User - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T10:08:38Z http://mathoverflow.net/feeds/user/15136 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130789/are-the-two-meanings-of-undecidable-related Are the two meanings of "undecidable" related? unknown (google) 2013-05-16T05:30:39Z 2013-05-18T20:01:24Z <p>I am usually confused by questions of the type "could such and such a problem be undecidable", because as far as I know there are two distinct possible meanings of "undecidable". I regard the following as the standard definition:</p> <blockquote> <p>Let $P(n)$ be a statement like "$n$ has at least three prime factors" or "the $n$th Turing machine halts" (we can also forget the statement and consider just its "truth value" function <code>$P:\mathbb N\to\{0,1\}$</code>). We say that <strong>the problem $P(n)$ is undecidable</strong> iff there is no Turing machine which answers "Is $P(n)$ true" correctly for all $n$ (equivalently, iff the function <code>$P:\mathbb N\to\{0,1\}$</code> is non-recursive).</p> </blockquote> <p>On the other hand, there is also the following alternative notion:</p> <blockquote> <p>Let $P$ be a statement like <code>"$\forall x,y,z,n\in\mathbb Z,x^n+y^n+z^n=0\;\&amp;\;n\geq 3\implies xyz=0$"</code>. We say that <strong>the problem $P$ is undecidable</strong> iff it is independent of some particular chosen set of axioms (like PA or ZFC).</p> </blockquote> <p>I would prefer to call this notion "independent of PA" or "independent of ZFC", since it makes explicit reference to a particular choice of axiomatic system.</p> <p>Is there a good reason to use "undecidable" for both of these notions? Has this collision of terminology bothered anyone else? Is there a deeper connection between these notions which justifies using the same word for both?</p> http://mathoverflow.net/questions/131016/terminology-complex-of-sheaves-with-cohomology-sheaves-concentrated-in-degree-ze Terminology: complex of sheaves with cohomology sheaves concentrated in degree zero unknown (google) 2013-05-18T02:30:36Z 2013-05-18T07:00:22Z <p>What is the proper terminology for a complex of sheaves $\mathcal F^\bullet$ whose homology sheaves $\mathcal H^i\mathcal F^\bullet$ vanish for $i\ne 0$?</p> http://mathoverflow.net/questions/130884/reference-request-sheaves-on-closed-sets Reference request: sheaves on closed sets unknown (google) 2013-05-17T00:12:52Z 2013-05-17T19:49:21Z <p>I am faced with a context in which the most natural notion of a sheaf $\mathcal F$ is as a functor on the category of <em>compact</em> subsets of a (locally compact Hausdorff) space $X$. Specifically, I say a presheaf $\mathcal F$ is a sheaf iff it satisfies the three axioms: <code>\begin{align} \mathcal F(\varnothing)&amp;=0\\ 0\to\mathcal F(K_1\cup K_2)\to\mathcal F(K_1)\oplus\mathcal F(K_2)\to\mathcal F(K_1\cap K_2)&amp;\text{ is exact }\forall\text{ }K_1,K_2\subseteq X\\ \varinjlim_{\begin{smallmatrix}K\subseteq U\cr U\textrm{ open}\end{smallmatrix}}\mathcal F(\overline U)\to\mathcal F(K)&amp;\text{ is an isomorphism }\forall\text{ }K\subseteq X \end{align}</code></p> <p>I would very much like to not have to rewrite the entire foundations of sheaf theory in this context! I'm not sure how much the reader would appreciate such an exposition either. Granted, I would only need to write the foundations I am using, but even this turns out to require many proofs which in retrospect are mostly trivial applications of compactness and direct limit arguments.</p> <p>Is there a good reference for sheaves on the category of compact subsets somewhere in the literature? I hope that there is a book which develops the relevant foundations (perhaps by comparing this notion to the more familiar notion of sheaves on open subsets) so I don't have to include too much baggage explaining the basics of such sheaves.</p> <p>To be more specific about what I mean by "the basics of sheaf theory", I mean something like:</p> <ul> <li>Sections are determined by their stalks</li> <li>Functors $i_\ast$ and $j_!$ (for inclusions of closed and open subsets respectively)</li> <li>A theory of Cech cohomology</li> </ul> <p>I don't need anything about injective resolutions or making the category of sheaves into an abelian category.</p> http://mathoverflow.net/questions/129321/does-a-topological-manifold-have-an-exhaustion-by-compact-submanifolds-with-bou Does a *topological* manifold have an exhaustion by compact submanifolds with boundary? unknown (google) 2013-05-01T15:56:20Z 2013-05-02T08:31:15Z <p>If $M$ is a connected smooth manifold, then it is easy to show that there is a sequence of connected compact smooth submanifolds with boundary $M_1\subseteq M_2\subseteq\cdots$ such that $M=\bigcup_{i=1}^\infty(M_i)^\circ$.</p> <p>I would guess it should also be true that if $M$ is a connected <strong>topological</strong> manifold then there is a sequence of locally tame connected compact submanifolds with boundary $M_1\subseteq M_2\subseteq\cdots$ such that $M=\bigcup_{i=1}^\infty(M_i)^\circ$. How would one try to prove such a statement? The only proof I know of the statement in the smooth category is to start with any exhaustion by open sets with compact closure and then "smooth" their boundaries. However, modifying an open set in a topological manifold so that its boundary is a tamely embedded codimension 1 submanifold seems much more delicate (and perhaps there is even an obstruction to doing it!).