotes]: Countable models of PA fall into two categories: the standard one $(\omega, S)$ and the nonstandard ones (all the rest). The only way I've seen to construct a nonstandard model is through taking an ultraproduct or, equivalently, using the compactness theorem. My question is wether or not these are all the models there are? There are continuum many ultrafilters and continuum many nonstandard, countable models, but I don't know if there's a surjective correspondence. - -REPLY [4 votes]: It could be mentioned perhaps that Skolem's non-standard model of arithmetic (1933-1934) is countable and is a kind of a "definable" version of the ultraproduct construction. Namely, Skolem only uses definable sequences in his construction. The advantage of his model is that it is constructed without using the axiom of choice.<|endoftext|> -TITLE: The Quaternion Moat Problem -QUESTION [21 upvotes]: "One cannot walk to infinity on the real line if one uses steps of bounded -length and steps on the prime numbers. This is simply -a restatement of the classic result that there are arbitrarily -large gaps in the primes." -So begins the paper by -Gethner, Wagon, and Wick, -"A Stroll Through the Gaussian Primes" -(American Mathematical Monthly -105(4): 327-337 (1998).) -They explain that it is unknown if one can walk to infinity on the Gaussian primes -with steps of bounded length. -Paul Erdős was reported to have conjectured this is possible -("A conjecture of Paul Erdős concerning Gaussian primes." -Math. Comp 24: 221-223 (1970); -PDF). -Later Erdős is reported to have conjectured the opposite: -that no such walk-to-$\infty$ is possible [GWW98, p.327]. -This has become known as the Gaussian Moat Problem, apparently still unresolved. -My question is: - -Is there an analogous Quaternion Moat Problem? - Is it solved? Open? Is it easier or harder than the Gaussian Moat Problem? - -Define a nonzero quaternion $q = a + bi + cj + dk$ as prime -prime iff (a) it is a Hurwitz quaternion (all components integer, or all components half-integer) -and (b) its norm $a^2 + b^2 +c^2 + d^2$ is prime. -(Part (b) is a consequence of the inability to factor $q$; see, e.g., -Theorem 15 in "A Proof of Lagrange's Four Square Theorem Using Quaternion Algebras." -Drew Stokesbary, 2007; PDF). - -Can one "walk-to-$\infty$" on the quaternion primes using steps of bounded length? - -Perhaps relevant here is -Langrange's four-square theorem, which states -that any natural number can be represented as the sum of four squares. -I ask this question in relative naïveté, and appreciate being enlightened. - -REPLY [7 votes]: Having an infinite walk of bounded step length in the quaternions (or in $\mathbb Z^k$ in Gerry's version), gives us a sequence of primes $p_1,p_2\dots$ with $p_{k+1}-p_k=O(\sqrt{p_k})$. However the best unconditional result we have so far on prime gaps is $O(p_k^{0.525})$ by Baker, Harman and Pintz. So these problems are all open in general. -That said the heuristics that work for Gaussian primes can almost always be translated to a more general setting. One famous article on the topic is Vardi's paper "Prime percolation". There it is mentioned that the percolation model can be extended to the general case of primes represented by quadratic forms, quaternion primes etc. (where one can make the same predictions), though this is not written anywhere.<|endoftext|> -TITLE: A bestiary of topologies on Sch -QUESTION [67 upvotes]: The category of schemes has a large (and to me, slightly bewildering) number of what seem like different Grothendieck (pre)topologies. Zariski, ok, I get. Etale, that's alright, I think. Nisnevich? pff, not a chance. There are various ideas about stacks I would like to test out, but the sites I am most familiar with have few application-rich topologies. (Smooth, finite-dimensional manifolds are particularly boring in this respect, and topological spaces are not much better) -What I'm after is a table listing the well-known/common topologies on $Sch$ and their relative 'fineness'. Or, if you like, containment. We of course have the canonical topology - is there a characterisation of that in terms of schematic properties, as opposed to the obvious categorical definition? -And furthermore, one expects that for nice schemes, various topologies will coalesce, say one sort of covers becoming cofinal in another, when restricted to a subcategory of $Sch$. Say those schemes which are Noetherian, smooth or even just varieties. -Then there are things like when categories of sheaves, or 2-categories of stacks, are equivalent. But maybe this is asking too much. -Maybe I'm after something like 'Counterexamples in Grothendieck topologies'. Does such a thing exist, all in one place? I'm sure it is all there in SGA, or the stacks project, or in Vakil's Foundations of Algebraic Geometry, but I'm after the distilled essence. -PS I am interested in things which are (pre)topologies even if they are not usually used as such for the purposes of sheaves. - -EDIT: I'm not merely after examples of Grothendieck topologies on $Sch$, even though that is handy. I want a reference, if there is one, or just a straight-out answer, that compares the various topologies on $Sch$, and under which circumstances (restricting $Sch$ to a subcategory) they coincide. -For example, does an fppf cover of a variety have local sections over an etale cover? Do the fppf and fpqc topologies give rise to the same sheaves over a nicely behaved scheme? Is the etale topology strictly 'weaker' than some other topology no matter what schemes one looks at? Does one get the same Deligne-Mumford stacks for topology A and topology B? -(Grumble over) - -Figure 1 on page 7 of these notes gives a few more of the less common (pre)topologies: cdp, fps$\ell'$ etc. - -REPLY [36 votes]: I have just discovered a chart comparing topologies on Sch/S, made by Pieter Belmans. It includes all the topologies discussed above, and some more I haven't even heard of. It's even interactive and includes definitions and other information.<|endoftext|> -TITLE: A formal definition of Scaling Limits? -QUESTION [5 upvotes]: I'm looking for a formal definition of scaling limit in a rigorous math sense, also, if somebody knows a good translation to spanish. A good bibliography could be helpful. - -REPLY [8 votes]: This is the idea. Suppose that you have a family or sequence of structures of growing complexity (long random walks, realizations of a random field on a large piece of a lattice or in a large continuous domain, large random trees, etc.). You want to understand the behavior of the large structures of your family. Often you want to say that your large random object is similar to a simpler object that you can describe precisely. Since the random walk consisting of 1000 steps is quite different from the "same" random walk consisting of 100000 steps, but you still want to find similarities between them, it makes sense to normalize or rescale your objects appropriately. If you manage to find the right rescaling (it is given by shrinking the time by $n$ and space by $n^{1/2}$ for a standard simple symmetric random walk), then you might discover that thus rescaled (and appropriately embedded into the space of continuous functions or the Skorokhod space) random walk converges in distribution to the Wiener process. -So, scaling limits provide approximative descriptions of what your objects look like when "you look at them from a large distance" or "zoom out". -At a more formal level, suppose $\xi_n$ is a sequence of random objects in some space $X$ . Suppose $\phi_n$ is a (carefully chosen) sequence of scaling tansformations in $X$. It is hard to say precisely what a scaling transformation is, often it is a linear map depending on $n$ with coefficients decaying in $n$. Often, a (time)-reparametrization of the random objects involved is a part of $\phi_n$. A scaling limit is the distributional limit of $\phi_n(\xi_n)$. -Some more comments: - -One point of view is understanding scaling limits as limiting points for renormalization group. -Papers by P.Major from around 1980 on self-semilarity and renormalization are useful in understanding the concept. -I will take this chance to advertise my own paper on scaling limits for random trees, where I describe what large random trees look like if drawn on the plane and looked at from a large distance. It appeared this year in Markov Processes and Related Fields and is also available at http://arxiv.org/abs/0909.2283 The construction and scaling used is different from Aldous's continuum trees (and there are strong connections to superprocesses).<|endoftext|> -TITLE: Notation in Frege's Grundgesetze der Arithmetik: The U with a flourish -QUESTION [17 upvotes]: In the Grundgesetze der Arithmetik, Frege used a number of strange characters for notation. I would be most interested to know anything about the typography (origin, usage and so on) of the strange U with a flourish which occurs in the following. -I am no logician, but I am given to understand that the symbol (U in the following) is used as "a function-name ‘Ux’ in such a way that if y is the extension of a relation, then Uy is the extension of its inverse". -Context: - -Detail: - -Thanks in advance! - -[edit] -In response to some of the comments as to the relevance of this question to mathematics, I add my motivation for it. I have heard it said (by a rather famous Frege scholar) that Frege chose his notation by taking whatever was available in the [type] box. I have come to the view that this is not the case, and that Frege often chose his notation rather carefully. This rather obscure issue leads me to seek the typographic origins of these symbols. I know the origins of most of those in the Grundgesetze (which are surprisingly diverse: phonetics, commerce, German, Greek, ...) but a few remain unidentified, hence the question. - -REPLY [18 votes]: The symbol stands for the currency ``Mark.'' -It is an old symbol developed in handwritten manuscripts. -As far as I know it is a lowercase m with an abbreviation symbol to -indicate that letters are dropped. The lowercase m has changed to -a simple horizontal bar. -See the OLD FLOURISH MARK SIGN on page 146 of -http://folk.uib.no/hnooh/mufi/specs/MUFI-Alphabetic-2-0.pdf<|endoftext|> -TITLE: Well-balanced covering of transpositions in $n$ elements -QUESTION [7 upvotes]: Let me denote $X_n$ the set of transpositions in $n$ elements. Equivalently, $X_n$ is the set of doubletons in $[1,n]\times[1,n]$. The cardinality of $X_n$ is $N=\frac{n(n-1)}{2}$. -If $f:{\mathbb Z}/N{\mathbb Z}\rightarrow X_n$ is a bijection, let us denote -$$r(f):=\min\{|\ell-m|;\ell\ne m\quad\hbox{and}\quad f(\ell)\cap f(m)\ne\emptyset\}.