User nicolas schmidt - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T03:15:12Z http://mathoverflow.net/feeds/user/3824 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/131490/langlands-product Langlands product Nicolas Schmidt 2013-05-22T17:49:18Z 2013-05-22T20:59:43Z <p>In his 'Märchen' Langlands considers for a local field $F$ a certain abelian category $\Pi(F)$ whose objects are given by isomorphisms classes of irreducible admissible representations of $GL_n(F)$, where $n \in \mathbf{N}$ runs over all natural numbers. For $[\pi],[\pi']$ represented by cuspidal reps $\pi,\pi'$ of $GL_n(F)$ and $GL_{n'}(F)$ he introduces the sum $[\pi] \boxplus [\pi]$ as the class of the unique irreducible quotient of the parabolic induction of $\pi\otimes \pi'$ where $GL_n(F)\times GL_{n'}(F)$ is viewed as a Levi component of the obvious standard parabolic of $GL_{n+n'}(F)$. Since he wants to recognize $\Pi(F)$ as the category of representations of some proalgebraic group via the Tannaka formalism he poses the problem of defining a tensor product $[\pi]\boxtimes [\pi']$ as a class of a representation of $GL_{n\cdot n'}(F)$. Moreover $\boxplus$, $\boxtimes$ and $\oplus$, $\otimes$ should correspond to each other under the Langlands correspondence. Of course since it has now been proven one could define $\boxtimes$ via the Local Langlands correspondence.</p> <p>My question is: Is there any known elementary construction of $[\pi]\boxtimes [\pi']$ for any example with $n,n' > 1$ ?</p> http://mathoverflow.net/questions/123358/when-is-this-braiding-not-a-symmetry When is this braiding not a symmetry? Nicolas Schmidt 2013-03-01T17:16:02Z 2013-03-01T22:58:31Z <p>Given a topological space $X$ instead of forming the fundamental groupoid $\pi(X)$ which is the category whose objects are the points and morphisms the homotopy classes of paths one can also form the fundamental 2-groupoid which is the bicategory with objects = points, 1-morphisms = paths and 2-morphisms = homotopy classes of homotopies between paths. Composition of 1-morphisms is the "standard" composition $a\otimes b$ of paths given by $(a\otimes b)(t) = a(2t)$ for $0 \leq t \leq \frac{1}{2}$ and $(a\otimes b)(t) = b(2t-1)$ for $\frac{1}{2} \leq t \leq 1$.</p> <p>By restricting to loops around a fixed point $e \in X$ one therefore gets a bicategory with one object, i.e., a monoidal category. When moreover $X$ has the structure of a topological monoid it is easy to see that the structure map $\mu: X \times X \rightarrow X$ induces a braiding $\gamma_{a,b}: a\otimes b \stackrel{\sim}{\rightarrow} b\otimes a$ on this monoidal category. </p> <p>My question is: Is there an (elementary) example where this braiding is not a symmetry? One obvious necessary condition is that $X$ has to be noncommutative in order to give such an example.</p> http://mathoverflow.net/questions/103213/when-do-sheaves-which-are-constant-along-the-fibers-come-from-the-base When do sheaves which are constant along the fibers come from the base? Nicolas Schmidt 2012-07-26T15:58:55Z 2012-08-15T19:24:35Z <p>Assume we are given a map $f: X \rightarrow Y$ between topological spaces which is open, surjective and has (pathwise) connected fibers. Consider categories $\text{Sh}(X),\text{Sh}(Y)$ of sheaves (with values in any given fixed sensible category) on $X$ and $Y$. The inverse image functor $f^\ast: \text{Sh}(Y) \rightarrow \text{Sh}(X)$ will then be fully faithful and hence we can consider $\text{Sh}(Y)$ up to equivalence as a full subcategory of $\text{Sh}(X)$. Obviously a necessary condition for a sheaf to lie in this subcategory is that it is constant when restricted to the fibers of $f$.</p> <p>My question is: Is this condition also sufficient? Is it even sufficient for trivial fibrations $F\times Y \rightarrow Y$ ?