User s. carnahan - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T04:28:59Z http://mathoverflow.net/feeds/user/121 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130564/references-for-period-matrix-of-abelian-variety/130608#130608 Answer by S. Carnahan for References for period matrix of abelian variety S. Carnahan 2013-05-14T17:06:25Z 2013-05-14T17:06:25Z <p>You won't get $A$ as a quotient of a vector space in general without some kind of strange transcendentality. For example, if $V$ is defined over $\mathbb{F}_p$, then any $S$-valued point of $V$ is $p$-torsion for any test object $S$. In particular, you can't possibly obtain any of the prime-to-$p$ torsion in $A$ without resorting to extraordinary means (whose existence I doubt).</p> <p>There are some interesting treatments for certain complete topological fields other than $\mathbb{C}$, but "most" fields do not fit this description.</p> http://mathoverflow.net/questions/129935/what-happens-to-virasoro-at-c25/130539#130539 Answer by S. Carnahan for What happens to Virasoro at c=25? S. Carnahan 2013-05-14T03:13:16Z 2013-05-14T03:13:16Z <p>I have an incomplete understanding of this, but I will try to say what I know.</p> <p>For each $c \in \mathbb{C}$, we define $Verma_c$ to be the category whose objects are Verma modules $V_{h,c}$ of central charge $c$ and highest weight $h$ for some $h \in \mathbb{C}$, and whose maps are Virasoro-module maps. Feigin and Fuks worked out the morphisms in all of these categories (they are embeddings), and they found that the categories $Verma_c$ and $Verma_{26-c}$ are antiequivalent. In particular, when $V_{h,c}$ has an embedded Verma submodule $V_{h+x,c}$, the corresponding module $V_{-1-h,26-c}$ embeds into $V_{-1-h-x,26-c}$.</p> <p>When you consider representations of Virasoro with central charge $0 &lt; c &lt; 1$ and positive $h$, the corresponding objects will have central charge $25 &lt; 26-c &lt; 26$, and highest weight that is negative. In particular, unitary representations don't seem to make an appearance anywhere in the complementary picture.</p> <p>According to <a href="http://mathoverflow.net/questions/6923/complexes-of-representations-with-complementary-central-charges" rel="nofollow">this MathOverflow question</a>, Positselski has developed the antiequivalence into a derived category statement, where modules are resolved by complexes of Vermas. However, I have been unable to extract an explicit theorem about Virasoro from <a href="http://arxiv.org/abs/0708.3398" rel="nofollow">his monograph</a>.</p> http://mathoverflow.net/questions/130333/a-possible-consequence-of-dirichlets-theorem-about-primes-in-arithmetic-progress/130346#130346 Answer by S. Carnahan for A possible consequence of Dirichlet's theorem about primes in arithmetic progression S. Carnahan 2013-05-11T15:14:05Z 2013-05-11T15:14:05Z <p>You seem to be using certain variable inputs in two different ways. First, $r$ is assumed to be less than $n-3$. Then, you define a notion of potential typical primality radius of $r$, which depends only on remainders of division by primes less than $\sqrt{2n-3}$. It is not clear in this definition that you are assuming $r$ is bounded by $n-3$. Finally, you discuss asymptotics of potential typical primality radii $x$ (or at least I assume that is the meaning of the symbol $\sim$ in the expression $\mathcal{N}_n(x) \sim k_n x$), and this requires $x$ to increase without bound.</p> <p>If we assume you meant to ignore the bound, then the asymptotic estimate is correct, but it doesn't depend on Dirichlet's theorem. Instead, it only depends on the Chinese remainder theorem, because you are making a statement about counting residue classes modulo the product of all primes less than $\sqrt{2n-3}$. Also, it's not clear how much benefit you will derive from this, since you haven't eliminated the possibility that $k_n = 0$.</p> http://mathoverflow.net/questions/3237/japanese-chinese-for-mathematicians/130315#130315 Answer by S. Carnahan for japanese/chinese for mathematicians? S. Carnahan 2013-05-11T01:23:35Z 2013-05-11T01:23:35Z <p>I am at a Japanese university, and I teach mathematics in Japanese, but I don't consider myself an expert yet. You asked for recommendations for technical Japanese textbooks, but I don't really have anything to offer, since I don't study from language texts. However, I can give you some tips from my own experience.</p> <p>If you want to progress reasonably quickly through a paper, it will help to have a basic understanding of grammar and the alphabet(s). This can come from most general-purpose introductions (e.g., software, textbooks, websites, classes).</p> <p>For words that involve difficult characters, I use an electronic dictionary to look up characters by their pieces. My particular tool of choice is the free "imiwa" (formerly "kotoba!") app on iOS, which has a "multi-radical" option. It also has a decent amount of technical vocabulary, but is not comprehensive. Since I have it on my phone, it is very convenient when I am away from a library.</p> <p>If an article is online and I can't understand a passage, I copy/paste into an online translator. My Japanese colleagues tell me that Yahoo is better than Google for technical language in physics and mathematics. I've also found that Wikipedia is not bad for this - you can search for keywords, and find corresponding articles in both languages.</p> <p>The standard reference for technical words is "the Iwanami", meaning <a href="http://en.wikipedia.org/wiki/Encyclopedic_Dictionary_of_Mathematics" rel="nofollow">The Encyclopedic Dictionary of Mathematics</a> (also mentioned by Iker). It has about 2000 pages, and weighs a lot. I almost never use it, and I don't own a copy. However, if I'm really stuck, I go to the library at work to find words in it.</p> <p>Regarding Neil Strickland's comment: I have yet to encounter a research article in Japanese, but there are expository articles, textbooks, conference slides, and grant applications that can be useful reading.