$$ -Finally, let us define -$$R_n:=\max\{r(f);\hbox{bijections}\quad f:{\mathbb Z}/N{\mathbb Z}\rightarrow X_n\}.$$ - -What is the asymptotics of $R_n$ as $n\rightarrow+\infty$. Is it $R_n\sim cn$ for some $c\in(0,\frac12)$? Or do we have $R_n=o(n)$? - -My motivation comes from a numerical algorithm due to Jacobi for the calculation of the spectrum of Hermitian matrices. Each step operates on a pair of rows/columns, with the effect of settong the entry $a_{ij}$ to zero. Once one has act on a row, it seems better to avoid coming back to it too soon. On an other hand, one needs to visit every pairs $(i,j)$ every $N$ steps. - -REPLY [11 votes]: Close to $n/2$ is possible. I'll do odd $n$ and leave even $n$ for someone else's pleasure. -Let $m=(n-1)/2$. For $i=0,\ldots,n-1$ and $j=1,\ldots,m$, let $M(i,j)$ be the pair $\{i-j,i+j\}$ (all values taken mod $n$, of course). The solution is -$$ M(0,1).\ldots,M(0,m),M(1,1),\ldots,M(1,m),\ldots,M(n-1,1),\ldots,M(n-1,m).$$ -Graph theorists will note that this is a standard 1-factorization of $K_n$ listed one factor at a time. -$M(i,j_1)$ and $M(i,j_2)$ are disjoint for $j_1\ne j_2$, so the only chance of two overlapping pairs being closer than $m$ positions is two pairs of the form $M(i,j_1)$ and $M(i+1,j_2)$. A little thought shows that $M(i,j)$ overlaps $M(i+1,j-1)$ and $M(i+1,j+1)$ and no other pairs $M(i+1,j')$. Thus the minimum separation is $m-1=(n-3)/2$. -There are $n-1$ pairs $\{0,j\}$, so two of them must be at most distance $\lfloor N/(n-1)\rfloor = (n-1)/2$, still assuming $n$ is odd. This shows that the solution above is at most 1 worse than the optimum. -EDIT: For even $n$, $(n-2)/2$ is achievable and is optimal. The remaining loose end is whether $(n-1)/2$ is possible for odd $n$.<|endoftext|> -TITLE: (3,2,1)-TQFTs and Verlinde algebras -QUESTION [13 upvotes]: Given a modular category $\mathcal{C}$ there are two natural ways to get a Frobenius algebra out of $\mathcal{C}$. One is to take the Verlinde algebra (or `fusion algebra') of $\mathcal{C}$. The other consist in considering the $(3,2,1)$-dimensional TQFT associated with $\mathcal{C}$, and to get out of it a $(2,1)$-dimensional TQFT by multiplication by $S^1$ (and a $(2,1)$-dimensional TQFT is the same thing as the datum of a Frobenius algebra). It is well known in fully extended TQFT folklore that these two constructions coincide. Is anyone aware of a reference I could cite as a source for this statement? (I know Dan Freed's -The Verlinde algebra is twisted equivariant K-theory, where this can be read between the lines) - -REPLY [7 votes]: A somehow detailed answer could be as follows (thanks to Alessandro Valentino, who is a coauthor of this answer (but me alone is to blame for mistakes and inaccuracies in it)). Kevin Walker's notes have been an essential source of inspiration. -Let $\mathcal{C}$ be a modular tensor category, and let $\{X_i\}$ a set of representatives for the isomorphism classes of its simple objects. We write $X^i$ for the dual of $X_i$. Then the element associated by the (3,2,1)-TQFT associated with $\mathcal{C}$ to the cylinder with two outgoing $S^1$'s is the element -$$ -\mathrm{coev}_{\mathcal{C}}=\bigoplus_{i\in I} X^i\boxtimes X_i -$$ -of $\mathcal{C}\boxtimes \mathcal{C}$, while the element -$$ -\mathrm{ev}_{\mathcal{C}} \colon \mathcal{C}\boxtimes \mathcal{C}\to Vect -$$ -associated to the cylinder with two ingoing $S^1$'s is -$$ -\mathrm{ev}_{\mathcal{C}}(A\boxtimes B) = \mathrm{Hom}(A^*,B) -$$ -We then have -$$ -Z(T^2)\equiv\mathrm{ev}_{\mathcal{C}}\circ \mathrm{coev}_{\mathcal{C}}= \bigoplus_{i\in I} \mathrm{End}(X_i)=\bigoplus_{i\in I} \mathbb{K} = \mathbb{K}^I. -$$ -In other words $Z(T^2)$ is isomorphic to the vector space with basis given by isomorphism classes of simple objects of $\mathcal{C}$. In particular we have -$$ -\dim Z(T^2) = \#\{\text{isomorphism classes of simple objects of $\mathcal{C}$}\} -$$ -Let us now describe a canonical basis for $Z(T^2)$. To do so we look at the boundary conditions for the TQFT associated with $\mathcal{C}$. Any object $X$ in $\mathcal{C}$ defines a boundary condition, that is, a $Vect$-linear functor $Vect\to \mathcal{C}$ which we will denote by the same symbol, i.e., we write $X\colon Vect\to \mathcal{C}$. The TQFT assigns the functor $X$ to a cylinder with the incoming copy of $S^1$ decorated by the colour $X$. Multplying this cylinder by $S^1$ we get the 3-dimensional cylinder over the basis $T^2$, with the incoming basis decorated by the colour $X$. This cylinder is a morphism $\mathbb{K}\to Z(T^2)$, i.e., an element of $Z(T^2)$. We will denote this element by $v_X$. Since $\mathcal{C}$ is semisimple, every short exact sequence $0\to X\to Y\to W\to 0$ in $\mathcal{C}$ splits, i.e., $Y\cong X\oplus W$. This implies that $v_Y=v_X+v_W$ and so $v$ defines a linear map -$$ -v\colon K(\mathcal{C})\to Z(T^2) -$$ -We are going to show that this amp is a linear isomorphism of vector spaces. - The cylinder over the basis $T^2$, with both copies of $T^2$ incoming defines the inner product on $Z(T^2)$; let us denote this inner product by $\langle \,|\,\rangle$. Then $\langle v_X | v_Y\rangle$ is the element in $k$ that the TQFT associates to the cylinder over the basis $T^2$, with both copies of $T^2$ incoming, the first one decorated by $X$ and the second one decorated by $Y$. This cylinder is obtained by multiplying by $S^1$ the cylinder over the basis $S^1$, with both copies of $S^1$ incoming, the first one decorated by $X$ and the second one decorated by $Y$. By the TQFT rules, this cylinder over the base $S^1$ is evaluated to $\mathrm{ev}_{\mathcal{C}}(X\boxtimes Y)=\mathrm{Hom}(X^*,Y)$. Therefore -$$ -\langle v_X | v_Y\rangle=\dim \mathrm{Hom}(X^*,Y). -$$ -In particular, if we write $v^i$ for $v_{X^i}$ and $v_i$ for $v_{X_i}$ we see that -$$ -\langle v^i | v_j\rangle=\dim \mathrm{Hom}(X_i,X_j)=\delta^i_j, -$$ -and so the vectors $\{v_i\}_{i\in I}$ are linearly independent vectors of $Z(T^2)$. Since their number equals the dimension of $Z(T^2)$, we see that the collection $\{v_i\}_{i\in I}$ is a linear basis of $Z(T^2)$. Since $\{v_i\}_{i\in I}$ is a basis, mapping $v_i$ to the equivalence class of $X_i$ in $K(\mathcal{C})\otimes k$ defines a linear map $Z(T^2)\to K(\mathcal{C})\otimes k$ which is manifestly the inverse to $v$. -The tensor product on $\mathcal{C}$ induces a multiplication on $K(\mathcal{C})\otimes k$ by setting -$$ -[X_i]\cdot [X_j]= [X_i\otimes X_j] -$$ -If we write -$$ -X_i\otimes X_j\cong \bigoplus_{k\in I} \mathbb{K}^{n_{ij}^k}\otimes X_k -$$ -then we see that the multiplication induced by the tensor product on $K(\mathcal{C})\otimes \mathbb{K}$ is given by -$$ -[X_i]\cdot [X_j]= \sum_{k\in I} n_{ij}^k [X_k]. -$$ -Notice that from $X_i\otimes X_j\cong \bigoplus_{k\in I} \mathbb{K}^{n_{ij}^k}\otimes X_k$ we see that -$$ -n_{ij}^k=\dim\mathrm{Hom}(X_k,X_i\otimes X_j). -$$ -On the other hand we have a multiplication induced on $Z(T^2)$ by the TQFT; namely, by the pair of pants times $S^1$. Since the pair of pants corresponds to the tensor product of $\mathcal{C}$ it is quite natural to expect that the multiplication on $Z(T^2)$ will coincide with the multiplication induced by the tensor product on $K(\mathcal{C})\otimes \mathbb{K}$, i.e., that $v$ is actually an isomorphism of commutative algebras. To see that it is indeed so, we only have to compute the product $v_i\cdot v_j$ in $Z(T^2)$ and check that this is -$$ -v_i\cdot v_j= \sum_{k\in I} n_{ij}^k v_k. -$$ -This is equivalent to showing that -$$ -\langle v^k | v_i\cdot v_j\rangle = n_{ij}^k -$$ -The element on the left hand side is what the TQFT assigns to a pair of pants times $S^1$ with all the three copies of $T^2$ incoming, decorated by the colours $X_i$, $X_j$ and $X_k$ respectively. This is the same as a pair of pants with all the three copies of $S^1$ incoming, decorated by the colours $X_i$, $X_j$ and $X^k$ respectively, multiplicated by $S^1$. Since the pair of pants with all the three copies of $S^1$ incoming, decorated by the colours $X_i$, $X_j$ and $X^k$ respectively is mapped by the TQFT to $\mathrm{Hom}(X_k,X_i\otimes X_j)$ we see that when we multiply it by $S^1$ we get $\dim \mathrm{Hom}(X_k,X_i\otimes X_j)$. Hence we have -$$ -\langle v^k | v_i\cdot v_j\rangle=\dim \mathrm{Hom}(X_k,X_i\otimes X_j)=n_{ij}^k, -$$ -which is precisely the sought for identity.<|endoftext|> -TITLE: Is the set of all curves that have a Galois map to the projective line Zariski closed in M_g? -QUESTION [5 upvotes]: Not all Riemann surfaces have branched Galois maps to the Riemann sphere. One way to see this is that if $C\rightarrow\mathbb{P}^1$ is Galois, this implies that $C$ is defined over its field of moduli (as a curve). -What is the shape of all genus $g$ Riemann surfaces that have a Galois map to $\mathbb{P}^1$? Is it Zariski closed in the coarse moduli space of genus $g$ curves $M_g$? - -REPLY [10 votes]: Here is a sketch of an argument that should prove that the set of curves of genus $g$ with automorphisms is closed. -As pointed out in the comments, by the Hurwitz bound there are at most finitely many groups that can act on a curve of genus $g$. Fix such a $G$ and fix a representation $V$ of $G$ of dimension $5g-5$. Again there are only finitely many such $V$'s up to isomorphism. Denote by $X_{V,G}$ the subset of curves in the moduli with a faithful $G$-action such that the representation of $G$ on $H^0(3K)$ is isomorphic to $V$. Denote by $H_{G,V}$ the subset of the Hilbert scheme of ${\mathbb P}(V)$ whose points correspond to the smooth curves $C$ of genus $g$ such that $C$ is $G$-invariant and the hyperplane section restricts to $3K_C$. -The set $H_{G,V}$ is quasi-projective variety and it maps onto $X_{G,V}$, so $X_{G,V}$ is constructible. -So, to prove that the set $X_G$ of curves with a faithful $G$-action is closed, it is enough to show that if a curve $C_0$ is the special fiber of a $1$-parameter family $\{C_t\}$ of smooth curves with a faithful $G$-action, then $C_0$ also has a faithful $G$-action. Up to base change we may assume that the base of the family is a small disk $D\subset {\mathbb C}$, hence that the total space $S$ of the family is a smooth surface. By assumption $G$ acts birationally on $S$. Hence it is known that it is possible to blow up $S$ and obtain a smooth surface $S'$ on which $G$ acts biregularly. The components of the special fiber $C_0'$ of $S'$ are the strict transform of $C_0$ and some rational curves. Since an automorphism cannot map a rational curve to a curve of positive genus, it follows that $G$ acts biregularly on $C_0$ (in fact by a similar argument one can show that $G$ acts biregularly on $S$). -The last thing to show is that the $G$-action on $C_0$ is faithful. So assume that $g\in G$, $g\ne 1$ fixes $C_0$ pointwise and let $x\in C$ be a general point. Then there are analytic coordinates $(z,t)$ near $x$ such that $C_0$ is given by $t=0$ and the $G$-action is given by $(z,t)\mapsto (z, ut)$, with $u\ne 1$ a root of unity. So locally $t=0$ is a section of the map $S\to D$ and $g$ acts nontrivially on this section. But we have a contradiction, since the action of $G$ preserves the fibers $C_t$ by construction. -Finally, the question whether $C/G$ is rational. -The genus of the quotient curve is equal to the number of times the trivial representation appears in the $G$-representation on $H^0(K_C)$. -EDIT: As above, let $f\colon S\to B$ be a smooth family of curves over a smooth connected curve $B$ such that $G$ acts on $S$ preserving the fibration. Then $f_*\omega_{S/B}$ is a rank $g$ vector bundle with a $G$-action and the fiber of $f_*\omega_{S/B}$ over $b\in B$ is $H^0(K_{C_b})$. The number of times the trivial representation appears in the decomposition of $H^0(K_{C_b})$ is the scalar product of the character of this representation with the trivial character. Hence it is a regular function of $b$ and, being an integer, it is constant. So the locus where $C/G$ is rational is closed, too.