# Corepresentations of Tensor Products of Hopf Algebras

Given two cosemisimple Hopf algebras $H,G$ over ${\mathbb C}$, denote their usual (not braided) tensor product by $G \otimes H$. What conditions do we need to impose on the Hopf algebras to ensure that the corepresentations of $G \otimes H$ are direct sums of corepresentations of the form $V \otimes W$, where $V$ is a corepresentation of $G$ and $W$ is a corepresentation of $H$?

I have heard a rumour that this is true when $G$ and $H$ can be completed to compact quantum groups. However, I can't find a reference, and I believe that it should be true for a larger class of Hopf algebras.

• Surely as written this will almost never be true. Let $M$ and $N$ be nontrivial corepresentations of $H$ and $G$ respectively, and let $1$ denote (for either Hopf algebra) the trivial representation. Then $M \otimes 1 \oplus 1 \otimes N$ has a hard time being of the form $V \otimes W$. But perhaps you mean that a general corepresentation should be a direct sum of products? I would have thought that if $A$ and $B$ are cosemisimple coalgebras over an algebraically closed field, then $A \otimes B$ would be cosemisimple with (simples of $A\otimes B$) =(simples of $A$) $\times$ (simples of $B$).... Jul 15 '16 at 17:56
• Is that not the case? I need the field to be algebraically closed to avoid the following situation. Consider the coalgebra $\mathbb H^*$, the linear dual to the quaternions. Its comodules are just $\mathbb H$-modules, of course. But $\mathbb H \otimes \mathbb H \cong \mathrm{Mat}_2(\mathbb R)$. Let $\mathbb H_{\mathbb H}$ denote the unique irrep of $\mathbb H$. Under the above isomorphism, $\mathbb H_{\mathbb H} \otimes \mathbb H_{\mathbb H}$ is the regular representation of $\mathrm{Mat}_2(\mathbb R)$, which breaks as a direct sum of two copies of the unique irrep. Jul 15 '16 at 18:00
• @Theo: What you have said is just what I meant. I have edited the question to remove my imprecisions. Jul 15 '16 at 18:14
• A few times in my comment from yesterday, I wrote "2" when I meant "4". $\mathbb H \otimes_{\mathbb R} \mathbb H =\mathrm{Mat}(4,\mathbb R)$ is the algebra of $4\times 4$ matrices, and the regular module is the direct sum of four copies of the unique irrep. I got confused by the related fact $\mathbb H \otimes_{\mathbb R} \mathbb C = \mathrm{Mat}(2,\mathbb C)$. Sorry for the error. Jul 16 '16 at 17:27

Let $A$ be a coalgebra over a field $\mathbb K$ and $\mathcal A = \mathrm{Comod}^A$ its category of comodules. A well-known result of Sweedler writes $A = \mathrm{colim}_i A_i$ where $A_i$ are finite-dimensional subcoalgebras of $A$. Let $A_i^*$ denote the linear dual to $A_i$ and $\mathcal A_i = \mathrm{Mod}_{A_i^*} = \mathrm{Comod}^{A_i}$ the category of $A_i^*$-modules, equivalently the category of $A_i$-comodules. As observed in my paper with Brandenburg and Chirvasitu, $\mathcal A = \mathrm{colim} \mathcal A_i$ is the colimit (over the same diagram) in the bicategory of $\mathbb K$-linear locally presentable categories. Here and throughout I sloppily write "$=$" for "there is a canonical equivalence".

Tensor products of coalgebras distribute over colimits of coalgebras, and tensor products of categories distribute over colimits of categories. Moreover, it is well-known that if $R$ and $S$ are $\mathbb K$-algebras, then $\mathrm{Mod}_{R \otimes S} = \mathrm{Mod}_R\boxtimes \mathrm{Mod}_S$. (This can be checked directly from the universal property of the tensor product.)

