User damienc - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T04:58:44Z http://mathoverflow.net/feeds/user/7031 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/127589/does-the-vanishing-of-the-poisson-bracket-on-s-mathfrakg-mathfrakg-ins/130091#130091 Answer by DamienC for Does the vanishing of the Poisson bracket on $S(\mathfrak{g})^{\mathfrak{g}}$ inspire the disover of Duflo's isomorphism theorem? DamienC 2013-05-08T16:10:41Z 2013-05-11T09:10:31Z <blockquote> <p>My question is: Does the vanishing of the Poisson bracket plays an important role in finding and proving Duflo's isomorphism theorem? Or it is just an literally first step?</p> </blockquote> <p>Let $A_0$ be a Poisson algebra and $A$ a deformation quantization of $A_0$ (assume we are in a context when it exists). </p> <p>Assume you have a quantization map $Q:A_0\to A$, by which I mean a section of the classical limit map $A\to A/(\hbar)=A_0$. </p> <p>Then for any two elements $a,b\in A_0$, <code>$[Q(a),Q(b)]=\hbar\{a,b\}+O(\hbar^2)$</code>. </p> <p>Hence if you want to have $Q(ab)=Q(a)Q(b)$ you must at least assume that <code>$\{a,b\}=0$</code>. </p> <p>My (non-)answer to your question is then: </p> <blockquote> <p>the vanishing of the Poisson bracket is a necessary requirement if you want a statement of Duflo-type. It is just a first step. </p> </blockquote> <p>The actual history comes from the Harish-Chandra isomomorphism. Duflo noticed that the original formula could be written for any Lie algebra, without any use of roots and similar stuff specific to the semi-simple case. </p> http://mathoverflow.net/questions/119991/is-the-quantum-algebra-unique-up-to-isomorphism-in-deformation-quantization/120089#120089 Answer by DamienC for Is the quantum algebra unique (up to isomorphism) in deformation quantization ? DamienC 2013-01-28T08:38:33Z 2013-01-28T08:38:33Z <p>Let me give a conjectural answer. </p> <p>Point 1. $GRT$ acts non-trivially on poly-vector fields by $L_\infty$-isomorphisms (see T.Willwacher, M. Kontsevich’s graph complex and the Grothendieck-Teichm¨uller Lie algebra, arXiv:1009.1654 for an explicit description of this actions). </p> <p>Point 2. I guess that $GRT$ actually acts non-trivially on the set of equivalence classes of Maurer-Cartan elements (I haven't checked this). </p> <p>Point 3. composing Kontsevich's formality map with the action of GRT gives a negative answer to your question if Point 2 is correct. . </p> <p>Point 4. concerning the symplectic case, the main reason why the classification map doesn't depend on any choice is because the quantization is unique. Let me explain further: the choices involved in the formality isomorphism appear in the local case. By itself, the globalization procedure does not require any additional choice. </p> http://mathoverflow.net/questions/47885/trace-of-the-atiyah-class-equals-chern-class/115613#115613 Answer by DamienC for trace of the atiyah class equals chern class DamienC 2012-12-06T14:27:44Z 2012-12-06T14:27:44Z <p>I might be wrong but it seems to me that the $p$-th Atiyah class does not have any reason to agree with the usual $p$-th Chern class unless the manifold under consideration is Kahler. </p> <p>Namely, if $X$ is not Kahler then for a holomorphic vector bundle $E\to X$, $c_p(E)\in H^{2p}(X)$ and $at_p(E)\in H^{p}(X,\Omega^p_X)$ live in different spaces. </p> <p>The point is that $c_p(E)$ can be defined as the class of $tr(R^p)$, where $R$ is the curvature of an hermitian connection on $E$, while $at_p(E)$ can be defined as the class of $tr(R_{1,1}^p)$, where $R_{1,1}$ is the $(1,1)$-part of the curvature of a $(1,0)$-connection on $E$. </p> <p>The point is that if $X$ is Kahler then there exists an Hermitian $(1,0)$-connection with curvature being of type $(1,1)$. The relation between the Atiyah classes and the Chern classes can be made through the Hodge-to-de Rham spectral sequence. </p> <hr> <p>So, I think that the Chern classes you are talking about are not the usual (i.e. topological) ones, but the Chern classes in Hodge cohomology. Then they coincide with the Atiyah classes almost by definition (by the way, there is a <a href="http://archive.numdam.org/ARCHIVE/BSMF/BSMF_1958__86_/BSMF_1958__86__137_0/BSMF_1958__86__137_0.pdf" rel="nofollow">very nice paper of Grothendieck</a> on Chern classes in Hodge cohomology). </p> http://mathoverflow.net/questions/31972/todd-class-and-baker-campbell-hausdorff-or-the-curious-number-12/115488#115488 Answer by DamienC for Todd class and Baker-Campbell-Hausdorff, or the curious number $12$ DamienC 2012-12-05T10:05:26Z 2012-12-05T10:05:26Z <p>The answer to your question is the following: given two non-commutative variables $x$ and $y$ one has $$log(e^xe^y)=x+e^{ad_x}\frac{ad_x}{e^{ad_x}-1}(y)+O(y^2)$$</p> <p>It is not the appearance of $12$ that is intriguing, but the appearance of the Todd series in algebraic geometry. It suggests that there is a group hidden somewhere... and this is indeed the case. This group is the derived loop space of your favorite algebraic variety $X$, and its tangent Lie algebra is the shifted tangent sheaf $T_X[-1]$, with Lie bracket given by the Atiyah class (the fact that the Atiyah class gives rize to a Lie structure was <a href="http://arxiv.org/abs/alg-geom/9704009" rel="nofollow">discovered by Kapranov</a>). </p> <p>The universal enveloping algebra of this Lie algebra is the Hochschild complex of $X$. One then gets a nice dictionnary between the Lie side and the algebraic geometry side. E.g.: </p> <ul> <li><p>any object in the derived category of $X$ turns out to be a representation of this Lie algebra. </p></li> <li><p>Poincare-Birkhoff-Witt is Hochschild-Kostant-Rosenberg. </p></li> <li><p>the Duflo isomorphism is the Kontsevich-Caldararu isomorphism between the Harmonic and Hochschild structures. </p></li> <li><p>there is also an relation between closed embeddings in algebraic geometry and inclusions of Lie algebras. </p></li> <li><p>...