# How to compute the abelianization of the representation theory of a Hopf algebra?

I will ask two versions of my question, which probably aren't precisely the same, and I am also interested in hearing about nuances between the two.

Version 1:

Let $(C,\otimes)$ be any monoidal category, and I will suppress all associators. By definition, the abelianization (or Hochschild homology) of $C$ is the (non-monoidal) category $C_{\rm ab}$ (along with a functor $C \to C_{\rm ab}$) formed as follows. The objects of $C_{\rm ab}$ are the same as the objects of $C$. For every pair of objects $X,Y \in C$, you freely include a new morphism $f_{X,Y} : X \otimes Y \overset\sim\to Y\otimes X$ in $C_{\rm ab}$. Let $X,Y,Z$ be any triple of objects in $C$. Then you have new morphisms $f_{X,Y\otimes Z} : X \otimes Y \otimes Z \to Y \otimes Z \otimes X$ and $f_{Y,Z\otimes X} : Z\otimes X \otimes Y$ and $f_{X\otimes Y,Z} : X\otimes Y \otimes Z \to Z\otimes X \otimes Y$; you impose for any such triple the equality $f_{X\otimes Y,Z} = f_{Y,Z\otimes X} \circ f_{X,Y \otimes Z}$. (For example, it follows from this that $f_{X,1} = \operatorname{id}_X = f_{1,X}$ and $f_{X,Y} = f_{Y,X}^{-1}$.) (Edit, following Victor's comment: For example, it follows from this that $f_{1,X} = \operatorname{id}_X$, and $f_{Y,X} \circ f_{X,Y} = f_{X\otimes Y,1}$, which is not required to be the identity.)

One probably should add a few more magic words, and I'm not sure how much follows for free. In general, it's nice to work with categories that are closed under colimits; perhaps we should add those in to $C_{\rm ab}$. And in every monoidal category that I care about, the tensor functor $\otimes$ is cocontinuous in each variable, so probably we should assert that $f_{X,Y}$ is a colimit if $X$ or $Y$ is.

Anyway, modulo these issues, my question is:

Let $H$ be a Hopf algebra and $C = H\text{-mod}$ the monoidal category of its left modules. Is there any reasonable description of $C_{\rm ab}$ in terms of $H$? For example, is $C_{\rm ab}$ the module theory of an algebra $R$ that can be computed directly from $H$?

Version 2:

Let $H$ be a Hopf algebra. Following the usual Sweedler notation, denote the comultiplication by $\Delta(a) = \sum a_{(1)}\otimes a_{(2)}$. Then $H \otimes H$ is a left $H$-module by $a \triangleright (h_1\otimes h_2) = \sum a_{(1)}h_1 \otimes a_{(2)}h_2$. Denote this structure by $_H(H\otimes H)$. There is also another left $H$-module structure given by $a \triangleright (h_1\otimes h_2) = \sum a_{(2)}h_1 \otimes a_{(1)}h_2$, which I will denote by $_H^{\rm op}(H\otimes H)$.

What is the universal (in the Morita category of algebras, bimodules, and intertwiners) triple $(R,B,f)$, where $R$ is an algebra and $_R B _H$ is a left $R$- right $H$- bimodule, and $f$ is an isomoprhism of $R$-$(H\otimes H)$-bimodules $$f : ({_R B_H}) \underset H \otimes ({_H^{\ } (H\otimes H) _{H\otimes H}}) \overset\sim\to ({_R B_H}) \underset H \otimes ({_H^{\rm op} (H\otimes H) _{H\otimes H}})$$ satisfying a certain equation of the form "$f = f\circ f$" between $R$-$H^{\otimes 3}$-bimodules?

This question is made slightly more complicated by the fact that it's posed in the Morita category of algebras. There might be some abstract nonsense that guarantees that the bimodule $B$ can be assumed to be, say, a rank-one free $R$-module, in which case it's the same as an algebra homomorphism $H \to R$; or perhaps it can be assumed to be a rank-one free $H$-module, in which case it's the same as an algebra homomorphism $R \to H$.

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Note that Z(C) is $\mathrm{Fun}_{C\text{-mod-}C}(C,C)$ while your H(C) is $\mathrm{Fun}_{C\text{-mod-}C}(C,S)$ where S is C where one of the actions is twisted by the double dual. (I can send you details about that calculation if you need them.) So I'd guess that you could somehow twist the Drinfel'd double of the Hopf algebra by inserting some double duals into the definition to get the algebra that you're looking for. – Noah Snyder Jun 8 '13 at 1:30
Hi Noah: $H(C)$ is a better notation than $C_{\rm ab}$, thanks! Some version of your comment would probably make a perfect answer, and I'd of course be happy to see details of calculations if you have them readily available. I assume you mean that $S$ is the category $C$, though of as a $C$-$C$-bimodule, there the left action is $X \triangleright = X\otimes$, and the right action is $\triangleleft X = \otimes X^{\ast\ast}$? This makes sense if $C$ is the finite-dimensional representation theory of my Hopf algebra, which usually I do not mean but in this case is perfectly fine with me... – Theo Johnson-Freyd Jun 8 '13 at 1:45
Dear Theo, it seems to me that your category is related with (universal receptacle of) commutator functors discussed in my paper arxiv.org/pdf/1108.3060.pdf One important remark: definition does imply that $f_{X,1}=id$ but does not imply that $f_{1,X}=id$; in a sense this is main observation of my paper. – Victor Ostrik Jun 8 '13 at 2:53
A priori, you have to be a little careful about whether those are left duals or right duals and whether you're twisting on the left or on the right, but once you know Radford's theorem you can canonically identify all the variations (nonetheless one of them is the right one to be Hochschild homology). That description works for any finite tensor category (in the sense of Etingof-Ostrik) not just Rep(H). But I don't actually know where to put in the duals to modify the definition of the Drinfel'd double which is why I hadn't written a full answer. – Noah Snyder Jun 8 '13 at 3:45
Another name is Dim(C), since it's the value of the (1-framed) circle in a 1-dimensional TFT. Or Tr(C) to avoid confusion with global dim. In 3 dimensions (which is the right dimension for tensor cats) there are two 3-framed circles: one is Hochschild cohomology (aka Z(C)) and the other is Hochschild homology (your $C_{ab}$). In our upcoming paper (should be this month?) with CD and CSP we give the above mentioned formula (all the techniques are already in ENO's papers but you have to be careful to not assume pivotally). If you want an unfinished draft earlier than that send me an email. – Noah Snyder Jun 8 '13 at 4:18