<|endoftext|> -TITLE: Is the composition of two bundle projections necessarily a bundle projection? -QUESTION [9 upvotes]: That is, if $f: X \rightarrow Y$ and $g:Y \rightarrow Z$ are bundle projections, is $g \circ f: X \rightarrow Z$ a bundle projection? Assume $X$, $Y$ and $Z$ are manifolds. -Here is what I know. The answer is affirmative when (1) $f$ is a covering map and $g$ is bundle projection; (2) $f$ is a bundle projection and $g$ is a covering map of finite degree. -What can we say about the most general situation? -Thanks. - -REPLY [8 votes]: In the category of finite dimensional manifolds, I proved this in a very long paper (p. 23): the composition of smooth fiber bundle maps is a smooth fiber bundle map. Spanier's counterexample is not a finite dimensional manifold.<|endoftext|> -TITLE: Continuously selecting elements from unordered pairs -QUESTION [16 upvotes]: The symmetric square of a topological space $X$ is obtained from the usual square $X^2$ by identifying pairs of symmetric points $(x_1,x_2)$ and $(x_2,x_1)$. Thus, elements of the symmetric square can be identified with unordered pairs $\{x_1,x_2\}$ from $X$ (including the degenerate case $x_1 = x_2$). With this identification, the following question is very natural: - -When is there a continuous selector from the symmetric square to the original space? - -This is always possible if $X$ is a generalized ordered space (a subspace of a linear order with the order topology) since min and max are both continuous selectors. I don't think this is a necessary condition since I can construct continuous selectors (on Baire space, for example) that aren't equal to min or max for any ordering of the underlying space. -A necessary condition is that removing the diagonal disconnects $X^2$. More precisely, as Adam Bjorndahl pointed out in the comments, the complement of the diagonal should be the union of two disjoint open sets such that $(x_1,x_2)$ is in one iff $(x_2,x_1)$ is in the other. So, for example, there are no continuous selectors for the symmetric square of the plane nor the circle. -I'm mostly interested in the case when $X$ is Polish, so it's fine to assume that $X$ is very nice: Hausdorff, normal, perfect, etc. - -REPLY [6 votes]: In this paper Van Mill and Wattel proved that the existence of a continuous selection characterizes orderability in the class of compact Hausdorff spaces.<|endoftext|> -TITLE: Intersection between category theory and graph theory -QUESTION [18 upvotes]: I'm a graduate student who has been spending a lot of time working with categories (model categories, derived categories, triangulated categories...) but I used to love graph theory and have always had a soft space in my heart for it. The semester I'm taking a graph theory course and for the first time seeing connections to category theory in the form of the underlying directed graph of a category (objects become vertices, and we put an edge $(A,B)$ if there's a morphism $A\rightarrow B$). - -I'd like to learn more about places where graph theory has informed category theory. - -One question that can be asked in this vein is whether a given graph could be the underlying graph of a category. This has been answered for finite graphs by Samer Allouch in his thesis. Sadly, the thesis is in French and I can't really understand it. Here is a nice n-Category Cafe post about this. In that post someone asked for an explanation of the results or ideas in English, but that was never given. I'd love to hear about this if anyone here knows about it, but my question is more general. Since most categories I come across and understand are not finite (nor even small), I'm wondering about the existence of work connecting large categories with graph theory. -Because this is very vague, I came up with a few specific questions. However, if all I get is reference requests which have nothing to do with these questions I'll still be happy. In particular, I'd love to hear about times questions similar to those below have been asked and answered, or to get a recommendation for a textbook on category theory which focuses on the underlying graph more than Maclane does. - -1) Has anyone classified (sub)graphs which correspond to replete subcategories? For instance, is there a purely graph theoretic test for whether or not a subgraph $\mathcal{H}$ of $G$ underlies a replete subcategory $\mathcal{D}$ of $\mathcal{C}$? - -Here's what I can tell so far about this problem: for each non-isomorphic object in $\mathcal{D}$ you will get a complete graph in $H$. Composition morphisms ensure that if there is any arrow from one vertex to another vertex in a different complete subgraph then there will be arrows from all vertices in the first complete subgraph to all in the second. So $H$ will need to look like a disjoint union of complete graphs, where complete here means for every two vertices there's an edge at least one way between them. One reason this is hard for me to wrap my brain around is that we're dealing with subgraphs of uncountable graphs in which vertices can have uncountable degree. I can solve this by restricting to isomorphism classes of objects, but I'm not sure this is a good idea. I'd love to hear from those more versed in category theory about the pros/cons of restricting to isomorphism classes. Note that the version of this question for full subcategories is answered by "induced subgraphs" and so it seems like you have no hope of classifying thick subcategories or other interesting full subcategories. I originally asked this question for dense subcategories and the helpful comment of David Roberts below shows you won't have success here because going to the underlying graph loses composition data. -Another sample question, about products of categories: - -2) The underlying graph of $\mathcal{C}\times \mathcal{D}$ is the strong product $G\boxtimes H$ of the two underlying graphs. Is there a category corresponding to the cartesian product of two underlying graphs? The tensor (direct) product? - -REPLY [7 votes]: There is an interaction between category theory and graph theory in -F.~W.~Lawvere. Qualitative distinctions between some toposes of generalized graphs. In {\em Categories in computer science and logic (Boulder, CO, 1987)/}, volume~92 of {\em Contemp. Math./}, 261--299. Amer. Math. Soc., Providence, RI (1989). -which we have exploited in -R. Brown, I. Morris, J. Shrimpton and C.D. Wensley, `Graphs of Morphisms of Graphs', Electronic Journal of Combinatorics, A1 of Volume 15(1), 2008. 1-28. -But that is actually about possible categories of graphs, which may be the opposite of the question you ask. -If you look at groupoid theory, then "underlying graphs" are fundamental, for example in defining free groupoids. See for example -Higgins, P.~J. Notes on categories and groupoids, Mathematical Studies, Volume~32. Van Nostrand Reinhold Co. London (1971); Reprints in Theory and Applications of Categories, No. 7 (2005) pp 1-195. -Groupoids are kind of "group theory + graphs".<|endoftext|> -TITLE: Invariants and base change -QUESTION [7 upvotes]: Suppose $R$ is a Noetherian commutative ring, and $M$ a finite free $R$-module, with an action of a finitely generated discrete group $G$ by $R$-linear maps. -Is there any homological condition on this data which would ensure that taking $G$-invariants commutes with base change? -That is, for any finite type map $R \rightarrow S$, we have -$$(M \otimes S)^G = (M^G)\otimes S.$$ -For instance, suppose $M^G = 0$? - -REPLY [2 votes]: Choose generators $g_1,\dots,g_n$ of $G$ and let $f\colon M\to M^n$, $f(m)=(g_im-m)_{i=1,\dots,n}$. This gives an exact sequence $$0\to M^G\to M\to M^n\to\mathrm{coker}(f)\to0$$ which can be interpreted (EDITED) as part of a flat resolution of $\mathrm{coker}(f)$. Since $\ker(f\otimes1_S)=(M\otimes S)^G$, the quotient $(M\otimes S)^G/\mathrm{im}(M^G\otimes S)$ may be identified with $\mathrm{Tor}_1(\mathrm{coker}(f),S)$, so they both vanish for all $S$ iff $\mathrm{coker}(f)$ is flat. If $\mathrm{coker}(f)$ is flat, we also get that $M^G\otimes S\to(M\otimes S)^G$ is injective, so we get the equivalence $M^G\otimes S\to(M\otimes S)^G$ bijective iff $\mathrm{coker}(f)$ is flat.