Suppose that $B = \mathrm{colim}_j B_j$ is another coalgebra over $\mathbb K$ with $B_j$ its finite-dimensional subcoalgebras, and $\mathcal B = \mathrm{Comod}^B$ and $\mathcal B_j = \mathrm{Mod}_{B_j^*}$ as above. We have \begin{aligned}\mathcal A \boxtimes \mathcal B & = (\mathrm{colim}_{i} \mathcal A_i) \boxtimes (\mathrm{colim}_j \mathcal B_j) \\ & = \mathrm{colim}_{i,j} (\mathcal A_i \boxtimes \mathcal B_j) \\ & = \mathrm{colim}_{i,j} (\mathrm{Mod}_{A_i}^* \boxtimes \mathrm{Mod}_{B_j^*}) \\ & = \mathrm{colim}_{i,j} (\mathrm{Mod}_{A_i^* \otimes B_j^*}) \\ & = \mathrm{colim}_{i,j} (\mathrm{Comod}^{A_i \otimes B_j}) \\ & = \mathrm{Comod}^{\mathrm{colim}_{i,j} A_i \otimes B_j} \\ & = \mathrm{Comod}^{A\otimes B}. \end{aligned}

On the other hand, the tensor product of categories can be presented as follows. For arbitrary $\mathbb K$-linear locally presentable categories $\mathcal A$, $\mathcal B$, objects of $\mathcal A \boxtimes \mathcal B$ are colimits over objects of the form $V \boxtimes W$ subject to the relation that $\mathrm{colim}_m (V_m \boxtimes W) = (\mathrm{colim}_m V_m) \boxtimes W$, and similarly in the $W$-variable, where on the LHS the colimit is computed in $\mathcal A \boxtimes \mathcal B$ and on the RHS the colimit is computed in $\mathcal A$.

It follows from this presentation that if $\mathcal A$ and $\mathcal B$ are both semisimple categories, then so is $\mathcal A\boxtimes \mathcal B$. Indeed, suppose that $\mathcal A$ is semisimple and let $\mathcal I = \{I_1,\dots\}$ be a complete set of simples. By "semisimple" I mean that every object of $\mathcal A$ is a direct sum of objects from $\mathcal I$. Let $\mathbb K_m = \mathrm{End}(I_m)$; by Schur's lemma, it is a division ring. Then $\mathcal A = \bigoplus_m \mathrm{Mod}_{\mathbb K_m}$. If $\mathcal B$ is also semisimple with simples $\{J_1,\dots\}$ and $\mathrm{End}(J_n) = \mathbb L_n$, then $$\mathcal A \boxtimes \mathcal B = \bigoplus_{m,n} \mathrm{Mod}_{\mathbb K_m} \boxtimes \mathrm{Mod}_{\mathbb L_n} = \bigoplus_{m,n} \mathrm{Mod}_{\mathbb K_m \boxtimes \mathbb L_n}.$$

Suppose now that $\mathbb K$ is algebraically closed and $\mathcal A = \mathrm{Comod}^A$ and $\mathcal B = \mathrm{Comod}^B$. Then $\mathbb K_m = \mathbb L_n = \mathbb K$ for all $m,n$. It follows that every object of $\mathcal A \boxtimes \mathcal B$ is a direct sum of objects of the form $I_m \boxtimes J_n$, as you wanted. (Under the equivalence $\mathcal A \boxtimes \mathcal B = \mathrm{Comod}^{A\otimes B}$, $I \boxtimes J$ corresponds to the $A\otimes B$-comodule $I \otimes J$.)

If $\mathbb K$ is not algebraically closed, then one can find many examples where $\mathbb K_m \boxtimes \mathbb L_n$ is not a division ring. It is always, however, a semisimple algebra, and so its module theory is semisimple with simples given by direct summands of the regular module. Thus in this case every object of $\mathrm{Comod}^{A\otimes B}$ is a direct sum of direct summands of objects of the form $I_m \boxtimes J_n$.

So as you can see, the answer to your question correctly interpreted is "yes", and it has nothing to do with Hopf algebras or quantum groups.