</p> http://mathoverflow.net/questions/130050/covering-of-a-group-by-seven-proper-subgroups-counterexample/130051#130051 Answer by S. Carnahan for Covering of a group by seven proper subgroups: Counterexample S. Carnahan 2013-05-08T04:59:00Z 2013-05-08T04:59:00Z <p>The only theorem of Tomkinson I could find about this was in</p> <blockquote> <p>Tomkinson, M.J.. "Groups as the union of proper subgroups." Mathematica Scandinavica 81.2 (1998): 191-198.</p> </blockquote> <p>In section 3, he proves that any group which is the union of seven proper subgroups can be given as the union of fewer than seven proper subgroups.</p> <p>The main theorem of the paper is that if you define $\sigma(G)$ to be the smallest $n$ such that $G$ is the union of $n$ proper subgroups, then there is no finite $G$ satisfying $\sigma(G) = 7$.</p> <p>In summary, I agree that your construction can yield seven proper subgroups whose union is $G$, but it is not a counterexample of the theorem, when the theorem is stated correctly.</p> http://mathoverflow.net/questions/129345/n-in-a-row-game-on-mathbbr2/129610#129610 Answer by S. Carnahan for $n$-in-a-row game on $\mathbb{R}^2$ S. Carnahan 2013-05-04T03:56:36Z 2013-05-05T00:36:57Z <p>Okay, I'm rewriting my old answer - I thought I could extract the necessary estimates from <a href="http://arxiv.org/abs/0910.3926" rel="nofollow">Polymath's paper</a>, but it looks like such an effort would take more time than I have in the near future.</p> <p>I'd like to modify the game as follows: Suppose we fix a positive integer $r$, and specify that player 2 is allowed to make $r$ moves for each move by player 1. The original game corresponds to the case $r=1$.</p> <p>As ARupinski proposed, a possible winning strategy for player 1 is to play on a high-dimensional cubic lattice $[1,\ldots,n]^d$ embedded in $\mathbb{R}^2$. To prove this works, it suffices to show that any arrangement of $\frac{n^d}{r+1}$ points in the lattice contains enough "lines" that player 2 cannot block all of them.</p> <p>The Furstenburg-Katznelson theorem (also known as Density Hales-Jewett) gives a way to solve this. The literal result only produces the existence of a single line of a special form, with $d$ bounded by $A_{n+1}(r+1)$ (where $A_n$ denotes a modified Ackerman function defined by $A_1(r) = 2r$, $A_n(1) = 2$, and $A_{n+1}(r+1) = A_n(A_{n+1}(r))$). Since a single line can be blocked by a single intervening point, we really need a result that gives lots of lines. There is a multidimensional DHJ result in Polymath's paper, producing subspaces instead of lines, but the asymptotics are to be much too weak for this purpose.</p> <p>A straightforward-sounding method is to show that if the dimension $d$ is large enough, then there are lots of subspaces $V$ of dimension close to $d'$ for which our set has density that isn't much less than $1/(r+1)$ on $V$, where $d'$ is large enough that we are guaranteed a line on each $V$. I suspect we could bound it by an Ackerman function with slightly larger parameters, but since I haven't worked out any details, we know how much that's worth.</p> http://mathoverflow.net/questions/129536/where-do-the-product-expansions-of-modular-forms-come-from/129596#129596 Answer by S. Carnahan for Where do the product expansions of modular forms come from? S. Carnahan 2013-05-04T01:51:29Z 2013-05-04T01:56:53Z <p>Many "natural" examples of automorphic infinite products (also known as Borcherds products) can be explained using the singular theta lift of Harvey-Moore and Borcherds. These examples have the property that the exponents of the product are coefficients of a modular form.</p> <p>For example, the 24 in the exponents of the product formula for $\Delta$ can be matched with the positive-power coefficients of $12 \theta(0;\tau) = 12 + 24q^{1/2} + 24 q^2 + 24 q^{9/2} + \cdots$ in the following way: Shimura gave a correspondence between forms of weight $k + 1/2$ on $\Gamma_0(4)$ (satisfying some conditions) and forms of weight $2k$ for $SL_2(\mathbb{Z})$ (for some character), taking $f(\tau) = \sum_n c(n)q^n$ to $-c(0)B_k/2k + \sum_n q^n \sum_{d|n} d^{k-1} c(n^2/d^2)$. While this really works best for $k$ positive and even, you can remove the infinite constant term when $k=0$ to get something almost modular. Applying this to $12 \theta$, you naturally get $\log (\Delta/q)$.</p> <p>In other words, products like $\Delta$ arise by exponentiation of a Howe theta lift (of which Shimura's correspondence is a special case), although the lift may need to be regularized. For example, the Koike-Norton-Zagier formula: $$j(\sigma) - j(\tau) = (p^{-1} - q^{-1}) \prod_{m,n>0} (1-p^m q^n)^{c(mn)}$$ arises as a lift of $j(\tau) - 744 = \sum_n c(n) q^n$ to $O(2,2)$, and the usual method of lifting involves a divergent integral of $(j-744)\theta$ over a fundamental domain of $SL_2(\mathbb{Z})$.</p> <p>In general, I don't think these products are very naturally related to Galois representations. It is easy to lift forms like $-12\theta$ to get products like $1/\Delta$ of negative weight, which are somewhat invisible to the Langlands program (as far as I know). Instead, the products tend to show up naturally in subjects related to string theory, like the representation theory of infinite dimensional Lie algebras. For example, the product expansion of $1/\Delta$ gives the partition function of free bosons propagating in 24-dimensional space, and the Koike-Norton-Zagier formula is the Weyl denominator formula for the <a href="http://en.wikipedia.org/wiki/Monster_Lie_algebra" rel="nofollow">Monster Lie algebra</a>. Borcherds has some expository overviews of this subject on <a href="http://math.berkeley.edu/~reb/papers/index.html" rel="nofollow">his web page</a>, e.g., number 28: "Automorphic forms and Lie algebras".</p> http://mathoverflow.net/questions/128831/infinity-categories-vs-kan-complexes/129009#129009 Answer by S. Carnahan for Infinity-categories vs Kan complexes S. Carnahan 2013-04-28T15:27:45Z 2013-04-28T15:27:45Z <p>I suspect your confusion arises in part because homotopies of paths are continuous maps $I^2 \to X$, while 2-morphisms in $\pi_{\lt \infty} X$ are continuous maps $\Delta^2 \to X$. That is, 2-morphisms are not strictly the same as homotopies of paths.