<|endoftext|> -TITLE: Given 2 towers of fields, when are these fields isomorphic? -QUESTION [5 upvotes]: Let $F_0 \subset F_1 \subset F_2 \subset \cdots$ and $K_0 \subset K_1 \subset K_2 \subset \cdots$ be two towers of fields. Also, let $F = \cup_{i=0}^\infty F_i$ and $K = \cup_{i=0}^\infty K_i$. -Now suppose for each $i$ we have injective homomorphisms from $F_i$ to $K_{\sigma(i)}$ and from $K_i$ to $F_{\mu(i)}$ where $i \leq \sigma(i)$ and $i \leq \mu(i)$. In other words, each field $F_i$ is isomorphic to a subfield of some $K_j$ where $j \geq i$ and each field $K_i$ is isomorphic to a subfield of some $F_j$ where $j \geq i$. [Think of the two towers sitting next to each other with arrows pointing diagonally upward.] -My question, can we conclude that $F \cong K$? -A colleague asked me this question some time ago. I came up with a sketch of a proof for the case when $F_{i+1}$ is an algebraic extension of $F_i$ and $K_{i+1}$ is an algebraic extension of $K_i$ for each $i$. I suspect it's false in general [something to do with the fact that injective and surjective aren't equivalent for maps between infinite dimensional spaces.] -Does anybody know a counterexample for the general case? -I would also appreciate a reference for the algebraic case (where I'm 99% sure it's true). -Thanks! - -REPLY [12 votes]: Here is a counterexample which, unlike JSpecter's, does not rely on transcendence degrees and Zorn's lemma (used to create an embedding of ${\mathbf C}(X)$ into $\mathbf C$). It comes from a pair of isogeneous but nonisomorphic elliptic curves. Let $E_1$ and $E_2$ be elliptic curves over ${\mathbf Q}$ admitting an isogeny $E_1 \rightarrow E_2$. There is a (dual) isogeny $E_2 \rightarrow E_1$ and these maps provide one with homomorphisms between the function fields ${\mathbf Q}(E_2) \rightarrow {\mathbf Q}(E_1)$ and ${\mathbf Q}(E_1) \rightarrow {\mathbf Q}(E_2)$. So each function field embeds into the other. Choosing $E_1$ and $E_2$ to be isogeneous but not isomorphic over ${\mathbf Q}$ -- say their $j$-invariants are not equal to assure non-isomorphism -- the function fields ${\mathbf Q}(E_1)$ and ${\mathbf Q}(E_2)$ are not isomorphic as fields. This is an absolute statement since we work over ${\mathbf Q}$ rather than, say, ${\mathbf C}$ (where one would conclude the function fields are not isomorphic over ${\mathbf C}$, i.e., fixing ${\mathbf C}$). The same argument goes through using isogenous but non-isomorphic elliptic curves over ${\mathbf F}_p$ for any prime $p$. -By the way, this question is similar to -Schroeder-Bernstein for Rings and my answer is similar to one given there.<|endoftext|> -TITLE: Suppose $\Gamma_m$ is a principal congruence subgroup of level m contained in a finite index subgroup $\Gamma$ of $SL(n,\mathbb Z)$. Is $\Gamma_m$ characteristic in $\Gamma$? -QUESTION [5 upvotes]: We know that principal congruence subgroups are characteristic in $SL(n,\mathbb Z)$. -Suppose $\Gamma$ is a finite index subgroup of $SL(n,\mathbb Z)$ and $\Gamma_m$ is a principal congruence subgroup of level m contained in $\Gamma$. Will it be characteristic in $\Gamma$? - -REPLY [6 votes]: This is false. Consider the case $n=2$, and let $p$ be a prime. Let $A=\left[\begin{array}{cc}0 & -1 \\\ p & 0\end{array}\right] -$ be an Atkin-Lehner involution (considered as an element of $PGL_2(\mathbb{Q})$), and consider the subgroup -$\Gamma_0(p) = \{ \left[\begin{array}{cc}a & b \\\ c & d\end{array}\right]\in SL_2(\mathbb{Z}) | c\equiv 0(\mod p) \}$. Then one may check that for $B\in \Gamma_0(p)$, $A^{-1} B A = \left[\begin{array}{cc}d & -c/p \\\ -pb & a\end{array}\right] -\in \Gamma_0(p)$, so $A\in Aut(\Gamma_0(p))$. Also, the principal congruence subgroup $\Gamma(p) \leq \Gamma_0(p)$. However, consider the matrix $C=\left[\begin{array}{cc}1 & 0 \\\ p & 1\end{array}\right] \in \Gamma(p)$. Then one has $A^{-1} C A = \left[\begin{array}{cc}1 & -1 \\\ 0 & 1\end{array}\right] \notin \Gamma(p)$. Thus, the subgroup $\Gamma(p)\leq \Gamma_0(p)$ is not characteristic. This extends to all $n$, taking the appropriate congruence subgroup by extending trivially $\mod p$.<|endoftext|> -TITLE: If I want to study Jacob Lurie's books "Higher Topoi Theory", "Derived AG", what prerequisites should I have? -QUESTION [59 upvotes]: I've been told that it's important to know modern physics, Differential Geometry and Algebraic Topology for understanding higher structures. Is there any other prerequisite for understanding Lurie's work? Since the title of the book indicates, I guess Algebraic Geometry is also important. Please tell me if I'm wrong. Moreover how deep should I known on those subjects and others I do not mention? - -REPLY [187 votes]: To read Higher Topos Theory, you'll need familiarity with ordinary category theory -and with the homotopy theory of simplicial sets (Peter May's book "Simplicial Objects in Algebraic Topology" is a good place to learn the latter). Other topics -(such as classical topos theory) will be helpful for motivation. -To read "Higher Algebra", you'll need the above and familiarity with parts of "Higher Topos Theory". Several other topics (stable homotopy theory, the theory of operads) will be helpful for motivation. -To read the papers "Derived Algebraic Geometry ???", you need all of the above plus familiarity with Grothendieck's theory of schemes, along with some more recent ideas in algebraic geometry (stacks, etcetera). -Since no knowledge of modern physics was required to write any of these books and papers, I can't imagine that you need any such knowledge to read them.<|endoftext|> -TITLE: Is there a name for the class of metric spaces such that the closure of the open ball of radius $r$ around each point $x$ is the set of elements $y$ such that $d(x,y)\leq r$ ? -QUESTION [5 upvotes]: Let $(X,d)$ be a metric space, let $B(x,r)$ be the open ball of radius $r$ about $x$ and $N(x,r)$ be the set of elements $y\in X$ such that $d(x,y)\leq r$. It is well-known that it is not always true that $N(x,r)$ is the closure of $B(x,r)$. -I need, for some research, to restrict my attention to metric spaces for which that property is true, i.e. $N(x,r)$ is the closure of $B(x,r)$. Do they have a particular name in literature? -Thanks in advance, -Valerio - -REPLY [5 votes]: I'm not sure what they're called, but according to this site an equivalent characterization of spaces $X$ where $\overline{B(x,r)} = N(x,r)$ is: for all $p\in X$, the -only local minimum of the function $x \rightarrow d(x,p)$ is at $x=p$. The proof is also there.<|endoftext|> -TITLE: How badly can strong multiplicity one fail in the theory of automorphic representations? -QUESTION [22 upvotes]: Let $G$ be a connected reductive group over a global field $k$, and let $\pi=\otimes_w\pi_w$ and $\pi'=\otimes_w\pi'_w$ be two automorphic representations for $G$, where here of course $w$ is ranging over all the places of $k$. -Assume now that $\pi_w\cong\pi'_w$ for all but finitely many places $w$ of $k$. I think people say "$\pi$ and $\pi'$ are nearly equivalent". -If ($G=GL(n)$ and $\pi$ is cuspidal), or if ($G=GL(n)$ and $\pi$ and $\pi'$ occur discretely in $L^2$), then this would force $\pi_v\cong\pi'_v$ for all places $v$. But in general this "strong multiplicity one" phenomenon does not occur. Indeed even if $G=GL(2)$ we can have $\pi_v\not\cong\pi'_v$ for a non-zero finite set of $v$: if $\pi$ is 1-dimensional then $\pi'$ can be Steinberg at $v$, for example. For groups other than $GL(2)$ we can even have $\pi$ and $\pi'$ cuspidal, with $\pi_v\not\cong\pi'_v$ -- this even happens if $G=SL(2)$: "strong multiplicity one" can fail here. -So here's a vague question. We've established that $\pi$ and $\pi'$ nearly equivalent does not imply $\pi_v\cong\pi'_v$ for all $v$. But can we say anything about the relationship between $\pi_v$ and $\pi'_v$? -But I am not a fan of vague questions so here are some more precise ones, together with some guesses for answers. Say $\pi_w\cong\pi'_w$ for almost all $w$, but $\pi_v\not\cong\pi'_v$. -0) Do $\pi_v$ and $\pi'_v$ necesarily have the same central character? [this should be an easy warm-up. It's just the question of whether tori satisfy some sort of strong mult 1. I feel a bit lame not being able to figure this out :-/] -1) Are $\pi_v$ and $\pi'_v$ necessarily in the same Bernstein component? [my guess is "no"; I half-suspect that for $G=GSp(4)$ one can have $\pi_v$ supercuspidal and $\pi'_v$ not, but my source is "I think someone once told me this" and it would be nice to have a more concrete one]. -2) If $v$ is infinite, do $\pi_v$ and $\pi'_v$ have the same infinitesimal character? [My guess is "this is known for $GL(n)$, and might follow from a super-optimistic version of Langlands functoriality for general $G$ but perhaps I am being a bit too optimistic."] -3) If $v$ is infinite, are $\pi_v$ and $\pi'_v$ in the same local $L$-packet as defined by Langlands? [I have very little understanding of local $L$-packets at infinity and daren't hazard a guess.] -4) Back to general $v$. Should one expect that $\pi_v$ and $\pi'_v$ are in the same "packet" in some way? I write this in quotes because I don't know that I can give a definition of $L$-packet or $A$-packet in this generality. So here I daren't even have an opinion. -I'd be interested to know in anything that is proved or conjectured. - -REPLY [5 votes]: I confirm what "someone once told you" about question 1 (so now "two people -once told you" or perhaps "someone twice told you"). This phenomenon -($\pi_\nu$ supercuspidal, and $\pi'_\nu$ principal series, even unramified) -occurs for example when $\pi$ and $\pi'$ are the non-tempered endoscopic representation in the discrete (or even cuspidal) spectrum that some people like to deform, for $U(3)$ and $GSP(4)$ and their inner forms. There is an article by Rogawski, "The multiplicity formula for A-packets" in the book "the Zeta Function of Picard Modular Surfaces", where he describes in details such an example for each of the two inner forms of $U(3)$ attached to a quadratic imaginary field $E$. You have analog examples for $GSP_4$, where $\pi$ is a Saito-Kurokawa lift of a modular form.<|endoftext|> -TITLE: Some weird "system" of inequalities in nonnegative integers. -QUESTION [5 upvotes]: Suppose I have a bunch of nonnegative integers $(a_{ijkl})_{1 \leq i \leq j \leq k \leq l \leq 17}$ such that for all 17-tuples nonnegative integers $w_t$ (for $1 \leq t \leq 17$) we have that $$\min_{1\leq i\leq j\leq k \leq l\leq 17}(a_{ijkl} + w_i + w_j + w_k+w_l) \leq \frac{4}{17}\sum_{1 \leq t \leq 17} w_t.$$ What can we say about the $a_{ijkl}$? Are they bounded? What are the optimal bounds? I would like to get as much information on the $a_{ijkl}$ as possible. This problem comes from a question in number theory I'm trying to answer, but as you can see, this has a smell of linear programming, of which I know nothing at all. Also, maybe some things can be done with a computer, but I'm not good with computers... -(EDIT) If you suppose moreover that for all $t$, at least one of the inequalities $a_{ijkl} \geq n$, where $n$ is the number of indices from $(i,j,k,l)$ which are equal to $t$ ($n \geq 1$), is false, can we get something more? -(MOTIVATION) This comes from a semistability condition in geometric invariant theory. - -REPLY [7 votes]: Take $a_{iiii}=0$ for all $i=1,\dots,17$ and let the other $a$'s be arbitrary large. Then the inequality is satisfied. -Indeed, let $w_1, \dots, w_{17}$ be arbitrary nonnegative integers. Without loss of generality assume that -$$\min \{ w_1, \dots, w_{17} \} = w_1.