</p> http://mathoverflow.net/questions/98777/moore-penrose-inverse-as-an-adjoint Moore-Penrose Inverse as an adjoint Nicolas Schmidt 2012-06-04T15:58:59Z 2012-06-09T19:47:48Z <p>A Moore-Penrose pseudoinverse of a morphism $f: V \rightarrow W$ between Euclidean vector spaces is a map $g: W \rightarrow V$ in the other direction satisfying the identities</p> <p>$fgf = f$</p> <p>$gfg = g$</p> <p>$(fg)^\ast = fg$</p> <p>$(gf)^\ast = gf$</p> <p>where $\phi^\ast$ denotes the adjoint of a linear map $\phi$.</p> <p>Now the first two identities obviously resemble the triangle equalities of an adjunction. My question is: Can one actually understand the Moore-Penrose inverse as an adjoint? One possibility would be to find a "nice" (compatible with composition) partial order on Hom-Sets $\text{Hom}(V,W)$ making the category of Euclidean vector spaces into a 2-category, where the notion of adjunction is defined and the triangle equality in fact would imply $fgf = f$. So a more precise question would be: Does there exist such an order?</p> http://mathoverflow.net/questions/96984/example-of-dynamical-system-m-such-that-m-rightarrow-mathbfr-backslash-m Example of dynamical system $M$ such that $M \rightarrow \mathbf{R}\backslash M$ is not locally trivial? Nicolas Schmidt 2012-05-15T10:06:13Z 2012-05-15T15:40:10Z <p>Let $\Phi: M \times \mathbf{R} \rightarrow M$ be a smooth dynamical system having no periodic orbits, i.e. such that the canonical map $\pi:M \rightarrow \mathbf{R}\backslash M$ is a principal $\mathbf{R}$-bundle. Is $\pi$ always locally trivial? If not, are there any nice (not too contrived/complicated) counterexamples? Which conditions ensure $\pi$ to be locally trivial?</p> http://mathoverflow.net/questions/59843/quotient-surface-of-a-hyperelliptic-involution/88470#88470 Answer by Nicolas Schmidt for Quotient Surface of A Hyperelliptic Involution Nicolas Schmidt 2012-02-14T23:32:51Z 2012-02-14T23:32:51Z <p>I think the confusion stems from a common misconception, that orbifold points should somehow be considered as singular. The quotient in question - whether as a complex variety or as an orbifold - is <em>not</em> singular. In fact you will never be able to produce something singular by taking the quotient of a compact Riemann surface (i.e. a smooth complex algebraic <em>curve</em>) wrt the action of a finite group (because smoothness in dimension 1 is equivalent to normality, which is preserved by quotients under finite group actions).</p> http://mathoverflow.net/questions/83027/what-is-ricardo-perez-marcos-ene-product-does-it-explain-his-statistical-resul/83237#83237 Answer by Nicolas Schmidt for What is Ricardo Pérez-Marco's eñe product? Does it explain his statistical results on differences of zeta zeros? Nicolas Schmidt 2011-12-12T13:02:19Z 2011-12-12T13:02:19Z <p>To my understanding the "eñe product" is the same as the usual product of Dirichlet series considered as elements in the big ring of Witt vectors over the complex numbers say. So as a reference you might want to take a look at Hazewinkels encyclopedic article <a href="http://arxiv.org/abs/0804.3888" rel="nofollow">http://arxiv.org/abs/0804.3888</a> .</p> http://mathoverflow.net/questions/62516/holomorphic-line-bundles-on-a-punctured-disc Holomorphic line bundles on a punctured disc Nicolas Schmidt 2011-04-21T10:30:56Z 2011-04-21T13:07:23Z <p>Is every holomorphic line bundle on the - say - punctured unit disc $\dot{\Delta} \subseteq \mathbf{C}$ trivial? Griffiths-Harris (p. 39) prove that $H^{p,q}_{\overline{\partial}}(\Delta) = 0$ (for $q \geq 1$), and mention that by replacing discs by annuli that one could prove also <code>$H^{p,q}_{\overline{\partial}}(\dot{\Delta}^k \times \Delta^\ell) = 0$</code>. They seem to imply that in particular <code>$H^{p,q}_{\overline{\partial}}(\dot{\Delta}) = 0$</code>. If this were true, using Dolbeault's theorem and the Kummer sequence one could conclude <code>$H^1(\dot{\Delta},\mathcal{O}^\times_{\dot{\Delta}}) = 0$</code>, hence that every holomorphic line bundle on $\dot{\Delta}$ is trivial.