</p> http://mathoverflow.net/questions/98915/are-there-maass-forms-where-the-expected-galois-representation-is-ell-adic Are there Maass forms where the expected Galois representation is $\ell$-adic? unknown (google) 2012-06-05T22:22:31Z 2013-04-05T02:45:05Z <p>Recall that by theorems of Deligne and Deligne--Serre, there is the following dichotomy:</p> <ol> <li><p>Modular forms on the upper half plane of level $N$ and weight $k\geq 2$ correspond to representations $\rho:\operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q)\to\operatorname{GL}(2,F_\lambda)$ for some number field $F$ and prime $\lambda$ over $\ell$.</p></li> <li><p>Modular forms on the upper half plane of level $N$ and weight $k=1$ correspond to representations $\rho:\operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q)\to\operatorname{GL}(2,\mathbb C)$ satisfying $\det\rho(\sigma)=-1$ ($\sigma$ is complex conjugation).</p></li> </ol> <p>Now I hear that Maass forms of eigenvalue $\frac 14$ are conjectured to correspond with representations $\rho:\operatorname{Gal}(\bar{\mathbb Q}/\mathbb Q)\to\operatorname{GL}(2,\mathbb C)$ satisfying $\det\rho(\sigma)=1$ ($\sigma$ is complex conjugation). Is this still true (that is, conjectured) for Maass forms of higher weight? Or do they "turn $\ell$-adic" in higher weight?</p> http://mathoverflow.net/questions/126350/how-to-call-a-simplicial-set-where-every-boundary-of-a-simplex-can-be-filled How to call a simplicial set where every boundary of a simplex can be filled? unknown (google) 2013-04-03T05:27:12Z 2013-04-03T06:00:23Z <p>What is the correct terminology for the following property of a simplicial set $X_\bullet$:</p> <blockquote> <p>For every $k\geq 0$, every map $\partial\Delta^k\to X_\bullet$ can be extended to a map $\Delta^k\to X_\bullet$.</p> </blockquote> http://mathoverflow.net/questions/123198/can-the-sl-2-character-variety-of-a-three-manifold-be-nonreduced Can the SL_2 character variety of a three-manifold be nonreduced? unknown (google) 2013-02-28T06:45:33Z 2013-03-18T20:07:15Z <p>Let $M^3$ be a three-manifold and consider the representation variety and the character variety of $M$: $$Y=\operatorname{Hom}(\pi_1(M^3),\operatorname{SL}(2,\mathbb C))$$ $$X=\operatorname{Hom}(\pi_1(M^3),\operatorname{SL}(2,\mathbb C))//\operatorname{SL}(2,\mathbb C)$$ I would like to think of these as schemes. They are usually singular; in fact the trivial representation is almost always a singular point.</p> <p>Are there any known examples where $Y$ (or $X$) are nonreduced as schemes?</p> <p>I really want to know the answer for $\pi_1(M^3)$; examples with just any finitely presented group would be less interesting.</p> http://mathoverflow.net/questions/124155/if-x-is-a-haussdorf-topological-space-and-r-and-equivalence-relation-on-x-when-i/124163#124163 Answer by unknown (google) for If X is a Haussdorf topological space and R and equivalence relation on X, when is X/R Haussdorf? unknown (google) 2013-03-10T17:01:45Z 2013-03-10T17:01:45Z <p>If $X$ is Hausdorff <em>and</em> the quotient map $X\to X/R$ is open, then $X/R$ is Hausdorff if and only if $R\subseteq X\times X$ is closed (see <a href="http://math.stackexchange.com/questions/91639/x-sim-is-hausdorff-if-and-only-if-sim-is-closed-in-x-times-x" rel="nofollow">http://math.stackexchange.com/questions/91639/x-sim-is-hausdorff-if-and-only-if-sim-is-closed-in-x-times-x</a>).</p> http://mathoverflow.net/questions/122605/do-the-solutions-of-the-maurer-cartan-equation-form-a-simplicial-set Do the solutions of the Maurer--Cartan equation form a simplicial set? unknown (google) 2013-02-22T05:53:07Z 2013-03-10T13:30:57Z <p>The Maurer--Cartan equation is the equation: $$d\gamma+\frac 12[\gamma,\gamma]=0$$ where $\gamma$ represents a degree one element in a differential graded Lie algebra $\mathfrak g^\ast$. Let's denote the set of solutions by $MC(\mathfrak g^\ast)$. I need to have some notion of (higher) homotopies between elements of the set $MC(\mathfrak g^\ast)$. One way of doing this would be to define a simplicial set $\mathsf{MC}(\mathfrak g^\ast)$ whose zero simplices are the set $MC(\mathfrak g^\ast)$. I have indeed seen the phrase "simplicial set of solutions to the Maurer--Cartan equation" in papers. Is there a standard construction of this simplicial set? If so, how should I think about it, and what are some good references?</p> <p>In fact, it seems that the $n$-simplices in $\mathsf{MC}(\mathfrak g^\ast)$ should be $MC(\mathfrak g^\ast\otimes\Omega^\ast(\Delta^n))$ where $\Omega^\ast(\Delta^n)$ is the differential graded algebra of differential forms on the standard simplex $\Delta^n$. Can I use something smaller (hopefully finite dimensional) instead of $\Omega^\ast(\Delta^n)$? Perhaps just the simplicial cochain complex $C^\ast(\Delta^n)$?</p> http://mathoverflow.net/questions/123694/is-the-free-abelian-group-on-an-affine-scheme-represented-by-an-ind-scheme Is the free abelian group on an affine scheme represented by an ind-scheme? unknown (google) 2013-03-06T02:15:24Z 2013-03-06T17:06:25Z <p>Let $X/\mathbb C$ be an affine scheme of finite type, and let $\mathbb Z[X(\mathbb C)]$ be the free abelian group generated by the $X(\mathbb C)$.