</p> <p>The dictionary between the two structures is not too bad:</p> <ol> <li><p>A 2-morphism in $\pi_{\lt \infty} X$ is a homotopy of paths, where either the beginning or the end is fixed (or perhaps it is a homotopy to or from a constant path).</p></li> <li><p>A continuous map $I^2 \to X$ can be viewed as a composite of two 2-morphisms in $\pi_{\lt \infty} X$. You end up using the diagonal $(0,0) \to (1,1)$ in the square to separate the two 2-morphisms, because the simplicial structure of $\Delta^1 \times \Delta^1$ has a diagonal 1-simplex.</p></li> </ol> <p>I personally find it rather magical that among the sixteen 2-simplices in $\Delta^1 \times \Delta^1$, a pair of them pops out as non-degenerate - it is a rewarding computation.</p> <p>More generally, higher dimensional cubes, viewed as products of the simplicial set $\Delta^1$, have canonical decompositions into nondegenerate simplices. The corresponding higher homotopies are composites of homotopies with various pieces held constant.</p> <p>I think this should account for the discrepancy you see in the number of face maps.</p> http://mathoverflow.net/questions/117415/old-books-still-used/128165#128165 Answer by S. Carnahan for Old books still used S. Carnahan 2013-04-20T11:38:34Z 2013-04-20T11:38:34Z <p>When I was an undergrad, at the turn of the millenium, I took a complex analysis class that used (an English translation of) Knopp's 1936 <em>Funktionentheorie</em>.</p> http://mathoverflow.net/questions/127275/what-is-the-purpose-of-section-3-of-bbd/127307#127307 Answer by S. Carnahan for What is the purpose of section 3 of BBD? S. Carnahan 2013-04-12T02:12:03Z 2013-04-12T02:12:03Z <p>I'm not an expert at BBD, but the introduction explains that chapter 3 is supplementary technical information that can be skipped. For example, if you want to work with $\mathbb{Z}$-coefficients, or the filtered derived category, you might find chapter 3 useful. As far as a "plan" of BBD is concerned, the introduction has a brief description of the contents of each chapter, but if you are looking for some kind of "tree" of lemmata leading to main results like the decomposition theorem, I haven't seen one.</p> <p>I only found 3 references to chapter 3 in the rest of the book:</p> <ol> <li><p>The proof of Proposition 2.1.23 uses section 3.2.</p></li> <li><p>Part 2.2.19 gives a proof of Proposition 2.1.23 that does not use results of chapter 3.</p></li> <li><p>The first page of chapter 4 mentions section 3.3 in an inessential way.</p></li> </ol> http://mathoverflow.net/questions/126595/algebraic-p-adic-integers-mod-p/126686#126686 Answer by S. Carnahan for Algebraic $p$-adic integers mod $p$ S. Carnahan 2013-04-06T01:57:33Z 2013-04-06T01:57:33Z <p>This was answered in the comments, so I'll just sum up:</p> <p>$(p)$ is the maximal ideal in $\mathbf{Z}_p^{nr}$, so we have an isomorphism $\mathbf{Z}_p^{nr}/(p) \to \overline{\mathbf{F}_p}$. The reverse isomorphism is given by the Teichmüller lift (a multiplicative monoid map to $\mathbf{Z}_p^{nr}$).</p> <p>$\mathbf{Z}_p^t \cong \mathbf{Z}_p^{nr}[\{p^{1/n}\}_{(n,p)=1}]$, so $\mathbf{Z}_p^t/(p)$ is isomorphic to $\overline{\mathbf{F}_p}[\{x_n\}_{(n,p)=1}]/(x_1, \{x_{nk}^k-x_n\}_{(n,p)=1,k>1})$. This is a local non-reduced non-Noetherian ring. Its nilradical, generated by all $x_n$, is maximal, idempotent, and nil, but not nilpotent. Any finite set of elements is contained in a principal ideal.</p> <p>I do not know an explicit structure for $\overline{\mathbf{Z}_p}/(p)$, but it has basically the same general ring-theoretic properties as the previous case (except for the fact that every element is a $p$th power).</p> http://mathoverflow.net/questions/125555/geometric-description-of-the-deligne-mumford-stacks/126163#126163 Answer by S. Carnahan for Geometric description of the Deligne-Mumford stacks S. Carnahan 2013-04-01T08:30:13Z 2013-04-01T08:30:13Z <p>For question 2, I've never seen a definition of DM 2-stack, so I don't know where to start. For question 1, I have a tentative counterexample even for the case where both the stack and the coarse moduli space are smooth.</p> <p>The étale local picture near a stacky point is that we take an affine space, quotient by a finite group $G$, and take the coarse moduli space to get an affine space. By the Chevalley-Shepard-Todd theorem, if you want your coarse moduli space to be smooth, it is necessary and sufficient that $G$ be a complex reflection group, acting by complex reflections. In order to find a pair of suitable non-isomorphic stacks for which labelings of the coarse space by orders of groups does not distinguish them, we need to find two complex reflection groups satisfying the following conditions:</p> <ol> <li><p>They have the same order.</p></li> <li><p>The orders of stabilizers of affine subspaces of a given dimension form identical sets of positive integers.</p></li> <li><p>There is an isomorphism of the coarse moduli spaces taking the images of suitably labeled subspaces to each other.</p></li> </ol> <p>The relevant information is in the big table at <a href="http://en.wikipedia.org/wiki/Complex_reflection_group" rel="nofollow">the Wikipedia page on complex reflection groups</a>.</p> <p>I will confine myself to rank 2 groups, because then the nontrivial subspace stabilizers are precisely the reflections. For some reason, I'm unable to think clearly right now (this may have something to do with the awkward wording of condition 3), so I'm going to make an assumption about transitivity that might not be true.</p> <p><strong>Assumption:</strong> For each reflection order in an irreducible complex reflection group $G$, all reflection hyperplanes of that order in the source affine space are mapped to a single hyperplane in the target. Equivalently, irreducible complex reflection groups act transitively by conjugation on the quotient of the set of reflections of any fixed order by the corresponding hyperplane stabilizers.