</p></li> </ul> http://mathoverflow.net/questions/54430/video-lectures-of-mathematics-courses-available-online-for-free/115474#115474 Answer by DamienC for Video lectures of mathematics courses available online for free DamienC 2012-12-05T07:18:37Z 2012-12-05T07:18:37Z <p>The courses of the summer school <strong>Poisson 2012</strong> (that took place in Utrecht), as well as lectures of the conference that followed, are available online: <a href="http://www.youtube.com/user/poissonutrecht" rel="nofollow">http://www.youtube.com/user/poissonutrecht</a></p> <p>The courses are: </p> <ul> <li><a href="http://www.youtube.com/watch?v=uYDi1NTJ930" rel="nofollow">Poisson and Symplectic Geometry of Moduli Spaces of Flat Connections</a>, by Anton Alekseev </li> <li><a href="http://www.youtube.com/watch?v=oYSE7p9fRBE" rel="nofollow">Poisson Geometry</a>, by Rui Loja Fernandes </li> <li><a href="http://www.youtube.com/watch?v=Yai8BIUfTuY" rel="nofollow">Lie Groupoids and Multiplicative Structures</a>, by Henrique Bursztyn </li> <li><a href="http://www.youtube.com/watch?v=_uiNqQIOzK4" rel="nofollow">Cluster Algebras and Compatible Poisson Structures</a>, by Michael Gekhtman </li> </ul> http://mathoverflow.net/questions/114701/one-more-question-about-pbw One more question about PBW DamienC 2012-11-27T21:03:43Z 2012-11-28T18:20:41Z <p>Let $k$ be a commutative ring with unit and $L$ be a Lie $k$-algebra. </p> <p>Let $U(L)$ be the universal enveloping $k$-algebra of $L$ (one can define it as a quotient of the tensor algebra, as it is explained <a href="http://mathoverflow.net/questions/61954/pbw-theorem-over-a-q-algebra-without-freeness-or-flatness" rel="nofollow">in this MO question</a>, or one can say that $U(-)$ is left adjoint to the forgetful functor sending an associative $k$-algebra to the Lie $k$-algebra obtained by taking the same underlying $k$-module and with Lie bracket being the commutator). </p> <p>The associative $k$-algebra $U(L)$ is filtered as a $k$-algebra, and there is a canonical epimorphism $S(L)\to gr\big(U(L)\big)$. </p> <p>If this epimorphism is an isomorphism, then we say that <em>$L$ has the PBW property</em>. </p> <p>All the examples of Lie $k$-algebras not satisfying the PBW property I am aware of are constructed in the following way: one first finds an example of a Lie algebra for which the map $L\to U(L)$ (the unit of the adjunction) is not injective, and then it is quite clear that the PBW property can't hold. </p> <p>My question is then: </p> <blockquote> <p>Is there any example of a Lie $k$-algebra $L$ such that the map $L\to U(L)$ is injective which does not satisfy the PBW property ?</p> </blockquote> <p>Or is it that the PBW property is just equivalent to $L\to U(L)$ being injective (it would be great, but I have no idea why this would be true - EDIT: one might want to use that $L\to U(L)$ is injective to reduce to the cas when $k\supset\mathbb{Q}$)?</p> http://mathoverflow.net/questions/99689/existence-of-dg-realization-for-6-functors/100570#100570 Answer by DamienC for Existence of dg realization for 6 functors DamienC 2012-06-25T07:45:08Z 2012-11-27T07:53:39Z <p>If I understand correctly your are looking for a dg enhancement of the six operation formalism. </p> <p>There seem to be a paper that does something very close to it: Yifeng Liu and Weizhe Zheng, <em>Enhanced six operations and base change theorem for sheaves on Artin stacks</em> (available at <a href="http://math.columbia.edu/~liuyf/sixi.pdf" rel="nofollow">http://math.columbia.edu/~liuyf/sixi.pdf</a>). </p> <p>They use the language of $(\infty,1)$-categories, but I think one can adapt it to dg-categories (assuming that one is working over a field of characteristic zero). </p> <p>EDIT Nov. 27, 2012: the above preprint has been posted on the arXiv: <a href="http://arxiv.org/abs/1211.5948" rel="nofollow">http://arxiv.org/abs/1211.5948</a></p> http://mathoverflow.net/questions/112069/a-fibrant-objects-structure-on-top/114412#114412 Answer by DamienC for A fibrant-objects structure on Top DamienC 2012-11-25T11:20:08Z 2012-11-25T12:40:06Z <p>This might be a naive answer but here is a suggestion for the definition of $\pi_n$-fibrations: maps having the RLP with respect to the the map $\Delta^k\to\Delta^k\times I$ for any $k\leq n$. </p> <p>In the limit you will get the "obvious" fibrant-object structure on $Top$ which comes from its usual model structure (recall that the full subcategory of fibrant objects in any model category is a category of fibrant objects.. and that all objects are fibrants in $Top$).</p> http://mathoverflow.net/questions/110020/algebraic-proof-of-hodge-decomposition-theorem/110024#110024 Answer by DamienC for algebraic proof of Hodge decomposition theorem DamienC 2012-10-18T17:04:36Z 2012-10-18T17:04:36Z <p>Deligne and Illusie proved the degeneration of Hodge to de Rahm (see e.g. <a href="http://math.bu.edu/people/potthars/writings/HdRSS.pdf" rel="nofollow">http://math.bu.edu/people/potthars/writings/HdRSS.pdf</a> for a brief review): </p> <blockquote> <p>Relevements modulo $p^2$ et decomposition du complexe de De Rham, Inv. Math. <strong>89</strong> (1987), 247-270.</p> </blockquote> <p>BUT the Deligne-Illusie approach does only says that the spectral suequence degenerates at $E_1$ (with $E_1$ terms being $H^q(X,\Omega^p_X)$). But is does not give an explicit isomorphism between $H^n(X)$ and $\oplus_{p+q=n}H^q(X,\Omega^p_X)$. </p> <p>I heard about a work-in-progress by Dima Arinkin, Andrei Caldararu and Marton Hablicsek, where they have a new approach to Deligne-Illusie <em>via</em> derived geometry. It might be that there approach gives the Hodge decomposition (I actualy don't know if this is the case or not). </p> http://mathoverflow.net/questions/109001/exact-dg-poisson-algebra/109122#109122 Answer by DamienC for Exact DG Poisson algebra DamienC 2012-10-08T07:51:27Z 2012-10-08T17:32:24Z <p>It seems to me that there is a (quasi-)isomorphism between the de Rham algebra and the dg algebra of polyvector fields equipped with the differential $[\pi,-]$ (where $\pi$ is the Poisson structure corresponding to the symplectic form). </p> <p>Through this isomorphism the equation $d\omega=0$ is sent to $[\pi,\pi]=0$, and the equation $\omega=d\lambda$ is sent to $\pi=[\pi,V]$, where $V$ is a vector field. </p> <p>On the level of the Poisson algebra of functions it tells you that for any two functions $f,g$, we have (up to a sign) <code>$$\{f,g\}=V(\{f,g\})-\{V(f),g\}+\{f,V(g)\}$$</code></p> <p>Algebraically you can say that there is a derivation $V$ for the product such that the Poisson bracket is its own derived bracket w.r.t. $V$. </p> http://mathoverflow.net/questions/63381/do-i-need-to-know-what-an-infinity-gerstenhaber-algebra-is-and-if-so-what-is-it/105937#105937 Answer by DamienC for Do I need to know what an infinity-Gerstenhaber algebra is, and if so, what is it? DamienC 2012-08-30T12:39:03Z 2012-08-30T12:39:03Z <blockquote> <p>Question 1: Does there necessarily exist a resolution of S that computes the derived $S\otimes_R$ and that is Gerstenhaber in a compatible way?</p> </blockquote> <p>Yes. As pointed out in the comments, the category of dg Gerstenhaber algebra admits a model structure in which the weak equivalences are the quasi-isomorphisms, fibrations are degreewise surjections, and cobifrant obects are those dg Gerstenhaber algebras that are free as graded algerbas. </p> <p>This actually wors with dg algebras over any given operad $\mathcal O$ (in place of Gerstenhaber). </p> <p>This is proved in Hinich's paper (quoted by the nLab: <a href="http://ncatlab.org/nlab/show/model+structure+on+dg-algebras+over+an+operad" rel="nofollow">http://ncatlab.org/nlab/show/model+structure+on+dg-algebras+over+an+operad</a>). </p> <p>Then there is also a natural model structure on the category of dg Gerstenhaber $R$-algebras (there is a more general statement about existence of a model structure on the category of objects under a given one $X$ in a model category $\mathcal C$). </p> <p>So, the answer to the title of your question is that you don't "need" to know what a $G_\infty$-algebra is. </p> <blockquote> <p>Question 2 if the answer to 1 is YES: How do I construct it?</p> </blockquote> <p>Shortly, bar-cobar. You can have a look at <a href="http://www.institut.math.jussieu.fr/%257Eginot/papers/gerstenhaber.pdf" rel="nofollow">Homologie et model minimal des algèbres de Gerstenhaber</a> in order to see how it works in details. </p> <p>Btw, the above paper also tells you what is the definition of a $G_\infty$-algebra. </p> <blockquote> <p>Question 2 if the answer to 1 is NO: Certainly my homotopy equivalence $S\leftrightarrow\widetilde{S}$ allows me to move the Gerstenhaber structure on $S$ to something on $\widetilde{S}$. What structure on $\widetilde{S}$ does it move to?</p> </blockquote> <p>Even though the answer to Question 1 is YES, there is still something to say here. There is on $\widetilde{S}$ a $G_\infty$-structure. This is "just" homotopy transfer formula (and the use of the explicit minimal model for the Gerstenhaber operad). </p> <p>The homotopy transfer for algebras over operad $\mathcal O$, w.r.t. to a cofibrant resolution $\widetilde{\mathcal O}\to\mathcal O$ is proved in the appendix A.2 of <a href="http://arxiv.org/abs/0710.4510" rel="nofollow">my paper with Van den Bergh</a> (see also Theorem 10.3.6 in <a href="http://math.unice.fr/~brunov/Operads.pdf" rel="nofollow">Loday-Vallette's Algebraic Operads</a> for the Koszul case). </p> http://mathoverflow.net/questions/105904/how-to-define-the-equivalence-of-maurer-cartan-elements-in-an-l-infty-algebr/105905#105905 Answer by DamienC for How to define the equivalence of Maurer-Cartan elements in an $L_{\infty}$-algebra? DamienC 2012-08-30T06:03:00Z 2012-08-30T06:03:00Z <p>This is explained in Section 4.5.2 of "deformation quantization of poisson manifolds" by Kontsevich (http://arxiv.org/abs/q-alg/9709040). </p> <p>The way you wrote the homotopy between two Maurer-Cartan elements is not enough : as it is explained in the above reference you also need a 1-parameter family of infinitesimal gauge equivalences. </p> <p>A quick reformulation of Kontsevich definition is the following. An equivalence between two Maurer-Cartan elements $a$ and $b$ in $\mathfrak g$ is a Maurer-Cartan element $c$ in $DR([0,1])\otimes\mathfrak g$ such that $a=c(0)$ and $b=c(0)$. </p> <p>Note that $DR(...)$ stands for the de Rham algebra of "...". </p> http://mathoverflow.net/questions/105806/what-structure-on-the-second-order-cotangent-bundle What structure on the second order cotangent bundle ? DamienC 2012-08-29T09:05:15Z 2012-08-29T12:56:19Z <p>It is well-known that the total space of the cotangent bundle $T^*X$ of a given smooth manifold $X$ admits a <a href="http://en.wikipedia.org/wiki/Symplectic_manifold" rel="nofollow">symplectic form</a> $\omega$. It is actually exact: $\omega=d\lambda$. The $1$-form $\lambda$ is called the <a href="http://en.wikipedia.org/wiki/Liouville_form" rel="nofollow">Liouville form</a> and can be defined in a quite tautological way: given an element $p=(x,\xi)\in T^*X$ (i.e. $x\in X$ and <code>$\xi\in T^*_xX$</code>) and a vector $v\in T_p(T^*X)$, we define <code>$\lambda_p(v):=\xi(\pi_*(v))$</code>, where <code>$\pi:T^*X\to X$</code> (and thus <code>$\pi_*(v)\in T_xX$</code>). </p> <blockquote> <p>Does the total space of the second order cotangent bundle $T^*_2X$ also admits a "natural" geometric structure ? It seems that there is similarly a tautological "second order form" on $T^*_2X$ (in place of the Liouville tautological $1$-form $\lambda$). But then I don't see any analog for $\omega$... </p> </blockquote> <p><strong>Recall that</strong> the second order cotangent bundle can be defined as the dual of the second order tangent bundle $T^2X$, which is the bundle of order $2$ differential operators on $X$ that vanish on constants. </p> <p>In case you would be an algebraic geometer, the fiber of <code>$T^*_2X$</code> at a point $x$ ($X$ is now a smooth algebraic variety) is just <code>$I_x/I_x^3$</code> if $I_x$ is the maximal ideal corresponding to $x$ (w.r.t. to an affine open neighbourhood of it).</p> http://mathoverflow.net/questions/30536/t-structures-and-higher-categories/105825#105825 Answer by DamienC for t-structures and higher categories? DamienC 2012-08-29T12:25:39Z 2012-08-29T12:25:39Z <blockquote> <p>So I'd like to ask: is there a higher categorical analog of a t-structure?</p> </blockquote> <p>As Mike Skirvin pointed out in a comment, higher categorical analog of t-structures have been introduced by Lurie. A more up-to-date reference might be <a href="http://www.math.harvard.edu/~lurie/papers/HigherAlgebra.pdf" rel="nofollow">Higher Algebra</a> (<code>$\S$</code> 1.2.1). </p> <blockquote> <p>More generally, how does the higher categorical viewpoint help one understand the set of all (or maybe all "nice" in an appropriate sense) t-structures on a given trangulated category, provided it is the homotopy category of a stable (∞,1) category?</p> </blockquote> <p>I guess the answer can be found at the same place. There, Lurie says that "there is a bijective correspondence between $t$-localizations of $\mathcal C$ (a stable $\infty$-category) and $t$-structures on the triangulated category $h\mathcal C$. </p> <p>The higher categorical point-of-view also seems to be useful to understand the yoga of derived functors in a more conceptual way. In Section 1.3 of the same reference (Higher Algebra) it is explained that if $\mathcal A$ is an abelian category with enough injectives, then its derived $\infty$-category $\mathcal D^-(\mathcal A)$ is stable, admits a $t$-structure, has homotopy category the standard derived category, and satisfies the following universal property: there is a canonical equivalence of abelian categories <code>$\mathcal A\to \mathcal D^-(\mathcal A)^\heartsuit$</code>, and if $\mathcal C$ is a stable $\infty$-category with a left-complete $t$-structure then any right exact functor $\mathcal A\to\mathcal C^\heartsuit$ extends (in an essentially unique way) to an exact functor $\mathcal D^-(\mathcal A)\to \mathcal C$. </p> http://mathoverflow.net/questions/102730/the-differential-of-the-exponential-map-reductive-homogeneous-space/103192#103192 Answer by DamienC for The differential of the exponential map: reductive homogeneous space DamienC 2012-07-26T13:23:46Z 2012-07-26T13:23:46Z <p>I think the formula will be essentially the same. </p> <p>The formula you wrote is valid in general for the exponential map of analytic manifolds equipped with an analytic affine connection. It is stated and proved in <a href="http://www.mscand.dk/article.php?id=1601" rel="nofollow">this paper</a> by Helgason (see pages 6-7 of the linked .pdf): <em>Some remarks on the exponential mapping of an affine connection. Math. Scand. 9 (l961), l29-l46.</em></p> <p>The existence of such connections on reductive homogeneous spaces, with an additional invariance property (that you might probably need at some point) is stated and proved in <em>K.Nomizu, Invariant affine connections on homogeneous spaces, Amer. J. Math. 76 (1954), 33-65</em>. </p> http://mathoverflow.net/questions/102597/history-question-autobiography-of-mathematicians/102618#102618 Answer by DamienC for History Question: AUTObiography of Mathematicians DamienC 2012-07-19T05:54:34Z 2012-07-19T10:11:57Z <p>Récoltes et semailles, by Alexander Grothendieck (available at the Grothendieck circle), might be considered as an autobiography. </p> http://mathoverflow.net/questions/101619/what-is-the-definition-of-the-l-infty-part-of-a-g-infty-morphism/101656#101656 Answer by DamienC for What is the definition of "the $L_\infty$ part of a $G_\infty$ morphism"? DamienC 2012-07-08T12:45:08Z 2012-07-08T12:45:08Z <p>A $G_\infty$-morphism $\phi$ is determined by structure maps $\phi^{k_1,\dots,k_n}$, $n\geq1$, $k_1,\dots,k_n\geq1$. </p> <p>The $L_\infty$-part of $\phi$ is the $L_\infty$-morphism $\ell$ with structure maps $\ell^k=\phi^{\overbrace{1,\dots,1}^{k~times}}$. </p> <p>In order to make this precise you might have to put (de)suspensions at the appropriate places.</p> http://mathoverflow.net/questions/75588/why-bi-module-bi-bundle-etc/75591#75591 Answer by DamienC for Why bi-module, bi-bundle, etc? DamienC 2011-09-16T11:35:28Z 2012-07-06T15:21:17Z <p>If you work with a single associative algebra $A$, then there is not so much sense to try to define a notion of $A$-tri-module. Namely, $A$-bimodules appear naturally as operadic $A$-modules. </p> <hr> <p>To give you a picture, an associative algebra $A$ is in particular an algebra over the colored operad given by intervals of the real line and their disjoint union inclusions (a-k-a <a href="http://math.northwestern.edu/~costello/factorization_public.html" rel="nofollow">prefactorization algebras</a> on the real line). </p> <p>It has the property to be locally constant: the space associated to any interval is the same ($A$ itself). </p> <p>Let me now relax a bit this locally constant property. Assume that you have such a (pre)factorization algebra over the real line such that it is locally constant outside the origin. Then you will get two associative algebras (one from $(0,+\infty)$ and the other from $(-\infty,0)$) together with a bunch of bimodules (One for each interval containing the origin). </p> <p>There is nothing against the idea of having a factorization algebra on the graph consisting of three half-edges attached to a single vertex. Things like this might have appeared in $2d$ lattice