User pavol s. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T07:52:22Z http://mathoverflow.net/feeds/user/9390 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108553/smooth-dependence-of-stable-manifold-on-parameters smooth dependence of stable manifold on parameters Pavol S. 2012-10-01T15:43:18Z 2012-10-01T19:00:23Z <p>Let $v$ be a $C^r$ vector field on a Banach space $V$ such that $0$ is its hyperbolic fixed point, and let $X_v\subset V$ be its local stable manifold. Does $X_v$ depend on $v$ smoothly?</p> http://mathoverflow.net/questions/91927/compatibility-of-the-kz-connection-with-operadic-composition Compatibility of the KZ connection with operadic composition Pavol S. 2012-03-22T15:48:09Z 2012-04-22T20:22:00Z <p>In what sense is the Knizhnik-Zamolodchikov connection compatible with the operadic composition in the little discs operad and/or the operad of $\overline{M}_{0,n}$'s?</p> <p>Here are (some) details, motivation, and a more precise question.</p> <p>The KZ connection $A_n\in\Omega^1(C_n)\otimes\mathfrak{t}^n$ is given by $$A_n =\sum_{i&lt; j} t^{ij}\,d\log(z_i-z_j).$$ Here $C_n$ is the configuration space of $n$ different points in $\mathbb{C}$ and $\mathfrak{t}^n$ is the Lie algebra with generators $t^{ij}$ ($1\leq i,j\leq n$, $i\neq j$, $t^{ij}=t^{ji}$) and relations $[t^{ij},t^{kl}]=0$ if all $i,j,k,l$ are different and $[t^{ij},t^{ik}+t^{jk}]=0$.</p> <p>The "regularized holonomy" of $A_3$, when $z_1$ stays at $0$, $z_3$ at $1$, and $z_2$ moves from $0$ to $1$, is the KZ Drinfeld associator $\Phi_{KZ}$.</p> <p>In general, an associator $\Phi$ (together with its coupling constant $\mu$) is equivalent to a morphism of operads of groupoids $F:PaB\to T$. Here $PaB_n$ is the groupoid of parenthesized braids, and $T_n=\exp \mathfrak{t}^n$ ($\mathfrak{t}^n$'s form an operad of Lie algebras). In the case of $\Phi_{KZ}$, $F_{KZ}$ is given by the "regularized holonomy" of $A_n$'s. The natural explanation of the fact that $F_{KZ}$ is a morphism of operads would be that $A$ is compatible with the operad structure. <strong>In what sense it is true?</strong> (there are two questions - in what sense is $A$ compatible with operad, and how it implies that $F_{KZ}$ is a morphism of operads)</p> <p><em>I'll mention some random stuff that might appear in the answer, but you can ignore it.</em></p> <p>There is a flat connection on moduli spaces of rational curves with marked points, closely related to $A$. If we define $\hat{\mathfrak{t}}^n$ by imposing one more relation $\sum_i t^{ij}=0$ ($[t^{ij},t^{ik}+t^{jk}]=0$ is then a consequence), the KZ connection descends to $M_{0,n}$; let us call it $\hat{A}_n\in\Omega^1(M_{0,n})\otimes\hat{\mathfrak{t}}^n$. By putting one of the points to $\infty$ we can identify $M_{0,n+1}$ with $C_n/\{az+b\}$ ($a\in\mathbb{C}^*,b\in\mathbb{C}$), and via the isomorphism $\hat{\mathfrak{t}}^{n+1}\cong\mathfrak{t}^n/\text{center}$ we can identify $\hat A_{n+1}$ with $A_{n}/\text{center}$. The center of $\mathfrak{t}$ is lost in this way. The Lie algebras $\hat{\mathfrak{t}}^{n}$ form a cyclic operad. The compactified moduli spaces $\overline{M}_{0,n}$ also form a cyclic operad.</p> <p>As $F_{KZ}$ is (modulo the center) given by the parallel transport of $\hat{A}$ between certain tangential base points of $M_{0,n+1}\subset\overline{M}_{0,n+1}$, I would imagine the operadic compositions to be (roughly) maps $$M_{0,n+1}\times M_{0,m+1}\times\text{formal punctured disc}\to M_{0,n+m}$$ coming from the operadic composition of $\overline{M}_{0,k}$'s. But I'm not sure in what category it would be an operad (of nice families over the punctured discs?) and how to make it technically work (something should be also said about maps of the trivial $\hat T$-bundles over these spaces).