</p> <p>If the assumption is true, we only need to find two rank 2 complex reflection groups of the same order, whose reflections have the same order. For example, the exceptional groups of order 144 listed as numbers 7 and 14 in the Wikipedia table work, as well as several other exceptional groups matched with the semidirect products $G(m,p,2)$.</p> http://mathoverflow.net/questions/125116/3d-rotation-representation-for-multiple-turns/125132#125132 Answer by S. Carnahan for 3D Rotation Representation for Multiple Turns S. Carnahan 2013-03-21T07:00:37Z 2013-03-21T07:00:37Z <p>Since you haven't described what you plan to do with this representation, I'm not sure what method would work well.</p> <p>One of the problems with representing "turns" in more than two dimensions is that you don't have much in the way of discrete invariants. This is because the fundamental group of $SO(n)$ only has two elements when $n \gt 2$. Indeed, a double rotation in any direction can be continuously deformed to the identity rotation. This means that two turns are indistinguishable from zero turns, if you consider paths that can be deformed to each other to be equivalent.</p> <p>If you want to keep track of how your rotation came about, here are two suggestions:</p> <ol> <li>You can use based paths in the group of rotations, i.e., continuous maps from a real interval into rotation matrices.</li> <li>If your paths are smooth, you can take the derivative, and use paths in velocity space, i.e., the Lie algebra of the rotation group.</li> </ol> <p>Both methods give you points in an infinite dimensional path space.</p> http://mathoverflow.net/questions/124864/representability-of-sheaves-of-groups/124872#124872 Answer by S. Carnahan for Representability of sheaves of groups S. Carnahan 2013-03-18T15:58:42Z 2013-03-18T15:58:42Z <p>I disagree with the premise of your question. There are many natural non-representable sheaves of groups. For example, the formal additive group $\widehat{\mathbb{G}_a}$ is a sheaf, as the colimit of (the sheaves represented by) spectra of $\mathbb{Z}[x]/(x^n)$, but it is not representable as a scheme.</p> <p>I suspect your experience is a result of mathematicians being generally more likely to encounter group sheaves that are given by quasicoherent sheaves of finite type Hopf algebras, than by weird moduli functors.</p> http://mathoverflow.net/questions/124490/smooth-connection-on-exterior-power/124603#124603 Answer by S. Carnahan for smooth connection on exterior power S. Carnahan 2013-03-15T10:52:59Z 2013-03-15T10:52:59Z <p>Yes, connections can be transported along any functor $F$ from vector spaces to vector spaces. Exterior powers are standard examples of such functors.</p> <p>Say $E$ is a vector bundle on $M$. The connection $\nabla$ is equivalent to an isomorphism $p_1^*E \to p_2^*E$ (where $p_1, p_2: M^{(2)} \to M$ are the projections from the first infinitesimal neighborhood of the diagonal) that restricts to the identity on the diagonal. The functor $F$ induces a corresponding functor on vector bundles over $M^{(2)}$, and hence a corresponding connection.</p> <p>The equivalence is explained in the beginning of section 2 (page 5) of <a href="http://www.math.ucdavis.edu/~osserman/math/connections.pdf" rel="nofollow">Osserman's notes</a> for schemes, but it holds for anything smooth (e.g., manifolds as ringed spaces).</p> http://mathoverflow.net/questions/122394/spectrum-and-scheme-of-the-commutative-group-algebra-of-an-abelian-group/122414#122414 Answer by S. Carnahan for Spectrum and scheme of the commutative group-algebra of an abelian group. S. Carnahan 2013-02-20T15:56:01Z 2013-02-20T15:56:01Z <p>The spectrum of the group algebra of a commutative group is called a diagonalizable group scheme. This is defined in <a href="http://www.math.jussieu.fr/~polo/SGA3/" rel="nofollow">SGA 3</a> Exposé VIII Section 1. Several geometric characterizations of group-theoretic properties are given in Proposition 2.1. A lot more is written in later sections, such as material on principal homogeneous spaces, quotients of affine schemes by diagonalizable group schemes, and representability of restriction of scalars.</p> <p>If that isn't enough for you, Exposés 9-11 are about group schemes that are locally-on-the-base isomorphic to diagonalizable group schemes.</p> http://mathoverflow.net/questions/121162/math-behind-databases-management-and-sql/122204#122204 Answer by S. Carnahan for Math behind databases management and SQL ? S. Carnahan 2013-02-18T17:53:27Z 2013-02-18T17:53:27Z <p><a href="http://arxiv.org/find/cs/1/au%3A+Spivak_D/0/1/0/all/0/1" rel="nofollow">David Spivak</a> (who is an occasional contributor here) has done some work on categorical aspects of database management, and the creation of dictionaries between math world and the DB world that allow one to prove that certain operations are well-behaved in a robust sense. The link gives a list of his ArXiv papers.</p> http://mathoverflow.net/questions/114595/can-we-promote-to-a-lie-group-isomorphism/114621#114621 Answer by S. Carnahan for Can we promote to a Lie Group Isomorphism? S. Carnahan 2012-11-27T02:18:01Z 2013-01-20T15:21:10Z <p>The answer to your question is "no". I've rewritten my previous answer to include details.</p> <p>In dimension $d$ at least 7, there are continuous positive-dimensional (non-isotrivial) families of nilpotent Lie algebras in characteristic zero. I had previously heard this as folklore, but some searching yields <a href="http://etd.uwaterloo.ca/etd/mpgong1998.pdf" rel="nofollow">Ming-Peng Gong's dissertation</a>, which has explicit presentations. By the exponential correspondence, this yields families of unipotent groups over subfields of $\mathbb{R}$, all of whose real-analytifications have underlying manifolds that are diffeomorphic to $\mathbb{R}^d$. </p> <p>I will choose the family (147E), with generating basis $x_1,\ldots,x_7$, and nonzero brackets $[x_1,x_2] = x_4$, $[x_1,x_3]=-x_6$, $[x_1,x_5] = -x_7$, $[x_2,x_3]=x_5$, $[x_2,x_6]=\lambda x_7$, $[x_3,x_4]=(1-\lambda)x_7$, as $\lambda$ varies over complex numbers satisfying $\lambda(\lambda-1) \neq 0$. Over the complex numbers, the Lie algebras are distinguished up to isomorphism by the value of $j(\lambda) = \frac{(1-\lambda+\lambda^2)^3}{\lambda^2(\lambda-1)^2}$. For each real $\lambda$, I will call the Lie algebra $L(\lambda)$, and the corresponding Lie group $G(\lambda)$.