</p> <blockquote> <p>Can the elements of $\mathbb Z[X(\mathbb C)]$ be identified with the $\mathbb C$-points of an ind-scheme $\mathcal X$ over $\mathbb C$ so that the natural map $(X^n\times X^m)(\mathbb C)\to\mathbb Z[X(\mathbb C)]$ given by $(x_1,\ldots,x_n,y_1,\ldots,y_m)\mapsto\sum x_i-\sum y_i$ lifts to a map $X^n\times X^m\to\mathcal X$?</p> </blockquote> <p>There is an obvious strategy to construct such an ind-scheme, but I don't know whether it works. For ease of notation, let us observe that there is an exact sequence: $$0\to\mathbb Z[X(\mathbb C)]_0\to\mathbb Z[X(\mathbb C)]\xrightarrow\epsilon\mathbb Z\to 0$$ where $\epsilon$ is simply the sum of all the coefficients. Thus it suffices to describe $\mathbb Z[X(\mathbb C)]_0$ as the $\mathbb C$-points of an ind-scheme over $\mathbb C$. Now we look at the map $(X^n\times X^n)(\mathbb C)\to\mathbb Z[X(\mathbb C)]_0$. If we are lucky (this is probably where $X$ being affine is helpful), then perhaps there is a scheme $Y_n$ and a surjection $X^n\times X^n\to Y_n$ so that the map above factors as: $$(X^n\times X^n)(\mathbb C)\to Y_n(\mathbb C)\to\mathbb Z[X(\mathbb C)]_0$$ where the second map is injective. Here the coordinate ring of $Y_n$ should be a subring of the coordinate ring of $X^n\times X^n$, and it is not immediately clear that the natural choice would be of finite type. Hopefully there are then natural inclusions $Y_1\hookrightarrow Y_2\hookrightarrow\cdots$ giving the desired ind-scheme.</p> http://mathoverflow.net/questions/121856/how-to-specify-a-finite-group-up-to-inner-automorphism How to specify a finite group up to inner automorphism? unknown (google) 2013-02-15T00:58:48Z 2013-03-01T06:02:14Z <p>I want some finite set of data to which I can canoically associate a "group up to inner automorphism", and which can be constructed canoically from a "group up to inner automorphism". I have a few answers which satisfy the first requirement, but not the second.</p> <ol> <li><p><strong>Give a topological space $X$.</strong> The fundamental group of $X$ is a "group up to inner automorphism": you can get a group by picking a basepoint $p$, but given two different basepoints $p$ and $q$, there is not a canonical isomorphism $\pi_1(X,p)\to\pi_1(X,q)$, but any two automorphisms coming from a path from $p$ to $q$ differ by an inner automorphism. This answer is unsatisfying because there are many topological spaces one can pick, even if we restrict them to be $K(\pi,1)$'s.</p></li> <li><p><strong>Give a groupoid $X$</strong> (this generalizes the first example by taking the fundamental groupoid). This answer is unsatisfying because again there are many (say, finite) groupoids corresponding to a given group up to inner automorphism.</p></li> <li><p><strong>Give a tensor category which is isomorphic to the category of finite dimensional representations of a finite group over an algebraically closed field of characteristic zero.</strong> This is almost a good answer. It looks at first that specifying such a category is a finite amount of combinatorial data: we specify the isomorphism classes of objects and specify how the tensor product of any two of them decomposes. However we also have to specify isomorphisms $(A\otimes B)\otimes C\to A\otimes(B\otimes C)$, and this means specifying some matrix which depends on the specific bases of the isotypic components of the tensor products that we chose when specifying how they decompose. So it is unsatisfying as well, although it is the best I have come up with. It is almost good enough since it is easy to say what an equivalence between two such finite categories is in terms of matrices.</p></li> <li><p><strong>Give an orbifold whose coarse space is a single point.</strong> This answer is unsatisfying because I am hoping the answer to this question will be useful for giving a nice definition of an orbifold. Also, by unraveling the definition of an orbifold, this answer really just reduces to (1) or (2) and is thus also unsatisfying.</p></li> </ol> http://mathoverflow.net/questions/122746/cech-cohomology-as-a-colimit-over-maps-to-a-cw-complex Cech cohomology as a colimit over maps to a CW complex unknown (google) 2013-02-23T19:07:20Z 2013-02-25T13:32:17Z <p>Given a topological space $X$, we consider the following category <code>$\mathsf{CW}_{X\to}$</code>. The objects are finite CW complexes $Y$ equipped with a continuous map $X\to Y$. The morphisms are continuous maps $Y\to Y'$ such that the composition $X\to Y\to Y'$ agrees with $X\to Y'$. Now there is a functor: <code>$$\mathsf{CW}_{X\to}\to\mathsf{AbGrp}$$</code> given by taking the cohomology of $Y$. I wish to consider the colimit of this functor: <code>$$\operatorname{colim}H^\ast(\mathsf{CW}_{X\to})$$</code></p> <p>Now let's assume that $X$ is compact Hausdorff. I believe that in this case <code>$\operatorname{colim}H^\ast(\mathsf{CW}_{X\to})$</code> is naturally isomorphic to the Cech cohomology of $X$. However, I am a bit unsure as to whether I should take the category <code>$\mathsf{CW}_{X\to}$</code> as I have defined it above, or "soften" it by taking only <em>homotopy classes</em> of maps $X\to Y$ and <em>homotopy classes</em> of maps $Y\to Y'$ (maybe we could call this category $\mathsf{CW}_{X\to}^h$).</p> <blockquote> <p>How should I think about the relationship betwee <code>$\operatorname{colim}H^\ast(\mathsf{CW}_{X\to})$</code> and <code>$\operatorname{colim}H^\ast(\mathsf{CW}_{X\to}^h)$</code>?