</p> <p>In fact, to get a parametrization of the punctured disc, it would be better to consider the moduli spaces $M_{0,n}'$ of rational curves with $n$ marked points and non-zero tangent vectors at those points. There is a flat connection on $M_{0,n}'$ which sees also the center. Let us replace the relations $\sum_i t^{ij}=0$ with $s^j:=\sum_i t^{ij}\textit{ is central}$. The Lie algebras $\check{\mathfrak{t}}^{n}$ that we obtain in this way still form a cyclic operad, and the connection $$\check{A}_n= \sum t^{ij}\,d\log(z_i-z_j)+ \sum s^i\,d\log(v_i)$$ is a flat connection on the configuration space of $n$ different points with chosen non-zero tangent vectors; this 1-form is again $SL(2,\mathbb{C})$-basic, and so it descends to the moduli space $M_{0,n}'$. By putting one of the points to $\infty$ and also normalizing its tangent vector, we can identify $M_{0,n+1}'$ with $(C_n/\text{translations})\times (\mathbb{C}^*)^n$. We have an isomorphism $\check{\mathfrak{t}}^{n+1}\cong\mathfrak{t}^n\oplus \mathbb{C}^n$, and so $\check{A}_{n+1}$ gets identified with $A_n$ plus a central part corresponding to the tangent vectors. It is the framed version of the KZ connection (and is my favorite).</p> <p><strong>edit:</strong> there is another (Alekseev-Torossian) connection $A^{AT}_n$ on $C_n$ with values is $\mathfrak{t}^n$. $A^{AT}_n$ is in fact on $FM_2(n)$ ($FM_2(n)$ is a compactification of $C_n/\{az+b\}$, where this time $a\in\mathbb{R}_+$). $FM_2(n)$'s form an operad (a version of the little discs operad) and $A^{AT}_n$'s are compatible with the operad structure in the obvious way: if $o_i:FM_2(m)\times FM_2(n)\to FM_2(m+n-1)$ is one of the compositions then $o_i^*A^{AT}_{m+n-1}$ is equal to the connection $A^{AT}_m\oplus A^{AT}_n$ on $FM_2(m)\times FM_2(n)$, after we apply the corresponding $o_i:\mathfrak{t}^m\oplus\mathfrak{t}^n\to\mathfrak{t}^{m+n-1}$. The operad $PaB$ is a sub-operad of the fundamental groupoid of $FM_2$. $\Phi_{AT}$ and $F_{AT}$ are defined as (ordinary, not regularized) parallel transport of $A^{AT}$, and $F_{AT}$ is obviously a morphism of operads. I would like to understand the corresponding picture for $A_{KZ}$.</p> http://mathoverflow.net/questions/55168/poincare-duality-with-boundary-conditions Poincare duality with boundary conditions Pavol S. 2011-02-11T21:04:06Z 2011-02-24T21:18:52Z <p>If $M$ is a compact oriented manifold with boundary then by Poincare duality the cohomology of $\Omega(M)$ (de Rham cohomology of $M$) is dual to the cohomology of $\Omega_0(M)$, where $\Omega_0(M)$ denotes differential forms vanishing on $\partial M$. This question is about a generalization of this fact to more complicated boundary conditions.</p> <p>Suppose $M$ is a compact oriented $C^\infty$ manifold with corners. Its boundary is decomposed (by the corners) to faces (of dimension $\dim M -1$).</p> <p>Let $V$ be a finite-dimensional vector space, and let us choose for every face $F\subset\partial M$ a subspace $V_F\subset V$. Let us consider the complex (of $V$-valued differential forms with boundary conditions given by $V_F$'s) $$\Omega(M)_{V,\{V_F\}}= \{\alpha\in\Omega(M)\otimes V;\quad \alpha|_F\in\Omega(F)\otimes V_F\text{ for all faces }F\}\subset\Omega(M)\otimes V.$$</p> <p>The "naive dual" of $\Omega(M)_{V,\{V_F\}}$ is $\Omega(M)_{V^*,\{\text{ann} V_F\}}$ with the pairing given by integration over $M$ ($\text{ann} V_F\subset V^*$ denotes the annihilator of $V_F$).</p> <p><strong>Is there a condition under which the pairing between the cohomologies of</strong> $\Omega(M)_{V,\{V_F\}}$ <strong>and of</strong> $\Omega(M)_{V^*,\{\text{ann} V_F\}}$ <strong>is perfect? What is the reason for the fact that the pairing is not always perfect?</strong></p> <p>Some remarks: </p> <ul> <li><p>The "dual complex" $\Omega(M)_{V^*,\{\text{ann} V_F\}}$ can be described as $$\{\alpha\in\Omega(M)\otimes V^*;\quad \int_M\langle\alpha\wedge d\beta\rangle= (-1)^{\deg\alpha +1}\int_M\langle d\alpha\wedge \beta\rangle\quad \forall\beta\in \Omega(M)_{V,\{V_F\}} \}.$$</p></li> <li><p>If $V_F=0$ for all $F$'s then we get the standard Poincare duality for manifolds with boundary. </p></li> <li><p>It is possible that manifolds with corners is not the right picture; the question might be about any "reasonable" division of $\partial M$ into "faces" (of dimension $\dim\partial M$).</p></li> </ul> http://mathoverflow.net/questions/49526/terminology-question-on-covering-spaces Terminology question on covering spaces Pavol S. 2010-12-15T13:31:37Z 2011-01-02T23:12:39Z <p>I'm teaching an elementary class about fundamental groups and covering spaces. It was very useful to use "fool's covering spaces" of a space $X$, defined as functors $\Pi_1(X)\to Sets$, where $\Pi_1(X)$ is the fundamental groupoids of $X$. In a more "covering space way", a fool's covering space can be described as a set $Y$, a map $p:Y\to X$, and a map $p^{-1}(x_1)\to p^{-1}(x_2)$ for any path between $x_1, x_2\in X$, satisfying the obvious properties.</p> <p>Is there a standard name for "fool's covering spaces"? Calling them "functors $\Pi_1(X)\to Sets$ " is a bit heavy for the class.</p> http://mathoverflow.net/questions/39366/is-a-quasi-iso-in-lie-algebra-cohomology-necessarily-an-iso/39389#39389 Answer by Pavol S. for Is a quasi-iso in Lie algebra cohomology necessarily an iso? Pavol S. 2010-09-20T14:33:32Z 2010-09-20T14:33:32Z <p>If $\mathfrak{g}$ and $\mathfrak{h}$ are say nilpotent and finite dimensional then the map $\mathfrak{g}\to\mathfrak{h}$ is an isomorphism. It follows from uniqueness of minimal Sullivan model, in this case $\bigwedge\mathfrak{g}^*$ (there should be a simpler reason, but I don't see it). As noted above by Tom Goodwillie, it is not true for solvable Lie algebras.</p> http://mathoverflow.net/questions/91927/compatibility-of-the-kz-connection-with-operadic-composition/91964#91964 Comment by Pavol S. Pavol S. 2012-03-23T10:37:21Z 2012-03-23T10:37:21Z Hi Adrien what do you mean by &quot;a well defined extension of the KZ equation on the compactified configuration space $\bar{C}_n$&quot;? Can you give an example? I understand that you obtained the connection in your last equation by applying one of the operadic maps $\mathfrak{t}_3\to\mathfrak{t}_4$. But I don't see a compatibility of the KZ connection w.r.t. the map in your next-to-last equation. http://mathoverflow.net/questions/64730/examples-of-lie-algebroids Comment by Pavol S. Pavol S. 2011-05-12T08:12:16Z 2011-05-12T08:12:16Z a small correction: Courant algebroids are in fact not Lie algebroids (depending on the definition, the bracket is either not skew-symmetric or doesn't satisfy Jacobi identity) http://mathoverflow.net/questions/55168/poincare-duality-with-boundary-conditions/56022#56022 Comment by Pavol S. Pavol S. 2011-02-26T20:40:27Z 2011-02-26T20:40:27Z Thanks a lot, now I believe you :) http://mathoverflow.net/questions/55168/poincare-duality-with-boundary-conditions Comment by Pavol S. Pavol S. 2011-02-22T14:12:36Z 2011-02-22T14:12:36Z If $V_F=V$ for all $F$'s we still get de Rham cohomology tensored with $V$, and de Rham with compact support in the interior tensored with $V^*$, so the duality holds. The motivation for the question was from symplectic form on the moduli space of flat connections on a surface, with various boundary conditions, and Lagrangian subspaces coming from cobordisms. http://mathoverflow.net/questions/55168/poincare-duality-with-boundary-conditions/56022#56022 Comment by Pavol S. Pavol S. 2011-02-22T14:05:22Z 2011-02-22T14:05:22Z I think this is a counterexample even though your condition is satisfied (if I understood it correctly): $M$ is tetrahedron, $\dim V =2$, we choose 3 different 1-dim subspaces of $V$ as $V_F$'s for 3 of the faces, and for the 4th $F$ we put (e.g.) $V_F=V$. http://mathoverflow.net/questions/55546/which-areas-of-arithmetic-algebraic-geometry-can-be-learned-as-black-boxes-and/55607#55607 