</p> http://mathoverflow.net/questions/13593/is-every-monomorphism-of-commutative-hopf-algebras-over-a-field-injective Is every monomorphism of commutative Hopf algebras (over a field) injective? Nicolas Schmidt 2010-01-31T22:30:25Z 2010-02-05T12:19:39Z <p>Is it true that any monomorphism of commutative Hopf algebras over a field is injective? Moreover, is it true that any epimorphism of commutative Hopf algebras over a field is surjective?</p> http://mathoverflow.net/questions/13593/is-every-monomorphism-of-commutative-hopf-algebras-over-a-field-injective/13639#13639 Answer by Nicolas Schmidt for Is every monomorphism of commutative Hopf algebras (over a field) injective? Nicolas Schmidt 2010-02-01T09:44:08Z 2010-02-01T09:44:08Z <p>Yes, the category of commutative Hopf algebras over k is antiequivalent to the category of (pro-)affine group schemes over k, but this equivalence does not respect the obvious functors to (Set). I really meant injective as maps between algebras, not between maximal spectrum, prime spectrum or rational points.</p> http://mathoverflow.net/questions/123358/when-is-this-braiding-not-a-symmetry/123380#123380 Comment by Nicolas Schmidt Nicolas Schmidt 2013-03-01T23:48:58Z 2013-03-01T23:48:58Z Very nice! Do you know a good reference for the relation between braided monoidal categories and homotopy 3-types? http://mathoverflow.net/questions/103213/when-do-sheaves-which-are-constant-along-the-fibers-come-from-the-base/103224#103224 Comment by Nicolas Schmidt Nicolas Schmidt 2012-07-26T20:37:42Z 2012-07-26T20:37:42Z If you start with any good section $s_y$ on $V_y$, I can extend it to a section $\widetilde{s_y}$ on $\widetilde{V_y}$ by assigning arbitrary values on the circles. Those sections $\widetilde{s}_y$ will never glue, and you can't eliminate these circles by intersecting $\widetilde{V}$ with opens of the form $f^{-1}(f(W))$. http://mathoverflow.net/questions/103213/when-do-sheaves-which-are-constant-along-the-fibers-come-from-the-base/103224#103224 Comment by Nicolas Schmidt Nicolas Schmidt 2012-07-26T20:37:37Z 2012-07-26T20:37:37Z Mmhh, still I don't see why the sections should have to be compatible on intersections. Clearly there has to be a restriction on the type of neighbourhood of the fiber you want to allow. For example let's take a trivial fibration over $Y = \mathbf{R}$ with fiber $F = \mathbf{R}$ and consider a &quot;Gaussian shaped&quot; neighbourhood $V_y$ of a fiber $\{y\}\times F$ which approaches the fiber as you run to $\pm \infty$. Then start adding circles with are disjoint from $V_y$ but come closer and closer to the fiber. Let $\widetilde{V_y}$ denote the resulting neighbourhood. http://mathoverflow.net/questions/103213/when-do-sheaves-which-are-constant-along-the-fibers-come-from-the-base/103224#103224 Comment by Nicolas Schmidt Nicolas Schmidt 2012-07-26T18:32:02Z 2012-07-26T18:32:02Z I'm not sure whether I understand you correctly, but this proof seems to be wrong. There is no reason why any given open neighbourhood of the fiber should contain an open neighbourhood of the form $f^{-1}(f(U_n))$, unless the fiber is compact. I considered an argument similar to yours but then got stuck on precisely this issue. http://mathoverflow.net/questions/98777/moore-penrose-inverse-as-an-adjoint/99181#99181 Comment by Nicolas Schmidt Nicolas Schmidt 2012-06-12T17:26:05Z 2012-06-12T17:26:05Z Oops ..., I didn't realize Moore-Penrose inverses don't compose! http://mathoverflow.net/questions/98777/moore-penrose-inverse-as-an-adjoint Comment by Nicolas Schmidt Nicolas Schmidt 2012-06-07T19:50:35Z 2012-06-07T19:50:35Z The equality $f = fgf$ is related to the triangle equations through the notion of &quot;idempotent adjunction&quot;, a particularly nice class of adjunctions. Whenever you have an idempotent adjunction, you get the familiar equations $f = fgf$ and $g = gfg$ (interpreted suitably) for free. For poset-categories any adjunction is automatically idempotent, hence my suggestion, but even outside this context many adjunctions &quot;in