14
$\begingroup$

Let $HH_*(A,N)$ (or $HH^*(A,N)$) be the Hochschild homology (or cohomology) of an associative algebra $A$ with coefficients in an $A$-bimodule $N$.

I was reading nlab's entry on Hochschild cohomology and I saw this term:

Hochschild homology object of any bimodule over an monoid in a monoidal ($\infty ,1$)-category

, which, when $N = A$, is also called the ($\infty ,1$)- or derived center of $A$.

In this paper (http://arxiv.org/pdf/0805.0157v5.pdf) they say that for an associative algebra object $A$ in a closed symmetric monoidal $\infty$-category $\mathcal{S}$, the derived center or Hochschild cohomology $Z(A) = HH^*(A) \in \mathcal{S}$ is the endomorphism object $End_{A \otimes A^{op}} (A)$ of $A$ as an $A$-bimodule.

Then I found this on page 45:

6.2. Deligne-Kontsevich conjectures for derived centers. The notion of Drinfeld center for monoidal stable categories is a categorical analogue of Hochschild cohomology of associative (or $A_{\infty}$ algebras)...

Honestly I had never heard of Hochschild cohomology objects (or derived centers?) or Hochschild (co)homlogy used in this type of context but mostly I wanted to know:

What is the relationship between Hochschild cohomology and Drinfeld centers?

$\endgroup$

2 Answers 2

14
$\begingroup$

Classically, Hochschild cohomology is an invariant defined for associative algebras while the Drinfeld centre is an invariant defined for monoidal categories. The latter is a categorification of the classical notion of centre of a monoid. In particular the Drinfeld centre of the category with a single object and endomorphism ring $A$ is nothing but $Z(A)$, the centre of $A$ in the usual sense (or more precisely the category with a single object and endomorphism ring $Z(A)$). The terminology originates from the fact that the Drinfeld centre can also be viewed as a categorification of the Drinfeld double construction for Hopf algebras, which goes back to [Vladimir Drinfeld, Quantum groups, Proceedings of the ICM, pp. 798–820, AMS, 1987]. It is easy to see that the definition of Drinfeld centre in fact gives a notion of centre for an associative algebra object in any symmetric monoidal category.

The relationship between these two constructions is that Hochschild homology is a derived version of the centre. That is, the Hochschild homology of an associative algebra $A$ is the centre of $A$ as an associative algebra object in $D(A \otimes A^{op})$, the derived category of $A \otimes A^{op}$-modules. However, to make this precise, one needs to introduce the centre in the setting of $(\infty,1)$-categories, since considering only the homotopy category of the $(\infty,1)$-category $D(A \otimes A^{op})$ loses too much information.

This is the subject of the paper of Ben-Zvi, Francis and Nadler: the generalization of centres to $(\infty,1)$-category theory. Thus they consider the notion of derived centre of an associative algebra object in any symmetric monoidal $(\infty,1)$-category. By specializing this gives a notion of derived centre of an associative algebra, i.e. Hochschild cohomology (by taking the derived $(\infty,1)$-category of bimodules over an associative algebra), and a notion of derived centre of a monoidal $(\infty,1)$-category, i.e. the derived Drinfeld centre (by taking the $(\infty,1)$-category of (presentable) $(\infty,1)$-categories).

$\endgroup$
5
  • $\begingroup$ Thank you, this is all still a bit foreign to me, but going by what you wrote and intuition from what I know about Hochschild (co)homology, do you mean that the Drinfeld center of the symmetric monoidal infinity-category of presentable infinity-categories and colimit-preserving functors is the same as the Hochschild cohomology of every associative algebra object in this same category? Does that make sense? $\endgroup$
    – Samuel M
    Commented Jan 13, 2015 at 0:40
  • $\begingroup$ Thank you again, when I read that Hochschild cohomology of the associative algebra $A$ is a "derived center" I thought it would be equivalent to something like the Drinfeld center of the derived category $D(Mod_{A \otimes A^{op}})$? Or is "derived" referring to something else? Sorry if that doesn't make sense, I might be a bit tired. $\endgroup$
    – Samuel M
    Commented Jan 14, 2015 at 10:26
  • $\begingroup$ So Hochschild cohomology of the associative algebra $A =$ Ext$^{\bullet}_{A \otimes A^{op}} (A,A)$ is End$_{D(A \otimes A^{op})} (A)$, I wonder what the (Drinfeld) center of $D(A \otimes A^{op})$ is and how it is related to this derived center of $A$? Thank you so much again. $\endgroup$
    – Samuel M
    Commented Jan 16, 2015 at 1:05
  • $\begingroup$ @SamuelM, according to Remark 1.5 in the paper, it seems that the Drinfeld centre of $D(A \otimes A^{op})$ will be the $\infty$-category of modules over the Hochschild chain complex of $A \otimes A^{op}$. $\endgroup$
    – AAK
    Commented Jan 16, 2015 at 18:51
  • $\begingroup$ Will look into that, all of this was very helpful, thanks. $\endgroup$
    – Samuel M
    Commented Jan 17, 2015 at 23:02
12
$\begingroup$

The key formula is, as you note, $End_{A\text{-mod-}A} (A)$. We can then interpret this same formula in lots of settings.

In the category of sets, $A$ is an ordinary algebra, and $End_{A\text{-mod-}A} (A)$ is the ordinary center of A. (The only things that commute with the left action are right multiplication by elements, and those only commute with right multiplication if the element is central.)

In the derived setting you get the derived version of the center, namely Hochschild cohomology.

In the $(2,1)$-category of categories, an associative algebra is exactly a monoidal category, and $End_{A\text{-mod-}A} (A)$ is the Drinfel'd center. To see this you need to be a bit careful unpacking what a functor of bimodule categories means. In particular instead of a condition saying $f(am) = af(m)$ there's a natural isomorphism $f(a \otimes m) \rightarrow a f(m)$ satisfying a coherence condition (see Ostrik for more details). Again left $A$-module endofunctors of $A$ correspond exactly to tensoring on the right with some object of $A$ (with the associator for $A$ giving the natural transformation). In order for tensoring on the right with an object to commute with the right action, we need to pick isomorphisms $\eta_b: a \otimes b \rightarrow b \otimes a$. In other words, $End_{A\text{-mod-}A} (A)$ consists of objects together with half-braidings and so is the Drinfeld center.

$\endgroup$
3
  • $\begingroup$ Thank you, that was very well-explained, it's the derived setting I was most interested in, so in the derived setting the formula $End_{A \otimes A^{op}}(A)$ is basically just Ext$^{\bullet}_{A \otimes A^{op}}(A,A)$ = Hochschild cohomology of $A$ with coefficients in $A%)$? Or am I using the term derived in the wrong context? $\endgroup$
    – Samuel M
    Commented Jan 14, 2015 at 10:23
  • $\begingroup$ @Samuel: yes (or one might instead mean Hochschild cochains). $\endgroup$ Commented Jan 14, 2015 at 12:09
  • $\begingroup$ @QiaochuYuan Thank you, it makes more sense now. $\endgroup$
    – Samuel M
    Commented Jan 16, 2015 at 1:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .