User mohammed abouzaid - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T04:08:53Z http://mathoverflow.net/feeds/user/6948 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/28948/gromovs-list-of-7-constructions-in-differential-topology Gromov's list of 7 constructions in differential topology Mohammed Abouzaid 2010-06-21T13:06:27Z 2010-09-23T09:54:30Z <p>At the <a href="http://www.ihp.jussieu.fr/Clay%20Institut/Inscriptions/Clay%20conference%207-8-9%20june%202010.html" rel="nofollow">2010 Clay Research Conference</a>, Gromov explained that we know of only 7 different methods for constructing smooth manifolds. Working from memory, and hence not necessarily respecting the order he used:</p> <ul> <li>Algebraic geometry (affine and projective varieties, ...)</li> <li>Lie groups (homogeneous spaces, ...)</li> <li>General position arguments (Morse theory, Pontryagin-Thom construction, ...)</li> <li>Solutions to PDE (Moduli spaces in gauge theory, Floer theory, ...)</li> <li>Surgery (Cut and paste techniques, ...)</li> <li>Markov processes</li> </ul> <p>I realise that I only gave 6 constructions; this was the number of separate items listed on his slides, and since he failed to discuss this part, I am left to guess that he either listed two different constructions on one line, which I interpreted to be variants of the same construction, or that failed to include one altogether.</p> <p><strong>Question</strong> How does one construct a smooth manifold from Markov processes?</p> <p>I asked Gromov after the talk for explanation, but due to the rudimentary nature of my Gromovian, I was unable to understand the answer. The only word I managed to parse is "hyperbolic," though I wouldn't put too much stock in that.</p> http://mathoverflow.net/questions/38919/generator-of-a-fukaya-category-with-certain-properties/38958#38958 Answer by Mohammed Abouzaid for Generator of a Fukaya category with certain properties Mohammed Abouzaid 2010-09-16T11:20:39Z 2010-09-17T23:34:35Z <p>Even the condition that you have a collection of Lagrangians which are categorically orthogonal and each with $HF^\ast(L)=H^\ast(L)$ as an $A_{\infty}$ algebra is unreasonable: There could a priori be symplectic manifolds with such Fukaya categories, but at the present state of knowledge, it is unlikely that we would be able to prove it since all methods for proving that a certain collection of Lagrangians generate the Fukaya category ultimately pass through a split-generation result for the diagonal (even the one used in Seidel's book can be interpreted in that language). On the other hand, the category you describe does not have such a resolution (you can see this by noting its Hochschild cohomology is a direct sum of homologies of free loop spaces and hence is of finite homological dimension).</p> http://mathoverflow.net/questions/36444/definition-of-an-e-infinity-algebra/36478#36478 Answer by Mohammed Abouzaid for Definition of an E-infinity algebra Mohammed Abouzaid 2010-08-23T18:40:24Z 2010-08-23T19:41:09Z <p>In characteristic 0, Kadeishvili has a notion of $C_{\infty}$ algebra which models rational homotopy theory. See the last paragraph of the introduction of his paper arXiv:0811.1655. His point of view is to simply consider $A_{\infty}$ algebras whose operations satisfy a certain property with respect to shuffle maps. So your computer doesn't have to remember any new operations, just check that the old ones are right.</p> <p>In characteristic $p$, things are probably hopeless.</p> <p>Added Remark: I just want to make clear that this does not give a "trivial proof" that a commutative dga is formal as a commutative dga if the underlying dga is formal in the "non-commutative" sense. The reason is that when you transfer from cochains from cohomology, you are restricted in the kind of morphisms allowed if you are interested in the commutative theory. So, just as in the answers to this <a href="http://mathoverflow.net/questions/9146/noncommutative-rational-homotopy-type" rel="nofollow">question</a>, there is some work to be done if you want results like that (to be completely honest, there is not yet a proof that I completely understand, so declare myself agnostic).</p> http://mathoverflow.net/questions/35900/when-is-a-symplectic-manifold-equivalent-to-a-cotangent-bundle/35901#35901 Answer by Mohammed Abouzaid for When is a symplectic manifold equivalent to a cotangent bundle? Mohammed Abouzaid 2010-08-17T20:12:33Z 2010-08-17T21:07:58Z <p>Using the h-principle, Gromov showed that there is a symplectic form on $\mathbb{R}^6$ which admits $S^3$ as a Lagrangian submanifold. Using holomorphic curves, he showed that the standard symplectic form on $T^* \mathbb{R}^3$ does not admit any such Lagrangian. There is now a whole industry of building exotic symplectic forms on non-compact manifolds (see papers of Seidel-Smith, Mark McLean, ...).</p> <p>Probably the only reasonable answer to characterising cotangent bundles uses the existence of a Lagrangian foliation by planes. If you have a foliation parametrised by a manifold which admits a Lagrangian section, then you have yourself an open subset of a cotangent bundle (this is just Weinstein's theorem). You can't drop the condition of the existence of a section precisely because you can add the pull back of a $2$-form on the base. If your symplectic form is "complete" then the existence of a Lagrangian section is a cohomological condition. Pick any section: If the pullback of $\omega$ doesn't vanish, then you don't have a cotangent bundle. If it vanishes in cohomology, you can use a primitive $1$-form to flow your section to a Lagrangian.