</p> </blockquote> <p>Is one easier to deal with theoretically (e.g. by virtue of being filtered)? Are they indeed isomorphic via the natural map: <code>$$\operatorname{colim}H^\ast(\mathsf{CW}_{X\to}^h)\to\operatorname{colim}H^\ast(\mathsf{CW}_{X\to})$$</code> induced by the forgetful functor <code>$\mathsf{CW}_{X\to}\to\mathsf{CW}_{X\to}^h$</code>?</p> http://mathoverflow.net/questions/121857/what-space-classifies-bundles-of-kpi-1s What space classifies bundles of K(pi,1)'s? unknown (google) 2013-02-15T01:08:58Z 2013-02-16T11:49:19Z <p>Given a discrete group $G$ (<strong>not</strong> assumed to be abelian), is there a nice construction of a topological space which classifies bundles of $K(G,1)$'s? I guess I should take something like $B\operatorname{Aut}(K(\pi,1))$ where $\operatorname{Aut}(K(\pi,1))$ is the monoid of self homotopy equivalences of some fixed $K(\pi,1)$. I don't know if this makes any sense, though.</p> <p>Morally speaking, things are defined as follows. A bundle of $K(\pi,1)$'s over a space $X$ is a fibration $Y\to X$ whose fibers are $K(\pi,1)$'s. Two bundles of $K(\pi,1)$'s $Y_1,Y_2\to X$ are said to be isomorphic iff there is a map $Y_1\to Y_2$ of spaces over $X$ which is a homotopy equivalence on each fiber.</p> http://mathoverflow.net/questions/68474/how-to-calculate-the-witten-reshetikhin-turaev-invariants-from-a-triangulation How to calculate the Witten-Reshetikhin-Turaev invariants from a triangulation? unknown (google) 2011-06-22T03:34:03Z 2012-12-12T21:13:51Z <p>I'm interested in the Witten-Reshetikhin-Turaev invariants for 3-manifolds, and in particular, how to calculate them from a <em>triangulation</em> 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?</p> <p>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.</p> <p>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.</p> <p>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 $M$.</p> http://mathoverflow.net/questions/115528/cobordism-of-orbifolds Cobordism of orbifolds? unknown (google) 2012-12-05T18:48:07Z 2012-12-05T23:24:18Z <p>Is it possible to setup classical cobordism theory in the context of orbifolds? For example, let's consider the free abelian group generated by oriented smooth orbifolds and quotient by those which are null cobordant (that is, they are the boundary of a smooth oriented orbifold with boundary). Is the resulting ring expressible as the homotopy groups of some Thom spectrum? Can we prove this just by smoothly embedding an orbifold in $\mathbb R^n/S_n$ and following the classical proof?</p> <p>My motivation is that in Gromov--Witten theory, the moduli space of stable maps $\bar M_{g,n}(X,A)$ is a smooth oriented orbifold (assuming it is cut out transversally, and assuming we have smooth charts for gluings) and it is defined up to cobordism. Thus instead of taking its fundamental class and pushing forward to $H_\ast(\bar M_{g,n}\times X^n)$ to get Gromov--Witten invariants, we could consider the class it represents in the generalized cohomology theory which we might call "oriented orbifold cobordism" of $\bar M_{g,n}\times X^n$, and get a slightly more refined invariant.</p> http://mathoverflow.net/questions/99413/journals-and-other-sources-with-easy-reading-papers/115404#115404 Answer by unknown (google) for Journals and other sources with "easy reading" papers ? unknown (google) 2012-12-04T15:01:39Z 2012-12-04T15:01:39Z <p>I am surprised that no one has mentioned the expository articles in the <a href="http://www.ams.org/publications/journals/journalsframework/bull" rel="nofollow">Bulletin of the AMS</a>, which are usually excellent.</p> http://mathoverflow.net/questions/114555/does-physics-need-non-analytic-smooth-functions/114568#114568 Answer by unknown (google) for Does Physics need non-analytic smooth functions? unknown (google) 2012-11-26T18:33:00Z 2012-11-26T18:33:00Z <p>In the field of mathematical general relativity, certain uniqueness results for black holes are known for real-analytic space-time but are open for smooth space-time. For example, the "No hair conjecture" was proven by Stephen Hawking for analytic space-time but is open in general. For more details, see slides by Klainerman:</p> <p>www.ihes.fr/~vanhove/Slides/Klainerman-ihes-fev2011.pdf</p> <p>(page 24 and on)</p> http://mathoverflow.net/questions/113794/non-cyclotomic-abelian-extensions/113796#113796 Answer by unknown (google) for Non-cyclotomic abelian extensions unknown (google) 2012-11-19T05:18:00Z 2012-11-19T17:34:42Z <p><strike>Pick some $\gamma_1\in L\setminus\mathbb Q$ which is not a square.</strike> Pick some $\gamma\in L^\times/(L^\times)^2$ which is not fixed by $\operatorname{Gal}(L/\mathbb Q)$ and fix a lift $\gamma_1\in L$. Let $\gamma_1,\ldots,\gamma_n$ be the orbit of $\gamma_1$ under $\operatorname{Gal}(L/\mathbb Q)$. Then it is an easy exercise in Galois theory to show that:</p> <ol> <li>$L(\sqrt{\gamma_1},\ldots,\sqrt{\gamma_n})/L$ is Galois with abelian Galois group $\subseteq(\mathbb Z/2\mathbb Z)^n$.</li> <li>$L(\sqrt{\gamma_1},\ldots,\sqrt{\gamma_n})/\mathbb Q$ is Galois with nonabelian Galois group $\subseteq\operatorname{Gal}(L/\mathbb Q)\ltimes(\mathbb Z/2\mathbb Z)^n$.