</p> <p>For any real quadratic extension $K/\mathbb{Q}$, there is a Galois-conjugate pair $(\lambda, \lambda')$ of irrational elements, such that $j(\lambda) \neq j(\lambda')$. One reason is that $j^{-1}j$ is not equivariant under translation by one - you can find the 6-element preimage of $j(\lambda)$ written out in <a href="http://en.wikipedia.org/wiki/Modular_lambda_function#Modular_properties" rel="nofollow">Wikipedia's Modular lambda article</a>. For such a pair, $L(\lambda)$ is not isomorphic to $L(\lambda')$, and $G(\lambda)$ is not isomorphic to $G(\lambda')$ as a real Lie group.</p> <p>However, the underlying groups of real points are isomorphic. The isomorphism is given by transporting the nontrivial Galois automorphism of $K$ through the functor $- \otimes_K \mathbb{R}$ on Lie algebras, followed by exponentiation. This is highly discontinuous. For example, each $\exp(x_i) \in G(\lambda)$ is taken to $\exp(x_i) \in G(\lambda')$, but $\exp (\lambda x_i) \in G(\lambda)$ is taken to $\exp(\lambda' x_i) \in G(\lambda')$.</p> <p>In summary, we have two Lie groups $G(\lambda)$ and $G(\lambda')$, we have an abstract group isomorphism $f$ (by transport of Galois) between them, and a diffeomorphism $g$ (because they are both diffeomorphic to $\mathbb{R}^7$) between them, but they are not isomorphic as Lie groups.</p> http://mathoverflow.net/questions/119069/is-any-g-set-a-coset-geometry-in-the-sense-of-tits-buekenhout/119076#119076 Answer by S. Carnahan for Is any $G$-set a coset geometry (in the sense of Tits-Buekenhout)? S. Carnahan 2013-01-16T14:23:28Z 2013-01-17T00:44:58Z <p>Any $G$-set with $n$ orbits has a rank $n$ coset geometry structure that is unique up to relabeling. Indeed, there is computer algebra software (e.g., GAP, Magma) that will produce such a structure from a list of stabilizers of orbits.</p> <p>Some Googling reveals that Tits came up with coset geometries, so it is likely that he was the first to realize your observation.</p> <p><strong>Edit:</strong> Regarding software, MAGMA is not free (although I have heard that they have some forgiving policies for students). The <a href="http://magma.maths.usyd.edu.au/magma/handbook/incidence_geometry" rel="nofollow">documentation on incidence geometry</a> has some discussion of the construction I outlined, especially in the introduction. In particular, they point out that a coset geometry is not in general a geometry in the sense of Buekenhout, due to an extra flag-transitivity condition. They reference: </p> <blockquote> <p>Tits, <em>Géométries polyédriques et groupes simples</em> Atti 2a Riunione Groupem. Math. Express. Lat. Firenze, (1962) pp66-88.</p> </blockquote> <p><a href="http://www.gap-system.org/" rel="nofollow">GAP</a> is free, and you can make coset geometries by <a href="http://cage.ugent.be/geometry/fining/chap13.html" rel="nofollow">following the documentation</a>.</p> http://mathoverflow.net/questions/117180/tannakian-fundamental-group-for-finitely-linear-representation-of-group/117217#117217 Answer by S. Carnahan for Tannakian fundamental group for finitely linear representation of group S. Carnahan 2012-12-26T00:35:37Z 2012-12-26T00:35:37Z <p>Following Niels's suggestion, I'm turning my comments into an answer.</p> <p>The first point I want to make is the observation that if $G$ is an infinite group, then the constant group scheme $G_k$ is not affine over $k$. As a scheme, it is an infinite disjoint union of copies of the spectrum of $k$, and the ring of global functions is an infinite product of copies of $k$. If you take the spectrum of that ring, the underlying topological space is not your original discrete space, first of all because it is quasi-compact. I think it has strictly larger cardinality, due to the existence of lots of exotic prime ideals. I imagine there is some relation with Stone-Cech compactification, but I am too lazy to look it up or try to figure it out.</p> <p>In general the group of automorphisms of the fiber functor is an object called the pro-algebraic hull of $G$. It is an initial object among all homomorphisms from $G$ to pro-algebraic groups, and exists by the completeness of the category of affine group schemes. There is a more concrete description given in the comments by ChrisLazda, and as you can see from his/her example, it can depend strongly on the base field.</p> <p>As far as I can tell, the notion of pro-algebraic hull was folklore (or perhaps an exercise for graduate students) for a while. Googling yields a description in <a href="http://arxiv.org/abs/math/0611521" rel="nofollow">this paper</a>, together with an application.</p> http://mathoverflow.net/questions/117019/quotient-of-a-compact-lie-group-by-maximal-torus/117069#117069 Answer by S. Carnahan for Quotient of a compact Lie group by maximal Torus S. Carnahan 2012-12-23T05:57:41Z 2012-12-23T05:57:41Z <p>The answer is yes. As Aakumadala mentioned in a comment, the quotient $K/\mathbb{T}$ is isomorphic to the flag variety of $K_{\mathbb{C}}$. If we let $N = N_{\mathbb{C}}$ denote the nilpotent radical of a Borel subgroup of $K_{\mathbb{C}}$, the flag variety has a dense $N$-orbit with a simply transitive $N$ action. Note that when $K = SU(3)$, we find that $N$ is a Heisenberg group, as you anticipated.</p> <p>The underlying manifold of the dense $N$-orbit is just an affine space, isomorphic to $\mathbb{C}^n = \mathbb{R}^{2n}$ for some $n$. If you make this identification, then the dense embedding $\mathbb{Z}[1/2] \subset \mathbb{R}$ gives you a dense set parametrized by $\mathbb{Z}[1/2]^{2n}$.</p> http://mathoverflow.net/questions/116847/intuitive-pictures-in-characteristic-p/116982#116982 Answer by S. Carnahan for Intuitive pictures in characteristic p S. Carnahan 2012-12-21T17:03:57Z 2012-12-22T03:21:24Z <p>I heard the following analogy when talking to some specialists in absolute de Rham theory. I think Deninger's name was mentioned at about the same time.