most recent 30 from http://mathoverflow.net 2013-05-21T17:04:44Z http://mathoverflow.net/feeds/question/33877 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/33877/hochschild-cohomology-of-a-and-of-mod-a Kevin Lin 2010-07-30T01:23:52Z 2010-07-31T01:39:39Z

<p>Let A be an algebra (or dg algebra). Where can I find a proof of HH_*(A) = HH_*(Mod_A) and HH^*(A) = HH^*(Mod_A)? (And does this hold for any A?) Here Mod_A is, e.g., the category of left A-modules.</p> <p>One reason why this is interesting/important/useful is because many categories which arise "in nature" are of the form Mod_A. For example, there is a theorem of Bondal and van den Bergh which states that derived categories of a large class of varieties (I forget their exact hypotheses) are equivalent to Mod_A for some A. Dyckerhoff also proved that categories of matrix factorizations are of this form. By mirror symmetry, Fukaya-type categories should be of this form as well...</p> <p>Anyway, so to compute HH of such a category, it suffices to find this A and then compute HH(A). I think that it generally(?) should be easier to compute HH of an algebra than HH of a category. (Of course finding this A can be a very nontrivial task.)</p>

http://mathoverflow.net/questions/33877/hochschild-cohomology-of-a-and-of-mod-a/33878#33878 Mariano Suárez-Alvarez 2010-07-30T01:43:29Z 2010-07-30T02:04:43Z

<p>I guess it follows from results in [Lowen, Wendy; Van den Bergh, Michel. Hochschild cohomology of abelian categories and ringed spaces. Adv. Math. 198 (2005), no. 1, 172--221. MR2183254 (2007d:18017)] </p> <p>For algebras $A$, at least, it follows more simply from the fact that the categories $\mathrm{Mod}(A)$ and $A$ are Morita equivalent. That must have been proved by Mitchell or Freyd...</p>

http://mathoverflow.net/questions/33877/hochschild-cohomology-of-a-and-of-mod-a/33889#33889 Aaron Bergman 2010-07-30T03:33:33Z 2010-07-30T03:33:33Z

<p>Basically this follows from the fact that the derived category of bimodules over two algebras is equivalent to the (suitably defined) functor category between the derived category of modules of each algebra. Say, Toen's paper on derived Morita equivalence. Then, the identity functor is given by the algebra itself interpreted as a bimodule, so the Hochschild cohomology is $\mathrm{Ext}^i_{A-A}(A,A)$. You can compute this using the bar resolution and a quick calculation gives you the usual definition of Hochschild cohomology.</p>