nature&quot; turn out to be idempotent. http://mathoverflow.net/questions/98777/moore-penrose-inverse-as-an-adjoint Comment by Nicolas Schmidt Nicolas Schmidt 2012-06-05T15:29:39Z 2012-06-05T15:29:39Z I guess the construction I suggested really is a dead-end. Maybe the answer to the weaker version of my question still is &quot;yes&quot;? Somehow I can't get this tempting idea out of my head. http://mathoverflow.net/questions/96984/example-of-dynamical-system-m-such-that-m-rightarrow-mathbfr-backslash-m/97012#97012 Comment by Nicolas Schmidt Nicolas Schmidt 2012-05-15T17:30:54Z 2012-05-15T17:30:54Z Nice answer, I was hoping for something along these lines. Could you explain why properness is necessary for existence of slices? I mean we only have a local condition on the action (which implies properness, locally), but a locally proper group action need not be proper. Regarding the terminology: Many people take principal $G$-bundles to be locally trivial, but many don't (e.g. Husem&#246;ller). http://mathoverflow.net/questions/62516/holomorphic-line-bundles-on-a-punctured-disc/62521#62521 Comment by Nicolas Schmidt Nicolas Schmidt 2011-04-21T13:05:32Z 2011-04-21T13:05:32Z Hey, thanks very much for your answer! Although I must say that I'm a bit disappointed, since for some reason I thought that there should be lots of nontrivial bundles. http://mathoverflow.net/questions/13593/is-every-monomorphism-of-commutative-hopf-algebras-over-a-field-injective/14234#14234 Comment by Nicolas Schmidt Nicolas Schmidt 2010-02-09T21:23:42Z 2010-02-09T21:23:42Z Ah, now I see what you mean. No, I think the argument is perfectly valid. http://mathoverflow.net/questions/13593/is-every-monomorphism-of-commutative-hopf-algebras-over-a-field-injective/14234#14234 Comment by Nicolas Schmidt Nicolas Schmidt 2010-02-07T13:37:45Z 2010-02-07T13:37:45Z Your argument basically depends on the fact, that an irreducible lie representation of sl(2) on a finite dimensional vector space $V$ is completely determined by the endomorphisms representing e and h: Take $v$ any eigenvector of $h$ with $ev = 0$, then we can construct canonically a basis of $V$, where the action of $f$ is a priori known. However splitting a given representation into irreducible ones, or even determining whether it is irreducible seems to depend on the whole action of sl(2). http://mathoverflow.net/questions/13593/is-every-monomorphism-of-commutative-hopf-algebras-over-a-field-injective/14234#14234 Comment by Nicolas Schmidt Nicolas Schmidt 2010-02-06T21:50:41Z 2010-02-06T21:50:41Z I'm not sure if this argument is 100% correct, but it seems to work under some mild technical assumptions. Either way I am convinced now that the answer is &quot;no&quot;. As I have understood the article of Chirvasitu mentioned below, the reason for this should be the lack of faithful coflatness. http://mathoverflow.net/questions/13593/is-every-monomorphism-of-commutative-hopf-algebras-over-a-field-injective/13639#13639 Comment by Nicolas Schmidt Nicolas Schmidt 2010-02-03T14:09:34Z 2010-02-03T14:09:34Z For some reason I cannot comment on his answer, I can only add comments to my own (?). http://mathoverflow.net/questions/13593/is-every-monomorphism-of-commutative-hopf-algebras-over-a-field-injective/13639#13639 Comment by Nicolas Schmidt Nicolas Schmidt 2010-02-03T11:45:19Z 2010-02-03T11:45:19Z The example of Ben cannot be correct, if I understand him correctly: The category of finite groups is embedded as a full subcategory of the category of affine group schemes over k via the constant group functor. Any epimorphism in the surrounding category remains of course an epimorphism in the smaller one. If the inclusion of a transposition into $S_3$ were an epimorphism of algebraic groups, it be so as abstract groups. But as was already mentioned, every epimorphism of finite groups is surjective (in fact this statement remains true without 'finite' and is an exercise in Saunders MacLane).