</p> <p>One possible way to imagine a variety over $\mathbb{F}_p$ is as a manifold equipped with a distinguished vector field, which we call "Frobenius". The usual discrete Frobenius that admits integer powers is the unit time evolution of the flow. Orbits of points on the variety that are defined over finite fields correspond to closed integral curves of the flow, and we assume that such curves only have integral periodicity. </p> <p>There are a few problems. If there are points defined over an imperfect field, you may have to consider flows that start somewhere, like a distinguished submanifold. Also, it is not clear to me how one connects such a picture to the set of solutions of a system of polynomial equations.</p> <p><strong>Edit:</strong> (response to Daniel Litt's comment) I must confess that I am not particularly familiar with the idea, so I can only fill in vague guesses. Also, I don't know why one would want to interpolate between different powers of Frobenius. The fundamental idea seems to be that if we examine the spectrum of a finite field using étale glasses rather than Zariski or Nisnevich glasses, it looks a lot like a circle, since the étale fundamental group is a completion of $\mathbb{Z}$. This suggests that if we were to propose some real geometric object as an analogue of a variety, $\mathbb{F}_q$-points should be distinguished circles, perhaps with some distinguished automorphism.</p> <p>The picture of finite fields as circles also shows up in "arithmetic topology" speculation for similar reasons. Here, the spectra of number rings are viewed as 3-manifolds, and the finite points are distinguished embedded circles, for which some kind of linking number may be defined homologically. As far as I know, this is another analogy that seems to be waiting for a substantial application.</p> http://mathoverflow.net/questions/116824/useful-notion-of-unramified-galois-representation/116835#116835 Answer by S. Carnahan for Useful notion of unramified Galois representation S. Carnahan 2012-12-20T02:08:26Z 2012-12-20T02:08:26Z <p>The standard definition of "unramified" applies to the function field case, so any finite branched cover of the projective line yields a Galois representation that is ramified in finitely many places. In particular, a representation of the absolute Galois group of $\mathbb{C}(t)$ that is unramified on $\mathbb{A}^1$ is just a representation of the fundamental group of $\mathbb{A}^1_{\mathbb{C}}$, i.e., the trivial group.</p> <p>However, there is a way to get a nontrivial object by following a conjectural dictionary between Galois representations on characteristic $p$ varieties and $D$-modules on complex varieties. In this dictionary, wild ramification (which doesn't occur in characteristic zero) corresponds to irregular singularities (which don't appear in the usual Riemann-Hilbert correspondence). In particular, the Artin-Schreier sheaf on a characteristic $p$ line is tied to the $D$-module on the complex line whose global solutions are exponential functions $k e^z$. They are similar in the sense that the Artin-Schreier representation is unramified away from infinity but has wild ramification there, and $e^z$ is entire but has an essential singularity at infinity.</p> <p>In conclusion, your smooth projective variety over $\mathbb{C}(t)$ may yield something interesting if you do some kind of $D$-module pushforward instead of looking at the Galois representation.</p> http://mathoverflow.net/questions/116519/is-an-eigenvector-of-a-hecke-operator-automatically-an-eigenform/116565#116565 Answer by S. Carnahan for Is an eigenvector of a Hecke operator automatically an eigenform? S. Carnahan 2012-12-17T06:01:57Z 2012-12-18T11:45:29Z <p>Following the comments, here is perhaps the simplest counterexample (once you know the Breuil-Conrad-Diamond-Taylor modularity theorem). The curves $y^2 + y = x^3$ and $y^2 + y = x^3 + 2x$ both reduce mod 2 to the same smooth curve, so their Hecke eigenforms have equal $T_2$ eigenvalues.</p> <p>However, the curves over $\mathbb{Q}$ are non-isomorphic, since only the first curve has vanishing <a href="http://mathoverflow.net/questions/105713/are-the-eigenvalues-of-hecke-operators-distinct" rel="nofollow">$j$-invariant</a>. This means the Hecke eigenforms have different coefficients for some prime $p$, and hence different eigenvalues for $T_p$ (in fact, by counting the mod 3 solutions, you can see that $p=3$ works).</p> <p>If you add the eigenforms, you get a (non-normalized) cusp form that is an eigenfunction for $T_2$, but not for $T_p$.</p> <p><strong>Edit</strong> I might as well say what I know about the level 1 case. It is "classically" known that Hecke operators are self-adjoint with respect to the Petersson inner product on cusp forms, and they commute with each other. By the spectral theorem, the Hecke operators are therefore simultaneously diagonalizable, i.e., there is a basis of the space of cusp forms made out of eigenforms.</p> <p>Now, let us assume all of the eigenvalues of all of the Hecke operators are distinct (in contrast to the counterexample above). If you have a form $f$ for which some $T_n$ acts by a scalar, then $f$ cannot be a non-trivial linear combination of some basis of eigenforms, i.e., it is necessarily an eigenform. This multiplicity one property is not known in level one in general, but it is known to be true for $T_2$, as I learned in Cardinal Wolsey's brilliant answer to <a href="http://mathoverflow.net/questions/105713/are-the-eigenvalues-of-hecke-operators-distinct" rel="nofollow">your previous question</a> (the answer was deleted, it seems, because you didn't accept it within 4 days).</p> <p>More generally, the multiplicity one property is implied by a conjecture of Maeda, which asserts that Hecke operators act irreducibly on $S_k(SL_2(\mathbb{Z}))$ for each $k$. In fact, David Loeffler mentioned, in a comment on the previous question, that there is a lot of computational evidence behind an even stronger assertion, namely that the eigenvalues generate as-big-as-possible Galois extensions over $\mathbb{Q}$.