</li> </ol> http://mathoverflow.net/questions/112024/how-does-one-go-from-chern-weil-to-cohomology-classes-on-bgln-c How does one go from Chern--Weil to cohomology classes on BGL(n,C)? unknown (google) 2012-11-10T21:06:34Z 2012-11-11T15:31:28Z <p>Let's assume we start with Chern--Weil theory in the following form:</p> <blockquote> <p>Given a manifold $M$ and a complex vector bundle $V$ over $M$, we can equip $V$ with a $\mathfrak g\mathfrak l_n(\mathbb C)$ connection and from this connection compute a closed differential $2k$-form (from the curvature of the connection) which thus determines an element $c_k(V)\in H^{2k}(M,\mathbb C)$ (by deRham theory). This value is independent of the connection chosen. If $f:N\to M$ is a smooth map, then $c_k(f^\ast V)=f^\ast c_k(V)$.</p> </blockquote> <p>I've often heard that Chern--Weil theory gives cohomology classes $c_k\in H^{2k}(B\operatorname{GL}_n(\mathbb C),\mathbb C)$. However, if we take Chern--Weil theory to mean the above boxed summary, then this does not seem obvious to me unless we use the fact that we can choose $B\operatorname{GL}_n(\mathbb C)$ to be a direct limit of manifolds (namely Grassmannians).</p> <p>My question is whether there is more abstract way of constructing the classes $c_k\in H^{2k}(B\operatorname{GL}_n(\mathbb C),\mathbb C)$ from Chern--Weil theory, using only the fact that $B\operatorname{GL}_n(\mathbb C)$ is the classifying space of complex vector bundles. I am fine with assuming that $B\operatorname{GL}_n(\mathbb C)$ is a direct limit of finite CW complexes, but of course, Chern--Weil theory does not obviously define Chern classes for vector bundles over CW complexes.</p> http://mathoverflow.net/questions/112024/how-does-one-go-from-chern-weil-to-cohomology-classes-on-bgln-c/112038#112038 Answer by unknown (google) for How does one go from Chern--Weil to cohomology classes on BGL(n,C)? unknown (google) 2012-11-10T22:40:20Z 2012-11-10T22:40:20Z <p>Expanding on Will Sawin's comment:</p> <p>Every finite CW complex is homotopy equivalent to a finite simplicial complex (by an approximation argument) and a simplicial complex is homotopy equivalent to a manifold (take a regular neighborhood of the natural geometric realization in <code>$\mathbb R^{\#\text{vertices}}$</code>). The naturality of the "very fine approximation of a CW complex by a simplicial complex" means that we get well-defined classes $c_k(V)\in H^{2k}(X,\mathbb C)$ for any complex vector bundle $V$ over a finite CW complex $X$ (alternatively, take two small manifold "extensions" of $X$; we can embed these in a third manifold extension, so they give the same cohomology class $c_k(V)$). These classes satisfy $f^\ast c_k(V)=c_k(f^\ast V)$ for $f:Y\to X$ where $Y$ and $X$ are both finite CW complexes because we can extend the map $f$ to the approximation/regular neighborhood.</p> <p>Now we have $c_k$ as characteristic classes of complex vector bundles over finite CW complexes which satisfy naturality (pullback), and so abstract nonsense about representability implies they come from unique $c_k\in H^{2k}(B\operatorname{GL}_n(\mathbb C),\mathbb C)$ (which we can construct by filtering $B\operatorname{GL}_n(\mathbb C)$ by finite CW complexes in any way we like).</p> http://mathoverflow.net/questions/109645/can-stabilizer-groups-in-an-orbifold-have-global-twisting Can stabilizer groups in an orbifold have global twisting? unknown (google) 2012-10-14T19:52:13Z 2012-10-15T13:38:18Z <p>Can stabilizer groups in an orbifold have global twisting?</p> <p>For example, consider the two groups $\mathbb Z/3\times\mathbb Z$ and $\mathbb Z/3\rtimes\mathbb Z$ (where $\mathbb Z\to\operatorname{Aut}(\mathbb Z/3)$ is the unique nontrivial map). Both groups act on $\mathbb R$ through their common quotient $\mathbb Z$, so there are two orbifolds here: $$M_1=[\mathbb R/(\mathbb Z/3\times\mathbb Z)]$$ $$M_2=[\mathbb R/(\mathbb Z/3\rtimes\mathbb Z)]$$ The coarse spaces of $M_1$ and $M_2$ are both $S^1=\mathbb R/\mathbb Z$, and in both $M_1$ and $M_2$, all points have stabilizer group isomorphic to $\mathbb Z/3$. Are $M_1$ and $M_2$ isomorphic as orbifolds?</p> <p><strong>Why they should be different:</strong> Over any orbifold $X$, we think of there being space $E$ with a "nice" map $\pi:E\to X$ such that <code>$\pi^{-1}(\{x\})$</code> is a $K(\Gamma_x,1)$ (see for example <a href="http://arxiv.org/abs/math/0112006" rel="nofollow">this paper</a> by André Henriques). In the case of $M_1$ and $M_2$, this just means we have a $K(\mathbb Z/3,1)$ bundle over $S^1$, and its monodromy is a homomorphism $\mathbb Z\to\operatorname{Aut}(\mathbb Z/3)$ which should recover the difference between $\mathbb Z/3\times\mathbb Z$ and $\mathbb Z/3\rtimes\mathbb Z$.</p> <p>This twisting is a subtle point that I don't see mentioned explicitly anywhere, and I didn't realize it could happen until now, when I actually have to do something with orbifolds and the precise definition becomes important.</p> <p>End note: The answers to these related questions have very good answers in regards to defining orbifolds.</p> <p><a href="http://mathoverflow.net/questions/1861/looking-for-an-introduction-to-orbifolds" rel="nofollow">http://mathoverflow.net/questions/1861/looking-for-an-introduction-to-orbifolds</a></p> <p><a href="http://mathoverflow.net/questions/19530/what-is-meant-by-smooth-orbifold" rel="nofollow">http://mathoverflow.net/questions/19530/what-is-meant-by-smooth-orbifold</a></p> http://mathoverflow.net/questions/108384/beta-mathbbn-vs-beta-mathbbz/108386#108386 Answer by unknown (google) for $\beta\mathbb{N}$ vs $\beta\mathbb{Z}$ unknown (google) 2012-09-29T05:20:26Z 2012-09-29T05:20:26Z <p>Since $\mathbb N$ and $\mathbb Z$ are homeomorphic, so are $\beta\mathbb N$ and $\beta\mathbb Z$, though of course the semigroup structure will be different.