models. </p> http://mathoverflow.net/questions/92988/can-duflo-type-map-be-defined-for-invariant-differential-operators-in-a-way-co/100578#100578 Answer by DamienC for Can Duflo type map be defined for invariant differential operators ? (In a way compatible with "Harish-Chandra" isomorphism defined by F. Knop) DamienC 2012-06-25T09:35:50Z 2012-06-25T09:35:50Z <p>Let $X=T^*M$, and apply Dolgushev's equivariant formality theorem (see <a href="http://arxiv.org/abs/math/0307212" rel="nofollow">http://arxiv.org/abs/math/0307212</a>) to get a $G$-equivariant $L_\infty$-quasi-isomorphism $$\mathcal U:T_{poly}X \longrightarrow D_{poly}X$$</p> <p>Let $\pi$ be the Poisson structure on $X$ which comes from the canonical symplectic form on $T^*M$. Then one gets a $G$-equiavriant quasi-isomorphism of complexes $\mathcal U_\pi^1$ from the Poisson cochain complex of $(X,\pi)$ to the Hochschild cochain complex of $(\mathcal O_X,\star_\pi)$, where $\star_\pi$ is the star-product the quantizes $\pi$ through $\mathcal U$. </p> <p>Moreover, one has a homotopy betwen the cup-product one both sides, and it is $G$-equivariant (compatibility with cup-products is sketched in the original paper of Kontsevich <a href="http://arxiv.org/abs/q-alg/9709040" rel="nofollow">http://arxiv.org/abs/q-alg/9709040</a>, is rigorously proved in <a href="http://arxiv.org/abs/math/0106205" rel="nofollow">http://arxiv.org/abs/math/0106205</a> by Manchon and Torossian, and the globalization is adressed in <a href="http://arxiv.org/abs/0805.3444" rel="nofollow">http://arxiv.org/abs/0805.3444</a> - in this last reference you will easily see that if you start with a $G$-invariant connection to perform globalization then your homotopy will be $G$-invariant too). </p> <p><strong>Conclusion</strong>. taking $G$-invariants and restricting yourself to the degree zero part of cohomology (Poisson and Hochschil, respectively), you'll find that the Poisson center of $\mathcal O_X^G$ is isomorphic, <em>as an algebra</em>, to the center of the subalgebra of $G$-invariants elements in the quantization. </p> <p><strong>Things I have been hiding under the carpet</strong>. 1. issues involving $\hbar$. In this case you have to prove that the quantization is actually convergent and can be specialized to $\hbar=1$. 2. You have to prove that the quantization really gives back differential operators... this is probably the more tricky part. </p> http://mathoverflow.net/questions/98291/are-fully-extended-tqfts-generalized-cohomology-theories/98298#98298 Answer by DamienC for Are fully extended TQFTs generalized cohomology theories? DamienC 2012-05-29T18:54:33Z 2012-06-18T12:14:04Z <p>A fully extended tqft is not quite a generalized homology theory... But almost. You can find a preliminary reference here (notes of a talk by Hiro Tannaka at the mit Talbot workshop): <a href="http://math.mit.edu/conferences/talbot/2012/notes/14_Tanaka_FactorizationHomology(hiro).pdf" rel="nofollow">http://math.mit.edu/conferences/talbot/2012/notes/14_Tanaka_FactorizationHomology(hiro).pdf</a> The precise statements you might be interested in are Theorems 2.16 and 2.20. </p> <p>(side remark: the notes of the whole workshop are worth reading: <a href="http://math.mit.edu/conferences/talbot/2012/notes/talbot_2012_notes(claudia).pdf" rel="nofollow">http://math.mit.edu/conferences/talbot/2012/notes/talbot_2012_notes(claudia).pdf</a>). </p> <p>EDIT : the work announced in the above talk is now partly available on John Francis' webpage: <a href="http://www.math.northwestern.edu/~jnkf/writ/" rel="nofollow">http://www.math.northwestern.edu/~jnkf/writ/</a> (see "Factorization homology of topological manifolds" and "Structured singular manifolds and factorization homology"). </p> http://mathoverflow.net/questions/93140/associative-yang-baxter-on-ug/93197#93197 Answer by DamienC for associative Yang-Baxter on U(g) DamienC 2012-04-05T07:42:15Z 2012-04-05T07:42:15Z <p>I don't know about full classification results for $U(\mathfrak{g})$ in general (this is probably out of reach), but there are a bunch of very interesting constructions (together with partial classification resultas) when $A=M_n(\textbf{k})$ and/or when $\mathfrak{g}$ is a semi-simple Lie algebra: </p> <ul> <li><p>in the work of Travis Schedler: see e.g. <a href="http://arxiv.org/abs/math/0212258" rel="nofollow">http://arxiv.org/abs/math/0212258</a> (this is very much in the spirit of Belavin-Drinfeld classification of solutions of the classical Yang-Baxter equation for a semi=simple $\mathfrak{g}$)</p></li> <li><p>in the work of Igor Burban: see e.g. <a href="http://www.mi.uni-koeln.de/~burban/AYBEnewsemistable.pdf" rel="nofollow">http://www.mi.uni-koeln.de/~burban/AYBEnewsemistable.pdf</a> (imo, this is beautiful geometric approach). </p></li> <li><p>in the original work of Alexander Polishchuk: <a href="http://arxiv.org/abs/math/0008156" rel="nofollow">http://arxiv.org/abs/math/0008156</a></p></li> </ul> http://mathoverflow.net/questions/92356/a-infty-basic-reference/92379#92379 Answer by DamienC for $A_\infty$ basic reference DamienC 2012-03-27T14:02:02Z 2012-03-27T14:02:02Z <p>I believe the thesis of Kenji Lefèvre-Hasegawa is a remarkable piece of work, and is very readable (references spotted by Samuel Tinguely are very good, but they are about $A_\infty$-algebras): </p> <p><a href="http://www.math.jussieu.fr/~keller/lefevre/TheseFinale/tel-00007761.pdf" rel="nofollow">http://www.math.jussieu.fr/~keller/lefevre/TheseFinale/tel-00007761.pdf</a></p> <p>Unfortunately it is in only available in French. </p> http://mathoverflow.net/questions/86414/pbw-theorem-and-multigraded-lie-algebras/86613#86613 Answer by DamienC for PBW-Theorem and multigraded Lie algebras DamienC 2012-01-25T10:37:42Z 2012-01-25T10:37:42Z <p>As $\mathfrak{a}[0]$-modules we have: <code>$$U(\mathfrak{a}_+)[k]=\bigoplus_{\substack{l\geq1 \\ c_1r_1+\cdots+c_lr_l=k}}S^{c_1}(\mathfrak{a}_+[r_1])\otimes\cdots\otimes