</p> http://mathoverflow.net/questions/116219/rational-orthogonal-matrices/116237#116237 Answer by S. Carnahan for Rational orthogonal matrices S. Carnahan 2012-12-13T04:23:08Z 2012-12-13T04:23:08Z <p>This is not a complete answer, but it's a start.</p> <p>In the $2 \times 2$ case, the act of choosing the first column and clearing denominators describes a two-to-one map from orthogonal matrices to primitive vectors in the lattice $\mathbb{Z} \oplus \mathbb{Z}$. In particular, there are twice as many orthogonal matrices with denominator exactly $q^2$ as there are primitive vectors of length $q$, and there are twice as many orthogonal matrices with denominator dividing $q^2$ as there are vectors of length $q$. The latter quantity is enumerated by <a href="http://en.wikipedia.org/wiki/Theta_function#Generalizations" rel="nofollow">the theta function of the lattice</a>, which is a modular form of weight 1, and its coefficients grow logarithmically.</p> <p>In higher dimension, I guess the generating function comes from successively choosing vectors in orthogonal complements, so it should be at least related to modular forms.</p> http://mathoverflow.net/questions/116023/automorphy-factors-and-bundles/116049#116049 Answer by S. Carnahan for Automorphy Factors and Bundles S. Carnahan 2012-12-11T06:19:06Z 2012-12-11T06:19:06Z <p>A cocycle for a modular group is precisely the same as a descent datum for the quotient map $\mathbf{H} \to [\mathbf{H}/\Gamma]$, where the target is the quotient orbifold. This is a special case of the fact that pullback induces an equivalence of categories between vector bundles on the quotient, and vector bundles $V$ on the upper half-plane equipped with a cocycle with coefficients in $\operatorname{Aut}(V)$.</p> <p>If you want to make a vector bundle on $X(1)$ or $X(\Gamma)$, your cocycle has to satisfy some properties, and you need to specify some additional data. In particular, you need triviality at elliptic points to descend from $[\mathbf{H}/\Gamma]$ to the affine coarse space $\mathbf{H}/\Gamma$ (which is often written $Y(\Gamma)$), and you need a gluing datum to describe the behavior at the cusps in order to define a vector bundle on the compact curve $X(\Gamma)$.</p> <p>We have a standard example in level 1. The stack $Ell$ has Picard group $\mathbb{Z}/12\mathbb{Z}$ (I think this may be a theorem of Fulton). The trivial bundles descend to the coarse space $Y(1)$, which is an affine line, and all vector bundles on $Y(1)$ are trivial. Adding a cusp yields $X(1)$, which is a projective line, and vector bundles on $X(1)$ are just sums of line bundles parametrized by degree. The upshot is that if you don't specify gluing data at infinity, your cocycle will not tell you anything about the weight of a modular form.</p> <p>Regarding your comment about level 1 forms of weight 6: they all vanish at $i$.</p> http://mathoverflow.net/questions/115743/an-algebra-of-integrals/115789#115789 Answer by S. Carnahan for An algebra of "integrals" S. Carnahan 2012-12-08T10:29:01Z 2012-12-10T10:38:46Z <p>I don't see why you want $A$ to be an algebra, since the integral of 1 doesn't seem like a reasonable unit. Did you want some compatibility with higher dimensional integrals using the Fubini theorem? Otherwise, if you follow Kähler's lead, it seems more natural to expect a real (or complex) vector space.</p> <p>Let $C^\infty(\mathbb{R})_{int}$ denote the subspace of $C^\infty(\mathbb{R})$ whose elements are integrable on $[0,\infty)$, and let $C^\infty(\mathbb{R})_{int}^0$ denote the codimension one subspace of functions whose integral is zero. Here's a rephrasing of the desired properties of $A$ and $T$:</p> <ol> <li><p>Linearity of $T$.</p></li> <li><p>The restriction of $T$ to $C^\infty(\mathbb{R})_{int}$ lands in a distinguished subspace $\mathbb{R} \subset A$, and is given by ordinary integration.</p></li> <li><p>Good behavior under the action of the group $\mathbb{R} \rtimes \mathbb{R}^\times_{>0}$ generated by translations and orientation-preserving dilations.</p></li> </ol> <p>[Edit:] Let $X$ is a space of smooth functions closed under addition by $C^\infty(\mathbb{R})_{int}$, such that $\mathbb{R} \rtimes \mathbb{R}^\times_{>0}$ acts freely on the quotient vector space $X/C^\infty(\mathbb{R})_{int}$. If a universal target $A$ for integration existed, then $X/C^\infty(\mathbb{R})_{int}^0$ should admit an injection to $A$, because your list of conditions specifies no further relations. The problem (as pointed out by Tao) is that there are lots of smooth functions with nontrivial stabilizer in $\mathbb{R} \rtimes \mathbb{R}^\times_{>0}$.</p> <p>I think a common method to removing such a difficulty is to ignore the requirement that integration be $\mathbb{R} \rtimes \mathbb{R}^\times_{>0}$-equivariant. Then your universal space is just $C^\infty(\mathbb{R})/C^\infty(\mathbb{R})_{int}^0$.</p> http://mathoverflow.net/questions/115567/covering-maps-in-real-life-that-can-be-demonstrated-to-students/115575#115575 Answer by S. Carnahan for Covering maps in real life that can be demonstrated to students S. Carnahan 2012-12-06T06:01:40Z 2012-12-06T06:01:40Z <p>As far as I can tell, you can cover any graph embedded in $\mathbb{R}^3$ with its universal cover using a homotopy that satisfies your conditions. This is because trees are contractible, and can snake themselves into tight spaces without intersecting themselves.</p> http://mathoverflow.net/questions/108978/has-any-attempt-been-made-to-classify-finite-groupoids/114756#114756 Answer by S. Carnahan for Has any attempt been made to classify finite groupoids? S. Carnahan 2012-11-28T10:57:25Z 2012-11-28T10:57:25Z <p>Todd Trimble basically answered the question you literally asked, but it sounds like you may be thinking about a slightly different question, e.g., is there a classification of objects like the Mathieu groupoid, where it may show up as an exceptional example?