</p> http://mathoverflow.net/questions/106717/sarkars-maslov-index-formula/107431#107431 Answer by unknown (google) for Sarkar's Maslov index formula unknown (google) 2012-09-18T02:11:35Z 2012-09-18T02:11:35Z <p>(1) Yes, $p_1$ and $p_n$ are treated different from other points. If you cyclically permute the indices <code>$\{1,\ldots,n\}$</code> you do of course get the same answer, but it's not immediately obvious from Sarkar's formula that this is the case.</p> <p>(2) The Euler measure of a region $R$ equals:<code>$$e(R):=\chi(R)-\frac 14\#\{\text{acute corners}\}+\frac 14\#\{\text{obtuse corners}\}$$</code>Usually one just talks about regions which are disks, so $\chi(R)=1$. Now every connected component of $\Sigma_g\setminus(\eta_1^1\cup\cdots\cup\eta_g^1\cup\cdots\cup\eta_1^k\cup\cdots\eta_g^k)$ has zero obtuse corners. With these two facts in hand one reduces to Sarkar's formula $1-\frac n4$ for an $n$-sided polygon.</p> <p>(3) The arcs in $\partial_iD$ all lie inside the $\boldsymbol\eta^i$ circles, and they are considered moved in "both directions" along the $\boldsymbol\eta^i$ circles. If you indulge me as I create ugly ascii art, I can give a few examples. Here $|$ denotes $\partial_iD$ and $=$ denotes $\partial_jD$.</p> <p>The following has intersection $1$: <code>$$\begin{matrix} &amp;|&amp;\cr &amp;|&amp;\cr =&amp;=&amp;=\cr &amp;|&amp;\cr &amp;|&amp;\cr \end{matrix}$$</code> The following both have intersection $\frac 12$: <code>$$\begin{matrix} &amp;|&amp;&amp;&amp;&amp;|\cr &amp;|&amp;&amp;&amp;&amp;|\cr =&amp;=&amp;&amp;&amp;=&amp;=&amp;=\cr &amp;|&amp;&amp;&amp;&amp;\cr &amp;|&amp;&amp;&amp;&amp;\cr \end{matrix}$$</code> The following all have intersection $\frac 14$: <code>$$\begin{matrix} &amp;|&amp;&amp;&amp;&amp;|\cr &amp;|&amp;&amp;&amp;&amp;|\cr =&amp;=&amp;&amp;&amp;&amp;=&amp;=\cr &amp;&amp;&amp;&amp;&amp;\cr\cr\cr &amp;&amp;&amp;&amp;&amp;\cr &amp;&amp;&amp;&amp;&amp;\cr &amp;&amp;&amp;&amp;&amp;\cr =&amp;=&amp;&amp;&amp;&amp;=&amp;=\cr &amp;|&amp;&amp;&amp;&amp;|\cr &amp;|&amp;&amp;&amp;&amp;|\cr \end{matrix}$$</code></p> http://mathoverflow.net/questions/105756/is-sl2-c-sl2-z-a-quasi-projective-variety Is SL(2,C)/SL(2,Z) a quasi-projective variety? unknown (google) 2012-08-28T20:27:22Z 2012-08-31T09:16:52Z <p>Consider the complex 3-fold $SL(2,\mathbb C)/SL(2,\mathbb Z)$ (just for clarity: note that $SL(2,\mathbb Z)$ acts without stabilizers, so this is a complex manifold, not a complex orbifold).</p> <blockquote> <p>Is $SL(2,\mathbb C)/SL(2,\mathbb Z)$ a quasi-projective variety?</p> </blockquote> <p>The natural generalization of this question seems to be the following. Let $G$ be a semisimple linear algebraic group over $\mathbb Q$. Then $G(\mathbb Z)$ is well-defined up to taking a finite-index subgroup. Thus we can ask the same question: is the complex manifold $G(\mathbb C)/G(\mathbb Z)$ a quasi-projective variety?</p> http://mathoverflow.net/questions/104965/non-regular-connected-hausdorff-banach-manifold/105075#105075 Answer by unknown (google) for Non-regular Connected Hausdorff Banach Manifold unknown (google) 2012-08-20T05:03:51Z 2012-08-20T16:21:03Z <p>I'm confused as to why (failure of) local compactness is an issue. I think one can just prove this directly as follows. It's rather trivial, though, so maybe I'm missing something.</p> <p>Let $x\in X$ be a point and $K\subseteq X$ a closed subset with $x\notin K$. Find some coordinate chart near $x$, that is, a map $f:E\to X$ which is a homeomorphism onto some open set containing $x$, where $E$ is a Banach space and $f(0)=x$. Now $f^{-1}(K)$ is a closed subset of $E$ which does not contain $0$. Thus there exists $\epsilon>0$ such that $\|k\|\geq\epsilon$ for all $k\in f^{-1}(K)$. Let $B(0,a)$ denote the open ball in $E$ centered at $0$ with radius $a$. Then we simply observe that the following two open sets are disjoint and "separate" $x$ and $K$: $$x\in f(B(0,\epsilon/3))$$ $$K\subseteq X\setminus f(\overline{B(0,2\epsilon/3)})$$ I'm using Hausdorffness to conclude that $f(\overline{B(0,2\epsilon/3)})$ is closed in $X$.</p> <p><strong>EDIT</strong>: As the comments point out, the point here is exactly to show that $f(\overline{B(0,2\epsilon/3)})$ is closed, which in finite dimensions (i.e. locally compact) follows easily from Hausdorfness. So, as of now this argument is incomplete.</p> http://mathoverflow.net/questions/104320/equivariant-handle-decompositions Equivariant handle decompositions unknown (google) 2012-08-09T02:52:49Z 2012-08-09T15:43:31Z <p>Suppose I have some smooth closed <strong>high-dimensional</strong> manifold $M$ acted on smoothly by a <strong>finite</strong> group $G$. By a metric averaging procedure, we can equip $M$ with a smooth Riemannian metric so that $G$ acts by isometries. I can't necessarily pick a $G$-invariant morse function $f:M\to\mathbb R$, but nevertheless, I can certainly pick a smooth function $f:M\to\mathbb R$ which, though perhaps not Morse, still has only isolated "nice" critical points in some precise sense. We therefore conclude:</p> <blockquote> <p>There is a "handle" decomposition of $M$ (where I haven't said what I mean by "handle") which is preserved by $G$. Thus $G$ just permutes (and/or acts on individually) the handles.