$$ -Then -$$\min_{1\leq i\leq j\leq k\leq l\leq 17} (a_{ijkl} + w_i + w_j + w_k + w_l) \leq a_{1111} + 4w_1 = 4w_1 \leq \frac{4}{17} \sum_{t=1}^{17} w_t$$ -as required. -Hence, $a_{ijkl}$ are not bounded in general.<|endoftext|> -TITLE: Motivating the Casimir element -QUESTION [49 upvotes]: Weyl's theorem states that any finite-dimensional representation of a finite-dimensional semisimple Lie algebra is completely reducible. In my mind, the "natural" way to prove this result is by way of Lie groups. However, as a student, I first encountered Weyl's theorem in the textbook by Humphreys, in which he gives a purely algebraic proof. I remember finding this proof very mysterious, and in particular it seemed that Humphreys pulled the Casimir element out of a hat. Looking at the proof again now, I still find it somewhat mysterious, and if I had to present the proof in a graduate class, I don't think I would be able to motivate it very well. -How would you motivate this purely algebraic proof of Weyl's theorem? - -REPLY [8 votes]: This is an old question, but I was reviewing this stuff and wanted to elaborate on Daniel Litt's answer to explain why we should expect the Casimir element to give a proof of Weyl's theorem. -General arguments reduce the problem to showing that every short exact sequence of the form $0 \to V \to W \to k \to 0$ splits, where $V$ is a nontrivial irreducible rep and $k$ has a trivial action. The argument for Weyl's theorem goes as follows: the Casimir element of the trace form on $V$ acts as a nonzero scalar on $V$ but as $0$ on $k$, so we can find a splitting via an element in its kernel. -The reason we should expect this to be so is that the Casimir element is the Laplacian of the compact Lie group $G$. Every irreducible representation of $G$ embeds into $L^2(G)$ with the translation action via matrix coefficients. -On $L^2(G)$, the Casimir element really acts as the Laplacian, and its kernel, the harmonic functions, are exactly the constants since $G$ is compact. Since the kernel is one-dimensional, this shows that for every nontrivial irreducible representation, the Casimir element acts by a nonzero scalar, hence the argument for Weyl's theorem should work.<|endoftext|> -TITLE: How does the conjectural Langlands group fit into the Tannakian point of view? -QUESTION [10 upvotes]: I've read that one way to formulate the Langlands program is the following: -Let $\mathcal{L}_ {\mathbb{Q}}$ be the conjectural Langlands group. Then the category of semi-simple (continuous) representations of $\mathcal{L}_{\mathbb{Q}}$ that are algebraic(!) is equivalent to the category of motives over $\mathbb{Q}$ with coefficients in $\overline{\mathbb{Q}}$. -This is peculiar to me, since the category of motives is Tannakian, and so is equivalent to the category of (all!) representations of some affine group scheme. -How does one think of the condition that the representations must be algebraic? Does this mean that the Langlands group is not meant to be the motivic Galois group (the group guaranteed by Tannakian formalism applied to the category of motives, using some fiber functor)? Is there a way to reconcile the two approached in an insightful way? - -REPLY [11 votes]: The Langlands group is not meant to be the motivic Galois group; rather, it is larger (in Langlands's original formulation), or alternatively not an algebraic group, but a locally compact group which has some kind of underlying algebraic avatar (this is the more recent, indeed current, formulation, due to Kottwitz), so that one can speak of both continuous and algebraic representations. -A toy model to think about is the group $\mathbb C^{\times}$, and the difference between representations of $\mathbb C^{\times}$ just as a topological group, as opposed to $\mathbb C^{\times}$ thought of as a real algebraic group (i.e. thought of as the restriction of scalars of $\mathbb G_m$ from $\mathbb C$ to $\mathbb R$). -To see this example arising in real life, one can think about the difference between arbitrary and algebraic (type $A_0$ in Weil's terminology) Hecke characters for some number field $F$. (This is the theory for $1$-dimensional reps. of the Langlands group $\mathcal L_F$.) The relevant topological group is the idele class group of $L$, while the corresponding algebraic group is what Langlands calls the Serre group in his Ein Maerchen paper (and which is studied, but with different notation and terminology, in Serre's Abelian $\ell$-adic reps. book). -A harder example can be obtained by comparing the global Weil group over a number field to the Taniyama group over this field. This is discussed in Ein Maerchen, and in the book of Deligne, Milne, Ogus, and Shih, Hodge cycles, motives, and Shimura varieties. -(This is the theory obtained by combining $1$-dimensional reps. with finite image reps. of arbitrary dimension.) (See this answer for more on Weil groups.)<|endoftext|> -TITLE: Mathematical habits of thought and action which would be of use to non-mathematicians -QUESTION [106 upvotes]: Once again I come to MO for help with something I'm writing for the public. -Which habits of mathematicians -- aspects of the way we approach problems, the way we argue, the way we function as a community, the way we decide on our goals, whatever -- would you recommend that non-mathematicians adopt, at least in certain contexts? -In other words: if you can imagine a situation in which someone came to you for advice, and you said, "Look, I think you should be a little more like a mathematician about this and...." what would be the end of the sentence? - -REPLY [7 votes]: If you have an idea that you think or hope might be true, then don't just look for reasons to believe it. (It is amazing how strong and how wrong is this instinct.) Look for reasons that your idea might be wrong. If it is wrong, you will save yourself a lot of time and possible embarrassment. If it is right, you will learn a lot about why and have probably found some new reasons to believe it anyway.<|endoftext|> -TITLE: What is the analog of a topos in quantum logic? -QUESTION [15 upvotes]: If I'm studying classical mechanics, we might start by viewing propositions as true/false valued questions on points of phase space. -Then, if I'm interested in a proposition-oriented view of things, I might flip things around and ask what points of phase space correspond to propositions, and observe a Boolean algebra of subsets of phase space. Then I might think about things that aren't propositions (like "x-coordinate of the 7th particle") which are clearly functions on phase space and I start turning the mathematical crank, and eventually decide that I ought to interpret classical mechanics in terms of the topos of sheaves on phase space. (with the discrete topology -- or maybe I consider the usual topology with interesting consequences) -If I want to carry out this procedure quantum physics, I somewhat get the idea that I should associate propositions with projection operators in a C*-algebra -- but where to go from here is unclear. I've only managed to find material talking about the very special case where we consider working with orthogonal projections -- and nothing on what higher structure should be built on top of this. -Just because it sounds natural, I imagine the answer to my question ought to be something like "The category of Hilbert-space representations of the C*-algebra", but I'm having difficulty seeing how one would go about interpreting quantum mechanics in this category. -Can anyone elaborate on how the interpretation would go or point me at references or give the right answer to my question? -(I hope this is the right place -- I'm asking here since physics.stackexchange.com didn't appear to have people knowledgeable about category theory) - -REPLY [10 votes]: While the other two answers referred to very interesting connections of topos theory and quantum physics I think the following is going more into the direction the OP was imagining: In non-commutative topology one considers quantales, which are, roughly, an axiomatization of what you get when you replace open sets with projection operators - in particular intersection of open sets becomes composition of operators and does no longer have to be commutative. Here a few approaches to this idea are listed. One short definition is: A quantale is a monoid object in the monoidal category of lattices. -One can define sheaves over quantales (e.g. as done by Miraglia and Solitro in this article, or by Mulvey and Nawaz in this nicely written paper) and the categories of such sheaves would be the analogon to the notion of Grothendieck topos (but: to my knowledge no Giraud type characterization of such categories is known, let alone an elementary one). One can interpret logic in such categories, as e.g. done by Coniglio here for the Miraglia/Solitro setting. -In some ways such categories of sheaves over quantales connect back to actual topos theory as for example seen in the last corollary of the above Mulvey/Nawaz paper and more impressively in this article by Pedro Resende. - -Edit: After having looked into the article pointed out by Urs Schreiber I would like to add a comment on what the authors see as drawbacks of quantum logic. They write that, on the logical side, quantum logic is not distributive and thus "difficult to interpret as a logical structure", that no satisfactory implication operator has been found "so that there is no deductive system in quantum logic" and that no satisfactory first order quantum logic has been found. -I cannot judge what would be satisfactory, but Coniglio's interpretation linked to above has an implication operator and first order quantifiers. In any case, deduction systems can be built without implication operators being present inside the language, one just has to devise rules which say when one can infer a formula from a set of hypotheses (this is done in many nonclassical logics). The non-distributivity presents undeniable and annoying technical difficulties, the difficulty to interpret a non-distributive system "as a logical structure" on the other hand may be a matter of reading it in an appropriate way, see the next point. -The authors also see a physical drawback: They see the law of excluded middle $x \vee x^\bot = 1$, valid in quantum logic, as not reflecting the probabilistic spirit of quantum physics, because there it is not the case that either a proposition $x$ or its complement $x^\bot$ are true - they both may have intermediate "degrees of truth". I think an answer to this is that the interpretation of $\vee$ should not be that one of its arguments is true, but rather that its arguments together span the space of all possibilities. Non-distributivity then makes some sense also. -In a similar vein many things in intuitionistic logic make more sense if one interprets it as talking not about the truth of assertions but about whether they are known or provable (and an observer-centered perspective on quantum physics seems very appropriate). So the choice of an underlying logic is the choice of a point of view. Maybe one can see the two topos approaches and the quantale approach as talking about quantum physics from different perspectives - one could even consider creating a bigger formal language containing the connectives from both interpretations and allowing to relate the several points of view (giving a formal semantics for this language might be quite a challenge, though). -Disclaimers: 1. My background in the things discussed above lies in categorical and non-classical logic - my knowledge of quantum physics is quite superficial. 