S^{c_l}(\mathfrak{a}_+[r_l])$$</code></p> <p>PS: I assume you work over a field of characteristic zero and I use the isomorphism of graded $\mathfrak a$-modules $S(\mathfrak a)\to U(\mathfrak a)$ given by the symmetrization map. </p> http://mathoverflow.net/questions/17182/is-an-irreducible-holomorphic-symplectic-manifold-a-simple-lie-algebra/82486#82486 Answer by DamienC for Is an irreducible holomorphic symplectic manifold a simple Lie algebra? DamienC 2011-12-02T16:54:54Z 2012-01-23T13:15:28Z <p>I believe this is a very interesting question, that I have been asking myself for quite a long time. </p> <p>Nevertheless, I have been told by Prof. Beauville that even in the irreducible case one does not have that $$Ext_X(\mathcal O_X,\mathcal O_X)=Ext_X(T_X,T_X)$$</p> <p>Namely, consider $X$ being the Hilbert scheme of two points on a $K3$ surface. </p> <p>Then $Ext_X(\mathcal O_X,\mathcal O_X)=\mathbb{C}\oplus\mathbb{C}[-2]\oplus\mathbb{C}[-4]$. </p> <p>But <code>$Ext_X(T_X,T_X)=Ext_X(\mathcal O_X,(T^*_X)^{\otimes 2})$</code> contains $Ext_X(\mathcal O_X,\Omega^2_X)$, which is huge ($h^{2,2}=232$). </p> <p>Anyway, I must say that this does not kill the question (this just tells we have to reformulate it). I hope to be able to write more about it soon. </p> <p>EDIT: it seems that the answer to the question is NO. The point is that 232 is also the dimension of $H^1(X,S^3(T_X))$ ($X$ is again a $K3$), therefore $Ext_X^1(S^2(T_X),T_X)=RHom_X(\wedge^2(T_X[-1]),T_X[-1]))$ has dimension $\geq232$. </p> http://mathoverflow.net/questions/85258/is-the-hochschild-chain-complex-c-a-a-a-b-infty-module-over-the-hochschi/85261#85261 Answer by DamienC for Is the Hochschild chain complex $C_*(A, A)$ a $B_\infty$-module over the Hochschild cochain complex $C^*(A, A)$? DamienC 2012-01-09T15:13:57Z 2012-01-11T07:49:52Z <p>This is the subject of Section 2 in <a href="http://arxiv.org/abs/0805.3444" rel="nofollow">that paper</a> (sorry for self-promotion). Chains actually have two $B_\infty$-module structures (over cochains). Those two module structures are moreover compatible (see Theorem 2.4 of the above paper for a precise statement). </p> http://mathoverflow.net/questions/32889/a-few-questions-about-kontsevich-formality/81891#81891 Answer by DamienC for A few questions about Kontsevich formality DamienC 2011-11-25T14:47:48Z 2012-01-10T13:26:01Z <p>Hi Kevin, even if the question is answered I would like to add a few remarks. </p> <p><strong>(0)</strong> the claim that </p> <blockquote> <p>this quasi-isomorphism $U$ is compatible with the dg algebra structures on $T$ and $D$</p> </blockquote> <p>is not exactly true. It is compatible only on tangent cohomology. </p> <p><strong>(1)</strong> I agree with Daniel and James that there exists a $G_\infty$-quasi-isomorphism between $T$ and $D$ (this implies compatibility at the level of tangent cohomology, and is strictly stronger). But until recently this was not known if Kontsevich's $L_\infty$-quasi-isomorphism can be upgraded to a $G_\infty$-one. <a href="http://arxiv.org/abs/1109.3520" rel="nofollow">A recent paper of Willwacher</a> solves (in a positive way) this question (EDIT: Willwacher makes the comparison with Tamarkin's $G_\infty$-quasi-isomorphism in Section 10). </p> <p><strong>(2)</strong> the proof for smooth affine varieties is essentially the same as the one for smooth differentiable manifolds. Both rely </p> <ul> <li><p>either on the exitence of a connection in the tangent bundle (see the papers of Dolgushev, e.g. <a href="http://arxiv.org/abs/math/0307212" rel="nofollow">this one</a>). </p></li> <li><p>or (equivalently) on acyclicity of sheaves of sections of bundles (see e.g. my <a href="http://arxiv.org/abs/0708.2725" rel="nofollow">paper with Michel Van den Bergh</a>). </p></li> </ul> <p><strong>(3)</strong> references are </p> <ul> <li><p><a href="http://arxiv.org/abs/math/0310399" rel="nofollow">Yekutieli's paper</a> and <a href="http://arxiv.org/abs/math/0603200" rel="nofollow">Van den Bergh'sone</a> for smooth algebraic varieties. </p></li> <li><p>my <a href="http://arxiv.org/abs/math/0504372" rel="nofollow">paper with Dolgushev and Halbout</a> for complex manifolds. </p></li> <li><p>the <a href="http://arxiv.org/abs/0708.2725" rel="nofollow">above paper</a> with Van den Bergh for a uniform approach to smooth, complex and algebraic settings (using Lie algebroids). </p></li> <li><p><a href="http://arxiv.org/abs/math/0605141" rel="nofollow">Dolgushev-Tamarkin-Tsygan paper</a> for the $G_\infty$ version (see also <a href="http://arxiv.org/abs/0710.4510" rel="nofollow">another paper with Van den Bergh</a>). </p></li> </ul> <blockquote> <p>For a general smooth variety, though, instead of taking the Hochschild cochain complex of $A=\Gamma(X;\mathcal O_X)$, presumably we should take the Hochschild cochain complex of the (dg?) derived category of $X$. Is this correct?</p> </blockquote> <p>One could do this, but one instead works on the sheaf level. Consider $T$ and $D$ as sheaves and prove that they are $L_\infty$- (or $G_\infty$-)quasi-isomorphic as sheaves of DG Lie (or $G_\infty$)-algebras. </p> <p><strong>(3+$\epsilon$)</strong> "Deformation quantization of Poisson manifolds, II" does not exist, but there is <a href="http://arxiv.org/abs/math/0106006" rel="nofollow">"Deformation quantization of algebraic varieties"</a> (quite sketchy). You might also be interested by the very inspiring paper <a href="http://arxiv.org/abs/math/9904055" rel="nofollow">"Operads and motives in deformation quantization"</a>. </p> http://mathoverflow.net/questions/71315/tamarkin-tsygan-formalism/84962#84962 Answer by DamienC for Tamarkin-Tsygan Formalism DamienC 2012-01-05T14:20:03Z 2012-01-05T14:43:44Z <p>Here is a sketch of topological description of a Tamarkin-Tsygan precalculus. </p> <p>Consider the compactified configuration spaces $C_n$ and $D_{1,n}$ of $n$ points on $\mathbb{R}^2$ and $\mathbb{R}^2-{(0,0)}$, respectively. The collection $(C_n,D_{1,n})_n$ form a topological (colored) operad. </p> <p><strong>Claim</strong>: the collection <code>$(H_{-\bullet}(C_n,\mathbb{Q}),H_{-\bullet}(D_{1,n},\mathbb{Q}))_n$</code> is the operad of Tamarkin-Tsygan precalculi. </p> <p>It is well-known that <code>$(H_{-\bullet}(C_n,\mathbb{Q}))_n$</code> is the Gerstenhaber operad. Then the operations $L$ and $i$ are the two classes (of respective degrees $1$ and $0$) in <code>$H_{\bullet}(D_{1,1},\mathbb{Q})=H_{\bullet}(S^1,\mathbb{Q})$</code>. </p> <p>The two compatibility conditions can be understood as identities in <code>$H_1(D_{1,2},\mathbb{Q})$</code>. </p> <p>If you want to get the operator $d$ (i.e. get a calulus rather than a precalculus) you'll have to replace $D_{1,n}$ by its semi-direct product with $S^1$. </p> <p>See also Section 11 (especially $\S 11.3$) of <a href="http://arxiv.org/abs/math/0606241" rel="nofollow">this paper by Kontsevich and Soibelman</a> for a similar point-of-view and its relation to a generalization of Deligne's conjecture. </p> <hr> <p>EDIT: there is also an algebraic motivation for the compatibility conditions, which is explained in <a href="http://arxiv.org/abs/math/0002116" rel="nofollow">the original paper of Tamarkin and Tsygan</a>. Namely, if $A$ is a Gerstenhaber algebra then $A[\epsilon]$, with $deg(\epsilon)=1$, is a Gerstenhaber algebra with modified product $a*b=a\cdot b+(-1)^{|a|}\epsilon[a,b]$. And if $(A,M)$ is a precalculus then $M$ becomes a Gerstenhaber module over $A[\epsilon]$ with $$(a+\epsilon b)*m=(-1)^{|a|}i_am\quad\textrm{and}\quad [a+\epsilon b,m]=L_am+i_bm$$</p> http://mathoverflow.net/questions/70881/duality-between-proper-homotopy-theory-and-strong-shape-theory Duality between proper homotopy theory and strong shape theory DamienC 2011-07-21T08:09:05Z 2011-12-13T17:13:57Z <p>In the <a href="http://ncatlab.org/nlab/show/shape+theory" rel="nofollow">n-lab entry about shape theory</a> one can read that </p> <blockquote> <p>Strong Shape Theory [...] has, especially in the approach pioneered by Edwards and Hastings, strong links to proper homotopy theory. The links are a form of duality related to some of the more geometric duality theorems of classical cohomology.</p> </blockquote> <p>I would be interested in any reference where I can find a precise formulation of this duality. </p> <p>EDIT: according to Gjergji Zaimi's answer the duality might be an improvement of Chapman's complement theorem. One can find it as Theorem 6.5.3 on page 230 of the book by Edwards and Hastings ("Cech and Steenrod Homotopy Theories with Applications to Geometric Topology"). Nevertheless, it seems to me that what was meant on the n-lab entry was more a cohomology type duality (like an instance of Verdier duality in the $(\infty,1)$-ctageorical context). Am I completely wrong?</p> http://mathoverflow.net/questions/76913/is-the-nc-torus-a-quantum-group/83343#83343 Answer by DamienC for Is the nc torus a quantum group? DamienC 2011-12-13T15:13:06Z 2011-12-13T15:13:06Z <p>Despite the negative result quoted by MTS, there have been some attempts to put a Hopf-like structure on the quantum torus. </p> <p>One of these attemps, which seems orthogonal to the one mentioned by Pierre in his answer, is <em>via</em> <a href="http://arxiv.org/abs/math/0510421" rel="nofollow">Hopfish algebras</a>. To be short, Hopfish algebras (after Tang-Weinstein-Zhu) are unital algebras equipped a coproduct, a counit and an antipode that are morphisms in the Morita category (they are bimodules,rather than actual algebra morphisms). </p> <p>The Hopfish structure on the quantum torus has been studied in details in <a href="http://arxiv.org/abs/math/0604405" rel="nofollow">this paper</a>. </p> <p>To be complete, let me emphazise the following point (taken from the above paper): </p> <blockquote> <p>It is important to note that, although the irrational rotation algebra may be viewed as a deformation of the algebra of functions on a 2-dimensional torus, our hopfish structure is not a deformation of the Hopf structure associated with the group structure on the torus. Rather, the classical limit of our hopfish structure is a second symplectic groupoid structure on $T^∗\mathbb{T}^2$ (...), whose quantization is the multiplication in the irrational rotation algebra. We thus seem to have a symplectic double groupoid which does not arise from a Poisson Lie group.</p> </blockquote> http://mathoverflow.net/questions/81209/galois-theory-for-polynomials-in-several-variables Galois theory for polynomials in several variables DamienC 2011-11-17T23:11:34Z 2011-12-04T13:37:46Z <p>I feel a bit ashamed to ask the following question here. </p> <blockquote> <p>What is (actually, is there) Galois theory for polynomials in $n$-variables for $n\geq2$?</p> </blockquote> <p>I am preparing a large audience talk on Lie theory, and decided to start talking about symmetries and take Galois theory as a "baby" example. I know that Lie groups are somehow to differential equations what discrete groups are to algebraic equations. But I nevertheless would expect Lie (or algebraic) groups to appear naturally as higher dimensional analogs of Galois groups. </p> <p>Namely, the Galois group $G_P$ of a polynomial $P(x)$ in one variable can be defined as the symmetry group of the equation $P(x)=0$ (very shortly, the subgroup of permutations of the solutions/roots that preserves any algebraic equation satisfied by them). </p> <p>Then one of the great results of Galois theory is that $P(x)=0$ is solvable by radicals if and only if the group $G_P$ is solvable (meaning that its derived series reaches <code>$\{1\}$</code>). </p> <p>I was wondering what is the analog of the story in higher dimension (i.e. for equations of the form $P(x_1,\dots,x_n)=0$. I would naively expect algebraic group to show up... </p> <hr> <p>I googled the main key words and found <a href="http://www.ucl.ac.uk/~ucahmki/sheffield.pdf" rel="nofollow">this presentation</a>: on the last slide it is written that </p> <blockquote> <p>the task at hand is to develop a Galois theory of polynomials in two variables</p> </blockquote> <p>This convinced me to anyway ask the question</p> <hr> <p><strong>EDIT: the first "idea" I had</strong></p> <p>I first thought about the following strategy. Consider $P(x,y)=0$ as an polynomial equation in one variable $x$ with coefficients in the field $k(y)$ of rational functions in $y$, and consider its Galois group. But then we could do the opposite...what would happen?