Comment by Pavol S. Pavol S. 2011-02-16T16:32:00Z 2011-02-16T16:32:00Z it might be simpler to notice that $b_2=0$, so it cannot be symplectic :) http://mathoverflow.net/questions/55168/poincare-duality-with-boundary-conditions Comment by Pavol S. Pavol S. 2011-02-14T21:58:24Z 2011-02-14T21:58:24Z @John Klein: Then $V$ is not important. One cohomology is de Rham's tensored with $V^*$ and the other one is (isomorphic to) de Rham's with compact support in the interior of $M$ tensored with $V$. We can safely put $V=\mathbb R$ in this case. More-dimensional $V$ is only needed for those more complex boundary conditions. For $V=\mathbb{R}$ we can only have $V_F=0$ or $V_F=\mathbb{R}$. In that case, if say $\partial M$ is divided by a hypersurface to two faces, one with $V_F=0$ and the other with $V_F=\mathbb{R}$, then we do get perfect pairing (at least I hope :) http://mathoverflow.net/questions/43445/baker-campbell-hausdorff-formula-prime-divisors-of-denominators Comment by Pavol S. Pavol S. 2011-02-14T21:36:54Z 2011-02-14T21:36:54Z The best I know is that the multiplicity of $p$ in the denominators of the terms of length $\leq n$ is at most $(n-1)/(p-1)$. The proof follows from the elementary observation that if $v_p(x)\geq1/(p-1)$ then $v_p(e^x-1)\geq1/(p-1)$ and $\log(1+x)\geq1/(p-1)$. See the above-mentioned paper of Lazard. I don't know any intelligent lower estimates of the multiplicity, however http://mathoverflow.net/questions/49526/terminology-question-on-covering-spaces/50965#50965 Comment by Pavol S. Pavol S. 2011-01-03T20:53:09Z 2011-01-03T20:53:09Z This is perhaps not exactly what I wanted: a local system should be a functor from the Cech fundamental groupoid, while I was asking about the usual (path) fundamental groupoid (e.g. the application to Van Kampen theorem was meant for the usual (path) fundamental group). http://mathoverflow.net/questions/49526/terminology-question-on-covering-spaces/49570#49570 Comment by Pavol S. Pavol S. 2010-12-15T22:56:20Z 2010-12-15T22:56:20Z I asked a terminology question, and instead I learned something very interesting. Thanks a lot! http://mathoverflow.net/questions/49526/terminology-question-on-covering-spaces Comment by Pavol S. Pavol S. 2010-12-15T16:13:57Z 2010-12-15T16:13:57Z @Ryan: here are 2 pedagogical reasons for giving &quot;fool' covering spaces&quot; independent life and name: 1. the fact that they are equivalent to ordinary covering spaces (if the base is nice) can be described by lifting the topology from $X$ to $Y$, which is an easy operation. 2. Using the classification of covering spaces (for nice base), van Kampen theorem (for nice spaces) is the trivial statement that if something is locally (in $X$) a covering space then it is a covering space. Without this classification and for arbitrary spaces it is the same locality statement for fool's covering spaces. http://mathoverflow.net/questions/49526/terminology-question-on-covering-spaces Comment by Pavol S. Pavol S. 2010-12-15T14:20:08Z 2010-12-15T14:20:08Z @Todd: &quot;fool's&quot; because the equivalence of &quot;fool's&quot; with ordinary is only true for nice spaces, and (given the level of the course) is considered as one of more difficult theorems. – Trial 5 http://mathoverflow.net/questions/49526/terminology-question-on-covering-spaces Comment by Pavol S. Pavol S. 2010-12-15T14:15:16Z 2010-12-15T14:15:16Z Sorry for being unclear. Any covering space is also a &quot;fool's covering space&quot;. For a locally path-connected and semilocally 1-connected spaces the two notions are equivalent (this is considered a &quot;difficult theorem&quot; in the cours). In a fool's covering space the set $Y$ is just a set, with no topology (if it's uclear, take &quot;functor $\Pi_1(X)\to Sets$ &quot; as a definition of fool's covering space, and forget the other description).