2. Although I jumped to the defense of quantales here, I really like both topos approaches...<|endoftext|> -TITLE: Examples of Non-algebraic Fibered Knots? -QUESTION [7 upvotes]: I am currently reading a monograph by Jose Seade, " On the topology of isolated singularities in analytic spaces". - I have following questions but before asking questions I recall the definition of algebraic knots/ links -Definition : Let $ f : (\mathbb{C}^2,0) \rightarrow (\mathbb{C},0)$ be a holomorphic function with an isolated critical point at $0$. Then for sufficiently small $\epsilon >0$ we have $ K= \mathbb{S}^3_{\epsilon} \cap f^{-1}(0)$. The knot/link $K$ is called algebraic knot/link. -By Milnor's fibration theorem for complex singularities, we get an open book decomposition $(\mathbb{S}^3,K) $ . - -Questions : - -Here we think of knot as an embedded copy of $\mathbb{S}^1$ in $\mathbb{S}^3 $. Suppose we have an algebraic knot $K$. Is there a fibered knot $K'$ in the isotopy class of $K$ which is not algebraic? -In general, how to detect whether a fibered knot $K$ is algebraic or not? - -REPLY [8 votes]: The algebraic knots are iterated cablings of the unknot, so have simplicial volume $=0$. -Thus, "most" fibered knots will not be algebraic. -There is an algorithm to find a minimal genus Seifert surface of a knot. -Moreover, one may determine if it is a fiber of a fibration, and compute -the conjugacy class in the mapping class group of the monodromy of the fiber. -There are various approaches, e.g. using sutured manifold theory. One may -determine whether a mapping class is completely reducible (having no -pseudo-Anosov components), and therefore -whether the mapping torus has simplicial volume $=0$. In fact, you -may use this to determine which iterated cabling of the unknot you have -algorithmically. Then you can apply the criterion of Eisenbud and Neumann -to see if it is algebraic. -This answer is interpreting "detect" as "does -there exist an algorithm", which I'm not sure if this is what you want. -I think there may be ways of doing this now using Heegaard Floer homology, -although I'm not certain if one can detect simplicial volume $=0$ using -that invariant. -I should remark also that the term "algebraic knot" can have other meanings -in topology.<|endoftext|> -TITLE: Unique limits of sequences plus what implies Hausdorff? -QUESTION [18 upvotes]: It is known that there are non-Hausdorff spaces which admit unique limits for all convergent sequence (see here) and it is also known that unique limits for nets implies Hausdorff. -What I am wondering is, if there is a (somehow weak) condition which one should add to "unique limits of sequences" to obtain a Hausdorff space. Would, for example, some countability help? -Somehow in the same direction: What is the central property which is needed for a space such that it can be non-Hausdorff but has unique sequence limits? Is there a whole class of non-Hausdorff spaces which admit unique limits for convergent sequence? - -REPLY [12 votes]: Here is an answer to Dirk's last question, "Is there a class of non_Hausdorff spaces in which convergent sequences have unique limits?" -Yes. The so called KC-spaces or maximal compact spaces. These are spaces such that every compact subspace is closed. -(The 1967 Monthly article of Wilansky, "Between $\mathrm T_1$ and $\mathrm T_2$" (MSN), subsumes, references, or implies all of the following.) -In a KC-space, convergent sequences have unique limits. -(Suppose $x_n\to x$ in the KC space $X$. The set $\{x,x_1,x_2,\dotsc\}$ is compact and hence closed. Thus, if $y$ is not in the set $\{x,x_1,x_2,\dotsc\}$, then the open set $X \setminus \{x,x_1,x_2,\dotsc\}$ shows it is false that $x_n\to y$. Thus, if $x_n\to y$, then $y=x$ or $y=x_n$ for some $n$. If $y=x_n$ for infinitely many indices $n$ then $y=x$ (since every KC space is $\mathrm T_1$ (since singletons are compact) and since constant sequences have unique limits in a $\mathrm T_1$ space). If $y=x_n$ for finitely many indices $n$ then (deleting $y$ from the sequence $x_1,x_2,\dotsc$) we are left with a subsequence $z_n\to x$, the knowledge that $y$ is not equal to any $z_n$, and the knowledge that $y$ is in the set $\{x,z_1,z_2,\dotsc\}$, and we conclude that $y=x$). -To exhibit a large class of non-Hausdorff KC spaces, let $X$ be a non-locally-compact metric space (for example, the rationals) and let $Y=X \cup \{y\}$ denote the Alexandroff compactification of $X$ (i.e., $V$ is open in $Y$ if $V$ is open in $X$ or if $Y\setminus V$ is a compact subspace of $X$). -The space $Y$ is a KC space, but $Y$ is not Hausdorff.<|endoftext|> -TITLE: Counting the (additive) decompositions of a quadratic, symmetric, empty-diagonal and constant-line matrix into permutation matrices -QUESTION [6 upvotes]: Dear community, -I have the following combinatorial question which I will explain in short first and then with some more detail. At the end you will find a very simple example. -Short version -Le $A \in \mathbb{N}_0^{n \times n}$ be a symmetric matrix with zeros on the diagonal, whose row- and column-sums add up to some fixed, positive integer $c$. In how many ways can we write $A$ as a sum of permutation matrices, ignoring the order of summation? -Detailed version -Suppose you have a quadratic matrix $A$ of dimension $n$ with non-negative integral entries whose row- and column-sums add up to some common number $c$. Then it is known that $A$ can be written as a sum of $c$ permutation matrices, i.e. we have that -$$A = P_{\sigma_1} + P_{\sigma_2} + \ldots + P_{\sigma_c},$$ -where each $P_{\sigma_i}$ is a permutation matrix representing a -permutation $\sigma_i \in S_n$, $S_n$ being the symmetric group of -order $n$. Let's call the set $\{ \sigma_1, \sigma_2, \ldots, \sigma_n -\}$ a decomposition for $A$. -If we add the property that the diagonal of $A$ vanishes -(i.e. contains only zeros), the permutation matrices $P_{\sigma_i}$ of -any decomposition as above will have vanishing diagonals, too, -i.e. the corresponding permutations $\sigma_i$ will have no fixed -points. -If we add another property to $A$, namely that it is symmetric, we get -decompositions with even more structure. Either all matrices of a -decomposition are symmetric themselves (which in this framework is the -case, if and only if the cycles of the corresponding permutations are -all of length two), or the non-symmetric matrices add up to something -which is symmetric. This is for example the case if for a given -permutation $\sigma_i$, the permutation $\sigma_i^{-1}$ (inverse in -$S_n$, i.e. with respect to composition) is also in the -decomposition, because $P_{\sigma_i^{-1}} = P_{\sigma_i}^T$ and -$P_{\sigma_i} + P_{\sigma_i}^T$ is symmetric, but this condition is not necessary. -My question is the following: How many decompositions are there in total? -Example -Let me give you a very simple example for the situation in the case -$n=3$. -If -$$A = -\begin{pmatrix} - 0 &1 &1 \\ - 1 &0 &1 \\ - 1 &1 &0 -\end{pmatrix} -$$ -the only possible way to write this as a sum of permutation matrices -(up to order of summands) is -$$ A = -\begin{pmatrix} - 0 &0 &1 \\ - 1 &0 &0 \\ - 0 &1 &0 -\end{pmatrix} -+ -\begin{pmatrix} - 0 &1 &0 \\ - 0 &0 &1 \\ - 1 &0 &0 -\end{pmatrix}, -$$ -and the corresponding decomposition is $\{ (1 3 2), (2 3 1) \}$. -Thanks for your help, -Simon - -REPLY [8 votes]: In general the number of decompositions depends on the structure of the matrix, not just on its size and row sum. This is even so in the case of 0-1 matrices, where the question is equivalent to 1-factorization of regular bipartite graphs. Even very simple-looking cases are difficult, for example if the matrix is full of ones the decompositions are the Latin squares (only counted exactly up to order 11). A lower bound for the 0-1 case follows from Schrijver's lower bound on the permanent, and an upper bound follows from the van der Waerden upper bound on the permanent. These bounds are far apart. Sharper upper bounds exist for the 0-1 case with row sums not too large. Symmetry and zero diagonal don't make a great difference as far as I know. There are some asymptotic results. In the symmetric case there are asymptotic results for the number of decompositions into symmetric permutation matrices (which corresponds to 1-factorisation of regular graphs). Let us know which aspects interest you.<|endoftext|> -TITLE: Simplices in convex polytopes -QUESTION [5 upvotes]: This question is a direct generalization of: -Counting the (additive) decompositions of a quadratic, symmetric, empty-diagonal and constant-line matrix into permutation matrices -Given a convex polytope $P \in \mathbb{R}^d,$ and a point $p\in P,$ how many simplices $S$ are there such that the $S$ is the convex hull of some $d+1$ vertices of $P$ and such that $p \in S?$ -I assume that this is $\#P$-complete... - -REPLY [5 votes]: There are two ways to state this question, and the answers differ. -1) Let the dimension $d$ is fixed, and the input is via $n$ vertices of the polytopes. In this case the total number of possible simplices is polynomial, and so is the counting problem. -2) If the dimension $d$ is arbitrary, everything falls apart. Think of a simplex with vertices $O=(0,\ldots,0)$ and $(0,\ldots,a,\ldots,0)$, where $a\in$ {$a_1,\ldots,a_m$}. Here $n=dm+1$. Let $z=(c,\ldots,c)$ and consider all (closed) simplices which contain $z$. Check that #simplices containing O is a variation on the #knapsack problem, several versions of which are known to be #P-complete. I omit the details.