</p> </blockquote> <p>I am interested in knowing to what extent this can be generalized to the case of an "action up to homotopy". More specifically, suppose we have $G\to\operatorname{Homeo}(M)/\text{homotopy}$. To what extent can we "decompose" M into simple pieces in a $G$-invariant way? If it helps, then it is OK to assume that the action of $G$ is "close to the identity" in a vague coarse sense.</p> <p>(I am essentially just interested on the case of high-dimensional $M$, but of course the question makes sense in any dimension).</p> http://mathoverflow.net/questions/104308/do-the-following-set-of-dehn-twists-generate-the-mapping-class-group Do the following set of Dehn twists generate the mapping class group? unknown (google) 2012-08-08T21:28:21Z 2012-08-09T10:22:23Z <blockquote> <p>If $S$ is the surface illustrated below, do the Dehn twists about the red curves generate the mapping class group $\operatorname{MCG}(S,\partial S)$?</p> </blockquote> <p><img src="http://math.stanford.edu/~pardon/img/dehntwists.png"/></p> http://mathoverflow.net/questions/104254/rank-two-vector-bundles-on-a-curve-of-genus-two Rank two vector bundles on a curve of genus two unknown (google) 2012-08-08T03:50:49Z 2012-08-08T06:41:04Z <p>I recently learned of an <a href="http://www.ams.org/mathscinet-getitem?mr=242185" rel="nofollow">interesting result</a> of Narasimhan and Ramanan from 1969, which says that moduli space of rank two vector bundles with trivial determinant on a curve $X$ of genus two is naturally isomorphic to $\mathbb PH^0(\operatorname{Pic}^1X,\mathcal L_\Theta^{\otimes 2})$ (this vector space has dimension four). I'd like to understand this isomorphism better in the context of the <a href="http://www.ams.org/mathscinet-getitem?mr=0184252" rel="nofollow">Narasimhan--Seshadri theorem</a>.</p> <p>First, fix a closed <strong>topological</strong> surface $X$ of genus two. Let: $$M_{X,SU(2)}=\operatorname{Hom}(\pi_1(X),SU(2))//SU(2)$$ denote the $SU(2)$ character variety of $X$. For a complex structure $\sigma$ on $X$, let $M_{X,\sigma,\operatorname{rk}2}$ denote the moduli space of rank two vector bundles on $X$ with trivial determinant (actually, there is a technical stability/equivalence relation which should be included, but I will ignore this). According to the <a href="http://www.ams.org/mathscinet-getitem?mr=0184252" rel="nofollow">Narasimhan--Seshadri theorem</a>, $M_{X,SU(2)}$ and $M_{X,\sigma,\operatorname{rk}2}$ are naturally diffeomorphic (again, there are some qualifications to this which I am ignoring; in particular both spaces are usually singular).</p> <p>Now I want to recall the isomorphism $M_{X,\sigma,\operatorname{rk}2}\to\mathbb PH^0(\operatorname{Pic}^1X,\mathcal L_\Theta^{\otimes 2})$. To any rank two vector bundle $E$, we consider the subvariety $C_E$ of $\operatorname{Pic}^1X$ consisting of bundles $\xi$ for which there is an exact sequence: $$0\to\xi\to E\to\xi^{-1}\to 0$$ (that is, $E$ is an extension of $\xi^{-1}$ and $\xi$). Then Narasimhan and Ramanan prove that $C_E$ is a divisor on $\operatorname{Pic}^1X$ and is linearly equivalent to $2\Theta$. This gives a map $M_{X,\sigma,\operatorname{rk}2}\to\mathbb PH^0(\operatorname{Pic}^1X,\mathcal L_\Theta^{\otimes 2})$, which Narasimhan and Ramanan go on to show is an isomorphism. (That was only a rough outline).</p> <p>OK, now let's reinterpret the map $M_{X,\sigma,\operatorname{rk}2}\to\mathbb PH^0(\operatorname{Pic}^1X,\mathcal L_\Theta^{\otimes 2})$ in terms of the Narasimhan-Seshadri theorem. Remember that $\operatorname{Jac}X$ is $\operatorname{Hom}(\pi_1(X),U(1))$. Thus for a homomorphism $\rho:\pi_1(X)\to SU(2)$ (corresponding to a vector bundle of rank two), we can define the subvariety $C_E$ as those $U(1)$ representations $\alpha:\pi_1(X)\to U(1)$ for which $\rho$ can be conjugated to have the form: $$\left(\begin{matrix}\alpha&amp;\beta\cr 0&amp;\alpha^{-1}\end{matrix}\right)$$ This is a <strong>subset</strong> of $\operatorname{Hom}(\pi_1(X),U(1))=\operatorname{Jac}X$. Now according to Narasimhan and Ramanan, it should be a <strong>subvariety</strong> of $\operatorname{Jac}X$ for <em>any</em> complex structure on $X$. This seems a bit unlikely to me, because there is a large moduli of complex structures on $X$. Also, somehow I've constructed naturally $C_E\subseteq\operatorname{Jac}X$, but according to the construction in Narasimhan and Ramanan I should be getting $C_E\subseteq\operatorname{Pic}^1X$, which is really not the same thing canonically.</p> <p>I suppose I'm getting confused in applying the Narasimhan-Seshadri theorem. Any assistance is appreciated!</p> http://mathoverflow.net/questions/67764/why-does-the-parameterization-ff1-happen Why does the parameterization (F:F':1) happen? unknown (google) 2011-06-14T14:09:10Z 2012-08-04T10:43:19Z <p>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 $\cos$).</p> <p>2) To parameterize an elliptic curve $y^2=4x^3-g_2x-g_3$, we can use $(x,y)=(\wp(t),\wp'(t))$.</p> <p>(I know how to prove (1) and (2), and that it is possible to parameterize curves of higher genus using the unit disc.)</p> <p>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?