</p> http://mathoverflow.net/questions/126705/sh-lie-algebra-cohomology Comment by DamienC DamienC 2013-04-07T19:52:25Z 2013-04-07T19:52:25Z @Jim Conant: higher sh Lie operations have arity in arbitrary positive degree, but they also have an inner cohomological degree which is precisely 1-arity. So that their total degree is still 1. http://mathoverflow.net/questions/124505/the-work-of-pierre-deligne Comment by DamienC DamienC 2013-03-14T14:26:19Z 2013-03-14T14:26:19Z Is this question appropriate for mathoverflow? I personnally believe not: what kind of answer can one expect? http://mathoverflow.net/questions/124404/phd-in-operator-algebras-and-non-commutative-geometry/124410#124410 Comment by DamienC DamienC 2013-03-13T23:06:38Z 2013-03-13T23:06:38Z Benameur is no longer in Metz. He is now in Montpellier. http://mathoverflow.net/questions/119991/is-the-quantum-algebra-unique-up-to-isomorphism-in-deformation-quantization/120089#120089 Comment by DamienC DamienC 2013-01-28T15:17:12Z 2013-01-28T15:17:12Z @Adrien. I think that point 2 is not the content of Dolgushev's paper (though related to it). But your last sentence is correct (and THIS is the main point of 2.). @Alexander 1. I was just saying that if you make the choice of a local universal formality morphism (i.e. given by weights associated to graphs) then the globalization is essentially unique. @Alexander 2. this is not what I am saying BUT in the context of the class of a star-product, the Poisson structures you are looking at are of the form $\hbar\pi+\cdots$. wiht a given fixed $\pi$. If $\pi$ is ND then they are all equivalent. http://mathoverflow.net/questions/116137/the-bch-series-in-terms-of-lyndon-words Comment by DamienC DamienC 2012-12-12T08:23:46Z 2012-12-12T08:23:46Z Is it really periodic or something more like a sturmian sequence? Namely, if one writes &quot;A&quot; whenever the even and odd guys coincide, and then &quot;B&quot; whenever they don't, then one gets a (semi-)infinite word in two letters &quot;A&quot; and &quot;B&quot;. Is this sequence ultimately periodic or sturmian (the later meaning that it is of minimal complexity among non-periodic words)? http://mathoverflow.net/questions/31972/todd-class-and-baker-campbell-hausdorff-or-the-curious-number-12/115488#115488 Comment by DamienC DamienC 2012-12-05T20:07:55Z 2012-12-05T20:07:55Z I actualy mean quasi-coherent sheaves (the action of $T_X[-1]$ on a given one $E$ is given by the Atiyah class of $E$). http://mathoverflow.net/questions/54430/video-lectures-of-mathematics-courses-available-online-for-free/54462#54462 Comment by DamienC DamienC 2012-12-05T09:28:08Z 2012-12-05T09:28:08Z I added the (hyper)link. http://mathoverflow.net/questions/54430/video-lectures-of-mathematics-courses-available-online-for-free/94390#94390 Comment by DamienC DamienC 2012-12-05T09:22:03Z 2012-12-05T09:22:03Z For me it says &quot;404 Not Found&quot;. http://mathoverflow.net/questions/54430/video-lectures-of-mathematics-courses-available-online-for-free/100559#100559 Comment by DamienC DamienC 2012-12-05T09:18:08Z 2012-12-05T09:18:08Z Would you mind giving a link ? :-) http://mathoverflow.net/questions/115369/detadetb-detabcorrection-capelli-identities-factorzied-representati Comment by DamienC DamienC 2012-12-04T08:54:05Z 2012-12-04T08:54:05Z I would suggest to see if this is true even when you consider non-commutative variables $E_{ij}$ that are not necessarily of the form $a_ib_j$. http://mathoverflow.net/questions/115032/non-rigorous-reasoning-in-rigorous-mathematics Comment by DamienC DamienC 2012-12-01T15:54:28Z 2012-12-01T15:54:28Z Why isn't it CW? http://mathoverflow.net/questions/114701/one-more-question-about-pbw/114781#114781 Comment by DamienC DamienC 2012-11-28T20:13:55Z 2012-11-28T20:13:55Z There is also has a relatively simple example in <a href="http://archive.numdam.org/ARCHIVE/ASNSP/ASNSP_1958_3_12_1-2/ASNSP_1958_3_12_1-2_1_0/ASNSP_1958_3_12_1-2_1_0.pdf" rel="nofollow">archive.numdam.org/ARCHIVE/ASNSP/&hellip;</a> (see also the discussion here: <a href="http://mathoverflow.net/questions/61954/" rel="nofollow">mathoverflow.net/questions/61954</a> ). http://mathoverflow.net/questions/114701/one-more-question-about-pbw Comment by DamienC DamienC 2012-11-28T07:52:15Z 2012-11-28T07:52:15Z I know that PBW holds whenever $k\supset\mathbb{Q}$. About my EDIT, this was just a dummy suggestion for a strategy. I was thinking about something like: if $L\to U(L)$ is injective and if PBW holds for $L\otimes_k\mathbb{K}$ (for some $\mathbb{K}$) then PBW holds for $L$ http://mathoverflow.net/questions/111433/symplectic-manifold-with-totally-geodesic-foliation Comment by DamienC DamienC 2012-11-04T14:39:00Z 2012-11-04T14:39:00Z @helge. As paolo ghiggini and robert bryant said, if a poisson manifold is symplectic then its symplectic foliation consists of a single leaf. Therefore there is no example of a poisson manifold which is symplectic and have codimension 2 symplectic leaves. http://mathoverflow.net/questions/110613/for-which-calabi-yau-threefolds-is-syz-conjecture-known-to-hold Comment by DamienC DamienC 2012-10-25T04:52:00Z 2012-10-25T04:52:00Z It seems that the OP is asking for a list... shouldn't it be community wiki then ?