<|endoftext|> -TITLE: Where can I find an explicit description of the pseudocolimit of a small pseudofunctor to Cat? -QUESTION [5 upvotes]: Given a functor from a small category to $Set$, we can describe the colimit set as a quotient of the disjoint union of image sets by an equivalence relation arising from morphisms in the source category (as seen in Wikipedia, or Kashiwara-Schapira's Categories and Sheaves). I am looking for an analogous description on the 2-categorical level: - -Is there an explicit description of "the" pseudo-colimit of a pseudo-functor $F$ from a small 2-category to $Cat$, and is it in the literature? In particular, who should I reference for the fact that such pseudo-colimits exist? - -Here, by pseudo-colimit, I mean "pseudo-bi-colimit" in the sense of Borceaux's Handbook of Categorical Algebra, i.e., a pair $(L, \pi)$, where $L$ is a small category, and $\pi: F \to \Delta(L)$ is a pseudo-natural transformation to the constant pseudo-functor, such that the functor $Fun(L,B) \to PsNat(F, \Delta(B))$ given by $f \mapsto \Delta(f) \odot \pi$ is an equivalence. -Presumably, this should be a small category whose set of objects is something like a quotient of a disjoint union of object sets, but when I tried to work it out by myself, all of the arrows gave me a headache. I also looked at several sources, e.g., Kelly's Elementary Observations on 2-categorical limits, the elephant, and Borceaux, without success (but I may have missed something). - -REPLY [8 votes]: An answer can more or less be extracted from Kelly's Elementary Observations on 2-categorical limits, at least if you already know that it's there. (-: -First, as Kelly notes in section 6, it would suffice to construct what we may call strict pseudo-colimits, that is pseudo-colimits in your sense for which the functor $f\mapsto \Delta(f)\odot \pi$ is an isomorphism (since any isomorphism of categories is a fortiori also an equivalence). -Second, in Proposition 5.1, Kelly shows how to construct strict pseudo-limits (which he calls merely "pseudo-limits") in any 2-category out of (strict) products, cotensor products, iso-inserters, and iso-equifiers. This is a 2-categorical version of the construction of limits out of products and equalizers. (In the correction to the paper Fibrations in bicategories, Street gives an equivalent construction of non-strict pseudo-limits in terms of products, cotensor products, and descent objects.) -Dually, of course, pseudo-colimits may of course be constructed from coproducts, tensor products, iso-coinserters, and iso-coequifiers. Coproducts and tensor products in $Cat$ are easy — they are disjoint unions and cartesian products — so it suffices to construct iso-coinserters and iso-coequifiers. -At this level, though, I think we do have to descend into writing down strings of composites of arrows modulo equivalence relations. Iso-coinserters and iso-coequifiers, being both particular Cat-weighted colimits, can be constructed in terms of cartesian products in Cat and ordinary unweighted colimits, so it would suffice to understand the latter. We know that Cat is cocomplete (as a 1-category) since it is the models of an essentially algebraic theory, so this settles the existence question. But an explicit description of colimits is going to be kind of messy.<|endoftext|> -TITLE: "Almost all" quantifier -QUESTION [7 upvotes]: Suppose I enlarge the first-order logic with an "almost all" quantifier, let's denote it by G, ie.: -$G_x P(x) \iff$ for all but finitely many x, P(x) -Syntax for G is the same as for other quantifiers. -Suppose I am working over the first-order theory of natural numbers. For every sentence $T$ using $G$, does there exist a sentence $S$ over "standard" first-order logic, such that $S \iff T$ in every countable model of natural numbers? - -REPLY [10 votes]: No. For example, Robinson’s Q + $\forall x\,G_y\,x -TITLE: Countable unions and the axiom of countable choice -QUESTION [19 upvotes]: Let us denote by ACC the axiom of countable choice, namely the assertion that the product of countably many non-empty sets is non-empty, and denote by UCC the assertion that a countable union of countable sets is countable. -UCC is a simple theorem of ZF+ACC. -Proof Suppose for every $i\in\omega$ we have $X_i$ a countable set, and $X_i\cap X_j=\varnothing$ for $j\neq i$. -Since $X_i$ is countable $O_i=\{f\colon X_i\to\omega\mid f\ \text{ injective}\}$ is non-empty. We can choose $f_i\in O_i$ by the axiom of countable choice, and define: $$F\colon\bigcup X_i\to \omega\times\omega\colon\qquad x\mapsto\langle n,f_n(x)\rangle$$ -Where $n$ is the unique $n\in\omega$ such that $x\in X_n$. -The Cantor pairing function shows that $\omega\times\omega$ is countable and we are done. - -Is the opposite assertion is true, namely ZF+UCC implies ACC? If the answer is negative, does that imply at least some other weaker form of choice? - -As noted by Emil Jeřábek below, UCC implies the axiom of countable choice for countable sets (the latter abbreviated as CCF). -Digging through the paper mentioned by godelian in the comments, I reached [1] in which Howard constructs a model of ZFA in which CCF holds and UCC does not, and by the transfer theorem of Pincus constructs this over ZF. Therefore we have: $$\text{ACC}\Rightarrow\text{UCC}\Rightarrow\text{CCF}$$ The first implication is irreversible in ZFA, by the comment of godelian, and the second irreversible in ZF by [1]. Both papers are two decades old, is there any known progress? - -Bibliography: - -Howard, P. The axiom of choice for countable collections of countable sets does not imply the countable union theorem. Notre Dame J. Formal Logic Volume 33, Number 2 (1992), 236-243. - -REPLY [7 votes]: The implication $ACC \implies UCC$ is irreversible in $ZF$. This follows again from the transfer theorem of Pincus: - -$UCC$ is an injectively boundable statement, see note 103 in "Consequences of the axiom of choice" by Howard & Rubin. Right after the theorem in pp. 285 and its corollary, there are examples of statements of this kind, one of which is form 31 (which is precisely $UCC$). The fact that this is the case follows in turn from the application of lemma 3.5 in Howard, P.-Solski, J.: "The strength of the $\Delta$-system lemma", Notre Dame J. Formal Logic vol 34, pp. 100-106 - 1993 - -It is known that $¬ACC$ is boundable, and hence injectively boundable. - - -Then we can apply the transfer theorem of Pincus to the conjunction $UCC \wedge ¬ACC$ and we are done. -SECOND PROOF: Browsing through the "Consequences..." book I've just found another less direct proof of the same fact. I'm adding it here to avoid the interested person from looking it up for himself (especially since the AC website is not working these days). It involves form 9, known as $W_{\aleph_0}$: a set is finite if and only if it is Dedekind finite. Now, $ACC \implies W_{\aleph_0}$ is provable in $ZF$ (in fact this was already proved by Dedekind), while $UCC$ does not imply $W_{\aleph_0}$. The latter follows from the fact that in the basic Fraenkel model, $\mathcal{N}1$, $UCC$ is valid while $W_{\aleph_0}$ is not, and such a result is transferable by considerations of Pincus that can be found at Pincus, D.: "Zermelo-Fraenkel consistency results by Fraenkel Mostowski methods", J. of Symbolic Logic vol 37, pp. 721-743 - 1972 (doi:10.2307/2272420)<|endoftext|> -TITLE: Complex Eigenvalues of Directed Graphs -QUESTION [5 upvotes]: I have been computing eigenvalues of adjacency matrices for several directed (not necessarily strongly connected) graphs and one remarkable property seemed to hold (each graph that I have examined contained at least one cycle, but this need not to be a necessary condition): -"If $\lambda$ is an eigenvalue of an adjacency matrix $A$, then it can be expressed as -$r\cdot z$, where $r$ is some real number and z is $n$-th root of unity for some $n \in \mathbb{N}$. Moreover, if some eigenvalue $\lambda$ can be expressed in this form, then for all $n$-th roots of unity $z'$, $r \cdot z'$ is an eigenvalue of $A$. As a consequence, if $\lambda$ is eigenvalue of $A$, then also $|\lambda|$ (absolute value) is an eigenvalue of $A$." -Since I am not an expert in the field of spectral graph theory, I was unable to proof or disproof the property by myself. Is it known if this property or any similar property holds? Has anyone proved any similar property (it is possible, that the property does not hold exactly in the form I had written it down, but it may hold for instance for some special class of digraphs)? Any reference would be welcomed. -Thank you in advance. - -REPLY [9 votes]: Let $D$ be the Paley tournament on seven vertices. Its vertices are the integers mod seven -and there is an arc from $i$ to $j$ is $j-i$ is a non-zero square mod seven. The characteristic polynomial of the adjacency matrix is $(x-3)(x^2+x+2)^3$. The only real eigenvalue is 3, the remaining eigenvalues are equal to $(-1\pm\sqrt{-7})/2$, with absolute value $\sqrt{2}$. This can be generalized to Paley tournaments on $q$ vertices, where -$q\equiv3$ mod 4. -According to my computations (in sage) random directed graphs with seven vertices and arc probability 0.5 do well as counterexamples too.