</p> http://mathoverflow.net/questions/103146/terminology-for-a-cyclically-ordered-set-of-objects Terminology for a cyclically ordered set of objects unknown (google) 2012-07-26T02:20:19Z 2012-07-26T14:25:12Z <p>If I have an ordered set of objects (for concreteness, say they're integers) $(x_1,\ldots,x_n)$, I might call it a <em>tuple of integers</em>.</p> <p>Perhaps, though, I have an set of integers $(x_1,\ldots,x_n)$ but the order is only defined up to cyclical permutation (imagine they're sitting at distinct points along a circle); so $(x_1,\ldots,x_n)$ is the same as $(x_2,\ldots,x_n,x_1)$. I thus can't call this a "tuple of integers" because that would imply there is a canonical ordering. Is there some standard term I can use, besides the unweildy phrase "cyclically ordered set of integers"?</p> http://mathoverflow.net/questions/131322/are-small-knots-generic Comment by 2013-05-22T15:55:27Z 2013-05-22T15:55:27Z @Misha and Julien: why not just condition on the random knot being hyperbolic (or prime, etc.)? [concretely: pick random knots until you get a hyperbolic/prime one]. http://mathoverflow.net/questions/131322/are-small-knots-generic Comment by 2013-05-21T15:56:14Z 2013-05-21T15:56:14Z For prime knots, I believe this result (<a href="http://www.ams.org/mathscinet-getitem?mr=1304395" rel="nofollow">ams.org/mathscinet-getitem?mr=1304395</a>) implies that (for a certain sense of random knots) most are satellite knots (have an incompressible torus). So I would guess you also want to restrict to hyperbolic knots. Even in that case my guess would be that with lots of crossings you get lots of incompressible surfaces generically. http://mathoverflow.net/questions/131076/complex-morse-function-on-a-four-manifold Comment by 2013-05-18T18:29:13Z 2013-05-18T18:29:13Z There are various notions of being Morse for a function $X\to\mathbb C=\mathbb R^2$. Could you say specifically which you mean? http://mathoverflow.net/questions/130884/reference-request-sheaves-on-closed-sets Comment by 2013-05-17T01:06:47Z 2013-05-17T01:06:47Z @David Carchedi: sure, I edited it to give the precise definition. http://mathoverflow.net/questions/130789/are-the-two-meanings-of-undecidable-related Comment by 2013-05-16T16:20:05Z 2013-05-16T16:20:05Z @Jason Rute: Thanks, I do mean quantified over all $x,y,z,n$ in the second example. http://mathoverflow.net/questions/129759/modern-mathematical-achievements-accessible-to-undergraduates/129767#129767 Comment by 2013-05-07T01:16:16Z 2013-05-07T01:16:16Z Correction: Sabitov's result is that the volume of a polyhedron can be expressed as <b>a root</b> of a polynomial whose <b>coefficients</b> are polynomials in the edge lengths. http://mathoverflow.net/questions/129321/does-a-topological-manifold-have-an-exhaustion-by-compact-submanifolds-with-bou/129330#129330 Comment by 2013-05-01T21:50:28Z 2013-05-01T21:50:28Z This answer doesn't seem to address the question asked . . . http://mathoverflow.net/questions/124470/is-it-possible-to-have-a-stable-3-form-in-dim-7-such-that-the-dimension-of-its-or Comment by 2013-03-14T01:56:24Z 2013-03-14T01:56:24Z Hint: if the orbit is open, then calculating its dimension should be easy . . . http://mathoverflow.net/questions/124292/solution-in-distinct-elements-for-a-system-of-n-equations-over-finite-fields Comment by 2013-03-12T17:16:37Z 2013-03-12T17:16:37Z Your equations specialize to $x_m^2-x_m^2=y_m$ mod $p$, but you want $y_m$ nonzero mod $p$. So there are never any solutions. http://mathoverflow.net/questions/124155/if-x-is-a-haussdorf-topological-space-and-r-and-equivalence-relation-on-x-when-i/124157#124157 Comment by 2013-03-10T17:04:15Z 2013-03-10T17:04:15Z Are you sure? I thought that one needed $X\to X/R$ to be an open map for this to hold. http://mathoverflow.net/questions/123482/is-there-a-constructive-proof-of-cantorbernsteinschroeder-theorem/123484#123484 Comment by 2013-03-03T18:48:46Z 2013-03-03T18:48:46Z This answer seems orthogonal to the original question. The OP states that the proof requires the law of excluded middle. http://mathoverflow.net/questions/122858/an-abelian-hodge-tate-representation-lands-in-a-torus Comment by 2013-02-25T06:38:36Z 2013-02-25T06:38:36Z which I guess in this case is then $\ell$-adic Hodge theory :) http://mathoverflow.net/questions/122092/equivalence-of-two-well-known-forms-of-rh-reference-request Comment by 2013-02-17T19:52:55Z 2013-02-17T19:52:55Z Have you tried writing $\pi(x)=\int_1^x\frac 1{\log x}d\theta(x)$, using integration by parts on the right hand side, and then plugging in (2) for $\theta(x)$? For the other direction, do something similar. http://mathoverflow.net/questions/4214/equivalence-of-grothendieck-style-versus-cech-style-sheaf-cohomology/18674#18674 Comment by 2013-02-15T21:13:59Z 2013-02-15T21:13:59Z @Li Zhan: No, because in Georges Elencwajg's answer, paracompact means also Hausdorff and the Zariski topology here is not Hausdorff http://mathoverflow.net/questions/121856/how-to-specify-a-finite-group-up-to-inner-automorphism Comment by 2013-02-15T07:50:31Z 2013-02-15T07:50:31Z @Dylan Wilson: well, it's not quite localization (since inner automorphisms are already isomorphisms in the category of groups). It's more like I set $\operatorname{Hom}(G_1,G_2)$ to be the set of group homomorphisms quotiented by the equivalence relation where two which differ by an inner automorphism of the target are considered the same.