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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.

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...

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.)

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  • $\begingroup$ Sorry for not providing any background. Perhaps this is bad etiquette. I just started a discussion about this at meta: tea.mathoverflow.net/discussion/564/… $\endgroup$
    – Kevin H. Lin
    Commented Jul 30, 2010 at 1:30
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    $\begingroup$ You absolutely don't have to write a tutorial on Hochschild (co)homology, but, at least, you should explain the notation, just like you would in a paper. $\endgroup$
    – Victor Protsak
    Commented Jul 30, 2010 at 6:00
  • $\begingroup$ I've now expanded the text of my question :-) $\endgroup$
    – Kevin H. Lin
    Commented Jul 30, 2010 at 23:23
  • $\begingroup$ sorry to ask for more, but how do you compute HH of a category? take some sort of cyclic nerve? $\endgroup$
    – Sean Tilson
    Commented Jul 31, 2010 at 5:13
  • $\begingroup$ $\mathcal{Nat}(\mathrm{Id},[n])$ $\endgroup$
    – Aaron Bergman
    Commented Jul 31, 2010 at 18:54

2 Answers 2

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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.

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  • $\begingroup$ Very nice. It seems like everything is answered very nicely by Toen's Morita theory. $\endgroup$
    – Kevin H. Lin
    Commented Jul 31, 2010 at 0:04
  • $\begingroup$ are the bar resolution and hochschild complex really the same? i thought that the differentials were a little different. One is a simplicial set and the other is cyclic. Or are you only using the bar resolution to compute the ext group? $\endgroup$
    – Sean Tilson
    Commented Jul 31, 2010 at 5:17
  • $\begingroup$ The latter. It'e Lemma 9.1.3 of Weibel, for example. $\endgroup$
    – Aaron Bergman
    Commented Jul 31, 2010 at 13:17
  • $\begingroup$ This is the answer for cohomology. What is the corresponding answer for homology? $\endgroup$
    – Kevin H. Lin
    Commented Apr 5, 2011 at 22:57
  • $\begingroup$ Presumably you're taking derived tensor products in the endofunctor category, but I'm not sure of a reference. It must be in Lurie or Ben-Zvi-Francis-Nadler. $\endgroup$
    – Aaron Bergman
    Commented Apr 6, 2011 at 4:11
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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)]

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...

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  • $\begingroup$ Despite having tagged this question with "morita-theory", I don't really know anything about Morita theory. In particular, I don't know what it means for an algebra to be Morita equivalent to a category. $\endgroup$
    – Kevin H. Lin
    Commented Jul 30, 2010 at 2:09
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    $\begingroup$ @Kevin: an algebra can be seen as a linear category with one object, so let's do that. Next, if $C$ is a linear category, its modules are the functors $C\to\mathrm{Vect}$, and they form a category ${}_C\mathrm{Mod}$. Now, wo linear categories $C$ and $C'$ are Morita equivalent if they have equivalent module categories ${}_C\mathrm{Mod}$ and ${}_{C'}\mathrm{Mod}$. Finally, under any sensible definition, Hochschild cohomology is invariant under Morita equivalences. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Jul 30, 2010 at 2:12
  • $\begingroup$ @Mariano: In your second paragraph, did you mean abelian algebras? If not, what about, say, the path algebra of the A_2 quiver? $\endgroup$
    – Kevin Walker
    Commented Jul 30, 2010 at 2:39

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