<|endoftext|> -TITLE: Unmathematical habits of thought and action which would be of use to mathematicians -QUESTION [21 upvotes]: In Question 74707, we ask what mathematical habits of thoughts are useful in other areas. It seems only fair to ask also what we can learn from them. It is also fair to ask what they should not learn from us. -The first question is - -What habit of thoughts in other areas - can be of use in mathematics. - -Here are a few suggestions to the first question: - -In many areas the quality of exposition is very important. -In some areas simplicity is considered as an advantage. In mathematics to some extent difficulty is a criterion for quality. -Other areas give more weight to heuristic and non rigourous arguments compared to pure mathematics. -In other areas there is much heavier use of computers. -In some areas discussions and debates are basic part of the academic discipline. This is not the case in mathematics. - -The second question is: - -What habit of thoughts in mathematics - should be avoided (even by mathematicians) outside mathematics. - -Here are a few suggestions (to the second question) for starters. - -Mathematicians (as a rule) avoid ambiguity. Nonmathematicians recognize the value (in appropriate circumstances) of ambiguity. -Mathematicians don't care what anybody thinks. We know when our assertions are facts, because we can prove them, and we know how important they are; we don't need anyone's opinion on that. There is a danger that this attitude carries over to our everyday lives. Nonmathematicians recognize the value (in appropriate circumstances) of the opinions of others. -Mathematicians, with all due respect to Godel, think simple declarative sentences are either true or false. Nonmathematicians are better able to deal with shades of gray. -Those of us who teach are constantly judging the mathematical abilities of others. If we are not careful, we start to judge the worth of others by their mathematical abilities. Nonmathematicians know that some of the best people alive can't add fractions. -Mathematicians (and theoretical physicists) consider a spherical cow. Nonmathematicians understand that conclusions based on unrealistically oversimplified models are untenable. - -REPLY [5 votes]: A long quote, from which one can extrapolate trivially a tentative answer. -"[T]echnical treatises in science do not generally receive such a license for explicitly personal expression. I believe that this convention in technical writing has been both harmful and more than a bit deceptive. Science, done perforce by ordinary human beings expressing ordinary motives and foibles of the species, cannot be grasped as an enterprise without some acknowledgment of personal dimensions in preferences and decisions – for, -although a final product may display logical coherence, other decisions, leading to other formulations of equally tight structure, could have been followed, and we do need to know why an author proceeded as he did if we wish to achieve our best understanding of his accomplishments, including the general worth of his conclusions. -Logical coherence may remain formally separate from ontogenetic construction, or psychological origin, but a full understanding of form does require some insight into intention and working procedure. Perhaps some presentations of broad theories in the history of science – Newton's Principia comes immediately to mind – remain virtually free of personal statement (sometimes making them, as in this case, virtually unreadable thereby). But most comprehensive works, in all fields of science, from Galileo's Dialogo to Darwin's Origin, gain stylistic strength and logical power by their suffusion with honorable statements about authorial intents, purposes, prejudices and preferences." -SJ Gould The structure of evolutionary theory p. 34.<|endoftext|> -TITLE: Is it true that all sphere bundles are boundaries of disk bundles? -QUESTION [16 upvotes]: Let $E$ be the total space of the sphere bundle $S^k\to E\to M$, is it true that there exists a disk bundle $D^{k+1}\to N\to M$ such that $E=\partial N$? (where $D^{k+1}$ is the unit disk in $\mathbb R^n$) - -REPLY [6 votes]: The question of whether or not a smooth sphere bundle fibrewise extends to a smooth disc bundle over a space $X$ boils down to whether or not the classifying map -$$ X \to BDiff(S^n) $$ -lifts up -$$ BDiff(D^{n+1}) \to BDiff(S^n)$$ -Where $Diff(D^{n+1})$ is the group of diffeomorphisms of the disc. -The map $Diff(D^{n+1}) \to Diff(S^n)$ splits as a product: -$$O_{n+1} \times PDiff(D^{n}) \to O_{n+1} \times Diff(D^n rel \partial)$$ -where $PDiff(D^{n})$ is the group of pseudo-isotopy diffeomorphisms of $D^{n}$. These are diffeomorphisms of $D^{n} \times [0,1]$ that are the identity on $(D^{n} \times \{0\}) \cup (S^{n-1} \times [0,1])$. -There is a fibre-bundle: -$$Diff(D^{n+1} rel \partial) \to PDiff(D^{n}) \to Diff(D^n rel \partial)$$ -so basically this is asking whether or not this bundle has a section. I think it can't have a section, in particular the map $\pi_1 Diff(D^n rel \partial) \to \pi_0 Diff(D^{n+1} rel \partial)$ is epic by Cerf's Pseudoisotopy theorem. -Okay, so this is now an answer. So this is saying that there are sphere bundles over $S^2$ which do not extend to smooth disc bundles over $S^2$.<|endoftext|> -TITLE: Finite field analogue of representations in same packet have equal central character -QUESTION [6 upvotes]: In Kevin Buzzard's recent question, a warm up question was: if two automorphic representations are nearly equivalent, then are the central characters of their local components equal? -Working my way up to local and later global automorphic representations, I am currently studying the situation over finite fields. - -What is the analogue of Buzzard's question in the finite field case? -Is it true? - - -Here are some of my own thoughts on this. -If $$(T_1,\theta_1)\sim (T_2,\theta_2)$$ are two geometrically conjugate pairs of torus+character, then I think the characters have to agree on the center of $G^F$ (elements in the center are norms [?], so the geometric conjugacy equation shows equality). Using 7.2 in [DL] we see that the value of $R_T^\theta$ on the center is $\theta$ (maybe up to sign), i.e. -$$\frac{R_{T_1}^{\theta_1}(z)}{R_{T_1}^{\theta_1}(1)} = \frac{R_{T_2}^{\theta_2}(z)}{R_{T_2}^{\theta_2}(1)}=\theta(z)$$ -for $z\in Z(G^F)$. -This might be considered an analogue as requested, but a naive one at that. A deeper analogue should consider the irreducible representations of $G^F$, and not the Deligne-Lusztig virtual characters, which can be reducible. -Consider the two cross sections $\rho_x$, $\rho_x'$ from the set of geometric conjugacy classes to irreducible representations (10.7.1/2 in [DL]): -$$\rho_x=\sum_{[(T,\theta)]=x} \frac{(-1)^{\sigma(G)-\sigma(T)}}{\langle R_T^\theta,R_T^\theta\rangle}R_T^\theta$$ -$$\rho_x'=(-1)^{\sigma(G)-\delta_x} \sum_{[(T,\theta)]=x} \frac{1}{\langle R_T^\theta,R_T^\theta\rangle}R_T^\theta.$$ - -Do these have the same central character? - -If the $\theta$'s are trivial on the center, then the answer is yes. Computations that I have done before show that this is the case for the $\theta$'s that are not in general position in $\mathrm{Sp}_4$. -We can divide in an obvious way the $[(T,\theta)]\in x$ into two sets, of "positive" and "negative", such that $\rho_x'$ is a sum and $\rho_x$ is a difference. We see that for the two representations to have equal central character, either the sum of "negative" terms, or "positive" terms (depending on $\delta_x$), must be zero on the center (minus the identity). -Note that $\rho_x$ appears in the Gelfand-Graev representation of $G^F$, whose character restricted to the center (minus the identity) is zero, so this supports in spirit the paragraph above. I'm not sure if it can be extended from spirit to an actual proof. -[DL] - this is, of course, the original Deligne-Lusztig paper from 1976. - -REPLY [5 votes]: This is quite an old question but I believe the answer to your question is given in Lemma 2.2 of Malle's paper "Height 0 characters of finite groups of Lie type" (2007) which is freely available online here. -His lemma states that any two characters in the same (geometric) Lusztig series have the same central character. It follows very simply from the fact you gave for Deligne-Lusztig characters. This is because the characteristic functions of semisimple elements are explicit uniform functions, in the sense that they are linear combinations of Deligne-Lusztig characters.<|endoftext|> -TITLE: Formal group law of unoriented cobordism -QUESTION [14 upvotes]: It is well known that the formal group law $F_U$ of complex cobordism, expressing the Euler class of a tensor product of complex line bundles, is universal. -Also, the formal group law $F_O$ of unoriented cobordism, expressing the Euler class of a tensor product of real line bundles, is universal among formal group laws in characteristic 2 with the property that $F(X,X)=0$. -There is a nice description of $F_U$ in terms of manifold generators, due to Buchstaber: -$$ -F_U(X,Y) = \frac{\sum_{i,j\geq 0} [H_{ij}]\;X^iY^j}{\left(\sum_{r\geq 0}[\mathbb{C}P^r] X^r\right) \left(\sum_{s\geq 0}[\mathbb{C}P^s] Y^s\right)} -$$ -where the $H_{ij}$ are Milnor hypersurfaces. Here I am quoting this page. -Is there a similar description of $F_O(X,Y)$? - -REPLY [8 votes]: I'm fairly sure you just get the same formula, with $\mathbb{C}P^k$ replaced by $\mathbb{R}P^k$, and $H_{ij}$ replaced by the corresponding real hypersurface in $\mathbb{R}P^i\times\mathbb{R}P^j$. The proof of the equivalent formula -$$ \left(\sum [\mathbb{R}P^r]\;X^r\right) - \left(\sum [\mathbb{R}P^s]\;Y^s\right) - F_O(X,Y) = - \sum H_{ij} X^i Y^j -$$ -is quite direct and geometric. (I might come back and write more tomorrow.)<|endoftext|> -TITLE: Finding a hyperplane that splits a convex polytope evenly -QUESTION [5 upvotes]: Say we have a convex polytope in standard form: -\begin{equation*} -\begin{array}{rl} -\mathbf{A}\mathbf{x} = \mathbf{b} \\\\ -\mathbf{x} \ge 0 -\end{array} -\end{equation*} -Are there any known methods for finding a hyperplane $\mathbf{d} \mathbf{x} +d_0= 0$ that splits the polyhedron in a way that the number of vertices on each side of the hyperplane is approximately the same? (i.e. a hyperplane that minimizes the absolute difference of vertex cardinalities on the two sides of the split). -Also, are there any known results regarding the computational complexity of this problem? -Addendum: Restricting the types of cuts: -Here is a variation of the original problem with the hope that it is easier to solve than the original one: -Is there a way to efficiently compute or estimate for which coordinate $i$ a hyperplane of the form $d_ix_i + d_0 = 0$ would yield the lowest absolute difference of vertex cardinalities on both sides of the split? By efficient I mean anything more efficient than the exhaustive enumeration of vertex cardinalities for all such possible splits. -Note: -I first asked this question in CSTheory.stackexchange.com last week. Since the question has not seen any significant progress since then, I thought I could try here. - -REPLY [2 votes]: This is not an answer, it's a warning of a dead end. -I considered using a Markov Chain to randomly sample the vertices. Just start at an arbitrary vertex, and at each time step, pick an edge-connected neighbor of the current vertex and step to that vertex. The distribution of points hit by this chain converges to the uniform distribution on the vertices. So we could use it to take a random sample of vertices and pick a splitting plane for that sample. This would be an approximation to the real splitting plane. Unfortunately I was able to construct a polytope where the convergence takes time exponential in the number of constraints. So as long as you're taking exponential time, you might as well just enumerate all the vertices. \ No newline at end of file