</p> <p>To elaborate, the Mathieu groupoid is not only a groupoid, but it is equipped with a distinguished representation on a <a href="http://en.wikipedia.org/wiki/Finite-state_machine" rel="nofollow">finite state machine</a>. The set of states is the set of reachable labelings of vertices, and the transition operations are described by the generators of the groupoid. In other words, you may be seeking not a classification of finite groupoids (which, as Todd Trimble mentioned, is equivalent to a classification of finite groups), but a classification of reversible finite state machines.</p> <p>That said, it seems unlikely that anyone has made a concerted attempt, even on the level of Hölder's 2-step program, due to a lack of structure.</p> http://mathoverflow.net/questions/114471/quotients-of-varieties-as-non-noetherian-schemes/114537#114537 Answer by S. Carnahan for quotients of varieties as non-noetherian schemes? S. Carnahan 2012-11-26T15:14:00Z 2012-11-26T15:14:00Z <p>The quotient exists as an algebraic space (if we ignore the literature that defines algebraic spaces to be separated). The equivalence relation is given by the standard action groupoid. This is étale, since both the projection and the action map from $X \times G$ to $X$ are locally finitely presented and formally étale (indeed, they are locally isomorphisms). I'm not sure what you are looking for by passing to non-Notherian schemes, beyond allowing the "morphism space" $X \times G$ to be used.</p> http://mathoverflow.net/questions/93437/math-blog-directory/102220#102220 Comment by S. Carnahan S. Carnahan 2013-05-18T02:38:08Z 2013-05-18T02:38:08Z This question was not intended as a place to promote your own blog. http://mathoverflow.net/questions/93437/math-blog-directory/130801#130801 Comment by S. Carnahan S. Carnahan 2013-05-18T02:37:58Z 2013-05-18T02:37:58Z This question was not intended as a place to promote your own blog. http://mathoverflow.net/questions/130969/derivation-of-bessel-functions Comment by S. Carnahan S. Carnahan 2013-05-18T00:20:01Z 2013-05-18T00:20:01Z As far as I can tell, the Bessel function only gives an approximate solution to the equation, so it is unlikely to pop out of a first-principles attempt at an exact solution. It might be best if you broke your question down into simpler pieces, and asked them at <a href="http://math.stackexchange.com" rel="nofollow">math.stackexchange.com</a> or one of the other sites listed in the FAQ. There are nice free materials on fluid mechanics and differential equations at MIT's OpenCourseware, but they require a serious commitment and investment of time to work well. http://mathoverflow.net/questions/130883/is-there-any-proof-that-you-feel-you-do-not-understand/130928#130928 Comment by S. Carnahan S. Carnahan 2013-05-17T16:09:53Z 2013-05-17T16:09:53Z Instead of cutting subintervals, you can define $x \in [0,1]$, for a given cover, to be the least upper bound of the set of $y$ for which there is a finite subcover of $[0,y]$. Then by the existence of an open set containing $x$, it is not hard to show that $x=1$. Plus: explicit use of least upper bound property. Minus: large quantifier load. http://mathoverflow.net/questions/130952/problem-estimating-peak-traffic-on-web-site Comment by S. Carnahan S. Carnahan 2013-05-17T14:56:14Z 2013-05-17T14:56:14Z I'm afraid your question is somewhat outside the scope of this web site. You may want to consider one of the other sites listed in the FAQ. http://mathoverflow.net/questions/119157/runs-in-coin-flips/119167#119167 Comment by S. Carnahan S. Carnahan 2013-05-17T00:01:56Z 2013-05-17T00:01:56Z This is not an answer. http://mathoverflow.net/questions/6166/is-there-a-name-for-the-matrix-equation-a-x-b-b-x-a-c-x-c-d/6211#6211 Comment by S. Carnahan S. Carnahan 2013-05-15T02:18:12Z 2013-05-15T02:18:12Z Chetan Tonde (in a deleted answer) suggests that you change $X$ to $vec(X)$ and $D$ to $vec(D)$ in your formula. http://mathoverflow.net/questions/6166/is-there-a-name-for-the-matrix-equation-a-x-b-b-x-a-c-x-c-d/130640#130640 Comment by S. Carnahan S. Carnahan 2013-05-15T02:17:14Z 2013-05-15T02:17:14Z I assume this was meant to be a comment on Ho Chung Siu's answer. http://mathoverflow.net/questions/130654/a-catalog-of-faithful-representations-of-finite-groups Comment by S. Carnahan S. Carnahan 2013-05-15T01:51:21Z 2013-05-15T01:51:21Z Have you tried GAP? <a href="http://www.gap-system.org" rel="nofollow">gap-system.org</a> http://mathoverflow.net/questions/130545/proof-of-the-weak-goldbach-conjecture/130559#130559 Comment by S. Carnahan S. Carnahan 2013-05-15T01:02:38Z 2013-05-15T01:02:38Z Perhaps I travel in different circles than pooper, but the theorems of Vinogradov and Chen were among the first things I learned after hearing of the existence of the Goldbach problem. Isn't it common, when introducing a famous open problem, to describe roughly how far we've come? http://mathoverflow.net/questions/117662/a-question-about-lp-integral-of-an-entire-function-on-mathbbc/128843#128843 Comment by S. Carnahan S. Carnahan 2013-05-14T13:58:25Z 2013-05-14T13:58:25Z You may want to look at the Wikipedia article on subharmonic functions. http://mathoverflow.net/questions/130574/why-do-i-get-estimated-error-1-ind-when-doing-bicgstab-linear-solver-using-il Comment by S. Carnahan S. Carnahan 2013-05-14T12:59:13Z 2013-05-14T12:59:13Z Sorry, this site is a bit more concerned with software-independent questions. http://mathoverflow.net/questions/130516/the-category-of-right-comodule-of-coalgebra-is-a-monoidal-category-why Comment by S. Carnahan S. Carnahan 2013-05-14T01:51:32Z 2013-05-14T01:51:32Z Your question seems to be missing the definitions of some of your notation. http://mathoverflow.net/questions/130533/finance-problem Comment by S. Carnahan S. Carnahan 2013-05-14T01:42:51Z 2013-05-14T01:42:51Z I'm afraid this website is not quote the right one for your question. http://mathoverflow.net/questions/53471/are-there-any-very-hard-unknots/130439#130439 Comment by S. Carnahan S. Carnahan 2013-05-13T08:22:16Z 2013-05-13T08:22:16Z I have transferred this comment to the appropriate place.