</p> <p>Added Remark:</p> <p>I want to point out that the methods we have for producing different symplectic forms do not proceed by writing down different $2$-forms on the same space. Rather, you find some construction of symplectic manifolds (using some general notion of symplectic surgery) which produces a large class of symplectic manifolds, then you prove that some of these result in the same smooth manifold. The existence of a diffeomorphism is obtained abstractly, so I do not know of examples where we can write down a Hamiltonian whose dynamics for two different symplectic forms can be compared.</p> http://mathoverflow.net/questions/84576/pseudocycle-definition-of-open-gromov-witten-invariants Comment by Mohammed Abouzaid Mohammed Abouzaid 2011-12-31T01:29:39Z 2011-12-31T01:29:39Z In the notes you're citing, the class $\beta$ has the minimal amount of energy among classes admitting holomorphic discs, so there is no bubbling issues. In particular, the moduli space is, for generic almost complex structure, a smooth manifold. The fact that evaluation maps at interior points can be made transverse to any cycle in $M$ can be extracted from work of Floer-Hofer-Salamon. For cycles on Lagrangians, you can use a doubling trick. http://mathoverflow.net/questions/68803/how-to-relate-equivariant-symplectic-cohomology-contact-homology-cyclic-homolog/69490#69490 Comment by Mohammed Abouzaid Mohammed Abouzaid 2011-07-05T01:06:49Z 2011-07-05T01:06:49Z It is only $HH^∗$ of the wrapped Fukaya category which has a chance to be isomorphic to symplectic cohomology (which is verified for cotangent bundles). There are two versions of the Calabi-Yau property: the one most people are familiar with concerns the existence of a pairing, and the fact that it holds for compact Lagrangians implies that $HH^∗$ of the ordinary Fukaya category is always, up to shift, dual to $HH_∗$. There is another version (due to Kontsevich) which is a statement about duality in a category of bimodules, and which implies an isomorphism $HH^∗ \cong HH_∗$ up to shift. http://mathoverflow.net/questions/38919/generator-of-a-fukaya-category-with-certain-properties/38958#38958 Comment by Mohammed Abouzaid Mohammed Abouzaid 2010-09-17T19:40:45Z 2010-09-17T19:40:45Z My point is that if you're interested in having a general theory, symplectic topology currently only produces (1) manifolds like cotangent bundles which are not compact, but where you can hope to have Fukaya categories like you originally asked for in your question (2) examples like toric varieties, where there are Lagrangian tori which are disjoint and generate, but have deformed cohomology rings. There is no general class in between that we know of an understand. The LG model for $x^3$ probably appears for toric varieties if you allow bulk deformations. http://mathoverflow.net/questions/38919/generator-of-a-fukaya-category-with-certain-properties/38958#38958 Comment by Mohammed Abouzaid Mohammed Abouzaid 2010-09-17T18:35:20Z 2010-09-17T18:35:20Z If you allow yourself to vary $m_2$, then you're essentially allowing an arbitrary proper Calabi-Yau algebra, and there are indeed going to be some that are smooth (though I'm not the expert on this subject). The question becomes too general at this stage to probably have a good answer. The example that you have in mind for the Landau-Ginzburg model requires changing $m_2$. http://mathoverflow.net/questions/38919/generator-of-a-fukaya-category-with-certain-properties/38958#38958 Comment by Mohammed Abouzaid Mohammed Abouzaid 2010-09-17T01:05:07Z 2010-09-17T01:05:07Z I intentionally answered the question for $A_{\infty}$ algebra structures as opposed to cohomology because I can't think of any reasonably geometric criterion which make the products agree, while the higher products to diverge. Moreover, I highly doubt that you can construct a smooth $A_\infty$ category by taking the $A_\infty$ structure on cohomology and changing the higher products. http://mathoverflow.net/questions/37963/lecture-notes-by-thurston-on-tiling/37973#37973 Comment by Mohammed Abouzaid Mohammed Abouzaid 2010-09-07T15:47:39Z 2010-09-07T15:47:39Z The arXiv accepts pdf or ps documents as long as they have not been produced by directly processing a (La)TeX file. If a (La)TeX document is ever found/produced, one can always replace the arXiv submission to reflect that. There is a maximum size of about 6 mb, so something like Thurston's notes on 3 manifolds would not work but these should. Of course, one can split a book into separate submissions. http://mathoverflow.net/questions/36444/definition-of-an-e-infinity-algebra/36478#36478 Comment by Mohammed Abouzaid Mohammed Abouzaid 2010-08-23T21:49:00Z 2010-08-23T21:49:00Z My understanding is that if one cares about quasi-isomorphism types of commutative dga's then the two theories ($E_\infty$ or $C_{\infty}$) give the same answer rationally. It would be great if we had a comment from experts confirming or denying this. http://mathoverflow.net/questions/35863/generating-the-derived-category-with-line-bundles Comment by Mohammed Abouzaid Mohammed Abouzaid 2010-08-17T14:51:42Z 2010-08-17T14:51:42Z I think you mean to say that these powers split-generate the derived category (otherwise, all varieties would have finite rank $K_0$ which fails already for curves).